Tutor MathsGuru: Ask me for your burning Maths questions!

[Daddy]

Thanks Vanilla Cake

[adhdadhd]

1. During an IT fair, for each book sold, Leena received a commission of $20. An additional bonus of $10 would be given to her for every 5 books sold.


a) How much would Leena receive if she sold a total of 15 books?

comm of 1 book = $20
comm of 15 books = $20x15= $300
bonus of 15 books = $10 x 15/5 = $30

Leena would rx $330.

b) How many books had she sold if she was given $8880?

For every 5 books she get $100 comm and $10 bouns, sum up to be $110.

For every 10 books, 10x$20 + 10/5x$10 = $220.

8800/220 = 40, thus 400 books
80/20=4

She sold 404 books.

Check: 400 books = 400x20 + 400/5x10 = 8000+800.

[firebird]

Dear maths guru


Good morning

Please help me on the following maths:

1) Susan saved $345 in $5, $10 and $50 notes. There were 35 notes althogether. She had 4 more $10 notes than $50 notes. How many $5 notes did she have?

2) Mr. Tan has some charity tickets. If he sends 4 tickets each to some companies, he will need 20 more tickets. However, if he sends 3 tickets each to the same companies, he will have a remainder of 50 tickets. How many tickets does he have.

3) Alex, Boon Kheng and Casmuri paid for a meal. Alex paid 1/5 of the amount Boon Kheng and Casmuri paid together, Boon Keng paid 1/4 of the amount Alex and Casmuri paid.

a) What fraction of the total cost of the meal did Casmuri pay?
b) If the meal cost $205.20, how much did Alex pay?

Thank you
firebird

[Vanilla Cake]

firebird:

1) Susan saved $345 in $5, $10 and $50 notes. There were 35 notes altogether. She had 4 more $10 notes than $50 notes. How many $5 notes did she have?

$5: 35-unit-unit-4 = 31-2units

$10: unit+4
$50: unit

5(31-2units)+10(unit+4)+50(unit) = 345
50units =150
1 unit = 3

$5: 31-2units = 31-2(3)=25
Sussan had 25 $5 notes.

firebird:

2) Mr. Tan has some charity tickets. If he sends 4 tickets each to some companies, he will need 20 more tickets. However, if he sends 3 tickets each to the same companies, he will have a remainder of 50 tickets. How many tickets does he have.

Instead of 3 tickets, each company is given 4 tickets, ie each company gets 1 more ticket.
With 50 tickets remain and 20 tickets short , total tickets issued = 50+20=70.

So each company receives 1 more ticket and 70 tickets are sent, number of companies =70 divided by 1 = 70.

Number of tickers = 4x70-20= 260

firebird:

3) Alex, Boon Kheng and Casmuri paid for a meal. Alex paid 1/5 of the amount Boon Kheng and Casmuri paid together, Boon Keng paid 1/4 of the amount Alex and Casmuri paid.

a) What fraction of the total cost of the meal did Casmuri pay?
b) If the meal cost $205.20, how much did Alex pay?

A : B+C
1 : 5
5 : 25

B : A+C
1:4
6:24

A:5
B:6
C:19

a) Fraction of the total cost of the meal that Casmuri paid = 19/30

30 units = $205.20
5 units = $34.20
b)Alex paid $34.20

Submitted by VC's mum.

[Dharma]

firebird:

Dear maths guru


Good morning

Please help me on the following maths:

Thank you
firebird

1) Susan saved $345 in $5, $10 and $50 notes. There were 35 notes althogether. She had 4 more $10 notes than $50 notes. How many $5 notes did she have? `

Quantity :
$50 : 1u
$10 : 1u + 4
$5 : 35 – 2u – 4 = 31 – 2u

Total = 1u($50) + (1u+4)$10 + (31-2u)$5 = $50u + $195 = $345

$50u = $345 - $195 = $150
1u = 3

No. of $5 notes = 31 - 6 = 25


2) Mr. Tan has some charity tickets. If he sends 4 tickets each to some companies, he will need 20 more tickets. However, if he sends 3 tickets each to the same companies, he will have a remainder of 50 tickets. How many tickets does he have.


If Mr Tan sends 1 less ticket to each of the companies, he will have extra (20 + 50 = 70) tickets
Therefore there must be 70 companies that he his sending his tickets to.

No. of tickets Mr Tan has
= (4 x 70) – 20 = 260
Or
= (3 X 70) + 50 = 260

3) Alex, Boon Kheng and Casmuri paid for a meal. Alex paid 1/5 of the amount Boon Kheng and Casmuri paid together, Boon Keng paid 1/4 of the amount Alex and Casmuri paid.

a) What fraction of the total cost of the meal did Casmuri pay?
b) If the meal cost $205.20, how much did Alex pay?

Alex paid 1/6 of total cost
Boon Keng paid 1/5 of total cost

a)\tFraction of total cost Casmuri paid = 1 – 1/6 – 1/5 = 30/30 – 5/30 – 6/30 = 19/ 30
b)\tAlex paid = 1/ 6 x $ 205.20 = $34.20

[firebird]

Dear Vanilla cake & Dharma


Good morning.

Thank you very much for solving the questions posted by me.

With best regards
firebird

[starlight1968sg]

Dharma:

starlight1968sg:

Hi mathsguru and others,

I need some help:
(1) P, T and V have 960 beads altogether. P gave some of her beads to T and the number of T's beads was doubled. Then T gave some of her beads to V and the number of V's beads was doubled. If the 3 girls had the same number of beads in the end, how many beads did P have at first.
(ans: 560)
How to solve it?


MTIA.

Since P, T and V had the same no. of beads at last ; each will have 960/3 = 320 beads at last

When T gave some of her beads to V and the number of V's beads was doubled
No. of beads T and V before T gave V some beads :
T : 320 + 160 = 480
V : 320 – 160 = 160

When P gave some of her beads to T and the number of T's beads was doubled.
No. of beads P and T before P gave T some beads :
T : 480 – 240 = 240
P : 320 + 240 = 560

Hi Dharma,
The number 320 is due to 960/3 = 320.
May I ask how to get the number 160?
MTIA.

[Brenda10]

Good afternoon Mathsguru and others,


I need help for the following P5 SA1 Q11. :?

Mrs. Sham had 4 times as many lollipops as chocolate bars. After giving 177 lollipops and 25 chocolate bars to her students, she had thrice as many chocolate bars as lollipops left. How many lollipops and chocolate bars did she have altogether at first?
(Given answer: 230)

Thank you for your help.

[Dharma]

starlight1968sg:

Dharma:

[quote=\"starlight1968sg\"]Hi mathsguru and others,

I need some help:
(1) P, T and V have 960 beads altogether. P gave some of her beads to T and the number of T's beads was doubled. Then T gave some of her beads to V and the number of V's beads was doubled. If the 3 girls had the same number of beads in the end, how many beads did P have at first.
(ans: 560)
How to solve it?


MTIA.

Since P, T and V had the same no. of beads at last ; each will have 960/3 = 320 beads at last

When T gave some of her beads to V and the number of V's beads was doubled
No. of beads T and V before T gave V some beads :
T : 320 + 160 = 480
V : 320 – 160 = 160

When P gave some of her beads to T and the number of T's beads was doubled.
No. of beads P and T before P gave T some beads :
T : 480 – 240 = 240
P : 320 + 240 = 560

Hi Dharma,
The number 320 is due to 960/3 = 320.
May I ask how to get the number 160?
MTIA.[/quote]We know that P, T and V had 320 at the end.

The strategy is to work backwards.

1. When T gave some of her beads to V and the number of V's beads was doubled

Before V gets some beads from T, V would have had ½ of the no. of beads he had at last.

So, 320 x ½ = 160 (V had 160 heads and received 160 beads from T, to have 320 beads at the end).

T on the other hand, must had (320 + 160) 480 beads at first before giving away 160 to V and up with 320 beads at last.

[Vanilla Cake]

starlight1968sg:

Dharma:

[quote=\"starlight1968sg\"]Hi mathsguru and others,

I need some help:
(1) P, T and V have 960 beads altogether. P gave some of her beads to T and the number of T's beads was doubled. Then T gave some of her beads to V and the number of V's beads was doubled. If the 3 girls had the same number of beads in the end, how many beads did P have at first.
(ans: 560)
How to solve it?

MTIA.

Since P, T and V had the same no. of beads at last ; each will have 960/3 = 320 beads at last

When T gave some of her beads to V and the number of V's beads was doubled
No. of beads T and V before T gave V some beads :
T : 320 + 160 = 480
V : 320 – 160 = 160

When P gave some of her beads to T and the number of T's beads was doubled.
No. of beads P and T before P gave T some beads :
T : 480 – 240 = 240
P : 320 + 240 = 560

Hi Dharma,
The number 320 is due to 960/3 = 320.
May I ask how to get the number 160?
MTIA.[/quote]Hi starlight1968sg,
Your question is from CHIJ 2009 P5 SA1 Paper 2 Q18. VC's younger sister had done this paper before and pls see whether you are able to understand her workings while waiting for Dharma's explanation.

Workings done by VC's P5 younger sister
Beginning
P-> 1 and 3/4u
T-> 3/4u
V-> 1/2u

Middle
P->u
T-> 1 and 1/2u
V-> 1/2u

End
P->u
T->u
V->u

3u->960
1 and 3/4u->560

Prema had 560 beads at first.

Submitted by VC's mum.