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- Question
- A has double the money of B and B has 50% more money then C. If average money of all the three persons is ? 12000, how money does A have?
Options- A. ? 211000/11
- B. ? 315000/11
- C. ? 216000/11
- D. ? 316000/11
- Correct Answer
- ? 216000/11
ExplanationLet money of C be ? N.
According to the question
Total money of B = N + N x 50%
= N + 50N/100 = 3N/2
Total money of A = 2 x 3N/2 = 3N
Average money of three persons = 12000
? Total money of three = 12000 x 3
? 3N + (3N/2) + N = 12000 x 3
? (6N + 3N + 2N)/2 = 36000
? N = (36000 x 2)/11 = 72000/11
Now, money of A = 3N = (3 x 72000)/11
= ? 216000/11 - 1. The average age of two boys and their father is greater then average age of those two boys and their mother by 3 yr. The average age of the four is 19 yr. If the average age of the two boys be 5 1/ 2, then find the age of the father and mother.
Options- A. 37 yr and 28 yr
- B. 47 yr and 38 yr
- C. 50 yr and 41 yr
- D. 35 yr and 32 yr Discuss
- 2. A cricketer whose bowling average is 12.4 runs per wicket, takes 5 wickets for 26 runs and there by decreases his average by 0.4. Find out the number of wickets taken by him till the last match?
Options- A. 75
- B. 81
- C. 85
- D. 79 Discuss
- 3. A cricketer has a certain average for 10 inning. In the eleventh innings. he scored 216 runs, there by increasing his average by 12 runs. Find out his new average?
Options- A. 96
- B. 84
- C. 97
- D. 87 Discuss
- 4. A car reached Raipur from Somgarh in 35 min With an average speed of 69 km/h. If the average speed is increased by 36 km/h, how much time will it take to cover the same distance?
Options- A. 24 min
- B. 27 min
- C. 23 min
- D. 29 min Discuss
- 5. The average runs scored by a cricketer is 42 innings, is 30. The difference between his maximum and minimum scores in an inning is 100. If these two innings are not taken into consideration, then the average score of remaining 40 inning is 28. Calculate the maximum runs scored by him in an innings?
Options- A. 125
- B. 120
- C. 110
- D. 100 Discuss
- 6. A cricketer played 80 innings and scored an average of 99 runs. His score in the last inning is zero run. To have an average of 100 at the end, his score in the last innings should have been.
Options- A. 10 runs
- B. 1 runs
- C. 60 runs
- D. 80 runs Discuss
- 7. The arithmetical mean of the scores of a group of students in a test was 52. The brightest 20% of them secured a mean score of 80 and the dullest 25%, a mean score of 31. The mean score of remaining 55% is
Options- A. 54.6%
- B. 45%
- C. 50'%
- D. 51.4% Discuss
- 8. The average weight of 4 men, A, B, C and D is 67 kg. The 5th man E is included and the average weight decreses by 2 kg. A is replaced by F. The weight of F is 4 kg more then E . Average weight decreases because of the replacement of A and now the average weight is 64 kg. Find the weight of A.
Options- A. 78 kg
- B. 66 kg
- C. 75 kg
- D. 58 kg Discuss
- 9. 5 yr ago, the average age of Rashmi and Surbhi was 20 yr, Now, the average age of Rashmi, Surbhi and Meeta is 30 yr. What will be the age of Meeta after 15 yr.?
Options- A. 35 yr
- B. 55 yr
- C. 39 yr
- D. 67 yr Discuss
- 10. The average weight of the students in four sections A, B, C and D is 60 kg. The average weight of the students of A, B, C and D individually are 45 kg, 50 kg, 72 kg and 80 kg, respectively. If the average weight of the students of section A and B together is 48 kg and that of B and C together is 60 kg. What is the ratio of the number of the students in sections A and D?
Options- A. 12 : 7
- B. 4 : 3
- C. 3 : 2
- D. 8 : 5 Discuss
Average problems
Search Results
Correct Answer: 37 yr and 28 yr
Explanation:
Let the ages of mother, father and boys be M, F, B1 and B2, respectively.
The total age of four member
= 19 x 4
= 76 yr.
Given (B1 + B1)/2 = 11/2
? B1 + B2 = 11
M + F + B1 + B2 = 76 - 11
? M + F = 65 .......(i)
According to the question,
(B1 + B2 + F) / 3 = (B1 + B2 + M) / 3 + 3
? B1 + B2 + F) = B1 + B2 + M + 9
? F = M + 9
? F - M = 9 ............. (ii)
From Eqs (i) and (ii), we get
F = 37 yr and M = 28 yr
Correct Answer: 85
Explanation:
Let x be the number of wickets taken till the last match
According to the question,
(12.4 x + 26) = 12(x + 5)
? 12.4 x + 26 = 12x + 60
? 0.4 x = 34
? x = 34/0.4 = 85
Correct Answer: 96
Explanation:
Let average after 10th innings = x.
According to the question.
10x + 216 = 11(x +12)
? x = 216 - 132 = 84
? New average = 84 + 12 = 96
Correct Answer: 23 min
Explanation:
Distance between Raipur and Somgarh
= Average speed x Time
= 69 x 35/60= 161/4 km
New speed = (69 + 36) km/h = 105 km/h
? Time = Distance/ Speed = 161/4 x 105 h
= 161 x 60/4 x 105min = 23 min
Correct Answer: 120
Explanation:
Let the minimum score = x
? Maximum score = x + 100
? x + (x + 100) = 30 x 42 - 40 x 28
? 2x + 100 = 1260 - 1120 = 140
? 2x = 140 - 100 = 40
? x = 20
Hence, the maximum score = x + 100
= 20 + 100 = 120
Correct Answer: 80 runs
Explanation:
Let the required runs be N.
According to the question,
(80 x 99 + N) / 80 = 100
? 7920 + N = 8000
? N = 80
Correct Answer: 51.4%
Explanation:
Let the total number of students be 100.
Also, let the mean score = N
According to the question,
20 x 80 + 25 x 31 + (100 - 20 - 25) x N = 100 x 52
? 1600 + 775 + (55) x N = 5200
? 55N = 5200 - 1600 - 775
? N = 2825/55 = 51.36 = 51.4%
So, mean score of remaining 55% is 51.4%
Correct Answer: 66 kg
Explanation:
Sum of the weight of A, B, C and D = 67 x 4 = 268 kg
and average weight of A, B , C , D and E = 67 - 2 = 65kg
? Sum of the weight of A, B, C, D and E = 65 x 5 = 325 kg
? Weight of E = 325 - 268 = 57 kg
? Weight of F = 57 + 4 = 61 kg
Now, average weight of F, B, C, D and E = 64 kg
? Sum of the weight of F, B, C, D and E = 64 x 5 = 320 Kg
? Sum of the weights of B, C and D = 320 - 57 - 61 = 202 kg
? Weight of A = 268 - 202 = 66 kg
Correct Answer: 55 yr
Explanation:
Total age of Rashmi and Surbhi (5 yr ago) 20 x 2 = 40 yr.
Total age of Rashmi and Surbhi (at present) = 40 + 5 + 5 = 50 yr.
Total age of Rashmi, Surbhi and Meeta (at present) = 30 x 3 = 90 yr.
? Present age of Meeta = 90 - 50 = 40 yr.
? Age of meeta after 15yr. = 40 + 15 = 55 yr.
Correct Answer: 4 : 3
Explanation:
Let number of students in the sections
A, B, C and D be a, b, c and d respectively.
Then, total weight of students of sections A = 45a
Total weight of students of section B = 50b
Total weight of students of section C = 72c
Total weight of students of section D = 80d
According to the question,
Average weight of students of sections of A and B = 48 kg
? (45a + 50b)/(a + b) = 48
? 45a +50b = 48a + 48b
? 3a = 2b
? 15a = 10b
And average weight of students of sections B and C = 60 kg
? 50b + 72c = 60(b + c)
? 10b = 12c
Now, average weight of students of A, B, C and D = 60 kg
? 45a + 50b + 72c + 80d = 60(a + b + c + d)
? 15a + 10b - 12c - 20d = 0
? 15a = 20d
? a : d = 4 : 3
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