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kitty2
KiasuGrandMaster
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Post by kitty2 » Wed May 05, 2010 11:46 am

Vanilla Cake wrote:
kitty2 wrote:1)Audrey and Belle have some money each.If Audrey spends $18 and Belle spends $24 each day.Audrey will have $25 left.when belle has spent all her money.If Audrey spends $13 and belle spends $30 each day,Audrey will still have $139 left.when Belle has spent all her money.How much money do they have altogether?

2)Figure No of cakes

1 1
2 4
3 10

a)How many cubes will ther be in Fig 6?
b)Which figure will have 120 cubes?
Hi kitty2,
Pls refer to the links below while waiting for Mathsguru's solutions:

Link 1
Link 2
Link 3
Link 4

Submitted by VC's mum.

:thankyou:

i-mum
YellowBelt
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Post by i-mum » Wed May 05, 2010 12:03 pm

Hi MathsGuru,

1. A, B & C shared some sweets. A received 1/11 of the sweets. B received 1/4 of the number of sweets C received. When C gave away 104 sweets to be shared between A and B, they found that all of them had the same number of sweets. How many sweets were there at first?

2. A, B and C shared a sum of money.
A's share was 1 1/2 timess that of B and B's share was 1 1/4 times that of C.
(a) what fraction of A's share was C?
(b) If A gave B $19 and C $22, A would have as much money as C. Find the sum of money shared.

Thank You.

i-mum

Dharma
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Post by Dharma » Wed May 05, 2010 1:15 pm

i-mum wrote:Hi MathsGuru,

1. A, B & C shared some sweets. A received 1/11 of the sweets. B received 1/4 of the number of sweets C received. When C gave away 104 sweets to be shared between A and B, they found that all of them had the same number of sweets. How many sweets were there at first?


i-mum
Q1)
A & B : 3u + 104 = 2p

C : 8u – 104 =1p

16u - 208 = 3u + 104
13u = 312
1u = 24

Total no. of sweets at first = 24 x 11 = 264


Q2)
A: 15u
B: 10u
C: 8u

(a) Fraction = 8/15b)

A : 15u - $41 = 1p
C : 8u + $22 = 1p

(b)
15u - $41 =8u + $22
7u = $63
1u = $9

Sum of money shared = 15u + 10u + 8u = 33u = 33 x $9 = $297
Last edited by Dharma on Wed May 05, 2010 3:06 pm, edited 1 time in total.

Vanilla Cake
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Post by Vanilla Cake » Wed May 05, 2010 2:21 pm

i-mum wrote:1. A, B & C shared some sweets. A received 1/11 of the sweets. B received 1/4 of the number of sweets C received. When C gave away 104 sweets to be shared between A and B, they found that all of them had the same number of sweets. How many sweets were there at first?
Hi i-mum,
For the reference of readers in this forum, your Q1 was from Nanyang Primary School P5 SA1 2009 Paper 2 Q18 [5 marks] set on 15 May 2009.
A stands for Aifang, B stands for Bala and C stands for Cindy in Q1.

You may wish to refer to these links for the solutions while awaiting Mathsguru's reply.Thanks
Solution 1
Solution 2

Submitted by VC's mum

happies
YellowBelt
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Post by happies » Wed May 05, 2010 3:19 pm

Hello Mathsguru,

Q1) A man is 45 years old when his son is t years old. How old is the man when his son is 15 years old?

Q2) 3/5 of May's mass is 3/4 of June's mass. If June's mass is 52 kg, what is their total mass?

Q3) A number when divided by 6 gives a remainder of 1 and gives a remainder of 2 when divided by 5. What is the number?

Q4) I am a 4-digit number. All my digits are different. My first digit is 1/6 of my last digit. My second digit is 1/4 of my third digit. What number am I?

Thank you.


happies
YellowBelt
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Post by happies » Wed May 05, 2010 3:54 pm

happies wrote:Hello Mathsguru,

Q1) A man is 45 years old when his son is t years old. How old is the man when his son is 15 years old?

Q2) 3/5 of May's mass is 3/4 of June's mass. If June's mass is 52 kg, what is their total mass?

Q3) A number when divided by 6 gives a remainder of 1 and gives a remainder of 2 when divided by 5. What is the number?

Q4) I am a 4-digit number. All my digits are different. My first digit is 1/6 of my last digit. My second digit is 1/4 of my third digit. What number am I?

Thank you.
Oops, pls ignore Q1 & Q4 questions:))

YLH88
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Post by YLH88 » Wed May 05, 2010 5:31 pm

Hi YLH88,
I am VC's mum and VC won't be back home today until 7 pm plus due to SMO training and dance practice.
Sorry for the inconvenience caused due to VC's inadequate explanation as she is still a Sec 2 girl.

Area of square : Area of rectangle+Area of circle
1 : 2
Multiply by 10
10 : 20

because 60% of the square is shaded.60% = 3/5.
Instead of 1x 3/5 which will be inconvenient as 3/5 unit, expand the ratio by multiplying by 10. So, 10x3/5 = 6 units which is easier to calculate.

Hope this helps.
Submitted by VC's mum
Hi VC's mum,

No worries, there is no need to apologies, really. In fact, I'm very very grateful for all the help the she, mathsguru and all other parents have given in this forum. Greatly appreciated!

Sorry, i'm still abit blur in this question. Is it possible to multiple by 5 instead of 10 ? ie 5 x 3/5 = 3 ?

Thank you.

i-mum
YellowBelt
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Post by i-mum » Wed May 05, 2010 6:22 pm

Thanks Vanilla Cake and Dharma.

But for Q2, the answers was given by the school as 49 1/2 and $330. This paper was set before the ratio is taught and teacher did not give any solutions due to limited time :? :cry:

i-mum

tisha
BrownBelt
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Post by tisha » Wed May 05, 2010 8:43 pm

CoffeeCat wrote:
tisha wrote:Is it possible to solve these without algebra? These are P5 Questions from past year exam papers for SA1.
I believe algebra is only taught in P6, so just curious how come these questions can be asked for SA1 of P5 :?

Ken and Patrick had $625 at first. After buying some books, the amount of money Ken has left is 4 times the amount of money he spent and the amount of money Patrick has left is 5 times the amount of money he spent. Find the amount of money Patrick had at first if they have a total of $510 left.


There are 85 plates of chicken rice for 80 people. Each adult eats 2 plates of chicken rice and every 3 children share 1 plate of chicken rice. Find the number of adults and children respectively.
For qns 2. One common non-algebra way to solve such algebraical questions is to use this heuristic called "make an assumption".
First assume there are 80 adults. Then these 80 adults will eat 160 plates of rice.
Now we must try to find the number of children.
Notice that if we interchange 3 adults for 3 children, there will be a decrease in (2*3 -1) = 5 plates of rice from the total.
(if you are not convinced you can try 77 adults, 75 adults to see the pattern)
For every group of interchange of 3 adults for 3 children, we are 5 plates nearer the true total of 85. therefore, the number of children is 3*(160-85)*5 =45.
Thank you very much coffeecat for the non algebric solution.
But I'm not so clear on this part
For every group of interchange of 3 adults for 3 children, we are 5 plates nearer the true total of 85. therefore, the number of children is 3*(160-85)*5 =45
I suppose it is a typo and should be 3*(160-85)/5, but I still don't understand it. Sorry to bother you so much.

tisha
BrownBelt
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Post by tisha » Wed May 05, 2010 8:54 pm

small wrote:Hi tisha,

Hope this clear enough. :)


Question:
Ken and Patrick had $625 at first. After buying some books, the amount of money Ken has left is 4 times the amount of money he spent and the amount of money Patrick has left is 5 times the amount of money he spent. Find the amount of money Patrick had at first if they have a total of $510 left

Total of $510 left -----> 4 units of Ken + 5 units of Patirck = 510 ---------------> (1)

The amount they spent -----> 1 unit of Ken + 1 unit of patrick = (625-510) = 115
4 units of Ken + 4 units of patrick = 115 x 4 = 460
4 units of Ken + 4 units of patrick = 460 ---------------> (2)

By substitute (2) in (1):
460 + 1 unit of Patrick = 510
1 unit of Patrick = 50

Patrick had 6 units at first.
6 x 50 = 300

Patrick had $300 at first.
Thank you small, it is very clear. At first I thought that unit of patrick and unit of Ken are nothing but x and y disguised in new clothes. But if we do solve the problem by algebra the equations and calculations are different from yours. Indeed a novel way to solve the problem. :udaman:

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