Question 1197826: Hi
There were 207 pears and 176 mangoes in crate A . There were 293 pears and 274 mango in crate B. Rob moved some pears and mangoes from B to A . In the end 40% of the fruit in A and 70% of the fruit in B were pears. How many pieces of fruit did he move from B to A. Thanks
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
p = number of pears moved from B to A
m = number of mangoes moved from B to A
These variables represent positive whole numbers {1,2,3,...}
After moving p of the pears and m mangoes from crate B to crate A, we have 207+p pears out of 207+176+p+m = 383+p+m total fruit in crate A.
The ratio (207+p)/(383+p+m) is set equal to 0.40 since the new percentage of pears in crate A is 40%.
Let's solve for the variable m.
(207+p)/(383+p+m) = 0.40
207+p = 0.40*383+0.40*p+0.40*m
207+p = 153.2+0.40p+0.40m
0.40m = 207+p-153.2-0.40p
0.40m = 0.6p + 53.8
m = (0.6p + 53.8)/0.40
m = (0.6p)/0.40 + (53.8)/0.40
m = 1.5p + 134.5
We'll use this in a substitution step later.
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After moving p of the pears and m of the mangoes from B to A we have 293-p pears left in crate B
This is out of 293+274-p-m = 567-p-m total fruit in crate B.
We want this ratio
(293-p)/(567-p-m)
to be set equal to 0.70 to reflect the fact 70% of the fruit left in crate B are pears.
So we form this equation
(293-p)/(567-p-m) = 0.70
Let's replace m with 1.5p+134.5
This is what I referred to in the substitution step.
(293-p)/(567-p-m) = 0.70
(293-p)/(567-p-(1.5p+134.5)) = 0.70
(293-p)/(567-p-1.5p-134.5) = 0.70
(293-p)/(-2.5p+432.5) = 0.70
We're left with one variable in which we can solve for p.
(293-p)/(-2.5p+432.5) = 0.70
293-p = 0.70(-2.5p+432.5)
293-p = 0.70(-2.5p)+0.70(432.5)
293-p = -1.75p+302.75
-p+1.75p = 302.75-293
0.75p = 9.75
p = 9.75/0.75
p = 13
Rob moved 13 pears from crate B to crate A.
Use this value of p to find m.
m = 1.5p + 134.5
m = 1.5*13 + 134.5
m = 19.5 + 134.5
m = 154
He also moved 154 mangoes from crate B to crate A.
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Let's check the answers.
| Crate A | Crate B | | # of pears | # of mangoes | # of pears | # of mangoes | | Before moving | 207 | 176 | 293 | 274 | | After moving | 207+13 = 220 | 176+154 = 330 | 293-13 = 280 | 274-154 = 120 |
which cleans up a bit to this
| Crate A | Crate B | | # of pears | # of mangoes | # of pears | # of mangoes | | Before moving | 207 | 176 | 293 | 274 | | After moving | 220 | 330 | 280 | 120 |
The table format is optional, but I find it helps to keep all the numbers organized.
After moving 13 pears and 154 mangoes from B to A, there are:
220 pears out of 220+330 = 550 fruit in crate A.
Then note how 220/550 = 0.40 = 40% to show that 40% of the fruit in crate A are pears.
So far so good.
After moving 13 pears and 154 mangoes from B to A, there are:
280 pears left out of 280+120 = 400 fruit left in crate B.
280/400 = 0.70 = 70%
Showing that 70% of the fruit left in crate B are pears.
The answers are fully confirmed.
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Answers:
He moved 13 pears and 154 mangoes from crate B to crate A.
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