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Maths Questions for DSSSB Exam 2017

Maths Questions for DSSSB Exam 2017_30.1
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