Q&A - P5 Math

[Jamesbond]

A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

Pl help.

[MathIzzzFun]

Jamesbond:

A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

Pl help.

possible approaches - cross-multiply/UP, units, MD.

Using units:

10 x $0.50 = 25 x $0.20
--> 10 fifty-cent coins were replaced by 25 twenty-cent coins

At first
number of 50c --> 20 units + 10
number of 20c (= 3/4 number of 50c) --> 15 units + 7.5

In the end,
number of 50c --> 20 units
number of 20c (=7/5 number of 50c) --> 28 units

since 25 twenty-cent coins were added, we have:

15 units + 7.5 + 25 = 28 units
1 unit --> 2.5

Sum of money in the box at first
= sum of money in the box in the end
= 20 x 2.5 x $0.50 + 28 x 2.5 x $0.20
= $39

cheers.

[snowball]

4/7 of Ming’s mass is equal to 2/5 of Rob’s mass.

If Ming puts on 6 kg and Rob loses 12kg, the two men will have the same mass.
a) Draw a clearly labelled model to solve the problem.
b) What is Rob’s mass?

[snowball]

In January, Mrs Wong spent 2/5 of her salary on rent and $120 on parking fees.

She spent 1/4 of her remaining salary on food and $480 on shopping.
She saved the remaining $1896.
How much was Mrs Wong’s salary for January ?

[MathsOlympiadtrainer]

snowball:

In January, Mrs Wong spent 2/5 of her salary on rent and $120 on parking fees.

She spent 1/4 of her remaining salary on food and $480 on shopping.
She saved the remaining $1896.
How much was Mrs Wong's salary for January ?

This is a working backwards qn.
3/4 of her remaining salary-> 480+1896= 2376
4/4 of her remaining salary-> 2376/3 * 4= 3168

3/5 of her salary-> 3168+120 = 3288
5/5 of her salary-> 3288/3 * 5 = 5480

Hope that helps.

[veraclari]

MathIzzzFun:

Jamesbond:

A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

Pl help.

possible approaches - cross-multiply/UP, lunits, MD.

Using units:

10 x $0.50 = 25 x $0.20
--> 10 fifty-cent coins were replaced by 25 twenty-cent coins

At first
number of 50c --> 20 units + 10
number of 20c (= 3/4 number of 50c) --> 15 units + 7.5

In the end,
number of 50c --> 20 units
number of 20c (=7/5 number of 50c) --> 28 units

since 25 twenty-cent coins were added, we have:

15 units + 7.5 + 25 = 28 units
1 unit --> 2.5

Sum of money in the box at first
= sum of money in the box in the end
= 20 x 2.5 x $0.50 + 28 x 2.5 x $0.20
= $39

cheers.

Hi Mathizzfun, can I ask.. how to get the 20units in the first place?
And why cross multiply? TIA!!!

[MathIzzzFun]

veraclari:

MathIzzzFun:

[quote=\"Jamesbond\"]A box contained a number of 20 cent and 50 cent coins in the ratio 3:4. When ten 50 cent coins were taken out and replaced by 20 cent coins, the ratio became 7:5. Find the sum of money in the box at first?

Pl help.

possible approaches - cross-multiply/UP, lunits, MD.

Using units:

10 x $0.50 = 25 x $0.20
--> 10 fifty-cent coins were replaced by 25 twenty-cent coins

At first
number of 50c --> 20 units + 10
number of 20c (= 3/4 number of 50c) --> 15 units + 7.5

In the end,
number of 50c --> 20 units
number of 20c (=7/5 number of 50c) --> 28 units

since 25 twenty-cent coins were added, we have:

15 units + 7.5 + 25 = 28 units
1 unit --> 2.5

Sum of money in the box at first
= sum of money in the box in the end
= 20 x 2.5 x $0.50 + 28 x 2.5 x $0.20
= $39

cheers.

Hi Mathizzfun, can I ask.. how to get the 20units in the first place?
And why cross multiply? TIA!!![/quote]
At first, number of 20c = 3/4 number of 50c,
In the end, number of 20c = = 7/5 number of 50c

4 x 5 = 20

cross multiply method is another way that can be used get the solution.

here are some questions that were solved using the cross multiply method:

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=951428&hilit=cross+multiply#p951428

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=960600&hilit=cross+multiply#p960600

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=961764&hilit=cross+multiply#p961764

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=915223&hilit=cross+multiply#p915223

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=974082&hilit=cross+multiply#p974082

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=974032&hilit=cross+multiply#p974032

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=878744&hilit=cross+multiply#p878744

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=68&t=25129&p=866815&hilit=cross+multiply#p866815

cheers.

[MathIzzzFun]

snowball:

4/7 of Ming's mass is equal to 2/5 of Rob's mass.

If Ming puts on 6 kg and Rob loses 12kg, the two men will have the same mass.
a) Draw a clearly labelled model to solve the problem.
b) What is Rob's mass?

http://i45.tinypic.com/2nret7p.png\">


cheers.

[chloecube]

there were as many men as women in a bus. 2/7 of the men and 1/3 of the women alighted the bus. what fraction of the people did not alight the bus?



two ribbon X and Y were cut into 5 equal pieces. each piece cut from ribbon y was 3/5m longer than each piece of ribbon x. if ribbon x was 4m long, what is the lenght of ribbon y?

what i did was:

3/5m x 100 = 60cm (each piece of ribbon y is 60cm longer than ribbon x)
ribbon y = 5u + (60x5)
= 5u + 300cm
5u = 4m
so ribbon y is = 4m + 300cm = 7m

correct?

[PapayaDad]

Assume 21 man and 21 woman. Total 42


2/7 men = 6
1/3 woment = 7
Total alighted = 13
So 29 did not alight. Answer 29/42