practice question on average (aptitude)

Square, Cube, indices and Surds

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Question no. 1

A school has 4 section of Chemistry in Class X having 40, 35,45 and 42 students. The mean marksobtained in Chemistry test are 50, 60, 55 and 45 respectively for the 4 sections. Determine the overall average of marks per student

looks_one 50.25

looks_two 52.25

looks_3 51.25

looks_4 53.25

Option looks_two 52.25

Solution :
average marks = total marks / total students
total marks = 50*40 + 60*35 + 55*45 + 45*42 = 8462
total students = 40 + 35 + 45 + 42 = 162
Average marks = 8462/ 162 = 52.25

Question no. 2

The mean of 30 values was 150. It was detected on rechecking that one value 165 was wrongly copied as 135 for the computation of the mean. Find the correct mean.

looks_one 151

looks_two 149

looks_3 152

looks_4 None of these

Option looks_one 151

Solution :

Question no. 3

A car owner buys petrol at $7.50, $8.00 and $8.50 per litre for three successive years. What approximately is his average cost per litre of petrol if he spends $4000 each year ?

looks_one $8

looks_two $9

looks_3 $7.98

looks_4 $8.50

option looks_3 $7.98

Solution :
Average cost per litre = total cost / total petrol
total cost = 12000
total petrol = 4000/ 7.5 + 4000/ 8 + 4000/ 8.5 = 1503.92
Average cost per litre = 12000/1503.92 = 7.98

Question no. 4

A batsman in his 12th innings makes a score of 65 and thereby increases his average by 2 runs. What is his average after the 12th innings if he had never been ‘not out’?

looks_one 42

looks_two 43

looks_3 44

looks_4 45

Option looks_two 43

Solution :
x is the avearge after 12nd
Total run = 12x
according to question
total run = 11(x-2) + 65
12x = 11(x-2) + 65
12x = 11x - 22 + 65
x = 43

Question no. 5

In the first 10 overs of a cricket game, the run rate was only 3.2. What should be the run rate in the remaining 40 overs to reach a target of 282 runs ?

looks_one 6.25

looks_two6.5

looks_3 6.75

looks_4 7.00

Option looks_one 6.25

Solution :
Run in the first ten over = 10 * 3.2 = 32
Run required in next 40 over = 282 - 32 = 250
run rate in the remaining 40 over =run needed/reamining over = 250/40 = 6.25

Question no. 6

The average attendance in a school for the first4 days of the week is 30 and for the first 5 days of the week is 32. The attendance on the fifth day is

looks_one 32

looks_two 40

looks_3 38

looks_4 36

Option looks_two 40

Solution :

Question no. 7

The average of six numbers is 3.95. The average of two of them is 3.4, while the average of the other two is 3.85. What is the average of the remaining two numbers?

looks_one 4.5

looks_two 4.6

looks_3 4.7

looks_4 4.8

Option looks_two 4.6

Solution :

Question no. 8

Nine persons went to a hotel for taking their meals. Eight of them spent $ 12 each on their meals and the ninth spend $ 8 more than the average expenditure of all the nine. What was the total money spent by them?

looks_one$ 115

looks_two $117

looks_3 $119

looks_4$ 122

Option looks_two 117

Solution :
x₁ + x₂ + ..... + x₈ = 12*8 = 96
x₉ = 8 + (x₁ + x₂ + ..... + x₉)/9
x₉ = 8 + (x₁ + x₂ + .....+x₈ + x₉)/9
x₉ = 8 + (96 + x₉)/9
9x₉ = 72 + 96 + x₉
8x₉ = 168
x₉ = 21
Total cost = 21 + 96 = $117

Question no. 9

The average expenditure of a labourer for 6 months was Rs. 85 and he fell into debt. In the next 4 months by reducing his monthly expenses to Rs. 60 he not only cleared off his debt but also saved Rs. 30. His monthly income is

looks_one 70

looks_two 75

looks_3 72

looks_4 78

Option looks_4 78

Solution :
monthly income = x
debt = total expenditure - total income = 85*6 - 6x
according to question
85*6 - 6x = 4x - 60*4 - 30
10x = 780
x = 78

Question no. 10

30 pens and 75 pencils were purchased for Rs. 510. If the average price of a pencil was Rs 2.00, find the average price of a pen.

looks_one 10

looks_two 11

looks_3 12

looks_4can't be determined

Option looks_3 12

Solution :
let the average cost of pencil is y and average cost of pen is x
According to Question
30x + 75y = 510
30x + 75*2 = 510
30x = 510 - 150 = 360
30x = 360
x = 12

Question no. 11

The average of 20 numbers is zero. Of them, at the most, how many may be greater than zero?

looks_one 0

looks_two 1

looks_3 10

looks_4 19

Option looks_4 19

Solution :

Question no. 12

The average of a batsman for 40 innings is 50 runs. His highest score exceeds his lowest score by 172 runs. If these two innings are excluded, his average drops by 2 runs. Find his highest score.

looks_one 172

looks_two 173

looks_3 174

looks_4 175

Option looks_3 174

Solution :
let highest run = H , lowest run = L
Total runs = 40*50 = 2000
H = L + 172
total run excluded by H and L = remaining matches * average = 38 * 48 = 1824
H + L = 2000 - 1824 = 176
Solve both the equation of H and L
H = L + 172 , H + L = 176
H = 174 , L = 2

Question no. 13

The average number of printing error per page in a book of 512 pages is 4. If the total number of printing error in the first 302 pages is 1,208, the average number of printing errors per page in the remaining pages is

looks_one 0

looks_two 4

looks_3 90

looks_4 840

Option looks_two 4

Solution :
Total no. of error = 4 * 512 = 2048
remaining error = 2048 - 1208 = 840
remaining pages = 512 - 302 = 210
Average of remaining page error = remaining error/ remaining pages = 840/ 210 = 4

Question no. 14

A batsman makes a scores of 98 runs in his 19th inning and thus increases his average by 4. What is his average after 19th inning?

looks_one 22

looks_two 24

looks_3 26

looks_4 28

Option looks_4 28

Solution :
Average after 19th inning is x
total marks = 19x
According to question
19x = (x-4)18 + 98
19x = 18x - 72 + 98
x = 26

Question no. 15

If a, b, c, d, e are five consecutive odd numbers, then the average in terms of ‘a’ will be

looks_one a+2

looks_two a+3

looks_3 a+4

looks_4 a

Option looks_3 a+4

Solution :
a , b = a + 2 , c = a+ 4 , d = a+6 , e =a+8
Average = (a + a + 2 + a + 4 + a + 6 + a + 8)/5
= (5a + 20 ) / 5 = a + 4

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