**Q1. A grocer has a sale of Rs. 6435, Rs. 6927, Rs. 6855, Rs. 7230 and Rs. 6562 for 5 consecutive months. How much sale must he have in the sixth month so that he gets an average sale of Rs. 6500?**

(a) Rs. 4991

(b) Rs. 5991

(c) Rs. 6001

(d) Rs. 6991

S1. Ans.(a)

Sol. Total sale for 5 month = Rs. (6435 + 6927 + 6855 + 7230 +

6562) = Rs. 34009.

6562) = Rs. 34009.

∴ Required sale = Rs. [(6500 × 6) – 34009] = Rs. (39000 – 34009) =

Rs. 4991.

Rs. 4991.

**Q2. 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 the target of 282 runs?**

(a) 6.25

(b) 6.5

(c) 6.75

(d) 7

**Q3. Of the three numbers, the average of the first and the second is greater than the average of the second and the third by 15. What is the difference between the first and the third of the three numbers?**

(a) 15

(b) 45

(c) 60

(d) None of these

**Q4. The average weight of a class of 24 students is 35 kg. If the weight of the teacher be included, the average rises by 400 g. The weight of the teacher is:**

(a) 45 kg

(b) 50 kg

(c) 53 kg

(d) 55 kg

S4. Ans.(a)

Sol. Weight of the teacher = (35.4 × 25 – 35 × 24) kg = 45 kg.

**Q5. The captain of a cricket team of 11 members is 26 years old and the wicket keeper is 3 years older then captain. 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?**

(a) 23 years

(b) 24 years

(c) 25 years

(d) None of these

S5. Ans.(a)

Sol. 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.

**Q6. The average weight of 8 persons increases by 2.5 kg when a new person comes in place of one of them weighing 65 kg. What might be the weight of the new person?**

(a) 76 kg

(b) 76.5 kg

(c) 85 kg

(d) None of these

S6. Ans.(c)

Sol. Total weight increased = (8 × 2.5) kg = 20 kg.

Weight of new person = (65 + 20) kg = 85 kg.

**Q7. The average age of 8 men is increased by 2 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:**

(a) 22 years

(b) 24 years

(c) 28 years

(d) 30 years

S7. Ans.(d)

Sol. 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

**Q8. 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:**

(a) 64

(b) 72

(c) 80

(d) 85

**Q9. The average weight of 3 men A, B and C is 84 kg. Another man D joins the group and the average now becomes 80 kg. If another man E, whose weight is 3 kg more than that of D, replaces A, then the average weight of B, C, D and E becomes 79 kg. Then weight of A is:**

(a) 70 kg

(b) 72 kg

(c) 75 kg

(d) 80 kg

S9. Ans.(c)

Sol. Let, A, B, C, D and E represent their respective weights.

Sol. Let, A, B, C, D and E represent their respective weights.

Then,

A + B + C = (84 × 3) = 252 kg, A + B + C + D = (80 × 4) = 320 kg.

A + B + C = (84 × 3) = 252 kg, A + B + C + D = (80 × 4) = 320 kg.

∴ D = (320 – 252) kg = 68 kg, E = (68 + 3) kg

= 71 kg.

B + C + D + E = (79 × 4) = 316 kg.

B + C + D + E = (79 × 4) = 316 kg.

Now, (A + B + C + D) – (B + C + D + E) = (320 – 316) kg

= 4 kg

∴ A – E = 4 ⇒ A = (4 + E) = 75 kg.

∴ A – E = 4 ⇒ A = (4 + E) = 75 kg.

**Q10. 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:**

(a) 19 years

(b) 23 years

(c) 28.5 years

(d) 29.3 years

S10. Ans.(a)

Sol. Sum of the present ages of husband, wife and child = (23 × 2 + 5 × 2) + 1 = 57 years.

∴ Required average =(57/3) = 19 years.

**Q11. 3 years ago, the average age of a family of 5 members was 17 years. A baby having been born, the average age of the family is the same today. The present age of the baby is**

(a) 1 year

(b) 1 1/2 years

(c) 2 years

(d) 3 years

S11. Ans.(c)

Sol. Total age of 5 members, 3 years ago = (17 × 5) years = 85 years.

Total age of 5 members now = (85 + 3 × 5) years = 100 years.

Total age of 6 members now = (17 × 6) years = 102 years.

∴ Age of the baby = (102 – 100) years = 2 years.

**Q12. The average age of 3 children in a family is 20% of the average age of the father and the eldest child. The total age of the mother and the youngest child is 39 years. If the father’s age is 26 years, what is the age of second child?**

(a) 15 years

(b) 18 years

(c) 20 years

(d) Cannot be determined

S12. Ans.(d)

Sol. Since the total or average age of all the family members is

not given, the given data is inadequate. So, the age of second child cannot be

determined.

not given, the given data is inadequate. So, the age of second child cannot be

determined.

**Q13. A certain factory employed 600 men and 400 women and the average wage was Rs. 25.50 per day. If a woman got Rs. 5 less than a man, then what are their daily wages?**

(a) Man : Rs. 25, Woman : Rs. 20

(b) Man: Rs. 27.50, Woman : Rs. 22.50

(c) Man : Rs. 30, Woman : Rs. 25

(d) Man: Rs. 32.50, Woman : Rs. 27.50

S13. Ans.(b)

Sol. Let the daily wage of a man be Rs. x

Then, daily wage of a woman = Rs. (x – 5)

Now, 600x + 400 (x – 5) = 25.50 × (600 + 400) ⇔ 1000x = 27500 ⇔ x = 27.50

∴ Man’s daily wages = Rs. 27.50; Woman’s daily wages = (x – 5) = Rs. 22.50

**Q14. The arithmetic mean of the scores of a group of 100 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:**

(a) 45

(b) 50

(c) 51.4 approx.

(d) 54.6 approx.

S14. Ans.(c)

Sol. Let the required mean score be x. Then,

20 × 80 + 25 × 31 + 55 × x = 52 × 100

⇔ 1600 + 775 + 55x = 5200 ⇔ 55x = 2825 ⇔ x=565/11=51.4.

**Q15. In an examination, a pupil’s average marks were 63 per paper. If he had obtained 23 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?**

(a) 8

(b) 9

(c) 10

(d) 11

S15. Ans.(d)

Sol. Let the number of papers be x. Then, 63x + 20 + 2 = 65x or 2x

= 22 or x = 11.

= 22 or x = 11.