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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?

 

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Answer

Why? 1u    -> 170÷20=8.5

slope game

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

Try again.

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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//

 

 

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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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