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

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?

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

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Why? 1u    -> 170÷20=8.5

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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.

You got it wrong.  The original answer provided by hwachie is correct.

Try again.