 # Arithmetic Questions for SSC CHSL and RRB NTPC

Q1. At a game of billiards, A can give B 15 points in 60 and A can give C 20 points in 60. How many points can B give C in a game of 90?
(a) 30 points
(b) 20 points
(c) 10 points
(d) 12 points
S1. Ans.(c)
Sol. A : B = 60 : 45 and A : C = 60 : 40.
∴ B/C=(B/A×A/C)=(45/60×60/40)=45/40=90/80 = 90 : 80
∴ B can give C 10 points in a game of 90.

Q2. A solid piece of iron of dimensions 49 × 33 × 24 cm is moulded into a sphere. The radius of the sphere is:
(a) 21 cm
(b) 28 cm
(c) 35 cm
(d) None of these

Q3. The volume of a sphere is 4851 cu. cm. Its curved surface area is:
(a) 1386 cm^2
(b) 1625 cm^2
(c) 1716 cm^2
(d) 3087 cm^2

Q4. The cost of painting the whole surface area of a cube at the rate of 13 paise per sq. cm is Rs. 343.98 Then the volume of the cube is:
(a) 8500 cm^3
(b) 9000 cm^3
(c) 9250 cm^3
(d) 9261 cm^3

Q5. The volume of a rectangular block of stone is 10368 dm^3. Its dimensions are in the ratio of 3 : 2 : 1. If its entire surface is polished at 2 paise per dm^3, then the total cost will be:
(a) Rs. 31.50
(b) Rs. 31.68
(c) Rs. 63
(d) Rs. 63.36

Q6. A sum of money amounts to Rs. 6690 after 3 years and to Rs. 10,035 after 6 years on compound interest. Find the sum.
(a) 4360
(b) 4460
(c) 4260
(d) 4560

Q7. The principal that amounts to Rs. 4913 in 3 years at 6(1/4)% per annum compound interest compounded annually, is:
(a) Rs. 3096
(b) Rs. 4076
(c) Rs. 4085
(d) Rs. 4096

Q8. Two taps A and B can fill a tank in 5 hours and 20 hours respectively. If both the taps are open then due to a leakage, it took 30 minutes more to fill the tank. If the tank is full, how long will it take for the leakage alone to empty the tank?
(a) 4(1/2) hrs
(b) 9 hrs
(c) 18 hrs
(d) 36 hrs

S8. Ans.(d)
Sol. Part
filled by (A + B) in 1 hour =

, A
and B together can fill the tank in 4 hours.
Work
done by the leak in 1 hour =

Leak will empty the tank in 36 hrs.

Q9. 12 buckets of water fill a tank when the capacity of each tank is 13.5 litres. How many buckets will be needed to fill the same tank, if the capacity of each bucket is 9 litres?
(a) 8
(b) 15
(c) 16
(d) 18
S9. Ans.(d)
Sol. Capacity of the tank = (12 × 13.5) litres = 162 litres.
Capacity of each bucket = 9 litres.
Number of buckets needed = (162/9) = 18.

Q10. A water tank is two-fifth full. Pipe A can fill a tank in 10 minutes and pipe B can empty it in 6 minutes. If both the pipes are open, how long will it take to empty or fill the tank completely?
(a) 6 min. to empty
(b) 6 min. to fill
(c) 9 min. to empty
(d) 9 min. to fill
S10. Ans.(a)
Sol. Clearly, pipe B is faster than pipe A and so, the tank will be emptied.
Part to be emptied = 2/5.
Part emptied by (A + B) in 1 minute = (1/6-1/10)=1/15.
∴ 1/15:2/5∷1∶x or x=(2/5×1×15) = 6 min.
So, the tank will be emptied in 6 min.

Q11. A sum of money is sufficient to pay A’s wages for 21 days and B’s wages for 28 days. The same money is sufficient to pay the wages of both for:
(a) 12 days
(b) 12(1/4) days
(c) 14 days
(d) 24(1/2) days
S11. Ans.(a)
Sol. Let total money be Rs. x.
A’s 1 day’s wages = Rs. x/21, B’s 1 day’s wages = Rs. x/28.
∴ (A + B)’s 1 day’s wages = Rs. (x/21+x/28) = Rs. x/12.
∴ Money is sufficient to pay the wages of both for 12 days.

Q12. A and B can do a piece of work in 30 days, while B and C can do the same work in 24 days and C and A in 20 days. They all work together for 10 days when B and C leave. How many days more will A take to finish the work?
(a) 18 days
(b) 24 days
(c) 30 days
(d) 36 days
S12. Ans.(a)
Sol. 2 (A + B + C)’s 1 day’s work = (1/30+1/24+1/20)=15/120=1/8.
⇒ (A + B + C)’s 1 day’s work = 1/16.
Work done by A, B and C in 10 days = 10/16=5/8. Remaining work = (1-5/8)=3/8.
A’s 1 day’s work = (1/16-1/24)=1/48.
Now, 1/48 work is done by A in 1 day.
So, 3/8 work will be done by A in (48×3/8) = 18 days.
Q13. A and B together can do a piece of work in 12 days, which B and C together can do in 16 days. After A has been working at it for 5 days and B for 7 days, C finishes it in 13 days. In how many days C alone will do the work?
(a) 16
(b) 24
(c) 36
(d) 48
S13. Ans.(b)
Sol. A’s 5 days’ work + B’s 7 days’ work + C’s 13 days’ work = 1
⇒ (A + B)’s 5 days’ work + (B + C)’s 2 days’ work + C’s 11 days’ work = 1
⇒ 5/12+2/16 + C’s 11 days’ work = 1
⇒ C’s 11 days’ work = 1-(5/12+2/16)=11/24.
⇒ C’s 1 day’s work = (11/24×1/11)=1/24.
∴ C alone can finish the work in 24 days.
Q14. A, B and C can do a piece of work in 36, 54 and 72 days respectively. They started the work but A left 8 days before the completion of the work while B left 12 days before the completion. The number of days for which C worked is:
(a) 4
(b) 8
(c) 12
(d) 24
S14. Ans.(d)
Sol. Suppose the work was finished in x days.
The, A’s (x – 8) days’ work + B’s (x – 12) day’s work + C’s x days’ work = 1
⇒ ((x – 8))/36+((x – 12))/54+x/72 = 1 ⇔ 6 (x – 8) + 4 (x – 12) + 3x = 216
∴ 13x = 312 or x = 24.
Q15. 10 men can complete a piece of work in 15 days and 15 women can complete the same work in 12 days. If all the 10 men and 15 women work together, in how many days will the work get completed?
(a) 6
(b) 6(1/3)
(c) 6(2/3)
(d) 7(2/3)
S15. Ans.(c)
Sol. 10 men’s 1 day’s work = 1/15; 15 women’s 1 day’s work = 1/12.
(10 men + 15 women)’s 1 day’s work = (1/15+1/12)=9/60=3/20.
∴ 10 men and 15 women will complete the work in 3/20=6 (2/3) days.

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