Exams

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y

Difference = 14
=> 7x - 7y = 14
=> x - y = 2

product of numbers = product of their hcf and lcm
=> 7x * 7y = 441 * 7
=> x * y = 63

Now, we have
x * y = 63 , x - y = 2
=> x = 9 , y = 7

The numbers are 7x and 7y
=> 63 and 49

Let the sum be Rs. x. Then,
C.I. = x ( 1 + ( 10 /100 ))^2 - x = 21x / 100 ,
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100
( x / 100 ) = 632 * x = 63100.
Hence, the sum is Rs.63,100.

If we count letters till the end of the alphabet including that letter in reverse order then
Here the values are nothing but the values corresponding to the alphabets when taken in a reverse order,i.e,A=26 ,B=25...,Z=1
G = 20 and O = 12 totaling 32
S = 8, H = 19 and E = 22 totaling 49

So
SOME = 8 + 12 + 14 + 22 = 56.