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- Question
Out of 9 persons, 8 persons spent Rs. 30 each for their meals. The ninth one spent Rs. 20 more than the average expenditure of all the nine. The total money spent by all of them was :
Options- A. Rs. 292.50
- B. Rs. 297.50
- C. Rs. 298
- D. Rs. 298.50
- Correct Answer
- Rs. 292.50
ExplanationLet the average expenditure be Rs. x Then,
9x=[(8*30)+(x+20)]<=>9x =x+260 <=> x =32.50
Total money spent = 9x = Rs. (9 x 32.5O) = Rs 292. 50
- 1. 10 years ago, the average age of a family of 4 members was 24 years. Two children having been born (with age diference of 2 years), the present average age of the family is the same. The present age of the youngest child is :
Options- A. 1
- B. 2
- C. 3
- D. 4 Discuss
- 2. Of the four numbers, whose average is 60, the first is one-fourth of the sum of the last three. The first number is :
Options- A. 17
- B. 29
- C. 36
- D. 48 Discuss
- 3. The mean of 50 observations was 36. It was found later that an observation 48 was wrongly taken as 23. The corrected new mean is :
Options- A. 35
- B. 36.5
- C. 40
- D. 42 Discuss
- 4. A pupil's marks were wrongly entered as 83 instead of 63. Due to that the average marks for the class got increased by half. The number of pupils in the class is :
Options- A. 45
- B. 40
- C. 39
- D. 37 Discuss
- 5. After replacing an old member by a new member, it was found that the average age of five members of a club is the same as it was 3 years ago. What is the difference between the ages of the replaced and the new member ?
Options- A. 12
- B. 13
- C. 14
- D. 15 Discuss
- 6. In an examination, a pupil's average marks were 63 per paper. If he had obtained 20 more marks for his Geography paper and 2 more marks for his History paper, his average per paper would have been 65. How many papers were there in the examination ?
Options- A. 8
- B. 9
- C. 10
- D. 11 Discuss
- 7. The average of 11 numbers is 10.9. If the average of the first six numbers is 10.5 and that of the last six numbers is 11.4, then the middle number is :
Options- A. 10.5
- B. 11.5
- C. 12.5
- D. 13.5 Discuss
- 8. The average age of 8 men is increased by years when two of them whose ages are 21 years and 23 years are replaced by two new men. The average age of the two new men is :
Options- A. 20
- B. 30
- C. 40
- D. 50 Discuss
- 9. The average age of a husband and his wife was 23 years at the time of their marriage. After five years they have a one-year old child. The average age of the family now is :
Options- A. 25
- B. 23
- C. 19
- D. 18 Discuss
- 10. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older. If the ages of these two are excluded, the average age of the remaining players is one year less than the average age of the whole team. What is the average age of the team ?
Options- A. 20 years
- B. 21 years
- C. 23 years
- D. 25 years Discuss
Average problems
Search Results
Correct Answer: 3
Explanation:
Total age of 4 members, 10 years ago = (24 x 4) years = 96 years.
Total age of 4 members now = [96 + (10 x 4)] years = 136 years.
Total age of 6 members now = (24 x 6) years = 144 years.
Sum of the ages of 2 children = (144 - 136) years = 8 years.
Let the age of the younger child be x years.
Then, age of the elder child = (x+2) years.
So, x+(x+2) =8 <=> x=3
Age of younger child = 3 years.
Correct Answer: 48
Explanation:
Let the first number be x,
Then, sum of the four numbers = x + 4x = 5x.
so, 5x/4 = 60 or x = (60 * 4) / 5 = 48.
Correct Answer: 36.5
Explanation:
Correct Sum = (36 * 50 + 48 - 23) = 1825.
Correct mean = = 1825/50 = 36.5
Correct Answer: 40
Explanation:
Let there be x pupils in the class.
Total increase in marks = (x * 1/2) = x/2.
x/2 = (83 - 63) => x/2 = 20 => x = 40.
Correct Answer: 15
Explanation:
i) Let the ages of the five members at present be a, b, c, d & e years.
And the age of the new member be f years.
ii) So the new average of five members' age = (a + b + c + d + f)/5 ------- (1)
iii) Their corresponding ages 3 years ago = (a-3), (b-3), (c-3), (d-3) & (e-3) years
So their average age 3 years ago = (a + b + c + d + e - 15)/5 = x ----- (2)
==> a + b + c + d + e = 5x + 15
==> a + b + c + d = 5x + 15 - e ------ (3)
iv) Substituting this value of a + b + c + d = 5x + 15 - e in (1) above,
The new average is: (5x + 15 - e + f)/5
Equating this to the average age of x years, 3 yrs, ago as in (2) above,
(5x + 15 - e + f)/5 = x
==> (5x + 15 - e + f) = 5x
Solving e - f = 15 years.
Thus the difference of ages between replaced and new member = 15 years.
Correct Answer: 11
Explanation:
Let the number of papers be x. Then, 63x + 20 + 2 = 65x or 2x = 22 or x = 11.
Correct Answer: 11.5
Explanation:
Middle numbers = [(10.5 x 6 + 11.4 x 6) - 10.9 x 11] = (131.4 - 119-9) = 11.5.
Correct Answer: 30
Explanation:
Total age increased = (8 * 2) years = 16 years.
Sum of ages of two new men = (21 + 23 + 16) years = 60 years
Average age of two new men = (60/2) years = 30 years.
Correct Answer: 19
Explanation:
Sum of the present ages of husband, wife and child = (23 * 2 + 5 * 2) + 1 = 57 years.
Required average = (57/3) = 19 years.
Correct Answer: 23 years
Explanation:
Let the average age of the whole team be x years.
11x - (26 + 29) = 9 (x - 1)
=> 11x - 9x = 46
=> 2x = 46
=> x = 23.
So, average age of the team is 23 years.
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