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Q:

Each question contains Quantity I and Quantity II. Read the contents carefully and answer your questions accordingly.

Quantity I: there are three numbers in the ratio 5:6:10. The sum of the largest and the smallest numbers is 126 more than the other number. Find the largest number?

Quantity II: 12% of first number is equal to 25 % second number. The difference of these two numbers is 78. Then find the largest number?

A) Quantity I > Quantity I	B) Quantity I ≥ Quantity II
C) Quantity I < Quantity II	D) Quantity I < Quantity II

Answer & Explanation Answer: C) Quantity I < Quantity II

Explanation:

Quantity I= Let the three numbers will be 5x, 6x and 10x respectively

ATQ

10x+5x−6x=126

9x=126 X=14

So largest number =10x=10×14=140

Quantity II=let x and y are two numbers

ATQ

12% of x=25%of y And their difference is 78

So first no.=150 2nd no.=72

Hence Quantity I < Quantity II

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Filed Under: Ratios and Proportions

Q:

Each question contains Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

Quantity I: 3 years ago the ratio of age of A and B is 3: 4 after 2 years the sum of their ages is 45. Then find the present age of A?

Quantity II: 5 years ago, the ratio of age of p and Q id 3: 4 P’s age after 6 years is equal to the present age of Q. the find the present age of P?

A) Quantity I > Quantity II	B) Quantity I ≥ Quantity II
C) Quantity I < Quantity II	D) Quantity I < Quantity II

Answer & Explanation Answer: C) Quantity I < Quantity II

Explanation:

Quantity I= Let 3 year ago the age of A and B be 3x and 4x respectively

Then ATQ 3x+5+4x+5=45 X=5

Hence present age of A=3x+3=18year

Quantity II== Let 5 year ago the age of p and q be 3x and 4x respectively

Then ATQ

3x+11=4x+5

X=6 Hence present age of P=3x+5=23

Hence Quantity I < Quantity II

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Filed Under: Problems on Ages

Q:

Each question contains Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

Quantity I: The Sl on a certain sum of money for 3 years at 5% per annum is Rs 4800. Then the principle is?

Quantity II: The Cl on a certain sum of money for 2 years at 6% per annum is Rs. 3708. Then the principle is?

A) Quantity I > Quantity II	B) Quantity I ≥ Quantity II
C) Quantity I < Quantity II	D) Quantity I < Quantity II

Answer & Explanation Answer: A) Quantity I > Quantity II

Explanation:

Quantity I=15% of Principle=4800

Principle=Rs.32000

Quantity II=12.36% of Principle =3708

Principle=Rs.30000

Hence Quantity I > Quantity II

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Filed Under: Simple Interest

Q:

Each question contains Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

Quantity I: Vipin can swim at 6 km/hr in still water. The river flows at 3 km/hr and it takes 8 hours more upstream then downstream for the same distance. How far is the place?

Quantity II: A man can row 25 km/hr in still water and the river is running at 15 km/hr. if the man takes 2 hr to row to a place and back, how far is the place?

A) Quantity I > Quantity II	B) Quantity I ≥ Quantity II
C) Quantity I < Quantity II	D) Quantity I < Quantity II

Answer & Explanation Answer: A) Quantity I > Quantity II

Explanation:

Quantity I: let d be the distance covered ATQ d/(6-3)-d/(6+3)=8 d=36km

Quantity II: ATQ d/(25-15)+d/(25+15)=2 D=16km

Hence Quantity I > Quantity II

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Filed Under: Time and Distance

Q:

$3 x^{2} - 14 x + 16 = 0 5 y^{2} - 16 y + 12 = 0$

A) x > y	B) x < y
C) x=y or relation cannot be established between x and y	D) x ≥ y

Answer & Explanation Answer: C) x=y or relation cannot be established between x and y

Explanation:

3x2 – 14x + 16=0

3x2 –6x – 8x + 16=0

3x (x-2) – 8 (x-2)

(3x-8) (x-2)=0

x=8/3 or 2

5y2 – 16y + 12=0

5y2 – 10y – 6y + 12=0

5y (y-2) – 6 (y-2)=0

(5y-6) (y-2)=0

y=6/5 or 2

Can't be determined.

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Filed Under: Simplification

Q:

In each of these questions, two equations (I) and (II) are given.

$I . 2 x^{2} + 19 x + 42 = 0 II . 4 y^{2} + 43 y + 30 = 0$

A) x > y	B) x < y
C) x ≥ y	D) x ≤ y

Answer & Explanation Answer:

Explanation:

I. 2x2 + 19x + 42 = 0

2x2 + 12x + 7x + 42 = 0

2x (x+6) + 7(x +6) = 0

(2x+7) (x+6) = 0 x = -7/2, -6 I

I. 4y2 + 43y + 30 = 0

4y2 +40y + 3y + 30 = 0

4y(y + 10) + 3(y + 10) (4y + 3)(y + 10) = 0

y = -3/4, -10

So answer is no relation.

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Filed Under: Simplification

Q:

In each of the following questions two equations I and II are given. Solve both the equations and give answer:
$I . x^{2} + 2 x - 195 = 0 II . y^{2} + 30 y + 225 = 0$

A) x = y or relation can’t be established between x and y	B) x > y
C) x < y	D) x ≥ y

Answer & Explanation Answer: D) x ≥ y

Explanation:

I. x2 + 2x - 195 = 0 = (x+15) (x-13) => x = -15, 13

II. y2 + 30y + 225 = 0 = (y+15) (y+15) => y = -15, -15

So x ≥ y

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Filed Under: Simplification

Q:

What should come in place of question mark (?) in the following question?

600 + 300 ÷ 15 = ? × 31

A) 18	B) 25
C) 40	D) 20

Answer & Explanation Answer: D) 20

Explanation:

As per the BODMAS rule, Addition and subtraction can be treated on same priority (from left to right) when they are in consecutive order. 600 + (300/15) = ? × 31 600 + 20 = ? × 31 ? = 620/31 = 20

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Filed Under: Number Series

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