A. 2bc
B. 8b – 10c
C. 8c – 10b
D. 10c -8b

A. X-2
B. A – 2
C. X + 12
D. None of these

A. 3(x – 5)
B. 5(x – 3)
C. 3(x + 5)
D. None of these

A. 7 days
B. 6 days
C. 5 days
D. 8 days

A. 24
B. 30
C. 120
D. 240

A. 11
B. 12
C. 9
D. 14

A. 4
B. 5
C. 8
D. 10

A. 9
B. 1
C. 3
D. 0

2x +9x = 11
11 x = 11
x = 11 / 11
x = 1

## A candidate, appearing for an examination, was asked to find 3/14 of a certain number but mistakenly, he found 3/4 of it. When he cross checked his answer, he found his answer was 150 more than the correct answer. Which number was given in the examination for this calculation?

A. 160
B. 180
C. 240
D. 280

Let the number be x.

The candidate was asked to find 3/14, i.e. 3/14 of x but mistakenly, he found 3/4 of x, and the difference between these answers = 150

3/4 of x – 3/14 of x = 150
⇒ 3/4 × x − 3/14 × x = 150
⇒ 3x/4 − 3x/14 = 150
⇒ 3x × (1/4 – 1/14) = 150
⇒ 3x (5/28) =150

x = 150 × 28 / 5 × 3 = 280

## In the following question, one or two equation(s) is/are given. On their basis you have to determine the relation between x and y and then give answer I. x^2 + 3x + 2 = 0 II. 2y^2 = 5y

A. x < y
B. x > y
C. x ≤  y
D. x = y

Correct Answer: A. x < y

We will separately solve both equations.

Equation 1:

x2 + 3x + 2 = 0
⇒ x2 + 2x + x + 2 = 0
⇒ x (x + 2) + 1 (x + 2) = 0
⇒ (x + 2) × (x + 1) = 0
⇒ x = -2 or, x = -1

Equation 2:

2y2 = 5y
⇒ 2y2 – 5y = 0
⇒ 2y × (y – 5/2) = 0
⇒ y = 0 or, y = -5/2
Both values of y are positive while both values of x are negative.

∴ y > x

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