The area of a rectangle is equal to the area of a circle of radius [1/sqrt(pi)]m. If length of the rectangle is twice the width. Then sum of length and width will be
according to the given condition
l*w = pi(r)^2 r is radius and is given as 1/sqrt(pi)
so l*w = 1
so l*l/2 = 1
or l = sqrt(2)
so w = 1/sqrt(2)
so sum = [sqrt(2)] + 1/[sqrt(2)]
= 3/sqrt(2)
A and B complete a work in 10 days, B and C complete the same in 15 days, C and A in 20 days. How many days are required to do the same work if A, B, C work together.
one day work of all the 3 pairs are 1/10, 1/15, 1/20 respectively
so one day work of A, B, C will be 1/2[(1/10)+(1/15)+(1/20)] = 13/120
so days required = 120/13
k,4,n,40,121,364
364-121 = 243 = 3^5
121-40 = 81 = 3^4
so
40-n will be 27 = 3^3
so n = 13
and 13-4 = 9 = 3^2
so 4-k will be 3 = 3^1
so k = 1
so k+n = 14
Arun and Tarun can do a work in 10 days. After 4 days Tarun went to his village. How many days are required to complete the remaining work by Arun alone. Arun can do the work alone in 30 days.
After 4 days Tarun went to his village.
then remaining work = 6/10 part
one day work of Arun = 1/30 part
so days required = 6/10/1/30 = 180/10 = 18 days
[2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]
on multiplying [2-1] in the numerator and denominator, we get
[2-1][2+1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]/[2-1]
now we know that (a+b)(a-b) = a^2-b^2 so next line will be
[2^2-1][2^2+1][2^4+1][2^8+1][2^16+1]...........[2^256+1]/1
=[2^4-1][2^4+1][2^8+1][2^16+1]...........[2^256+1]
=[2^8-1][2^8+1][2^16+1]...........[2^256+1]
=[2^16-1][2^16+1]...........[2^256+1]
=[2^32-1][2^32+1]...........[2^256+1]
solve this, final result will be 2^512-1