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
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.
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
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.
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.
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.
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.
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.
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.
What was the day of the week
on 12th January, 1979?
Ans: Friday
Sol: Number of odd days in (1600 + 300) years = (0 + 1) = 1 odd day.
78 years = (19 leap years + 59 ordinary years) = (38 + 59) odd days = 6 odd days
12 days of January have 5 odd days.
Therefore, total number of odd days= (1 + 6 + 5) = 5 odd days.
Therefore, the desired day was Friday.
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