SOLUTION: 1512 guavas were placed in three containers, X, Y and Z, at a supermarket.
The number of guavas in Container X is 2/7 of the total number of guavas.
190 guavas from Container Y a
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-> SOLUTION: 1512 guavas were placed in three containers, X, Y and Z, at a supermarket.
The number of guavas in Container X is 2/7 of the total number of guavas.
190 guavas from Container Y a
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Question 1205383: 1512 guavas were placed in three containers, X, Y and Z, at a supermarket.
The number of guavas in Container X is 2/7 of the total number of guavas.
190 guavas from Container Y and 1/4 of the guavas in Container Z were sold.
The number of guavas left in Container Y was twice the number of guavas left in Container Z. How many guavas were there in Container Y at first? Found 2 solutions by ikleyn, josgarithmetic:Answer by ikleyn(52884) (Show Source):
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1512 guavas were placed in three containers, X, Y and Z, at a supermarket.
The number of guavas in Container X is 2/7 of the total number of guavas.
190 guavas from Container Y and 1/4 of the guavas in Container Z were sold.
The number of guavas left in Container Y was twice the number of guavas left in Container Z.
How many guavas were there in Container Y at first?
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From the condition, the number of guavas in both containers Y and Z altogether
was 5/7 of 1512, or = 1080 originally.
Let y be the number of guavas in container Y originally (the value under
the problem's question). Then the number of guavas in container Y left is (y-190);
the number of guavas in container Z left is .
So, we can write this equation
y - 190 = .
Simplify and find y
y - 190 =
2(y-190) = 3*(1080-y)
2y - 380 = 3*1080 - 3y
2y + 3y = 3240 + 380
5y = 3620
y = 3620/5 = 724.
ANSWER. The number under the problem's question is 724.
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