**Q1. 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. The weight of A is:**

(a) 76

(b) 65

(c) 75

(d) 50

**Q2. If the product 4864 × 9 P 2 is divisible by 12, the value of P is:**

(a) 2

(b) 5

(c) 6

(d) 1

**Q3. A boy multiplied 987 by a certain number and obtained 559981 as his answer. If in the answer both 9 are wrong and the other digits are correct, then the correct answer would be:**

(a) 553681

(b) 555181

(c) 555681

(d) 554581

**Q4. Three numbers are co-prime to each other are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is:**

(a) 75

(b) 81

(c) 85

(d) 89

**Q5. Three different containers contain 496 litres, 403 litres and 713 litres of mixtures of milk and water respectively. What biggest measure can measure all the different quantities exactly?**

(a) 1 litre

(b) 7 litres

(c) 31 litres

(d) 41 litres

**Q6. A sum of Rs. 76 is divided among A, B and C in such a way that A gets Rs. 7 more than that B gets and B gets Rs. 6 more than that C gets. The ratio of their share is**

(a) 19 : 24 : 33

(b) 32 : 25 : 19

(c) 32 : 24 : 20

(d) 19 : 25 : 33

**Q7. The ratio of the ages of two persons is 4 : 7 and the age of one of them is greater than that of the other by 30 years. The sum of their ages (in years) is**

(a) 110

(b) 100

(c) 70

(d) 40

**Q8. There are 480 coins of half rupees, quarter rupees and 10 paise coins and their values are proportional to 5 : 3 : 1. The number of coins in each case are**

(a) 100, 200, 180

(b) 50, 30, 400

(c) 150, 180, 150

(d) 300, 90, 90

**Q9. Rs. 738 is divided among A, B, C so that their shares are in the ratio of 2 : 3 : 4. B’s share is**

(a) Rs. 328

(b) Rs. 246

(c) Rs. 264

(d) Rs. 164

**Q10. A mixture contains alcohol and water in the ratio of 4 : 3. If 5 litres of water is added to the mixtures the ratio becomes 4 : 5. The quantity of alcohol in the given mixture is**

(a) 3 litres

(b) 4 litres

(c) 15 litres

(d) 10 litres

**Solutions:**

S1. Ans.(c)

Sol. A + B + C = (3 × 84) kg = 252 kg

A + B + C + D = (4 × 80) kg = 320 kg

D = (320 – 252) kg = 68 kg

E = (68 + 3) kg = 71 kg

B + C + D + E = (4 × 79) kg = 316 kg

B + C + D = (316 – 71) kg = 245 kg

A = (320 – 245) kg = 75 kg

S2. Ans.(d)

Sol.

Clearly, 4864 is divisible by 4.

So, 9P2 must be divisible by 3. So, (9 + P + 2) must be divisible by 3.

∴ P = 1.

S3. Ans.(c)

Sol.

987 = 3 × 7 × 47

So, the required number must be divisible by each one of 3, 7, 47

553681 → (Sum of digits = 28, not divisible by 3)

555181 → (Sum of digits = 25, not divisible by 3)

555681 is divisible by each one of 3, 7, 47.

S4. Ans.(c)

Sol.

Since the numbers are co-prime, they contain only 1 as the common factor.

Also, the given two products have the middle number in common

So, middle number = H.C.F of 551 and 1073 = 29;

First number = (551/29) = 19; Third number = (1073/29) = 37

∴ Required sum = (19 + 29 + 37) = 85.

S5. Ans.(c)

Sol.

Required measurement = (H.C.F of 496, 403, 713) litres = 31 litres

S6. Ans.(b)

Sol. A = x + 13

B = x + 6

C = x

3x + 19 = 76

3x = 57

x = 19

A = 32

B = 25

C = 19

S7. Ans.(a)

Sol. 7x – 4x = 3x

3x = 30

x = 10

total = 11x

11 × 10 = 110

S8. Ans.(c)

Sol.

50p 25p 10p

5x 3x x

5x × 2 3x × 4 x × 10

10x + 12x + 10x = 480

32x = 480

x = 15

150 180 150

S9. Ans.(b)

Sol. B = 738 × 3/9

B = 246

S10. Ans.(d)

Sol. 4x/(3x + 5)=4/5

5x = 3x + 5

2x = 5

x = 5/2

alcohol = 4 × 5/2

= 10