## GRE: Number System

Natural Numbers:- These are the numbers (1,2,3 etc) that are used for counting. In other words, all positive integers are natural numbers.

Prime Numbers:- A natural number larger than unity is a prime number if it does not have other divisors except for itself & unity.

Whole Numbers:- The set of numbers that includes all natural numbers and the number zero are called whole numbers.

Composite Numbers:- It is a natural number that has at least one divisor different from unity and itself.

Real Numbers:- All numbers between negative infinity & positive infinity that can be represented on a number line are called real numbers.

Rational Numbers:- A rational numbers is defined as a number of the form (a/b) where a and b are integers and b ≠ 0.

Rational numbers that are not integers have decimal values, can be of two types:

1.Terminating Decimal Fraction

2.Non- Terminating Decimal Fractions

Terminating decimal fractions :- For example  17/4 = 4.24  ,  21/5 = 4.2

Non-Terminating decimal fractions: Non-terminating decimal fractions are of two types:

Non-Terminating Repeating Fractions: These are a non terminating repeating fraction. For Example 5.333…..,   6.161616…..

Neither terminating nor repeating fractions: For example 5.273612…..

## Concept of GCD (Greatest Common Divisor or Height Common Factor)

Consider two natural numbers n1 and n2.

If the numbers n1 and n2 are exactly divisible by same number x, then x is a common divisor of n1 and n2.

The highest of all the common divisors of n1 and n2 is called as the GCD or the H.C.F:-

### Steps for calculating HCF

Suppose we have to find H.C.F of 150,210 and 375.

Step 1:- Writing down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

Step 2:- Writing prime factors common to all the three numbers are taking their product is 51  ×  31

Step 3:- H.C.F = 5×3= 15

## Concept of LCM (Least common multiple)

Let n1 and n2 be two natural numbers distinct from each other. The smallest natural number n that is exactly divisible by n1 and n2 is called the least common multiple (LCM)

### Steps for calculation of LCM:-

Suppose we have to find LCM of 150,210,375.

Step 1:- Write down the standard form of numbers

150=5×5×3×2

210=5×2×7×3

375=5×5×5×3

Step 2:-  Write down all the prime factors that appear at least once in any of the numbers: 5,3,2,7.

Step3: -Raise each of the prime factor to their highest available power

The LCM= 2 × 3 × 53 × 7 = 5270

HCF OF TWO OR MORE FRACTION IS GIVEN BY :

=HCF of Numerators Divided By LCM Of Denomination

LCM OF TWO OR MORE FRACTIONS IS GIVEN BY:

=LCM of Numerators  Divided By HCF of Denominators

Problem 1) Three traffic signals change after 36,42 and 72 seconds respectively.If the lights are first switched on at 9:00 sharp. At What time will they change simultaneously?

Sol. In order to distribute $4000 and 180 pencils evenly, the numbers of employees must be a factor of each of these two numbers. We have to find the greatest number of employees possible therefore, we have to find HCF of$4000 and 180.

4000=2×2×2×5×2×5×2×5 =25  ×  53

180= 2×9×2×5 = 22  ×  32 × 5

HCF = 22 × 5  = 20

Correct Answer = 20

Problem 2) The number of students who attended a school could be divided among 10,12 or 16 buses, such that each bus transports an equal number of students that could attend the school?

Sol.  No, of students must be divisible by 10,12 and 16. We have to find a minimum number of students that could attend the school or we have to find a minimum number that is divisible by 10,12 and 16.

LCM of 10,12,16                            10= 2×5

=24 × 3×5                                       12=2×2×3

=16×15                                          16=2×2×2×2

ANS:-  (240)

Problem 3) There are 576 boys and 448 girls in a school that are to be divided into sections containing equal no. of students of either boys or girls alone. Find the minimum total number of sections thus formed

a) 24

b) 32

c) 16

d) 20

Sol.  We have to find HCF of 576 and 448

576= 32  × 26

448= 26  × 7

HCF= 26 = 64

Number of sections = 576/64 + 448/64  = (9+7) = $${ 16 }^{ Ans }$$

Divisibility:- Any integer I (Commonly called dividend)

When divided by a natural number N (Called Divisor), there exist a unique pair of number Q and R ,Which are called the quotient and remainder respectively.

Dividend=Divisor × Quotient + Remainder

Suppose We divide 42 by 5. The result has a quotient of 8 and remainder of 2.

$$Let\quad N={ p }_{ 1 }^{ { \alpha }_{ 1 } }.{ p }_{ 2 }^{ { \alpha }_{ 2 } }.{ p }_{ 3 }^{ { \alpha }_{ 3 } }—–{ p }_{ k }^{ { \alpha }_{ k } }\quad Where\quad { p }_{ 1 },{ p }_{ 2 },{ p }_{ 3 }—–{ p }_{ k }\\ are\quad different\quad primes\quad and\quad { \alpha }_{ 1 },{ \alpha }_{ 2 },{ \alpha }_{ 3 }——{ \alpha }_{ k }\quad are\quad natural\quad numbers.\\ Total\quad number\quad of\quad divisors\quad of\quad N\quad =\\ =\quad (1+{ \alpha }_{ 1 })(1+{ \alpha }_{ 2 })(1+{ \alpha }_{ 3 })——-(1+{ \alpha }_{ k })\\ Sum\quad of\quad these\quad divisors\quad =\\ =\quad ({ p }_{ 1 }^{ 0 }+{ p }_{ 1 }^{ 1 }+{ p }_{ 1 }^{ 2 }—–{ p }_{ 1 }^{ { \alpha }_{ 1 } })({ p }_{ 2 }^{ 0 }+{ p }_{ 2 }^{ 1 }+{ p }_{ 2 }^{ 2 }——{ p }_{ 2 }^{ { \alpha }_{ 2 } })——-({ p }_{ k }^{ 0 }+{ p }_{ k }^{ 1 }—-{ p }_{ k }^{ { \alpha }_{ k } })\\$$

Problem 4) Find the sum of factors and the number of factor of 240?

Sol.  240 = 24 × 41 × 51

Total Number of divisor = (1+4) (1+1) (1+1)

=5×2×2= 20

Sum of these divisors

= (20 + 21 + 22 + 24 ) (30 + 31) (S0 + S1)

= (1+2+4+8+16) (4) (6)

=24 (31) = (744)

Problem 5)

Quantity A  The number of positive factor of 20.

Quantity B  The number of positive factor of 32.

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.  32 = 25                                        ,                               20= 22  51

Number of distinct positive factors of 32 = (6)

Number of distinct positive factors of 20 = 3×2 = 6

The two quantities are equal.

The highest power of a prime number p, which divides n! exactly is given by

= [n/p] + [n/p2] +[n/p3]……..

Where [x] denotes the greatest integer less than or equal to x

Remainder Theorem:-

The remainder of the expression

[A1 × B1 × C1 ………………+ A2 × B2 × C2] / m

will be the same as the remainder of the expression

[A1R × B1R × C1R ………………+ A2R × B2R × C2R] / m

Where A1R  is the remainder when A1 is divided by M. B1R is the remainder when B1 is divided by M and so on.

Problem 6) Find the remainder when 17 ×23×126×38 divided by 8

Sol.     ( 17 ×23×126×38 / 8 )

= (1×7×6×6 / 8 ) = ( 42×6 / 8) = (2*6 / 8 )     =    ( 12 / 8 )

Remainder=   ( 4 )

Problem 7) If a=16 b and b is a prime number greater than 2, How many positive distinct factors does a have?

Sol.   a =16b = 24. b1

No. of positive distinct factors = ( 4 + 1 ) ( 1 + 1 )    = (10)

Problem 8) If x is a positive integer, which one of the following could be the remainder. When 73x  is divisible by

a) 0

b) 1

c) 2

d) 3

e) 4

f) 5

g) 6

h) 7

i) 8

j) 9

Indicate all such remainders

Sol.  We have to find the unit digit of 73x

As with multiplication, when an integer is raised to a power, the unit digit is determined solely by the product of units digits.

Those products will form a repeating pattern.

31= 3. 32 =9. 33 =27. 34 = 81 and 35 =243

Here  the pattern returns to its original value of 3 and we have got a repeating pattern

3,9,7,1,3,7,1,3 —–

Thus the unit digit of  73x  must be 1,3,7,9

### Divisibility Rules:

Divisibility by 2-A number is divisible by 2 if the last digit is divisible by 2.

Divisibility by 5 – A number is divisible by 5 if the last digit is 5 or 0

Divisibility by 4 – A number is divisible by 4 if the last  two digit  are divisible by 4

Divisibility by 8 – A number is divisible by 8 if the last  3 digit  are divisible by 8

Divisibility by 11 – A number is divisible by 11 if the difference of the sum of the digits in the odd places and the sum of the digits in the even places is zero or divisible by 11.

Problem 9) w, x, y and z are consecutive odd integers such that $$w<x<y<z.$$ which of the following statements must be true?

Indicate all such statements.

a) wxyz is odd.

b) w+x+y+z is odd

c) w+z=x+y

Sol.  Odd×odd = odd

wx is odd. yz is odd and wxyz is also odd.

Odd+odd = even

Even+even=even

w+x=even, y+z=even

Therefore,w+x+y+z is also even.

w,x,y and z are consecutive odd integers, all can be defined in terms of w.

w=w, y=w+4

x=w+z, z=w+6

w+z=2w+6, x+y=2w+6

Therefore, w+z=x+y

Problem 10)

Quantity A Remainder when $${ (67) }^{ 99 }$$ is divided by 7.

Quantity B Remainder when $${ (75) }^{ 80 }$$ is divided by 7.

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.  Using remainder theorem

$$(\frac { { 67 }^{ 99 } }{ 7 } )\rightarrow (\frac { { 4 }^{ 99 } }{ 7 } )=\frac { { { (4 }^{ 3 }) }^{ 33 } }{ 7 } =\frac { { 64 }^{ 33 } }{ 7 } \\ Remainder\quad =\quad 1\\ \frac { { 75 }^{ 80 } }{ 7 } \rightarrow \frac { { 5 }^{ 80 } }{ 7 } \rightarrow \frac { { ({ 5 }^{ 6 }) }^{ 13 }\times { 5 }^{ 2 } }{ 7 } \rightarrow \frac { { 1 }^{ 13 }\times 25 }{ 7 } \\ Remainder\quad =\quad 4$$

Quantity B is greater.

Problem 11) If set S consists of all positive integers that are multiples of both 2 and 7, how many numbers in set S are between 140 and 240, inclusive?

Sol.  Correct answer is 8.

A positive integer that is a multiple of both 2 and 7 is a multiple of 14. Since 140 is a multiple of 14, start listing there and count the terms in the range:

140, 154, 168, 182, 196, 210, 224, 238

Problem 12) If then which of the following must be true?

a) $${ a }^{ 2 }>a>b>{ b }^{ 2 }$$

b) $$b>a>{ a }^{ 2 }>{ b }^{ 2 }$$

c) $${ b }^{ 2 }>a>{ a }^{ 2 }>b$$

d) $${ b }^{ 2 }>{ a }^{ 2 }>b>a$$

e) $${ b }^{ 2 }>b>a>{ a }^{ 2 }$$

Sol.  The correct answer is $$({ b }^{ 2 }>b>a>{ a }^{ 2 })$$

The goal in this question is to order $$a,{ a }^{ 2 },b\quad and\quad { b }^{ 2 }$$ by magnitude. Based on the original inequality

$$0<a<\frac { 1 }{ b } <1$$, several things are true.

$$\frac { 1 }{ b } <1\quad \& \quad b>1$$

Therefore, $${ b }^{ 2 }>b$$

Second, note that $$a<1$$ in the given inequality.

a is a positive number less than 1,

$${ a }^{ 2 }<a<1\\ { b }^{ 2 }>b>1\\ { ({ b }^{ 2 }>b>a>{ a }^{ 2 }) }^{ Ans }$$

Problem 13) On a number line, the distance from A to B is 4 and the distance from B to C is 5.

Quantity A  The distance from A to C

Quantity B   9

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.  If the points A, B and C are in alphabetical order from left to right, then the distance from A to C will be 9. However, alphabetical order is not required. If the points are in the order C, A and B from left to right,then the distance from A to C is 5-4=1. Therefore, the relationship cannot be determined.

Problem 14) The integer a is even and the integer b is odd. Which of the following statements must be true?

Indicate all such statements

a) a+2b is even

b) 2a+b is even

c) ab is even

d) $${ (a) }^{ b }$$ is even

e) $${ (a+b) }^{ 2 }$$ is even

f) $${ a }^{ 2 }-{ b }^{ 2 }$$ is odd

Sol.  If b is odd, then 2b must be even.

even+even = even

(a+2b) must be even

even+odd = odd

2a+b must be odd.

even×odd = even

ab must be even.

$${ (even) }^{ odd }$$ must be even.

$${ (a) }^{ b }$$ must be even

$${ (odd) }^{ even }$$ must be odd

$${ (odd) }^{ odd }$$ must be odd

(a+b) is odd

Therefore, $${ (a+b) }^{ 2 }$$ must be odd.

$${ (a) }^{ 2 }$$ must be even and $${ (b) }^{ 2 }$$  must be odd.

$$({ a }^{ 2 }-{ b }^{ 2 })$$ must be odd.

Problem 15) Find the number of zeros in 137!?

Sol.  The highest power of a prime number p, which divides n! exactly is given by

$$\left[ \frac { n }{ p } \right] +\left[ \frac { n }{ { p }^{ 2 } } \right] +\left[ \frac { n }{ { p }^{ 3 } } \right] ——-$$

Where $$\left[ x \right]$$ denotes the greatest integer less than or equal to x.

$$\left[ \frac { 137 }{ 5 } \right] +\left[ \frac { 137 }{ 25 } \right] +\left[ \frac { 137 }{ 125 } \right] \\ =1+5+27=\quad 33\quad zeros$$

Since the restriction on the number of zeroes is due to the number of fives.

Therefore, 137! Has 33 zeros.

Problem 16) Find the number of numbers from 1 to 100 which are not divisible by any one of 2 & 3.

a) 16

b) 17

c) 18

d) 33

Sol.  Number of numbers not divisible by 2 & 3 = Total no. of numbers – number of numbers divisible by either 2 or 3.

Number of numbers divisible by 2 = 50

Numbers divisible by 3 (but not by 2, as it has already been counted) are 3,9, 15, 21, 27, 33—–93, 99.

Number of numbers divisible by 3 (but not by 2)

$$=\frac { (99-3) }{ 6 } +1=17$$

Hence, number of numbers divisible by either 2 or 3

= 50+17 = 67

Number of numbers not divisible by any of 2 & 3 =

Total number of numbers – no. of numbers divisible by either 2 or 3

= 100 – 67

= $${ 33 }^{ Ans }$$

## Conditions of similarity:-

• AAA Similarity:- If in two triangles, corresponding angles are equal, then triangles are similar.

If ∠A = ∠P , ∠B = ∠Q , ∠C = ∠R

Then △ABC ~ △ PQR

• SSS Similarity:- If the corresponding sides of two triangles are proportional then they are similar.

If AB/PQ = BC/QR = AC/PR

Then   △ABC ~ △ PQR

• SAS Similarity: If in two triangles, one pair of corresponding sides are proportional and the included angles are equal then the two triangles are similar.

If AB/PQ = BC/QR  &  ∠B = ∠Q

Then △ABC ~ △ PQR

The similarity of Triangles:- Similarity of triangles is a special case where if either of the conditions of similarity holds, The other will hold automatically.

Congruency of Triangles:-If two triangles are congruent, then corresponding angles are equal.

## Conditions of Congruency:-

1.SAS Congruency:- If two sides and an included angle of one triangle are equal to two sides and an included angle of another, the two triangles are congruent.

In  △ABC and △ PQR

IF AB=PQ,

BC=QR,

∠ABC = ∠PQR

Then △ABC~△ PQR

2.ASA Congruency:- If two angles and the included side of one triangle is equal to two angles and the included side of another, The two triangles are congruent.

In  △ABC and △ DEF

If  ∠A =∠D

AB = DE

∠B = ∠E, Then △ABC ≅ △ PQR

3. AAS Congruency:- If two angles and the side opposite to one of the angles is equal to the corresponding angles and the side of another triangle, the triangles are congruent.  If  In △ABC and △ DEF

∠A =∠D

∠B=∠E

AC=DF

Then △ABC≅△ DEF

4.SSS Congruency:- If three sides if one triangle is equal to three sides of another triangle, the two triangles are congruent.  In △ABC and △ DEF

If AB=DE, BC=EF and AC=DF

THEN  △ABC ≅ △ DEF

5.SSA Congruency:- If two sides and the angle opposite to any of the sides are equal to corresponding sides and the angle opposite to any of the sides, Then the triangles are congruent.

In △ABC and △ DEF

AB=DF

BC= EF

∠BAC = ∠EDF

Then △ABC ≅ △ DEF

## Right Angled Triangle:

Pythagoras theorem: In the case of a right angled triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. In right △ABC, AB2 + BC2 = AC2

## Special figure: Note: A lot of questions are based on the figure

In this figure, △ABC ~△ ADB ~ △BDC

• △ABC ~△ ADB

AB/AD   =  BC/DB  =  AC/AB

• △ADB ~△ BDC

AD/BD   =  DB/DC  = AB/BC

• △ABC ~△ BDC

AB/BD = BC/DC = AC/BC

## Area Theorem: According to area theorem:The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

If △ABC ~ △ DEF

Ar(ABC): Ar(DEF) = (AB/DE)2 = (BC/EF)2 = (AC/DF)2

Problem 1) For similar triangles, the ratio of their corresponding sides is 2:3. What is the ratio of their areas?

Sol. According to area theorem :The ratio of areas of two similar triangles is the square of the ratio of corresponding sides.

Ratio of areas= (2/3)2  =  4/9

Problem 2) What is the area of the triangle shown? Sol. ∠B = ∠C = 60

∠A= 60

(Sum of angles of a triangle is equal to 180°)

∠A = ∠B = ∠C = 60°

(Angles of an equilateral triangle are equal to 60° & triangle having each angle equal to 60° is an equilateral triangle )

:. △ABC is an equilateral △

In △ABD

BD2 +AD2 = AB2 (Pythagoras theorem)

Area of Triangle = ½ × Base × Height

=1/2 × 10 × 5√3

= 25√3  sq. unit

Problem 3) What is the area of the triangle Shown? Sol.

x2 +( x + 1)2 =25 (by Pythagoras theorem)

2x2 +2x =24

2x2 + 2x -24 =0

x2 + x – 12 =0

x = 3 or -4

x= 3 (Because length cannot be negative)

Area = ½ (base) x (height) = ½(7)(8)   = 28 sq. units

Problem 4) What is the value of x in the triangle shown? Sol.      In △ABC

4 + 3 = AC  (BY Pythagoras theorem)

AC = 5

△ABC~ △BDC

AB/BD =BC/DC = AC/BC

4/x  =  5/3

12 = 5x          ,      x =(12/5)  = Ans

## Polygons

A polygon is a closed figure whose sides are straight line segments.

The perimeter of the polygon is the sum of the lengths of the sides. A vertex of a polygon is the point where two adjacent sides meet.

A diagonal of a polygon is a line segment connecting two nonadjacent vertices.

A regular polygon has sides of equal length and interior angles of equal measure.

Sum of all angles of a polygon with n sides

= (n-2) π radians = (n-2) (180°)

Area of Regular polygon = (ns2/4) x Cot (180/n)

Where     s = length of side

n = No. of sides

## Parallelogram (IIgm)

A quadrilateral with two pairs of parallel sides. Properties of parallelogram:

Area of Parallelogram = Base x Height

Diagonals of Parallelogram bisect each other.

The opposite angles in a Parallelogram are equal.

## Rectangle

A Parallelogram with four equal angles, each at right angle is a rectangle. AB = CD, BC =AD

∠ABC = ∠BCD = ∠CDA = ∠BAD

Properties of a Rectangle:

Diagonals of a rectangle are equal and bisect each other.

## Rhombus

A Parallelogram having all the sides equal is a rhombus. Area of Rhombus = ½ x product of diagonals

Properties:-

Diagonals of a rhombus bisect each other at right angle.

## Square

A Square is a rectangle with adjacent sides equal or rhombus with angles equal to 90°.

Properties:

Diagonals of square are equal and bisect each other.

Area = ½ x (diagonal) 2

Diameter of circle circumscribing a square also acts as a diagonal of square. BD=Diameter of Circle

= Diagonal of Square = a √2

## Trapezium

A Trapezium is a quadrilateral with only two sides parallel to each other. Area of Trapezium = ½ (Sum of Parallel sides)x height

## Rectangular Hexagon

A Rectangular Hexagon is a actually a combination of 6 equilateral triangles all of side “a” Area of hexagon = 6(3/4 a2) = 3√3 / 2 a2

Problem 5) Right △ABC and rectangle EFGH have the same perimeter. What is the value of x? Sol. In △ABC

AB2 + BC2 = AC2 (BY Pythagoras theorem)

AC=5

Perimeter of △ABC = 3+4+5 = 12

Perimeter of rectangle EFGH =2(2+x)

12 = 2 (2+x)

2 + x = 6

x = 4

Problem 6) What is the area of the square in the figure above? Sol. (Side2 + Side2 = Diagonal2) (by Pythagoras theorem)

2 Side2  = 10= 100

Side = 5 √2

Area = Side2 = 50 sq units

Problem 7) The area of a rectangle can be represented by the expression 2x2+9x+10. The length is 2x+5 and the width is 6. What is the value of the area of the rectangle?

Sol. Area = 2x2 +9x 10 = Length x Breadth

2x2 +9x 10 = (2x +5)6

2x2  – 3x – 20 = 0

x = 4 or -2.5

If x = -2.5, then length = 2(-2.5) + 5 = 0 (Length can’ t be zero,∴ Negative value of x is discarded)

x = 4

Area = 2 (16) +9 (4) +10

= 32 +36+ 10   = 78 sq Units

Problem 8) In the figure shown, ABCD is a rectangle. AB=8 and BC=6. R,S,T, AND Q are midpoints of the sides of rectangle ABCD. What is the perimeter of RSTQ? Sol. AB = CD = 8, BC = AD = 6

AR = RB = DT = CT

AQ = QC = BS = SD = 3

AR2 + AQ2 = RQ2  (By Pythagoras theorem)

32 + 42 = RQ2

RQ = 5

RQ=RS=ST=QT=5

Perimeter of RSTQ = 20

Problem 9) In the figure shown ABCD is a square with a side length of 16. R,S,T AND Q are midpoints of the sides of ABCD .

What is the area of RSTQ? Sol. AB=16

AR=RB=BS=SC=TC=DT=QD=AQ=8

AR2 + AQ2 = RQ2  (By Pythagoras Theorem)

RQ=8 √2

Similarly RQ=RS=QT=ST= 8 √2

In △ARQ, ∠QAR = 90°, AR = AQ

∠AQR = ∠DQT (Angles opposite to equal sides are equal)

∠AQR = ∠DQT = 45°

∠RQT=90(Linear pair)

Therefore, RSTQ is square

Area (8 √2)2= 128 sq units

Problem 10) In the given figure, AD||BC. Find the value of x.

a) x=8, $$\frac { 15 }{ 3 }$$

b) x=8, $$\frac { 11 }{ 3 }$$

c) x=8, $$\frac { 13 }{ 3 }$$

d) x=7, $$\frac { 13 }{ 3 }$$ Sol.

$$In\quad \triangle ODA\quad and\quad \triangle OBC,\quad AD\parallel BC\\ \angle ODA=\angle OBC$$ (Alternate interior Angles are equal)

$$\angle OAD=\angle OCB$$ (Alternate interior angles are equal)

$$\triangle ODA\sim \triangle OBC\\ \frac { OD }{ OB } =\frac { DA }{ BC } =\frac { OA }{ OC } \\ \frac { OD }{ OB } =\frac { AO }{ OC } \\ \frac { x-5 }{ x-3 } =\frac { 3 }{ 3x-19 } \\ (x-5)(3x-19)=3x-9\\ 3{ x }^{ 2 }-19x-15x+95-3x+9=0\\ 3{ x }^{ 2 }-37x+104=0\\$$

(solving this quadratic equation)

$$x=\frac { 37\pm \sqrt { { 37 }^{ 2 }-4(3)(104) } }{ 6 } =\frac { 37\pm 11 }{ 6 } \\ =8,\frac { 11 }{ 3 }$$

c) is correct.

Problem 11) In the right angle $$\triangle PQR$$, find RS? Sol.

$$In\quad \triangle QRP\quad and\quad \triangle RSP\\ \angle QRP=\angle RSP\quad (both\quad are\quad right\quad angles)\\ \angle QPR=\angle SPR\quad (common)\\ Therefore,\quad \triangle QRP\sim \triangle RSP\\ \frac { QR }{ RS } =\frac { RP }{ SP } =\frac { QP }{ RP } \\ \frac { 5 }{ RS } =\frac { QP }{ 12 } \\ In\quad \triangle QRP\\ { QR }^{ 2 }+{ RP }^{ 2 }={ QP }^{ 2 }(By\quad pythagoras\quad theorem)\\ { 12 }^{ 2 }+{ 5 }^{ 2 }={ QP }^{ 2 }\\ QP=13\\ \frac { 5 }{ RS } =\frac { 13 }{ 12 } \\ RS={ (\frac { 60 }{ 13 } ) }^{ Ans }$$

Problem 12) The area of semicircle O is $$16\pi$$.

$$\angle CDE=\angle ABE={ 30 }^{ 0 }\\ AC=\sqrt { 2 }$$

Quantity A:  DE

Quantity B:  $$3\sqrt { 5 }$$

Sol.

$$In\quad \triangle EDC\quad and\quad \triangle EBA\\ \angle EDC=\angle EBA\\ \angle CED=\angle AEB={ 90 }^{ 0 }\\ \triangle EDC\sim EBA\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ Let\quad EC\quad be\quad x\\ In\quad \triangle EDC\\ \angle CED={ 90 }^{ 0 },\quad \angle CDE={ 30 }^{ 0 },\quad \angle DCE={ 60 }^{ 0 }$$

Recall that the sides of a 30-60-90 triangle are in the proportion

$$x:x\sqrt { 3 } :2x$$.

.Or if you are familiar with trigonometry, just recall that

$$\sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { perpendicular }{ hypotenuse } \\ Therefore,\quad \frac { CE }{ CD } =\frac { 1 }{ 2 } =\frac { x }{ CD } ,\quad CD=2x\\ \frac { EC }{ AE } =\frac { DC }{ BA } \\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ BA }$$

Area of semicircle = $$\frac { \pi { r }^{ 2 } }{ 2 } =16\pi$$ $${ r }^{ 2 }=36,\quad r=4\sqrt { 2 } =OB\\ \frac { x }{ x+\sqrt { 2 } } =\frac { 2x }{ 8\sqrt { 2 } } \\ x+\sqrt { 2 } =4\sqrt { 2 } \\ x=3\sqrt { 2 } \\ Therefore,\quad CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 }$$

Side of $$\triangle CDE$$ are in ratio $$x:x\sqrt { 3 } :2x$$ $$CE=3\sqrt { 2 } ,\quad CD=6\sqrt { 2 } \\ ED=3\sqrt { 6 } \\ Therefore,\quad DE=3\sqrt { 6 }$$

(Quantity A is greater than Quantity B).

Problem 13) What is the area of a triangle that has two sides that each have a length of 10, & whose perimeter is equal to that of a square whose area is 81?

a) 30

b) 48

c) 36

d) 60

e) 42

Sol.

Area of square = 81

Side of square = $$\sqrt { 81 } =9$$

Perimeter of square = 4×9 = 36

Perimeter of triangle = 36

Two sides of triangle have a length of 10.

Third side of triangle = 36-10-10 = 16 Any triangle that has equal sides is an isosceles triangle. Drop a perpendicular to divide it into two individual right triangles.

$$In\quad \triangle ABC,\quad AD\bot BC\\ In\quad \triangle ABD,\quad { AB }^{ 2 }={ BD }^{ 2 }+{ AD }^{ 2 }\quad (by\quad pythagoras\quad theorem)\\ { 10 }^{ 2 }={ 8 }^{ 2 }+{ AD }^{ 2 }\\ AD=6\\ Area\quad of\quad \triangle ABC=\frac { 1 }{ 2 } \times Base\times Height=\frac { 1 }{ 2 } \times 16\times 16=48\quad Sq.units$$

Problem 14) Michael wants to split his rhombus-shaped garden into two triangular plots, one for planting strawberries & one for planting vegetables, by erecting a fence from one corner of the garden to the opposite corner. If one corner of the garden measures $${ 60 }^{ 0 }$$ & one side of the garden measures $$x$$ meters, which of the following could be the area of the vegetable plot?

Indicate all such value

a) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 2 }$$

b) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }$$

c) $$\frac { { x }^{ 2 } }{ 2 }$$

d) $$\frac { { x }^{ 2 }\sqrt { 3 } }{ 4 }$$

e) $$\frac { { x }^{ 2 } }{ 4 }$$

Sol.  Whenever a geometry problem comes without a figure, start by drawing one yourself: Michael wants to split this garden into two triangles.

Try it this way first: $$\triangle ACD$$ is an equilateral. (Because all angles are equal to $${ 60 }^{ 0 }$$)

Area of triangle ACD = Area of vegetable plot =

$$=\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }$$

This is one of the choices, but the garden might be split the other way: $$Draw\quad AE\bot BD,\quad In\triangle AEB\\ \angle AEB={ 90 }^{ 0 },\quad \angle ABE={ 30 }^{ 0 },\quad \angle EAB={ 60 }^{ 0 }$$

Recall that the sides of a 30-60-90 triangle are in the proportion $$x:\sqrt { 3 } x:2x$$

Or if you are familiar with basic trigonometry

$$In\quad \triangle AEB,\quad \sin { { 30 }^{ 0 } } =\frac { 1 }{ 2 } =\frac { AE }{ AB } =\frac { AE }{ x } \\ AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }$$

Therefore, we can easily derive that

$$AE=\frac { x }{ 2 } ,\quad EB=\frac { x\sqrt { 3 } }{ 2 }$$ (by using properties of triangle or by using basic trigonometry)

Similarly, we can find that $$ED=\frac { \sqrt { 3 } x }{ 2 } ,\quad BD=\sqrt { 3 } x$$ $$Ar(\triangle ABD)=\frac { 1 }{ 2 } \times base\times height\\ =\frac { 1 }{ 2 } \times \sqrt { 3 } x\times \frac { x }{ 2 } \\ =\frac { \sqrt { 3 } { x }^{ 2 } }{ 4 }$$

This is same as the previous area, so only one answer choice is B.

## GRE: Quadratic Equation

If you set the polynomial $$a{ x }^{ 2 }+bx+c$$ equal to zero, where a, b and c are constants and

a ≠ 0 , there is a special name for it. It is called a quadratic, You can find the values for x that make the equation true.

Example:     $${ x }^{ 2 }-3x+2=0$$

To find the solutions, also called roots of an equation, start by factoring whenever possible. You can factor $${ x }^{ 2 }-3x+2$$ into (x-2)  (x-1) , making the quadratic equation:

(x-2) (x-1) = 0

To find the roots, set each binomial equal to 0. That gives you:-

(x-2) = 0      or     (x-1) =0

Solving for x, you get x=2 or x=1

The solutions to a quadratic in the form $$a{ x }^{ 2 }+bx+c=0$$ can also be found using the quadratic formula provided a, b and c are real numbers and a ≠ 0 then:

$$x\quad =\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$$

EXAMPLE:- Find the Solution of 2x2 + 9x +9 = 0

Solution :- a=2 , b=9 , c=9

$$x\quad =\quad \frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a }$$ $$x\quad =\quad \frac { -9\pm \sqrt { 81-4(2)(9) } }{ 2(2) }$$

=  (-9±3) / 4

ANS= -3     or -3/2

Sum of the roots of a quadratic equation =  -b/a

Product of the roots of a quadratic equation = c/a

• $$D\quad =\quad { b }^{ 2 }-4ac$$ is the discriminant of a quadratic equation.

If D < 0 (i.e. the discriminant is negative) then the equation has no real roots.

If D= 0 then the quadratic equation has two equal roots.

If D> 0 (i.e. the discriminant is positive) then the quadratic equation has two distinct roots.

Graph of a Quadratic Expression:-

Let f(x)= ax2 + bx +c , Where a, b, c are real and a ≠ 0, Then y=f(x) represents a parabola, Whose axis is parallel to y- axis. This gives the following cases:

1) a> 0 and D = (b2 -4ac) < 0 (Roots are imaginary).

$$f(x)>0\quad \forall \quad x\in R$$ 2) a > 0 and D = 0 (The roots are real and equal)

f(x) will be positive for all values of x except at the vertex where f(x) = 0 3) When a> 0 and D > 0 (The roots are real and distinct)

F(x) will be equal to zero when x is equal to either of $$\alpha \quad or\quad \beta$$.

$$f(x)>0\quad \forall x\in (-\infty ,\alpha )\bigcup (\beta ,\infty )\\ and\quad f(x)<0\quad \forall x\in (\alpha ,\beta )$$ 4) When a < 0 and D=0 (Roots are imaginary)

$$f(x)<0\quad \forall x\in R$$ 5) When a<0 and D=0 (Roots are real and equal)

F(x) is negative for all values of x except at the vertex Where f(x)= 0 6) When a<0 and D>0 (Roots are real and distinct)

$$f(x)<0\quad \forall x\in (-\infty ,\quad \alpha )\bigcup (\beta ,\quad \infty )\\ f(x)>0\quad \forall x\in (\alpha ,\quad \beta )$$ ## Sample Problems

Problem 1)  If -4 is a solution for x in the equation $${ x }^{ 2 }+kx+8=0$$, what is k?

Sol.  If -4 is a solution for x in the equation.

Then,$${ (-4) }^{ 2 }+k(-4)+8=0\\ 16-4k+8=0\\ -4k=-24\\ k={ (6) }^{ Ans }$$

Problem 2) Given that $${ x }^{ 2 }-13x=30$$, what is x?

Sol.

$${ x }^{ 2 }-13x=30\\ { x }^{ 2 }-13x-30=0\\ x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ =(\frac { 13\pm \sqrt { 169-4(-30) } }{ 2 } )\\ =\frac { 13\pm \sqrt { 289 } }{ 2 } =\frac { 13\pm 17 }{ 2 } \\ =\frac { -4 }{ 2 } \quad or\quad \frac { 30 }{ 2 } \\ =-2\quad or\quad 15$$

Problem 3) If the area of a certain square (expressed in square meters) is added to its perimeter (expressed in meters), the sum is 77. What is the length of a side of the square?

Sol.  Let side of square be x.

Area if square $$={ x }^{ 2 }$$

Perimeter of square = 4x

$${ x }^{ 2 }+4x=77\\ { x }^{ 2 }+4x-77=0\\ \alpha ,\beta =\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2 } \\ \alpha =\frac { -4+\sqrt { 16+4(77) } }{ 2 } ,\quad \beta =\frac { -4-\sqrt { 16+4(77) } }{ 2 } \\ \alpha =\frac { -4+18 }{ 2 } ,\quad \beta =\frac { -4-18 }{ 2 } \\ \alpha =7,\quad \beta =-11$$

Since the ride of a square must be positive, negative value (-11) has been discorded.

Side of the square = 7

Problem 4)  If $${ x }^{ 2 }-6x-27=0$$ and $${ y }^{ 2 }-6y-4=0$$, what is the maximum value of x+y?

Sol.

$${ x }^{ 2 }-6x-27=0\\ \alpha =\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } ,\quad \beta =\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \alpha =\frac { 6+\sqrt { 36-4(-27) } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 36-4(-27) } }{ 2 } \\ \alpha =\frac { 6+\sqrt { 144 } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 144 } }{ 2 } \\ \alpha =\frac { 6+12 }{ 2 } ,\quad \beta =\frac { 6-12 }{ 2 } \\ \alpha =9,\quad \beta =-3$$

x may be 9 or -3.

$${ y }^{ 2 }-6y-40=0\\ \alpha =\frac { -b+\sqrt { { b }^{ 2 }-4ac } }{ 2a } ,\quad \beta =\frac { -b-\sqrt { { b }^{ 2 }-4ac } }{ 2a } \\ \alpha =\frac { 6+\sqrt { 36-4(-40) } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 36-4(-40) } }{ 2 } \\ \alpha =\frac { 6+\sqrt { 196 } }{ 2 } ,\quad \beta =\frac { 6-\sqrt { 196 } }{ 2 } \\ \alpha =\frac { 6+14 }{ 2 } ,\quad \beta =\frac { 6-14 }{ 2 } \\ \alpha =10,\quad \beta =-4$$

y may be 10 or -4.

For maximum value of (x+y),

x should be maximum & y should be maximum.

Maximum value of (x+y) = 9+10 = 19

Problem 5)  xy>0

Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

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.  Quantity A: $${ (x+y) }^{ 2 }$$

Quantity B: $${ (x-y) }^{ 2 }$$

Now subtract $${ x }^{ 2 }+{ y }^{ 2 }$$ from both columns.

Then, Quantity A:       2xy

Quantity B:         -2xy

xy>0

xy is positive, so the value in quantity A will be positive, regardless of the values of x and y. Similarly, the value in Quantity B will always be negative, regardless of the values of x & y.

Quantity A is larger.

Problem 6)  Find the value of $$\sqrt { 6+\sqrt { 6+\sqrt { 6+——— } } }$$

a) 4

b) 3

c) 5

d) 5

Sol.  Let

$$y\quad =\quad \sqrt { 6+\sqrt { 6+\sqrt { 6+——— } } } \\ y\quad =\quad \sqrt { 6+y } \quad (Squaring\quad both\quad sides)\\ { y }^{ 2 }=6+y\\ { y }^{ 2 }-y-6=0\\ \alpha =\frac { 1+\sqrt { 1-4(-6) } }{ 2 } ,\quad \beta =\frac { 1-\sqrt { 1-4(-6) } }{ 2 } \\ \alpha =\frac { 1+5 }{ 2 } ,\quad \beta =\frac { 1-5 }{ 2 } \\ \alpha =3,\quad \beta =-2\\ Therefore,\quad y=3\quad (Since\quad \sqrt { 6+\sqrt { 6+\sqrt { 6—— } } } cannot\quad be\quad negative)$$

Problem 7)  If the roots of the equation , differ by 2, then which of the following is true?

a) $${ c }^{ 2 }=4(c+1)$$

b) $${ b }^{ 2 }=4(c+1)$$

c) $${ c }^{ 2 }=b+4$$

d) $${ b }^{ 2 }=4(c+2)$$

Sol.

$${ x }^{ 2 }-bx+c=0$$

Let $$\alpha \quad and\quad \beta$$ be the roots of equation.

$$\alpha +\beta =b\quad \quad \alpha \beta =c\\ \alpha -\beta =2\\ { (\alpha +\beta ) }^{ 2 }={ \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta \\ (Subtract\quad 4\alpha \beta \quad from\quad both\quad sides)\\ { (\alpha +\beta ) }^{ 2 }-4\alpha \beta ={ \alpha }^{ 2 }+{ \beta }^{ 2 }+2\alpha \beta -4\alpha \beta \\ { \alpha }^{ 2 }-2\alpha \beta +{ \beta }^{ 2 }\\ ={ (\alpha -\beta ) }^{ 2 }\\ \\ { b }^{ 2 }-4c={ 2 }^{ 2 }=4\\ { b }^{ 2 }=4c+4\\ Therefore,\quad { ({ b }^{ 2 }=4(c+1)) }^{ Ans }$$

## GRE: Solving Inequalities

Inequalities may be written with the symbols Shown here:-

 < Less than > Greater than ≤ Less than or equal to ≥ Greater than or equal to

A range of values is often expressed on a number line.

Two ranges are shown below:- a) Represents the set of all numbers between -4 and 0 excluding the endpoints -4 and 0 or -4< x < 0 b) Represents the set of all numbers greater than -1, upto and including 3, or -1 < x ≤ 3

Rules for working with inequalities:-

• Treat inequalities like equations when adding or subtracting terms or when multiplying/Dividing by a positive number on both sides of the inequality.
• Flip the inequality sign if you multiply or divide both sides of an equality by a negative number.,

Problem 1) Solve for x and represent the solution set on a number line.

$$3-\frac { x }{ 4 } \ge 2$$

Sol.     Multiply both sides by 4                              12-x  ≥ 8

Subtract 12 from both sides                        -x ≥ -4

Multiply both sides by -1 and flip the inequality symbol      x ≤ 4

Problem 2) Solve the inequality

$$\frac { 3x }{ 2 } -\frac { 3x }{ 4 } <\frac { 7x }{ 4 } -1$$

Sol.

$$\frac { 3x }{ 2 } -\frac { 3x }{ 4 } -\frac { 7x }{ 4 } <-1\quad (Subtract\quad \frac { -7x }{ 4 } \quad from\quad both\quad sides)\\ \frac { 3x }{ 4 } -\frac { 7x }{ 4 } <-1\\ \frac { -4x }{ 4 } <-1\\ -x<-1\quad (Multiply\quad both\quad sides\quad by\quad -1)\\ x>1$$

Absolute Value: The absolute value of a number describes how far that number is away from 0 on a number line. The symbol of absolute value is |number|.

STEPS TO SOLVE ABSOLUTE VALUE EQUATION:-

• Take what’s inside the absolute value sign and set up two equations.First sets the positive value equal to the other side of the equation, and the second sites the negative value equal to the other side.
• Solve both equations:

Problem 3)        $$|2x+4|=30$$

Sol.

$$2x+4=30\quad or\quad -(2x+4)=30\\ x=13\quad \quad \quad \quad \quad \quad \quad +2x+4=-30\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=-17\\ x=13\quad or\quad (-17)$$

Inequalities and absolute value:-

### Steps for solving questions involving both inequalities and absolute value

• Set up Two Equations:- The first inequality replaces the absolute value with the positive of what’s inside and the second replaces the absolute value with the negative of what’s inside.

Problem 4)  | x | ≥ 4

Sol.

+( x ) ≥ 4 or –x ≥ 4

x ≥ 4 or x ≤ -4 Problem 5) | x+3 | ≤ 5

Sol.

+( x+3) < 5 and        -( x+3 ) < 5

x < 2                        x > -8 The only numbers that make the original inequality true are those that are true for both inequalities. X should be greater than -8 and less than 2.

NOTE:-   In conclusion, for solving absolute value expressions that are greater than some quantity set up two equations & after sol; having two equations.We have to take real numbers that are satisfied by either equation. However, for solving absolute value expressions that are less than some quantity set up two equations and after solving two equations, We have to take real numbers that are satisfied by both the equations.

Problem 6) If |3x + 7|  ≥ 2x + 12 , then which of the following is TRUE?

a) $$x\le \frac { -19 }{ 5 }$$

b) $$x\ge \frac { -19 }{ 5 }$$

c) $$x\ge 5$$

d) $$x\le \frac { -19 }{ 5 } \quad or\quad x\ge 5$$

Ans:- ( 3x + 7 ) ≥ 2x +12                               -( 3x + 7 ) ≥ 2x + 12

x ≥ 5                                                    -3x ≥ 2x +19

-5x ≥ 19

x ≤  -19/5

$${ (x\le \frac { -19 }{ 5 } \quad or\quad x\ge 5) }^{ Ans }$$

a), c) and d) are correct answers.

Using extreme values: Extreme values are helpful where the questions involve the potential range of value for variables in the problem.

Problem 7) If  -7  ≤ a  ≤ 6   and  -7  ≤ b  ≤ 8, What is the maximum possible value for ab?

Sol. Let us consider the different extreme value scenarios for a, b and ab.

abab
Min -7Min -749
Min -7Max 8-56
Max 6Min -7-42
Max 6Max 848

We can easily reckon that ab is maximized when we take the negative extreme values for both a and b. Therefore maximum value of  ab = 49

Problem 8) If $$|y|\le -4x\quad and\quad |3x-4|=2x+6$$,what is the value of x?

Sol.      $$|y|\le -4x$$

Any absolute value cannot be negative. |y| must be positive & positive value cannot be lesser than any negative value. Therefore, -4x must be positive.

For -4x to be positive, x must be negative.

Now solve the absolute value equation by using the identity that |a|=a when a is positive or zero & |a|=-a when a is negative.

$$|3x-4|=2x+6\\ 3x-4=2x+6\quad \quad or\quad -(3x-4)=2x+6\\ 3x-4-2x-6=0\quad \quad \quad \quad -3x+4=2x+6\\ x=10\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad -5x=2\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x=(\frac { -2 }{ 5 } )$$

x must be negative. Therefore, x can only be $$\frac { -2 }{ 5 } $. Problem 9) If [latex]2{ (x-1) }^{ 3 }+3\le 19$$, which of the following must be true? a) $$x\ge 3$$ b) $$x\le 3$$ c) $$x\ge -3$$ d) $$x\le -3$$ e) $$x<-3\quad or\quad x>3$$ Sol. $$2{ (x-1) }^{ 3 }+3\le 19\\ 2{ (x-1) }^{ 3 }\le 16\\ { (x-1) }^{ 3 }\le 8$$ Taking the cube root of an inequality is permissible here, because cubing a number, while squaring it, does not change its sign. $$x-1\le 2\\ { (x\le 3) }^{ Ans }$$ Problem 10) Which of the following could be the graph of all values of $$x$$ that satisfy the inequality $$4-11x\ge (\frac { -2x+3 }{ 2 } )$$ Sol.First solve the inequality, $$4-11x\ge \frac { -2x+3 }{ 2 } \\ 8-22x\ge 20x\\ \frac { 1 }{ 4 } \ge x\\ x\le \frac { 1 }{ 4 }$$ Therefore, correct answer is (a). Thus, the correct choice should show the gray line beginning to the right of zero (in the positive zone), & continuing indefinitely into the negative zero. Problem 11) $$p+|k|>|p|+k$$ Quantity A: p Quantity B: k 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. In general, there are four cases for the signs of p & k, some of which can be ruled out by the constraints of this equation:  p k p+|k|>|p|+k + + In this case, both sides should be positive. Therefore, this case is not true. + - This case is true. (positive value)+absolute value>(positive value)+(negative value) - - It may between. Both sides are positive plus a negative. For this condition to the true (|k|>|p|) - + This case is not true. k+(negative value) Additionally, check whether p or k could be zero. If p=0, p+|k|>|p|+k is equivalent to |k|>k. This is true when k is negative. If k=0, p+|k|>|p|+k (when k=0) p>|p| This is not true for any p value. So, there are three possible cases for p & k. Let’s interpret this more: – Use identity that |a|=-a, when a is negative.  p k Interpret + - p=positive >k=negative - - p+|k|>|p|+k p-k>-p+k 2p>2k Therefore, p>k 0 - p=0>k=negative Therefore, p>k In all cases, p is greater than k. Therefore, Quantity A is greater. Problem 12) $$|x|y>x|y|$$ Quantity A: $${ (x+y) }^{ 2 }$$ Quantity B: $${ (x-y) }^{ 2 }$$ 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. In general, there are four cases for the signs of x & y, some of which can be ruled out by the constraint in the question using identity that |a|=-a when a is negative.  x y |x|y>x|y| is equivalent to True or False? + + xy>xy False: xy=xy + - xy>x(-y) False: xy is negative & (-xy) is negative. - + (-x)y>(-xy) True: (-xy) is positive & xy is negative - - (-x)y>(-xy) False: -xy=-xy Note that if either x or y equals 0, that case would also fail the constraint. Quantity A: $${ (x+y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }+2xy$$ Quantity B: $${ (x-y) }^{ 2 }={ x }^{ 2 }+{ y }^{ 2 }-2xy$$ X is negative & y is positive. Therefore, xy is negative. $${ x }^{ 2 }+{ y }^{ 2 }-2xy>{ x }^{ 2 }+{ y }^{ 2 }+2xy\\ { (x-y) }^{ 2 }>{ (x+y) }^{ 2 }$$ Therefore, Quantity B is greater. Problem 13) If y<0 & 4x>y, which of the following could be equal to $$\frac { x }{ y }$$ ? a) 0 b) $$\frac { 1 }{ 4 }$$ c) $$\frac { 1 }{ 2 }$$ d) 1 e) 4 Sol. 4x>y (divide both sides by y) $$\frac { 4x }{ y } <1$$ Remember to switch the direction of the inequality sign when multiplying or dividing by a negative Quantity. $$\frac { 4x }{ y } <1$$ (Now divide both sides by 4) $$\frac { x }{ y } <\frac { 1 }{ 4 }$$ The only answer choice less than $$\frac { 1 }{ 4 }$$ is 0. Problem 14) If |1-x|=6 & |2y-6|=10, which of the following could be the value of xy? Indicate all such value? a) -40 b) -10 c) -14 d) 56 Sol. Remember that to solve the absolute value equation, First set the positive value equal to the other side of the equation & the second set the negative value equal to the other side of the equation. $$|1-x|=6,\quad -(1-x)=6\\ -1+x=6\\ x=7\\ \\ |1-x|=6,\quad +(1-x)=6\\ 1-x=6\\ x=-5\\ Therefore,\quad x=-5\quad or\quad 7\\ \\ |2y-6|=10,\quad 2y-6=10\\ 2y=16,\quad y=8\\ \\ 2y-6=-10,\quad 2y=-10+6=-4\\ y=-2\\ Therefore,\quad y=8\quad or\quad y=-2$$ Now calculate all four possible combinations for xy: (-5) (8) =-40 (-5) (-2) =10 7(8) =56 7(-2) =-14 The correct answer are -40, -14 & 56 only. Problem 15) If & x<0, which of the following could be the value of x? Indicate all such values: a) -6 b) -14 c) -18 d) -22 e) 36 Sol. Remember that for solving absolute value expressions that are greater than some quantity, set up two equations (the first inequality replaces the absolute value with the positive of what’s inside & the second replaces the absolute value with the negative of what’s inside) & we have to take real numbers that are satisfied by either equation. $$\frac { |x+4| }{ 2 } >5,\quad |x+5|>10\\ x+4>10\quad \quad \quad -(x+4)>10\\ x>6\quad \quad \quad \quad \quad \quad -x>14\\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad x<-14$$ x>6 is not a valid solution range, as the other inequality indicates that x is negative. x<-14, note that this fits the other inequality which states that x<0. If x<-14, only -18 & -22 are correct answers. ## GRE: Permutation AND Combination Permutation and Combination In this chapter, we shall learn about some basic counting techniques which will be useful in determining the number of different ways of arranging and selecting objects without actually listing them. Multiplication principle- Suppose student Mohan has 3 pants and 2 shirts. How many different pairs of a pant and shirt can he dress up with? There are three ways in which a pant can be chosen. For 1st pant————-any of the two shirts can be chosen———2 ways For 2nd pant————-any of the two shirts can be chosen———2 ways For 3rdpant————–any of the two shirts can be chosen———-2 ways For every pant, there are two shirts, therefore, he can dress up with 3×2 = 6. Let the three pants be P1, P2 and P3 and two shirts be S1, S2 and S3. Above can be easily understood by using this diagram: For every pant, there are two shirts, therefore, he can dress up in 3×2 = 6 ways. Number of pair of pants and shirts he can choose = (No. of pants he can choose) x (No. of shirts he can choose) Problems of above type are solved by using multiplication principle or fundamental principle of multiplication which states that: If an event can occur in p ways and another event can occur in n ways, number of ways in which both events p & q can be performed is equal to p x q. Addition principle- If an event can occur in p different ways and another event can occur in q different ways, number of ways in which event p or q can be performed is equal to p + q. Problem: In a debate competition, there are 5 candidates from science side, 4 from commerce side and 3 from humanities side. In how many ways a winner of competition can be selected? Sol. A winner out of science students can be selected in 5 ways. A winner out of commerce students can be selected in 4 ways. A winner out of humanities students can be selected in 3 ways. An overall winner of competition can be selected in 5 + 4 + 3 = 12 ways. Problem 1) In a shop, there are 10 shirts and there are 10 pants • In how many ways can a pair of pant and a pair of shirt can be selected. • In how many ways can a pair of cloth (pant or shirt) can be selected. Sol. Number of ways in which a pair of pant and a pair of shirt can be selected = 10 x 10 = 100 Number of ways in which a pair of cloth (pant or shirt) can be selected = 10 + 10 = 20 Factorial notation- The notation n! (factorial n) represents the product of first n natural numbers. Factorial of only natural numbers is defined. For example, 5! = 1 x 2 x 3 x 4 x 5 0! = 1 n! = n (n – 1)! = n (n-1) (n-2)! = n (n-1) (n-2) (n-3)! (provided n ≥ 3) n! tool for arrangement- Number of ways in which, we can palace N things in N places = N! or No. of ways in which we can arrange n things = N! Problem 2) How many 4-digits numbers can be formed from digits 1, 2, 3 and 4? (Repetition is not allowed) Sol. Number of 4-digits numbers that can be formed from digits 1, 2, 3 and 4 = 4! = 24 Problem: How many 7 letter words with or without meaning can be formed from letters M, N, O, P, Q, R and S? Sol. Number of 7 letter words with or without meaning formed from letters M,N,O,P,Q, R and S = 7! Problem 3) Of the different words that can be formed from the letters of the word BEGINS how many words begins with B and end with S? Sol. Number of words beginning with B and ending with S formed from letters of word BEGINS = 4! (Number of arrangements of letters E, G, I, N) Question based on combination of fundamental principle of multiplication and n! principle Problem 4) In how many ways can the 7 letters M, N, O, P, Q, R and S be arranged so that P and Q occupy continuous position? Sol. We should assume PQ as a single word. No. of arrangements of words M, N, O, PQ, R and S = 6! PQ may act as a single word and QP may also act as a single word. No. of ways in which letters M, N, O, P, Q, R and S can be arranged so that P and Q occupy continuous position = 6! X 2! = 1440 Combination (nCr) tool for arrangement- This tool is used for specific situation of counting, the number of ways of selecting r things out of n distinct things = [ nCr]. The number of ways of selecting r things out of n identical things = 1. nCr = n! /r! x (n-r)! important points regarding nCr • nCr + nCr – 1 = n + 1Cr • nCx = nCy It implies that (x = y) or (x + y = n) • When n is constant and r is variable, for nCr to be greatest 1. If n is even, r = n/2 2. If n is odd, r = (n + 1)/2 or (n-1)/2 Problem 5) A committee of 5 members is to be formed from group of 8 members. In how many ways can this be done? Sol. 5 members out of 8 members can be selected in 8C5ways. Number of ways in which 5 members out of 8 members can be selected = 8C5 = 56 Problem 6) How many chords can be drawn through 21 points on a circle? Sol. Drawing a chord is equivalent to selecting two specific points. Number of chords that can be drawn through 21 points = 21C2 Problem 7) In how many ways can a student choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory for every student? Sol. 2 specific courses are compulsory. Therefore, student has to choose 3 courses out of 7 courses. Number of ways in which a student can choose a program of 5 courses if 9 courses are available and 2 specific courses are compulsory = 7C3 Question based on combination of fundamental principle of multiplication and Combination Problem 8) A committee of 3 persons is to be constituted from a group of 2 men and 3 women. In how many ways can this be done? How many of these committees would consist of 1 man and 2 women? Sol. We have to select 1 man out of 3 men and 2 women out of 3 women. No. of ways in which this can be done = 2C1 X 3C2 Problem 9) In how many ways can one select a cricket team of eleven from 17 players in which only 5 players can bowl if each cricket team of 11 must include exactly 4 bowlers? Sol. We have to select 4 bowlers out of 5 bowlers and we have to select 7 players out of remaining 12 players. This can be done in 5C4 X 12C7 Problem 10) A bag contains 5 black and 6 red balls. Determine the number of ways in which 2 black and 3 red balls can be selected? • If balls of same color are distinct. • If balls of same color are identical. Sol. If balls of same color are distinct. No. of ways in which 2 black ball out of 5 black ball can be selected = 5C2 X 6C3 If balls of same color are same. No. of ways in which 2 black ball out of 5 black ball can be selected = 1 Permutation (arrangement) nPr Number of ways in which r things out of N distinct things can be selected and can be arranged at r different places = NCr x r! = NPr Problem 11) In how many ways 5 fruits out of group of 8 fruits can be selected and arranged at 5 different places? Sol. Number of ways in which 5 fruits out of 8 fruits can be selected and arranged at 5 different places = 8C5 x 5 = 56 Arrangement of objects not all distinct Number of arrangements of N things out of which P1 are of one type, P2 are of second type, P3 are of third type and the rest are all different = (N! / (P1! P2! P3!)) Problem 12) How many words with or without meaning can be formed with the letters of following words: – 1) ALLAHABAD 2) MISSISSIPPI 3) APPLE Sol. Number of words that can be formed with the letters of word ALLAHABAD = (9!)/ (4!2!) = 7560 Number of words that can be formed with the letters of word MISSISSIPPI = (11!)/ (4!4!2!) = 34650 Number of words that can be formed with the letters of word APPLE = (5!)/ (2!) = 60. Circular permutations There are three objects A, B and C. We can linearly arrange these three objects in 3! = 6 possible ways. There are following linear arrangements of these three objects: – ABC, ACB, CAB, CBA, BAC, BCA.  If we talk about circular arrangements, we can observe from figure 1 that three arrangements ACB, CBA and BAC are same. We can observe from figure 2 that three arrangements ABC, BCA and CAB are same. We conclude that there are two ways of circular arrangements of these three objects. This happens because there is no point of a starting point on a circular arrangement. Three objects can be arranged in ((3!)/3) ways = 2! Ways. Generalizing the whole concept, on circular table, N objects can be arranged in (N – 1)! Ways or (N!)/N. Problem 13) In how many ways five guests can sit on a round table? Sol. If we talk about necklace or garland, where we can turn necklace or garland and clockwise and anti-clockwise arrangements are not different. No. of circular arrangements of N different beads of necklace = (N – 1)! /2. Problem 14) How many different garlands can be formed from eight different beads? Sol. Number of garlands formed from eight different beads = 7!/2 = 2520. Selection of any number of things out of N distinct things If there are N distinct things, 1 thing out of N distinct thing can be chosen in NC1. m things out of N distinct things can be chosen in NCm ways. Any number (0 or 1 or 2) of things out of N can be chosen in (NC0 + NC1 + NC2 + NC3 + NC4 + NC5 . . . . . . . . . . . .. NCN) ways. (By using combination and fundamental principle of addition) NC0 + NC1 + NC2 + NC3 + NC4 + NC5 . . . . . . . . . . . .. NCN = 2N The number of selections of 1 or more things out of N different things = NC1 + NC2 + NC3 + NC4 + NC5 . . . . . . . . . . . .. NCN = 2N– 1 Problem 15) A boy has gone to a library and there are 200 different books in library. In how many ways he can pick one or more books from library? Sol. Number of ways in which he can pick one or more book = $${ 2 }^{ 200 }-1$$ Number of diagonals of N sided polygon Number of diagonals of N sided polygon = (Number of straight lines formed by joining N points) – (Number of sides) = NC2 – N Number of straight lines formed by N points of which l are collinear NC2 straight lines are formed by joining N points which are not collinear and only 1 straight line is formed by joining R collinear points. Therefore, number of straight lines formed by joining N points of which l are collinear = NC2 – lC2 + 1 Number of Triangles formed by N points of which l are collinear NC3 triangles are formed by joining N points which are not collinear and no triangle is formed by joining R collinear points. Therefore, number of triangles formed by joining N points of which l are collinear = NC3 – lC3 Number of parallelograms formed by intersection of m parallel lines with n parallel lines Two lines out of m parallel lines can be selected in mC2 ways and two lines out of n parallel lines can be selected in nC2 ways. Therefore, parallelogram is formed by selecting two lines out of m parallel lines and two lines out of n parallel lines. Number of parallelograms formed by intersection of m parallel lines with n parallel lines = mC2 x nC2 Problem 16) Find the number of diagonals & triangles formed in a decagon? a) 35, 130 b) 25, 120 c) 35, 120 d) 45, 120 Sol. No. of diagonals formed by decagon = $${ _{ }^{ 10 }{ C } }_{ 2 }-10\\ =\frac { 9\times 10 }{ 2 } -10\\ ={ (35) }^{ Ans }$$ No. of triangles formed in a decagon = $$=_{ }^{ 10 }{ { C }_{ 3 } }\\ =\frac { 8\times 9\times 10 }{ 2\times 3 } \\ ={ (120) }^{ Ans }$$ Problem 17) Out of 18 points in a plane, no three are in a straight line except 5 which are collinear. How many straight lines can be formed. a) $$_{ }^{ 18 }{ { C }_{ 2 } }$$ b) $$_{ }^{ 18 }{ { C }_{ 2 } }-_{ }^{ 5 }{ { C }_{ 2 } }$$ c) $$_{ }^{ 18 }{ { C }_{ 2 }-_{ }^{ 5 }{ { C }_{ 2 }+1 } }$$ d) $$_{ }^{ 18 }{ { C }_{ 2 }-_{ }^{ 5 }{ { C }_{ 2 }-1 } }$$ e) $$_{ }^{ 18 }{ { C }_{ 2 }-_{ }^{ 5 }{ { C }_{ 2 }+2 } }$$ Sol. No. of straight lines that can be formed = $$=_{ }^{ 18 }{ { C }_{ 2 }-_{ }^{ 5 }{ { C }_{ 2 }+1 } }$$ Problem 18) How many numbers of 3-digits can be formed with the digits 1, 2, 3, 4, 5 (repetition of digits not allowed) a) 125 b) 120 c) 60 d) 180 Sol. The number of numbers formed would be given by $$=5\times 4\times 3=60$$ (First digit can be filled in 5 ways, second can be filled in 4 ways & third can be formed in 3 ways.) Problem 19) How many nos. between 2000 & 3000 can be formed with the digits 0, 1, 2, 3, 4, 5, 6, 7 (repetition of digits not allowed)? a) 42 b) 210 c) 336 d) 440 Sol. The first digit can only be 2 (1 way), 2nd digit can filled only in 7 ways, 3rd digit can be filled only in 6 ways & 4th digit can be filled only in 5 ways. No. of object = $$=7\times 6\times 5=210\quad ways$$ Problem 20) In how many ways can a person send invitation cards to 6 of his friends if he has four servants to distribute the cards? a) 42 b) 210 c) 24 d) 120 Sol. Each invitation can be sent in 4 ways.ways. Thus $$4\times 4\times 4\times 4\times 4\times 4={ (4) }^{ 6 }$$ Problem 21) In how many ways can 7 Indians, 5 Pakistanis & 6 Dutch be seated in a row so that all persons of the same nationality sit together? a) 3! b) 7!5!6! c) 3!7!5!6! d) 182 Sol. We should assume 7 Indians as one person, 5 Pakistanis as one person, 6 Dutch as one person. 5 Pakistanis can be arranged in 5! Ways. 7 Indians can be arranged in 7! Ways & 6 Dutch can be arranged in 6! Ways Pakistanis, Dutch & Indians can be arranged in 5! Ways. Therefore, 7 Indians, 5 Pakistanis and 6 Dutch can sit in (7!×5!×6!×3!) ways. Problem 22) In how many ways five chocolates can be chosen from an unlimited number of dairy milk, five-star & perk chocolates? a) 81 b) 243 c) 21 d) 31 Sol. We have to choose five chocolates. We can choose first chocolate in three ways (either dairy milk or five-star or perk). Similarly, we can choose another chocolate in three way. Therefore, five chocolates can be choosen in $$3\times 3\times 3\times 3\times 3={ (3) }^{ 5 }$$ ways. Problem 23) There are 6 pups & 4 cats. In how many ways can they be seated in a row so that no cats sit together? Sol. 1st arrange 6 pups in 6 places in 6! ways. This will leave us with 7 places for 4 cats. Assume = $$_{ }^{ 7 }{ { P }_{ 4 }(6!) }$$ ways Problem 24) A state issues automobile license plates that begin with two letters selected from a 26-letter alphabet, followed by four numerals selected from the digits 0 through 9 inclusive. Repeats are permitted. For example, one possible license plate combination is GF3352. Quantity A: The number of possible unique license plate combinations. Quantity B: 60,00,000 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 license plate have 2 letters followed by 4 numbers. No. of possible license plate = $$26\times 26\times { 10 }^{ 4 }={ (6760000) }^{ Ans }$$ There are 26 letters in the alphabet and 10 digits to pick from. Problem 25) A 10-student class is to choose a president, vice president & security from the group. If no person can occupy more than one post, in how many ways can this be accomplished? Sol. There are 10 students from whom president can be chosen & after that there are 9 students from which vice president can be chosen, 8 students from which secretary can be chosen. Whole process can be accomplished in = (8×9×10) = 720 ways. Problem 26) A history exam features five questions. There of the questions are multiple-choice with four options each. The other two questions are true or false. If Caroline selects one answer for every question, how many ways can she/he answer the exam? Sol. No. of ways in ways she/he answer the question $$=4\times 4\times 4\times 2\times 2=64\times 4={ (256) }^{ Ans }$$ Problem 27) In a school of 150 students, 75 study Latin, 110 study Spanish & 11 study neither. Quantity A: The no. of students who study only Latin Quantity B: 46 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 A be the no. of students who study only Latin. Let B the no. of students who study both Latin & Spanish. Let C be the no. of students who study only Spanish. Let D be the no. of students who studies neither. $$A+B+C+D=150\\ A+B=75\\ B+C=110\\ D=11\\ A=29,\quad B=46,\quad C=64$$ No. of students who study only Latin = 29 Quantity B is greater. Problem 28) Mario’s pizza has two choices of crust. The restaurant also has 5 choices of toppings. Finally Mario’s offers every pizza in extra-cheese as well as regular. If Linda’s volleyball team decides to order a pizza with four different toppings, how many different choices do the teammates have at Mario’s pizza? Sol. Consider the topics first, four the toppings are on the pizza & one isn’t. Four out of five can be chosen in $$_{ }^{ 5 }{ { C }_{ 4 } }$$ ways. If each of these pizzas can be offered in 2 choices of crust, 2 choices of cheese & 5 choices of toppings. Total no. of choices do the teammates have at Mario’s pizza = 5×2×2 = 20 ways Problem 29) Country X has a four-digit postal code assigned to each town & none of the digits is repeated. Quantity A: The numbers of possible portal codes in country X Quantity B: 4536 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. No. of possible postal code in country X = 10×9×8×7 = 5040 Quantity A is bigger. Problem 30) 8 athletes compete in a race in which gold, silver & a bronze medal will be awarded to the top three finishers, in that order. Quantity A: The number of ways in which the medals can be awarded. Quantity B: 8×3! 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. No. of ways in which the medals can be awarded$=8\times 7\times 6=336\quad ways

Quantity A is bigger.

(Gold medal winner can be selected in 8 ways, silver medal winner can be selected in 7 ways & bronze medal winner can be selected in 6 ways.

# DISQUIET (Verb)

Meaning- to make (someone) worried or nervous.

Synonym- DISCOMPOSE, PERTURB

Uses-

• In the exam, I was disquieted by the strange noises we herd outside our school.

DISQUIET (Noun)

Meaning- A feeling of worry (or Nervousness) or lack of peace (or tranquility).

Synonym- ANXIETY

Uses-

• There is increasing public disquiet about the number of crimes against women in the city.
• A period of disquiet before the results of the elections was confirmed.

# DISSEMBLE (Verb)

Meaning- 1) to hide your true feeling, opinion, nature etc.

2) to make a false show of.

Mnemonic trick- The word dissemble contains word ‘ssemble’. ‘ssemble’ sounds similar to assemble. Director dissembled after ‘de-assembling’ the company. Therefore, dissemble means to hide your true feeling, opinion, nature etc. or to make a false show of.

Synonym- COUNTERFEIT, FEIGN, PRETEND, SHAM

Uses-

• He accused the government of dissembling the facts about our economy.
• Children learn to dissemble at a surprisingly early age.
• Hayden’ s friend dissembled his real feeling regarding his wife’ s leaving him for another man.
• He dissembled his feelings at the news that his old girlfriend was getting married to someone else.

Meaning- 1) lack of agreement; especially: inconsistency between the beliefs.

2) mingling of discordant or inharmonious sounds.

Mnemonic trick- The word dissonance is opposite in meaning to word resonance. The word resonance means the quality of sound that stays loud, clear and pleasant for a long time or quality that makes something personally meaningful or important to see. Therefore, dissonance means mingling of inharmonious sounds or conflict of opinion.

Synonym- DISHARMONY, DISACCORD, DISCORDANCE

Uses-

• Western composers often use dissonance -clashing or unresolved chords for special effects in their musical work.
• Ending as an unsolved dissonance is a peculiar quality of his stories.

Meaning- Inattentive or distracted especially because of anxiety.

Synonym- DELIRIOUS, DISTRACTED, DISTRAUGHT

Uses-

• He grew more and more distrait as hours passed without confirmation that there were survivors of the plane crash.
• When Rahul did not show up for the meeting, her employer was uneasy and distrait.
• I appeared to be distrait before the exam because I lost admit card before the exam.

# DOCILE (OR PLIABLE) (Verb)

Meaning- 1) Easily, taught, led or controlled.

2) Easily managed or obedient.

Synonym- AMENABLE, TRACTABLE, OBEDIENT

Uses-

• I had a docile, well-behaved pet.
• Docile people sometimes wonder why they are not more popular; after all, they never make any trouble for anyone.
• When a child becomes unusually pliant, obedient and manageable, he or she may, in short, become docile.
• Other people usually find nothing admirable or necessarily likable about a docile person who is willing to be pushed around, or who does what he or she is told out of fear of offending others.

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# DISABUSE (Verb)

Meaning- Free someone from a mistake in thinking.

Uses-

• Do you really believe that toilets flush one way in the Northern hemisphere and another way in the Southern hemisphere? Any physicist would be happy to disabuse you of that silly notion.

# DISCERN (Verb)

Meaning-  1) to see, recognize or understand something that is not clear.

2) to see or understand the difference.

Mnemonic trick- The word discern contains word ‘ern’. ‘ern’ sounds similar to earn. We normally try to discern how much our competitors earn. Therefore, discern means to see, recognize or understand something that is not clear.

Synonym- PERCEIVE, DISTINGUISH, OBSERVE

Uses-

• The reason behind this sudden change is difficult to discern.
• I was barely able to discern anything in the darkness.

# EMANCIPATE (OR DISENTHRALL) (Verb)

Meaning- to free somebody especially from legal, political or social restrictions.

Synonym- FREE, LIBERATE, MANUMIT

Uses-

• Slaves were not emancipated until 1863 in the United States.
• The women’ s liberation movement would like to emancipate women from being restricted to their homes.
• Mildred was quite smart and received her high school diploma while she was just 15, enabling her to start her university education in another city and being independent; therefore, being emancipated from control by her mother and father.

# DISJOINT (Verb)

Meaning- 1) to separate something at the joints.

2) to force or to move something out of its usual position, or be moved out of the usual position.

3) to destroy the unity or coherence of something.

Synonym- DECOUPLE, DISJOINT, DISUNITE, SEPARATE

Uses-

• He is disjointing himself from the organization because he realizes his organization id advocating illegal practices.
• The author deliberately disjoints his narrative in favor of a more impressionistic account of the war.

Meaning- Different from each other.

Mnemonic trick- The word disparate contains word ‘sparate’. ‘Sparate’ sounds similar to separate. If we combine separated things, then we get a disparate combination of things. Therefore, disparate means different from each other.

Synonym- HETEROGENEOUS, DISSIMILAR, INCONGRUENT, CONTRASTING, DISTINCT

Uses-

• This wine has more disparate flavor than that one.
• When I discussed the cashless economy with my relatives, I got disparate ideas from adults and children.
• The school was composed of students from disparate cultures.
• Even though the boys were twins, their personalities were disparate.

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## CHAPTER 28 on Vocabulary

Meaning- bitter scolding or piece of writing that strongly criticizes someone or something.

Mnemonic trick- The word diatribe contains word ‘tribe’. If you speak something against any tribe or quota for schedule tribe. Definitely, tribe will launch a diatribe against you. Therefore, diatribe means bitter scolding or piece of writing that strongly criticizes someone or something.

Uses-

• His novel was a strong and more outspoken diatribe against society and marriage laws.
• The article is a diatribe against mainstream media.
• During the lengthy diatribe delivered by his opponent, he remained calm and self-controlled.
• Channel launched an open diatribe against government policies.

Meaning- 1) lack of confidence or not feeling comfortable around certain people.

2) very careful about acting or speaking.

Mnemonic trick- The word diffident is opposite in meaning to word confident. The word diffident is used for a person having lack of self-confidence and who does not feel comfortable around certain people.

Synonym- UNASSERTIVE, SHY, INTROVERTED, HESISTANT, RESERVED, MODEST

Uses-

• When John is called on to answer a question that his teacher is asking, he usually has a diffident way of expressing himself even when he is certain that he is correct.
• When the reporter asked the politician, if he would be running for a third term, she made a diffident response because she wanted to avoid questions on his governance.

# DIGRESS (Verb)

Meaning- to speak or write about something that is different from the main subject being discussed.

Synonym- SWERVE, DEVIATE

Uses-

• Mayor did not take a stand on corruption, he digressed from the topic.
• If I can digress for a moment, I want to briefly mention her earlier films.

# DIGRESSION (Noun)

Meaning- A spoken topic or written document about something which is different from the main subject being discussed.

Uses-

• The speaker at the conference presented many humorous digressions even though it was supposed to be a serious financial report.
• On TV, the political commentator was revealing the social and political digressions that the senator had committed while he was a member of the U.S Congress.

# DILETTANTE (OR DABBLER) (Noun)

Meaning- person whose interest in an art or in an area of knowledge is not very deep or serious

Mnemonic trick- dilettante contains two words ‘dil’ and ‘let’. We are not letting our dil or heart or heart to completely open for something. It means that we are not deeply engaged in or concerned with something. Therefore, dilettante means a person whose interest in an art or in an area of knowledge is not very deep or serious.

Synonym- SUPERFICIAL, AMATEUR

Uses-

• Michael flit like a butterfly from interest to interest, never concentrating for any length of time on one. He is a true dilettante.
• AS a Salesman, Erwin was still a dilettante because he was really not interested in that type of work.
• He is not a struggling artist. Just like a dilettante, he follows the art as a pastime.
• You can tell the difference between a true expert and a scientist who is just a dilettante.

# DIRGE (Noun)

Meaning- A slow song or piece of music that expresses sadness or sorrow.

Mnemonic trick- The word dirge sounds similar to Hindi word ‘dird’. Dirge expresses ‘dird’. Therefore, dirge means a slow song or piece of music that expresses sadness or sorrow.

Synonym- ELEGY, THRENODY, LAMENT, REQUIEM

Uses-

• The choir will sing dirge from Bollywood movie.
• The funeral dirge stirred us to tears.

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## CHAPTER 27 on Vocabulary

Meaning- 1) to write or to talk about another person, or something in a very critical or insulting way.

2) to say that someone or something is of no value or is ridiculous.

Mnemonic trick- The word deride can be memorized by thinking that it is made up of two words ‘de’ and ‘ride’. When I ask him for ride by saying ‘de de ride’, he derided me. When a person fell off a ride in an amusement park. He was derided by his friends. Therefore, deride means to write or to talk about another person, or something in a very critical or insulting way.

Synonym- RIDICULE, SCOFF, MOCK, SCORN, TAUNT

Uses-

• My friends derided my efforts, but were forced to eat their words when we won first place.
• Deriding other politicians is a contemporary fashion in Indian politics.
• A newspaper critic derided David’ s book as dull and worthless.
• His boss derided his ideas as ludicrous.

# SACRILEGE (OR BLASPHEMY OR DESECRATION OR IRREVERENCE) (Noun)

Meaning- The act of treating a holy thing or place without respect.

Synonym- IMPIETY, DEFILEMENT

Uses-

• Hindu consider it sacrilege to wear shoes inside a temple.
• They accused him of committing a sacrilege.
• In 2010, Advertisement agency, produced images of Prophet Mohammad in an episode that featured the Prophet as a bear mascot. Angered by the blasphemy, one Islamic website threatened the producer for what they deemed as a high level of disrespect for the Prophet.

# DESICCATE (Verb)

Meaning- 1) to dry up.

2) to preserve (something such as food) by removing the moisture.

3) to drain the power or ability of something to continue to live, be successful etc.

Mnemonic trick- The word desiccate can be memorized by thinking that it is made up of two words ‘des’ and ‘cat’. ‘DESERT KI CAT’ died because of scarcity of water or because of desiccation. Therefore, desiccate means to dry up.

Synonym- PARCH, SHRIVEL, DEHYDRATE, WITHER, EXSICCATE

Uses-

• The surgeon thoroughly desiccated the tissue with a needle electrode.
• While making sweets for the party, Elizabeth used a cup of desiccated coconuts.
• This movie revolves around the married life and desiccated romance after marriage.
• Historian’ s prose desiccates what is actually an exciting period in European history.
• Desiccated land or drought is a result of environmental degradation.

# DESPONDENT (OR DEJECTED OR DISCONSOLATE) (Adj.)

Meaning- very sad and without hope because of loss, failure etc.

Synonym- DEPRESSED, MELANCHOLIC, GLOOMY

Uses-

• Karan was very despondent about losing his job after fifteen years of work in the company; however, the business was having economic problems and laying off many of its employees.
• He became despondent after working very hard and failing to achieve desired results.
• She looked despondent when she was told that she had n’ t got the job.

# DESPOT (OR TYRANT) (Noun)

Meaning- a person especially a ruler who has unlimited power over other people and often uses it unfairly and cruelly.

Synonym- DICTATOR

Uses-

• The king was regarded as having been an enlightened despot.
• Persians installed despots to rule their Greek subjects, but such rulers were unpopular.
• There are some despots, even in these modern times, who are ruling their people in cruel and inhumane ways.
• My uncle was a good football coach; however, a lot of people considered him to be a despot because he was always dictating what the players had to do during the games.

# DESTITUTE (OR IMPECUNIOUS OR PENURIOUS OR INDIGENT OR PENNILESS) (Adj.)

Meaning- very poor or having little or no money.

Synonym- PAUPERIZED

Antonym- AFFLUENT, PLUSH, PROSPEROUS, OPULENT

Explanation- Destitute, impecunious, penurious, indigent and penniless means poor. Impoverish has two meanings. It means to make someone poor or to deplete the quality or richness of something. For example –

• poor farming practices impoverished the soil.
• Dictators enriched themselves but impoverished the public.

Uses of Destitute-

• He got fired from the job and now he is destitute.
• Many families were left destitute by the second world war.

Meaning- not having a plan or purpose or done without making a serious effort.

Mnemonic trick- The word desultory contains word desult. Desult sounds similar to result. Something done without aiming to produce a good result is desultory. Therefore, desultory means something done just perfunctorily without making a serious effort or without aiming to produce a good result.

Synonym- AIMLESS, UNFOCUSSED, HAPHAZARD

Uses-

• We usually make a desultory search on the internet for entertainment.
• Michael was searching for something of interest on television desultorily.
• Your essay seems desultory and unclear. You should write to the point.
• She made a desultory attempt at conversation.

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# DELINEATE (Verb)

Meaning- to describe, show or explain something in detail.

Mnemonic trick- The word delineate contains word deli and eat. Deli sounds similar to Delhi. My parents clearly delineated what to eat and what not to eat in the crowded market of Delhi. Therefore, delineate means to explain or describe something in detail.

Synonym- SPECIFY, OUTLINE, DELIMIT, DEFINE

Uses-

• Our objectives need to be properly delineated.
• The ship’ s route is clearly delineated on the map.
• The politician was delineating his development plan in his election rally.
• In his novel, the story was carefully delineated.

Meaning- A political leader who wins support by exciting people’ s emotions rather than by having good ideas. Demagogue tries to get support by making false claims and promises.

Explanation- “dem” is a root word for people. For example- Democracy means the government of the people. Greek word “agogos” means “leading”. Therefore, demagogue means leader of the mob who appeals to the passions and prejudices of the mob in order to get political advantage.

Synonym- RABBLE-ROUSER

Uses-

• His opponent called him a bigoted demagogue.
• Here’s what demagogues like Trump do to their countries when they take power.
• He was an enthusiastic, but a fickle and ambitious demagogue, and he achieved a better reputation as a writer.

Meaning- School teacher especially who gives attention to formal rules or traditional method of teaching.

Explanation- ped- is a word-forming element. It means “child or boy”. Greek word “agogos” means “leading”. Therefore, pedagogue means teacher or schoolmaster.

Synonym- PEDANTIC TEACHER, EDUCATIONIST

Uses-

• The teaching of a pedagogue consisted entirely of reading directly from the textbook in the same tone.
• Teaching and giving direction to children is what a pedagogue does.

# DENIZEN (OR INHABITANT) (Noun)

Meaning- a person or animal that lives in a particular place.

Mnemonic trick- The word denizen sounds similar to citizen. Therefore, citizen or denizen means a person who lives in particular place.

Synonym- DWELLER, NATIVE

Uses-

• The denizens of this island don’ t like the tourist.
• Lion-tailed macaque, koala bears are among the denizens of the forest.
• The polar bear is an iconic denizen of the snowy Arctic.

# DEPRECIATE (Verb) and DEPRECATE (Verb)

## DEPRECIATE(Verb)

Meaning- to decrease in value or to describe something as having little value

Synonym- REDUCE, ATTENUATE

Uses-

• The value of the property has depreciated greatly.
• There was a slight period during which the stocks seemed to depreciate but they rallied after a short time.
• He made unworthy attempt to depreciate the learning of great historians.
• Michael felt that he had to depreciate Seema’ s effort to help her.

## DEPRECATE (Verb)

Meaning- to criticize or express disapproval of something

Synonym- BELITTLE, DISPARAGE

Uses-

• He looks down on the acts of others or He deprecates.
• He showed contempt and disapproval of younger generation. He deprecated the younger generation.

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