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Aptitude Interview Questions And Answers

16. 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 .

17. 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

18. Finding the wrong term in the given series

7, 28, 63, 124, 215, 342, 511
Ans:28
Sol:Clearly, the correct sequence is
       2^3 – 1, 3^3 – 1, 4^3 – 1, 5^3 – 1, ……….
       Therefore, 28 is wrong and should be replaced by (3^3 – 1) i.e, 26.

19. If i walk with 30 miles/hr i reach 1 hour before and if i walk with 20 miles/hr i reach 1 hour late.Find the distance between 2 points and the exact time of reaching destination is 11 am then find the speed with which it walks.

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

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

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.
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