I think experience because if experience brings pay. If someone having experience in any work then he/she can work easily. Although pay is important but not more then experience
C. 8 kg
Let the money borrowed by Nitin=Rs.P
According to the question,
(P×6×3)/100+(P×9×5)/100+(P×13×3)/100=Rs.8160
ans=8000
Cousin
9 and 16
Phillips
1, 5, 14, 30, 50, 55, 91
C. 50
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
Question is not completed
The letters A, B, C, D, E, F and G, not necessarily in that order, stand for seven consecutive integers from 1 to 10
D is 3 less than A
B is the middle term
F is as much less than B as C is greater than D
G is greater than F
1. The fifth integer is
(a) A
(b) C
(c) D
(d) E
(e) F
ans:a
2.A is as much greater than F as which integer is
less than G
(a) A
(b) B
(c) C
(d) D
(e) E
ans:a
3. If A = 7, the sum of E and G is
(a) 8
(b) 10
(c) 12
(d) 14
(e) 16
4. An integer T is as much greater than C as C is
greater than E. T
can be written as A + E. What is D?
(a) 2
(b) 3
(c) 4
(d) 5
(e) Cannot be determined
ans:a
CASE 1: First we should take six balls divided equally and
then it is placed on the two pans.three on one and three on
other..
if the two pans are balanced then the defective ball is not
in the six..then we should the two and keep them one ball
on each.
CASE2: Again We should take any of the six balls and
divided equally and then it is placed on the two pans.. if
any of the pan weighs less than the other.. We should take
the three balls seperately..Now from that three we should
take any two and placed one on each.. fi both the pan
balances the ball which is left over is the defective.. if
one ball weighes less than the other,while keeping one on
each,then it is the defective one….
Correct Deven,