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

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The following question based on average topic of quantitative aptitude

Questions : A student finds the average of 10, 2 – digit numbers. If the digits of one of the numbers is interchanged, the average increases by 3.6. The difference between the digits of the 2-digit numbers is

(a) 3

(b) 2

(c) 4

(d) 5

The correct answers to the above question in:

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

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Read more excluded numbers Based Quantitative Aptitude Questions and Answers

Question : 1

In a class, the average score of girls in an examination is 73 and that of boys is 71. The average score for the whole class is 71.8. Find the percentage of girls.

a) 50%

b) 55%

c) 40%

d) 60%

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%

Question : 2

Average weight of 25 students of a class is 50 kg. If the weight of the class teacher is included, the average is increased by 1 kg. The weight of the teacher is

a) 77 kg

b) 74 kg

c) 76 kg

d) 75 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

Question : 3

The average of five numbers is 7. When three new numbers are included, the average of the eight numbers becomes 8.5. The average of the three new numbers is

a) 10.5

b) 11

c) 9

d) 11.5

Answer: (b)

Sum of the three new numbers

= 8 × 8.5 – 5 × 7 = 68 – 35 = 33

∴ Required average = $33/3$ = 11

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