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decrease two times
Very exited
130 years
441=21^2
Let ‘N’ is the smallest number which divided by 13 and 16 leaves respective remainders of 2 and 5.
Required number = (LCM of 13 and 16) – (common difference of divisors and remainders)
= (208) – (11) = 197.
s1(1+x/100)+s2(1+x/100)+s3(1+x/100)
Let ‘x’ be the number of perons in the group and let ‘y’ be
amount per head which they have to pay.
Then xy = 2400.
or y = 2400/x
Since two friends have forget the purse, x-2 persons should
share the total amount(2400).
If they share, they have to make an extra contribution of
Rs 100 to pay up the bill
That is x-2 persons should pay y+100 each
or (x-2)(y+100) = 2400
or y+100 = 2400/(x-2)
or y = 2400/(x-2) – 100
Therefore we have got two equations namely,
y= 2400/x and y = 2400/(x-2) – 100
Comparing these two, we get
2400/x = 2400/(x-2) – 100
Solving this we get
x^2 – 2x -48 = 0
or (x-8) (x+6) = 0
or x = 8 or x = -6
Since ‘x’ denote the number of persons, it should be
positive.
So, x = 8.
75/8 days
Solution # 15-18
Sitting solution be- D T M B H R
Ans 15 :Mr. D
ans 16 : 1 2 3
Ans 17 : 3
Ans 18 : Mr B Suffering from anemia.
pass English =80%
fail=100-80=20%
pass Math = 70%
fail in Math=100-70=30%
fail in both=10%
total fail students= fail in Eng+fail in Math-common
= 20+30-10=40%
if 40% fail then 60% will pass
let total students=x
hence
60% 0f (total students)=144
60/100 of (x)= 144
x=(144×100)/60
X=240
total students=240
Greatest number of 5 digits=99, 999
Smallest number of 5 digits=10, 000
And their sum=99, 999+10, 000=109, 999
=36*84/ lcm
= 36 * 84 / (12 * 3 * 7 )
HCF= 12
D
Suppose that x glassses are supplied safely , therefore
(100-x) glases are damage. Then equation is :
3x – 3(100-x) = 270
on solving this equation value of x comes out to be 95.
21
38 / 2 = 19 – 5 = 14 years old
answer = 16.5Rs
A=3*9=27Rs
B=7*15=105Rs
132 R for 10 kg
therefore,
1kg amount = 132/10=13.2 so,
5kg mixture price = 13.2*5=66Rs
mixture will sell 25%profit then the price =66*(25/10)=33/2=16.5Rs