# 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: ?

If there were 28u pears in the beginning, there were ( 200 + 28u ) apples in the beginning.

There were ( 1 – 1/4) = 3/4 apples left in the end and there were ( 1 – 2/7 ) = 5/7 pears left in the end.

Given

3( 200 + 28u )/4 – 5(28u)/7 = 130

150 + 21u – 20u = 130

u = 20

200 + 28u = 760

There were 760 apples in the beginning

20u = 400

There were 400 pears in the end.

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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//

Thanks.  I’m still not very clear if simultaneous equations are taught in Primary school and if we are actually allowed to use them instead of using some “models”.  Can you clarify?

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The original number of apples: x

The original number of pears: y

x-y=200 (1)

3/4 x-5/7 y=170 (2)

(1)*20: 20x-20y=4000 (3)

(2)*4*7: 21x-20y=4760 (4)

(4)-(3): (21x-20y)-(20x-20y)=4760-4000

So x=760, then according(1), y=560

(a)How many apples were there at the fruit stall at first?

x=760

(b) How many pears were there at the fruit stall in the end?

(1-2/7)y=5/7 y=5/7*560=400

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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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```You were right in recognizing that 1/4 = 7/28 and 2/7 = 8/28.
Identifying the 28 was quite important. Good observation.

This is how I would do it:

Step 1: Finding a "friendly" unit
As you have noted, there is division by 4 and 7. So 28 is probably useful.

Step 2: Modeling the initial condition
Pears  = 28u
Apple  = 28u + 200   (there were 200 more apples as given)

Step 3: Modeling the final condition
Pears  = 20u        (2/7 sold means 8u sold; so 20u left)
Apple  = 21u + 150  (1/4 sold means 7u + 50 sold; so 21u + 150 left)
Difference = (21u + 150) - 20u = u + 150
That means 170 = u + 150 and so u = 20.

a) There were 28 x 20 + 200 = 760 apples at first.
b) There were 20 x 20 = 400 pears in the end.

This answer is similar to the earlier ones. But it uses less fractions
and only 1 unknown. I find that helps to avoid careless mistakes.```
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You got it wrong.  The original answer provided by hwachie is correct.

Try again.

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