Q&A - P5 Math

[Maths Hub]

ajmum:

Need some help for RGPS SA2 2011 question.


1. Becky bought 7 basketballs with 3/7 of the money she had. She bought another 2 basketballs and 11 baseballs with her remaining money. If Becky used the amount of money she had at first to buy only baseballs, what was the maximum number of baseballs she could buy ?

2. The mass od the sand in sack B was 1/4 the mass of sand in sack A. After 52.5 kgs of the sand in sack A and 10kg 150g of the sand in sack B was used, the mass of sand in sack A was 1/2 the mass of sand in sack B. What was the total mass of and in sack A and sack B at first?

3. The average number of sit-ups performed by 15 girls and some boys was 50. The average number of sit-ups performed by the girls was 80% of the average number od sit-ups performed by all the pupils. The average number of sit-ups performed by the boys was 30% more than the average number of sit-ups performed by the girls.
b) what percentage of the pupils were boys ?

Thank you.

1. Becky bought 7 basketballs with 3/7 of the money she had. She bought another 2 basketballs and 11 baseballs with her remaining money. If Becky used the amount of money she had at first to buy only baseballs, what was the maximum number of baseballs she could buy ?

Let Becky has 7 Units of money at first.
3 Units --> 7 basketballs__________________________ (1)
4 Units --> 2 basketballs + 11 baseballs_____________ (2)
Using common multiple of 3 and 4,
12 Units --> 28 basketballs
12 Units --> 6 basketballs + 33 baseballs
Therefore
28 basketballs --> 6 basketballs + 33 baseballs
22 basketballs --> 33 baseballs
So 2 basketballs --> 3 baseballs
And adding (1) and (2), we have 7 Units --> 9 basketballs + 11 baseballs
Therefore 7 Units --> 13.5 baseballs + 11 baseballs = 24.5 baseballs
Maximum number of baseballs she could buy --> 24

2. The mass od the sand in sack B was 1/4 the mass of sand in sack A. After 52.5 kgs of the sand in sack A and 10kg 150g of the sand in sack B was used, the mass of sand in sack A was 1/2 the mass of sand in sack B. What was the total mass of and in sack A and sack B at first?

Let sack A has 4 parts of sand at first and 1 unit of sand in the end.
Let sack B has 1 part of sand at first and 2 units of sand in the end.
A: 1 unit --> 4 parts – 52.5
B: 2 units --> 1 part – 10.15
Therefore 2 units --> 8 parts – 105 --> 1 part -10.15
7 parts --> 94.85
1 part --> 13.55
5 parts --> 67.75
Total mass in sack A and B at first --> 67.75 kg

3. The average number of sit-ups performed by 15 girls and some boys was 50. The average number of sit-ups performed by the girls was 80% of the average number od sit-ups performed by all the pupils. The average number of sit-ups performed by the boys was 30% more than the average number of sit-ups performed by the girls.
b) what percentage of the pupils were boys ?

Let the average of all the pupils be 100 units.
The average of girls --> 80 units
Average of boys --> 130% × 80 = 104 units
Using comparison, the average of girls is short of 20 units to become 100 units while boys have excess of 4 units. In order to make the combined average 100 units, there must be 5 boys for every girl.
Therefore percentage of boys --> 5/6 × 100% = 83 1/3%

[lapapillion]

After editing the question, it's okay now


I'll do it this way, which is similar to yours. But I think you should just skip the first step and go straight to the second step. The first step does not logically lead to the second step

Let x be the no. of green apples:

The average cost of all the red+green apples = 0.55
=> [Total cost of green + red apples]/total no. of apples = 0.55
=> [(x-4)(0.45)+4(0.75)]/x] = 0.55
=> x = 12

[KP2]

Hi , need help with this one . Many thanks in advance


Jim bought some chocolates and gave half to them to Ken. Ken bought some sweets and gave half of them to Jim.

Jim ate 12 sweets and Ken ate 18 chocolates. After that, the number of sweets and chocolates Jim had were in the ratio 1:7 and the number of sweets and chocolates Ken had were in the ratio 1:4.

How many sweets did Ken buy?

[Maths Hub]

KP2:

Hi , need help with this one . Many thanks in advance


Jim bought some chocolates and gave half to them to Ken. Ken bought some sweets and gave half of them to Jim.

Jim ate 12 sweets and Ken ate 18 chocolates. After that, the number of sweets and chocolates Jim had were in the ratio 1:7 and the number of sweets and chocolates Ken had were in the ratio 1:4.

How many sweets did Ken buy?

The yellow boxes refer to the unchanged quantity.
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Cheers! :salute:

[zombie]

Hello, please help me to check if my answer and solution are acceptable.

I'm a parent teaching my P5 boy maths. :scratchhead: Thanks !

If Mr Ibrahim buys 3 plates and 5 cups, he will have $6 left. If he wants to buy 5 plates and 3 cups, he will need another $4. Given that a cup costs $1.50, how much does he have?

My solution

Let x be the total amount he has
3p + 7.50 = x - 6 (x5)
5p + 4.50 = x + 4 (x3)

15p + 37.50 = 5x - 30
15p + 13.50 = 3x + 12
24.00 = 2x – 42

24 + 42 = 2x

66 ÷ 2 = x
X = 33

Answer :he has $33

[Maths Hub]

zombie:

Hello, please help me to check if my answer and solution are acceptable.

I'm a parent teaching my P5 boy maths. :scratchhead: Thanks !

If Mr Ibrahim buys 3 plates and 5 cups, he will have $6 left. If he wants to buy 5 plates and 3 cups, he will need another $4. Given that a cup costs $1.50, how much does he have?

My solution

Let x be the total amount he has
3p + 7.50 = x - 6 (x5)
5p + 4.50 = x + 4 (x3)

15p + 37.50 = 5x - 30
15p + 13.50 = 3x + 12
24.00 = 2x – 42

24 + 42 = 2x

66 ÷ 2 = x
X = 33

Answer :he has $33

Hi zombie,

Depending on your child's school, algebraic method may or may not be accepted, even though it is correct.

Allow me to offer another non-algebraic way:
By drawing models,
2 plates will cost $10 more than 2 cups.
So each plate cost $5 more than each cup,
which means the cost cost of each plate is $6.50.
Therefore, he has $6.50 X 3 + 5 X $1.50 + $6 = $33. :snuggles:

Cheers.

[YumYum]

Hi, need help with qn:


In may, the average number of participants in one pottery class was 24. In June, 5 new pottery classes were added. The average number of participants in all the pottery classes became 26. If there were 32 participants in each of the new pottery classes, find the total number of participants in all the pottery classes in June.

Thanks!

[Maths Hub]

YumYum:

Hi, need help with qn:


In may, the average number of participants in one pottery class was 24. In June, 5 new pottery classes were added. The average number of participants in all the pottery classes became 26. If there were 32 participants in each of the new pottery classes, find the total number of participants in all the pottery classes in June.

Thanks!

Total extra members from the new classes that contribute to the rise in old average --> (32 - 26) X 5 = 30
Rise in old average --> 26 - 24 = 2
Original number of classes --> 30 / 2 = 15
New number of classes --> 15 + 5 = 20
Total number of participants --> 20 X 26 = 520

Cheers.

[zombie]

This question is taken from Nanyang SA2, 2011 question 17 paper 2.

Can someone help me to solve ? Thanks !

At first, the ratio of the number of men to the number of women in a
cinema was 2:5. When 48 women and some men entered the cinema
the number of women increased by 12% and the total number of people
in the cinema increased by 20%.
(a) How many men were there at first?
(b) What percentage of the people in the cinema were men in the end?
Express your answer as a mixed number.

[dazzlego]

savoury sweet:

Hi everyone,kindly help to solve the following question. Thanks so much.


A grp of pupils claculated their average score for Science test. They noted that if anyone scored 14 marks and more, their average would be 90. If anyone scored 4 marks fewer, their average score would be 88. How many pupils are there in the grp?

Hi savoury sweet,

You can assume that the number of pupils is 1 unit
case 1: the total score -> 90 units - 14
case 2: the total score -> 88 units + 4
2 units -> 14 + 4 = 18
1 unit -> 18 ÷ 2 = 9

Cheers :celebrate: