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MOST IMPORTANT quantitative aptitude - 9 EXERCISES

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The following question based on average topic of quantitative aptitude

Questions : If average of 20 observations $x_1 , x_2 , ....., x_20$ is y, then the average of $x_1 – 101, x_2 – 101, x_3 – 101, ....., x_20 –101$ is

(a) 20y

(b) 101y

(c) y – 101

(d) y – 20

The correct answers to the above question in:

Answer: (c)

Using Rule 2,

If the given observations (x) are occuring with certain frequency (A) then,

$Average = {A_1x_1+A_2x_2+...............+A_nx_n}/{x_1+x_2+......+x_n}$

where, $A_1, A_2, A_3, .......... A_n$ are frequencies

Required average

${x_1+x_2+…+x_20}/20$ = ${101×2}/20$

= y-101

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Read more simple average Based Quantitative Aptitude Questions and Answers

Question : 1

The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. The weight of the new person is

a) 76 kg

b) 76.5 kg

c) 85 kg

d) 84 kg

Answer: (c)

Weight of new person

= (65 + 8 × 2.5) kg

= (65 + 20) kg

= 85 kg

Aliter : Using Rule 23,

Here, x = 2.5, n = 8

Weight of new person

= weight of replaced boy + x × n

Question : 2

The average marks of 32 boys of section A of class X is 60 whereas the average marks of 40 boys of section B of class X is 33. The average marks for both the sections combined together is

a) 46$1/2$

b) 45$1/2$

c) 45

d) 44

Answer: (c)

Required average

${32×60+33×40}/72$

${1920+1320}/72$ = $3240/72$ = 45

Question : 3

The average of 20 numbers is 15 and the average of first five is 12. The average of the rest is

a) 14

b) 13

c) 15

d) 16

Answer: (d)

If the average of remaining numbers be x, then

20 × 15 = 5 × 12 + 15x

⇒ 300 = 60 + 15x

⇒ 15x = 300 – 60 = 240

⇒ x = $240/15$ = 16

Aliter : Using Rule 13,

Here, m = 20, x = 15

n = 5, y = 12

Average of remaining Numbers = $({mx-ny}/{m-n})$

= $({20×15-5×12}/{20-5})$

= $({300-60}/15)$ = $240/15$ = 16

Question : 4

The average of 30 results is 20 and the average of other 20 results is 30. What is the average of all the results ?

a) 25

b) 50

c) 48

d) 24

Answer: (d)

Using Rule 10,

If the average of '$n_1$' numbers is $a_1$ and the average of '$n_2$' numbers is $a_2$, then average of total numbers $n_1$ and $n_2$ is

$Average = {n_1a_1 + n_2a_2}/{n_1 + n_2}$

Required average

= ${20×30+20×30}/{30+20}$

= ${600+600}/50$

= $1200/50$ = 24

Question : 5

The average weight of 3 men A, B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E whose weight is 3 kg more then that of D, replaces A, then the average weight of B, C, D and E becomes 79 kg. Then weight of A is

a) 75 kg.

b) 76 kg.

c) 74 kg.

d) 72 kg.

Answer: (a)

D’s weight = 80 × 4 – 84 × 3

= 320 – 252 = 68 kg.

E’s weight = 68 + 3 = 71 kg.

Total weight of (A + B + C + D + E)

= 84 × 3 + 68 + 71

= 252 + 68 + 71 = 391 kg.

Total weight of (B + C + D + E)

= 79 × 4 = 316 kg.

∴ A’s weight= 391 – 316 = 75 kg.

Question : 6

The average of 1, 3, 5, 7, 9, 11, -------- to 25 terms is

a) 625

b) 50

c) 25

d) 125

Answer: (c)

Sum of first n odd natural numbers = $n^2$

∴ Their average = ${n^2}/n$ = n

∴ Required average = 25

because n = 25

Aliter : Using Rule 7,

Average = 25

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