Kano Applications Interview Questions, Process, and Tips

( e ) Apple

A = (B*H)/2
H = 2B
formula becomes A = (B*2*B)/2
this can be rewritten as A = (2*B^2)/2
the 2 in the numerator and denominator cancel out and you get:
A = B^2

3×8=24=2xt ===> t=12hours

C

Let’s assume the length of each train is ‘L’ and the speeds of the two trains are ‘V₁’ and ‘V₂’ respectively.

When the trains are moving in the opposite direction, their relative speed is the sum of their individual speeds. The total distance they need to cover is the sum of their lengths. Since they cross each other completely in 5 seconds, we can set up the following equation:

(V₁ + V₂) × 5 = 2L

When the trains are moving in the same direction, their relative speed is the difference between their individual speeds. The total distance they need to cover is the difference between their lengths. Since they cross each other completely in 15 seconds, we can set up the following equation:

(V₁ – V₂) × 15 = 2L

Now, let’s solve these equations to find the ratio of their speeds.

From the first equation, we have:
(V₁ + V₂) × 5 = 2L
V₁ + V₂ = (2L) / 5

From the second equation, we have:
(V₁ – V₂) × 15 = 2L
V₁ – V₂ = (2L) / 15

Let’s add these two equations together:
V₁ + V₂ + V₁ – V₂ = (2L) / 5 + (2L) / 15
2V₁ = (6L + 2L) / 15
2V₁ = (8L) / 15
V₁ = (4L) / 15

So, the speed of the first train is (4L) / 15.

Now, let’s substitute this value back into the first equation to find V₂:
(4L) / 15 + V₂ = (2L) / 5
V₂ = (2L) / 5 – (4L) / 15
V₂ = (6L – 4L) / 15
V₂ = (2L) / 15

Therefore, the speed of the second train is (2L) / 15.

The ratio of their speeds is given by:
(V₁ / V₂) = ((4L) / 15) / ((2L) / 15)
(V₁ / V₂) = 4L / 2L
(V₁ / V₂) = 2

So, the ratio of their speeds is 2:1.

Question is not completed
The letters A, B, C, D, E, F and G, not necessarily in that order, stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F

1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F
ans:a
2.A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E
ans:a

3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16

4. An integer T is as much greater than C as C is
greater than E. T
can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined
ans:a

ANS = 24.
16 = 2^4
24 = 2^3 x 3
H.C.F of 16 and 24 = 8
L.C.M will be 48

Might be using dual E1 in that case will be much more helpful.

4.5

let x be sum.
SI of 18% for 1 year= x*18/100=0.18x;
SI of 18% for 2 years is 0.36x;

SI of 12% for 1 year= x*12/100=0.12x;
SI of 12% for 2 years is 0.24x;

Given, 0.36x-0.24x=840
0.12x=840
x=840/0.12=7000.

c

30

First write equations from info:
(A) (Mon + Tue + Wed)/3 = 111 Rearrange as ——–> Tue + Wed = 111 – Mon
(B) (Tue + Wed + Thu)/3 =102 Rearrange as ——–> Tue + Wed = 102 – Thu
(C) Thu = 0.8(Mon)

Substitute equation C into B:
(B) Tue + Wed = 102 – 0.8(Mon)

At this point I changed the values for clearer algebra:
Mon = x
Tue + Wed = y

Re-write equations A & B with new values:
(A) y = 111 – x
(B) y = 102 – 0.8x

Solve simultaneous equations:

111 – x = 102 – 0.8x
111 – 102 = x – 0.8x (Re-arraged)
9 = 0.2x
x = 45

Thus, Mon = 45C

Thu = 0.8(45)
Thu = 36C

So the answer is it was 36C on Thursday

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