Q&A - PSLE Math

[wanderlust-alm]

CloudeeDaz:

Hi

need help withthis question

Mdm Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal no of beads in each small box and equal no of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box?

thanks

Each big box has 5 more beads ==> extra beads over the small box ==> 75 beads.

14 small box + 15 small box + 75 beads = total amount of beads ( Y beads )

each small box contain ( 3/8 y ) / 14 beads ==> 3/112 y
Beads in big box = 3/8 y + 3/112 y + 75 beads

hence total number of beads = 3/8 y + ( 3/8 y + 3/112 y + 75 )

Work out to common denominator

42/112 + 42/112 + 3/112 + 75 beads = total number of beads

87/112 + 75 beads = total number of beads

==> 25 /112 = 75 beads
hence 3 / 8 = 126 beads

hence each small box has 126/14 = 9 beads

[wanderlust-alm]

wanderlust-alm:

CloudeeDaz:

Hi

need help withthis question

Mdm Zhang packed some beads into 14 small boxes and 15 big boxes. There were equal no of beads in each small box and equal no of beads in each big box. Each big box contained 5 more beads than each small box. 3/8 of the beads were packed in small boxes. How many beads were there in each small box?

thanks

Each big box has 5 more beads ==> extra beads over the small box ==> 75 beads.

14 small box + 15 small box + 75 beads = total amount of beads ( Y beads )

each small box contain ( 3/8 y ) / 14 beads ==> 3/112 y
Beads in big box = 3/8 y + 3/112 y + 75 beads

hence total number of beads = 3/8 y + ( 3/8 y + 3/112 y + 75 )

Work out to common denominator

42/112 + 42/112 + 3/112 + 75 beads = total number of beads

87/112 + 75 beads = total number of beads

==> 25 /112 = 75 beads
hence 3 / 8 = 126 beads

hence each small box has 126/14 = 9 beads

Thanks for link too. The illustration by units is much clearer . My method is more like the old style although eventually answer still the same but definately easier to view through units . Thanks Too !!

[pixiedust]

Need some help, thanks in advance :


A box contained 50-cent and 20-cent coins in the ratio 2:3.
Some 50-cent coins were taken out and exchanged for 20-cent coins.
The ratio of the number of 50-cent coins to the number of 20-cent coins became 2:7.
Find (a) the total value of money in the box (b) the number of 50-cent coins which were taken to exchange for 20-cent coins.

[acehkr3009]

Let the number of coins taken out (50 cents) and exchange (20 cents) be U numbers.


Before exchange,

50 cents : 20 cents
2 : 3
After the exchange of U numbers of 50 cents with 20 cents,

50 cents = 2 - U
20 cents = 3 + U

New ratio becomes,
2 : 7

So 7 times number of 50 cents = 2 times number of 20 cents

2 X (3 + U) = 7 X (2 - U)
6 + 2U = 14 - 7U
9U = 8
U = 8/9 (fraction)

Now we know each of ratio unit of 50 cents & 20 cents is subdivided into 9 smaller units. Of which, 8 units (or 8 numbers of coins) of 50 cents coins is used to exchange for 8 units of 20 cents coins.

Before the exchange,
Based on the ratio (50 cents: 20 cents) 2:3,
There will be 2 x 9 = 18 number of 50 cents coins
Value of 50 cents coins = 18 x 50 cents = $9
There will be 3 x 9 = 27 number of 20 cents coins
Value of 20 cents coins = 27 x 20 cents = $5.40
Total value at first will be $14.40

After exchanging 8 number of 50 cents with 20 cents,
There will be 18 - 8 = 10 number of 50 cents &
10 number of 50 cents coins: 10 x 50 = $5
There will be 27 + 8 = 35 number of 20 cents.
35 number of 20 cents coins: 35 x 20 = $7
Total value after the exchange = $12.

Hope the above lenghtly solution is correct & acceptable.
Pls advise if there is simpler solutions.
Thanks in advance.

[tianzhu]

pixiedust:

Need some help, thanks in advance :


A box contained 50-cent and 20-cent coins in the ratio 2:3.
Some 50-cent coins were taken out and exchanged for 20-cent coins.
The ratio of the number of 50-cent coins to the number of 20-cent coins became 2:7.
Find (a) the total value of money in the box (b) the number of 50-cent coins which were taken to exchange for 20-cent coins.

Hi

You may use a heuristics called \"Systematic Listing\".

We assume that some 50 cents were taken out and exchanged by a number of 20 cents of the same value so that the total sum of money in the box stays the same.

Due to time constraints, I shall not present the answer in slide but give some pointers for you to work it out.

1)Make a table of equivalent ratios of the number of 50 cents coins and 20 cents coins before and after the coins are replaced.

2)Calculate the total value of coins in the box before and after the replacement.

Before: (2*50)+(3*20) ----- 160
After: (2*50)+(7*20) ----- 240

The total value of coins remains the same before and after the replacement of coins.

From the table, the total sum of money ------ 480 cents.

Before:(6*50)+(9*20) ----- 480
After:(4*50)+(14*20) ----- 480

The number of 50 cents being exchanged ------ 2

Best wishes

PS ----- In this question, we are calculating the least possible amount of money in the box.Usually, for such question, the range of money in the box is usually specified, otherwise, there'll be multiple answers.

[Tang]

tianzhu:

pixiedust:

Need some help, thanks in advance :


A box contained 50-cent and 20-cent coins in the ratio 2:3.
Some 50-cent coins were taken out and exchanged for 20-cent coins.
The ratio of the number of 50-cent coins to the number of 20-cent coins became 2:7.
Find (a) the total value of money in the box (b) the number of 50-cent coins which were taken to exchange for 20-cent coins.

Hi

You may use a heuristics called \"Systematic Listing\".

We assume that some 50 cents were taken out and exchanged by a number of 20 cents of the same value so that the total sum of money in the box stays the same.

Due to time constraints, I shall not present the answer in slide but give some pointers for you to work it out.

1)Make a table of equivalent ratios of the number of 50 cents coins and 20 cents coins before and after the coins are replaced.

2)Calculate the total value of coins in the box before and after the replacement.

Before: (2*50)+(3*20) ----- 160
After: (2*50)+(7*20) ----- 240

The total value of coins remains the same before and after the replacement of coins.

From the table, the total sum of money ------ 480 cents.

Before:(6*50)+(9*20) ----- 480
After:(4*50)+(14*20) ----- 480

The number of 50 cents being exchanged ------ 2

Best wishes

PS ----- In this question, we are calculating the least possible amount of money in the box.Usually, for such question, the range of money in the box is usually specified, otherwise, there'll be multiple answers.

Hi,

There are multiple answers.

The number of 50-cent coins which was exchanged for 20-cent coins is a multiple of 2, i.e. it can be 2, 4, 6, 8, etc.

[Tang]

acehkr3009:

Let the number of coins taken out (50 cents) and exchange (20 cents) be U numbers.


Before exchange,

50 cents : 20 cents
2 : 3
After the exchange of U numbers of 50 cents with 20 cents,

50 cents = 2 - U
20 cents = 3 + U

New ratio becomes,
2 : 7

So 7 times number of 50 cents = 2 times number of 20 cents

2 X (3 + U) = 7 X (2 - U)
6 + 2U = 14 - 7U
9U = 8
U = 8/9 (fraction)

Now we know each of ratio unit of 50 cents & 20 cents is subdivided into 9 smaller units. Of which, 8 units (or 8 numbers of coins) of 50 cents coins is used to exchange for 8 units of 20 cents coins.

Before the exchange,
Based on the ratio (50 cents: 20 cents) 2:3,
There will be 2 x 9 = 18 number of 50 cents coins
Value of 50 cents coins = 18 x 50 cents = $9
There will be 3 x 9 = 27 number of 20 cents coins
Value of 20 cents coins = 27 x 20 cents = $5.40
Total value at first will be $14.40

After exchanging 8 number of 50 cents with 20 cents,
There will be 18 - 8 = 10 number of 50 cents &
10 number of 50 cents coins: 10 x 50 = $5
There will be 27 + 8 = 35 number of 20 cents.
35 number of 20 cents coins: 35 x 20 = $7
Total value after the exchange = $12.

Hope the above lenghtly solution is correct & acceptable.
Pls advise if there is simpler solutions.
Thanks in advance.

Hi,

The total value before and after should be the same.
Two 50-cent coins would be exchanged for five 20-cent coins (and not 1 to 1), etc.
There are multiple answers.

[Daddy]

MathIzzzFun - Thanks very much

[pixiedust]

Thank you acehkr3009, tang, tianzhu.

The question is from school’s ws under ‘listing’ heuristics. I am still digesting your explanations but I think I see some light. Thanks again.

[cimman]

pixiedust:

Thank you acehkr3009, tang, tianzhu.

The question is from school's ws under 'listing' heuristics. I am still digesting your explanations but I think I see some light. Thanks again.

this category of problem is classified as \"Internal Transfer with Unchanged Total Concept\". Basically, the total number of 50 cents and 20 cents coins remains the same, before and after an internal transfer (ie. between the 2 coins, no coins were taken out of the system or added to the system).


Before Transfer
50 cents::2 units x 9->18 units
20 cents::3 units x 9->27 units
Total::::::5 units------>45 units

After Transfer
50 cents::2 units\tx 5->10 units
20 cents::7 units\tx 5->35 units
Total::::::9 units------>45 units

Changes in 50 cent coin: 18 units to 10 units = - 8 units, thus 8 50cent coins were transferred.
Total value of money in the box -> 10units * 50 cents + 35units * 20 cents = $5 + $7 = $12

As to why we need to multiply the first table by 9 and the second table by 5, is that we need to ensure that the total number of units in both table are the same. I would call this step normalizing the totals so that both totals (before and after) are the same, and that is the crux for this type of problems . \"Normalizing\" is very similar to having the same denominators when adding 2 fractions together, ie. 1/5 + 1/9 = (9 + 5)/45

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This concept and others are very well covered in this assessment book: Challenging Maths Problems Made Easy by Ammiel Wan Chee Hong. Publisher: Marshall Cavendish. Can be found in Popular Bookstore. Do have a look at the book, it is very useful.

The book teaches using the Unit Transfer Method to solve the above problem. It is very effective for certain types of problems, especially those where either the total number of units remains the same before and after the transfer or one of the units remain the same. The book is really good in exposing students to the more difficult problems. A number of problem categories not covered in OnSponge are covered here. I like the Gap and Difference topic. Very useful for those kinds of problems.