Question

Rohit had 48 sweets .he arrange them in three heaps having unequal numbers .if he take six sweets from first heap and put two sweets in second heap and four in third heap ,all heap have equal number of sweets what was the original no of sweets in each heap

Answer 1

Original number of sweets in each heap is 22, 14, 12

- let the three heaps original number be x , y and z respectively

-Since total no of sweets are 48,

we have x + y + z = 48.................(1)

-According to the condition,

   taking out 6 from 1st heap => x-6 in 1st heap

   adding 2 to 2nd heap => y+2 in 2nd heap

   adding 4 to 3rd heap => z+4 in 3rd heap

-We have x-6 = y+2 = z+4

from here we can get equations,

  x=y+8..........(2)

  x=z+10.........(3)

-Using equations 1, 2 and 3, we can find the value of x, y and z