model 1 basic average questions Practice Questions Answers Test with Solutions & More Shortcuts

Question : 21

The average income of 40 persons is Rs. 4200 and that of another 35 persons is Rs. 4000. The average income of the whole group is :

a) Rs.4106$2/3$

b) Rs.4108$1/3$

c) Rs.4106$1/3$

d) Rs.4100

Answer: (a)

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

Average income of whole group

= ${4200×40+4000×35}/75$

= ${168000+140000}/75$

= $308000/75$ =Rs. 4106$2/3$

Question : 22

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

Question : 23 [SSC CAPFs SI, CISF 2015]

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 : 24 [SSC CGL Prelim 2003]

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 : 25

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

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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