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?
+128707 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
+128707 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
