Hey Angie,

Following pls find my worked solution :

Total (X + Y) = 1300

X = 800

A(X) = 30% x 800 = 240

P(X) = 800 - 240 = 560

Y = 500

P(Y) = 40% x 500 = 200

A(Y) = 500 - 200 = 300

** - for alignment purpose

<Before>

*** X ************* Y

A **** P ****** A *** P

240 : 560 *** 300 : 200

<Process>

some fruits in both baskets were transferred from one to another

<After>

** X ********* Y

A ** P ***** A ** P

25 : 75 *** 75 : 25

* 1 : 3 ******* 3 : 1

Using <Unchanged Total> concept, total number of A in X,Y (After> = total number of A in X,Y <Before>. Likewise for number of P in X,Y.

Using Simultaneous method, let the <After> ratio in X be part (p) qtys, and the <After> ratio in Y be unit (u) qtys, and we have ...

1p + 3u --> 240 + 300 = 540 -- (1)

3p + 1u --> 560 + 200 = 760 -- (2)

To find number of fruits in basket Y after transfer, we need to find the unit (u) qty, while eliminating the part (p) qty.

Eliminating (p), we have ...

(1)x3: 3p + 9u --> 1620 -- (3)

Comparing (3) and (2):

(3)-(2): 9u - 1u --> 1620 - 760

8u --> 860

1u --> (860/8)

Total number of fruits in Y (after transfer) = 4u

4u --> 4 x (860/8) = 430

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Do let me know again if the answer does not match your Answerkey or if there's further clarification.

Btw which sch prelimpaper is this??

Cheers,

Edward