# Quant Questions – Average Problems with Solution Set 2

Dear Readers, Important Average Problems for upcoming competitive Exams with Solutions. Average is an important topic from banking exam point of view. You could solve Average problems easily if you understand the basic concepts. Aspirants preparing for SBI Clerk, SBI PO, IBPS Clerk, IBPS PO, Insurance, RBI and other competitive examination can make of use these average problems. We have included some important average problems that are repeatedly asking in competitive exams. Practice lot of questions to become an expert in solving average problems and learn to use shortcuts.

## Average Problems – Set 2

1).Three different qualities of wheat flour  – A, B& C are mixed together in the ratio of 7 : 5 : 4. The price of A, B and C are respectively Rs. 35, Rs. 40 Rs. 42 per kg. Find the average price of the new mixture
a)  Rs. 38.31
b)  Rs. 42.5
c)  Rs.36.15
d)  Rs. 40.5
e)  None of these

2).The average expenditure of P, Q,& R is Rs. 5000 per month. Also, the average expenditure of Q, R& S is Rs. 7000 per month. If the average expenditure of S is thrice of that of P then the average expenditure of Q and R is :
a)  Rs. 6000
b)  Rs. 3000
c)  Rs. 5000
d)  Rs. 4000
e)  None of these

3).The average of four consecutive odd numbers P, Q, R and S respectively (in increasing order) is 104. What is the sum of P and S ?
a)  204
b)  208
c)  206
d)  212
e)  None of these

4).Average weight of 19 men is 74 kgs, and the average weight of 38 women is 63 kgs. What is the average weight (rounded off to the nearest integer) of all the men and the women together?
a)  59 kgs
b)  65 kgs
c)  69 kgs
d)  67 kgs
e)  71 kgs

5).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. What might be the weight of the new person ?
a)  76 kg
b)  76.5 kg
c)  85 kg
e)  None of these

6).The average age of a husband and wife, who were married 7 years ago, was 25 years at the time of their marriage. Now the average age of the family, including husband, wife and a child (born during the interval) is 22 years, what is the present age of the child?
a)  4 years
b)  2 years
c)  5 years
d)  None of these
e)  None of these

7).In a group of 5 friends, the sums of ages (in years) of each group of 4 of them are 124, 128, 130, 136 and 142. The age (in years) of the youngest of them is
a)  18
b)  21
c)  23
d)  27
e)  None of these

8).The mean of 15 different natural numbers is 13. The maximum value of the second largest of these numbers is
a)  53
b)  52
c)  51
d)  50
e)  None of these

9).The average weight of M, N and O is equal to 65 kg. The average weight of M and N is equal to 69.5 kg. The average weight of N and O is 66.5 kg. What is the weight of N ?
a)  74 kg
b)  75 kg
c)  77 kg
d)  80 kg
e)  None of these

10).The present age of Romila is one fourth of that of her father. After 6 years the father’s age will be twice the age of Kapil. If Kapil celebrated fifth birth day 8 years ago, What is Romil’s present age?
a)  7 years
b)  7.5 years
c)  8 years
d)  8.5 years
e)  None of these
1).a) 2).a) 3).b) 4).d) 5).c) 6).b) 7).c) 8).c) 9).c) 10).c)

Solution:

1).Let the Quantity of A, B& C be 7x, 5x, and 4x. So,
Average Price of Mixture = [7x(35) + 5x(40) + 4x(42) ] / (7x + 5x + 4x)
= (245x + 200x + 168x) / 16x = 613 / 16 = Rs. 38.31 per kg

2).Total expenditure of P, Q& R = 3 × 5000 = 15000
Total expenditure of Q, R& S = 3 × 7000 = 21000
S – P = 21000 – 15000 = 6000
Also, we are given S = 3P.
So 3P – P = 6000& P = 3000
Total expenditure of Q& R = 15000 – 3000 = 12000
Average expenditure of Q& R = 12000 / 2 = Rs. 6000

3).Sum of four consecutive odd numbers = 104 × 4 = 416
Let P, Q, R and S be 2n – 3, 2n – 1, 2n + 1, 2n + 3 respectively
2n – 3 + 2n – 1 + 2n + 1+ 2n + 3 = 416
8n = 416
n = 52
Hence, P = 2 × 52 – 3 = 10 4 – 3 = 101& S = 2n + 3
= 2 × 52 + 3 = 104 + 3 = 107. Required sum = 101 + 107 = 208

4).Required average weight = (19 × 74 + 38 × 63) / (19 + 38)
=(1406 + 2394) / 57
= 3800 / 57 = 66.66 = 67 kg

5).Weight of new person = 65 + 8 × 2.5 = 65 + 20 = 85 kg

6).7 years ago, Husband + wife
2 × 25 = 50 years
Sum of present ages of husband and wife = 50 + 14 = 64 years
Sum of present ages of husband, wife and child = 22 × 3 = 66 years
Present age of child = 66 – 64 = 2 years

7).a + b + c + d = 124
b + c + d + e = 128
c + d + e + a = 130
d + e + a + b = 136
e + a + b + c = 142
on adding, 4(a + b + c + d + e) = 660
a + b + c + d + e = 165
Age of the youngest of them = d = 165 – 142 = 23 years
Subtract the highest sum, because in that the friend with lowest age must be dropped, then only the sum can be max.

8).Sum of 15 numbers = 15 × 13 = 195
For maximum value, the other 13 numbers will be least.
1 + 2 + ……. + 13
= (13 × 14) / 2 = 91
Sum of last two numbers = 195 – 91 = 104
Numbers = 51 and 53

9).M + N + O = 3 × 65 = 195 kg
M + N = 2 × 69.5 = 139 kg
N + O = 2 × 66.5 = 133 kg
Weight of N = (139 + 133 – 195 ) kg = 77 kg 