1. A car owner buys petrol at Rs.7.50, Rs. 8 and Rs. 8.50 per litre for three successive years. What approximately is the average cost per litre of petrol if he spends Rs. 4000 each year?
a) Rs. 7.98
b) Rs. 8
c) Rs. 8.50
d) Rs. 9
Explanation: Total quantity of petrol consumed in 3 years
$$\eqalign{ & = \left( {\frac{{4000}}{{7.50}} + \frac{{4000}}{8} + \frac{{4000}}{{8.50}}} \right){\text{litres}} \cr & = 4000\left( {\frac{2}{{15}} + \frac{1}{8} + \frac{2}{{17}}} \right){\text{litres}} \cr & = {\frac{{76700}}{{51}}} {\text{ litres}} \cr & {\text{Total}}\,{\text{amount}}\,{\text{spent}} = Rs.\,\left( {3 \times 4000} \right) \cr & = Rs.\,12000 \cr & {\text{Average}}\,{\text{Cost}} = Rs.\, {\frac{{12000 \times 51}}{{76700}}} \cr & = Rs.\,\frac{{6120}}{{767}} \cr & = Rs.\,7.98 \cr} $$
2. In Arun's opinion, his weight is greater than 65 kg but less than 72 kg. His brother doest not agree with Arun and he thinks that Arun's weight is greater than 60 kg but less than 70 kg. His mother's view is that his weight cannot be greater than 68 kg. If all are them are correct in their estimation, what is the average of different probable weights of Arun?
a) 67 kg
b) 68 kg
c) 69 kg
d) Data inadequate
Explanation: Let Arun's weight by X kg.
According to Arun, 65 < X < 72
According to Arun's brother, 60 < X < 70
According to Arun's mother, X < 68
The values satisfying all the above conditions are 66, 67 and 68
$$\eqalign{ & {\text{Required}}\,{\text{average}} = {\frac{{66 + 67 + 68}}{3}} \cr & = {\frac{{201}}{3}} \cr & = 67\,kg \cr} $$
3. The average weight of A, B and C is 45 kg. If the average weight of A and B be 40 kg and that of B and C be 43 kg, then the weight of B is:
a) 17 kg
b) 20 kg
c) 26 kg
d) 31 kg
Explanation: Let A, B, C represent their respective weights.
A + B + C = (45 x 3) = 135 .... (i)
A + B = (40 x 2) = 80 .... (ii)
B + C = (43 x 2) = 86 ....(iii)
Adding (ii) and (iii), we get: A + 2B + C = 166 .... (iv)
Subtracting (i) from (iv), we get : B = 31
B's weight = 31 kg.
4. The average weight of 16 boys in a class is 50.25 kg and that of the remaining 8 boys is 45.15 kg. Find the average weights of all the boys in the class.
a) 47.55 kg.
b) 48 kg.
c) 48.55 kg.
d) 49.25 kg.
Explanation:
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{50.25 \times 16 + 45.15 \times 8}}{{16 + 8}}} \cr & = {\frac{{804 + 361.20}}{{24}}} \cr & = \frac{{1165.20}}{{24}} \cr & = 48.55 {\text{ kg.}} \cr} $$
5. A library has an average of 510 visitors on Sundays and 240 on other days. The average number of visitors per day in a month of 30 days beginning with a Sunday is:
a) 250
b) 276
c) 280
d) 285
Explanation: Since the month begins with a Sunday, to there will be five Sundays in the month.
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{510 \times 5 + 240 \times 25}}{{30}}} \cr & = \frac{{8550}}{{30}} \cr & = 285 \cr} $$
6. There were 35 students in a hostel. If the number of the students is increased by 7, then the expenses of the mess increased by Rs. 42 per day, while the average expenditure per head diminishes by Rs. 1. The original expenditure of the mess per day was
a) Rs. 400
b) Rs. 420
c) Rs. 432
d) Rs. 442
Explanation: Let the original expenditure of the mess per day be Rs. x
Then, new expenditure
= Rs. (x + 42)
$$\eqalign{ & \therefore \frac{x}{{35}} - \frac{{\left( {x + 42} \right)}}{{42}} = 1 \cr & 6x - 5\left( {x + 42} \right) = 210 \cr & x - 210 = 210 \cr & x = 420 \cr} $$
7. The mean of 5 numbers is 18. If one number is excluded, their mean is 16. Find the excluded number.
a) 25
b) 26
c) 27
d) 28
Explanation: Excluded number = (18 × 5) - (16 × 4)
= 90 - 64
= 26
8. The average of the reciprocals of x and y is-
a) $$\frac{(x + y)}{(x - y)}$$
b) $$\frac{(x + y)}{2xy}$$
c) $$\frac{2(x + y)}{xy}$$
d) $$\frac{2xy}{(x + y)}$$
Explanation: Required average
$$\eqalign{ & = \frac{{\left( {\frac{1}{x} + \frac{1}{y}} \right)}}{2} \cr & = \frac{{x + y}}{{2xy}} \cr} $$
9. The arithmetic mean of 15 numbers is 41.4. Then the sum of these numbers is
a) 414
b) 420
c) 620
d) 621
Explanation: Sum of numbers = (41.4 × 15)
= 621
10. If the average of m numbers is n2 and that of n numbers is m2 , then the average of (m + n) numbers is
a) m - n
b) mn
c) m + n
d) $$\frac{m}{n}$$
Explanation: Sum of m numbers = mn2
Sum of n numbers = nm2
Average of (m + n) numbers
= $$\frac{mn(m + n)}{(m + n)}$$
= mn