Special Quant Quiz (Arithmetic) SSC CGL Tier-II 2016

Q1. Two trains, 130 m and 110 m long, while going in the same direction, the faster train takes one minute to pass the other completely. If they are moving in opposite direction, they pass each other completely in 3 seconds. Find the speed of each train.  
(a) 42 m/s 38 m/s
(b) 38 m/s 36 m/s
(c) 36 m/s 42 m/s
(d) None of these

Q2. A train leaves the station at 5 am at 60 km/h. Another train leaves the same station at 6.30 am at 75 km/h and travels in the direction of the first train. At what time and at what distance from the station will they meet?
(a) 12.30 am 450 km
(b) 1.30 pm 375 km
(c) 11.30 am 425 km
(d) None of these

Q3. A and B are two stations. A train goes from A to B at 64 km/h and returns to A at a slower speed. If its average speed for the whole journey is 56 km/h, at what speed did it return?  
(a) 48 km/h
(b) 49.77 km/h
(c) 30 km/h
(d) 47.46 km/h
S3. Ans.(b)
Sol. Let the required speed b x km/h
Then, (2 × 64 × x)/(64 + x)=56
⇒ 128x = 64 × 56 + 56x
∴x=(64 × 56)/72 = 49.77 km/h

Q4. A 100 metres long train completely passes a man walking in the same direction at 6 km/h in 5 seconds and also a car travelling in the same direction in 6 seconds. At what speed was the car travelling?   
(a) 18 km/h
(b) 48 km/h
(c) 24 km/h
(d) 30 km/h
S4. Ans.(a)
Sol. Let, x km/h be the speed of the train
⇒ (x-6)×5/18×5=100
⇒ x = 78 km/h
Let, the speed of the car be y km/h
⇒ (78-y)×5/18×6=100 ⇒ y = 18.

Q5. If a 110 metres long train passes a man walking in opposite direction at a speed of 6 km/h in 6 seconds, it will pass another man walking at the same speed in the same direction in time of: 
(a) 28/3 seconds
(b) 32/3 seconds
(c) 8 second
(d) 22/3 seconds
S5. Ans.(d)
Sol. Let, the speed of the train = x km/h
Relative speed = (x + 6) km/h
=(x+6)×5/18 m/s
∴(x+6)×5/18×6=110
∴ x = 60
∴ Speed of train = 60 km/h for the person,
Relative speed = (60 – 6)km/h
=54×5/18 m/s = 15 m/s
∴ Time taken the cross the speed person
=110/15=22/3 seconds

Q6. If a distance of 50 m is covered in 1 minute, 90 m in 2 minutes and 130 m in 3 minutes find the distance covered in 15 minutes.   
(a) 610 m
(b) 750 m
(c) 1000 m
(d) 650 m

S6. Ans.(a)
Sol. Distance covered in 2nd minute = 90 – 50 = 40 metres 
Distance covered in 3rd minute = 130 – 90 = 40 m 
∴ Required distance = 50 + 40 × 14 
= 50 + 560 = 610m 

Q7. If a train runs at 40 km/h, it reaches its destination late by 11 minutes, but if it runs at 50 km/h, it is late by 5 minutes only. Find, the correct time for the train to complete its journey.  
(a) 19 minutes
(b) 20 minutes
(c) 21 minutes
(d) 18 minutes
S7. Ans.(a) 
Sol. 40 km/h =40/60 km/h minute =2/3 km/minute
50 km/h =50/60=5/6 km/minute
Let, distance be x km and the actual time be t minutes,
Then, according to the question,
x/(2/3)=t+11 ….(i)
x/(5/6)=t+5 …(ii)
By equations (i) and (ii), we get
3x/2-6x/5=6
⇒ (15x – 12x)/10=6
⇒ 3x = 60 km
⇒ x = 20 km
From equation (i),
3x/2=t+11
⇒ (3 × 20)/2=t+11
⇒ t = (30 – 11) = 19 minutes

Q8. Two trains, A and B, start from stations X and Y towards Y and X respectively. After passing each other, they take 4 hours 48 minutes and 3 hours 20 minutes to reach Y and X respectively. If train A is moving at 45 km/h, then the speed of the train B is: 
(a) 60 km/h
(b) 64.8 km/h
(c) 54 km/h
(d) 37.5 km/h

Q9. A train covers a distance between station A and station B in 45 minutes. If the speed of the train is reduced by 5 km/h, then the same distance is covered in 48 minutes. The distance between stations A and B is:  
(a) 60 km
(b) 64 km
(c) 80 km
(d) 55 km


Q10. Two places A and B are 100 km apart on a highway. One car starts from A and another from B at the same time. If the cars travel in the same direction at a constant speed, they meet in 5 hours. If the cars travel towards each other, they meet in 1 hour. What is the speed of the car running faster?  
(a) 60 km/h
(b) 50 km/h
(c) 40 km/h
(d) 32 km/h
S10. Ans.(a)
Sol. Let the speeds of the cars starting from A and B be x km/h and y km/h respectively.
Now, according to the question,
Case I: When both the cars are running in the same direction
Relative speed = (x – y) km/h
∴ 5x – 5y = 100 …(i)
Case II: When both the cars are running in the opposite directions
Relative speed = (x + y) km/h
∴ (x + y) × 1 = 100
⇒ x + y = 100 …(ii)
Now, solving equation (i) and equation (ii), we have
x = 60 and y = 40
Therefore, speed of the faster car is 60 km/h

Q11. A bus started its journey from Ramgrah and reached Devgrah in 44 minutes at an average speed of 50 km/h. If the average speed of the bus is increased by 5 km/h, how much time will it take to cover the same distance?
(a) 40 minutes
(b) 38 minutes
(c) 36 minutes
(d) 31 minutes
S11. Ans.(a)
Sol. Distance =44/60×50=x/60×55
∴ x = 40 minutes

Q12. 0.05 is what % of 0.5?
(a) 1%
(b) 0.1%
(c) 10%
(d) 100%
S12. Ans.(c)
Sol.
Let x% of 0.5 is 0.05
x×.5/100=.05
x=(.05 × 100)/.5=10%

Q13. One liter of water evaporates from 6 liters of sea water containing 4% salt. Find the % of salt in the remaining solution?
(a) 7.5%
(b) 4.5%
(c) 4.8%
(d) 8%
S13. Ans.(c)
Sol. Salt in sea water = 4% × 6 = .24
After evaporation of 1L of sea water remaining water is = 5 liters
Required % of salt in the remaining solution =.24/5=4.8%

Q14. In an election with only two contestants the candidates who gets 30% of the votes polled is defeated by 15000 votes. What is the number of votes polled?
(a) 112350
(b) 10580
(c) 26250
(d) 37500
S14. Ans.(d)
Sol. Let the total vote polled = 100x
According to question,
70x – 30x = 15000
x =15000/40
x = 375
So, total vote polled = 37500

Q15. If in a fraction, numerator is increased by 200% and denominator by 500%, then the new fraction is what % of previous fraction?
(a) 25%
(b) 45%
(c) 60%
(d) 50%
S15. Ans.(d)
Sol.
Let the fraction is x/y
New fraction =(x (100 + 200))/(y (100 + 500))
=3x/6y=x/2y
% of new fraction of pervious fraction
⇒(x/2y×100)/(x/y)=50%

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