Q&A - P5 Math

[shawnlim88]

hi all,

need help on this:
how to solve this using model?

1. 25% of Browns money is the same as 20% of Charlie’s Money. Express Brown’s money as a fraction of their total amount of money.

2.
Figure : 1, 2, 3, 4, …,10,…
Sequence: 5, 12, 21, 32, …, 140,…
what figure has the number 10197?

many thanks
shawnlim88

THank you

[neowanz]

Hi shawnlim88


1) For this type of question, you have to make the numerator of Browns and Charlie the same. In this case, it is 1 unit.

http://i45.tinypic.com/14j52sl.jpg\">

Then total amount of money is = 9units.

Ans: Brown's money / total amt of money = 4/9

2) Figure 1 2 3 4 ..... 10
Sequence 5 12 21 32 ......140

Sequence = figure x (figure + 4)

Eg for figure 10:

Sequence = 10 x (10 + 4) = 140

For sequence = 10197,

it can be seen that 100 x 104 = 10400 (which is around 10197), the answer for the figure should be around this value.

Hence, by trial and error, figure is 99. (99 x 103 = 10197).

Thanks

[MathIzzzFun]

shawnlim88:

hi all,

need help on this:
how to solve this using model?

1. 25% of Browns money is the same as 20% of Charlie's Money. Express Brown's money as a fraction of their total amount of money.

2.
Figure : 1, 2, 3, 4, ....,10,..
Sequence: 5, 12, 21, 32, .., 140,.....
what figure has the number 10197?

many thanks
shawnlim88

THank you

Q1
1/4 Brown's money = 1/5 Charlie's money
Brown's money --> 4 units
Charlie's money --> 5 units
Total --> 9 units

Brown's money as a fraction of the total amount of money = 4/9

Q2.
Add 4 to each number --> 9, 16, 25, 36 ... which is
3²,4², 5², 6² …

so, the sequence for figure number N = (N+2)²-4

10197+4 = 10201 = 101 x 101
101-2=99
Figure 99 = 10197

cheers.

[shawnlim88]

Dear neowanz & MathIzzzFun


thanks for the help.

cheers

[zombie]

Hello, helping my son to solve using agebra,

can someone tell me if the answer is correct ?

Can someone suggest other method? i was taught using Algebra in the 1970s. lost touch. Thanks :roll:

Mrs Lim bought some green apples for an average cost of $0.45. However, if she exchanges 4 green apples for 4 red apples, which cost $0.75 each, the average price of each apple would be $0.55
How many apples are there altogether ?

Let the total number of apples be X
X (0.45)+ (4x 0.75) \t= 0.55 X
0.45 (X - 4) + 3.00\t= 0.55 X
0.45 X -18 + 3.00\t= 0.55 X - 0.45 X
1.2\t= 0.10 X
12\t= X
there are 12 apples altogether

[lapapillion]

Did you miss out some parts of the question? How come in your working there's a 0.75? :scared:

[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.

[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?

[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