Tutor MathsGuru: Ask me for your burning Maths questions!

[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.

[Dharma]

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.

Chocolates\t: 1u – 25 = 3p
Lollipops \t: 4u – 177 = 1p

12u – 531 = 1u - 25
11u = 531 – 25 = 506
1u = 46

Total no. of chocs & lollipops at first = 5u = 46 x 5 = 230

[Vanilla Cake]

Brenda10:

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?

This is a 4-mark Q11 from CHIJ P5 2009 SA1 Maths paper. You need to wait for Mathsguru and others to provide model solutions. Here's what VC's P5 younger sister did to solve. Sorry, she didn't use models.

Before
L : C
4 : 1

After
L : C
1: 3

(4u-177)/(1u-25)=1/3
3(4u-177)=1(1u-25)
12u-531=1u-25
11u=506
5u=230

Mrs Sham had 230 lollipops and chocolate bars at first.

Submitted by VC's mum.

[Brenda10]

Thank you Dharma and Vanilla Cake for your kind help while waiting for mathsguru’s model. I really feel headache to go through all these SA1 papers to prepare for coming SA1 Exam.

[chrisho]

2 simple questions that need to find better way to explain to my son. d


1. James had twice as many marbles as Bob. After James lost 240 marbles, Bob had 3 times as many marbles as James. How many more marbles did James have Than Bob at first.

2.Peter is 8 years old and his mother is 30 years older than he. how old will Peter be when his mother is thrice his age?

TQ