Tricky questions of (Arithmetic misc.) for SSC CGL Tier-II 2016

Q1. Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?   
(a) 7 hours 30 minutes 
(b) 8 hours 
(c) 8 hours 15 minutes 
(d) 8 hours 25 minutes 
S1. Ans.(c)
Sol. Number of pages types by Ronald in 1 hour =32/6=16/3 
Number of pages types by Elan in 1 hour =40/5=8 
Number of pages typed by both in 1 hour =(16/3+8)=40/3   
∴ Time taken by both to type 110 pages =(110×3/40)hrs=8 (1/4) hrs = 8 hrs 15 min


Q2. A is 30% more efficient than B. How much time will they, working together, take to complete a job which A alone could have done is 23 days?  
(a) 11 days 
(b) 13 days 
(c) 20 days 
(d) None of these 
S2. Ans.(b)
Sol. Ratio of times taken by A and B = 100 : 130 = 10 : 13 
Suppose B takes x days to do the work 
Then 10 : 13 :: 23 : x ⇒ x=((23 × 13)/10)⇒x=299/10 
A’s 1 day’s work =1/23; = B’s 1 days work =10/299 
(A + B)’s 1 day’s work =(1/23+10/299)=23/299=1/13 
∴ A and B together can complete the job in 13 days.  



Q3. X and Y can do a piece of work in 20 days and 12 days respectively. X started the work alone and then after 4 days Y joined him till the completion of the work. How long did the work last? 
(a) 6 days 
(b) 10 days 
(c) 15 days 
(d) 20 days 
S3. Ans.(b)
Sol. Work done by X in 4 days =(1/20×4)=1/5 
Remaining work =(1-1/5)=4/5 
(X + Y)’s 1 day’s work  =(1/20+1/12)=8/60=2/15
Now, 2/15 work in done by X and Y in 1 day
So, 4/5 work will be done X and Y in (15/2×4/5)= 6 days.
Hence, total time taken = (6 + 4) days = 10 days



Q4. A, B and C can complete a work separately in 24, 36 and 48 days respectively. They started together but C left after 4 days of start and A left 3 days before the completion of the work. In how many days will the work be completed?  
(a) 15 days 
(b) 22 days 
(c) 25 days 
(d) 35 days 
S4. Ans.(a)
Sol. (A + B + C)’s 1 day’s work =(1/24+1/36+1/48)=13/144  
Work done by (A + B + C) in 4 dyas =(13/144×4)=13/36 
Work done by B in 3 days =(1/36×3)=1/12 
Remaining work =[1-(13/36+1/12)]=5/9
(A + B)’s 1 day’s work =(1/24+1/36)=5/72 
Now, 5/72 work is done by A and B in (72/5×5/9) = 8 days 
Hence, total time taken = (4 + 3 + 8) days = 15 days. 



Q5. Sixteen men can complete a work in twelve days. Twenty-four children can complete the same work in eighteen days. Twelve men and eight children started working and after eight days three more children joined them. How many days will they now take to complete the remaining work?  
(a) 2 days 
(b) 4 days 
(c) 6 days 
(d) 8 days 
S5. Ans.(b)
Sol. 1 man’s 1 day’s work =1/192; 1 child’s 1 day’s work =1/432 
Work done in 8 days =8(12/192+8/432)=8(1/16+1/54)=35/54 
Remaining work =(1-35/54)=19/54 
(12 men + 11 children)’s 1 day’s work =(12/192+11/432)=19/216 
Now, 19/216 work is done by them in 1 day. 
∴ 19/54 work will be done by them in (216/19×19/54)= 4 days. 



Q6. If 6 men and 8 boys can do a piece of work in 10 days while 26 men and 48 boys can do the same in 2 days, the time taken by 15 men and 20 boys in doing the same type of work will be: 
(a) 4 days 
(b) 5 days 
(c) 6 days 
(d) 7 days 
S6. Ans.(a)
Sol. Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work = y 
Then, 6x + 8y =1/10 and 26x + 48y =1/2 
Solving these two equations, we get: x=1/100 and y=1/200 
(15 men + 20 boys)’s 1 day’s work =(15/100+20/200)=1/4 
∴ 15 men and 20 boys can do the work in 4 days.  



Q7. A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m^3. The emptying capacity of the tank is 10 m^3 per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?  
(a) 50 m^3/min
(b) 60 m^3/min
(c) 72 m^3/min
(d) None of these 
S7. Ans.(a) 
Sol. Let, the filling capacity of the pump be x m^3/min 
Then, emptying capacity of the pump =(x+10) m^3/min
So, 2400/x-2400/((x+10) )=8⇒x^2+10x-3000=0 
⇒(x-50)(x+60)=0⇒x=50 [Neglecting the -ve value of x)] 



Q8. A leak in the bottom of a tank can empty the full tank in 8 hours. An inlet pipe fills water at the rate of 6 litres a minute. When the tank is full, the inlet is opened and due to the leak, the tank is empty in 12 hours. How many litres does the cistern hold?  
(a) 7580
(b) 7960
(c) 8290
(d) 8640 
S8. Ans.(d)
Sol. Work done by the inlet in 1 hour =(1/8-1/12)=1/24  
Work done by the inlet in 1 min =(1/24×1/60)=1/1440 
∴ Volume of 1/1440 part = 6 litres 
∴ Volume of whole = (1440 × 6) litres = 8640 litres. 



Q9. A car travels from P to Q at a constant speed. If its sped were increased by 10 km/hr, it would have taken one hour lesser to cover the distance. It would have taken further 45 minutes lesser if the speed was further increased by 10 km/hr. What is the distance between the two cities?   
(a) 420 km 
(b) 540 km 
(c) 600 km 
(d) 650 km 
S9. Ans.(a)
Sol
. Let distance = x km and usual rate = y kmph
Then,
x/y-x/(y+10)=1 or y (y + 10) = 10x …(i)
And, x/y-x/(y+20)=7/4 or y(y + 20) =80x/7 …(ii)
On dividing (i) by (ii), we get y = 60
Substituting y = 60 in (i) we, get : x = 420 km

Q10. Three persons are walking from a place A to another place B. Their speeds are in the ratio of 4 : 3 : 5. The time ratio to reach B by these persons will be:  
(a) 4 : 3 : 5 
(b) 5 : 3 : 4 
(c) 15 : 9 : 20 
(d) 15 : 20 : 12 
S10. Ans.(d)
Sol. Ratio of speeds = 4 : 3 : 5 
∴ Ratio of times taken =1/4:1/3:1/5 = 15 : 20 :12 



Q11. A walks around a circular filed at the rate of one round per hour while B runs around it at the rate of six rounds per hour. They start in the same direction from the same point at 7:30 a.m. They shall first cross each other at: 
(a) 7.42 a.m.
(b) 7.48 a.m. 
(c) 8.10 a.m. 
(d) 8.30 a.m. 
S11. Ans.(a)
Sol. Since A and B move in the same direction along the circle, so they will first meet each other when there is a difference of one round between the two. 
Relative speed of A and B = (6 – 1) = 5 rounds per hour 
Time taken to complete one round at this speed =1/5 hr = 12 min  
Hence, option(a) is correct



Q12. Two guns were fired from the same place at an interval of 10 minutes and 30 seconds, but a person in the train approaching the place hears the second shot 10 minutes after the first. The speed of the train (in km/hr), supposing that speed travels at 330 metres per second is: 
(a) 19.8 
(b) 58.6
(c) 59.4
(d) 111.80 
S12. Ans.(c)
Sol. Let the speed of the train be x m/sec. Then, 
Distance travelled by train in 10 min = Distance travelled by sound in 30 sec.
⇒ x × 10 × 60 = 330 × 30 ⇒ x = 16.5 
∴ Speed of the train = 16.5 m/sec =(16.5×18/5) km/hr = 59.4 km/hr 

Q13. A and B walk around a circular track. They start at 8 a.m. from the same point in the opposite directions. A and B walk at a speed of 2 rounds per hour and 3 rounds per hour respectively. How many times shall they cross each other before 9.30 a.m.?
(a) 5 
(b) 6
(c) 7
(d) 8
S13. Ans.(c)
Sol. Relative speed = (2 + 3) = 5 rounds per hour
So, they cross each other 5 times in an hour and 2 times in half an hour.
Hence, they cross each other 7 times before 9 : 30 a.m.  



Q14. A jogger running at 9 kmph alongside a railway track is 240 metres ahead of the engine of a 120 metre long train running at 45 kmph in the same direction. In how much time will the train pass the jogger?  
(a) 3.6 sec 
(b) 18 sec 
(c) 36 sec 
(d) 72 sec 
S14. Ans.(c)
Sol. Speed of train relative to jogger =(45-9) km/hr = 36 km/hr 
=(36×5/18)sec = 10 m/sec
Distance to be covered = (240 + 120) m = 360 m.
∴ Time taken =(360/10)sec = 36 sec.  



Q15. A train 110 m long passes a man, running at 6 kmph in the direction opposite to that of the train, in 6 seconds. The speed of the train is:  
(a) 54 km/hr 
(b) 60 km/hr 
(c) 66 km/hr  
(d) 72 km/hr
S15. Ans.(b)
Sol. Speed of the train relative to man 
=(110/6) m/sec =(110/6×18/5)  km/hr = 66 km/hr
Let the speed of the train be x kmph. Then, relative speed = (x + 6) = kmph

∴ x + 6 = 66 or x = 60 kmph.    





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