3;4
189
6+8+10+10= 36
6²+8²+10²+12²=344
Form IV is the register of receipt and issue of raw materials which. With the replacement of private records in lieu of statutory records, no separate Form V is required provided you maintain private accounts for receipt, issue and closing stock of raw materials.
LET ACTUAL PRICE BE 100 RS(FOR CONVINIENCE)
THEREFORE IF PRICE IS CUT BY 20%
THEREFORE NEW PRICE=100-20=80 RS
THEREFORE WE NEED TO ADD X% TO 80 RS TO MAKE IT 20 RS AND ADD IT TO 80 TO MAKE IT 100RS
THEREFORE, 80X/100 = 20
I.E. 4X/5=20
THEREFORE x=20*5/4
=5*5
=25
HENCE 25% OF PRICE SHOULD BE ADDED TO MAKE THE PRICE EQUAL TO THE ACTUAL PRICE.
C
Ans is 10%
Printer price = x;
Computer price = 3x
Total overall price = 20x + 60 x 3x = 200x
Printer percent = (20x/200x ) x 100 = 10%
In the given series 5,6,7,8,10,11,14…
There are two series
First is 5,7,10,14…
Second is 6,8,11….
First series
5+2=7
7+3=10
10+4=14
Second series
6+2=8
8+3=11
11+4=15
Hence complete series is 5,6,7,8,10,11,14,15…
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
11
wake up guys and see modulo % can return only integer value.
So answer will be 1 and all other answers are wrong. You can
run your equation in C code and check the output.
9654
20000