SOLUTION: 40 biscuits are shared amongst three children,A,B and C ,so that B has four more than A and C has twice as many as B.how many do they each have
Algebra ->
Equations
-> SOLUTION: 40 biscuits are shared amongst three children,A,B and C ,so that B has four more than A and C has twice as many as B.how many do they each have
Log On
Question 842595: 40 biscuits are shared amongst three children,A,B and C ,so that B has four more than A and C has twice as many as B.how many do they each have Found 2 solutions by ewatrrr, thejackal:Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website! formulate an equation out of the description.
What do we know:
1) there are 40 biscuits
2) there are 3 children
3) B has four more than A which said another way whatever A has, B has 4 + A's
4) C has twice as many as B's which is the same as saying 2 * (4 + A's)
when we put all this together we have a simple equation with only one unknown. lets call A's number of biscuits x to be clearer so that
x + (4+x) + 2(4+x) = 40; this is saying A's biscuits + B's biscuits + C's biscuits = 40
solve for x by simplifying first
2x + 4 + 8 + 2x = 40
4x + 12 = 40
4x = 40 - 12 = 28
x = 7 = A's biscuits
go back to equation 3 so that B's biscuits = 7 + 4 = 11
go back to equation 4 so that C's biscuits = 11 * 2 = 22
check your answer is 11+22+7 = 40?
Q.E.D