Biopharmax Interview Questions, Process, and Tips

7/5 * 5/q = 7/10
35/5q = 7/10
q=10

Answer: z and u

Explanation:

Y is to the right of U and exactly in front of V. Therefore,

U Y
V

Z is behind W and W and X are at extreme ends. So, W has to be to the right of Y. The final arrangement is as follows.
U Y W
X V Z Therefore, Z and U are at extreme ends is true.

Let cost for apple be a Cost for banana be b and Orange be c

So by first value expression becomes 17a + 13b + 9c =130 ———-1 therefore if you further solve a = (130 – 13b – 9c)/17 ———- 2 the second expression becomes: 13c + 7a + 10b = 100 ———- 3 If you put value of a in second expression it becomes: 13c + 7[(130 – 3b – 9c)/17] +10b = 100

Further if you solve you get value of b:

b = 10 – 2c ———-4

put value of b from 4 in 1

17a + 13 [10 – 2c] + 9c = 130

Further if you solve you find value of a

a = c ———-5

Put 5 in 3

13c + 7c + 10b = 100

further solve you get: c = 1 ———-6

from 5 and 6

a = c = 1 ———-7

Substitute value of c in expression 4

b = 10 – 2c b = 10 -2 * 1 b = 8 ———-8

therefore a + b + c = 10

Kisan Baburao Hazare

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

54

20(85+0)/2=850

B. 32%

Kerala