1. A person purchase 1 kg of tomatoes from each of the 4 places at the rate of 1 kg, 2 kg, 3 kg, 4 kg per rupee respectively. On an average, he has purchased x kg of tomatoes per rupee. Then the value of x is
a) Rs. 1.92
b) Rs. 2
c) Rs. 2.50
d) None of these
Explanation: Total quantity purchased = 4 kg
Total Money paid
$$\eqalign{ & = {\text{Rs}}{\text{. }}\left( {1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}} \right) \cr & = {\text{Rs}}{\text{. }}\frac{{25}}{{12}} \cr} $$
Required average
$$\eqalign{ & = \left( {4 \times \frac{{12}}{{25}}} \right){\text{ kg/rupee}} = \left( {\frac{{48}}{{25}}} \right){\text{ kg/rupee}} \cr & = 1.92{\text{ kg/rupee}} \cr} $$
2. Out of 9 persons, 8 persons spent Rs. 30 each for their meals. The ninth one spent Rs. 20 more than the average expenditure of all the nine. The total money spent by all of them was
a) Rs. 260
b) Rs. 290
c) Rs. 292.50
d) Rs. 400.50
Explanation: Let the average expenditure be Rs. x
9x = 8 × 30 + (x + 20)
9x = x + 260
8x = 260
x = 32.50
Total money spent = Rs. 9x = Rs. (9 × 32.50)
= Rs. 292.50
3. Of the three numbers, the average of the first and the second is greater than the average of the second and the third by 15. What is the difference between the first and the third of the three numbers?
a) 15
b) 45
c) 60
d) None of these
Explanation: Let the numbers be x, y and z.
$$\eqalign{ & \left( {\frac{{x + y}}{2}} \right) - \left( {\frac{{y + z}}{2}} \right) = 15 \cr & \left( {x + y} \right) - \left( {y + z} \right) = 30 \cr & x - z = 30 \cr} $$
4. The average of runs of a cricket player of 10 innings was 32. How many runs must he make in his next innings so as to increase his average of runs by 4 ?
a) 2
b) 4
c) 70
d) 76
Explanation: Average after 11 innings = 36
Required number of runs = (36 × 11) - (32 × 10) = 396 - 320
= 76
5. There are 3 groups of students, each containing 25, 50 and 25 students respectively. The mean marks obtained by the first two groups are 60 and 55. The combined mean of all the three groups is 58. What is the mean of the marks scored by the third group ?
a) 52
b) 57
c) 58
d) 62
Explanation: Let the mean marks of the third group be x.
$$\eqalign{ & \frac{{25 \times 60 + 50 \times 55 + 25 \times x}}{{25 + 50 + 25}} = 58 \cr & 1500 + 2750 + 25x = 5800 \cr & 25x = 1550 \cr & x = 62 \cr} $$
6. If the average marks of three batches of 55, 60 and 45 students respectively is 50, 55, 60, then the average marks of all the students is:
a) 53.33
b) 54.68
c) 55
d) None of these
Explanation:
$$\eqalign{ & {\text{Required}}\,{\text{average}} \cr & = {\frac{{55 \times 50 + 60 \times 55 + 45 \times 60}}{{55 + 60 + 45}}} \cr & = {\frac{{2750 + 3300 + 2700}}{{160}}} \cr & = \frac{{8750}}{{160}} \cr & = 54.68 \cr} $$
7. A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half $$\frac{1}{2}$$ . The number of pupils in the class is:
a) 10
b) 20
c) 40
d) 73
Explanation:
$$\eqalign{ & {\text{Let}}\,{\text{there}}\,{\text{be}}\,x\,{\text{pupils}}\,{\text{in}}\,{\text{the}}\,{\text{class}}{\text{.}} \cr & {\text{Total}}\,{\text{increase}}\,{\text{in}}\,{\text{marks}} = {x \times \frac{1}{2}} = \frac{x}{2} \cr & \frac{x}{2} = \left( {83 - 63} \right) \cr & \frac{x}{2} = 20 \cr & x = 40 \cr} $$
8. The marks of six boys in a group are 48, 59, 87, 37, 78 and 57. What are the average marks of all six boys?
a) 61
b) 65
c) 69
d) None of these
Explanation: Total marks of six boys
= 48 + 59 + 87 + 37 + 78 + 57
= 366
Required Average = $$\frac{{366}}{6}$$ = 61
9. Six numbers are arranged in decreasing order. The average of the first five numbers is 30 and the average of the last five numbers is 25. The difference of the first and the last numbers is
a) 20
b) 25
c) 5
d) 30
Explanation: Numbers are
x > y > z > p > q > r
Average of first five numbers = 30
Sum of first five numbers
= a + y + z + p + q = 5 × 30 = 150.....(i)
Average of last five numbers
= y + z + p + q + r = 5 × 25 = 125.....(ii)
By equation (i) and (ii)
a - r = 150 - 125 = 25
10. The average of 12 numbers is 15 and the average of the first two is 14. What is the average of the rest?
a) 15
b) $$15\frac{1}{5}$$
c) 14
d) $$14\frac{1}{5}$$
Explanation: Average of 12 numbers = 15
Total of 12 numbers = 15 × 12 = 180
Average of first two number = 14
Total of first two number = 14 × 2 = 28
Total of remaining ten numbers = 180 - 28 = 152
Required average of remaining ten number
$$\eqalign{ & = \frac{{152}}{{10}} \cr & = \frac{{76}}{5} \cr & = 15\frac{1}{5} \cr} $$