Q&A - PSLE Math

[tianzhu]

chrisu:

The question is from MyPal workbook where a number of the parents' kids are using too. Those students from other schools are also given the 450 as answer and not too sure if there is any feedback from their parents.

Hi

Questions involving two cases or double if are common in primary maths.

Let your kid go through these questions, do we compare with the original quantity or remaining?

1) Wawa had some red and blue stickers. If 72 red stickers were removed, the ratio of the number of red stickers to that of the blue stickers would be 1 : 2. If 180 blue stickers were removed instead, the ratio would become 5 : 1. How many red stickers were there?

2) There are some red pins and blue pins in a box. If one of the red pin is removed, the number of red pins left will be 1/7 of the total number of pins left. If two blue pins are removed, the number of red pins will be 1/5 of the total number of pins left. How many pins are there in the box?

3) If Mark gives Lenny 14 stickers, he will have the same number of stickers as Lenny. If Lenny gives 10 stickers to Mark, the ratio of the number of stickers Lenny has to that of Mark will be 2: 5. How many stickers does Mark have?

Best wishes

[James Ang]

When Larry's age was twice Henry's age, Don's age was 27. When Don's age was twice Larry's age, Henry's age was 15. What was Henry's age when Don was 58 years old ?



Let me show how a P5 student would do to solve by drawing a model. (Remember that P5 students do not know any algebra!)


L [][]+[31]
H []+[31]
D [27]+[31] = 58

L [15][] + [][16]
H [15] +[][16]
D [15][][15][] + [][16] = 58

from above
D [15][][15][] + [][16] = 58
[][][] = 58-15-15-16 = 12
[]=4

Therefore Henry = [15] +[][16] = 15+4+16 = 35 years old.

A picture speaks a thousand words!

[Belle2011]

Hello,

Can somebody help me with this problem?

F, L and W each owned a collection of comics. The total collected owned by L and W was 3/2 as many comics as F owned. L owned 4/5 as many comics as the total collection owned by F and W. If W owned 169 fewer comics than L, how many comics must F and L each give to W in order for the 3 girls to have the same number of comics?

Thank-you.

Cheers,
Belle

[iFruit]

Belle2011:

Hello,

Can somebody help me with this problem?

F, L and W each owned a collection of comics. The total collected owned by L and W was 3/2 as many comics as F owned. L owned 4/5 as many comics as the total collection owned by F and W. If W owned 169 fewer comics than L, how many comics must F and L each give to W in order for the 3 girls to have the same number of comics?

Thank-you.

Cheers,
Belle

W = L - 169

F = 2/3 ( W + L) = 2/3 (L-169 + L) = 2/3 (2L -169)

5L/4 = W+F = (L - 169) + 2/3 (2L -169) = (3L - 3x169 + 4L -2x169)/3 = (7L -5x169)/3

15L = 28L -20x169----->13L = 20x169----> L = 20x13 = 260

So

W = L -169 = 260 -169 = 91

F = 2/3 (91+260) = 2/3 (351) = 234

(L+W+F)/3 = (260+234+91)/3 = 585/3 = 195

So,

L must give 260-195 = 65 to W

F must give 234 -195 = 39 to W

HTH.

[Dharma]

Belle2011:

Hello,

Can somebody help me with this problem?

F, L and W each owned a collection of comics. The total collected owned by L and W was 3/2 as many comics as F owned. L owned 4/5 as many comics as the total collection owned by F and W. If W owned 169 fewer comics than L, how many comics must F and L each give to W in order for the 3 girls to have the same number of comics?

Thank-you.

Cheers,
Belle

F : L + W = 2 : 3 = 18 : 27
L : F + W = 4 : 5 = 20 : 25

L => 20u
F => 18u
W => 7u

20u – 7u = 169
13u = 169
1u = 13

No. of comics that F gives to W = 18u – 15u = 3u = 3 x 13 = 39
No. of comics that L gives to W = 20u – 15u = 5u = 5 x 13 = 65

[Belle2011]

Dear iFruit and Dharma,

Thanks for helping.

Dharma,
F : L + W = 2 : 3 = 18 : 27
L : F + W = 4 : 5 = 20 : 25
L => 20u
F => 18u
W => 7u

I know you are converting 2:3 = 18:27 as equivalent ratios.
But how do you know you have to "stop" at 18:27 so that you can get L is 20u, F is 18u and W is 7u?
Thank-you.

Cheers,
Belle.

[Dharma]

Belle2011:

Dear iFruit and Dharma,

Thanks for helping.

Dharma,
F : L + W = 2 : 3 = 18 : 27
L : F + W = 4 : 5 = 20 : 25
L => 20u
F => 18u
W => 7u

I know you are converting 2:3 = 18:27 as equivalent ratios.
But how do you know you have to \"stop\" at 18:27 so that you can get L is 20u, F is 18u and W is 7u?
Thank-you.

Cheers,
Belle.

1. LCM of 5 and 9 is 45 (total of F, L and W) is the same in the 2 ratios
2. From ratios, we know F =18u and L = 20u
3. W = 45u - 18u - 20u = 7u

[Mum1113]

Hi, may I get help to solve this and in case it been posted and answered before , from Nanyang 2010 papers :-


Tanks X and Y are each filled with some water. If water from Tank Y is poured into Tank X until the water in Tank X reaches the brim, there will be 8 litres of water left in Tank Y. If water from Tank X is poured into Tank Y until the water in Tank Y reaches the brim, there will be 26 litres of water left in Tank X. The ratio of the volume of Tank X to the volume of Tank Y is 5 : 3 . How many more litres of water are needed to fill both tanks to their brim ?

TIA

[iFruit]

Mum1113:

Hi, may I get help to solve this and in case it been posted and answered before , from Nanyang 2010 papers :-


Tanks X and Y are each filled with some water. If water from Tank Y is poured into Tank X until the water in Tank X reaches the brim, there will be 8 litres of water left in Tank Y. If water from Tank X is poured into Tank Y until the water in Tank Y reaches the brim, there will be 26 litres of water left in Tank X. The ratio of the volume of Tank X to the volume of Tank Y is 5 : 3 . How many more litres of water are needed to fill both tanks to their brim ?

TIA

Let Water in Tank X = x
Water in Tank Y = y

Capacity of Tank X = x+y -8
Capacity of Tank Y = x+y -26

(x+y-8): (x+y-26) = 5:3 ----> 3(x+y-8) = 5(x+y-26)

3x + 3y -24 = 5x+5y-130----> x+y = 53

Capacity of Tank X = x+y -8 = 53 -8 = 45
Capacity of Tank Y = x+y -26 = 53 -26 = 27


Water needed to fill the tanks to the brim = 45 +27 -53 = 19 liters

HTH.

[Yu Xuan]

Mum1113:

Hi, may I get help to solve this and in case it been posted and answered before , from Nanyang 2010 papers :-


Tanks X and Y are each filled with some water. If water from Tank Y is poured into Tank X until the water in Tank X reaches the brim, there will be 8 litres of water left in Tank Y. If water from Tank X is poured into Tank Y until the water in Tank Y reaches the brim, there will be 26 litres of water left in Tank X. The ratio of the volume of Tank X to the volume of Tank Y is 5 : 3 . How many more litres of water are needed to fill both tanks to their brim ?

TIA

Ratio of X : Ratio of Y = 5 : 3
Difference in volume --> 5u -3u = 2u

There is some existing water in both tanks. The total volume of the existing water remains the same whether it is poured into Tank X or Y. Hence the difference in the 'excess' water is the difference in the volume of the two tanks. i.e.

2u --> 26 - 8 = 18
1u --> 9

Volume of Tank X --> 9 * 5 = 45
Volume of Tank Y --> 9 * 3 = 27

To find the existing volume of water in both tanks, we simply add the 8 litres of water to the volume of X ;

45 + 8 = 53

So the amount of watere required to fill up both tanks is
45 + 27 - 53 = 19 litres