model 6 find average of excluded number Section-Wise Topic Notes With Detailed Explanation And Example Questions

Answer: (a)

Total weight of 40 children = 40 × 36.2 kg = 1448 kg

Total weight of 43 children = 1448 + 42.3 + 39.7 + 39.5 = 1569.5 kg

∴ Required average weight = $1569.5/43$ = 36.5 kg

Answer: (b)

Weight of the student who left = 50 × 45 – 49 × 44.9

= 2250 – 2200.1 = 49.9 kg.

Answer: (d)

Weight of the teacher = 42 kg. + ${35 × 400}/1000$kg

= 42 + 14 = 56 kg.

Aliter : Using Rule 24,

Here, x = $400/100$ = 0.4, n = 34

Weight of teacher = Average + x (n + 1)

= 42 + 0.4 (34 + 1)

= 42 + 14 = 56 kg

Answer: (c)

Weight of teacher = 50 + 26 × 1 = 76 kg

Aliter : Using Rule 24,

( If a new member is added in a group then. age (or income) of added member = Average (or income) ± x (n + 1)

where x = increase (+) or decrease (–) in average age (or income) n = Number of members. )

Here, N = 50, T = 45, t = $1/10$ = 0.1

Weight of teacher = Average + × (n + 1)

= 50 + 1 (25 + 1)

= 50 + 26 = 76 kg

Answer: (c)

Let the number of boys be x and that of girls be y.

∴ 71x + 73y = 71.8 (x + y)

⇒ 71.8x – 71x = 73y – 71.8y

⇒ 0.8x = 1.2y

⇒ $\text"x"/ \text"y"$ = ${1.2}/{0.8}$ = $12/8$ = $3/2$

∴ $\text"x"/\text"y"$+1 = $3/2$ + 1⇒ $\text"x+y"/ \text"y"$ = $5/2$

∴ Percentage of girls

= $\text"y"/ \text"x+y"$ × 100= $2/5$ × 100= 40%

Answer: (c)

Total increase = 3.6 × 10 = 36

∴ If the number be 10x + y,

then Number obtained after reversing the digits = 10y + x

∴ 10y + x – 10x – y = 36

⇒ 9y – 9x = 36

⇒ 9 (y – x) = 36

⇒ y – x = $36/9$ = 4