Jane addams hull house association Interview Questions, Process, and Tips

13. One group of each = 2+3+5 = 10 and 1 remainder for each = 1+1+1 = 3. 10+3 = 13

75%

Let the firt number be n,
The largest number will be n+1
Their sum
n+n+1=55
2n+1=55
2n=54
n=27
Largest number =27+1=28

15

Answer:

Since the car has met the person 20 minutes before hand, it has actually saved 10 mins of journey (to and fro)

since the man has started 1.30 hrs before and car has met him 10 mins before actual time he takes to reach daily is 1hr and 20 mins

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

let,first num be x and second num be y
hcf =264
lcm=44
given equation
x/2=44 i.e x=88
product of 2 numbers is equal to lcm*hcf
88*y=44*264
y=132

perimeter = pi(r) + 2r
= 19.782 + (12.6)
= 32.382 cm

(3)^ 7.5 ÷ (27)^1.5 x (9)^2 = 3?
⇒(3)^7.5 ÷ {(3)^(3 x 1.5)} x {(3)^(2 x 2)} = 3 ?
⇒ 3^(7.5 – 4.5 + 4) = 3?
⇒ 3^7 = 3?
⇒ ? = 7

22

Therefore, the number of possible 13 digit numbers using 1, 2, 3, 4, 5 which are divisible by 4

= x 3 + x 2

= (3 + 2)

=

= 244140625 ways

I have seen a few say 25% as an answer to this. But looks
like their math isn’t good and is extremeley complex to
have an asnwer of 25% to this question. Consider – Rs.1.00/-
per mango; so 40 mangoes = Rs.40.00/-, implies Rs.4.00/-
extra at the end of the deal. So he’s left with Rs.44.00/-
after selling 40 mangoes implies a profit of Rs.4.00/- over
Rs.40.00 = 44/40 = 110% which means his profit is 10%.
Hope this helps…
Thanks.