## Coordinate Geometry Parabola

It is unlikely that you will get a problem on Coordinate geometry parabola on the new GRE. Its an uncommon topic but you are supposed to go through our article on Coordinate geometry parabola. Our article is definitely helpful in tackling problem on coordinate geometry parabola.

A parabola is the set of all points in a plane that are equidistant  from a fixed line & a fixed point (not on the line) in the plane.

The fixed line is called the directrix of the parabola & the fixed point F is called the focus.

$${ B }_{ 1 }{ P }_{ 1 }={ B }_{ 2 }{ P }_{ 2 }$$

A line through the focus & perpendicular to the directrix is called the axis of the parabola. The point of intersection of parabola with the axis is called the vertex of the parabola.

### Standard equations of parabola:

The equation of a parabola is simplest if the vertex is at the origin and the axis of symmetry is along the x-axis or y-axis. The four possible such orientation of parabola are shown in fig (a), fig (b), fig (c) & fig (d).

Fig(a)

Fig (b)

Fig (c)

Fig (d)

Let F be the focus & l be the directrix. In fig(a), the equation for the parabola with focus at (a,0) (where a>0) & directrix at x = a is

$${ y }^{ 2 }=4ax$$

In fig (b), the equation for the parabola with focus at (-a,0)  & directrix at x = +a  is

$${ y }^{ 2 }=-4ax$$

In fig (c), the equation for the parabola with focus at (0,a) & directrix at y = -a  is

$${ x }^{ 2 }=4ay$$

In fig (d), the equation for the parabola with focus at (0,-a) & directrix at y = a is

$${ y }^{ 2 }=4ax$$…… (1)

$${ y }^{ 2 }=-4ax$$…… (2)

$${ x }^{ 2 }=4ay$$…… (3)

$${ x }^{ 2 }=-4ay$$…… (4)

These four equations are known as standard equation of the parabola, derivation is beyond the scope here.

For any positive number c,

• The graph of parabola $${ y }^{ 2 }=4ax$$ when shifts upward by c units, the equation of parabola becomes
$${ (y-c) }^{ 2 }=4ax$$
• The graph of parabola $${ y }^{ 2 }=4ax$$ when shifts downward by c units, the equation of parabola becomes
$${ (y+c) }^{ 2 }=4ax$$
• The equation of parabola $${ y }^{ 2 }=4ax$$ when shifts to the left by c units, the equation of parabola becomes
$${ y }^{ 2 }=4a(x+c)$$
• The equation of parabola $${ y }^{ 2 }=4ax$$ when shift to the right by c units, the equation of parabola becomes
$${ y }^{ 2 }=4a(x-c)$$

Similar rules apply to all the standard parabolas.

Problem   The equation $$y={ x }^{ 2 }-2x-3$$ has the following graph. Find the following:

• Coordinates of the vertex
• The y-intercept

Solution:  The equation

$$y={ x }^{ 2 }-2x-3\\ y={ x }^{ 2 }-2x+1-4\\ (y+4)={ (x-1) }^{ 2 }\\ { (x-1) }^{ 2 }=4(\frac { 1 }{ 4 } )(y+4)$$

Therefore, the point (1,-4) is the vertex of parabola.

The y-intercept is the y-coordinate of the point on the parabola at which  x=0, which is

$$y=0-2(0)-3={ (-3) }^{ Ans }$$

Problem

Which of the following could be the equation of the parabola in the coordinate plane above?

a) $$y={ x }^{ 2 }+3$$

b) $$y={ (x-3) }^{ 2 }+3$$

c) $$y={ (x+3) }^{ 2 }-3$$

d) $$y={ (x-3) }^{ 2 }-3$$

e) $$y={ (x+3) }^{ 2 }+3$$

Sol.

This graph has been obtained by shifting standard parabola (having vertex at (0,0)  & focus at (0,a)  k units upward & h units toward the left.

$${ x }^{ 2 }=4ay$$—— (equation of standard parabola)

Equation after shifting vertex at (-h,k)

$${ (x+h) }^{ 2 }=4a(y-k)\quad \quad (h>0,K>0)\\ { (x+h) }^{ 2 }+4ak=4ay$$

After analysing answer choices, we can easily reckon that choice $$y={ (x+3) }^{ 2 }+3$$ places the vertex in the correct quadrant.

Problem

Which of the following could be the equation of the parabola in the coordinate place above?

a) $$y=-x-1$$

b) $$y={ x }^{ 2 }+1$$

c) $$y={ -x }^{ 2 }-1$$

d) $$y={ -x }^{ 2 }+1$$

e) $$y=-{ (x-1) }^{ 2 }$$

Sol.   This graph has been obtained by shifting standard parabola (having vertex at (0,0) & focus at (0, –a) 1 unit upward.

$${ x }^{ 2 }=-4ay——-$$(Equation of parabola having vertex at (0,0) & focus at (0, –a)).

Equation after shifting vertex at (0,1):

$${ x }^{ 2 }=-4a(y-1)\quad \quad (a>0,\quad let\quad a=\frac { 1 }{ 4 } )\\ { x }^{ 2 }=-4(\frac { 1 }{ 4 } )(y-1)\\ { x }^{ 2 }=-(y-1)=-y+1\\ ({ y=-{ x }^{ 2 }+1) }^{ Ans }$$

Problem

If the equation of the parabola in the coordinate plane above is $$y={ (x-h) }^{ 2 }+k$$ and (-3,n) is a point on the parabola, what is the value of n?

Sol.   This graph has been obtained by shifting standard parabola (having vertex at (0,0)  & focus at (0,a)) 2 units rightward.

$${ x }^{ 2 }=4ay$$— — (Equation of standard parabola having vertex at (0,0) and focus at (0,a)).

$${ (x-2) }^{ 2 }=4ay$$— — (Equation of standard parabola by shifting vertex at (2,0)).

$${ (x-2) }^{ 2 }=4ay$$

Let’s compare this equation with given equation

$$y={ (x-h) }^{ 2 }+k\\ 4ay={ (x-2) }^{ 2 }+k\\ Therefore,\quad a=\frac { 1 }{ 4 } ,\quad h=2,\quad k=0\\ y={ (x-2) }^{ 2 }$$

Point (-3,n) lies on this parabola

$$n={ (-3-2) }^{ 2 }\\ n=25\\ Therefore,\quad { (n=25) }^{ Ans }$$

## GRE Sequences and Series

A GRE quant problem varies from vapid arithmetic problem to maverick second-degree algebra problem. In our article on GRE Sequence and series we have almost everything beginning from formulas to tedious problems. After going through our article on GRE Sequences and series, solving a GRE Sequences and series will be an easy task for you.

A sequence is a collection of number in a set order. Each term of a sequence is determined by a RULE. Here are examples of sequence RULES:

$${ A }_{ n }=9n+3$$

The $${ n }^{ th }$$ term of this sequence is defined by the rule 9n+3, for integer . For example, the fourth term in this sequence = 9(4) + 3 = 39.

Let’s take one more example:

$${ Q }_{ n }={ n }^{ 2 }+4$$

The $${ n }^{ th }$$ term of this sequence is defined by the rule $${ n }^{ 2 }+4$$, for integers $$n\ge 1$$. The first term in this sequence is

$${ 1 }^{ 2 }+4=5$$

In the above cases, each term of the sequence is defined as a function of n, the place in which the term occurs in the sequence. This is a direct definition of a sequence formula.

Questions in GRE mostly appear from Arithmetic propagation from Geometric progression.

## Arithmetic progressions:

Number are said to be in arithmetic progression when they increase or decrease by a common difference. Thus each of the following series forms an arithmetic progression:

3, 7, 11, 15………

8, 2, -4, -10………

a, a+d, a+2d, a+3d………

The common difference is found by subtracting any term of the series from the next term.

Common difference of an A.P. =

$${ N }^{ th }term\quad -\quad { (N-1) }^{ th }term$$

If we examine the series a, a+d, a+2d, a+3d……… we notice that in any term the coefficient of d is always less by one than the position of that term in the series.

Thus the $${ N }^{ th }term$$ term of an arithmetic progression is given by $${ N }^{ th }term$$ term = a+(N-1)d.

To find the sum of the given number of terms in an Arithmetic progression:

Let a denote the first term, d be the common difference,

Sum of term upto N terms =

$$=\quad \frac { N }{ 2 } (2a\quad +\quad (N-1)d)$$

## Geometric Progression:

Quantities are said to be in Geometric Progression when they increase or decrease by a constant factor.

The constant factor is also called the common ratio & it is founded by dividing any term by the term immediately preceding it.

If we examine the series

$$a,\quad ar,\quad a{ r }^{ 2 },\quad a{ { r }^{ 3 } }\_ \quad \_ \quad \_ a{ r }^{ n-1 }\quad$$

We notice that in any term the index of r is always less by one than the number of terms in the series.

$${ n }^{ th }term\quad$$ term of the sequence =

$$a{ r }^{ n-1 }\quad$$

To find the sum of a Number of terms in a Geometric Progression:

Let a be the first term, r be the common ratio,

Sum of numbers upto n terms

$$=\quad \frac { a(1-{ r }^{ n }) }{ (1-r) } \quad (If\quad r>1)\quad \\ =\quad \frac { a }{ (1-r) } \quad (If\quad r<1)$$

The GRE also uses recursive formulas to define sequences with direct formulas, the values of each term in a sequence is defined in term of its term number in the sequence. With recursive formulas, each item number of a sequence is defined in terms of the value of PREVIOUS ITEM in the sequence.

A recursive formula looks like this:

$${ A }_{ n }={ A }_{ n-1 }+9$$

This formula simply means “THIS term $${ A }_{ n }$$ equals the PREVIOUS term $${ A }_{ n-1 }$$ plus 9”.

Whenever you look at a recursive formula, articulate its meaning in words in your mind.

As we have discussed, sequence problems generally involve finding patterns among the terms in a sequence.

Let’s take an example:

If $${ A }_{ n }={ 3 }^{ n }$$, what is the unit digit of $${ A }_{ 65 }$$?

Clearly, you cannot be expected to multiply out $${ 3 }^{ 65 }$$ on the GRE, even with a calculator. Therefore, you must look for a patterns in the powers of three.

$${ 3 }^{ 1 }=3\\ { 3 }^{ 2 }=9\\ { 3 }^{ 3 }=27\\ { 3 }^{ 4 }=81\\ { 3 }^{ 5 }=243\\ { 3 }^{ 6 }=729\\ { 3 }^{ 7 }=2187\\ { 3 }^{ 8 }=6561$$

You can see that the unit’s digits of powers of 3 follow the pattern “3, 9, 7, 1”. As 64 is divisible by 4. Unit digit of $${ 3 }^{ 64 }$$ is 1 and Unit digit of $${ 3 }^{ 65 }$$ is 3.

Problem 1) The sequence A is defined by

$${ A }_{ n }\quad =\quad { A }_{ n-1 }+{ A }_{ n-2 }+{ A }_{ n-3 }-5$$ for each integer $$n\ge 4$$. If $${ A }_{ 1 }=4,\quad { A }_{ 2 }=0\quad and\quad { A }_{ 4 }\quad =\quad -4$$, what is the value of $${ A }_{ 6 }$$?

Sol.

$${ A }_{ n }\quad ={ \quad A }_{ n-1 }+{ A }_{ n-2 }+{ A }_{ n-3 }-5\quad (let\quad n=4)\\ { A }_{ 4 }\quad =\quad { A }_{ 3 }+{ A }_{ 2 }+{ A }_{ 1 }-5\\ -4\quad =\quad { A }_{ 3 }+0+4-5\quad =\quad { A }_{ 3 }-1\\ -4\quad =\quad { A }_{ 3 }-1\\ { A }_{ 3 }\quad =\quad -4+1\quad =\quad -3\\ { A }_{ n }\quad =\quad { A }_{ n-1 }+{ A }_{ n-2 }+{ A }_{ n-3 }-5\quad (let\quad n=5)\\ { A }_{ 5 }={ A }_{ 4 }+{ A }_{ 3 }+{ A }_{ 2 }-5\\ \quad \quad =\quad -4+(-3)+0-5\quad =\quad -4-3-5\quad =\quad -12\\ { A }_{ 6 }\quad =\quad { A }_{ 5 }+{ A }_{ 4 }+{ A }_{ 3 }-5\\ { A }_{ 6 }\quad =\quad -12\quad +\quad (-4)\quad +\quad (-3)\quad -\quad 5\quad =\quad -24\\ Therefore,\quad { A }_{ 6 }\quad =\quad { (-24) }^{ Ans }$$

Problem 2) The sequence P is defined by

$${ P }_{ n }\quad =\quad 10({ P }_{ n-1 })-2$$ for each integer $$n\ge 2$$. If $${ P }_{ 1 }=2$$, what is the value of $${ P }_{ 4 }$$?

Sol.

$${ P }_{ n\quad }=\quad 10({ P }_{ n-1 })-2\quad (let\quad n=2)\\ { P }_{ 2\quad }=\quad 10{ P }_{ 1 }-2\quad =\quad 10(2)-2\quad =\quad 18\\ { P }_{ n }\quad =\quad 10{ P }_{ n-1 }-2\quad (let\quad n=3)\\ { P }_{ 3 }\quad =\quad 10{ P }_{ 2 }-2\\ \quad \quad \quad =\quad 10(18)-2\quad =\quad 180-2\quad =\quad 178\\ { P }_{ n }\quad =\quad 10({ P }_{ n-1 })-2\quad (let\quad n=4)\\ { P }_{ 4 }\quad =\quad 10({ P }_{ 3 })-2\\ \quad \quad \quad =\quad 10(178)-2\\ \quad \quad \quad =\quad { (1778) }^{ Ans }$$

Problem 3) The sequence S is defined by

$${ S }_{ n }\quad =\quad 2({ S }_{ n-1 })-4$$ for each integer $$n\ge 2$$. If $${ S }_{ 1 }=6$$, what is the value of $${ S }_{ 5 }$$?

Sol.

$${ S }_{ n }=2{ S }_{ n-1 }-4\quad (let\quad n=2)\\ { S }_{ 2 }=2{ S }_{ 1 }-4\\ { S }_{ 2 }=2{ S }_{ 1 }-4\quad ({ S }_{ 1 }=6)\\ \quad =\quad 2(6)-4\quad =\quad 8\\ { S }_{ n }=2{ S }_{ n-1 }-4\quad (let\quad n=3)\\ { S }_{ 3 }=2{ S }_{ 2 }-4\\ \quad \quad \quad =2(8)-4\quad =\quad 12\\ { S }_{ n }\quad =\quad 2{ S }_{ n-1 }-4\quad (let\quad n=4)\\ { S }_{ 4\quad }=\quad 2{ S }_{ 3 }-4\quad =\quad 2(12)-4\quad =\quad 24-4\quad =\quad 20\\ { S }_{ n }\quad =\quad 2{ S }_{ n-1 }-4\quad (let\quad n=5)\\ { S }_{ 5 }=\quad 2{ S }_{ 4 }-4\\ \quad \quad =\quad 2(20)-4\quad =\quad { (36) }^{ Ans }$$

Problem 4) Find the value of the expression 1-6+2-7+3-8……… to 100 terms

a) -250

b) -500

c) -450

d) -300

Sol.  The series (1-6+2-7+3-8……… to 100 terms) can be written as:

$$=\quad (1+2+3+4———to\quad 50\quad terms)\quad +\\ (-6-7-8———to\quad 50\quad terms)\\$$

For sequence (1+2+3 ……… to 50 terms)

Using formula for sum of Airthmetic Progression

$${ S }_{ n }=\frac { n }{ 2 } (2a+(n-1)d)\\ { S }_{ 50 }=\quad \frac { 50 }{ 2 } (2(1)+(50-1)1)\\ \quad \quad =\quad 25(2+49)\\ \quad \quad =\quad 51\times 25\quad =\quad { (1275) }^{ Ans }\\$$

For sequence (6+7+8 ……… to 50 terms)

Using formula for sum of n terms of Arithmetic progression

$${ S }_{ n }=\frac { n }{ 2 } (2a+(n-1)d)\\ { S }_{ 50 }=\quad \frac { 50 }{ 2 } (2(6)+(50-1)1)\\ \quad \quad =\quad 25(12+49)\\ \quad \quad =\quad 61\times 25\quad =\quad { (1525) }^{ Ans }\\$$

Value of expression

$$=\quad (1+2+3——-to\quad 50\quad terms)\\ -(6+7+8——–to\quad 50\quad terms)\\ =\quad 1275\quad -\quad 1525\\ =\quad { (-250) }^{ Ans }$$

Problem 5) The first three terms in an arithmetic sequence are 30, 33 & 36. What is the 80th term?

Sol.  For arithmetic progression: 30, 33, 36, ………

a=30

Common difference = 33-30=3

Using formula for nth term of an arithmetic progression:

$${ a }_{ N }\quad =\quad a\quad +\quad (N-1)d\\ \quad =\quad 30\quad +\quad (N-1)(3)\\ { a }_{ 80\quad }=\quad 30\quad +\quad (80-1)(3)\\ \quad \quad \quad =\quad 30\quad +\quad 79\times 3\\ \quad \quad \quad =\quad { (267) }^{ Ans }$$

Problem 6) In a certain sequence, the term $${ a }_{ n }$$ is defined by the formula $${ a }_{ n }\quad =\quad { a }_{ n-1 }+10$$

for each integer $$n\ge 2$$. What is the positive difference between $${ a }_{ 10 }$$ and $${ a }_{ 15 }$$?

Sol.

$${ a }_{ n }\quad =\quad { a }_{ n-1 }+10\quad (let\quad n=2)\\ { a }_{ 2 }\quad =\quad { a }_{ 1 }+10$$

Therefore, this is an arithmetic sequence where the difference between successive terms is always +10.

The difference between, for example,$${ a }_{ 10 }$$  & $${ a }_{ 11 }$$, is exactly 10, regardless of the actual values of the two terms.

Starting from $${ a }_{ 10 }$$, there is a sequence of 5 terms to get to $${ a }_{ 15 }$$.

Therefore, the difference between $${ a }_{ 10 }$$ & $${ a }_{ 15 }$$ is 10×5 = 50.

Problem 7) The sequence A is defined by $${ A }_{ n }={ A }_{ n-1 }+2$$ for each integer $$n\ge 2$$, & $${ A }_{ 1 }=45$$. What is the sum of the first 100 terms in sequence A?

Sol.

$${ A }_{ 1 }=\quad first\quad term\quad =\quad 45\\ { A }_{ n }\quad =\quad { A }_{ n-1 }+2\quad (let\quad n=2)\\ { A }_{ 2 }\quad =\quad { A }_{ 1 }+2\quad =\quad 45+2\quad =\quad 47\\ { A }_{ 3 }\quad =\quad { A }_{ 2 }+2\quad =\quad 47+2\quad =\quad 49$$

The first few terms of the sequence are 45, 47, 49, 51 ………

This is an arithmetic progression with a=45 & common progression (d) = 47-45 = 2.

Using formula for sum of n terms of Arithmetic progression:

$${ S }_{ n }=\frac { n }{ 2 } (2a\quad +\quad (n-1)d)$$

Sum of first 100 terms in sequence A

$$=\frac { 100 }{ 2 } (2(45)\quad +\quad (100-1)2)\\ =50(90\quad +\quad 99\times 2)\\ =50(90+198)\\ ={ (14400) }^{ Ans }$$

Problem 8) The sequence S is defined by $${ S }_{ n-1 }=\frac { 1 }{ 4 } { S }_{ n }$$ for each integer $$n\ge 2$$. If $${ S }_{ 1 }=-4$$, what is the value of $${ S }_{ 4 }$$?

a) -256

b) -64

c) $$\frac { -1 }{ 16 }$$

d) $$\frac { 1 }{ 16 }$$

Sol.

$${ S }_{ n-1 }=\frac { 1 }{ 4 } { S }_{ n }$$

(let n=2)

$${ S }_{ 1 }=\frac { 1 }{ 4 } { S }_{ 2 }\\ { S }_{ 2\quad }=\quad 4{ S }_{ 1 }=\quad 4(-4)\quad =\quad -16\\ { S }_{ 2 }=\frac { 1 }{ 4 } { S }_{ 3 }\\ { S }_{ 3 }=4{ S }_{ 2 }=4(-16)=-64$$

First few terms of sequence are: -4, -16, -64

We can easily infer that this is a geometric progression with a=-4,

r (common ratio) =4

$${ S }_{ n }({ n }^{ th }\quad term\quad of\quad sequence)\quad =\quad a{ r }^{ n-1 }\\ Value\quad of\quad { S }_{ 4 }\quad =\quad (-4){ (4) }^{ n-1 }\\ =\quad (-4){ (4) }^{ 4-1 }=\quad (-4){ (4) }^{ 3 }={ (-256) }^{ Ans }$$

Problem 9) $${ A }_{ n }={ 2 }^{ n }-1$$ for all integer $$n\ge 1$$

Quantity A:         The unit digit of $${ A }_{ 26 }$$

Quantity B:         The unit digit of $${ A }_{ 34 }$$

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Ans.

$${ A }_{ 26 }={ 2 }^{ 26 }-1,\quad { A }_{ 34 }={ 2 }^{ 34 }-1$$

Clearly, we are not expected to multiply out $${ 2 }^{ 26 }\quad or{ \quad 2 }^{ 34 }$$ on the GRE, even with a calculator.

Therefore, we must look for a pattern in the powers of 2.

$${ 2 }^{ 1 }=2\\ { 2 }^{ 2 }=4\\ { 2 }^{ 3 }=8\\ { 2 }^{ 4 }=16\\ { 2 }^{ 5 }=32\\ { 2 }^{ 6 }=64\\ { 2 }^{ 7 }=128\\ { 2 }^{ 8 }=256$$

We can see that the unit’s digits of powers of 2. follow the pattern “2, 4, 8, 6, 2, 4, 8, 6, ………”

Unit digit of $${ 2 }^{ 26 }=4$$

Unit digit of $${ 2 }^{ 34 }=4$$

Unit digit of $$({ 2 }^{ 26 }-1)=3$$

Unit digit of $$({ 2 }^{ 34 }-1)=3$$

Therefore, two quantities are equal.

## GRE Functions

Our chapter on GRE functions enables you to learn each and every minuscule detail about GRE functions and covers problems ranging from easy to difficult.

An algebraic expression of only one variable may be defined as a function, usually f or g, of that variable.

Example:   What is the value of the function $$f(x)\quad =\quad { x }^{ 2 }-1$$

When x = 1?

Solution:   In the function $$f(x)\quad =\quad { x }^{ 2 }-1$$, if x = 1, then f(1)=1-1=0. In other words, when the input to the function is 1, the output is 0.

The set of input numbers for a function is the domain & the set of output values is the range of function. Every input value of a function has exactly one output value.

However, more than one input value may have the same output value.

Consider the function $$f(x)\quad =\quad { x }^{ 2 }-1$$ again.

If $$x\quad =\quad -1,f(-1)\quad =\quad { (-1) }^{ 2 }-1\quad =\quad 0$$

If $$x\quad =\quad 1,\quad f(1)\quad =\quad { (1) }^{ 2 }-1\quad =\quad 0$$

Therefore,$$f(1)=f(-1)=0$$ .

Restricted domain of a function:

A function may be defined for all real numbers or it may be only for a subset of the real number.

For example: $$g(x)\quad =\quad -{ x }^{ 2 }+2$$ , where $$-3\le x\le 3$$

The domain of the function $$g(x)$$ is restricted to values of $$x$$ between -3 and +3 inclusive.

The maximum value of $$g(x)\quad =\quad -{ x }^{ 2 }+2$$ occurs when $$x=0,\quad g(0)=\quad -{ x }^{ 2 }+2\quad =\quad 0+2\quad =\quad 2.$$

To find the minimum value of $$g(x)\quad =\quad -{ x }^{ 2 }+2$$, where $$-3\le x\le 3$$

We will find g(x) at extreme values.

$$g(3)\quad =\quad -9+2\quad =\quad -7,\quad g(3)\quad =\quad -9+2\quad =\quad -7$$

Therefore, minimum value occurs when x = $$-3$$ or x = 3. That value is -7.

The domain of a function may also be restricted to avoid having a zero in the denominator of a fraction or to avoid taking the square root of a negative number.

Example: $$f(x)\quad =\quad \frac { 2x-5 }{ x-3 } \quad$$, where $$x\quad \neq \quad 3$$

Here, To find the domain of the function, we are supposed to restrict domain of f to avoid x-3=0.

Example:  $$h(x)\quad =\quad \sqrt { x+10 } ,\quad where\quad x\ge -10$$

Here, To find the domain of this function, we are supposed to restrict domain of h to avoid x+10<0.

## Graphs of functions:

To graph, a function in the xy-plane, use the x-axis for the input & the y-axis for the output. You can represent every input value, x & its corresponding output value y, as an ordered pair (x,y).

Problem 1) 1) Plot function $$f(x)\quad =\quad 2x-1\quad and\quad g(x)\quad =\quad { x }^{ 2 }-1$$

2) Find points of intersection of two graphs.

Sol.   We can easily see that three graph intersect at two points. These are the points for which f(x)=g(x).

$$2x\quad -1\quad =\quad { x }^{ 2 }-1\\ { x }^{ 2 }-2x\quad =\quad 0\\ x(x\quad -\quad 2)\quad =\quad 0\\ x=0\quad or\quad x=2.\\ f(0)\quad =\quad -1,\quad f(2)\quad =\quad 3\\$$

Therefore, the point of intersection is (0,1) & (2,3).

## Piecewise function:

A piecewise function is defined by more than one equation, where each equation applies to a different part of the domain of the function.

The absolute value function is an example of a piecewise of a piecewise function.

The absolute value function is defined as

$$f(x)\quad =\quad |x|\quad =\\ \{ \quad x\quad for\quad x\ge 0\\ \{ \quad x\quad for\quad x<0$$

## Shifting of graphs: –

The ability to visualize how graphs shift when the basic analytical expression is changed is a very important skill.

In order to be able to do so, you first need to understand the following points clearly.

• The graph of f(x)+c is the graph of f(x) shifted upward c units or spaces.
• The graph of f(x)-c is the graph of f(x) shifted downward c units or spaces.
• The graph of f(x+c) is the graph of f(x) shifted to the left c units or spaces.
• The graph of f(x-c) is the graph of f(x) shifted to the right c units or spaces.

Problem 2) The graph of $$f(x)\quad =\quad { x }^{ 2 }$$ is shown. How is the graph of the function $$g(x)\quad =\quad { x }^{ 2 }-4$$  related to the graph of f(x)?

Sol.

Graph of $$f(x)\quad =\quad { x }^{ 2 }$$

Graph of $$g(x)\quad =\quad { x }^{ 2 }-4$$

As we have studied that the graph of f(x)-c is obtained by shifting graph of f(x) c units downward.

Problem 3) The graph of $$h(x)\quad =\quad |x|$$ is shown. How is the graph of the function $$k(x)\quad =\quad |x-1|$$ related to the graph of $$h(x)$$ ?

Sol.

As we have studied that the graph of f(x-c) is the graph of f(x) shifted to the right c units or spaces.

The graph of k(x)=|x-1| is obtained by shifting graph of h(x)=|x| 1 unit to the right.

The graph of a function may also be vertically stretched away or compressed toward the x-axis by a factor of c, where c is a positive number. You can stretch the graph by making the slope larger & steeper, or you can compress the graph by making the slope smaller & less steep. To change the vertical slope of the graph of f(x), follow these rules:

• The graph of c×f(x) is the graph of f(x) stretched away from the x-axis by a factor of c.
• The graph of (1/c)×f(x) is the graph of f(x) compressed away from the x-axis by a factor of c.

Problem 4) describe the relationship between the graphs of $$f(x)\quad =\quad { x }^{ 2 }$$  & $$g(x)\quad =\quad \frac { 1 }{ 2 } { x }^{ 2 }$$.

Sol.  As we have studied that the graph of  is the graph of (1/c)×f(x) is the graph of f(x) compressed toward the x-axis by a factor of c.

The graph of $$g(x)\quad =\quad \frac { 1 }{ 2 } { x }^{ 2 }$$ is obtained by compressing the graph of f(x) toward the x-axis by a factor of 2.

## PROBLEMS

Problem 5) For what values should be the domain be restricted for this function?

$$f(n)\quad =\quad \frac { n-6 }{ { n }^{ 2 }-6n }$$

Sol.  Factor the denominator: $${ n }^{ 2 }-6n\quad =\quad n(n-6)$$

As a denominator of 0 is not permissible, the value of n cannot be 6 or 0.

Problem 6) If $$g(a)\quad =\quad { a }^{ 2 }-1$$ & f(a)=a+4, what is g(f(-4))?

Sol.

$$f(a)\quad =\quad a+4\\ f(-4)\quad =\quad a\quad +\quad 4\\ -4\quad +\quad 4\quad =\quad 0.\\ g(0)\quad =\quad 0-1\quad =\quad -1$$

Problem 7) If $$f(x)\quad =\quad \frac { 2x+6 }{ 4 }$$ and [/latex]g(x)\quad =\quad 2x-1[/latex], what is f(g(x))?

Sol.

$$f(x)\quad =\quad \frac { 2x+6 }{ 4 } \\ g(x)\quad =\quad 2x-1\\ f(g(x))\quad =\quad \frac { 2(2x-1)+6 }{ 4 } \\ (Replace\quad x\quad with\quad 2x-1)\\ =\quad (\frac { 4x-2+6 }{ 4 } )\\ =\quad \frac { 4x+4 }{ 4 }$$

Problem 8) If $$g(x)\quad =\quad \frac { { x }^{ 2 }(4x+9) }{ (3x-3)(x+2) }$$, for which of the following x values is g(x) undefined?

Indicate all such values of x.

a) $$-\frac { 9 }{ 4 }$$

b) -2

c) 0

d) 1

e) 2

Sol.  Factor the denominator:$$\quad \quad (3x-3)(x+2)\\ =\quad 3(x-1)(x+2)$$

As a denominator of 0 is not permissible, the value of x cannot be 1 or -2.

The correct answers are -2 & 1.

Problem 9) If $$f(x)\quad =\quad x+5$$ & $$f(2g)\quad =\quad 19$$. What is the value of f(3-g)?

Sol.

$$f(x)\quad =\quad x+5\\ f(2g)\quad =\quad 2g+5\quad =\quad 19\\ g\quad =\quad 7\\ f(3-g)\quad =\quad f(3-7)\quad =\quad f(-4)\quad =\quad -4+5\quad =\quad { 1 }^{ Ans }$$

Problem 10) If $$f(x)\quad =\quad { x }^{ 2 }-1$$, what is the value of f(y)+f(-1)?

Sol.

$$f(x)\quad =\quad { x }^{ 2 }-1\\ f(y)\quad =\quad { y }^{ 2 }-1\\ f(-1)\quad =\quad { (-1) }^{ 2 }-1\quad =\quad 0\\ f(y)\quad +\quad f(-1)\quad =\quad { y }^{ 2 }-1+0\\ =\quad { ({ y }^{ 2 }-1) }^{ Ans }$$

Problem 11) If $$\ast x$$ is defined as the square of one-half of x, what is the value of $$\frac { \ast 5 }{ \ast 3 }$$ ?

Sol.  $$\ast x$$ is defined as the square of one-half of x, $$\ast x\quad =\quad { (\frac { x }{ 2 } ) }^{ 2 }$$

Therefore,

$$\ast 5\quad =\quad { (\frac { 5 }{ 2 } ) }^{ 2 }=\quad \frac { 25 }{ 4 } \\ \ast 3\quad =\quad { (\frac { 3 }{ 2 } ) }^{ 2 }=\quad \frac { 9 }{ 4 } \\ \frac { \ast 5 }{ \ast 3 } \quad =\quad (\frac { \frac { 25 }{ 4 } }{ \frac { 9 }{ 4 } } )\quad =\quad \frac { 25 }{ 9 }$$

Problem 12) If $$h(x)\quad =\quad 5{ x }^{ 2 }+x$$, then which of the following is equal to h(a+b)?

a) $$5{ a }^{ 2 }+5{ b }^{ 2 }$$

b) $$5{ a }^{ 3 }+5{ b }^{ 3 }$$

c) $$5{ a }^{ 2 }+5{ b }^{ 2 }+a+b$$

d) $$5{ a }^{ 3 }+10ab+5{ b }^{ 3 }$$

e) $$5{ a }^{ 2 }+10ab+5{ b }^{ 2 }+a+b$$

Sol.

$$h(x)\quad =\quad 5{ x }^{ 2 }+x\\ h(a+b)\quad =\quad 5{ (a+b) }^{ 2 }+(a+b)\\ =\quad 5{ a }^{ 2 }+5{ b }^{ 2 }+5(2ab)+a+b\\ =\quad 5{ a }^{ 2 }+10ab+5{ b }^{ 2 }+a+b$$

Problem 13) $$\ast x$$ is defined as the least integer greater than x for all odd values of x, & the greatest integer less than x for all even values of x. What is the value of $$\ast (-2)-\ast 5$$?

Sol. $$\ast x$$ is defined as the least integer greater than x for all odd values of x, & the greatest integer less than x for all even values of x.

If x is odd, $$\ast x$$ equals the least integer greater than x (e.g., if x=3, then the “least integer greater than 3” is equal to 4).

If x is even, $$\ast x$$ equals the greatest integer less than x (e.g., if x=6, the “greatest integer less than x” is equal to 5).

Since -2 is even, $$\ast (-2)$$ = the greatest integer less than -2, or -3.

Since 5 is odd, $$\ast (5)$$ = the least integer greater than 5, or 6.

$$\ast (-2)-\ast (5)\quad =\quad -3-6\quad =\quad { (-9) }^{ Ans }$$

## GRE Speed, Distance Problems

Our chapter on GRE Speed, distance problems will make solving a question on GRE Speed, distance an easy task for you. We have just focused on how to use simple formula Speed × Time = Distance. You are supposed to go through our chapter and solve all the problems on GRE Speed, distance problems.

The mathematical model that describes motion has three variables, speed, time & distance. The interrelationship between these three is also the most important formula for this chapter, namely:

Speed × Time = Distance

Problem 1) A car moves for 2 hours at a speed of 25 km/h & another car moves for 3 hours at the same speed. Find the ratio of distance covered by the two cars?

Sol.  Distance covered by first car = 2 × 25 = 50 km

Distance covered by second car = 3 × 25 = 75 km

Ratio of distance covered by the two cars =

$$=\quad (\frac { 50 }{ 75 } )\quad =\quad { (\frac { 2 }{ 3 } ) }^{ Ans }$$

## Conversion between km/h to m/sec

1km/h = 1000m/h = $$\quad (\frac { 1000 }{ 3600 } )$$ m/sec = $$\quad (\frac { 1000 }{ 3600 } )$$ m/sec

Hence, to convert y km/h into m/sec multiply by $$\quad (\frac { 5 }{ 18 } )$$

Thus, y km/h = $$\quad (\frac { 5y }{ 18 } )$$ m/sec

And vice-versa: y m/sec = $$\quad (\frac { 18y }{ 5 } )$$ km/h.

To convert from m/sec to km/h, multiply by $$\quad (\frac { 18 }{ 5 } )$$.

Relative movement, therefore, can be viewed as the movement of one body relative to another moving body.

• When two bodies are moving in opposite direction at speeds
$${ S }_{ 1 }\quad$$ and $${ S }_{ 2 }\quad$$ respectively.

The relative speed is defined as $${ S }_{ 1 }\quad$$ + $${ S }_{ 2 }\quad$$.

• When two bodies are moving in the same direction.

Relative speed is $$|{ S }_{ 1 }-{ S }_{ 2 }|$$.

Suppose a car goes from A to B at an average speed of $${ S }_{ 1 }$$  & then comes back from B to A at an average speed of $${ S }_{ 2 }$$. If you had to find out the average speed of the whole journey, what would you do?

Let distance from A to B is d.

Time taken by car to travel from A to B = $$\frac { d }{ { S }_{ 1 } }$$

Time taken by car to travel from B to A = $$\frac { d }{ { S }_{ 2 } }$$

Average speed of whole journey =

$$=\frac { 2d }{ \frac { d }{ { S }_{ 1 } } +\frac { d }{ { S }_{ 2 } } } \\ =\quad \frac { Total\quad distance }{ Total\quad time } \\ =\quad \frac { 2{ S }_{ 1 }{ S }_{ 2 } }{ { S }_{ 1 }+{ S }_{ 2 } }$$

Average speed =

$$\frac { 2{ S }_{ 1 }{ S }_{ 2 } }{ { S }_{ 1 }+{ S }_{ 2 } }$$

The following thing needs to be kept in mind before solving questions on trains:

• When the train is crossing a moving object, the speed has to be taken as the relative speed of the train with respect to the object. All the rules for relative speed will apply for calculating the relative speed.
• The distance to be covered when crossing an object whenever a train crosses an object will be equal to:

Length of train + Length of object.

Problem 2) Davis drone from Amityville to Betel town at 50 miles per hour & returned by the same rate at 60 miles per hour.

Quantity A:         Davis’s average speed for the round trip, in miles/hour.

Quantity B:         55

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.  Let’s use formula for the average speed

Average speed = $$\frac { 2{ S }_{ 1 }{ S }_{ 2 } }{ { S }_{ 1 }+{ S }_{ 2 } } =\quad \frac { 2(50)(60) }{ 50+60 } \\ =\quad 54.54\quad miles/hour$$

Quantity A = 54.54miles/hour

So, Quantity B is greater.

Problem 3) Brenda walked a 12-mile scenic loop in 3 hours. If she then reduced her walking speed by half, how many hours would it take Brenda to walk the same scenic loop two more times?

Sol.  Brenda walked a 12-mile scenic loop in 3 hours.

Speed =

$$\frac { Distance }{ Speed } =\frac { 12 }{ 3 } =\quad 4\quad miles/hour$$

She reduced her peed to 2miles/hour.

Time taken by Brenda to walk the same scenic loop two more times =

$$\frac { Total\quad distance }{ Speed } \\ =\quad \frac { 24 }{ 2 } \quad =\quad 12\quad hour$$

Problem 4) if a turtle traveled $$\frac { 1 }{ 30 }$$ of a mile in 5 minutes, what was its speed in miles per hour?

Sol. Turtle traveled $$\frac { 1 }{ 30 }$$  of a mile in 5 minutes.

Distance traveled by turtle in 1 minute = $$(\frac { 1 }{ 30\times 5 } )$$ miles

Speed of turtle = $$(\frac { 1 }{ 150 } )$$ miles/minute

1 minute = $$(\frac { 1 }{ 60 } )$$ hour

Therefore, speed of turtle (in miles/hour)

$$=\quad (\frac { 1 }{ 150 } \times 60)\\ =\quad (\frac { 2 }{ 5 } )\\ =\quad 0.4\quad miles/hour$$

Problem 5) Running on a 10-mile loop in the same direction, Sue ran at a constant rate of 8 miles/hour & Rob ran at a constant rate of 6 miles/hour. If they began running at the same point on the loop. How many hours later did Sue complete exactly 1 more lap than Rob?

Sol.  Sue ran at a constant rate of 2 miles/hour

Rob ran at a constant rate of 6miles/hour

Relative speed = 2 miles/hour

Time taken by Sue to complete exactly 1 more loop than Rob

$$=\quad (\frac { Relative\quad distance }{ Relative\quad speed } )\\ =\quad (\frac { 10\quad mile }{ 2\quad miles/hour } )\\ =\quad 5\quad hour$$

Problem 6) Svetlana ran the first 5 kilometers of a 10 kilometer race at a constant rate of 12 kilometer per hour, if she completed the entire 10 kilometer race in 55 minutes, at which constant rate did she run the last 5 kilometers of the race, in Kilometers per hour?

Sol.  Svetlana ran the first 5 kilometers of a 10-km race at a constant rate of 12 kilometers per hour.

Time taken by Svetlana to complete first 5 km

$$=\quad \frac { Distance }{ Speed } \\ =\quad \frac { 5 }{ 12 } hour\\ =\quad \frac { 5 }{ 120 } hour$$

Time taken by Svetlana to complete entire race $$=\quad \frac { 55 }{ 60 } hour$$

Time taken by Svetlana to complete last 5 km

$$=\quad (\frac { 55 }{ 60 } -\frac { 25 }{ 60 } )\quad hour\\ =\quad (\frac { 30 }{ 60 } )\quad hour$$

Speed at which she ran last 5 km

$$=\quad (\frac { Distance }{ Time } )\quad =\quad (\frac { 5 }{ \frac { 30 }{ 60 } } )\\ =\quad 10\quad km/hour$$

Problem 7) Alina took a non-stop car trip that encompassed three different sections of roadways. Each of three sections covered the same distance. Alina averaged 45 miles/hour over the first section, 60 miles/hour over the second section, & 54 miles/hour for the entire trip. What was her average speed for the third section to the nearest miles/hour?

Sol.  Let distance covered by one section be ‘d’ miles.

Let average speed for the third section be V miles/hour.

Total distance = 3d

Average speed in first section = 45 miles/hour

Time taken to complete first section

$$=\quad (\frac { d }{ 45 } )\quad hour$$

Time taken to complete second section

$$=\quad (\frac { d }{ 60 } )\quad hour$$

Time taken to complete third section

$$=\quad (\frac { d }{ V } )\quad hour$$

Average speed for entire trip =

$$=\quad \frac { Total\quad distance }{ Total\quad time } \\ 54\quad =\quad (\frac { 3d }{ \frac { d }{ 45 } +\frac { d }{ 60 } +\frac { d }{ V } } )\quad =\quad (\frac { 3 }{ \frac { 1 }{ 45 } +\frac { 1 }{ 60 } +\frac { 1 }{ V } } )\\ \frac { 1 }{ 45 } +\frac { 1 }{ 60 } +\frac { 1 }{ V } \quad =\quad \frac { 3 }{ 54 } \quad =\quad \frac { 1 }{ 18 } \\ \frac { 1 }{ V } \quad =\quad \frac { 1 }{ 18 } -\frac { 1 }{ 60 } -\frac { 1 }{ 45 } =\quad \frac { 1 }{ 3 } (\frac { 1 }{ 6 } -\frac { 1 }{ 20 } -\frac { 1 }{ 15 } )\quad =\quad \frac { 1 }{ 3 } (\frac { 10-3-4 }{ 60 } )\quad =\quad \frac { 1 }{ 3 } (\frac { 3 }{ 60 } )\\ V\quad =\quad { (60\quad miles/hour) }^{ Ans }$$

 Distance Speed Time Section 1 d 45 m/hour d/45 Section 2 d 60 m/hour d/60 Section 3 d V = ? d/V Total 3d 54 m/hour 3d/54

## GRE Time and Work

Problem on GRE Time and Work may appear in any of the quantitative reasoning format and these problems on GRE time and work may cause anxiety for test takers, but by going through our chapter on GRE Time and Work with a slightly clever ability to understand you can attempt each and every problem accurately.

The concept of time & work is another important topic for the quantitative aptitude exam.

We have to understand the following basic concepts of this chapter:

If A does a work in a days, his rate = $$\frac { 1 }{ a }$$ work/day

If B does a work in b days, his rate = $$\frac { 1 }{ b }$$ work/day

If A & B work together, then their combined rate = $$\frac { 1 }{ a } +\frac { 1 }{ b }$$  work/day

= $$(\frac { a+b }{ ab } )$$ Work/day

The equation that applies to Time & Work problems is

Work Rate × Time = Work Done

Problem 1) A can build a wall in 10 days & B van build it in 5 days, if they both start working at the same time, in how many days will the wall be built?

Sol.  A can build a wall in 10 days, his rate = $${ (\frac { 1 }{ 10 } ) }^{ th }$$ wall/days

B can build a wall in 5 days, his rate = $${ (\frac { 1 }{ 5 } ) }^{ th }$$ wall/days

If A & B work together, than their combined rate = $${ (\frac { 1 }{ 10 } +\frac { 1 }{ 5 } ) }^{ th }$$ wall/day

= $${ (\frac { 3 }{ 10 } ) }^{ th }$$ wall/day

Work Rate × Time = Work Done

$${ (\frac { 3 }{ 10 } ) }^{ th }$$ Wall/day × Time = 1 wall

Time = $$(\frac { 10 }{ 3 } )$$ days

When numbers of workers are working, equation becomes

(Individual work rate × Number of workers × Time = Work done)

Problem 2) To service a single device in 12 seconds, 700 Nano robots are required with all Nano robots working at the same constant rate. How many hours would it take for a single Nano robot to service 12 devices?

Sol.   700 Nano robots can service 1 device in 12 seconds.

Here ‘work’ is 1 devic.

Rata at which Nano robot is working =

$$(\frac { 1 }{ 700\times 12 } )$$ Nano robots/sec

Using this equation:

Individual work rate × Number of workers × Time = Work

$$(\frac { 1 }{ 700\times 12 } )\times 1\times Time\quad =\quad 12\\ Time\quad =\quad (700\times 12\times 12)\quad seconds\\ \quad \quad \quad \quad =\quad (\frac { 700\times 12\times 12 }{ 60\times 60 } )\quad hours\\ \quad \quad \quad \quad =\quad (\frac { 700 }{ 25 } )\quad =\quad 28\quad hours$$

Problem 3) Twelve workers pack boxes at a constant rate of 60 boxes in 9 minutes. How many minutes would it take 27 workers to pack 180 boxes, if all workers pack boxes at the same constant rate?

Sol.   Twelve workers boxes at a constant rate of 60 boxes in 9 minutes.

1 worker can pack box at rate of $$(\frac { 60 }{ 12 } )$$ boxes in 9 minute.

Rate at which workers work = $$(\frac { 60 }{ 12\times 9 } )$$  boxes/minute

Using this equation:

Individual work rate × Number of workers × Time = Work

$$(\frac { 60 }{ 12\times 9 } )\times 27\times time\quad =\quad 180\\ time\quad =\quad (\frac { 180\times 12\times 9 }{ 60\times 27 } )\\ =\quad 12\quad minutes$$

.Problem 4) Machine A, which produces 15 golf clubs per hour, fills a production lot in 6 hours. Machine B fills the same production lot in 1.5 hours. How many golf clubs does machine B produce per hour?

Sol.   First, calculate the size of a production lot machine A works at a rate of 15 golf clubs per hour & fills a production lot in 6 hours.

Using, Rate × Time = Work

15 golf clubs per hour × 6 hours = 90 golf clubs

Machine A produces 90 golf clubs in 6 hours & fills a production lot.

Therefore, 90 golf clubs are required to fill a production lot.

Machine B produces 90 golf clubs in 1.5 hours.

Using equation:

Rate × Time = Work

Rate × 1.5 hours = 90 golf clubs

Rate = $$\frac { 90 }{ 1.5 } =(\frac { 90 }{ 15 } \times 10)=60\quad$$ golf clubs/hour

Problem 5) A standard machine fills paint cans at a constant rate of 1 gallon every 4 minutes. A deluxe machine fills gallons of point at twice the rate of a standard machine. How many hours will it take a standard machine & a deluxe machine, working together, to fill 135 gallons of paint?

a) 1

b) 1.5

c) 2

d) 2.5

e) 3

Sol.   Rate at which standard machine fill paint cans = 1 gallon/4 minutes

=  $$\frac { 1 }{ 4 } \quad$$ gallon/minute

Rate at which deluxe machine fills = $$\frac { 1 }{ 2 } \quad$$ gallon/minute

Combined rate = $$(\frac { 1 }{ 2 } +\frac { 1 }{ 4 } )\quad$$  gallon/minute = $$\frac { 1 }{ 2 } \quad$$  gallon/minute

Using equation,

Rate × Time = Work

$$(\frac { 3 }{ 4 } )\quad \times \quad Time\quad =\quad 135\\ Time\quad =\quad (\frac { 135\times 4 }{ 3 } )minute\\ =\quad (\frac { 135\times 4 }{ 3\times 60 } )\quad hour\quad =\quad (3)\quad hour$$

Problem 6) With 4 identical servers working at a constant rate, a new internet search provider processes 9,600 search requests per hour. If the search provider adds 2 more identical servers & server work rate never varies, the search provider can process 216,000 search requests in how many hours?

Sol.   With 4 identical servers working at a constant rate, a new internet search provider processes 9,600 search requests per hour.

1 sever work at a rate of $$\frac { 9600 }{ 4 }$$ search request per hour.

Using equation:

Individual rate × Number of servers × Time = Work

2400 × 6 × Time = 216,000

Time =

$$(\frac { 216,000 }{ 2400\times 6 } )\quad =\quad 15\quad hours$$

Problem 7) A can do a piece of work in 10 days & B can do in 12 days. Both simultaneously start working.

Quantity A:         Time taken to complete the work, if B leaves 2 days before the actual completion of work.

Quantity B:         Time taken to completes the work, if B leaves 2 days before the scheduled completion of the work.

a) Quantity A is greater

b) Quantity B is greater

c) The two quantities are equal

d) The relationship cannot be determined from the information given.

Sol.   A can do a piece of work in 10 days & B can do in 12 days.

If B leaves 2 days before the actual completion of the work: In this case, the actual completion of the work is after 2 days of B’s leaving. This means that A has worked alone for the last 2 days to complete the work.

Rate at which A works =  $$(\frac { 1 }{ 10 } )$$/day

Rate at which B works =  $$(\frac { 1 }{ 12 } )$$/day

Work done by A in last two days = $${ (\frac { 2 }{ 10 } ) }^{ th }$$ of work

A & B work together at a rate of

$$(\frac { 1 }{ 10 } +\frac { 1 }{ 12 } )/day\\ =\quad (\frac { 11 }{ 60 } )/day$$

Work done by combined by A & B

$$(1-\frac { 2 }{ 10 } )\\ =\quad { (\frac { 8 }{ 10 } ) }^{ th }$$ of work

Rate × Time = Work

$$(\frac { 11 }{ 60 } )\times time\quad =\quad (\frac { 8 }{ 10 } )\\ time\quad =\quad (\frac { 8 }{ 10 } \times \frac { 60 }{ 11 } )\\ =\quad (\frac { 48 }{ 11 } )days\quad =\quad 4.3636\quad days$$

Total time taken to complete the work = Quantity A = 6.3636 days

Now let’ s calculate time taken to complete the work if B leaves 2 days before the scheduled completion of work.

A & B work together at a rate of $$\quad (\frac { 11 }{ 60 } )$$/days

Rate × time = work

$$(\frac { 11 }{ 60 } )\times time\quad =\quad 1\\ time\quad =\quad (\frac { 60 }{ 11 } )\quad days\quad =\quad 5.45\quad days$$

A & B work together for $$(\frac { 60 }{ 11 } -2)\quad days\quad =\quad (\frac { }{ } )\quad days$$

Let’s use equation:

Rate × time = work

$$\frac { 11 }{ 60 } \times \frac { 38 }{ 11 } ={ (\frac { 38 }{ 60 } ) }^{ th }of\quad work$$

Work done by A alone =

$$(1-\frac { 38 }{ 60 } )\quad =\quad { (\frac { 22 }{ 60 } ) }^{ th }of\quad work$$

A work at a rate of $${ (\frac { 1 }{ 10 } ) }^{ th }of\quad work/day\\ (\frac { 1 }{ 10 } )\times time\quad =\quad \frac { 22 }{ 60 } \\ time\quad =\quad (\frac { 22 }{ 6 } )\quad days\\$$

Total time taken to complete the project =

$$=\quad (\frac { 22 }{ 6 } +\frac { 38 }{ 11 } )\\ =\quad 7.12\quad days\\$$

Therefore, Quantity B is greater.

Problem 8) Phil collects virtual gold in an online computer game & then sells the virtual gold for real dollars. After playing 10 hours a day for 6 days, he collected 540,000 gold pieces. If he immediately sold this virtual gold at a rate of $1 per 1000 gold pieces, what were his average earnings per hour in real dollars? Sol. After playing 60 hour, he collected 540,000 gold pieces. Rate × time = work Rate × 60 = 540,000 Rate = $$(\frac { 540,000 }{ 60 } )\\$$ gold pieces /hour = 9000 gold piece /hour He sold his virtual gold at a rate of$1 per 1000 gold pieces.

1 gold pieces = $$(\frac { 1 }{ 1000 } )\\$$ dollars

His average earning per hour = $$(\frac { 9000 }{ 1000 } )\\$$ dollars/hour = 9 dollars/ hour.

## GRE Percentage

Percentage. We utter this word daily in our real life. You will absolutely get one or more GRE percentage problem in your exam. GRE percentage problems are slightly more tricky than our real life problems. By going through our chapter you will learn how to attack on a GRE percentage problem.

The concept of percentage mainly applies to ratios, & the percentage value of ratio is arrived at by multiplying by 100 the decimal value of the ratio.

For example, a student scores 20 marks out of a maximum possible 30 marks. His marks can then  be denoted as 20  out of 30

= $$(\frac { 20 }{ 30 } )\quad or\quad (\frac { 20 }{ 30 } )\quad \times \quad 100%\quad =\quad 66.66%$$

# GRE PERCENTAGE PROBLEMS

Problem 1)   A hostess at an art gallery makes $100 for each exhibit that she works. She also receives $$2\frac { 1 }{ 2 } %$$ of the art sales. If she earned$900 for a single exhibit, how much were the art sales?

Sol.  Money earned by hostess of the art sales = $800 $$(Art\quad Sales)\quad (\frac { 100 }{ 10000 } \times \frac { 5 }{ 2 } )\quad =\quad 800\\ (\frac { 1 }{ 100 } )(Art\quad Sales)(\frac { 5 }{ 2 } )\quad =\quad 800\\ Art\quad Sales\quad =\quad (800\quad \times \quad 100)(\frac { 2 }{ 5 } )\quad =\quad (800)(20)(2)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad (800)(40)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 32000/-$$ Problem 2) Gasoline at a certain station has increased from$2.98 to $3.07. If it then decreases by half the percent of the percent increase, what is the new price? Sol. Gasoline increased from$2.98 to $3.67. Let’s percentage increase be x. $$2.98(1+\frac { x }{ 100 } )\quad =\quad 3.07\\ \quad \quad \quad \quad \quad \frac { x }{ 100 } =\quad (\frac { 3.07 }{ 2.98 } -1)\\ \quad \quad \quad \quad \quad \quad \quad x\quad =\quad (\frac { 3.07 }{ 2.98 } -1)(100)$$ New Price $$=\quad (1-\frac { 50 }{ 100 } (\frac { 3.07 }{ 2.98 } -1))\\ =\quad 3.07(1-0.015)\\ =\quad { (3.02395) }^{ Ans }$$ Problem 3) Which of the following is the largest number a) 20% of 200 b) 7% of 500 c) 1300% of 3 d) 700% of 9 Sol. 700% of 9 = 63 is the highest number. (d) is correct. Problem 4) What is 20% of 50% of 75% of 70? Sol. $$(\frac { 20 }{ 100 } )(\frac { 50 }{ 100 } )(\frac { 75 }{ 100 } \times 70)\\ =\quad \frac { 1 }{ 5 } \times \frac { 1 }{ 2 } \times \frac { 3 }{ 4 } \times 70\\ =\quad { 5.25 }^{ Ans }$$ Problem 5) The length, breadth and height of a room in the shape of a cuboid are increased by 10%, 20% and 50% respectively. Find the percentage change in the volume of the cuboid? Sol. $$lbh\quad =\quad V\\ l(\frac { 110 }{ 100 } )b(\frac { 120 }{ 100 } )h(\frac { 150 }{ 100 } )\quad =\quad V(1+\frac { x }{ 100 } )\\ (100\quad +\quad x)\quad =\quad \frac { (110)(120)(150) }{ 10000 } \quad =\quad 198\\ x\quad =\quad 98%\quad$$ Problem 6) The salary if A it is 30% more than of Varun. Find by what percentage is the salary of Varun less than that of Amit? a) 12% b) 23.07% c) 21.23% d) 27.27% Sol. Let Salary of Varun is x% less than that of Amit. $$Salary\quad of\quad Amit\quad =\quad (Salary\quad of\quad Varun)(\frac { 130 }{ 100 } )\\ Salary\quad of\quad Varun\quad =\quad (Salary\quad of\quad Amit)(1-\frac { x }{ 100 } )\\ (1-\frac { x }{ 100 } )\quad =\quad (\frac { 100 }{ 130 } )\\ 1\quad -\quad \frac { 100 }{ 130 } \quad =\quad \frac { x }{ 100 } \\ x\quad =\quad 100(\frac { 30 }{ 130 } )\quad =\quad 23.07%$$ (b) is correct. Problem 7) In the recent, climate conference in New York, out of 700 men, 500 women, 800 children present inside the building premises, 20% of the men, 40% of the women & 10% of the children were Indians. Find the percentage of people who were not indian. a) 73% b) 77% c) 79% d) 83% Sol. Percentage = $$(\frac { 700(\frac { 80 }{ 100 } )\quad +\quad 500(\frac { 60 }{ 100 } )\quad +\quad 800(\frac { 90 }{ 100 } ) }{ 700\quad +\quad 500\quad +\quad 800 } )\times 100\\ =(\frac { 560\quad +\quad 300\quad +\quad 720 }{ 2000 } )\times 100\\ =79%$$ (c) is correct. Problem 8) Out of the total production of iron from hematite. An ore of iron, 20% of the ore gets wasted, & out of the remaining ore, only 25% is the pure iron. If the pure iron obtained in a year from a mine of hematite was 80,000kg, then the quantity of hematite mined from that mine in the year is? a) 5,000 kg b) 4,00,000 kg c) 4,50,000 kg d) None of these Sol. Let Quantity of hematite mixed from that mine in the year is x. $$x(\frac { 80 }{ 100 } )(\frac { 25 }{ 100 } )\quad =\quad 80,000\\ x\quad =\quad (\frac { 80,000\times 10000 }{ 80\times 25 } )\\ =\quad 4,00,000\quad Kg$$ (b) is correct. Problem 9) A man buys a truck for ¥2,50,000. The annual repair cost comes to 2% of the price of the purchase. Besides, he has to pay an annual tax of ¥2,000. At what monthly rent must be rent out the truck to get a return of 15% on his net investment of the first year? a) 3,350 b) 2,500 c) 4,000 d) 3,212.50 Sol. Monthly rent = $$(2,50,000(1\quad +\quad \frac { 2 }{ 100 } )\quad +\quad 2000)\times (\frac { 15 }{ 100 } )\times (\frac { 1 }{ 12 } )\\ =\quad (2,50,000(\frac { 102 }{ 100 } )\quad +\quad 2000)\times (\frac { 15 }{ 1200 } )\\ =\quad (3212.5)$$ (d) is correct. Problem 10) $${ (\frac { 4 }{ 5 } ) }^{ th }$$ of the voters in Bellary promised to vote for Sonia & the rest promised to vote for Sushi. Of these voters, 10% of the voters who had promised to vote for Sonia, didn’t vote on the election day, while 20% of the voters who had promised to vote for Sushi didn’t vote on the election day. What is the total number of voters polled if Sonia got 216 votes? a) 200 b) 300 c) 264 d) 100 Sol. Sonia got 216 votes. Let voters be x. $$216\quad =\quad (\frac { 4 }{ 5 } )x(\frac { 90 }{ 100 } )\\ x\quad =\quad (\frac { 216\times 5\times 100 }{ 4\times 90 } )$$ Total number of votes polled = $$=\quad x\quad -\quad x(\frac { 4 }{ 5 } )(\frac { 10 }{ 100 } )\quad -\quad x(\frac { 1 }{ 5 } )(\frac { 20 }{ 100 } )\\ =\quad (\frac { 216 }{ 4\times 90 } )(500-40-20)\\ =\quad (\frac { 216 }{ 4\times 90 } )\times (440)\\ =\quad (\frac { 216\times 44 }{ 4\times 9 } )\\ =\quad (\frac { 216\times (11) }{ 9 } )\\ =\quad 264$$ (c) is correct. Problem 11) Ravana spends 30% of his salary on house rent, 30% of the rest he spends on his children’s education & 24% of the total salary he spends on clothes. After his expenditure, he is lift with ₹2500. What is Ravana’s salary? a) 11,494.25 b) 20,000 c) 10,000 d) 15,000 Sol. Let salary be x. $$2500\quad =\quad x(1-\frac { 30 }{ 100 } -\frac { 70 }{ 100 } (\frac { 30 }{ 100 } )-\frac { 24 }{ 100 } )\\ 2500\quad =\quad x(1-\frac { 30 }{ 100 } -\frac { 21 }{ 100 } -\frac { 24 }{ 100 } )\\ 2500\quad =\quad x(1-\frac { 75 }{ 100 } )\\ 2500\quad =\quad x(\frac { 25 }{ 100 } )\\ x\quad =\quad 10,000$$ (c) is correct. Problem 12) The entrance ticket at the Minerva theatre in London is worth £ 250. When the price of the ticket was lowered, the sale of tickets increased by 50% while the collection recorded a decrease of 17.5% . Find the deduction in the ticket price. a) 150 b) 112.5 c) 105 d) 120 Sol. Let number of ticket sold previously be x. Number of ticket sold now be y. $$y\quad =\quad x(\frac { 150 }{ 100 } )$$ Let new price be P. $$250(x)(1-\frac { 17.5 }{ 100 } )\quad =\quad yp\quad =\quad (\frac { 150 }{ 100 } )xp\\ (250)(1-\frac { 17.5 }{ 100 } )\quad =\quad (\frac { 150 }{ 100 } )p\\ p\quad =\quad \frac { 25\times (825) }{ 150 }$$ Deduction = $$250\quad -\quad \frac { 25(825) }{ 150 } \\ =\quad 25(10\quad -\quad \frac { 825 }{ 150 } )\\ =\quad 112.5$$ (b) is correct. Problem 13) In 1970, company X had 2,000 employees, 15% of whom were women & 10% of these women were executives, what was the percent increase in the number of women executives from 1970 to 2012? Sol. No. of women executives in 1970 = $$=\quad (\frac { 15 }{ 100 } \times 2000)\quad \times \quad (\frac { 10 }{ 100 } )\\ =\quad 30$$ No. of women executives in 2012 = $$=\quad (12000)\times ()\times ()\\ =\quad 12\times 45\times 4\\ =\quad 12\times 180\\ =\quad 2160$$ Percent increase in the number of women executive from 1970 to 2012 = $$=\quad (\frac { Difference }{ Original } )\times 100\\ =\quad (\frac { 2130 }{ 30 } \times 700)\\ =\quad 7100%$$ Problem 14) Raymond borrowed$450 at 0% interest. If he pays back 0.5% of the total amt. every 7 days, beginning exactly 7 days after the loan was disbursed & has thus far paid back $18, with the most recent payment made today. How many days ago did he borrow the money? Sol. 0.5% of the total = $$(\frac { 5 }{ 1000 } \times 450)\\ =\quad \frac { 45 }{ 20 } \\ =\quad 2.25$$ 2.25$ × 8 = 18$8 × 7 = 56 He borrowed money 56 days ago. Problem 15) If Ken’s salary were 20% higher, it would be 20% less than Lorena’s. If Lorena’s salary is$60,000. What is Ken’s salary?

Sol.  Let Ken’s salary be x

$$x\quad \times \quad (\frac { 120 }{ 100 } )\quad =\quad 60.000\quad \times \quad (\frac { 80 }{ 100 } )\\ x\quad =\quad 60000\quad \times \quad \frac { 80 }{ 100 } \quad \times \quad \frac { 100 }{ 120 } \\ =\quad 60,000\quad \times \quad \frac { 4 }{ 6 } \\ =\quad 40,000$$

Ken’ s Salary is  40,000. Problem 16) Aloysius spends 50% of his income on rent, utilities & insurance & 20% on food. If he spends 30% of the remainder on video games & has no other expenditures. What percent of his income is lift after all of the expenditures? Sol. After spending 70% money on rent, utilities, insurance & food. He spends 30% of his remaining 30% salary on video games. After all expenditure 70% of his 30% salary remains with him. Let x% of his income left after all of the expenditures. $$(\frac { 70 }{ 100 } )\times (\frac { 30 }{ 100 } \times Salary)\quad =\quad \frac { x }{ 100 } \times Salary\\ =\quad 21%$$ Problem 17) Roselba’s annual income exceeds twice Jane’s annual income & both pay the same positive percent of their respective incomes in transportation fees. Quantity A: The annual amt. Jane pays in transportation fees. Quantity B: Half the annual amt. Roselba pays in transportation fees. a) Quantity A is greater b) Quantity B is greater c) The two quantities are equal d) The relationship cannot be determined from the information given. Sol. The correct answer is “Quantity B is greater”. Roselba’s income is more than twice as great as Jane’s income. If both pay the same percent of income in transportation fees, that means Roselba must pay more than twice as much as Jane in transportation fees. Quantity B is greater. Problem 18) At the end of April, the price of fuel was 40% greater than the price at the beginning of the month. At the end of May, the price of fuel was 30% greater than of the price at the end of April. Quantity A: The price increase in April. Quantity B: The price increase in May. a) Quantity A is greater. b) Quantity B is greater. c) The two quantities are equal. d) The relationship cannot be determined from the information given. Sol. Let price of fuel at the starting of April be x. Price increase in April = 0.40x Price of fuel increases to 1.40x Price increase in May = $$=\quad (1.40x)\times (\frac { 30 }{ 100 } )\\ =\quad 0.42x$$ “Quantity B is greater” price increase in May is greater than price increase in April. Problem 19) 75% of all the boys & 48% of all the girls at Smith high school take civics. If there are 20% flower boys then there are girls in the school. What percent of all the students take civics? Sol. No. of boys took civics = $$(\frac { 75 }{ 100 } \times (Total\quad number\quad of\quad boys))$$ No. of girls took civics = $$\frac { 48 }{ 100 } \times (Total\quad No.\quad of\quad girls)$$ No. of boys = $$(\frac { 80 }{ 100 } \times (No.\quad of\quad girls))$$ Percent of all the students who took civics = $$(\frac { \frac { 75 }{ 100 } \times (Total\quad no.\quad of\quad boys)\quad +\quad \frac { 48 }{ 100 } \times (Total\quad no.\quad of\quad girls) }{ Total\quad no.\quad of\quad boys\quad +\quad Total\quad no.\quad of\quad girls } )\times 100\\ =\quad (\frac { \frac { 75 }{ 100 } \times (\frac { 80 }{ 100 } \times Total\quad no.\quad of\quad girls)\quad +\quad \frac { 48 }{ 100 } \times (Total\quad no.\quad of\quad girls) }{ \frac { 80 }{ 100 } \times Total\quad no.\quad of\quad girls\quad +\quad Total\quad no.\quad of\quad girls } )\times 100\\ =\quad (\frac { \frac { 48 }{ 100 } +(\frac { 75 }{ 100 } \times \frac { 80 }{ 100 } ) }{ 1\quad +\quad \frac { 80 }{ 100 } } )\times 100\\ =\quad (\frac { \frac { 48 }{ 100 } \quad +\quad (\frac { 60 }{ 100 } ) }{ \frac { 180 }{ 100 } } )\times 100\\ =\quad (\frac { 108 }{ 180 } \times 100)\\ =\quad 60%\\ =\quad ()$$ 60% of all students took civics. ## GRE Exponents We are absolutely sure that you will get problem on GRE exponents on test day. In our chapters on GRE exponents we have covered from each and every property to difficult problems on GRE exponents. If you go through our article carefully problems can’ t trick you if they meant to trick you. Problem 1) $${ (\frac { 3x }{ 4 } ) }^{ 2 }\quad +\quad { (\frac { { y }^{ 2 } }{ 2 } ) }^{ 4 }$$ a) $$(\frac { 9{ x }^{ 2 }+{ y }^{ 8 } }{ 4 } )$$ b) $$(\frac { 9{ x }^{ 2 }+{ y }^{ 8 } }{ 16 } )$$ c) $$(\frac { 9{ x }^{ 2 }+{ y }^{ 4 } }{ 16 } )$$ d) $$(\frac { 3{ x }^{ 2 }+{ y }^{ 8 } }{ 16 } )$$ Sol. $$\frac { 9{ x }^{ 2 } }{ 16 } +\frac { { y }^{ 8 } }{ 16 } \\ =(\frac { 9{ x }^{ 2 }+{ y }^{ 8 } }{ 16 } )$$ (b) is correct. Problem 2) $$(\frac { { x }^{ 4 }{ y }^{ -2 } }{ { x }^{ -3 }{ y }^{ 4 } } )$$ a) $$\frac { { x }^{ 7 } }{ { y }^{ 4 } }$$ b) $$\frac { { x }^{ 7 } }{ { y }^{ 6 } }$$ c) $$\frac { { x }^{ 3 } }{ { y }^{ 6 } }$$ d) $$\frac { { x }^{ 4 } }{ { y }^{ 6 } }$$ Sol. $$(\frac { { x }^{ 4 }{ y }^{ -2 } }{ { { x }^{ -3 }{ y }^{ 4 } } } )\\ =\quad (\frac { { x }^{ 4 }{ x }^{ 3 } }{ { y }^{ 6 } } )\\ =\quad (\frac { { x }^{ 7 } }{ { y }^{ 6 } } )$$ (b) is correct. Problem 3) $$(\frac { { r }^{ 0 }{ s }^{ 4 } }{ t } )\div { (\frac { 3s }{ t } ) }^{ 2 }$$ a) $$(\frac { 3{ s }^{ 2 }t }{ 9 } )$$ b) $$(\frac { { s }^{ 2 }t }{ 9 } )$$ c) $$(\frac { { s }^{ 6 } }{ { t }^{ 3 } } )$$ d) $$(\frac { { 3s }^{ 6 } }{ { t }^{ 3 } } )$$ Sol. $$(\frac { { r }^{ 0 }{ s }^{ 4 } }{ t } )\div (\frac { 9{ s }^{ 2 } }{ { t }^{ 2 } } )\\ =(\frac { { s }^{ 4 } }{ t } )(\frac { { t }^{ 2 } }{ 9{ s }^{ 2 } } )\\ =(\frac { 1 }{ 9 } )({ s }^{ 2 }t)$$ (b) is correct. Problem 4) $${ (\frac { -3 }{ { x }^{ 2 } } ) }^{ 3 }{ (6xy) }^{ 0 }(\frac { { x }^{ 5 } }{ 9 } )$$ a) $$-3{ x }^{ 3 }$$ b) $$\frac { 3 }{ x }$$ c) $$\frac { -3 }{ x }$$ d) $$3{ x }^{ 3 }$$ Sol. $$(\frac { -27 }{ { x }^{ 6 } } )(\frac { { x }^{ 5 } }{ 9 } )\\ =\quad (\frac { -3{ x }^{ 3 } }{ { x }^{ 4 } } )\\ =\quad (\frac { -3 }{ x } )$$ (c) is correct. Problem 5) 80 is divisible by $${ 2 }^{ x }$$ Quantity A: x Quantity B: 3 a) Quantity A is greater b) Quantity B is greater c) The two quantities are equal d) The relationship cannot be determined from the information given. Sol. 80 is divisible by $${ 2 }^{ 4 }$$ & therefore also by $${ 2 }^{ 3 }{ ,\quad 2 }^{ 2 }\quad { ,2 }^{ 1 }$$ and $${ 2 }^{ 0 }$$. Thus x could be 0, 1, 2, 3 or 4 & could therefore be less than, equal to or greater than 3. Thus relationship cannot be determined. Problem 6) For which of the following positive integers is the square of the integer divided by the cube root of the same integer equal to nine times that integer? a) 4 b) 8 c)16 d) 27 e) 125 Sol. The correct answer is 27. Let integer be a $${ \frac { { a }^{ 2 } }{ { a }^{ \frac { 1 }{ 3 } } } \quad =\quad 9a\\ \\ { a }^{ 2-\frac { 1 }{ 3 } -1 }\quad =\quad { 3 }^{ 2 } }\\ { a }^{ \frac { 2 }{ 3 } }\quad =\quad { 3 }^{ 2 }\\ a\quad =\quad { 3 }^{ 3 }\quad =\quad { 27 }^{ Ans }$$ (d) is correct. Problem 7) If $${ 2 }^{ k }-{ 2 }^{ k+1 }+{ 2 }^{ k-1 }\quad =\quad { 2 }^{ k }m$$, what is the value of m? a) -1 b) $$-\frac { 1 }{ 2 }$$ c) $$\frac { 1 }{ 2 }$$ d) 1 e) 2 Sol. $${ 2 }^{ k }-{ 2 }^{ k+1 }+{ 2 }^{ k-1 }={ 2 }^{ k }m\\ { 2 }^{ k }(1-2+\frac { 1 }{ 2 } )={ 2 }^{ k }m\\ m\quad =\quad (-1+\frac { 1 }{ 2 } )\quad =\quad -\frac { 1 }{ 2 }$$ (b) is correct. Problem 8) If the hash marks above are equally spaced, What is the value of p? a) $$\frac { 3 }{ 2 }$$ b) $$\frac { 3 }{ 2 }$$ c) $$\frac { 3 }{ 2 }$$ d) $$\frac { 3 }{ 2 }$$ e) $$\frac { 512 }{ 125 }$$ Sol. $$(\frac { 2 }{ 5 } )(4)\quad =\quad { p }^{ \frac { 1 }{ 3 } }$$ To determine the distance between Harsh marks, divide 2 (the distance from 0 to 2) by 5 (the number of strength the number line has been divided into). The result is $$\frac { 2 }{ 5 }$$. Therefore, $$(\frac { 2 }{ 5 } )(4)\quad =\quad { p }^{ \frac { 1 }{ 3 } }\\ (\frac { 8 }{ 5 } )\quad =\quad { p }^{ \frac { 1 }{ 3 } }\\ p\quad =\quad { (\frac { 8 }{ 5 } ) }^{ 3 }\quad =\quad (\frac { 512 }{ 125 } )$$ (e) is correct. Problem 9) What is the greatest prime factor of $${ 2 }^{ 99 }-{ 2 }^{ 96 }$$ ? Sol. $${ 2 }^{ 99 }-{ 2 }^{ 96 }={ 2 }^{ 96 }(8-1)={ 2 }^{ 96 }(7)\quad$$ Greatest prime factor = 7 Ans. Problem 10) Which of the following is equal to $$(\frac { { 10 }^{ -8 }{ 25 }^{ 7 }{ 2 }^{ 16 } }{ { 20 }^{ 6 }{ 8 }^{ -1 } } )\quad$$ ? a) $$(\frac { 1 }{ 5 } )\quad$$ b) $$(\frac { 1 }{ 2 } )\quad$$ c) 2 d) 5 e) 10 Sol. $$(\frac { { 10 }^{ -8 }{ 25 }^{ 7 }{ 2 }^{ 16 } }{ { 20 }^{ 6 }{ 8 }^{ -1 } } )\quad =\quad (\frac { { 5 }^{ -8 }{ 2 }^{ -8 }{ 5 }^{ 14 }{ 2 }^{ 16 } }{ { 2 }^{ 12 }{ 5 }^{ 6 }{ 2 }^{ -3 } } )\\ =\quad { 5 }^{ -8+14-6 }{ 2 }^{ 16-8+3-12 }\\ =\quad { 2 }^{ 19-20 }\\ =\quad { 2 }^{ -1 }\\ =\quad \frac { 1 }{ 2 }$$ (b) is correct. Problem 11) Solve $$\sqrt { 4+\sqrt { 131+\sqrt { 154+\sqrt { 225 } } } }$$ ? Sol. $$\sqrt { 4+\sqrt { 131+\sqrt { 154+\sqrt { 225 } } } } \\ =\quad \sqrt { 4+\sqrt { 131+\sqrt { 154+15 } } } \\ =\quad \sqrt { 4+\sqrt { 131+13 } } \\ =\quad \sqrt { 4+\sqrt { 144 } } \\ =\quad \sqrt { 4+12 } \\ =\quad { 4 }^{ Ans }$$ ## GRE Averages Understanging GRE averages is pretty easy and problems on GRE averages are pretty easy. Just bu going through our chapter on GRE averages you can solve each and every problem efficiently and effectively. The average of a number is a measure of the central tendency of a set of numbers. In other words, it is an estimate of where the Centre point of a set of number lies. The basic formula for the average of n numbers $${ x }_{ 1, }{ \quad x }_{ 2 },\quad { x }_{ 3—— }{ x }_{ n }$$ is $$Average\quad =\quad \frac { { x }_{ 1 }+{ x }_{ 2 }+{ x }_{ 3 }—-{ x }_{ n } }{ n } \\ \quad \quad \quad \quad \quad \quad \quad =\quad \frac { Sum\quad of\quad n\quad numbers\quad of\quad a\quad set }{ n }$$ ## Concept of weighted average: When we have two or more groups whose individual averages are known, then to find the combined average of all the elements of all the groups we use weighted average. Thus if we have k groups with average & having elements than the weighted average is given by the formula: $${ A }_{ w }\quad =\quad \frac { { A }_{ 1 }{ n }_{ 1 }+{ A }_{ 2 }{ n }_{ 2 }+{ A }_{ 3 }{ n }_{ 3 }——{ A }_{ k }{ n }_{ k } }{ { n }_{ 1 }+{ n }_{ 2 }+{ n }_{ 3 }———-{ n }_{ k } }$$ Problem 1) If the average of 5 consecutive integers is 15, what is the sum of the least & greatest of the 5 integers? Sol. Average of 5 consecutive integers is 15. Five consecutive integers are 13, 14, 15, 16, and 17. Sum of the least & greatest of the 5 integers = 30 Ans. Problem 2) The average of five numbers is 30. After one of the number is removed, the average arithmetic mean of the remaining numbers is 32. What number was removed? Sol. Average = $$\frac { Sum\quad of\quad Quantities }{ Number\quad of\quad Quantities }$$ $$30\quad =\quad \frac { Sum\quad Of\quad Quantities }{ 5 }$$ Sum of quantities = 150 After one number is removed, $$\frac { Sum\quad Of\quad Remaining\quad numbers }{ 4 } \quad =\quad 32$$ Sum of remaining number = 128 Removed number was 22. Ans. Problem 3) Two airplanes leave the same airport at the same time, one traveling west & the other traveling to east. Their average speeds different by 10 miles/hr. After 1.5 hours, they are 520 miles apart. What is the approximate average speed of each plane over the 1.5 hours? Round to the nearest tenth. Sol. Let average speed of one airplane be $$x$$ miles/hr. Let average speed of another airplane be $$x+10$$ miles/hr. $$1.5x\quad +\quad 1.5(x+10)\quad =\quad 520\\ 1.5(2x+10)\quad =\quad 520\\ 2x\quad +\quad 10\quad =\quad \frac { 520 }{ 1.5 } \\ x\quad =\quad 168.3$$ Average speed of one airplane is 168.3 miles/hr. Average speed of another airplane is 178.3miles/hr. Problem 4) If the average of 10 consecutive odd integers is 224, what is the least of these integers? Sol. Average of 10 consecutive odd integers is 224. Therefore, five consecutive odd integers are less than 224 & five consecutive integers are more than 224. 10 consecutive odd integers are:- 215, 217, 219, 221, 223, 225, 227, 229, 231 & 233 Least of these integers = 215 Ans. Problem 5) A man travels at 60 kmph on the journey from A to B & returns at 100 kmph. Find the average speed for the journey? Sol. Let distance be d. Time taken by man from A to B = $$\frac { d }{ 60 }$$ Time taken by man in returns from B to A = $$\frac { d }{ 100 }$$ Average speed of the journey = $$\frac { Total\quad distance }{ Total\quad time }$$ $$=\frac { 2d }{ \frac { d }{ 60 } +\frac { d }{ 100 } } \\ =\frac { 2d }{ d(\frac { 1 }{ 60 } +\frac { 1 }{ 100 } ) } \\ =\frac { 2d }{ d(\frac { 5+3 }{ 300 } ) } \\ =\frac { 2(300) }{ 8 } \\ =\frac { 300 }{ 4 } \\$$ Average speed of journey = 75 kmph Problem 6) A school has only 3 classes that contain 10, 20 & 30 student respectively. The pass percentages of these classes are 20%, 30% & 40% respectively. Find the pass percentage of the entire school. a) $$5\frac { 3 }{ 16 } \\$$ b) $$3\frac { 3 }{ 16 } \\$$ c) $$16\frac { 5 }{ 3 } \\$$ d) $$3\frac { 16 }{ 5 } \\$$ Sol. Total students = 60 Total students who passed the examination are $$=\frac { 20 }{ 100 } (10)\quad +\quad \frac { 30 }{ 100 } (20)\quad +\quad \frac { 40 }{ 100 } (30)\\ =\quad 2\quad +\quad 6\quad +\quad 12\\ =\quad 20$$ Pass percentage of the entire school = $$\frac { 20 }{ 60 } \times 100\quad =\quad 33.33%$$ Problem 7) Find the average of four numbers $$2\frac { 3 }{ 4 } ,\quad 5\frac { 1 }{ 3 } ,\quad 5\frac { 1 }{ 6 } ,\quad 8\frac { 1 }{ 2 }$$ ? Sol. Four numbers are $$\frac { 11 }{ 4 } ,\quad \frac { 16 }{ 3 } ,\quad \frac { 25 }{ 6 } ,\quad \frac { 17 }{ 2 }.$$ Average $$=\quad (\frac { \frac { 11(3)\quad +\quad 16(4)\quad +\quad 25(2)\quad +\quad 17(6) }{ 12 } }{ 4 } )\\ =\quad (\frac { 33\quad +\quad 64\quad +\quad 50\quad +\quad 102 }{ 48 } )\\ =\quad \frac { 249 }{ 48 } \\ =\quad \frac { 83 }{ 16 } \\ =\quad 5\frac { 3 }{ 16 }$$ Problem 8) Find the average increase rate if increase in the population in the first year is 30% & that in the second year is 40%? Sol. Let population be p. Population after the end of first year = $$p(\frac { 130 }{ 100 } )$$ Population after the end of second year = $$p(\frac { 130 }{ 100 } )(\frac { 140 }{ 100 } )$$ Let average increase rate be x. $$p(1\quad +\quad \frac { x }{ 100 } )\quad =\quad p(\frac { 130 }{ 100 } )(\frac { 140 }{ 100\\ } )\\ p(100\quad +\quad x)\quad =\quad p(13)(14)\\ x\quad =\quad 13\times 14\quad -\quad 100\quad =\quad 82%$$ Problem 9) A man covers half of his journey by train at 60 km/hr. half of the remainder by bus at 30 km/hr. & the rate by cycle at 10 km/hr. Find his average speed during the entire journey. a) 36 kmph b) 30 kmph c) 14 kmph d) 18 kmph Sol. Let total distance be d. Distance covered by train = $$\frac { d }{ 2 }$$ Distance covered by bus = $$\frac { d }{ 2 }$$ Distance covered by cycle = $$\frac { d }{ 2 }$$ Average speed during the entire journey = $$=\quad (\frac { d }{ \frac { d }{ 2(60) } +\frac { d }{ 4(30) } +\frac { d }{ 4(10) } } )\\ =\quad \frac { 1 }{ \frac { 1 }{ 4(30) } +\frac { 1 }{ 30(4) } +\frac { 1 }{ 4(10) } } \\ =\quad \frac { 4 }{ \frac { 1 }{ 30 } +\frac { 1 }{ 30 } +\frac { 1 }{ 10 } } \\ =\quad \frac { 4 }{ \frac { 2 }{ 30 } +\frac { 3 }{ 30 } } \\ =\quad \frac { 4(30) }{ 5 } \\ =\quad 4(6)\\ =\quad 24Kmph$$ Problem 10) The average of 71 results is 48. If the average of the 59 results is 46 & that of the last 11 is 52. Find the 60th result. Sol. Sum of Quantity= $$71\quad \times \quad 48$$ $$71\times 48\quad =\quad 59\times 46\quad +\quad 11\times 52\quad +\quad { (60 }^{ th }\quad result)\\ { (60 }^{ th }\quad result)\quad =\quad 71\times 48\quad -\quad 59\times 46\quad -\quad 11\times 52\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 3408-2714-572\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 122$$ Ans Problem 11) The average age of a group of 14 persons is 27 years & 9 months. Two persons, each person years old, left the group. What will be the average age of the remaining persons in the group? Sol. 9 months = 0.75 year Average age = 27.75 year $$Average\quad =\quad \frac { Sum\quad of\quad Quantities }{ No.\quad of\quad Quantities } \\ 27.75\quad =\quad \frac { Sum\quad of\quad Quantities }{ 14 } \\ Sum\quad of\quad Quantities\quad =\quad 27.75\times 14\\$$ Average age of remaining persons = $$(\frac { 27.75\times 14\quad -\quad 42(2) }{ 12 } )\\$$ = 25.375 years Problem 12) Find the average of f(x), g(x), h(x), d(x) at x=10. $$f(x)\quad =\quad { x }^{ 2 }+2\\ g(x)\quad =\quad 5{ x }^{ 2 }-3\\ h(x)\quad =\quad log({ x }^{ 2 })\\ d(x)\quad =\quad (\frac { 4 }{ 5 } )\quad { x }^{ 2 }\\$$ a) 170 b) 170.25 c) 70.25 d) 70 Sol. Average of f(x), g(x), h(x) and d(x) at x = 10 $$=(\frac { { x }^{ 2 }+2+5{ x }^{ 2 }-3+\log { { x }^{ 2 } } +(\frac { 4 }{ 5 } ){ x }^{ 2 } }{ 4 } )\\ =(\frac { { { x }^{ 2 } }(1+\frac { 4 }{ 5 } +5)-1+\log { { x }^{ 2 } } }{ 4 } )\\ =(\frac { 100(1+\frac { 4 }{ 5 } +5)-1+2 }{ 4 } )\\ =(\frac { 100(\frac { 25+4+5 }{ 5 } )+1 }{ 4 } )\\ =100(\frac { 34 }{ 20 } )\quad +\quad \frac { 1 }{ 4 } \\ =5(34)\quad +\quad 0.25\\ =170.25$$. Therefore, Average = 170.25 Problem 13) There are five boxes in a cargo hold the weight of the 1st box is 200kg & the weight of the 2nd box is 20% higher than the weight of the 3rd box; whose weight is 25% higher than the 1st box’s weight. The 4th box at 350kg is 30% lighter than the 5th box. Find the difference in the average weight of the four heaviest boxes & the four lightest boxes. Sol. 1st box = 200kg 2nd box = $$Third(\frac { 120 }{ 100 } )=200(\frac { 125 }{ 100 } )(\frac { 120 }{ 100 } )=300kg$$ 3rd box = $$200(\frac { 125 }{ 100 } )=250Kg$$ 4th box = 350kg 5th box = 500kg $$350=(\frac { 70 }{ 100 } )(Fifth\quad box)$$ Difference in the average weight of the four heaviest boxes & the four lightest boxes. $$=\frac { 1 }{ 4 } (500+350+250+300)-\frac { 1 }{ 4 } (200+250+300+350)\\ =\frac { 1 }{ 4 } (300)\\ =75Kg$$ Problem 14) With an average speed of 40kmph, a train reaches its destination in time. If it goes with an average speed of 35kmph, it is late by 15 minutes. The length of the total journey is a) 40km b) 70km c) 30km d) 80km Sol. Let distance be d & time taken be t. $$(\frac { d }{ 40 } )\quad =\quad t\\ (\frac { d }{ 35 } )\quad =\quad t\quad +\quad \frac { 15 }{ 60 } \\ d(\frac { 1 }{ 35 } -\frac { 1 }{ 40 } )\quad =\quad \frac { 1 }{ 4 } \\ \frac { d }{ 5 } (\frac { 1 }{ 56 } )\quad =\quad \frac { 1 }{ 4 } \\ d\quad =\quad \frac { 56\times 5 }{ 4 } \\ d\quad =\quad 70Km$$ ## GRE Text Completion Text Completion questions are designed to further test your ability to understand what you read. They will also test your vocabulary, particularly your ability to apply sophisticated vocabulary in context. You will be asked to read a short passage of 1-5 sentences. There will be 1-3 blanks in the passage where a crucial word is missing. You will need to fill in each blank using several options provided. All Text Completion questions will be in multiple choice format. Each passage will have 1-3 blanks, and you will be given several options to choose from to fill in the blanks. Generally, passages with only one blank will have five options to choose from, while those with 2-3 blanks will have only three options per blank. ### What does a Text Completion question look like? Text Completion questions will always be based off a short passage of one or more sentences. The passage will have 1-3 blanks. You will be asked to read the passage and get a feel of what words will best complete it using several options given. Below are examples of what Text Completion questions will look like on the GRE revised General Test. ### For each blank select one entry from the corresponding column of choices. Fill all blanks in the way that best completes the text. 1. Harry’ s ________ performance on the project both failed to impress his superiors and helped to lose the company an important client.  a) rapacious b) desultory c) indomitable d) arcane e) indefatigable 2. While far from the bane that some scholars have declared them to be, _________ versions of novels and essay do indeed excise essential elements; students would have to supplement their reading with ________sources to fully understand the intent of the original.  Blank I Blank II a) Annotated a) Complementary b) Abridged b) Complimentary c) Antedated c) Compelling 3. If the candidate’ s speech was intended to stir up _______ feelings, he must have been solely disappointed by the almost ________ effect it had on the audience. Although it was refreshing to see a politician _________ inflammatory taglines in favor of reasoned argument, his speech was simply too long to hold the audience’ s attention.  Blank I Blank II Blank III a) Rancorous a) Incendiary a) Eschew b) Partisan b) Noxious b) Preclude c) Indigenous c) Soporific c) Abhor ## Answer key 1 Desultory 2 (i) Abridged (ii) Complementary 3 (i) Partisan (ii) soporific (iii) eschew ## Text Completion Tips In the first place, the passages used for Text Completion questions will generally be much shorter than those used for Reading Comprehension questions. This means that it will take you less time to carefully read a passage for a Text Completion. Moreover, Text completion questions are much more focused, and to answer them effectively, you may need to grasp subtle aspects of sentence structure. When answering Text Completion questions, one of the key skills you will need is the ability to grasp the overall meaning of the passage even with certain words missing. One of the key strategy while answering text completion question is to read the text completely to grasp the overall meaning of the passage. After doing this you will have an idea which the easiest blank is. Starting with the easiest blank first, fill all the blanks with a word of your own before you look at the answer options. Don’ t try to plug-in the answer choices because most of the times answer choices are smartly constructed. After filling the blank with a word of your own look for the closest synonym of the word out of options. If you are unsure of what some of the options mean and cannot identify closest synonym, try to eliminate words with opposite meanings to your own word. Don’ t commit the mistake of filling the blank as you read. You have to read the entire passage and grasp the overall meaning of the passage first. After reading the text completely. Don’ t assume that you should always fill in the first blank first. Always fill the easiest blank with a word of your own first. For some questions, it is difficult to fill in one blank without first correctly filling in another blank. Often there are overt indications of which word will best complete the passage; either the word you need to complete the passage will be essentially defined in another clause, or the context will strongly suggest that you use a word with the opposite meaning to what has already been said. These contextual indicators are quite valuable, so keep a close eye out for them. Be especially alert for words like “although” and “despite”. Also keep an eye out for words like “therefore”, “moreover” and “thus” which suggest summation or additional support for points that have already been made. Always re-read the passage with the answer(s) you have selected and filled in to see whether the passage is grammatical and makes sense with the options you have chosen. Even if your choice seemed right, if it does not produce a coherent, grammatically correct passage, it is n’ t the correct answer. ## Academic structures to know for Text Completions In both text completion sections, there are certain phrases that may show up that can give the sentence a spin. He was _______, always giving to those in need. He was anything but ________, always giving to those in need. ### What exactly does “anything but” mean? We should know the meaning of “anything but” to attempt this question correctly and how these idiomatic phrases can be highly misleading if you’ re not paying close attention. Below are some of the most common phrases you can expect to see on the GRE. Keep an eye out for them when answering text completions, and be sure you feel comfortable with how they’ re used in sentences. ### Nothing but In most cases, the phrase nothing but means “only (something).” When we went to her house, she was nothing but kind, showering us with gifts. In his book critiques, Jones was nothing but fair, always judging an author on the merits of his or her latest novel, regardless of previous flops. ### Anything but In most cases, the phrase anything but means “not.” He was anything but ________, always giving to those in need. It’ s an expression that implies that he’ s many things, A, B, C and D…….but he’ s definitely not E. In this case, E would be the opposite of the second part of the sentence. A simple way to think about it is to replace anything but with not. As in, “he was (not) ________, always giving to those in need.” ### All but The phrase all but is identical to “.almost” It can also mean “everything except the ones mentioned” Contrast the two sentences below to see the differences in how the phrase is used. All but the most famous actors of our day will likely not be remembered fifty years from now. At the end of the marathon, Charles was all but dead; he stumbled across the finish line, mentioning something about his pet iguana. ### At once X and Y The phrase at once X and Y is a tricky structure! First off, X and Y are words or phrases that are opposite in meaning. Second, that’ s an “and” you see, and not an “or.” So this phrase is used to suggest an element of surprise because a person/thing has these opposing qualities. At once melodious and dissonant, Perkins symphony is full of beautiful melodies that are suddenly interrupted by a burst of clashing gongs and screeching sopranos. Melodious = X; dissonant = Y He was at once hysterically funny, making people roll on the floor in laughter, and overly serious as soon as the conversation turned to politics. Hysterically funny = X; overly serious = Y At once forward-thinking and traditionalist, the mayor’ s new plan will usher in unprecedented changes while using approaches that have shown enduring efficacy in the civic share. Forward-thinking = X; traditionalist = Y ### Nothing more than The phrase nothing more than is used to show that somebody is n’ t very good at something. The word that follows “than” should be a negative description. He is nothing more than a second-rate musician, busking at a bus stops, his friends are always happy to escape his warbling falsetto. Harry is nothing more than a seasoned Hollywood hack; his scripts are as numerous as they are contrived. ### All the more so If you want to add emphasis but need an entire phrase to do so, you can use all the more so. Quentin’ s sudden termination was shocking- all the more so because he helped build the company as many know it today. ### For all The phrase for all is another way of saying “despite.” For all his hard work, Michael was passed over for a promotion. For all their talk on purging the environment of toxins, the two brothers can’ t do without their hourly smoke break. ### If anything The phrase if anything means “if at all.” It’ s meant to suggest that somebody is disagreeing with something and wants to prove that the other case is actually true. Bob: It seems like this city is getting more dangerous every day. Steve: Actually, it does n’ t seem that much worse from when I first moved here. If anything, the crime rate has actually dropped since the city’ s population has almost doubled in the last ten years. ### As such The phrase as such can be confusing because it is often misinterpreted as “therefore.” However, as such must refer to something that came before it. Correct: The CEO walked around the office as though he was King Kong, walking over anyone who came in his way. As such, anyone who was n’ t upper management tried to avoid him. Incorrect: We missed our train to Brussels. As such, we will have to take another one. ### Not so much A as B The phrase not so much A as B implies that to describe a situation, B is a better word or phrase than A. The scholar was not so much insightful as he was patient: he would peruse texts far longer than any of his peers. He was not so much jealous as downright resentful of his sister’ s talents, believing that their parents had put little interest in his education. ### But for The phrase but for is just another way of saying “except for.” But for her eloquence, she had little aptitude as an attorney. His contribution to cinema has been mostly forgotten but for his Oscar-winning role. ### Save (for) Similarly, the phrase save (for) means the same as “except(for).” Watching TV was Mama’ s favorite activity, save for eating chocolate cream puffs. Randy did not consider any of the class ruffians friends, save for Donald, who once came to his defense in a playground scuffle. ### Stem from To stem from just means to “come from” or “be caused by.” His insecurity stems from his lack of friends in grade school. The current crises stem from the former administration’ s inability to rein in spending. ## GRE Text Completion ## Sample Questions Q 1) Ms. Llewellyn is known to gently ___________ students who do not do their homework, but because of her generally amiable demeanor, she refuses to punish anyone, and seldom even raises her voice.  a) Pillory b) Detest c) Malign d) Penalize e) chide Ans 1) The sentence describes what the teacher “gently” does to “students who do not do their homework”. Her demeanor is generally amiable to punish them, so the blank must contain something like lightly criticize or “chide”. Malign or Pillory means to criticize, it is impossible to gently “detest” or “pillory” them. “detest” means to loathe or hate. Demeanor means person’s behavior or outward manner. Seldom means “almost never”. Q 2) Far too ________ to consider a career in the political limelight, the unassuming aide contented herself with a career behind the scenes, ________ supporting the political heavyweights of her day.  a) Pillory b) Detest c) Malign d) Penalize e) chide Ans 2) “Diffident” means lacking confident. “Apathetic” means showing no emotion or having no feeling. “implicit” means suggested but not stated directly. It has second meaning also. “implicit” means something stated directly without any doubt. “skeptical” means having or expressing doubt about something (such as a claim or statement). “unassuming” means shy or not having or showing a desire to be noticed, praised etc. The initial “Far too” indicates that the first blank will oppose “limelight” (an old theatrical expression meaning to be in the spotlight) and agree with “unassuming” (Shy). “Diffident” (shy or reserved) is a good match and is also a clue for the second blank, which must match the first one; “quietly” is the best choice. She is supporting the political heavyweights of her day. Because of shyness, she does not want her to be in limelight. Focus is on her “shyness” and “diffidence”. Therefore, quietly is the best choice. “Implicitly”, “Skeptically” do not make sense. ## GRE Argument Essay # Analyze an Argument The Analyze an Argument task gives you 30 minutes to plan and write a critique of an argument presented in the form of a short passage. A critique of any other argument will receive a score of zero. To score well, you need to do three things. First, analyze the line of reasoning in the argument (which will always be faulty). Then, explain the logical flaws and assumptions that underlie that reasoning. Finally, you must discuss what the author could add in order to make the conclusion of the argument more logically sound. It is absolutely critical that you recognize that you are not being asked to present your own views on the subject matter of the argument. You are being asked only to discuss how well the author made his argument. # Argument Essay Ground Rules In the Official Guide, ETS lists eight possible sets of instructions that could accompany an Argument essay prompt. However, the eight of them are even less interesting than the six provided for the Issue essay! Not one of them demands anything that would n’ t be featured in any successful Argument essay on the given prompt. Here they are, from The official Guide to the GRE revised General Test. You would be given an argument followed by one of these: 1. Write a response in which you discuss what specific evidence is needed to evaluate the argument and explain how the evidence would weaken or strengthen the argument. 2. Write a response in which you examine the stated and/or unstated assumptions of the argument. Be sure to explain how the argument depends on these assumptions and what the implications are if the assumptions prove unwarranted. 3. Write a response in which you discuss what questions would need to be answered in order to decide whether the recommendation and the argument on which it is based are reasonable. Be sure to explain how the answers to these questions would help to evaluate the recommendation. 4. Write a response in which you discuss what questions would need to be answered in order to decide whether the advice and the argument on which it is based are reasonable. Be sure to explain how the answers to these questions would help to evaluate the advice. 5. Write a response in which you discuss what questions would need to be answered to decide whether the recommendation is likely to have the predicted result. Be sure to explain how the answers to these questions would help to evaluate the recommendation. 6. Write a response in which you discuss what questions would need to be answered in order to decide whether the prediction and the argument on which it is based are reasonable. Be sure to explain how the answers to these questions would help to evaluate the prediction. 7. Write a response in which you discuss one or more alternative explanation that could rival the proposed explanation and explain how your explanation(s) can plausibly account for the facts presented in the argument. 8. Write a response in which you discuss what questions would need to be addressed in order to decide whether the conclusion and the argument on which it is based are reasonable. Be sure to explain how the answers to the questions would help to evaluate the conclusion. A well-written essay in which you locate logical flaws in the argument and then explain how they could be fixed will likely score highly. That said, you should, of course, read the specific instructions and make sure that they are addressed, just to be on safe side. Once you have properly read the argument, it’ s time to brainstorm the flaws. How do you find them? Fortunately, most of the mistakes have been made before…… # Argument Essay: Flaws to Watch Out For The following is a list of common fallacies found in GRE arguments. Unjustified Assumption-The argument is based on a questionable assumption. That is, in order for the argument to be true, the author is depending on a premise that he or she did n’ t write down and has n’ t proven. Thus, the conclusion can’ t be validated unless the assumption(s) can be proved to be true. The Urban Apartment Towers complex has seen a number of police visits to the property recently, resulting in the police breaking up loud parties held by young residents and attended by other young people. These police visits and the reputation for loud parties are hurting Urban Apartment Towers’ reputation and ability to attract new residents. To reduce the number of police visits and improve profitability. Urban Apartment Towers plans to advertise its vacant apartments in a local publication for people age 50 and up. What is this argument assuming but not proving? That people age 50 and up are less likely to have loud parties or attract police visits. That does n’ t sound like a totally unreasonable assumption, but it is an assumption nevertheless, and it is the job of the arguer to prove it (and your job to point out that the arguer has n’ t done so). Perhaps older residents would attract visits of another type (e.g., healthcare personnel) that could also impact the reputation of the complex. Skill & Work– The argument assumes that people have the ability (skill) to do something or the motivation (will) do it, when this has not been proven to be the case. The recommendation that “Everyone should exercise two hours per day,” and “Children should be offered green vegetables three times daily” run into problems regarding the ability of people to exercise that much (what about people who are already ill?) and the desire o children to eat the vegetables. The Urban Apartment Towers argument above also has both a “skill” problem and a “will” problem. Maybe over-50 people in the local area are largely on a fixed income and cannot afford to live in the Towers. And why would they want to? It’ s not clear that people over 50 have much motivation to live in an apartment complex where the police are always raiding loud parties. Extreme Language- The argument (usually the conclusion) uses language so extreme that the premises cannot justify the conclusion. People who jog more than 10 miles per week have a lower incidence of heart disease than people who exercise the same amount on stationary bicycles. Therefore, jogging is the best method of exercise for reducing heart disease. The conclusion is the final sentence: Jogging is the best method of exercise heart disease. The word “best” is quite extreme! The best method ever? Better than swimming, tennis, and a million other things? Proving that jogging is better than stationary bicycling (and there are some problems with that as well) just proves that jogging is “better” than one thing, not the “best.” Other extreme words to watch out for include the following: only, never, always, cannot, certainly Terms are Too vague- Just as you are on the lookout for language that is too extreme, you’ re also on the lookout for language that is too vague. The people who jog argument above has this problem. What on earth does it mean to “exercise the same amount” as someone who is jogging 10 miles? Does it mean biking for the same amount of time or the same distance? The same number of calories burned? Since it’ s much faster to ride 10 miles on a stationary bike than to jog 10 miles, if the arguer means that the distances are the same, then there’ s another reason (besides the author’ s conclusion) that the joggers have less heart disease: they are exercising more hours per week. Predicting the future– There’ s nothing wrong with trying to predict the future, of course; it’ s hard to run a government (or anything) without doing so. However, whenever an argument tries to predict the future, that’ s your opportunity to point out that the future could actually turn out some other way. Anyone who tries to predict the future is automatically introducing a level of uncertainty into his or her argument. The police chief in Rand city, a major urban metropolis, has proposed cutting down on speeding by doubling the fines levied on those who are caught. Speeding has been a major problem in Rand City, where over 5,000 tickets are issued each month. Of those who are issued tickets, over 95% mail in the fines, while less than 1% contest the charges in court, thus indicating the offender’ s admission of guilt. Doubling the fines for speeding will substantially reduce speeding in Rand City. The arguer is trying to predict the future: Doubling the fines will substantially reduce speeding. To find a weak link in this chain of events, ask yourself what could happen in between “the fines double” and “people speed less.” What else could happen? What about “the fines double” and then “people speed just the same but don’ t pay their tickets”? What if the fines are so low already (hence the lack of motivation to contest the charges) that doubling them won’ t make a difference? You can think of lots of ways that the first part of the conclusion could lead to something other than the second part of the conclusion. What’ s their Motivation? – whenever an argument is in the form of an advertisement or company announcement, you get to ask, “What’ s the Speaker’ s motivation?” Is the speaker trying to promote a medication, make a company look good, sell something, or get elected? The Police Chief in Rand City argument above potentially has this problem. What motivation does the police have in doubling traffic fines? Probably an honest desire to reduce speeding-but maybe a desire to increase the police budget by increasing what has historically been a reliable source of funding. The Troubled Analogy- There’ s nothing wrong with a good analogy, of course, but analogies in GRE arguments are never good. Every time you make an analogy, you’ re saying that something is like something else- except that it is n’ t exactly like that, or you’ d just be talking about the original topic. It’ s your job to find and exploit the dissimilarities. Bowbridge University, a prestigious institution with a long history of educating great scholars and national leaders, launched a distance learning program five years ago. Bowbridge students were very happy with the flexibility afforded to them by the program: for instance, they could continue studying with professors on the Bowbridge campus while conducting, research, traveling, or volunteering anywhere in the world. A study showed that the quality of education, as measured by students’ grades, did not decrease. Thus, if the tuition-free Local City College implements a distance learning program, student satisfaction will increase without compromising quality of education. Is Bowbridge University similar to Local City College? There are a lot of assumptions there. You’ re told that Bowbridge is prestigious, and that its students travel, volunteer, and conduct research around the world. They sound like a wealthy bunch! The students at the free Local City College? Probably not as wealthy. Maybe they don’ t even own computers. Do they need distance learning? It’ s not clear that someone who attends a “local” college would want – or have the means – attend that college from halfway around the world. In the end, you don’ t know that much about Local City College. It’ s not your job to prove that distance learning won’ t work there; it’ s your job to point out that the arguer has not established enough similarities to make a good analogy between the two institutions. Confusing Signs of a Thing for the Thing Itself- This effect is especially acute when people have an incentive (such as money) to over-report something, or an incentive (such as fear or laziness) to under-report something. For instance, reports of crimes such as littering and jaywalking are extremely low, but that does n’ t mean people are n’ t committing those crimes all the time. Reports of whiplash from car accidents tend to be highly inflated, since victims are often in a position to gain money from insurance companies. Reports of workplace harassment may be lower than actual incidents of harassment because workers fear losing their jobs or worsening the problem. Another common variation on this problem assumes that, because a law exists, people must be following it. A law is not the same as compliance with a law. Short Term vs. Long Term- Something that’ s good in the short term, under certain circumstances (antibiotics, for instance) may not be good for you in the long term. Similarly, something that’ s good or possible long term may not be good or possible short term. Sample Isn’ t Representative- If the GRE mentions a study, chances are that the sample is not representative. One in the argument pool refers to “French women in their eighties who were nursing-home residents.” Wow, what a very specific group! It’ s your job to point out that what works for French female octogenarians (A person who is between 80 and 89 years old) might not work for non-French people, men and people under 80. Sample Is Too Small- If a GRE argument mentions how many people were in a study, it’ s your job to say that the study should have been bigger or sample size should be bigger. No control group– A good study should have a control group- that is, a group of people who are as similar as possible in every way, and differ from the test group by only one variable. You can’ t just give people a new medicine and measure whether their condition improves; you have to get together a big enough group of people who meet certain conditions (such as having a particular illness at a particular stage), divide the people into two groups (balanced by gender, age and a host of other factors), and give the drug to only one group. It’ s important to make sure that the people receiving the drug do not just get better, but better than the other group. After all, what if it’ s the sort of illness that goes away on its own? Maybe some outside force (the changing seasons?) will cause improvement in both groups. It’ s your job to point out when a study lacks a control group, and what impact this might have on the study’ s findings. The Ever-Changing Pool- Most groups of people have a rotating cast of members. If a civic club voted in favor of something yesterday and against it 20 years ago, you would n’ t automatically conclude that people in the club changed their minds over time; it’ s pretty likely that the club includes different people than it did back then. Correlation Does Not Equal Causation- Just because two things are happening at the same time does n’ t mean one cause the other. Researchers have noted that cats that eat Premium Cat Food have healthier coats and less shedding. While Premium Cat Food costs more, the time saved cleaning up pet hair from furniture and rugs makes Premium Cat Food a wise choice. Two things are happening at the same time: cats are eating Premium food, and they are shedding less. Does that mean the food causes the reduced shedding? A third factor could be causing both A and B. In this case, perhaps a pet owner who is willing to pay for Premium Cat Food is also willing to pay for regular grooming, or for a dietary supplement that helps create healthy skin and fur, or… let your imagine run wild! Perhaps people who pay for Premium Cat Food are also more likely to own special breeds of cats that naturally shed less. This kind of setup-a third factor that could cause both A and B- is very common when as an argument makes a causation claim based solely on the fact that two things are correlated. Nothing in Quantified- Sometimes, you can get away with failing to attach numbers to things. Most people would be happy to be “healthier” or “richer,” even if you can’ t measure that exactly. The argument Premium Cat Food has this problem. “Healthier coats and less shedding” sounds like a nice enough benefit without needing to have numbers attached, but you run into problems with “the time saved cleaning up pet hair from furniture and rugs makes Premium Cat Food a wise choice.” Really? To validate this claim, you would need to know 1) how much more the cat food costs than the cat food the pet owner currently buys, 2) how much time the pet owner spends cleaning up cat hair, and 3) the monetary value of the pet owner’ s time. Of course, all of these factors vary from pet owner to pet owner to pet owner, so even if you could get all the facts and figures, it would certainly not be true that the premium food would be a “wise choice” for everyone. How Was It Before? A youth group applied for and received a permit to use the city park for a Culture Festival, which took place last weekend. On Wednesday, the Environmental Club, a group of local volunteers, visited the park and picked up 435 pieces of trash. The presence of such a quantity of rubbish signals a clear lack of respect for the park. Clearly, the youth group should be denied permits to use the park for any future events. Here, it is unclear whether the 435 pieces of trash were left by the youth group, or whether they were there beforehand. Who counts trash like that anyway? (At least they’ re quantifying). Just because two things happened in a certain order does n’ t mean one caused the other. Could some outside force be the cause? The 435 pieces of trash argument above has this problem. Maybe the trash was left by other groups that used the park (perhaps on Monday or Tuesday before the Environmental Club arrived?). There are many possible scenarios. Perhaps the trash was blown in by the wind. Alike Does n’ t Mean Identical – People who (or things that) are alike in some ways are undoubtedly different in others. Cetadone, a new therapy for the treatment of addiction to the illegal drug tarocaine, has been proven effective in a study centered around Regis Hospital in the western part of the state of New Portsmouth. The study involved local tarocaine addicts who responded to a newspaper and offering free treatment. Participants who received cetadone and counseling were 40% more likely to recover than were patients assigned to a control group, who received only counseling. Conventional therapies have only a 20% recovery rate. Therefore, the best way to reduce deaths from tarocaine overdose throughout all of New Portsmouth would be to fund cetadone therapy for all tarocaine addicts. Are tarocaine addicts in western New Portsmouth the same as tarocaine addicts in the rest of the state? Perhaps one area is rural and one is urban, or the demographics of different parts of the state vary. Furthermore, the addicts in this study seem pretty functional and motivated- they managed to successfully respond to a newspaper ad, and apparently were n’ t paid, so their motivation seems to have been to recover from addiction. Maybe the addicts who do well on cetadone are not the same addicts in danger of a fatal overdose. While drug addiction may seem to be a defining feature, the only thing that you can assume is uniform about tarocaine addicts is that they are addicted to tarocaine- anything else is up to the speaker to prove. Percents vs. Real Numbers (and Other Mathematical Confusion) – If David pays 28% of his income in taxes and Marie pays 33% of her income in taxes, who pays more money to the government? Without knowing how much the two people make, it’ s impossible to say. Don’ t confuse percents with actual numbers of dollars, people etc. In sum, if any numbers are presented in an Argument topic, see whether they are being cited in a logical way. Don’ t Forget to strengthen the Argument: Just Flip the Flaw Some sets of Argument essay instructions ask you to strengthen the argument. To discuss in your essay how the argument might be strengthened, just flip the flaw around. For instance: Nothing is quantified? This argument could be improved by quantifying X, Y, and Z . . . . . . Possible alternate causes? This argument could be improved by investigating and ruling out alternate causes such as . . . . . . Correlation does not equal causation? This argument could be improved by proving that X causes Y through a controlled study . . . . . No control group, non-representative sample, too-small sample? This argument could possibly be validated by a new study having the following qualities . . . . . . Brainstorming the Argument Essay Look back at the list of flaws and try to find several that apply to the following argument: Invoice Regulators, Inc. (IRC) can make your company more profitable. IRI examines our client firms’ outgoing invoices and vendor receipts to help clients recoup money owed and refunds due. One client, a family firm with a 100-year history, discovered75,000 worth of uncashed checks in an employee’ s desk drawer, and others have also made large gains. 80% of our client firms have experienced an increase in sales during the quarter our services were acquired. Hire IRI to improve your firm’ s profitability.

Did you make your own list of flaws? Jot some down before you keep reading.

Here’ s an example for this argument:

Correlation does not equal Causation: so, 80% of client firms had a sales increase around the time IRI was hired. So, what? Firms often have sales increases; one thing did not necessarily cause the other.

Alike does not Mean Identical/Unjustified Assumptions – The argument assumes that other businesses have outgoing invoices in the first place, and that, quite frankly, the business owners are a bit incompetent. It does not seem likely that the family firm with a 100-year history and a drawer full of forgotten money is representative of other companies. There are Small Sample/Unrepresentative Sample issues here as well.

Short Term vs. Long Term- The promise to make your company more profitable implies an ongoing financial improvement. The two cases cited seem temporary- the \$75,000 is a one-shot deal, and the increase in sales during the quarter makes no mention of some improved, systematic way to enhance ongoing profitability.

Terms Are Too Vague/Nothing Is Quantified- Other clients have made large gains. How large? Big enough to offset the cost (which was never mentioned) of IRI’ s services? What percent of clients experience the large gains?

What’ s Their Motivation? – obviously, this is an advertisement. But it does n’ t hurt to point out that IRI clearly has its own financial interests in mind here.

There is also another big problem that is n’ t named in this chapter, but is specific to this argument:

Confusing Sales with Profitability – Here, the argument confuses increased profitability (which is at least temporarily achievable by cashing a drawer full of checks or chasing refunds) with sales. Perhaps IRI costs more than the sales increase, and would thus hurt profitability.

Once you have identified the flaws, make a quick outline. (Don’ t use the names for the flaws- just write down what you’ re going to say.)

If you have more than four or so flaws to write about, you may wish to group any that are very similar, or simply omit the weakest.

You also want to put your ideas in a logical order so that your argument is persuasive and so that you can write nice transitions from one idea to the next.

Here is one sample outline:

the “checks in the drawer” client ≠ representative of other potential clients idiots!

even if not idiots, one biz is unsufficient evidence

not all biz even having outgoing invoices

claims of “other biz” are vague, nothing quantified

gains big enough to outweigh costs of IRI? (what ARE costs of IRI?)

claims of enhanced profitability even for existing clients are suspect

sales ≠ profitability

sales “in same quarter” – not even clear it’ s AFTER IRI

even if it were, correlation ≠ causation!

“profitability” implies ongoing

Make sure that you’ re not just throwing disconnected ideas on the page. Make sure that, in deconstructing a bad argument, you yourself are making a good argument.

Your outline should reflect a coherent argument that you’ have formed mentally before you begin to write:

• Just because hiring IRI has been profitable for some clients does n’ t mean it would be profitable for others.
• However, it’ s not even clear that IRI has been profitable for anyone, since we don’ t have any actual numbers to quantify most of the firms’ gains, and we don’ t know what it costs to hire IRI.
• The claims of profitability for existing clients are also suspect because IRI has confused sales with profitability, taking credit for something that is irrelevant to IRI’ s services and that possibly even began before IRI was hired.
• Profitability implies an ongoing financial improvement. IRI fails to define the period. A one-shot cash infusion is not the same thing as enhanced profitability.

The above statements are the parts of the argument each body paragraph will make. Note how the order seems “right”-its goes from arguing that IRI won’ t be profit-enhancing for everyone, to questioning whether it’ s profit enhancing for anyone at all. Statements II and III make the same point from different angles and clearly should come one after the other.

Note that there is no thesis written down. The thesis for an Argument essay will pretty much always be something like, “The argument rests on questionable assumptions, suffers from vaguely-defined terms, and contains numerous logical flaws that make it impossible to validate the conclusion.”

About Timing: On the real test, you should spend 2-3 minutes on the entire process of diagramming, brainstorming flaws, and organizing your thoughts into a coherent and persuasive outline. However, for now, it would be reasonable to take a bit more time (say, 5 or 6 minutes), knowing that with practice you’ ll get better and faster at spotting flaws.

# Argument Essay Outline

Introduction: In the intro, summarize the argument at hand and give your “take.” Do not repeat the argument; the grader is already very familiar with it.
You are then going to establish your “take” or thesis. It’s pretty much always going to say that the argument has some serious problems.

Body: Explain one main point in each of 2-4 paragraphs.
Each of the flaws you decided to write about should become the main point of a body paragraph. Or, if you decide to group more than one flaw into a paragraph, make sure that the two flaws are very closely and logically related; for instance, “the sample size is too small” and “the sample is not a representative” are good candidates to be grouped into a single paragraph. Generally, though, keep it to one main point per paragraph. GRE graders have given high scores to Argument essays that include as many as six body paragraphs- in such cases, many of the body paragraphs are quite short.

Arrange your main points in a logical way, and use transitions to segue from paragraph to paragraph. Transitions are usually located in the first sentence of a new body paragraph. For instance, if you have just written a paragraph about how a study’ s sample size was too small and not representative, you might begin the next paragraph with something like:

Not only should it be apparent that a study based on a sample of 80 Korean women is not necessarily applicable to humanity at large, it is also the case that, due to the lack of a control group, we are unable to evaluate the results of the study for even this extremely limited sample.This is a logical progression of ideas; the use of such transitions throughout an essay creates a sense of coherence and fluency.

Don’ t forget to improve the argument. There are (at least) three possible ways to arrange your argument to incorporate this component:

1. Each time you mention a flaw, follow up with how to fix it. The “improve the argument” component would therefore be part of each body paragraph.
2. Write two or more body paragraphs about the argument’ s flaws, and follow up with one body paragraph on how to fix those flaws.
3. Use the body paragraphs entirely to discuss the flaws, and save the discussion of how to fix those flaws for the conclusion. This may be the best plan for anyone frequently stuck for a conclusion. Keeping the “improve the argument” component brief is also a good way to keep from sounding repetitive.

Don’ t spend too much time making a single point or you will run out of time!

Conclusion: In the conclusion, resummarize your critique. The conclusion does not have to be lengthy: restate your thesis or main idea in different words, and state or restate what would need to be done to improve the argument. Ending with ideas for improvement give a nice, positive note at the end.
Ideally, the conclusion should sum things up while offering some special perspective or insight. In any case, try to avoid having your conclusion sound repetitive. If in doubt, keep it short.

## Style Points

Debate team persuasion tactics: It’ s possible to say something in a way that is not very persuasive, or in a way that is. Say you are trying to argue against the school superintendent’ s plan for year-round school:

1. The superintendent’ s has not proven that her plan will achieve the goal of improving academic performance. However, it may serve the function of reducing crime.
2. While the superintendent’ s plan may indeed reduce crime, she has not proven that her plan will achieve the stated goal of improving academic performance.

Which version sounds verse for the superintendent? The last one, right? If you have two opposing things to say, put the one that’ s on your side last. This makes the one that is not on your side seem less important. The order should be 1) Concession, then 2) Your assertion.
If you have a fairly weak point, use that point’ s weakness to your advantage to emphasize how strong your next point is. For instance, say you were only able to come up with three flaws for a particular argument, and one of them is pretty weak, but you can’ t toss it out because then you won’ t have enough to write about. Put the weakest point in the middle (if that won’ t disrupt the flow of the argument), and use it to underscore the final, biggest point.

Tone: There’ s no rule against saying “I,” but don’ t be too informal. Avoid conversational asides, and don’ t try to be funny. Keep the tone serious and academic. When you’ re referring to an argument and it’ s not clear who’ s talking, you can refer to that person as “the speaker.”

Varied direction: Throughout the essay, you will say the same thing several times. Don’ t use the exact same words. That is, paraphrase yourself. If in the introduction, you wrote:

While it is indisputable that a new train line would create some new jobs in Arrin City, the mayor’ s argument that the train line will improve the city’ s overall financial health is flawed due to a variety of counterfactors, including financial losses, that the mayor has neglected to take into account.

Then in your conclusion, you might write:

The mayor’ s contention that a new train line would improve the city’ s financial health is sadly misguided; while undoubtedly there would be some benefits, such as new jobs directly serving the train line, the financial benefit of those jobs would likely be dwarfed by other financial losses sustained in the wake of the train line’ s implementation.

Note that “undoubtedly” has been switched in for “indisputable,” and the three ideas in the sentence have been shuffled (“some new jobs, mayor is wrong, other factors” vs. “mayor is wrong, some new jobs, other factors”).

However, while you do want to avoid saying “indisputable” over and over when there are so many other good words (undeniable, unquestionable, irrefutable, incontrovertible, indubitable) you could use in its place, don’ t worry about repeating words such as “train” and “mayor.”
You must have learnt words in groups for GRE sentence equivalence questions. If you have learnt words in groups, you can use different words for varying diction. You can download android app on GRE vocabulary using word groups for learning words in groups. For downloading app, Just click here.

Varied sentence structure: Aim for a mix of long and short sentences. Throw in an occasional semicolon, hyphen, colon, or rhetorical question. For example:
Is it the case that sacrifice is the noblest of all virtues? Even a cursory analysis ought to indicate that it is not; the greatest of all virtues can hardly be said to be the one with, typically the least utilitarian value.
Make sure you know how to correctly use any punctuation you decide to include, of course.

Vocabulary: Use GRE-type words in your writing (but only if you are sure you can use them correctly). Some good vocab words to think about are those about arguments themselves, since those will work in nearly any essay. Some examples are:
Aver, extrapolate, contend, underpin, hypothesize, rebuttal, postulate, propound, concur
Transitions: A top-scoring essay has body paragraphs that lead logically into one other. You can create this chain of logic by arranging your examples or reasons in a progressive way, and by using transition phrases and similar signals. The simplest transitions involve phrases such as “” or “” A more sophisticated transition might take the form:

The obstacles towards international cooperation include not only [the stuff I discussed in my last paragraph], but also [the stuff I’ m about to discuss in this paragraph].
Transitions are usually located in the first sentence of a new body paragraph.

Finally, as a reminder: length on the GRE essay is highly correlated with scores. Write as much as you can in the time allotted. If you had a choice between painstakingly checking your spelling and writing another paragraph, it would probably be best to write another paragraph.

Trouble Getting started?
Remember, you are writing on a computer. If you “freeze” when trying to start your introduction, write something else first! Just pick whichever example seems easiest to write and dive in! You can certainly cut and paste as needed.
Just keep an eye on the clock and make sure you leave enough time for both an intro and a conclusion.

# Sample Essays

Sample Essay
Last month, National Issues ran an article about the decline- as measured by shrinking populations and the flight of young people-of small towns in Ganadia. Here in Lemmontown, a small resort town on the ocean, we are seeing just the opposite: citizens from the neighboring towns of Armontown and Gurdy City are moving here at a record rate. Furthermore, greater than ever number of high school graduates in Lemmontown are choosing to stay in Lemmontown, as the building of new hotels has created a significant number of jobs. All along the eastern seaboard are similar stories. Small towns in Ganadia are not in decline.

A letter to the editor of National issues magazine takes issue with the magazine’ s claim that small towns in Ganadia are declining. Of course, the magazines’ s contention was almost certainly that small towns, on average, are declining; a single counterexample does not disprove that claim. The arguments’ other flaws stem from the same central problem: Lemmontown is just one town, and not necessarily a very representative one.
The writer explains that Lemmontown is a resort town on the ocean. Resort towns depend on income flowing in from visitors, and the seaside (or whatever else visitors are there to see) is an asset that most towns do not have. These atypical resort assets are directly cited as the driver behind the jobs that are keeping young people in Lemmontown. Non-resort towns would not likely experience a similar effect. To set the argument on more sound footing, the writer would need to demonstrate that Lemmontown is typical of other Ganadian towns.
Of course, the writer does mention two other towns: Armontown and Gurdy City. While the writer means to cite those towns as evidence that Lemmontown is doing well, he or she inadvertently weakens the argument by giving two counterexamples: both Armontown and Gurdy City are losing residents, in accordance with the trend cited by National Issues. In fact, of the three towns the writer references, two of them are losing people. To strengthen the argument, the writer would have to prove that there are more Lemmontowns (so to speak) than Armontowns and Gurdy Cities, or that Armontown and Gurdy City are not small towns.
Finally, the writer points out that “all along the eastern seaboard are similar stories.” This assertion is vague. Are there enough stories of non-declining small towns to outweigh accounts of declining small towns? The claim lacks quantification. Also, the eastern seaboard is not necessarily representative of the rest of Ganadia. Perhaps the seaboard is full of thriving resort towns, but the bulk of Ganadia’ s small towns exist in the interior and on the west coast, where conditions are worse. To validate his or her claims, the writer would need to quantify the claim that eastern seaboard success stories are more numerous than accounts of small towns in decline.
The letter to the editor takes exception to a general claim by providing a specific exception. One anecdote does not make an argument. The argument as written fails to establish that Lemmontown’ s happy situation is representative of Ganadian towns at large.