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- Question
- 125.009 + 69.999 + 104.989 =?
Options- A. 420
- B. 300
- C. 285
- D. 415
- Correct Answer
- 300
Explanation125.009 + 69.999 + 104.989 = ?
Lets assume, each value is approximated to nearest whole number
? ? ? 125 + 70 + 105
? ? ? 300 - 1. 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
- 2. 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
- 3. 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
- 4. 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
- 5. 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
- 6. 16.003 x 27.998 - 209.010 =?
Options- A. 150
- B. 200
- C. 75
- D. 240 Discuss
- 7. 840.003 ÷ 23.999 =?
Options- A. 47
- B. 8
- C. 35
- D. 18 Discuss
- 8. 6885.009 - 419.999 - 94.989 =?
Options- A. 6650
- B. 6830
- C. 6370
- D. 6200 Discuss
- 9. 503 x 201=?
Options- A. 101100
- B. 1000000
- C. 110000
- D. 100003 Discuss
- 10. (2/3) x (6/8) x (2/3) x (3/4) =?
Options- A. 0.25
- B. 0.5
- C. 1.45
- D. 11.8 Discuss
Approximation problems
Search Results
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
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: 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: 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: 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: 240
Explanation:
? = 16.003 x 27.998 - 209.010
? ? 16 x 28 - 210 = 448 - 210
? = 238 ? 240
Correct Answer: 35
Explanation:
? = 840.003 ÷ 23.999
Here one number is increased and other is decreased to their nearest whole number
= 840 ÷ 24 = 35
Correct Answer: 6370
Explanation:
? = 6885.009 - 419.999 - 94.989
= 6885 - 420 - 95 ? 6370
Correct Answer: 101100
Explanation:
530 x 201 = 101103 ? 101100
Correct Answer: 0.25
Explanation:
(2/3) x (6/8) x (2/3) x (3/4) = 1/4 = 0.25
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