0
886

# SNAP Average Questions PDF [Most Important]

Average is an important topic in the Quant section of the SNAP Exam. You can also download this Free Average Questions for SNAP PDF (with answers) by Cracku. These questions will help you to practice and solve the Average questions in the SNAP exam. Utilize this PDF practice set, which is one of the best sources for practicing.

Question 1:Â The average age of three friends is 20 years and their ages are in the proportion 2:3:5, the age of the oldest friend is :

a)Â 12 years

b)Â 18 years

c)Â 30 years

d)Â 40 years

Solution:

Given average age of three friends is 20 years

Sum of their ages =Â $3\times\ 20$ = 60 years

Given, their ages are in ratio 2:3:5. Let their ages be 2x, 3x and 5x

2x + 3x + 5x = 60

10x = 60

x = 6

Their ages are 12 years, 18 years and 30 years

Age of oldest friend = 30 years

Question 2:Â In a batch of 50 students has ratio of Boys to Girls as 4:6. Average marks of Girls are 60 and Average marks of the batch is 62. Average Marks of Boys will be:

a)Â 61

b)Â 63

c)Â 62

d)Â 65

Solution:

Let the number of boys be 4x and number of girls be 6x

Given, 4x+6x = 50

x = 5

Number of boys = 20 and number of girls = 30

Average marks of girls = 60

Total marks scored by girls = 60$\times\$30 = 1800

Average marks score by batch = 62

Total marks scored by batch = 62$\times\$50 = 3100

Marks scored by boys = 3100 – 1800 = 1300

Average marks of boys =Â $\frac{1300}{20}$ = 65

Question 3:Â The average temperature of Tuesday, Wednesday and Thursday was $45^\circ C$. TheÂ average temperature of Wednesday, Thursday and Friday was $46^\circ C$. If the temperatureÂ on Friday was $47^\circ C$, what was the temperature on Tuesday?

a)Â $44^\circ C$

b)Â $48^\circ C$

c)Â $40^\circ C$

d)Â $46^\circ C$

Solution:

Sum of temperatures on Wednesday, Thursday and Friday= 46*3 = 138

The temperature on friday= 47

So, the sum of temperatures on Wednesday and Thursday= 138-47 = 91

Sum of temperatures on Tuesday, Wednesday and Thursday= 45*3 = 135

The temperature on Tuesday= 135-91= 44

Question 4:Â In a football tournament, a player has played a certain number of matches and 10 more matches are to be played. If he scores a total of one goal over the next 10 matches, his overall average will be 0.15 goals per match. On the other hand, if he scores a total of two goals over the next 10 matches, his overall average will be 0.2 goals per match. The number of matches he has played is

Solution:

Let Total matches played be n and in initial n-10 matches his goals be x
so we getÂ $\frac{\left(x+1\right)}{n}=0.15$
we get x+1 =0.15n Â  Â  Â  Â  Â  Â  Â Â  (1)
From condition (2) we get :
$\frac{\left(x+2\right)}{n}=0.2$
we get x+2 = 0.2nÂ  Â  Â  Â  Â  Â  Â  Â  Â  (2)
Subtracting (1) and (2)
we get 1 =0.05n
n =20
So initially he played n-10 =10 matches

Question 5:Â Onion is sold for 5 consecutive months at the rate of Rs 10, 20, 25, 25, and 50 per kg, respectively. A family spends a fixed amount of money on onion for each of the first three months, and then spends half that amount on onion for each of the next two months. The average expense for onion, in rupees per kg, for the family over these 5 months is closest to

a)Â 26

b)Â 18

c)Â 16

d)Â 20

Solution:

Let us assume the family spends Rs. 100 each month for the first 3 months and then spends Rs. 50 in each of the next two months.

Then amount of onions bought = 10, 5, 4, 2, 1, for months 1-5 respectively.

Total amount bought = 22kg.

Total amount spent = 100+100+100+50+50 = 400.

Average expense =Â $\frac{400}{22}=Rs.18.18\approx\ 18$

Question 6:Â A person goes from X to Y on a cycle at 20 kilometers per hour and returns at 24kilometers per hour. The method of Central Tendency most appropriate to calculate theaverage speed would be____________.

a)Â Geometric Mean

b)Â Harmonic Mean

c)Â Arithmetic Mean

d)Â Weighted Arithmetic Mean

Solution:

If the distance travelled is the same during all the speeds then the average speed for the whole journey is the harmonic mean of the speed.

Question 7:Â In a company, the average salary of the boys is Rs. 50000. The average salary of all the employees is Rs. 48000. There are 80 boys in the company and the average salary of the girl is Rs. 40000. What is the number of girls working in the company?

a)Â 12

b)Â 15

c)Â 20

d)Â 24

Solution:

Let girls be x
Total employees will be 80+x
now total value of money will be same :
i.e Average of 1employee (Number of employees ) will be equal to (Average of boys)* Number of boys +(Average of girls)*Number of girls
we get 48000(80+x) = 50,000(80)+40,000(x)
48(80+x) = 50(80) +40x
we get 8x =160
x=20

TakeÂ  SNAP mock tests here

Question 8:Â In a company there are 252 engineers, in which the ratio of the number of electronics engineers and computer engineers is 2 : 1. The ratio of the number of electronics engineers and computer engineers becomes 1:1 after recruitment of some more computer engineers. The average age of all the engineers is now 22 years and the average age of the computer engineers is 2 years less than the average age of electronics engineers.
(A) The average age (in years) of electronics engineers is 21
(B) The average age (in years) of computer engineers is 21
(C) The average age (in years) of electronics engineers is 23
(D) The average age (in years) of electronics engineers is 25
(E) The average age (in years) of computer engineers is 23
Choose the correct answer from the options given below:

a)Â (A) and (E)only

b)Â (B) and (C) only

c)Â (D) and (E) only

d)Â (E) only

Solution:

Number of electronics engineers initially =Â $\frac{2}{3}\times\ 252\ =\ 168$
So number of computer engineers initially = 252-168 =84
Now number of computer engineers recruited to make ratio 1:1 = 168-84 =84
Let average age of computer engineers be x
so total ageÂ  will be 168x
Now average age of electronics engineers will be : x+2
So total age will be : (x+2) 168
Now as per given
Total age = 336(22)
Equating we get :
336(22) = 336(x+1)
we get x=21
so B and C are correct

Question 9:Â Sumit stays in Noida in a joint family comprising of his parents, uncle, aunt, an elder sister and a
cousin two years younger than him. Sumit completed his MBA in 2013, from a reputed B-School.
That year, the average age of Sumit’s family was 41. Sumit got married in 2015 and two years after he
became father. If the average age of Sumit’s family in 2019 remains same as in 2013, and Sumit is
older than his wife by 3 years, at what age did Sumit graduate MBA ?

a)Â 31

b)Â 35

c)Â 38

d)Â 41

Solution:

Let Sumit’s age at the time of graduating MBA [in 2013] be $X$ years. Thus, his wife’s age that year will be $X-3$ years.

The average age of theÂ family in 2013 = 41 years.

Thus, the total age of the family in 2013 = 7 x 41 = 287

Sumit got married in 2015. Thus, his wife’s age in 2015 will added to the family’s total.

The total age of family in 2015 = 287 + 14 + $X$ – 3 + 2 = $X$ + 300

In 2017, Sumit had a daughter.

The total age of the family in 2017 = $X$ + 300 + 8(2) = $X$ + 316

The total age of the family in 2019 = $X$ + 316 + 9(2) = $X$ + 334

We are given that the average age of the family in 2019 is the same as that in 2013.

Thus, $X$ + 334 = 41 x 9 = 369

$X$ = 369 – 334 = 35 years

Thus, the age of Sumit at the time of graduating MBA is 35 years.

Hence, the answer is option B.

Question 10:Â A batsman played n + 2 innings and got out on all occasions. HisÂ average score in these n + 2 innings was 29 runs and he scored 38 andÂ 15 runs in the last two innings. The batsman scored less than 38 runsÂ in each of the first n innings. In these n innings, his average score wasÂ 30 runs and lowest score was x runs. The smallest possible value of xÂ is

a)Â 4

b)Â 3

c)Â 2

d)Â 1

Solution:

Given, $\frac{\text{sum of scoresÂ in n matches+38+15}}{n+2}=29$

Given,Â $\frac{\text{sum of scoresÂ in n matches}}{n}=30$

=> 30n + 53 = 29(n+2) => n=5

Sum of the scores in 5 matches = 29*7 – 38-15 = 150

Since the batsmen scored less than 38, in each ofÂ the first 5 innings. The value of x will be minimum when remaining four values are highest

=> 37+37+37+37 + x = 150

=> x = 2

Question 11:Â Five years ago, the average age of A,B,C and D was 45 years. By including X in the present lot their present average changes to 49 years. What is the present age of X?

a)Â 40 years

b)Â 45 years

c)Â 49 years

d)Â 48 years

Solution:

Sum of ages of A, B,Â C and D five years ago = $45\times4=180$

Sum of their present ages = $180+(5\times4)=200$

Sum of present ages of all five (including X) = $49\times5=245$

=> Present age of X = $245-200=45$ years

=> Ans – (B)

Question 12:Â The average of 5 consecutive numbers is n. If the next two numbers are also included the average will

a)Â remain the same

b)Â increase by 1

c)Â increase by 1.4

d)Â increase by 2

Solution:

Average of 5 consecutive number =Â $\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)}{5}$ = n+2

If 2 more consecutive number are added then its average =Â $\frac{\left(n\right)+\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+\left(n+4\right)\ +\left(n+5\right)+\left(n+6\right)}{7}$ = n+3

Hence average is increased by 1

Question 13:Â The average of nine numbers is M and the average of three of these is P. If the average of remaining numbers is N, then

a)Â M = N + P

b)Â 2M = N + P

c)Â 3M = 2N + P

d)Â 3M = 2P + N

Solution:

This can be solved easily by using allegations

$p=\frac{\left(p1q1+p2q2\right)}{q1+q2}$

Here p = M , p1=P , p2=N

q1=3 , q2=6

$\ \therefore\ M=\frac{\left(3P+6N\right)}{3+6}$

On solving 3M = P+2N

Question 14:Â In a retail outlet, the average revenue was Rs.10,000 per day over a 30 day period. During this period, the average revenue on weekends (total 8 days) was Rs. 20,000 per day. What was the average daily revenue on weekdays?

a)Â 6364

b)Â 5250

c)Â 6570

d)Â 8060

Solution:

Total revenue of the 30-day period = Rs.10000 x 30 days = Rs.300000

Total revenue of the weekends (8 days) = Rs.20000 x 8 days = Rs. 160000

Total revenue of weekdays = (300000-160000) = Rs. 140000

Avg. daily revenue on weekdays = 140000/22 = Rs.6363.63Â $\approx\$ Rs.6364

Question 15:Â Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is

a)Â 53

b)Â 51

c)Â 48

d)Â 49

Solution:

Assume the average of 21 students other than Ramesh = a

Sum of the scores of 21 students other than Ramesh = 21a

Hence the average of 22 students = a+1

Sum of the scores of all 22 students = 22(a+1)

The score of Ramesh = Sum of scores of all 22 students – Sum of the scores ofÂ 21 students other than Ramesh = 22(a+1)-21a=a+22Â  = 82.5Â  Â (Given)

=> a = 60.5

Hence,Â sum of the scores of all 22 students = 22(a+1) = 22*61.5 = 1353

Now the sum of the scores of students other than Gautam = 21*62 = 1302

Hence the score of Gautam = 1353-1302=51

Question 16:Â A cricket team has 11 players and each of them has played 20 matches till date. Virat, Rohit,Â Mahendra, Rahul and Shikhar have scored runs at an average of 60, 55, 50, 45 and 40Â respectively. Rest of the players have scored at an average of 25 each. In the next 10 matches,Â Virat and Rohit each scored 900 runs whereas Mahendra scored twice that of Rahul. AfterÂ 30 matches, if Viratâ€™s new average score is twice that of Rahul, what is the approximateÂ average score of Mahendra ?

a)Â 49

b)Â 41

c)Â 43

d)Â 45

Solution:

The average scores ofÂ Virat, Rohit, Mahendra, Rahul and Shikhar in the first 20 matches areÂ 60, 55, 50, 45,40 respectively.

In the next 10 matches, Virat and Rohit each scored 900 runs whereas Mahendra scored twice that of Rahul.

Virat’s score in 30 matches = 60*20+900=2100

Let us consider Rahul scored x runs in last 10 matches, then Mahendra scores 2x runs.

Rahul’s score in 30 matches =Â 45*20+x=900+x

Mahendra’s score in 30 matches = 50*20+2x = 1000+2x

It is given that

After 30 matches, Viratâ€™s new average score is twice that of Rahul.

2100=2(900+x)

x=150

Mahendra’s score in 30 matches =1300

Average score of Mahendra = 43