Finwizard Technology Interview Questions, Process, and Tips

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.

The answer is A)

y1 = 62 Rs/kg
y2 = 72 Rs/kg
y = 64.5 Rs/kg

y2 – y1 = 10 Rs/kg

The distance between the y and y1 is

y – y1 = 64.5 – 62 = 2.5

x1 = (y – y1)/(y2 – y1) = 2.5/10 = 0.25
x2 = 1 – x1 = 1 – 0.25 = 0.75

The target price is calculated by the lever method.

x1 * y2 + x2 * y1 = 0.25 * 72 + 0.75 * 62.5 = 64.5

The ratio is of y1 to y2 is

0.75 : 0.25

Divide by both by 0.25

3 : 1

1 through 24, divisible by 2 in descending order.

That leaves you with:
24, 22, 20, 18, 16, 14, 12, 10, 8, 6, 4, 2

8th place FROM the bottom means your answer will be 16.

b

2.5 sec

Let the number be x.

So x-2 =15/x, or

x^2–2x-15=0, or

(x-5)(x+3)=0

Hence x = 5 or -3.

Check: When x=5, 5–2=3 which is 15/5=3.

When x=-3, -3–2=-5 which is 15/-3=-5.

The number is 5 or -3.

option D

One day’s work of A, B and C = (1/24 + 1/30 + 1/40) = 1/10.

C leaves 4 days before completion of the work, which means only A and B work during the last 4 days.

Work done by A and B together in the last 4 days = 4 (1/24 + 1/30) = 3/10.

Remaining Work = 7/10, which was done by A,B and C in the initial number of days.

Number of days required for this initial work = 7 days.

Thus, the total numbers of days required = 4 + 7 = 11 days.

= 100 + 100 = 200 m

Time taken by train = 10 sec.

∵ Speed = Distance/Time

⇒ Speed of train = 200/10 = 20 m/sec.

∴ Speed of train = 20 × 18/5 = 72 kmph.

In question they left 10 seconds

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9 years
Bcz 3×3=9
5×3=15
7×3=21
Total = 45
Avg of this 45/3 = 15

600

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