Q&A - P5 Math

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?

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

chloecube

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

thks papayaDad

i do not quite understand, what i did was
i change the denominator to the same,
2/7 - 6/21
1/3 - 7/21
total of alighted pax = 13/21
so 8/21 did not alight

something wrong with my method?

MathIzzzFun

chloecube:
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
thks papayaDad

i do not quite understand, what i did was
i change the denominator to the same,
2/7 - 6/21
1/3 - 7/21
total of alighted pax = 13/21
so 8/21 did not alight

something wrong with my method?

total alighted should be 13/42
because there are equal number of men and women, so the total number of people is doubled ie 21 + 21 = 42.

29/42 did not alight

cheers.

chloecube

thks papayadad

veraclari

Hi all thanks to the sharing s here I know ratio must use cross multiply method sometimes. However can u kindly share how we determine what to cross multiply or the factor to multiply as I’m confused…Eg. In the qquestion and model answer attached, why X 6 & X 7? How to dderive 6&7? TIA!

Qn: ratio of no. Of girl to boy is 6:5. No. Of girl swimmers to girl non swimmers is 5:2. Ratio of boy swimmers to boy non swimmers is 3:2.
A) what fraction of children is non swimmers?
B) if there were 30 more boys non swimmers than girl non swimmers. How many swimmers were there altogether?
Model Answer

‘’‘’‘’‘’ 5 : 2 ---------3 : 2
((X6)) 30 : 12 ---------21 : 14 ((X7))
A) total units -> 42+35= 77
Total non swimmers -> 12+14 = 26

Kindly advise why they multiply by 6&7? How to get 6&7 in e first place?

Thanks! !!
P.s.I am new to this and am using mobile upload and I can’t rotate e picture which is turned when uploaded though original is upright Sso I didn’t Upload. How to rotate in tinypic?

Jamesbond

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.

Thank you MathIzzzFun for your help. Why this sum has to be soo difficult?

I managed to understand.

Jamesbond

In a classroom, when 5 boys left for the washroom, the ratio of the number of boys to the number of girls that remained became 1:1. On the other hand, when 5 girls left the classroom, the ratio of the number of boys to the number of girls that remained became 5:3. How many boys and girls were there in the class? :?

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