Let the number of cakes be a,b,c,d,e,f,g for the 7 children
a-(b+c+d+e+f+g) = 4
b+(2*6) = a-3
we want c+d+e+f+g. Call it x
a-b-x = 4
b+12 = a-3
x = (a-b)-4
a-b = 15
So, x = 11
there are 7 children standing in a line ,not all of whom have the same number of cakes with them, if the first child distributes his cakes to the remaining six children such that he doubles their respective no.of cakes ,then he will be left with four cakes. instead ,if the second child takes away 2 cakes from each of the remaining six ,then he will be left with 3 cakes less than the no.of cakes that the 1 child initially had.what is the total no.of cakes that are there with the 3 child to seventh child
1 answer
1 answer