Question

Raju took some some chocolates. He made groups with 5 of them and found 1 chocolate was left. He
also tried making groups of 6 and 8 each time 1 chocolate was left over. What is the smallest number

Answer 1

Step-by-step explanation:

1 chocolate= 1 whole

You had 1 chocolate at the beginning

Person 1: you gave him 1/4 of the chocolate

Now you have: 1-1/4 =3/4

Person 2: you gave him 1/2 of the remaining part of the chocolate .

Now you have: 3/4-1/2= 1/4 chocolate.

Therefore you have 1/4 of the chocolate you had at the beginning

plz mark me....

Answer 2

Answer:

The smallest possible number of chocolates Raju had = 121

Step-by-step explanation:

Given,

When Raju divided the number of chocolates he has into groups of 5,6 and 8 one chocolate was left over.

To find,

The minimum number of chocolate Raju has with this condition.

Solution:

Let 'X' be the number of chocolates Raju had

Then from the given conditions we have,

X = 5n+1

X = 6m+1

X = 8p+1

5n+1 = 6m+1 = 8p+1 = X

5n = 6m = 8p = X -1

The above condition will satisfy all common multiples of 5,6 and 8

The possible number of chocolates with Raju is 1 more than any common multiple of 5,6, and 8

The smallest possible number chocolates = LCM(5,6,8) + 1

LCM(5,6,8) = 2×2×2×3×5 = 120

The smallest possible number chocolates = 120+1  = 121

∴ The smallest possible number of chocolates Raju had = 121

#SPJ2