 # Question

There was 200 more apples than pears at a fruit stall. After 1/4 of the apples and 2/7 of the pears were sold, there were 170 more apples than pears left.
(a)How many apples were there at the fruit stall at first?
(b) How many pears were there at the fruit stall in the end?

Source: ?

1/4 = 7/28
2/7 = 8/28

Left for A -> 28-7=21
Left for P -> 28-8=20

20u -> 170
1u    -> 170÷20=8.5

(a) 28u -> 8.5×28=238

(b) 20u -> 8.5×20=170

I think it’s supposed to be solved like this but I don’t know.

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(A- A/4) – (P – 2P/7) = 170 <=> 3A/4 – 5P/7 = 170

tiny fishing

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At first:

Apples [—u—-]200

Pears—[— u—-]

In the end:

Apples [—–3/4 (u+200)]

Pears—[5/7 u ] <-170->

1/4 (u+200) – 2/7u = 200-170 =30     or    3/4(u+200) – 5/7u = 170  works out to u=560

7u + 1400 – 8u = 30 x 28 = 840

u = 1400-840 = 560

(a) Number of apples at first = u + 200 = 560+200 = 760

(b) Number of pears in the end = 5/7 x 560 = 5 x 80 = 400

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This question is testing you on forming algebraic equations and solving it.

Let 4 a be the number of apples initially. Let 7 p be the number of pears initially.

(I chose to use 4 a and 7 p because of the fractions involved in the question)

Initially,

4 a = 7 p + 200 — (1)

After the fruit were sold,

Remaining apples left = 4 a x 3/4 = 3 a

Remaining pears left = 7 a x 5/7 = 5 p

Hence, 3 a = 5 p + 170 — (2)

(Now, with the 2 equations, you can solve for a and p.)

(Since 12 is the LCM of 3 and 4, I choose to multiply equation 1 by 3 and equation 2 by 4.)

(1) x 3:

12 a = 21 p + 600 — (3)

(2) x 4:

12 a = 20 p + 680 — (4)

(Using the substitution method, I can substitute equation 3 into equation 4.)

Sub (3) into (4):

21 p + 600 = 20 p + 680

p = 80

Sub p = 80 into (1)

(a) 4 a = 7 (80) + 200

= 760//

(b) 5 p = 5  (80)

= 400//

Apple = A

Pear = P

Before:

A – P = 200

After:

(A- A/4) – (P – 2P/7) = 170 <=> 3A/4 – 5P/7 = 170

=>A= 650, P = 560

=> a) 650 Apples, b) 400 Pears