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

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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!
 

Hi,

A     7

B     11

C  22

 A+ B + C=  40

  4x + 12 = 40

     4x = 28

      x = 7 

CHECKING our answer***

  7 + 11 + 22 = 40

Wish You the Best in your Studies.

 

Answer by thejackal(72)   (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 

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