Jupiter infrastructure (bangalore) pvt ltd Interview Questions, Process, and Tips

4/5

Yellow Red Green
Blue Blue Green
Yellow Green Yellow

c

area is given by Area of Rectangle = Length of Rectangle*(sqrt((Diagonal of Rectangle^2)-(Length of Rectangle^2)))
so substituting we get area=4*(sqrt((25-16)))
area=4*(sqrt(9))
since area cannot be negative hence we take sqrt of 9 as 3 (as sqrt of 9 is 3 and -3 as well)
area=4*3=12m^2

(x**2 – 6* x + 5) = (x-1)*(x-5)
(x**2 + 2 * x + 1) = (x + 1) * (x+1) = (x+1)**2
For what x is (x-1)*(x-5)/( (x+1)**2) a minimum?
One way to answer this question is by using calculus.
Take the derivative, and set to zero.
Since this is a fraction of polynomials, and a fraction is
zero only if it’s numerator is zero, we need calculate only
the numerator of the derivative and set it to zero.
The numerator of the
Derivative of (x-1)*(x-5)/( (x+1)**2) is
( (x-1) + (x-5) ) ( x+1)**2 – (x-1)(x-5)( 2 (x+1) )
= (2 x – 6) (x+1)**2 – (2) (x-1)(x-5) (x+1)
= 0
Divide through by 2 (x+1)
(x-3)(x+1) – (x-1)(x-5) = 0
(x**2 – 2 x – 3 ) – (x**2 – 6 x + 5) = 0
x**2 – x**2 – 2 x + 6 x – 3 – 5 = 0
4 x – 8 = 0
x = 2
Plugging in x = 2 into the original
(x**2-6*x+5)/(x**2+2*x+1)
gives us (2**2 – 6 * 2 + 5)/(2**2 + 2*2 + 1)
= (4 – 12 + 5) / (4 + 4 + 1) = -3/9 = -1/3
Least value is -1/3

3000

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

(b) 16.66%

Earlier for ₹x we could purchase y gm of sugar.
Now we pay ₹1.2x for y gm of sugar

(As there was an increase in price so, x + 20%x = 1.2x)

At current rates for ₹x you can purchase y/1.2 gm of sugar

So the reduced consumption is y-(y/1.2)

Percentage change = (reduced consumption/ original consumption ) *100

That is (0.2/1.2) *100 = 16.66% (approx)

16

( a ) BQDCJCMF
“TERMINAL” split it into the half we get “TERM” “INAL”

Now the first half is decreased by one and the next half is
increased by one, so we get:

“SDQLJOBM” (pls note in the ques we have “SDQIJOBM” where
‘L’ shud have come instead of ‘I’)

so “CREDIBLE” is to “BQDCJCMF”

94

the pool will be half full at the 19th day .so that it doubles on the 20th day and becomes full