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Average Exercise 3

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Average 3

  • This online quiz will test your knowledge of Average in Quantitative Aptitude.
  • This Online Test is useful for academic and competitive exams.
  • Multiple answer choices are given for each question in this test. You have to choose the best option.
  • After completing the test, you can see your result.
  • There is no negative marking for wrong answers.
  • There is no specified time to complete this test.

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?

Total runs in the first 10 overs

= 10 × 3.2 = 32

Run rate required in the remaining 40 overs

average-28015.png runs per over

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’?

Let ‘x’ be the average score after 12 th innings

⇒ 12 x = 11 × (x – 2) + 65

∴ x = 43

A cricketer whose bowling average is 12.4 runs per wicket takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The number of wickets taken by him till the last match was:

Let the number of wickets taken till the last match be x. Then,

average-28009.png

average-28002.png

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?

Let the average score of 19 innings be x.

Then, average-28062.png

The average score after 20th innings

= x + 4 = 22 + 4 = 26

A batsman has scored an average of 46 runs for a certain number of innings played in England. When he came back to India, he played another two test matches of two innings each and scored at an average of 55 runs. For the innings in England and in India taken together, he has improved his average by 2 runs over the matches played in England. Find the number of innings played in England.

Let the number of innings played in England be x.

∴ Total runs scored in England = 46 x

Total runs scored for innings played in India = 55 × 4 = 220

(the number of innings played in India = 4)

Also, average-27996.png

⇒ 46x + 220 = 48 x + 192

⇒ 2x = 28

⇒ x = 14

N number of persons decide to raise ₹3 lakhs by equal contributions from each. If they contributed ₹50 each extra, the contribution would be ₹3.25 lakhs. How many persons are there?

Required persons

average-28192.png

= 500

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?

Let the average expenditure of all the nine be ₹x.

Then, 12 × 8 + (x + 8)

= 9x or 8x = 104 or x = 13.

∴ Total money spent

= 9x = ₹(9 × 13)

= ₹117.

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

Since average price of a pencil = ₹2

∴ Price of 75 pencils

= ₹150

∴ Price of 30 pens

= ₹(510 – 150) = ₹360

∴ Average price of a pen

average-28186.png

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

Remaining pages

= 512 – 302 = 210

Let average printing error in remaining pages = x

Then, average-28280.png

⇒ 210x = 840

⇒ x = 4

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

Income of 6 months

= ₹(6 × 85) – debt = ₹510 – debt

Income of the man for next 4 months

= ₹4 × 60 + debt + ₹30

= ₹270 + debt

∴ Income of 10 months = ₹780

Average monthly income

= ₹780 ÷ 10 = ₹78

In a coconut grove, (x + 2) trees yield 60 nuts per year, x trees yield 120 nuts per year and (x – 2) trees yield 180 nuts per year. If the average yield per year per tree be 100, then x is

average-28273.png

⇒ average-28267.png

⇒ average-28261.png

Nine men went to a hotel. 8 of them spent ₹3 each over their meals and the ninth spent ₹2 more than the average expenditure of all the nine. The total money spent by all of them was

Let the average expenditure of all the nine be ₹x

Then, 3 × 8 + x +2 = 9x

⇒ x = 3.25

∴ Total money spent

= 9x = 9 × 3.25 = ₹29.25

There were 35 students in a hostel. Due to the admission of 7 new students, the expenses of mess were increased by ₹42 per day while the average expenditure per head diminished by Re 1. What was the original expenditure of the mess?

Let the original average expenditure be ₹x. Then,

42(x – 1) – 35x = 42

⇒ 7x = 84

⇒ x = 12.

∴ Original expenditure

= ₹(35 × 12) = ₹420.

In an engineering college the average salary of all engineering graduates from Mechanical trade is ₹2.45 lacs per annum and that of the engineering graduates from Electronics trade is ₹3.56 lacs per annum. The average salary of all Mechanical and Electronics graduates is ₹3.12 lacs per annum. Find the least number of Electronics graduates passing out from this institute.

Let the number of Mechanical engineering graduates be M and Electronic engineering graduates be E. Then

average-28504.png

⇒ 2.45M + 3.56E = 3.12M + 3.12E

⇒ 0.44E = 0.67M

⇒ E average-28498.png

For E to be an integer, the least value will be 67.

The average monthly sales for the first eleven months of the year of a certain salesman were ₹12000, but due to his illness during the last month, the average monthly sales for the whole year came down to ₹11375. What was the value of sales during the last month?

Total sales for the first eleven months

= 12,000 × 11 = ₹132000

Total sales for the whole year

= ₹11375 × 12 = ₹136500

∴ Value of sales during the last month

= 136500 – 132000 = ₹4500.

Directions (Q. 63-65) : There are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D. What is the average weight of the students in group D?

Average weight of the students in group D cannot be determined since we do not know the average weight of each student. The given data is insufficient to compare its average with other groups.

Directions (Q. 63-65) : There are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D. If one student from Group A is shifted to group B, which of the following will be true?

If one student from group A is shifted to group B, still there is no effect on the whole class. In any case, the no. of students inside the class is same. Hence the average weight of the class remains same.

Directions (Q. 63-65) : There are 60 students in a class. These students are divided into three groups A, B and C of 15, 20 and 25 students each. The groups A and C are combined to form group D. If all the students of the class have the same weight, then which of the following is false?

Since all the students of the class have the same weight, then the average of weight of any group of any no. of students will be the same as that of each students weight. Hence, the average weight of D cannot be greater than average weight of A.

Directions (Q. 66 - 67) : Five heavy weight boxers measure their weights. Following results were obtained :

P is heavier than R by 14 LB. B is lighter than S by 10 LB. M’s weight is equal to the average weight of the four other boxers. P’s weight and B’s weight taken together equals the weight of M and S. The sum of the weights of all five boxers is 520 LB.

What is the average of the weights of M and R?

For Q. 66-67

Let p, m, r, s and b be the weights of boxers P, M, R, S and B respectively.

From data :

p = r + 14 ... (1)

b = s – 10 ... (2)

4m = p + b + r + s ... (3)

p + b = m + s ... (4)

p + b + m + r + s = 520 ... (5)

From (3) and (5),

5m = 520

⇒ m = 104 lb

From (1) and (2),

p + b = r + s + 4 ... (6)

From (4) and (6),

r + s + 4 = m + s

⇒ r = m – 4 = 100 lb

From (1),

p = r + 14 = 114 lb

From (5),

114 + b + 104 + s + 100 = 520

⇒ b + s = 202 ... (7)

From (7) and (2),

b – s = 10

and b + s = 202

⇒ b = 192/2 = 96 lb

and s = 106 lb

∴ Average of the weights of M and R

= (104 + 100)/2 = 102 lb

Average of the weights of P, S and B

= (114 + 106 + 96)/3 = 105.3 lb.

Directions (Q. 66 - 67) : Five heavy weight boxers measure their weights. Following results were obtained :

P is heavier than R by 14 LB. B is lighter than S by 10 LB. M’s weight is equal to the average weight of the four other boxers. P’s weight and B’s weight taken together equals the weight of M and S. The sum of the weights of all five boxers is 520 LB.

What is the average of the weights of P, S and B?

See the sol. of previous question

Now check your Result..

Your score is

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