Gain=(profit/c.p)*100
Gain =selling price-cost price =900-675
=225.
Gain=(225/675)*100.
=33.3%
2cm b)
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
6+8+10+10= 36
6²+8²+10²+12²=344
Ans = 8
Use simple box method
[1][2][2][2]= 1x2x2x2 =8
Logic >
once place have 2 number (2,6)
Tens place have also 2 number (7,3)
So that number is divisible by 4
Now, 2 numbers fixed at once and tens place left 2 numbers which is choice at hundredth place
Now 3 number fix left 1 number which is placed at thousands place
Total number form is 4×4 = 16
But divisible by 4 is 1x2x2x2 =8
let d distance and s be speed.
d/7 = x and d/5 = x+12
solving we get d=210.
The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
4:2
20/- per day
So, if he work one day
So,1*20=20
1/3*20=20/3
2/3*20=40/3
1/8*20=5/2
3/4*20=15
so,
20+20/3+40/3+5/2+15=57.5
So ans is 57.5 is the ans
Answer:
11 days.
Step-by-step explanation:
In the question,
Time taken by Ramesh to finish a piece of work = 20 days
Time taken by Sushil to finish a work = 25 days
Time for which they worked together = 5 days
Sushil left after = 5 days
So,
One day work of Ramesh is,
One day work of Sushil is,
So,
Work done in 5 days is given by,
Therefore, Remaining work is given by,
Now, as the Sushil left the remaining work was done by Ramesh,
Time taken by Ramesh for the remaining work is,
Therefore, the remaining work will be completed in 11 days by Ramesh.
4