A chocolate has 12 equal to pieces Manju gave 1/4 ^th of it to anju 1/3 ^th of it to raj and 1/6 ^th of to fiza The number of pieces of chocolate left with manju is

Let the number be a and b Then
a + b = 10 ..........(i)
ab = 20 ........(ii)

Sum of reciprocal = 1/a + 1/b
= (a + b) / ab
= 10/20
= 1/2

Total share given by the person = (1/4) + (1/2) + (1/5)
= (5 + 10 + 4) / 20
=19/20

Let the positive number be N.
According to the question,
N + 10 = 200 / N
? N² + 10N = 200
? N² +10N - 200 =0
? (N - 10)(N + 20) = 0
? N = 10 & N= -20
But N not equal to -20 since N is a positive number
So, the required number is 10

Let the number be N.
According to the question,
N - (N x 3/4) =163
? N/4 =163
? N=652

Given Exp. = 6/7 + [(y - x)/(y + x)]
= 6/7 + [1 - (x/y) / 1 + (x/y)]
= 6/7 + [1 - (3/4)] / [1 + (3/4)]
= 6/7 + 1/7 =1.

Factors of composite number = (P₁ +1)(P₂+1)(P₃+1)......(P_n+1).
Where P₁,P₂,P₃... P_n are the powers of respective prime factors.

So,
120= 2³ x 3¹ x 5¹
Factors=(3+1)(1+1)(1+1)=16.

Solving the question by taking two odd numbers greater than 1 i.e. 3 and 5,
for n =3, n( n² -1 ) = 3( 9 - 1) = 24
for n = 5 n( n² -1 ) = 5( 25 - 1) = 120

Using option we find that both the number are divisible by 24.

For n =1
7⁶ⁿ - 6⁶ⁿ = 7⁶ - 6⁶
? ( 7³ )² - ( 6³ )²
? ( 7³ - 6³ )( 7³ + 6³ )
= ( 343 - 216 )( 343 + 216 )
= 127 x 559

? it is clearly divisible by 127.

Sum of first n natural numbers = n(n+1)/2
= 25 ( 25 + 1 ) / 2
= ( 25 x 26 ) / 2
= 325

Sum of the square of first n natural numbers = n(n+1)(2n+1) / 6

Given n = 35
? Required sum = ( 35 x 36 x 71 ) / 6
= 14910