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Average Questions

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The average weight of 6 men decreases by 3 kg when one of them weighing 80 kg is replaced by a new man . The weight of the new man is :


A56kg

B58 kg

C62 kg

D76 kg

Answer: Option C

Explanation:

Total weight decreased = 6*3= 18 kg
New man weight = (80-18) = 62 kg.

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56
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A car averages 55 mph for the first 4 hours of a trip and averages 70 mph for each additional hour. The average speed for the entire trip was 60 mph. How many hours long is the trip?


A6 hrs.

B14 hrs.

C11 hrs.

D12 hrs.

Answer: Option A

Explanation:

Lets assume additional hours = x hrs.
So total No. hours in journey = (4+x)
[(55*4)+(70*x)]/(4+x)=60 =>
=>x=2
Therefore, Total No. of hrs. in Journey = (x+4) = 6 hrs.

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The average number of visitors of a library in the first 4 days of a week was 58. The average for the 2nd, 3rd, 4th and 5th days was 60.If the number of visitors on the 1st and 5th days were in the ratio 7:8 then what is the number of visitors on the 5th day of the library?


A64

B56

C45

D43

Answer: Option A

Explanation:

If number of visitors on 1st, 2nd, 3rd, 4th & 5th day are a, b, c, d & e respectively, then
a+b+c+d=58*4=232 ----(i)
b+c+d+e=60*4=240 ----(ii)
Subtracting eq. (i) from (ii)
e-a=8 ---(iii)
Given, a/e=7/8 ---(iv)
So from eq. (iii) & (iv)
a=56, e=64

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40
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64 / 94

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The average age of 10 members of a committee is the same as it was 4 years ago, because an old member has been replaced by a young member. Find how much younger is the new member?


A40 years

B35 years

C45 years

DNone of these

Answer: Option A

Explanation:

Lets assume the average of all 10 members 4 years ago = x years.
After 4 years, 10 members age increases by 10*4 = 40 years.
Conditon given: There is no change in the average as a person is replaced by an younger one.
=> The younger one age is 40 years less than the old one.

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In a test called ACSAT, the average marks of the 15 test takers is 240. If the marks of 5 test takers are subtracted , the average marks decreases by 40 . What is the average marks of 5 test takers ?


A1600

B320

C200

D40

Answer: Option B

Explanation:

Total of 15 students = 3600
Total of 10 students = [(240-40) * 10 ] = 2000
Total of 5 students = (3600 - 2000) = 1600
Avg of 5 students = 1600/5 = 320

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A man employs 20 men, 15 women and x children. he pays daily wages of Rs. 10 per men Rs. 8 per women and Rs. 4 per chil his daily average wages bill works out to Rs. 8.50 per person. What the value of x?


A5

B10

C15

DIndetermine

Answer: Option A

Explanation:

20*10 +15*8 +x*4/20+15+x = 8.5
x= 5 child

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consider the sequence 1,-2, 3, -4, 5 what is the average of first 200 terms


A-0.5

B0.5

C-0.9

D0.9

Answer: Option A

Explanation:

Series is divided into two terms 1,3,5….. will contain 100 terms
and second series is -2 ,-4,-6
S=n/2*(2a+(n-1)d)
First series give s=10000
Second series give s=-10100
so total = (10000 - 10100) = 100
so avg. = -100/200 = -0.5

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A student's average ( arithmetic mean) test score on 4 tests is 78. What must be the students score on a 5th test for the students average score on the 5th test to be 80?


A78

B88

C89

D98

Answer: Option B

Explanation:

Total scores of 4 test = 78 * 4 = 312
Total scores of 5 tests = 80 * 5 = 400

So, 5th test score = (400-312)=88

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The price of lunch for 15 people was 207 pounds, including a 15 percent gratuity of service. What was the average price per person, excluding the gratuity?


A18 pounds

B22 pounds

C16 pounds

D12 pounds

Answer: Option D

Explanation:

Let the net price excluding the gratuity of service = x pounds
Then, total price including 15% gratuity of service = x*(100+15100) = 1.15x pounds
So, 1.15x = 207 pounds
=> x = 207/1.15 = 180 pounds
Net price of lunch for each person = 180/15 = 12 pounds

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The average of run recorded by 8 players is 40. If the average of the runs by the first five players is 39, and that of the last four players is 34, what is the number of runs recorded by the fifth player?


A28

B29

C40

D11

Answer: Option D

Explanation:

Total run recorded by 8 players = 80*4 = 320
Total run by first five players = 5 * 39 = 195
Total run by four players = 4*34 = 136
Run record of fifth player = (195+136-320) = 11

Workspace

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