Stratapps Interview Questions, Process, and Tips

17and+-12=5-48=-43

Complete step-by-step answer:
Investment made by A for 1 year, IA=2000
Investment made by B for 2 year,
IB=2×3000IB=6000

Investment made by C for 2 year,
IC=2×4000IC=8000
The ratio of their investment is given by
IA:IB:IC=2000:6000:8000
To simplify the ratio divide it by 1000,
IA:IB:IC=2:6:8
Now again to simplify divide the ratio by 2,
IA:IB:IC=1:3:4
The total parts of this investment =1+3+4=8
A’s share in the investment is 1 out of the 8 parts; B’s share is 3 out of 8 while C’s share is 4.
Therefore, the profit share of A’s investment of Rs. 2000=18×3200=400.

72
Because it is not a square number

Let x and y be the two numbers. And x > y.
Then,
x + y = 80—–(1)
y=80-y
Also,
x = 4y – 5
Solving the two equations,
80 – y = 4y – 5
85 = 5y
Therefore, y = 17
put the value of y in equation(1)
then
x+y=80
x+17=80
x=63
the numbers are 17&63

10 a.m.

A certain sum amounts to Rs. 1725 in 3 years

and amounts to Rs.1875 in 5 years

so interest of 2 years = 1875 -1725

= 150

so interest of 1 year = 75

so interest of 3 years = 75 × 3 =225 rs

so , Principal = Amount – SI

= 1725 – 225

= 1500 rs

now ,

S.I. = P × N × R /100

75 = 1500 × 1 × R /100

R = 75 / 15

R = 5%

To determine how many consecutive zeros the product of S will end with, we need to find the highest power of 10 that divides the product. This is equivalent to finding the highest power of 5 that divides the product, since the number of factors of 2 will always be greater than the number of factors of 5.

The primes in S are {2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}.

There are 24 primes in S, so the product of S is:

2 x 3 x 5 x 7 x 11 x 13 x 17 x 19 x 23 x 29 x 31 x 37 x 41 x 43 x 47 x 53 x 59 x 61 x 67 x 71 x 73 x 79 x 83 x 89 x 97

We need to find the highest power of 5 that divides this product. To do this, we count the number of factors of 5 in the prime factorization of each number in S.

5 appears once: 5
5 appears once: 25
5 appears once: 35
5 appears once: 55
5 appears once: 65
5 appears once: 85
So, there are six factors of 5 in the product of S. However, we also need to consider the powers of 5 that arise from the factors 25, 35, 55, and 65.

25 = 5 x 5 appears once: 25
35 = 5 x 7 appears once: 35
55 = 5 x 11 appears once: 55
65 = 5 x 13 appears once: 65
Each of these numbers contributes an additional factor of 5 to the product of S. Therefore, there are 6 + 4 = 10 factors of 5 in the product of S.

Since each factor of 5 corresponds to a factor of 10, we know that the product of S will end with 10 zeros. Therefore, the product of S will end with 10 consecutive zeros

4:3
A—–>B (first train)
B——>A(second train)
A/B=Srureroot((time to B)/(time to A))

192 zeros in 1 to 1000

suppose
pipe:
A -30 hours A’s effeciency (60/30) =2
60( lcm of 30 and 20)
B- 20 hours B’s effeciency (60/20)=3

time taken by both to fill = 60/5 =12 as given in question (effeciencies of both a+b =2+3=5)
time taken by faster pipe i.e b = 60/3 =20

33

2

30 Sec

=12.05×5.4/0.6
=12.05×9.0
=108.45

c