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Question 1193893: Basket G and Basket H have a total of 284 lollipops. After 43 lollipops were transferred from Basket G to Basket H, there were 3 times as many lollipops in Basket H as Basket G. How many lollipops were there in (a) Basket G and (b) Basket H respectively at first?
Found 2 solutions by josgarithmetic, ikleyn:
Answer by josgarithmetic(39625)   (Show Source):
Answer by ikleyn(52855)  (Show Source):
You can put this solution on YOUR website! .
Basket G and Basket H have a total of 284 lollipops.
After 43 lollipops were transferred from Basket G to Basket H,
there were 3 times as many lollipops in Basket H as Basket G.
How many lollipops were there in (a) Basket G and (b) Basket H respectively at first?
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Since the transfer was between baskets G and H and nothing was transferred outside,
as well as nothing was transferred from outside, the total sum of 284 lollipops was unchangeable.
So after transfer, there were total 284 lollipops in basket G and basket H,
and were there 3 times as many lollipops in basket H as in basket G.
It implies that after transfer, were there 284/4 = 71 lollipops in basket G
and 71*3 = 213 lollipops in basket H.
Hence, BEFORE the transfer, were there 71 + 43 = 114 lollipops in basket G
and 213 - 43 = 170 lollipops in basket H.
ANSWER. At first, there were (a) 114 lollipops in basket G and (b) 170 lollipops in basket H.
Solved MENTALLY, without using any equations.
So, even a 4th grade young student may solve it in this way,
who even knows nothing about equations, but has living mind.
As a reward, he (or she) will get a piece of happiness for beautiful solution of the problem,
and will learn a beauty of Math.
The method is called " solving backward ".
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