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 Forum Index >  Test Paper Related >  Primary 6 Matters
 Primary 6 Math
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By: lee tong fong (offline)  Friday, December 07 2012 @ 02:23 AM CST (Read 1125 times)  
lee tong fong

Can you help me solve the following using model:

1) Joseph has some local and foreign stamps in 2 boxes. In Box A, the numbers of local and foreign stamps are in the ratio 3: 4. In Box B, the number of local stamps is twice the number of foreign stamps. Joseph transfers half of the foreign stamps from Box A to Box B. The numbers of stamps in Box A become 105 and the ratio of the number of local stamps to the number of foreign stamps in Box B becomes 6:5.
a) How many foreign stamps have been transferred from Box A to Box B?
b) What is the number of stamps in Box B at first?

2) Ravi had a total of 80 pieces of $10 notes and $50 notes. He used ½ of his $10 notes and withdrew another four pieces of $50 notes from the bank. After which, the number of $50 notes he had was 2/5 the number of $10 notes. Find the total value of the 80 pieces of notes he had at first.

3) 2/9 of Rani’s balloons were red and the rest were green. She gave away 15 green balloons and bought another 25 red balloons. She then had the same number of red and green balloons. How may green balloons did she have at first?

Thanks in advance.

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By: jo sarah (offline)  Friday, December 07 2012 @ 07:25 AM CST  
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Hi, if you insist on using models only, i am not sure if you might find the following useful.
But i shall share my solutions, hoping they be useful to any others reading this forum.
[Allow me also to comment that solutions that are mathematically correct should be allowed... i have heard kids saying their teachers "insist" on solving their problems certain ways only... my feeling is that these teachers are stifling the kids so affecting their interests in mathematics.]

1) Joseph has some local and foreign stamps in 2 boxes. In Box A, the numbers of local and foreign stamps are in the ratio 3: 4. In Box B, the number of local stamps is twice the number of foreign stamps. Joseph transfers half of the foreign stamps from Box A to Box B. The numbers of stamps in Box A become 105 and the ratio of the number of local stamps to the number of foreign stamps in Box B becomes 6:5.
a) How many foreign stamps have been transferred from Box A to Box B?
b) What is the number of stamps in Box B at first?

Solution:
..............................box A........................box B
........................local.....foreign.......local.....foreign
at 1st:................3u..........4u.............2p.........1p.........................[where 'u' represents 'unit' and 'p' represents 'part']
transfer:..........................-2u.........................+2u
in the end:........3u..........2u.............2p......1p+2u

total in box A = 5u = 105 stamps
therefore, .........1u = 105 / 5 = 21 stamps

(a) no. of foreign stamps transferred from A to B = 2u = 42. (ANS)

Now, in the end, in box B, the ratio of local : foreign = 6 : 5
so, the 2p = 6 pn.......................................['pn' represents 'portion']
thus, ....1p = 3 pn
so that, 2u = 2 pn [because 1p + 2u = 5 pn],
and so, 1 pn = 1u
At first, there are 3p stamps in box B, which is equal to 9u.

(b) therefore, there are 9 x 21 = 189 stamps in box B at first. (ANS)


2) Ravi had a total of 80 pieces of $10 notes and $50 notes. He used ½ of his $10 notes and withdrew another four pieces of $50 notes from the bank. After which, the number of $50 notes he had was 2/5 the number of $10 notes. Find the total value of the 80 pieces of notes he had at first.

Solution:
.......no. of...........$10................$50
at 1st:...................2u..................1p............total = 80, means that 2u + 1p = 80, or 1p = 80 - 2u........(1)
change:...............-1u.................+4
in end:.................1u................1p+4
ratio is:..................5........ : ........2
which means that 2u = 5p + 20
using (1), ............2u = 5(80 - 2u) + 20
therefore,.........12u = 420
that is, ..............1u = 35, and from (1), 1p = 80 -70 = 10

therefore, value of 80 pieces at first = 70 x $10 + 10 x $50 = $1200. (ANS)


3) 2/9 of Rani’s balloons were red and the rest were green. She gave away 15 green balloons and bought another 25 red balloons. She then had the same number of red and green balloons. How may green balloons did she have at first?

Solution:
.................................red................green
at first:.......................2u..................7u
change:...................+25.................-15
in end:...................2u + 25..........7u - 15

she had same no. of each in the end, so, 2u + 25 = 7u - 15
that is, .........5u = 40 balloons
so that, ........1u = 8 balloons

therefore, she had 7 x 8 = 56 green balloons at first. (ANS)

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By: echeewh (offline)  Friday, December 07 2012 @ 10:27 PM CST  
echeewh

Hey Fong Fong,

Following pls find alternative methods (with models ):

*** - for alignment purpose

Q1

Using ratio model (one can translate from ratio to a standard comparison model) , we have ...

<Before>
Box A ****** Box B
LA : FA **** LB : FB
3 : 4 ******** 2 : 1

<key statement>
Joseph transfers half of the foreign stamps from Box A to Box B.
This implies ...
(i) qty / number of local stamps in A, B remains unchanged before and after
(ii) total qty / number of foreign stamps in A and B remains unchanged before and after

Given that half the foreign stamps were taken out of A, the number of units of foreign stamps left in A is (4 - 2) units as shown below:

<After>
Box A ***** Box B
LA : FA *** LB : FB
3 : 2 ******* 6 : 5

Applying <Unchanged Qty> concept from (i) and comparing LB <before> and <after>, multiply Box B ratio <before> 3x. Hence the <before> ratios become ...

<Before>
Box A ***** Box B
LA : FA *** LB : FB
3 : 4 ******* 6 : 3

These models also agree with (ii).
FA + FB <before> = FA + FB <after>
4u + 3u = 2u + 5u

Given that (LA + FA) <after> = 105, we have ...

3u + 2u --> 105
5u --> 105
1u --> 21

(a)
Number of foreign stamps transferred from A:
FA <before> - FA <after> = 4u - 2u = 2u
2u --> 2 × 21 = 42

(b)
total number of stamps in B (at first) = 6u + 3u = 9u
9u --> 9 × 21 = 189


======================

Q2

<aft>
$50 |--|--|
$10 |--|--|--|--|--|

Apply <Work Backwards> method,
$50: -4
$10: x2

<bef>
$50 |-<4>
$10 |--|--|--|--|--|--|--|--|--|--|

Total Number/Qty <bef> = 80 pieces

12u - 4 --> 80
12u --> 80 + 4 = 84
1u --> 7

$50 (Qty): 2u - 4 --> (2 × 7) - 4 = 14 - 4 = 10
Val: 10 × 50 = $500

$10 (qty): 10u --> 10 × 7 = 70
Val: 70 × 10 = $700

Total val = $500 + $700 = $1200

================

Q3

<bef>
R : G
2 : 7

<aft>
R: |------------|
G: |------------|

Apply <work backwards> method,
R: -25
G: +15

<bef>
** <2u>
R: |----|<-25->
G: |------------<-15->|
** <-------- 7u ------->

Using <Gap & Difference> method, we have ...

7u - 2u --> 25 + 15
5u --> 40
1u --> 8

G (at first) = 7u --> 7 × 8 = 56

==============

Trust these help.

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

Cheers,
Edward

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