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5A=3B,
*3 SO,
15A=9B
Let’s say,
I have x coins of 50 paise and (80-x) coins of 100 paise,so the equation is like this ,
50x + (80-x)*100 = 64*100
x = 32
So ,I have 32 coins of 50 paise
Maximum number of edges = 9. Start from one corner. Select any face including that corner. Complete a square (4 edges) around the face to reach at the starting corner point . Now, move to the opposite face through the edge joining them and passing through the starting point(1 edge). Now, complete the square of edges around this face(4 edges). Total = 4+1+4 = 9 edges
d
zero – as one letter cant be wrong – atleast two are to be wrong
let the no be x
then given that the difference between the number and 3/5th
of the number is 50
therefor , x-3/5x=50
2x=250
x=125 is the answer
20- buy 15 books
15- buy 20 books
Original price is 20.
ans is C.
let H be present age of husaband.W be present age of wife.
H+W=91.
now let diff between their ages be x.that is H-W=x.
now when husband is W yaers old, his wife must be W-x years
old.and it is given that H=2(W-x). so 2W-H=2x. and H-
W=x.eliminating x we get 4W=3H. but H+W=91, so solving thse
two H=52 W=39.
c-> structured oriented lang
Pascal-> Procedure oriented lang
Days taken by A to complete whole work= 140 days
Days taken by B to complete whole work= 105 days
A’s 1 day work= 1/140
B’s 1 day work= 1/105
On adding these together we get: 1/60 (1 day work by working together)
So, together they will be able to complete it in 60 days
Dear friends i ofcourse confused like you when i come
across to solve this problem. Really friends its very
simple if you understand the question clearly. First thing
is what is mean by “as many as” means its called ‘idiom and
phrase’ in english and it means “the same number of”. now
read the question “how many pairs of letters in STAINLESS
which has same number of letters between them in the word
as they have in english alphabet”.
In the alphabetical order, A-Z can be numbered as 1-26.
In A(INL)E which is same as in the alphabetical order A
(BCD)E. In both the cases E is in the Fourth position. so
we got one pair.
And in ST, there are no letters between them in the word
stainless. In alphabetical orer from A-Z also there is no
letters between them..so we got the second pair…
In STAINLESS it has two pairs ST and AE
The last person covered 120.71 meters.
It is given that the platoon and the last person moved with
uniform speed. Also, they both moved for the identical
amount of time. Hence, the ratio of the distance they
covered – while person moving forward and backword – are
equal.
Let’s assume that when the last person reached the first
person, the platoon moved X meters forward.
Thus, while moving forward the last person moved (50+X)
meters whereas the platoon moved X meters.
Similarly, while moving back the last person moved [50-(50-
X)] X meters whereas the platoon moved (50-X) meters.
Now, as the ratios are equal,
(50+X)/X = X/(50-X)
(50+X)*(50-X) = X*X
Solving, X=35.355 meters
Thus, total distance covered by the last person
= (50+X) + X
= 2*X + 50
= 2*(35.355) + 50
= 120.71 meters
Note that at first glance, one might think that the total
distance covered by the last person is 100 meters, as he
ran the total lenght of the platoon (50 meters) twice.
TRUE, but that’s the relative distance covered by the last
person i.e. assuming that the platoon is stationary.
9 and 16
40
x – 30 = 1/4 x || *4
4x – 120 = x || -x + 120
3x = 120 || /3
x = 40