Aptitude Questions Y4

21. If a man walks at the rate of 5kmph, he misses a train by only 7min. However if he walks at the rate of 6 kmph he reaches the station 5 minutes before the arrival of the train.Find the distance covered by him to reach the station.



Ans:6km.

Sol: Let the required distance be x km.

Difference in the times taken at two speeds=12mins=1/5 hr.

Therefore x/5-x/6=1/5 or 6x-5x=6 or x=6km.

Hence ,the required distance is 6 km



22. Walking 5/6 of its usual speed, a train is 10min late. Find the usual time to cover the journey?

Ans:50 min



Sol: New speed = 5/6 of usual speed

New time = 6/5 of usual time

Therefore, (6/5 of usual time) – usual time = 10min

Therefore Usual time = 50min



23.A train running at 54 kmph takes 20 seconds to pass a platform. Next it takes 12 seconds to pass a man walking at 6 kmph in the same direction in which the train is going. Find the length of the train and the length of the platform.



Ans. length of the train=160m

length of the platform=140 m.



Sol: Let the length of the train be x meters and length of the platform be y meters.

Speed of the train relative to man=(54-6) kmph =48 kmph.

=(48*5/18) m/sec =40/3 m/sec.

In passing a man, the train covers its own length with relative speed.

Therefore, length of the train=(Relative speed *Time)

=(40/3 * 12) m =160 m.

Also, speed of the train=(54 * 5/18) m/sec=15 m/sec.



Therefore, x+y/2xy=20 or x+y=300 or y=(300-160 m=140 m.

Therefore, Length of the platform=140 m.



24. A man is standing on a railway bridge which is 180m long. He finds that a train crosses the bridge in 20seconds but himself in 8 seconds. Find the length of the train and its speed.



Ans: length of train=120m

Speed of train=54kmph



Sol: Let the length of the train be x meters

Then, the train covers x meters in 8 seconds and (x + 180) meters in 20 seconds.

Therefore x/8 = (x+180)/20 ó 20x = 8(x+180) ó x = 120

Therefore Length of the train = 120m

Speed of the train = 120/8 m/sec = 15 m/sec =15 * 18/5 kmph = 54kmph

25. A man sells an article at a profit of 25%. If he had bought it at 20 % less and sold it for Rs.10.50 less, he would have gained 30%. Find the cost price of the article?

Ans. Rs. 50.



Sol: Let the C.P be Rs.x.

1st S.P =125% of Rs.x.= 125*x/100= 5x/4.

2nd C.P=80% of x. = 80x/100 =4x/5.

2nd S.P =130% of 4x/5. = (130/100* 4x/5) = 26x/25.

Therefore, 5x/4-26x/25 = 10.50 or x = 10.50*100/21=50.

Hence, C.P = Rs. 50.

26. A grosser purchased 80 kg of rice at Rs.13.50 per kg and mixed it with 120 kg rice at Rs. 16 per kg. At what rate per kg should he sell the mixture to gain 16%?



Ans: Rs.17.40 per kg.



Sol: C.P of 200 kg of mix. = Rs (80*13.50+120*16) = Rs.3000.

S.P = 116% of Rs 3000= Rs (116*3000/100) = Rs.3480.

Rate of S.P of the mixture = Rs.3480/200.per kg. = Rs.17.40 per kg.



27. Two persons A and B working together can dig a trench in 8 hrs while A alone can dig it in 12 hrs. In how many hours B alone can dig such a trench?



Ans:24hours.



Sol: (A+B)’s one hour’s work =1/8, A’s one hour’s work =1/12

Therefore, B’s one hour’s work = (1/8-1/12) =1/24.

Hence, B alone can dig the trench in 24 hours.



28. A and B can do a piece of work in 12 days ; B and C can do it in 20 days. In how many days will A, B and C finishes it working all together?

Also, find the number of days taken by each to finish it working alone?



Ans:60 days



Sol: (A+B)’s one day’s work=1/12; (B+C)’s one day’s work=1/15 and (A+C)’s one day’s

work=1/20.

Adding, we get: 2(A+B+C)’s one day’s work = (1/12+1/15+1/20)=1/5.

Therefore, (A+B+C)’s one day’s work=1/10.

Thus, A, B and C together can finish the work in 10 days.

Now, A’s one day’s work

= [(A+B+C)’s one day’s work] – [(B+C)’s one day’s work]

= 1/10-1/15)

= 1/30.

Therefore, A alone can finish the work in 30 days.

Similarly, B’s 1 day’s work = (1/10 -1/20) = 1/20.

Therefore, B alone can finish the work in 20 days.

And, C’s 1 day’s work= (1/10-1/12) = 1/60.

Therefore, C alone can finish the work in 60 days.



29. A is twice as good a workman as B and together they finish a piece of work in 18

days.In how many days will A alone finish the work?



Ans:27 days.



Sol: (A’s 1 day’s work): (B’s 1 day’s work) = 2:1.

(A + B)’s 1 day’s work = 1/18.

Divide 1/18 in the ratio 2:1.

Therefore A’s 1 day’s work = (1/18 * 2/3) = 1/27.

Hence, A alone can finish the work in 27 days.



30. 2 men and 3 boys can do a piece of work in 10 days while 3 men and 2 boys can do the same work in 8 days. In how many days can 2 men and 1 boy do the work?



Ans: 12 ½ days.



Sol: Let 1 man’s 1 day’s work = x and 1 boy’s 1 day’s work =y.

Then, 2x+3y=1/10 and 3x+2y=1/8.

Solving, we get: x=7/200 and y=1/100.

Therefore (2 men +1 boy)’s 1 day’s work = (2*7/200 + 1*1/100) = 16/200 = 2/25.

So, 2 men and 1 boy together can finish the work in 25/2 =12 ½ days.