Tutor MathsGuru: Ask me for your burning Maths questions!

[livewith_vanilla]

maths6a:

livewith_vanilla:

[quote=\"super star\"]please help me with this problem

Tank p with 40cm by 30cm base is filled with water till it reaches the height 55cm .water was then poured from tank p to tank q with base 100cmby20cm, untill the height of the water level in tank p became twice that of the water level in tank q.find the volume of water in tank q. give your ans in liters.

55x40x30=66000

40x30=1200
100x20=12000
12000+1200=13200
66000/13200=5
5x2=10
10x100x20=20000

20000cm3 = 20L

Ans:20L

others please double confirm answer.

You made a calculation error with \"100x20=12000\"
it should be 2000

so answer should be 30 litres

Amount of water you have in tank p which you want to pour into p and q in such a way that the height in p is twice q so let the height in q be 1 unit and the height in p be two units.

55x40x30=66000 (amt of water in p)

vol to be filled in p + vol to be filled in q = 66000 ml

40x30x2units +100x20x 1 unit = 66000 ml
2400 units + 2000 units = 66000 ml
4400 units = 66000 ml
1 unit = 15 cm
2000 x 15 = 30000ml = 30 litres[/quote]Thanks for correcting.

[kwcllf]

livewith_vanilla:

kwcllf:

Hi,


Please help with the following question:

Andy, Benny and Calvin have 150 buttons altogether.
If Andy gives 3/7 of his buttons to Benny and Benny gives 32 buttons to Calvin and Calvin gives 26 buttons to Andy, they have the same number of buttons.

How many buttons do they have at first?

(I got the answers as Andy=42, Benny=64 and Calvin=44 by using algebra. But I am afraid my P4 son may not understand. Hope someone can help with an alternative solution).
Thanks!

Since the three boys give each other and not give away, the TOTAL number of marbles stay the same(eg. 150)

Hence, 150/3=50

Thus each boy has the same number of marbles left.

Andy:

Calvin gives Andy 26 marbles.

Thus, 50-26=24

In the first place, Andy gave Benny 3/7 of his marbles, before Calvin gave him any marbles.

Thus, Andy has (1-3/7=4/7) marbles left.

4u->24
1u->24/4=6
7u->6x7=42

Benny:
Andy gave 3u of his marbles to Benny.

Thus,

1u(of Andy)->6
3u(of Andy)->6x3=18

Andy gave Benny 18 marbles.

50+32=82

This is the number of marbles Benny would have if he did not give Calvin any marbles.

However, since Andy gave Benny some marbles, we have to substract 18 marbles from 82 marbles.

82-18=64

Benny has 64 marbles.

Calvin:

The Long Way:

Since Calvin gave Andy 26 marbles,

50+26=76

Calvin would have 76 marbles if he did not give any marbles to Andy.

However, Benny gave Calvin 32 marbles.

Thus,

76-32=44

Calvin would have 44 marbles.


Short way:

Since we found out Andy has 42 marbles and Benny has 64 marbles, we just need to take the total marbles subtract the sum of Andy's and Benny's marbles.

42+64=106
150-106=44

Calvin would have 44 marbles.

Hi livewith_vanilla,

Thanks for the solution!

[livewith_vanilla]

kwcllf:

livewith_vanilla:

[quote=\"kwcllf\"]Hi,


Please help with the following question:

Andy, Benny and Calvin have 150 buttons altogether.
If Andy gives 3/7 of his buttons to Benny and Benny gives 32 buttons to Calvin and Calvin gives 26 buttons to Andy, they have the same number of buttons.

How many buttons do they have at first?

(I got the answers as Andy=42, Benny=64 and Calvin=44 by using algebra. But I am afraid my P4 son may not understand. Hope someone can help with an alternative solution).
Thanks!

Since the three boys give each other and not give away, the TOTAL number of marbles stay the same(eg. 150)

Hence, 150/3=50

Thus each boy has the same number of marbles left.

Andy:

Calvin gives Andy 26 marbles.

Thus, 50-26=24

In the first place, Andy gave Benny 3/7 of his marbles, before Calvin gave him any marbles.

Thus, Andy has (1-3/7=4/7) marbles left.

4u->24
1u->24/4=6
7u->6x7=42

Benny:
Andy gave 3u of his marbles to Benny.

Thus,

1u(of Andy)->6
3u(of Andy)->6x3=18

Andy gave Benny 18 marbles.

50+32=82

This is the number of marbles Benny would have if he did not give Calvin any marbles.

However, since Andy gave Benny some marbles, we have to substract 18 marbles from 82 marbles.

82-18=64

Benny has 64 marbles.

Calvin:

The Long Way:

Since Calvin gave Andy 26 marbles,

50+26=76

Calvin would have 76 marbles if he did not give any marbles to Andy.

However, Benny gave Calvin 32 marbles.

Thus,

76-32=44

Calvin would have 44 marbles.


Short way:

Since we found out Andy has 42 marbles and Benny has 64 marbles, we just need to take the total marbles subtract the sum of Andy's and Benny's marbles.

42+64=106
150-106=44

Calvin would have 44 marbles.

Hi livewith_vanilla,

Thanks for the solution![/quote]You are welcome. Hope your child understands my solution.

[super star]

thanks

maths6a:

livewith_vanilla:

[quote=\"super star\"]please help me with this problem

Tank p with 40cm by 30cm base is filled with water till it reaches the height 55cm .water was then poured from tank p to tank q with base 100cmby20cm, untill the height of the water level in tank p became twice that of the water level in tank q.find the volume of water in tank q. give your ans in liters.

55x40x30=66000

40x30=1200
100x20=12000
12000+1200=13200
66000/13200=5
5x2=10
10x100x20=20000

20000cm3 = 20L

Ans:20L

others please double confirm answer.

You made a calculation error with \"100x20=12000\"
it should be 2000

so answer should be 30 litres

Amount of water you have in tank p which you want to pour into p and q in such a way that the height in p is twice q so let the height in q be 1 unit and the height in p be two units.

55x40x30=66000 (amt of water in p)

vol to be filled in p + vol to be filled in q = 66000 ml

40x30x2units +100x20x 1 unit = 66000 ml
2400 units + 2000 units = 66000 ml
4400 units = 66000 ml
1 unit = 15 cm
2000 x 15 = 30000ml = 30 litres[/quote]

[super star]

Please help me with this problem

cinema admission tickets for children were priced at $8 each.there were 20 more than girls inthe cinema for the movie. For the second movie ,the number of boys decreased by 10% but the number of girls increased by 5%.if there were 408 children at the second movie, how much money was collected from the sales of admission ticket for the first movie?

[super star]

Please help me with this problem

A librarian counted the number of adults in the library and found that 2/5 of the number of women was equal to 2 times the number of men.when another 12 men entered the library and 45 women left the library, the ratio of the number of women to the number of men became 5:2.

(A) what was the ratio of the number of women at first to the number of men at first?give youir answer in its simplest form.

(B)find the number of women in the library at first.

[livewith_vanilla]

Hi Superstar,


Your questions seem to come from Henry Park Preliminary Paper 2 2009.

The solution for Q14 as stated on the answer key is:

105%G+90% G+18=408
195%G+18=408
100%G=390/195x100=200
200x2+20=420
420x8=3360

Ans: $3360

For reference, the question is as of below:

Cinema admission tickets for children were priced at $8 each.there were 20 more than girls inthe cinema for the movie. For the second movie ,the number of boys decreased by 10% but the number of girls increased by 5%.if there were 408 children at the second movie, how much money was collected from the sales of admission ticket for the first movie?

For Q15, the answer key did not provide any solutions. But I’m able to solve (A). The answers are (a)5:1 and (b)150

For reference, the question is as of below:
A librarian counted the number of adults in the library and found that 2/5 of the number of women was equal to 2 times the number of men.when another 12 men entered the library and 45 women left the library, the ratio of the number of women to the number of men became 5:2.

(A) what was the ratio of the number of women at first to the number of men at first?give youir answer in its simplest form.

(B)find the number of women in the library at first.

2/5W=2 M
1/5W=1 M

W:M
5:1

Ans: 5:1

However, I am too unable to solve (B). The answer I got was 105 and not 150. Thus my apologies.

Are you a primary 6 student or a parent?

[Dharma]

livewith_vanilla:

Hi Superstar,


Your questions seem to come from Henry Park Preliminary Paper 2 2009.

The solution for Q14 as stated on the answer key is:

105%G+90% G+18=408
195%G+18=408
100%G=390/195x100=200
200x2+20=420
420x8=3360

Ans: $3360

For reference, the question is as of below:

Cinema admission tickets for children were priced at $8 each.there were 20 more than girls inthe cinema for the movie. For the second movie ,the number of boys decreased by 10% but the number of girls increased by 5%.if there were 408 children at the second movie, how much money was collected from the sales of admission ticket for the first movie?

For Q15, the answer key did not provide any solutions. But I'm able to solve (A). The answers are (a)5:1 and (b)150

For reference, the question is as of below:
A librarian counted the number of adults in the library and found that 2/5 of the number of women was equal to 2 times the number of men.when another 12 men entered the library and 45 women left the library, the ratio of the number of women to the number of men became 5:2.

(A) what was the ratio of the number of women at first to the number of men at first?give youir answer in its simplest form.

(B)find the number of women in the library at first.

2/5W=2 M
1/5W=1 M

W:M
5:1

Ans: 5:1

However, I am too unable to solve (B). The answer I got was 105 and not 150. Thus my apologies.

Are you a primary 6 student or a parent?

Hi superstar and livewith_vanilla,

Hope this helps.

At first
Ratio => Women : Men = 5 : 1

At last
Ratio => Women : Men = 5u - 45 : 1u + 12 = 5 : 2

10u – 90 = 5u + 60
5u = 150
1u = 30

No. of women at first = 5u = 5 x 30 = 150

[maths6a]

Hi Can anyone help with this question?

How many 3-digit number combinations are there on a 3-digit number lock if all the 3-digits used are different and the numbers are greater than 800?

Thanks.

[Dharma]

maths6a:

Hi Can anyone help with this question?

How many 3-digit number combinations are there on a 3-digit number lock if all the 3-digits used are different and the numbers are greater than 800?

Thanks.

Digit in hundreds place => 8 or 9
Digit in tens place => 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9
Digit in ones place => 0, 1, 2, 3, 4, 5, 6, 7, 8 or 9

Since the 3 digits used are different, you have 2 choices for hundreds place, 9 choices for tens place(1 choice already selected for 100s place) and 8 choices for the ones place (2 choices already chosen for 100s and 10s place)

No. of combinations = 2 x 9 x 8 = 144