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Basket A contained 70 apples and 100 pears. Basket B contained 90 apples and 20 pears. After some apples and pears were transferred from Basket A to Basket B, 40 % of Basket A contained apples and 30% of Basket B contained pears. How many apples and pears were transferred from Basket A to Basket B altogether?

 Aug 3, 2023

Best Answer 

 #1
avatar+126971 
+1

Let the number of appes transferred for A to B =  A

Let the number of pears transferred from A to B = P

 

Now  A contains (70 - A) apples and (100 - P) pears

So

Number of apples now in A / ( total fruits in A)  =  .40.....in equation form we have

 

(70 - A) / ( 170 - A - P)  = .40    simplify

(70 - A) =  .40 (170 - A - P)

70 -A = 68 - .40A -.40P

70 - 68 = A - .40A - .40P

2 = .60A - .40P     (1)

 

And in B....number of pears in B / (total fruits in B)  =  .30.....in equation form...

(20 + P)  / (110 + A + P)  = .30   simplify

20 + P =  .30 ( 110 + A + P)

20+ P  = 33 + .30A + .30P

33-20 = P -.30P - .30A

13 = .70P - .30A    (2)

 

Combining (1) and (2)  in a system  we have

 

.60A - .40P  =  2

.70P -.30A = 13       simplify

 

.6A - .4P =  2

-.3A + .7P =  13         multiply this equation through by 2

 

.6A - .4P =  2

-.6A + 1.4P  = 26       add these

 

P =  28 =  the number of pears transferred

 

Using either equation

 

.6A - .4(28)  = 2

.6A - 11.2 = 2

.6A = 2 + 11.2

,6A = 13.2

A = 13.2 / .6 =  22  =  the number of apples transferred

 

Proof    after   22 appples and 28 pears are transferred from A to B

 

( 70 - 22)  / ( 170 - 28 - 22)

48 / 120 =  40%  of the fruit in A  are apples

 

And in B

(20 + 28)  /( 110 + 22 + 28)

48 / 160  = 30% are pears

 

cool cool cool

 Aug 3, 2023
 #1
avatar+126971 
+1
Best Answer

Let the number of appes transferred for A to B =  A

Let the number of pears transferred from A to B = P

 

Now  A contains (70 - A) apples and (100 - P) pears

So

Number of apples now in A / ( total fruits in A)  =  .40.....in equation form we have

 

(70 - A) / ( 170 - A - P)  = .40    simplify

(70 - A) =  .40 (170 - A - P)

70 -A = 68 - .40A -.40P

70 - 68 = A - .40A - .40P

2 = .60A - .40P     (1)

 

And in B....number of pears in B / (total fruits in B)  =  .30.....in equation form...

(20 + P)  / (110 + A + P)  = .30   simplify

20 + P =  .30 ( 110 + A + P)

20+ P  = 33 + .30A + .30P

33-20 = P -.30P - .30A

13 = .70P - .30A    (2)

 

Combining (1) and (2)  in a system  we have

 

.60A - .40P  =  2

.70P -.30A = 13       simplify

 

.6A - .4P =  2

-.3A + .7P =  13         multiply this equation through by 2

 

.6A - .4P =  2

-.6A + 1.4P  = 26       add these

 

P =  28 =  the number of pears transferred

 

Using either equation

 

.6A - .4(28)  = 2

.6A - 11.2 = 2

.6A = 2 + 11.2

,6A = 13.2

A = 13.2 / .6 =  22  =  the number of apples transferred

 

Proof    after   22 appples and 28 pears are transferred from A to B

 

( 70 - 22)  / ( 170 - 28 - 22)

48 / 120 =  40%  of the fruit in A  are apples

 

And in B

(20 + 28)  /( 110 + 22 + 28)

48 / 160  = 30% are pears

 

cool cool cool

CPhill Aug 3, 2023

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