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In 3500 g rod contains 74% of silver = 3500 * 74/100 = 2590 g
Then 3500 g + 500 g of rod contains 84% of silver
Let x be the silver contained in 500 g of silver
(2590/3500 * 100) + (x/500 * 100) = 84
74 + x/5 = 84
(370 + x) /5 = 84
370 + x = 420
x = 50
Then the percentage of silver contained in 500 g of rod = 50/500 *100 =10%
4/5
only if the results are greater than my input
50
In the given series 5,6,7,8,10,11,14…
There are two series
First is 5,7,10,14…
Second is 6,8,11….
First series
5+2=7
7+3=10
10+4=14
Second series
6+2=8
8+3=11
11+4=15
Hence complete series is 5,6,7,8,10,11,14,15…
C. Rs. 6000
D
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.
4400+(97 leap year )=4497
Let Suvarna, Tara, Uma and Vibha be S,T,U,V respectively
initially in the beginning each persons share be
V = x U = y T = z
S = w = (x+y+z+32) Reason: She has to double others share, so she should have each and everyone’s share and still should be left out with 32
after 1st Round of game
S loses and is out with 32 and doubles the others share
V = 2x U = 2y T = 2z
After 2nd Round of game
T loses and is out with 32 and doubles the others share
V = 4x U = 4y
This means T had 2z = 2x + 2y + 32
After 3rd round of game
U looses and is out with 32 and doubles others share
V = 8x
This means U initially has 4y = 4x + 32
In the end V = 8x = 32
Solving this we get x = 4, y = 12, z = 32 and w = 80
There fore Suvarna had highest share in the beginning
c
Each PAIR of stations means a PAIR of tickets (A to B and B to A)
2(old stations)(new stations) + 2(new stations)(new stations – 1) = 46
(N * X) + (X * (X – 1)) = 23
factoring ___ X (N + X – 1) = 23
23 is a prime number with only two factors, 1 and 23
so N = 23 and X = 1