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?

Guest Aug 3, 2023

#1**+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

CPhill Aug 3, 2023

#1**+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

CPhill Aug 3, 2023