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 P6 Maths
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By: eqkt (offline)  Sunday, August 04 2013 @ 10:40 PM CDT (Read 1181 times)  
eqkt

Hi! Pls help me with these question. Thank you!

1. A fruit seller had some oranges, pears and mangoes. After selling some of them, there were 6 pears for every 5 oranges left and 7 oranges for every 3 mangoes left. He sold 168 oranges. The number of pears sold was the same as the number of mangoes sold. He had a total of 456 pears and mangoes left.
a) How many oranges had he at first?
b) Given that the number of pear was twice the number of mangoes at first, what was the total number of pears and mangoes sold?

2) Car A and Car B started travelling at 0945 from the opposite ends of a straight road. Car A travelled at an average speed of 40 km/h. Car B travelled 10 km/h faster and took 1.5 hours less than Car A to travel the entire road. Find the time when both cars met?

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By: echeewh (offline)  Monday, August 05 2013 @ 01:43 AM CDT  
echeewh

Hey <eqkt>,

Appreciate that you can provide the <Answerkey> for verification , whenever possible.


Q1.

*** - for alignment purpose

<After>

P : Or : M
6 : 5 ****** -- (1)
** 7 : 3 ** -- (2)

Using <Repeated Id> method { as Or is compared to P and M separately }, multiply (1) 7x, (2) 5x

P : Or : M
42 : 35
**** 35 : 15

42 : 35 : 15

P + M (left) = 456
42u + 15u = 456
57u --> 456
1u --> 456 ÷ 57 = 8

Or (left) = 35u
35u --> 35 × 8 = 280

(a)
Given that 168 oranges (Or) sold,
Or (at first) = 280 + 168 = 448

(b)
Given that number of pears was twice the number of mangoes at first,

<before>
P : M ** Diff
2 : 1 **** 1 **** -- (1)

<after>
42 : 15
14 : 5 *** 9 *** -- (2)

Since number of pears (P) sold was the same as the number of mangoes (M) sold, we use the <Unchanged Diff> concept { i.e. difference between P,M <before> and <after> remains the same }.

Multiply (1) 9x.

<before>
P : M ** Diff
18 : 9 *** 9 *** -- (3)

Note that 4u of P and 4u of M were sold.

P + M (left) = 456
14u + 5u = 456
19u --> 456
1u --> 456 ÷ 19 = 24

P + M (sold) = 8u
8u --> 8 × 24 = 192

==========

Q2.

Using <Common Distance> method, the time ratio is the reverse of the speed ratio { i.e. as one travels at a faster speed, it will take a shorter time (duration) to reach the same distance. }

s(A) : s(B )
40 : 50
4 : 5

t(A) : t(B )
5 : 4

Given that B took 1.5 hours less than A to travel the entire road,

1u --> 1.5

Hence, B took 4u to travel the entire distance.
4u --> 4 × 1.5 = 6 hrs

Distance covered = 50 × 6 = 300 km

Time taken (to meet up)
= 300 ÷ (50 + 40)
= 300 ÷ 90
= 3 1/3 hrs = 3 hrs 20 mins

Time (to meet up)
= 09 45 + 3 20
= 13 05

=========

Trust this helps.

Do let me know again if this is different from your <Answerkey> or if there's further clarification.

Cheers,
Edward

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By: eqkt (offline)  Monday, August 05 2013 @ 10:17 AM CDT  
eqkt

Hi Edward,
The answer provided is the same as yours. Thanks!
Cheers,
KJ

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By: eqkt (offline)  Monday, August 05 2013 @ 10:54 AM CDT  
eqkt

Hi Edward,
For Q1 part 2, I did not quite catch on the 2:11 part. Could you explain it further? Thanks!
Cheers,
KJ

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By: echeewh (offline)  Monday, August 05 2013 @ 12:03 PM CDT  
echeewh

Hey KJ,

About your clarification, I think the original text in my post was pretty close so much so that you took it as 2 : 11. There 's supposed to be some spaces separating the two ones. The ratio is 2 : 1 , while its difference is 1 unit. This difference is indicated beside the ratio to show how the <Unchanged Diff> concept is used.

Hope this clarifies.

Thank you,
Edward.

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By: eqkt (offline)  Tuesday, August 06 2013 @ 02:19 AM CDT  
eqkt

Thank you very much! Big Grin

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