# Exercise : 1

(b)

^{Y}⁄

_{2}

(c) 2XY – X

(d) X(Y – 1)

(e) None of these

**Ans.c**

Let the other number is N.

Then , \(\frac{X + N}{2}\) = XY ⇒ N = 2XY - X

(b) 3

(c) 5

(d) Cannot be determined

(e) None of these

**Ans.a**

Let the four consecutive odd nos. be

2x – 3, 2x – 1, 2x + 1 and 2x + 3.

Now, 2x = 12 or, x = 6

Lowest odd no. = 2 × 6 – 3 = 9

(b) 50

(c) 60

(d) Data inadequate

(e) None of these

**Ans.c**\(\frac{E + H + G + M}{4} \; - \; \frac{E + S + M + H}{4} = 15\)

⇒ G - S = 60

(b) 45 years

(c) 40 years

(d) Data inadequate

(e) None of these

**Ans.c**

Age of the CT = 25 × 16 – 24 × 15 = 400 – 360 = 40 yrs.

(b) 42.9 kg

(c) 49.9 kg

(d) 39.9 kg

**Ans.c**

Here one boy is excluded and final average of the group decreases.

∴ change in average is (–)ve = – 0.1 kg.

Using the formula

Sum of the quantities excluded

= \(\begin{pmatrix} Change \; in \; no. \; of \; quantities\\ \times \\ Original \; Average\end{pmatrix} + \begin{pmatrix} Change \; in \; Average\\ \times \\ Final \; no. \; of \; quantities\end{pmatrix}\)

⇒ weight of the boy who left = (1 × 45) – (– 0.1 × 49)

= 49.9 kg

∴ weight of the boy who left the class is 49.9 kg.

(b) 36

(c) 51

(d) cannot be determined

**Ans.c**

Age of the teacher = (37 × 15 – 36 × 14) years

= 51 years.

(b) 77 kg

(c) 76.5 kg

(d) Data inadequate

(e) None of these

**Ans.b**

total weight increases = 8 × 1.5 = 12 kg

so the weight of new person = 65 + 12 = 77 kg

(b) 400

(c) 450

(d) Cannot be determined

(e) None of these

**Ans.e**

Required persons = \(\frac{325000 - 300000}{50}\) = 500

(b) 45

(c) 60

(d) Data inadequate

(e) None of these

**Ans.e**

Set the first, second and third no be F, S and T

Respectively \(\frac{F + S}{2} = \frac{S + T}{2} + 15\). Solving, we get F – T = 30.

(b) 60

(c) 20

(d) Data inadequate

(e) None of the above

**Ans.c**

E + H = (55 × 2) = 110;

E + S = (65 × 2) = 130

∴ Reqd difference = 130 – 110 = 20