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: Ratios and Proportions If general, the ratio of a number x to a number y is defined as the quotient of the number x & y. The numbers that form the ratio are called the terms of the ratio. The numerator of the ratio is called the antecedent & the denominator is called the consequent of the ratio. GRE: Ratios and Proportions Important properties: 1. If $$\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } =\frac { g }{ h } =k\quad$$ , then $$\frac { a }{ b } =\frac { c }{ d } =\frac { e }{ f } =\frac { g }{ h } =k=\frac { (a\quad +\quad c\quad +\quad e\quad +\quad g) }{ (b\quad +\quad d\quad +\quad f\quad +\quad h) }$$ 2. If $$\frac { { a }_{ 1 } }{ { b }_{ 1 } } ,\frac { { a }_{ 2 } }{ { b }_{ 2 } } ,\frac { { a }_{ 3 } }{ { b }_{ 3 } } ——-\frac { { a }_{ n } }{ { b }_{ n } }$$ are unequal fractions. Then the ratio: $$\frac { { a }_{ 1 }\quad +\quad { a }_{ 2 }\quad +\quad { a }_{ 3 }\quad ——-{ a }_{ n } }{ { b }_{ 1 }\quad +\quad { b }_{ 2 }\quad +\quad { b }_{ 3 }——–{ b }_{ n } }$$ lies between the lowest & highest of these percentage. GRE: Ratios and Proportions Problem 1) In a pet store, the ratio of the number of puppies to kittens is 4:7. When 7 more puppies more puppies are received, the ratio of the number of puppies to the number of the kittens charges to 5:7. How many pets does the pet store now have? Sol. Let the number of puppies be $$4x$$ & number of kitters be $$7x$$. $$\frac { 4x\quad +\quad 7 }{ 7x } =\quad \frac { 5 }{ 7 }$$ $$(4x\quad +\quad 7)(7)\quad =\quad (7x)(5)$$ $$28x\quad +\quad 49\quad =\quad 35x$$ $$7x$$ = 49 $$x$$ = 7 No. of puppies = $$4\times 7+7$$ = 35 No. of kittens = $$7\times 7$$ = 49 No. of pets = 35 + 49 = 84 Problem 2) Liquids A & B are in the ratio 2:1 in the 1st container, and 1:2 in the 2nd container. In what ratio should the contents of the two containers be mixed to obtain a mixture of A & B in the ratio 1:1? Sol. Let 1st container contains 1 litre of liquid. 1st container contains $$\frac { 2 }{ 3 }$$ litre of liquid A. 1st container contains $$\frac { 1 }{ 3 }$$ litre of liquid B. Let 2nd container contains 1 litre of liquid. second container contains $$\frac { 1 }{ 3 }$$ litre of liquid A. 2nd container contains $$\frac { 2 }{ 3 }$$ litre of liquid B. Let $$x$$ of liquid is taken from first container & $$1-x$$ of liquid is taken from 2nd container. New mixture contains $$\frac { 2 }{ 3 } x\quad +\quad \frac { 1 }{ 3 } (1-x)$$ litre of liquid A. New mixture contains $$\frac { 1 }{ 3 } x\quad +\quad \frac { 2 }{ 3 } (1-x)$$ litre of liquid B. $$\frac { 2 }{ 3 } x\quad +\quad \frac { 1 }{ 3 } (1-x)\quad =\quad \frac { 1 }{ 3 } x\quad +\quad \frac { 2 }{ 3 } (1-x)$$ $$x=\frac { 1 }{ 2 }$$ Litre $$\frac { 1 }{ 2 }$$ of liquid is taken from first container & $$\frac { 1 }{ 2 }$$ of liquid is taken from 2nd container. Ans=1:1 Problem 3) If $$\frac { 1 }{ 2 }$$ of the number of white roses in a garden is $$\frac { 1 }{ 8 }$$ of the total number of roses, and $$\frac { 1 }{ 3 }$$ of the number of red roses is $$\frac { 1 }{ 9 }$$ of the total number of roses, then what is the ratio of white roses to red roses? Sol. Let number of white roses be $$x$$. No. of red roses be $$y$$. Total roses be $$x + y$$. $$\frac { 1 }{ 2 } x\quad =\quad \frac { 1 }{ 8 } (x\quad +\quad y)$$ $$\frac { 1 }{ 3 } y\quad =\quad \frac { 1 }{ 9} (x\quad +\quad y)$$ $$4x\quad =\quad (x\quad +\quad y)$$ $$3y\quad =\quad (x\quad +\quad y)$$ $$4x\quad =\quad 3y$$ $$\frac { x }{ y } =\frac { 3 }{ 4 }$$ Ratio=3:4 Problem 4) The ratio between two numbers is 3:4 & and their L.C.M. is 180. Find the first number? Sol. Let 1st number be $$3x$$ & 2nd number be $$4x$$. L.C.M. of $$3x$$ and $$4x$$ = $$3\times 4\times x$$ = $$12x$$ $$12x$$ = 180 1st number = $$\frac { 180 }{ 12 } \times 3$$ = $$\frac { 180 }{ 4 }$$ = 45 Ans Problem 5) If various inversely as $${ y }^{ 2 }-1$$ & is equal to 24 when y=10. Find x when y=5? Sol. $$x\quad =\quad \frac { k }{ { y }^{ 2 }-1 }$$ ; (x=10, when y=10) $$24\quad =\quad \frac { k }{ 100-1 }$$ k = (99)(24) ; $$x\quad =\quad \frac { k }{ 25\quad -\quad 1 }$$ $$x\quad =\quad \frac { 99\times 24 }{ 24 }$$ x = 99 Problem 6) The total number of pupils in three classes of a school is 333. The number of pupils in class I & II are in the ratio 3:5 & those in classes II & III are in the ratio 7:11. Find the number of pupils in the class that had the highest number of pupils? a) 63 b) 105 c) 165 d) 180 Sol. Let number of pupils in classes I & II are 3x & 5x. Let number of students in class II & III are 7y & 11y. $$5x\quad =\quad 7y$$ $$y\quad =\quad \frac { 5x }{ 7 }$$ $$3x\quad +\quad 7y\quad +\quad 11y\quad =\quad 333$$ $$\frac { (3)(7y) }{ 5 } +7y+11y\quad =\quad 333$$ $$y\quad =\quad \frac { 333(5) }{ 111 }$$ $$y\quad =\quad 21$$ No. of student in class I = 21(3) = 63 No. of student in class II = 21(5) = 105 No. of student in class III = 33(5) = 165 Problem 7) If a:b=c:d, e:f=g:h, then (ae+bf)(ae-bf) = ? a) $$\frac { e+f }{ e-f }$$ b) $$\frac { cg\quad +\quad dh }{ cg\quad -\quad dh }$$ c) $$\frac { cg\quad -\quad dh }{ cg\quad +\quad dh }$$ d) $$\frac { e\quad -\quad f }{ e\quad +\quad f }$$ Sol. $$\frac { a }{ b } \quad =\quad \frac { c }{ d } ……..(1)$$ $$\frac { e }{ f } =\frac { g }{ h } ………….(2)$$ Multiply equation (1) & (2) $$\frac { a\quad \times \quad e }{ b\quad \times \quad f } \quad =\quad \frac { c\quad \times \quad g }{ d\quad \times \quad h }$$ $$\frac { ae }{ bf } \quad =\quad \frac { cg }{ dh }$$ By using componendo and dividendo $$\frac { ae\quad +\quad bf }{ ae\quad -\quad bf } \quad =\quad \frac { cg\quad +\quad dh }{ cg\quad -\quad dh }$$ Problem 8) A factory employs skilled workers, unskilled workers & clerks in the proportion 8:5:1 & the wages of a skilled worker, an unskilled workers & clerks are in the ratio 5:3:2. Total wages of all workers amount to Rs. 31,800. The wages paid to each category of workers are a) 24000, 6000, 1800 b) 15000, 10800, 6000 c) 20000, 9000, 2800 d) 25000, 5000, 1800 Sol. Let number of skilled workers, unskilled workers & clerks be $$8x,\quad 5x\quad \quad and\quad x$$. Let wages of a skilled workers, unskilled workers & clerks be $$5y,\quad 2y\quad \quad and\quad 3y$$. Total wages of skilled workers = $$8x\quad \times \quad 5y\quad =\quad 40xy$$ Total wages of unskilled workers = $$5x\quad \times \quad 2y\quad =\quad 10xy$$ Total wages of clerks = $$x(3y)\quad =\quad 3xy\quad$$ $$53xy\quad =\quad 31800\quad$$ $$xy\quad =\quad 600\quad$$ Total wages of skilled workers = 40(600) = 24000/- Total wages of unskilled workers = 10(600) = 6000/- Total wages of clerks = 3(xy) = 3(600) = 1800/- Problem 9) On his deathbed, Mr. Kalu called upon his three sons & told them to distribute all his assets worth Rs. 525,000 in the ratio of amongst themselves. Find the biggest share amongst the three portions. Sol. Let the share of three sons be $$\frac { x }{ 15 } ,\quad \frac { x }{ 21 } \quad and\quad \frac { x }{ 35 }$$. $$\frac { x }{ 15 } +\frac { x }{ 21 } +\frac { x }{ 35 } =525000\\ 7x\quad +\quad 5x\quad +\quad 3x\quad =\quad 525000\times 3\times 5\times 7\\ 15x\quad =\quad (525000)\quad \times \quad (15)\quad \times \quad (7)\\ x\quad =\quad (525000)\times (7)\\ Biggest\quad share\quad =\quad \frac { 525000\times 7 }{ 15 } \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad (7000)\times (35)\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad 245,000/-$$ Problem 10) The ratio of 16 to g is equal to the ratio of g to 49. Quantity A Quantity B g 28 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. $$\frac { 16 }{ g } =\frac { g }{ 49 } \\ { g }^{ 2\quad }=\quad 49\times (16)\\ g\quad =\quad \pm 28$$ “The relationship cannot be determined from the information given”. $${ g }^{ 2\quad }=\quad 16\quad \times \quad 49$$ Remember that when “unsquaring” an equation with a squared variable, you must account for the negative possibility. The value of g could be either positive 28 or negative 28. Problem 11) In a group of adults, the ratio of women to men is 5 to 6, while the ratio-handed people is 7 to 9. Everyone is either left or right-handed; no is both. Quantity A The number of women in the group. : Quantity B :The number of left-handed people in the group. 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 number of woman be $$5x$$ & number of men be $$6x$$. Let number of left-handed people be $$7y$$ & number of right-handed people be $$9y$$. Total number of people = $$5x\quad +\quad 6x\quad =\quad 9y\quad +\quad 7y\\ 11x\quad =\quad 16y\\ x\quad =\quad \frac { 16y }{ 11 } \\ y\quad =\quad \frac { 11x }{ 16 }$$ Number of woman in the group = $$5x$$ Number of left-handed people in the group = $$7\times \frac { 11x }{ 16 } \quad =\quad \frac { 77x }{ 16 } \\ 5\quad >\quad \frac { 77 }{ 16 }$$ Number of woman is the group are more than the number of left-handed people in the group. (a) is correct. Problem 12) A pantry holds x cars of bears, twice as many cars of soup & half as many cars of tomato paste as there are cars of bears. If there are no other cars in the pantry, which of the following could be the total number of cars in the pantry? Indicate two such numbers. a) 6 b) 7 c) 36 d) 45 e) 63 Sol. Let number of cans of soup be y.. Let number of cans of bears be x. $$y\quad =\quad 2x$$ Cans of tomato = $$\frac { x }{ 2 }$$ Cans of soup = $$2x$$ Total number of cars = $$x\quad +\quad 2x\quad +\quad \frac { x }{ 2 } \quad =\quad 3x\quad +\quad \frac { x }{ 2 } \quad =\quad \frac { 7x }{ 2 }$$ x must be an even integer. Total number of cars could be 7, 14, 21, 28, 35……..etc. 7 & 63 are correct. Problem 13) Party Cranberry is 3 parts cranberry juice & 1 party seltzer. Fancy Lemonade is 1 part lemon juice & 2 parts seltzer. An amount of Party Cranberry is mixed with an equal amount of Fancy Lemonade. Quantity A Quantity B: The fraction of the resulting The fraction of the mix that is cranberry juice. resulting mix that is seltzer. 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. Party Cranberry = 3(Cranberry Juice)+(seltzer) Fancy Lemonade = (Lemon Juice)+2(seltzer) For Party Cranberry, Cranberry: seltzer: whole = 3:1:4 For fancy Lemonade, Lemon: Seltzer: Whole = 1:2:3 Let 1 litre of party Cranberry & 1 litre of fancy lemonade was taken. Cranberry juice in final mixture = $$\frac { 3 }{ 4 } \quad =\quad \frac { 9 }{ 12 }$$ Seltzer in final mixture = $$\frac { 1 }{ 4 } +\frac { 2 }{ 3 } =\frac { 11 }{ 12 }$$ $$\frac { 11 }{ 12 } >\frac { 9 }{ 12 }$$ (Seltzer in final mixture) > (Cranberry juice in final mixture) Problem 14) In a parking lot, $$\frac { 1 }{ 3 }$$ of the vehicles are black & $$\frac { 1 }{ 5 }$$ of the remainder are white. How many vehicles could be parked on the lot? a) 8 b) 12 c) 20 d) 30 e) 35 Sol. Black vehicle = $$\frac { 1 }{ 3 }$$ (Total Vehicle) White vehicle = $$\frac { 2 }{ 3 } \times \frac { 1 }{ 5 }$$(Total Vehicle) = $$\frac { 2 }{ 15 }$$(Total Vehicle) Number of white cars must be countable with whole numbers. Number of total vehicles must be divisible by 15 of the answer choices, only 30 are divisible by 15. Problem 15) In model town, the ratio of school going children to non-school going children is 5:4, if in the next year 20% of school going children turned into non-school going children making it to 35,400. What is the new ratio of school going children to non-school going children? a) $$\frac { 4 }{ 5 }$$ b) $$\frac { 5 }{ 4 }$$ c) $$\frac { 3 }{ 4 }$$ d) $$\frac { 5 }{ 3 }$$ Sol. Number of school going children = $$5x$$ Number of non-school going children = $$4x$$ New number of school going children = $$5x(\frac { 80 }{ 100 } )\quad =\quad \frac { 400x }{ 100 }$$ New number of school going children = $$4x\quad +\quad 5x(\frac { 20 }{ 100 } )\quad =\quad x(\frac { 500 }{ 100 } )$$ New ratio = $$\frac { 4 }{ 5 }$$ Problem 16) In the famous island of Italy, there are four men for every three women & four children for every three me. How many children are there in the island if it has 531 women? Sol. $$\frac { Men }{ Women } =\frac { 4 }{ 3 }$$ $$\frac { Children }{ Men } =\frac { 5 }{ 3 }$$ $$\frac { Men }{ 531 } =\frac { 4 }{ 3 }$$ Men = $$\frac { 4 }{ 3 } \quad (531)$$ Children = $$\frac { 5 }{ 3 } (\frac { 4 }{ 3 } )(531)\quad =\quad 1180$$ Ans. GRE: COORDINATE GEOMETRY Two real number lines that are perpendicular to each other and that intersect at their respective zero points define a rectangular coordinate system. Often called the x-y coordinate system or x-y plane. The horizontal number line is called the x-axis and the vertical number line is called the y-axis. The point where the two axes intersect is called the origin, denoted by O. The positive half of the x-axis is to the right of the origin, and the positive half of the y-axis is above the origin. The two axes divide the plane into four regions called quadrants IIIIII, and IV, as shown in the figure below. Each point P in the x-y plane can be identified with an ordered pair (x, y) of real numbers and is denoted by P (x, y). The first number is called the x-coordinate, and the second number is called the y-coordinate. A point with coordinates (x, y) is located lxl units to the right of the y-axis if x is positive or to the left of the y-axis if x is negative. Also, the point is located lyl units above the x-axis if y is positive or below the x-axis if y is negative. If x = 0, the point lies on the y-axis, and if y = 0, the point lies on the x-axis. The origin has coordinates (0, 0). In the figure above, the point P (4, 2) is 4 units to the right of the y-axis and 2 units above the x-axis, and the point P’’’ (-4, -2) is 4 units to the left of the y-axis and 2 units below the x-axis. Distance Formula: If Two points P and Q are such that they are represented by the points (x1, y1) and (x2, y2) on the x-y plane, then the distance between two pints P and Q = ((x– x2)2 + (y– y2)2)0.5 Section Formula: If any point C (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n internally, X = (mx2 + nx1)/ (m + n) Y = (my+ my2)/ (m + n) If any point C (x, y) divides the line segment joining the points A (x1, y1) and B (x2, y2) in the ratio m: n externally, X = (mx2 – nx1)/ (m + n) Y = (my– my2)/ (m + n) Slope of a line: The slope of a line joining two points A (x1, y1) and B (x2, y2) is denoted by m and is given by m = (y– y1)/ (x– x1) = tan , where is the angle that the line makes with the positive direction of x-axis. Equation of line: Slope Intercept form: The equation of a straight line passing through the point A (x1, y1) and having slope m is given by (y – y1) = m (x – x1) Intercept form: The equation of a straight line making intercepts a and b on the axes of x and y respectively is given by x/a + y/b = 1 Perpendicularity and parallelism: Conditions for two lines to be parallel: Two lines are said to be parallel if their slopes are equal. Conditions for two lines to be perpendicular: Two lines are said to be perpendicular if the product of slopes of two lines is equal to -1. Area of triangle: The area of a triangle whose vertices are $$A({ x }_{ 1 },{ y }_{ 1 }),\quad B({ x }_{ 2 },{ y }_{ 2 }),\quad C({ x }_{ 3 },{ y }_{ 3 })$$ is given by $$=\frac { 1 }{ 2 } ({ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }))$$ Since the area can’t be negative, we have to take the modulus value given by the above equation. Problem 1) Find the area of the triangle whose vertices are (1. 3), (-7, 6) & (5, 1)? solution. Vertices are A(1,3), B(-7,6) and C(5,-1) Area of triangle = $$=\frac { 1 }{ 2 } ({ x }_{ 1 }({ y }_{ 2 }-{ y }_{ 3 })+{ x }_{ 2 }({ y }_{ 3 }-{ y }_{ 1 })+{ x }_{ 3 }({ y }_{ 1 }-{ y }_{ 2 }))$$ $$=\frac { 1 }{ 2 } (1(6+1)-7(-1-3)+5(3-6))\\ =\frac { 1 }{ 2 } (7-7(-4)+5(-3))\\ =\frac { 1 }{ 2 } (7+28-15)\\ =\frac { 1 }{ 2 } (28-8)\\ =10\quad Sq.\quad units$$ Problem 2) What is the area of a triangle with vertices (-2, 4), (2, 4) & (-6, 6) in the coordinate plane? solution. Let point A(-6, 6), point B(-2, 4) & C(2,4). Line BC is parallel to x-axis. Slope of BC = $$(\frac { 4-4 }{ 2+2 } )=0$$ We can use line BC as the Base of triangle. Drop a height vertically from (-6, 6). Height = different of y coordinates of points A & B = 6-4 = 2 Length of Base (BC) = difference between the $$x$$ coordinates of point B & C = 2 – ( -2) = 4 Area of triangle = $$\frac { 1 }{ 2 } \times 4\times 2=4\quad Sq.\quad units$$ Problem 3) What will be the reflection of the point (4, 5) in the 3rd quadrant? solution. Point (-4,-5) is reflection of the point (4,5) in the third quadrant. Problem 4) If the origin gets shifted to (2, 2), then what will be the new coordinates of the point (4, -2)? a) (-2, 4) b) (2, 4) c) (4, 2) d) (2, -4) solution. If origin gets shifted to (2, 2), horizontal distance of point A from y-axis becomes (2) & vertical distance of point. A from x-axis becomes (4) in the downward direction. Therefore, new co-ordinate of the point will be (2, -4) Problem 5) What will be the equation of the straight line that passes through the intersection of the straight lines $$2x-3y+4=0$$ & $$3x+4y-5=0$$ and is perpendicular to the straight lime $$3x-4y-5=0$$? solution. Line $$2x-3y+4=0$$ & $$3x+4y-5=0$$ intersect at point $$(\frac { -1 }{ 17 } ,\frac { 22 }{ 17 } )$$ Slope of straight line $$3x-4y-5=0$$ is $$\frac { 3 }{ 4 }$$. Slope of line perpendicular to this line = $$\frac { -4 }{ 3 }$$ Equation of line having slope $$\frac { -4 }{ 3 }$$ & passing through the point $$(\frac { -1 }{ 17 } ,\frac { 22 }{ 17 } )$$ is $$(y-\frac { 22 }{ 17 } )=\frac { -4 }{ 3 } (x+\frac { 1 }{ 17 } )\\ 3y-\frac { 66 }{ 17 } =-4(x+\frac { 1 }{ 17 } )\\ 51y-66=-4(17)(x+\frac { 1 }{ 17 } )\\ 51y-66=-68x-4\\ 68x+51y=-4+66=62\\ Equation={ (68x+5y=62) }^{ Ans }$$ Problem 6) Line m & n are perpendicular, neither line is vertical or line m passes through the origin. Quantity A: The product of the slopes of lines m & n Quantity B: The product of the x-intercepts of lines m & n 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. solution. The product of slope of perpendicular lines = -1 (The only exception is when one of the lines has an undefined slope because its vertical, but that case has been specifically excluded). If line m passes through the origin, its x-intercept is 0, So regardless of the x-intercept of line n. Quantity B is 0. Therefore, quantity B is greater. Problem 7) Which of the following could be the slope of a line that passes through the point (-2, -3) & crosses the y-axis above the origin? a) $$-\frac { 2 }{ 3 }$$ b) $$\frac { 3 }{ 7 }$$ c) $$\frac { 3 }{ 2 }$$ d) $$\frac { 5 }{ 3 }$$ e) $$\frac { 9 }{ 4 }$$ f) 4 solution. Slope of the line m passes through the origin & through the point (-2,-3) = $$\frac { -3 }{ -2 } =\frac { 3 }{ 2 }$$ Let angle formed by line m with positive x-axis be $$\theta$$, $$\tan { \theta =\frac { 3 }{ 2 } }$$ Angle formed by all the lines passing through the point (-2,-3) & crossing the y-axis above the origin lines between $$\theta \quad and\quad { 90 }^{ o }$$. $$\tan { \theta =\frac { 3 }{ 2 } }$$ $$\tan { { 90 }^{ 0 }=\infty }$$ Therefore, slope of all lines passing through the point (-2,-3) & crossing the y-axis above the origin lies between $$\frac { 3 }{ 2 } \quad and\quad \infty$$. Therefore, $$\frac { 5 }{ 3 } ,\quad \frac { 9 }{ 4 }$$ & 4 could be the slope of such lines. GRE: PROBABILITY Probability means the chance of the occurrence of an event. In layman terms, we can say that it is the likelihood that something- that is defined as the event will or will not occur. In probability, our first approach is to define the event and then we approach to find out the probability of an event or we approach to find out the chance of the occurrence of an event. Sample space This is defined with respect to a random experiment and denotes the set representing all the possible outcomes of the random experiment. For example: – Sample space when a coin is tossed is S = (Head, Tail). Sample space when a dice is thrown is S = (1, 2, 3, 4, 5, 6). Sample space when two dices are thrown is S = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Event The set representing the desired outcome of a random experiment is called the event. Consider the experiment of tossing a coin two times. An associated sample space is S = {HH, TT, TH, HT}. Suppose, getting two heads is the desired outcome. We find that (HH) is the only element corresponds to the desired outcome. Therefore, event E = {HH} Consider the experiment of throwing a dice. An associated sample space is S = {1, 2, 3, 4, 5, 6}. Suppose, we want to get a number less than 4. We find that (1, 2, 3) are the only element of S corresponds to our requirement. Therefore, event E = {1, 2, 3}. Event (E) is always a subset of sample space (S). When the sets A and B are two events associated with a sample space, then event ‘either A or B or both’ is represented by A ∪ B. Event ‘A and B’ is represented by A ∩ B. Mutually exclusive events A set of events is mutually exclusive when the occurrence of any one of them means that the other event cannot occur. Consider the experiment of throwing a dice. An associated sample space is S = {1, 2, 3, 4, 5, 6}. Consider events A ‘an odd number appears’ and B ‘an even number appears’ A = {1, 3, 5} B = {2, 4, 6} Occurrence of event A means that the other event B cannot occur. A ∩ B = f (null) Therefore, A and B are mutually exclusive events. Equally likely events If two events have the same probability they are called equally likely events. In a toss of a coin, the chance of getting head is equal to chance of getting tail. Therefore, getting head and getting tail are equally likely events. Exhaustive set of events A set of events that includes all the possibilities of the sample space is said to be an exhaustive set of events. Let us define the following events A: ‘a number less than 3 appears’ B: ‘3 appears’ C: ‘a number more than three appears’ A = {1, 2} B = {3} C = {4, 5, 6} A U B U C = {1, 2} U {3} U {4, 5, 6} = {1, 2, 3, 4, 5, 6} = S Therefore, A, B and C are called exhaustive set of events. Independent events An event is described as such if the occurrence of an event has no effect on the probability of the occurrence of another event. (If the first child of a couple is a boy, there is no effect on the chances of the second child being a boy.) Formula for finding probability Let A be an event. Probability of event A is denoted by P(A). Probability of an event A = P(A) = (Number of outcomes favorable to event)/ (Total number of outcomes possible) Probability of an event ‘not A’ is denoted by P(A’). P(A’) = 1 – P(A) Some other Formulae: Let A and B be two events associated with a random experiment. Then, P (either A or B) = P(A) + P(B) – P (A and B) or P (A U B) = P(A) + P(B) – P (A ∩ B) P (A U B)’ = P (A’ ∩ B’) Let A, B and C be three events associated with a random experiment. Then, P (A U B U C) = P(A) + P(B) + P(C) – P (A ∩ B) – P (A ∩ B) – P (A ∩ B) + P (A ∩ B ∩ C) Problem 1) In a throw of two dice, find the probability of getting one prime and one composite number? Sol. Sample space when two dices are thrown is S = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) Let A denote the event of getting a prime and a composite number A = (1, 2) (1, 4) (1, 6) (2, 1) (2, 3) (2, 5) (3, 2) (3, 4) (3, 6) (4, 1) (4, 3) (4, 5) (5, 2) (5, 4) (5, 6) (6, 1) (6, 3) (6, 5) No. of outcomes favorable to A = 18 Total no. of outcomes = 36 Probability = (18)/ (36) = 1/ 2 Problem 2) The letters of the word LUCKNOW are arranged among themselves. Find the probability of always having NOW in the word? Sol. Number of arrangements of word LUCKNOW = 7! Number of arrangements of word LUCKNOW having NOW = 5! Probability = (Total number of outcomes favorable to event)/ (Total number of outcomes) = (5!)/ (7!) Use of conjunction AND If A AND B are two independent events, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A and B occur is equal to P(A) x P(B) Use of conjunction OR If A AND B are two independent events, and if the probability of their occurrence is P(A) and P(B) respectively, then the probability that A or B occur is equal to P(A) + P(B) Problem 3) If we have the probability of A winning a race is 1/3 and that of B winning the race is 1/ 2, then the probability that either A or B win a race is given by? Sol. Probability that either A or B wins a race is given by P(A) + P(B) = 1/3 + 1/ 2 = 5/6 Problem 4) One card is drawn from a well-shuffled deck of 52 cards. If each outcome is equally likely, calculate the probability that the card will be a) a diamond b) not an ace c) a black card d) not a diamond e) not a black card Sol. Let A be the event ‘the card drawn is a diamond’ = (Number of outcomes favorable to event)/ (Total number of outcomes possible) = P (A) = 13/52 = 1/ 4 b) Let B be the event ‘the card drawn is an ace’ P(B) = (Number of outcomes favorable to event)/ (Total number of outcomes possible) = 4/ 52 P(B)’ = 1 – 4/52 = 48/52 c) Let C be the event ‘the card drawn is a black card’ P(C) = (Number of outcomes favorable to event)/ (Total number of outcomes possible) = 26/52 = 1/2 d) Let D be the event ‘the card drawn is a diamond’ P(D) = (Number of outcomes favorable to event)/ (Total number of outcomes possible) = 13/52 = 1/ 4 P(D)’ = 1 – 1/4 = 3/4 e) Let E be the event ‘the card drawn is a black card’ P(E) = (Number of outcomes favorable to event)/ (Total number of outcomes possible) = 26/52 P(E)’ = 1 – 1/2 = 1/2 Problem 5) Find the probability that a year chosen at random will have 53 Sundays provided that year was not leap year? Sol. A Year has 365 days. Year has 52 complete weeks and one day. One day maybe a Sunday, Monday, Tuesday, Wednesday, Thursday, Friday, Saturday or Sunday. Probability that this day will be a Sunday = 1/7 Therefore, probability that a year chosen at random will have 53 Sundays = 1/7 Problem 6) If two dice are thrown, what is the probability that the sum of the numbers is not less than 10? Sol. Let A be an event ‘the sum of numbers is equal to 10’ Let B be an event ‘the sum of numbers is equal to 11’ Let C be an event ‘the sum of numbers is equal to 12’ Probability that the sum of numbers is not less than 10 = P(A) + P(B) + P(C) Sample space when two dices are thrown is S = (1, 1) (1, 2) (1, 3) (1, 4) (1, 5) (1, 6) (2, 1) (2, 2) (2, 3) (2, 4) (2, 5) (2, 6) (3, 1) (3, 2) (3, 3) (3, 4) (3, 5) (3, 6) (4, 1) (4, 2) (4, 3) (4, 4) (4, 5) (4, 6) (5, 1) (5, 2) (5, 3) (5, 4) (5, 5) (5, 6) (6, 1) (6, 2) (6, 3) (6, 4) (6, 5) (6, 6) A = {(4, 6), (6, 4), (5, 5)} B = {(5, 6), (6, 5)} C = {(6, 6)} P(A) = 3/36 P(B) = 2/36 P(C) = 1/36 Probability that the sum of numbers is not less than 10 = 1/36 + 2/36 + 3/36 = 6/36 = 1/6 Problem 7) There are two bags containing white and black balls. In the first bag, there are 8 white and 6 black balls and in the second bag, there are 4 white and 7 black balls. One ball is drawn at random from any of these two bags. Find the probability of this ball being black? Sol. Let A be an event ‘Selection of the first bag’ Let B be an event ‘Selection of second bag’ One bag is selected at random out of two bags. Therefore, P(A) = P(B) = 1/2 Let C be an event ‘Black ball drawn from the first bag’ P(C) = 6/14 Let D be an event ‘Black ball drawn from the first bag’ P(D) = 7/14 Probability of this ball being black = P(A). P(C) + P(B). P(D) = (1/2). (6/14) + (1/2). (7/14) = (1/2). (13/14) = 13/28 Problem 8) The probability that Ajeet will solve a problem is 1/5. What is probability that a) he does not solve a single problem out of ten problems b) he solves at least one problem out of ten problems Sol. Let A be an event ‘Ajeet will solve a problem’ P(A) = 1/5 P(A)’ = 4/5 He solves no problems i.e. he does not solve the first problem and he does not solve the second problem……….. and he does not solve the tenth problem. Probability that he does not solve a single problem out of ten problems = (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) x (1/5) = (1/5)10 b) Let B be an event ‘Ajeet does not solve a single problem out of ten problems’ P(B) = (1/5)10 P(B)’ = 1 – (1/5)10 Problem 9) In a four-game series between Radha and Anand, the probability that Anand wins a particular game is 2/5 and that of Radha winning a game is 3/5. Assuming that there is no probability of a draw in an individual game, what is the chance that the series is drawn (2-2)? Sol. Let R be an event ‘Radha will win a particular match’ Let A be an event ‘Anand will win a particular match’ P(A) = 2/5 and P(R) = 3/5 The event definition for the series to end in a draw can be described as (R AND R AND A AND A) OR (R AND A AND R AND A) OR (R AND A AND A AND R) OR (A AND A AND R AND R) OR (A AND R AND R AND A) OR (A AND R AND A AND R) = (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)+ (2/5)2(3/5)2 = (2/5)2(3/5)x (6) = (2/5)2(3/5)x (4C2) Where 4C2 gives us the number of ways in which Radha can win two games and Anand can win two games. If E1, E2, E3 ……..En are mutually exclusive events of an experiment. Then, P (EU EU E3………En) = P(E1) + P(E2) + P(E3)…………….+ P(En) If E1, E2, E3 ……..En are mutually exclusive and exhaustive events of an experiment. Then, P(E1) + P(E2) + P(E3)…………….+ P(En) = 1 Problem 10) If P(A) = 1/3, P(B) = 1/2, P (A∩ B) = 1/4 then find the P(A’ U B’) Sol. P (A U B)’ = P (A’ ∩ B’) Replacing A with A’ and B with B’ P (A’ U B’)’ = P (A ∩ B) P (A ∩ B) = 1/4 P (A’ U B’) = 1 – 1/4 = 3/4 Problem 11) A and B are two mutually exclusive events of an experiment. If P(A’) = 0.65 and P (A U B) = 0.65 and P(B) = p, find the value of p. Sol. P(A’) = 0.65 P(A) = 0.35 P (A U B) = 0.65 A and B are mutually exclusive events. Therefore, P (A U B) = P(A) + P(B) 0.65 = 0.35 + p P = 0.30 Problem 12) A committee of two persons is selected from two men and two women. What is the probability that the committee will have a) no man b) one man c) two men? Sol. Total number of persons = 4. Out of these 4 persons, two can be selected in 4Cways. a)There will be two women in the committee. Out of two women, two can be selected in 2C= 1 way Probability (no man) = 2C2/4C2 b)One man and one woman can be selected in 2C1 x 2C1 Probability (One man and one woman) = (2C1 x 2C1)/ (4C2) c)Out of two men, two can be selected in 2C= 1 way Probability (two-man) = 2C2/4C2 Problem 13) A carton contains 25 bulbs, 8 of which are defective. What is the probability that if a sample of 4 bulbs is chosen, exactly two of them will be defective? Sol. 4 bulbs out of 25 bulbs can be drawn in 25Cways. Two bulbs out of 8 defective bulbs can be drawn in 8Cways. Two bulbs out of 17 non-defective bulbs can be drawn in 17Cways. Two defective and two non-defective bulbs out of 25 bulbs can be drawn in (8C2) x (17C2) Probability that if a sample of 4 bulbs is chosen, exactly two of them will be defective = ((8C2) x (17C2))/ (25C4) Problem 14) From a bag containing 8 green and 5 red balls, three are drawn one after the another. Find the probability of all three balls being green if balls are drawn without replacement? Sol. 3 balls out of 13 balls can be drawn in 13Cways 3 green balls out of 8 green balls can be drawn in 8Cways probability of all three balls being green if balls are drawn without replacement = (8C3)/(13C3) Conditional Probability It is the probability of the occurrence of an event A given that the event B has already occurred. This is denoted by P(A/B). P(A/B) = (Number of elementary events favorable to A ∩ B)/ (Number of elementary events which are favorable to B) P(A’/B) = 1 – P(A/B) Problem 15) If P(A) = 7/ 13, P(B) = 9/13 and P (A∩ B) = 4/13, evaluate P(A/B). Sol. P(A) = 7/13 P(B) = 9/13 P (A ∩ B) = 4/13 P(A/B) = P (A ∩ B)/ P(B) = (4/13)/ (9/13) = 4/9 Problem 16) Ten cards numbered 1 to 10 are placed in a box, mixed up thoroughly and then one card is drawn randomly. If it is known that the number on the drawn card is more than 3, what is the probability that it is an even number? Sol. Let A be ‘number on the drawn card is even number’ Let B be ‘number on drawn card is more than 3’ P(A/B) = P (A ∩ B)/ P(B) = (4/10)/ (7/10) = 4/7 GRE Quantitative reasoning format The quantitative reasoning measure has four types of questions: • Quantitative Comparison questions • Multiple-choice questions-Select One Answer Choice • Multiple-choice questions-Select One or More Answer Choices • Numeric Entry questions Each question appears either independently as a discrete question or as part of a set of questions called a Data Interpretation set. All of the questions in a Data Interpretation set are based on the same data presented in tables, graphs, or other displays of data. In the computer-delivered test, you are allowed to use a basic calculator-provided on-screen-on the Quantitative Reasoning measure. Information about using the calculator appears later in this chapter. Quantitative Comparison Questions Description Questions of this type ask you to compare two quantities-Quantity A and Quantity B- and then determine which of the following statements describes the comparison. • Quantity A is greater. • Quantity B is greater. • The two quantities are equal. • The relationship cannot be determined from the information given. Sample Question Instructions Compare Quantity A and Quantity B, using additional information centered above the two quantities if such information is given, and select one of the following four answer choices: 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. A symbol that appears more than once in a question has the same meaning throughout the question. Quantity A Quantity B The least prime number greater than 24 2) The greatest prime number less than 28. 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) For the integers greater than 24, note that 25, 26, 27 and 28 are not prime numbers, but 29 is a prime number, as are 31 and many other great integers. Thus, 29 is the least prime number greater than 24, and Quantity A is 29. For the integers, less than 28, note that 27, 26, 25 and 24 are not prime numbers, but 23 is a prime number, as are 19 and several other lesser integers. Thus 23 is the greatest prime number less than 28, and Quantity B is 23. Therefore, Quantity A is greater. (a) is correct. Multiple-Choice Questions –Select One Answer Choice Description These questions are multiple-choice questions that ask you to select only one answer choice from a list of five choices. Sample Question Select a single answer choice. Q 1) If 5x + 32 = 4 – 2x, what is the value of x? A) -4 B) -3 C) 4 D) 4 E) 12 Ans) Solving the equation for x, you get 7x = -28, and so x = -4. The correct answer is Choice A, -4 Multiple-Choice Questions –Select One or More Answer Choice Description These questions are multiple-choice questions that ask you to select one or more answer choices. A question may or may not specify the number of choices to select. These questions are marked with square boxes beside the answer choices, not circles or ovals. Instructions Select one or more answer choices according to the specific question directions. If the question does not specify how many answer choices to select, select all that apply. 1. The correct answer may be just one of the choices or as many as all of the choices, depending on the question. 2. No credit is given unless you select all of the correct choices and no others. If the question specifies how many answer choices to select, select exactly that number of choices. Sample Question Select one or more answer choice. Q 1) Which of the following integers are multiple of both 2 and 3? A) 8 B) 9 C) 12 D) 18 E) 21 F) 36 Ans) We can easily identify multiple of 2 and multiple of 3. Here, multiple of 2 are 8, 12, 18 and 36. Multiple of 3 are 9, 12, 18, 21 and 36. Therefore, 12, 18 and 36 are multiples of both 2 and 3. Alternatively, if you realize that every number that is a multiple of 2 and 3 is also a multiple of 6. You can identify the choices that are multiple of 6. The correct answer consists of Choices C (12), D (18) and F (36). Numeric Entry Questions Description Questions of this type ask you either to enter your answer as an integer or decimal in a single answer box or to enter it as a fraction in two separate boxes- one for the numerator and one for the denominator. In the computer-delivered test, use the computer mouse and keyboard to enter your answer. Instructions Enter your answer as an integer or a decimal if there is a single answer box OR as a fraction if there are two separate boxes-one for the numerator and one for the denominator. To enter an integer or a decimal, either type the number in the answer box using the keyboard or use the Transfer Display button on the calculator. 1. First, click on the answer box- a cursor will appear in the box- and then type the number. 2. To erase a number, use the Backspace key. 3. For a negative sign, type a hyphen. For a decimal point, type a period. 4. To remove a negative sign, type the hyphen again and it will disappear; the number will remain. 5. The transfer Display button on the calculator will transfer the calculator display to the answer box. 6. Equivalent forms of the correct answer, such as 2.5 and 2.50, are all correct. 7. Enter the exact answer unless the question asks you to round your answer. To enter a fraction, type the numerator and the denominator in the respective boxes using the keyboard. 1. For a negative sign, type a hyphen; to remove it, type the hyphen again. A decimal point cannot be used in a fraction. 2. The Transfer Display button on the calculator cannot be used for a fraction. 3. Fractions do not need to be reduced to lowest terms, though you may need to reduce your fraction to fit in the boxes. Sample Question Q1) One pen costs$0.25 and one marker costs $0.35. At those prices, what is the total cost of 18 pens and 100 markers? Ans) Cost of one pen =$0.25

Cost of one marker = $0.35 Cost of 18 pens = 0.25 x 18 =$4.5

Cost of 100 markers = 0.35 x 100 = $35 Total cost of 18 pens and 100 markers =$39.5