The Brand Experience Company Interview Questions, Process, and Tips

A=P+SI——(1)
sum of money after 30 years = Double the money
A=2P———-(2)

Equate 1 and 2
P+SI=2P
SI=2P-P
SI=P————-(3)

WKT, SI= PTR/100

Equate 3 and SI

P=PTR/100
P=P*30*R/100
R=P*100/P*30
R=100/30 or 10/3 or 3(1/3) %

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.

we know,
area=b*h
b gets increased by 20% i.e (b+0.20b)
h gets decreased by 20% i.e (h-0.20h)
rewriting the equation(area=b*h),
area=(b+0.20b)*(h-0.20h)
area=b(1+0.20)*h(1-0.20)
area=b(1.20)*h(0.80)
area=b*h*(1.20)*(0.80)
area=b*h*(0.96)
i.e new area=0.96 times the original area
if 100% was the original area,it has decreased to 96%
so,100%-96%= 4%

Cooking

RABBIT

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

500 = Total+50
Total(450) = only one paper(p) + 29+20+35 + all three (g)
285+212+127 = p + 2( 29+20+35 )+ 3g
solve above .. to get g = 45 …
( small corrctn .. i think .. questn shud be 20 read ONLY
hindu and
times of India and 29 read ONLY hindu and Indian express
and 35
read ONLY times of India and Indian express)

57 and 3
12 x 57 = 684
-8 x 3 = -24
684 – 24 = 660
660/ 60 = 11

8

When A runs 1000 meters, B runs 900 meters and when B runs 800 meters, C runs 700 meters.

Therefore, when B runs 900 meters, the distance that C runs = (900 x 700)/800 = 6300/8 = 787.5 meters.

So, in a race of 1000 meters, A beats C by (1000 – 787.5) = 212.5 meters to C.

So, in a race of 600 meters, the number of meters by Which A beats C = (600 x 212.5)/1000 = 127.5 meters.

180*2/3 = 60degree

let x=speed
t=time taken when speed is x so…
xt=4/5x(t+40)
t=160 minutes
2 hr 40 minutes