# Math Questions for SSC CHSL and RRB NTPC EXAM

Q1. The value of x satisfying the equation √(2x+3)+√(2x-1) = 2 is
(a) 3
(b) 2
(c) 1
(d) 1/2
S1. Ans.(d)

Sol.  x=1/2 satisfies the given equation.

Q2. If A : B : C = 2 : 3 : 4, then A/B ∶B/C ∶C/A= ?
(a) 8 : 9 : 16
(b) 8 : 9 : 12
(c) 8 : 9 : 24
(d) 4 : 9 : 16
S2. Ans.(c)
Sol. A : B : C = 2 : 3 : 4 Let A = 2k, B = 3k and C = 4k. Then,
A/B=2k/3k=2/3=B/C=3k/4k=3/4 and C/A=4k/2k=2/1
(No need to write this help)

A/B:B/C:C/A=2/3:3/4 ∶2/1=8∶9∶24

Q3. Two numbers are in the ratio 3/2 ∶8/3. When each one of these in increased by 15, their ratio becomes 5/3 ∶5/2. The larger of the numbers is:
(a) 27
(b) 36
(c) 48
(d) 64
S3. Ans.(c)
Sol. Ratio of the given numbers =3/2 ∶8/3=9∶16
Let the numbers be 9x and 16x. Then,
(9x + 15)/(16x + 15)=((5/3))/((5/2))=2/3
⇒ 3(9x + 15) = 2(16x + 15)
⇒ 5x = 15
⇒ x = 3

Larger number = (16 × 3) = 48

Q4. The average age of 30 students is 9 year. If the age of the teacher is included, the average age becomes 10 years. What is the age of teacher?
(a) 27 years
(b) 31 years
(c) 35 years
(d) 40 years
S4. Ans.(d)
Sol. Age of teacher = n(b – a) + b
Where n = 30
Initial average, a = 9
Last average, b = 10
Required age = 30(10 – 9) + 10

= 30 + 10 = 40 years

Q5. In an examination A got 25% marks more than B, B got 10% less than C and C got 25% more than D. If D got 320 marks out of 500, the marks obtained by A were
(a) 405
(b) 450
(c) 360
(d) 400
S5. Ans.(b)
Sol. D got : 320 marks
C got : 320×125/100=400
B got : 400×90/100=360
A got : 36×125/100=450

A got marks = 450

Q6. From a container of wine, a thief has stone 15 litres of wine and replaced it with same quantity of water. He again repeated the same process. Thus in three attempts the ratio of wine and water became 343 : 169. The initial amount of wine in the container was:
(a) 150 litres
(b) 100 litres
(c) 120 litres
(d) 180 litres

Q7. A jar full of whisky contains 40% alcohol. A part of this whisky is replaced by another containing 19% alcohol and now the percentage of alcohol was found to be 26%. The quantity of whisky replaces is
(a) 1/3
(b) 2/3
(c) 2/5
(d) 3/5

Q8. In a farm there are cows and hens. If heads are counted there are 180, if legs are counted there are 420. The number of cows in the farm is
(a) 150
(b) 30
(c) 130
(d) 50
S8. Ans.(b)
Sol. Let the number of cows and hens in the farm be x and y, respectively
Then,  x + y = 180 …(i)
⇒ 4x + 2y = 420
⇒ 2x + y = 210  …(ii)
Subtracting equation (i) from equation (ii)
x = 30

Hence, there are 30 cows in the farm

Q9. A man covers half of his journey at 6 km/h and the remaining half at 3 km/hr. his average speed is:
(a) 3 km/hr
(b) 4 km/hr
(c) 4.5 km/hr
(d) 9 km/hr
S9. Ans.(b)
Sol. Average speed =2xy/((x+y) )km/hr
=(2 × 6 × 3)/((6 + 3) ) km/hr = 4 km/hr

Q10. The speed of a train 150 m long is 50 kmph. How much time will it take to pass a platform 600 m long?
(a) 50 sec
(b) 54 sec
(c) 60 sec
(d) 64 sec
S10. Ans.(b)
Sol. Speed =(50×5/18)=125/9 m/sec
Required time =((600 + 150))/((125/9)) sec

=(750×9/125) = 54 sec

Q11. A train travelling at a speed of 55 km/h, travels from X place to Y in 4 hours. If its speed is increased by 5 km/h, then the time of journey is reduced by
(a) 30 minutes
(b) 25 minutes
(c) 35 minutes
(d) 20 minutes
S11. Ans.(d)
Sol. Distance between station X and Y =speed ×Time
55 × 4 = 220 km
New speed = 55 + 5 = 60 kmph
∴ Required time =220/60=11/3 hours
3 hours 40 minutes

∴ Required answer = 4 hours – 3 hours 40 minutes = 20 minutes

Q12. P, Q and R can do a piece of work in 5, 8 and 10 days respectively, they start working together but R leaves after working 2 days and Q, 1 days before the completion of work. In how many days the work was finished?
(a) 2 (11/13) days
(b) 2 (12/13) days
(c) 2 (11/12) days
(d) None of these
S12. Ans.(a)
Sol. x/5+(x – 1)/8+2/10=1

Or, x=2 11/13 days

Q13. 6 men and 8 women can do a job in 10 days. In how many days can 3 men and 4 women finish the same job working together?
(a) 20 days
(b) 22 days
(c) 24 days
(d) 28 days
S13. Ans.(a)
Sol. Notice here we don’t know the relation of efficiencies but we can solve the problem due to clear relation between the work force
Since 6M + 8W = 10 days
2(3M + 4W) = 10 days
⇒ (3M + 4W) = 20 days

Since, the work force has become half of the original force so number of days must be double. Thus required number of days = 20

Q14. If 5 men and 3 boys can reap 23 acres in 4 days, and 3 men and 2 boys can reap 7 acres in 2 days, how many boys must assist 7 men in order than they may reap 45 acres in 6 days?
(a) 1
(b) 2
(c) 3
(d) 4

Q15. 8 taps are fitted to a water tank. Some of them are water taps to fill the tank and the remaining are outlet taps used to empty the tank. Each water tap can fill the tank in 12 hours and each outlet tap can empty it in 36 hours. On opening all the taps, the tank is filled in 3 hours. Find the number of outlet taps.
(a) 3
(b) 5
(c) 4
(d) 6
S15. Ans.(a)
Sol. Let water taps = x
x/12-(8-x)/36=1/3
⇒ x = 5

Number of outlet taps=3

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