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 P6 Maths Problem
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By: teapot (offline)  Saturday, February 27 2016 @ 09:49 PM CST (Read 841 times)  
teapot

Hi All

Please help. My daughter has 2 maths problem cannot solve. Thank you.

1. Sam and Peter had a total of 900 cards. Sam gave 1/5 of his cards to Peter. Peter then gave 1/4 of his cards to Sam. In the end, each of them had the same number of cards. How many cards did Sam at first?

2. Joys, Molly and Chris shared some sweets. The ratio of the total number of sweets received by Joys and Molly to the number of sweets received by Chris was 2:5. When Chris gave 25 sweets to Joys and 31 sweets to Molly, and Joys gave 12 sweets to Molly, each of them had the same number of sweets. Find the total number of sweets Joys had at first.

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By: echeewh (offline)  Sunday, February 28 2016 @ 11:44 PM CST  
echeewh

Hey <teapot>,

Appreciate you can help provide the <anskey> next time whenever possible. This will help to ease verification. Thank you.

My worked solutions are as follows:

Q1

*** - for alignment purpose

As this is an <Internal Transfer> process, the total number of cards that Sam (S) and Peter (P) had in the end is the same as before (apply <Unchanged Total> concept).

Given each of them had the same number of cards in the end, we have ...

900 ÷ 2 = 450

S ********** P
450 ******* 450

Apply <working backwards> method, we have ...

<After P gave (1/4) of his cards to S>

(3/4)P = 450
(4/4)P = 450 × 4 ÷ 3 = 600

S ********** P
300 ******* 600

<After S gave (1/5) of his cards to P>

(4/5)S = 300
(5/5)S = 300 × 5 ÷ 4 = 375

S ********** P
375 ******* 525

S (at first) = 375 cards

========

Q2

*** - for alignment purpose

<before>

(J + M) : C
** 2 *** : 5

<process>
C gave (25 + 31) sweets to J + M.
J gave 12 to M. {internal transfer between J,M so total remains unchanged}

(J + M): +56
C: -56

<after>

(J + M) : C
** 2 *** : 1

Apply <Changed Qtys> concept and we can use <Cross Multiply> method, we have ...

[(2u + 56) / 2] = [(5u - 56) / 1]
2u + 56 = 2 × (5u - 56)
2u + 56 = 10u - 112
10u - 2u --> 56 + 112
8u --> 168
1u --> 168 ÷ 8 = 21

C (after) = 5u - 56 --> (5 × 21) - 56 = 49

Hence, J (after) = 49 { given each of them had same number of sweets in the end }

<Working backwards>, we have ...

J (before) = 49 + 12 - 25 = 36



<< Alternative Method >>

<before>

(J + M) : C Total
** 2 *** : 5 *** 7 ----- (1)

<process>
C gave (25 + 31) sweets to J + M.
J gave 12 to M. {internal transfer between J,M so total remains unchanged}

(J + M): +56
C: -56

<after>

(J + M) : C Total
** 2 *** : 1 *** 3 ----- (2)

Given that this is an <Internal Transfer> process, we can use <Unchanged Total> concept. hence, we have ...

Lowest common multiple of 7,3 is 21.

<before>

(1)x3:
(J + M) : C Total
** 6 *** : 15 ** 21 ----- (3)

<after>

(2)x7:
(J + M) : C Total
** 14 ** : 7 *** 21 ----- (4)

Given C gave away 56 sweets, we have ...

8u --> 56
1u --> 56 ÷ 8 = 7

C (after) = 7u --> 7 × 7 = 49

Hence, J (after) = 49 { given each of them had same number of sweets in the end }

<Working backwards>, we have ...

J (before) = 49 + 12 - 25 = 36

======

Trust these help. Do let me know again if these are different from your <anskey> or if there's further clarification.

Cheers,
Edward

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