# Exercise : 5

^{4}⁄

_{7}years

(b) 31

^{5}⁄

_{7}years

(c) 32

^{1}⁄

_{7}years

(d) 27

^{1}⁄

_{2}years

(e) None of these

**Ans.b**

Required average = \(\left ( \frac{67 \times 2 + 35 \times 2 + 6 \times 3}{2 + 2 + 3} \right )\)

= \(\left ( \frac{134 + 70 + 18}{7} \right ) = \frac{222}{7}\)

= 31^{5}⁄_{7} years

(b) 68 kg

(c) 69 kg

(d) 66.5 kg

(e) None of the above

**Ans.d**

Let Arun’s weight be X kg.

According to Arun, 65 < X < 72.

According to Arun’s brother, 60 < X < 70.

According to Arun’s mother, X < 68.

The values satisfying all the above conditions are 66 and 67.

∴ Required average = \(\left ( \frac{66 + 67}{2} \right ) = \left ( \frac{133}{2} \right ) = 66.5 \; kg.\)

(b) 26 years

(c) 28 years

(d) 27 years

(e) None of the above

**Ans.a**

Let the new man was younger than the director = x years and 3 years ago, the sum of ages of board of directors

= S – 8 × 3 = S – 24

Then, 3 years ago, average age of board of directors

= \(\frac{S - 24}{8}\)

Now, \(\frac{S - 24}{8} = \frac{S - x}{8}\)

⇒ x = 24 years

**Shortcut Method :** If the new young director would have been not substituted, then total age would have increased at present by 8 × 3 = 24 years.

Therefore, the new man is 24 years younger keeping the average at present same as 3 years ago.

(b) 24

(c) 28

(d) 26

(e) None of the above

**Ans.d**

Let the average score of 19 innings be x.

Then, \(\frac{18x + 98}{19} = x + 4\)

The average score after 20th innings

= x + 4 = 22 + 4 = 26

^{1}⁄

_{3}

(b) 52

^{2}⁄

_{3}

(c) 52

^{1}⁄

_{3}

(d) 43.42

(e) None of these

**Ans.b**

Total weight of 45 students

= 45 × 52 = 2340 kg

Total weight of 5 students who leave

= 5 × 48 = 240 kg

Total weight of 5 students who join

= 5 × 54 = 270 kg

Therefore, new total weight of 45 students

= 2340 – 240 + 270 = 2370

⇒ New average weight = ^{2370}⁄_{45} = 52^{2}⁄_{3} kg

(b) 40.4

(c) 40.6

(d) 40.8

(e) None of the above

**Ans.a**

Sum of 10 numbers = 402

Corrected sum of 10 numbers

= 402 – 13 + 31 – 18 = 402

Hence, new average = ^{402}⁄_{10} = 40.2