On sports day,if 30 children were made to stand in a
column,16 columns could be formed. If 24 children were made to stand in a column
, how many columns could be formed?
Ans. 20
Sol: Total number of children=30*16=480
Number of columns of 24 children each =480/24=20.
Two trains 200mts and 150mts are running on the parallel rails at this rate of
40km/hr and 45km/hr.In how much time will they cross each other if they are
running in the same direction.
Ans: 252sec
Sol: Relative speed=45-40=5km/hr=25/18 mt/sec
Total distance covered =sum of lengths of trains =350mts.
So, time taken =350*18/25=252sec.
5/9 part of the population in a village are males. If 30% of the males are
married, the percentage of unmarried females in the total population is:
Ans: (250/9)%
Sol: Let the population =x Males=(5/9)x
Married males = 30% of (5/9)x = x/6
Married females = x/6
Total females = (x-(5/9)x)=4x/9
Unmarried females = (4x/9 – x/6) = 5x/18
Required percentage = (5x/18 * 1/x * 100) = (250/9)%
From height of 8 mts a ball fell down and each time it bounces half the distnace
back. What will be the distance travelled
Ans.: 24
Sol. 8+4+4+2+2+1+1+0.5+0.5+ and etc .. =24
First day of 1999 is sunday what day is the last day
Ans.: Monday
Increase area of a square by 69% by what percent should the side be incresed
Ans.: 13
Sol:Area of square=x2
Then area of increase=100+69=169
square root of 169 i.e 13 .
Ten years ago, chandrawathi’s mother was four times older than her daughter.
After 10years, the mother will be twice older than daughter. The present age of
Chandrawathi is:
Ans.20 years
Sol: Let Chandrawathi’s age 10 years ago be x years.
Her mother’s age 10 years ago = 4x
(4x+10)+10=2(x+10+10)
x=10
Present age of Chandrawathi = (x+10) = 20years
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
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