Tutor MathsGuru: Ask me for your burning Maths questions!

[Almighty]

mathsguru:

Almighty:

[quote=\"mathsguru\"]
Hi math monster,
I used yr algebra method.I am stuck..Pl correct me.. This is my solution:
Berfore:
A : 3m + 55
B : 5m - 55
After A gave B $55/-
A : B = 3m : 5m
B gives 20 % = (53 - 55) X 20% + 10
= m-1

So, A gets = (3m +55) + m-1
= 4m + 54
B left with = (5m -55) - m -1
= 4m -56
HOw can 4m +54 = 4m -56 n solve for m??Where did i go wrong??

If MM not active,Can MAth guru help me out? Pl.refer to P :165 for the method taught by MM.
I am not comfortable using Model...So, Pl.advise!!!

Hi Almighty,

Think u misread the question. It says \"If Brian gave Andrew 20% of THIS money\" not \"his money\". I think it's referring to 20% of $55 and not his entire sum of money.


MathsGuru

[/quote]Thanks MAthguru.I got it now!!!!

Tang, Here is the algebra solution ...
\tAt First :\t\t\t
\t\t\t\t
\tA = 3m+ 55 \t\t\t
\tB = 5m - 55\t\t\t
\t\t\t\t
\tAfter \t\t\t
\t 20% of 55 = 11\t\t\t
\t\t\t\t
\t5m - 55 -11 -10 = 5m -76\t\t\t
\t3m + 55 + 11 +10 = 3m + 76\t\t\t
\t\t\t\t
\t5m - 76 + 108 = 3m + 76\t\t\t
\t5m + 32 = 3m + 76\t\t\t
\t5m = 3m + 76 - 32\t\t\t
\t2m = 44\t\t\t
\tm = 22\t\t\t
So,\tA + B = 3m +55 + 5m -55 \t\t\t
\t = 8M\t\t\t
\t8m = 22 X 8 = 176

[Almighty]

Edmund drove Andrew from Jurong Point to Woodlands at a constant

speed of 75 Km/h. Andrew rested for 20 mins before walking 3km home at a speed of
6km/h.Edmund took 15 minutes to return to Jurong Point to pick Ben up. HE then drove
to Orchard at 75Km/h.The three people reached their destinations at the same time.
What was the distance Edmund drove from Jurong Point to Orchard?

Almighty:

singmathstutor:

We should be looking at the question in perspective from the time Andrew alighted at Woodlands.

In Andrew's case, to reach his destination (home) after alighting at Woodlands,
Total time he needs is 20min for resting and
30min for walking home at a speed of 6km/h in a distance of 3km.
T= 3km/6km/h
=1/2h
=30min

Therefore, the 20min+30min=50min is as such.

In Ben and Edmund's case, we should start comparing the time just when Edmund has started his return trip to JP to pick Ben. Since 15min is used to make that trip, and both the boys would have reached Orchard at the same as Andrew reaching home, therefore time left for their journey to Orchard is 50min-15min=35min

That 15min is only applicable to Ben and Edmund's journey, so you should not have added 15min to Andrew's return trip.

Hope this would have helped to clear your doubts

Thankyou Singmathstutor
Still trying to understand. Will come back to u if i dont get it...
:roll:

We should not take 15 mins as the time taken by Edmund to drop Andrew to Woodlands because:
Edmund's speed is only given and not the time taken .On his return from Woodlands to Jurong, his time is only given which means the speed need not be the same as 75km/h.
Am i right?pl.clarify :?

[singmathstutor]

Hi Almighty,


Edmund did not use 15min to alight Andrew, he used that time to get back to JP to fetch Ben. I think you have interpreted wrongly here.

[Almighty]

Thankyou singmathstutor…I got it!!!

[super star]

Hi Mathsguru

thanks for ur help.I have another problem pls help to solve this
Aurdrey & belle have some money each.If audrey spends $18 &belle spends $24 each day,audrey will still have $25 left when belle has spent all her money.If audrey spends $13 &bellespends $30each day,audrey will have $139 left when belle has spent all her money.How much money do they have altoghter?

mathsguru:

super star:

Maths guru
please help me in this problem.

Alan,Ben&charles had 864 trading cards.Ben won some of the cards from alan &as a result,ben's cards increased by 50%.charles then won some cards from ben & charles'cards increased by 40%.finally charles lost some of his cards to alan & alan's cards increased by 20%.In the end ,they realised that they each had an equal number of cards.how many cards did alan have at first?

Hi Super Star,

This question was posted quite some time back by another forum member. Here's my solution anyway...trick here is to work backwards systematically.

Cheers,
MathsGuru

http://www.postimage.org/image.php?v=gxHh7s0

[singmathstutor]

Almighty:

Thankyou singmathstutor..I got it!!!!

You're welcome, Almighty

[Daddy]

Hi


Need help on ratio question.

1) The ratio of the number of girls in 2A to the number of girls in 2B is 2:3.
The ratio of the number of boys in 2A to the number of boys in 2B is 5:3.
a) There are 5 more girls in 2B than 2A. Find the total number of girls in the 2 classes.
b) There are 6 more boys in 2A than 2B. Find the total enrolment of the pupils in the 2 classes.


2) Yvonne is 4 years older than Ken. The ratio of Yvonne’s age to Kenny’s age is 3:2. In how many years’ time will the ratio of Yvonne’s age to Kenny’s age be 9:7.

Thanks…

[Jolyn]

Hi


I have a question which needs your help:

1. 4 guavas and 5 managoes cost $12.50. A guava and a manago cost $3. Find the cost of 2 mangoes.

Thks

[singmathstutor]

Jolyn:

Hi


I have a question which needs your help:

1. 4 guavas and 5 managoes cost $12.50. A guava and a manago cost $3. Find the cost of 2 mangoes.

Thks

since a guava and a mango cost $3
4 guavas and 4 mangoes will cost $3 x 4 = $12
therefore since 4 guavas and 5 mangoes cost $12.50,
1 mango will cost $12.50-$12 = $0.50
2 mangoes cost $1

Hope this helps

[Dharma]

Daddy:

Hi


Need help on ratio question.

1) The ratio of the number of girls in 2A to the number of girls in 2B is 2:3.
The ratio of the number of boys in 2A to the number of boys in 2B is 5:3.
a) There are 5 more girls in 2B than 2A. Find the total number of girls in the 2 classes.
b) There are 6 more boys in 2A than 2B. Find the total enrolment of the pupils in the 2 classes.

a)\t3u – 2u = 1u = 5
No. of girls in the 2 classes = 5u = 5 x 5 = 25

b)\t5p – 3p = 2p = 6
1p = 3
Total enrolment = 5u + 8p = 25 + 8(3) = 49

Daddy:

2) Yvonne is 4 years older than Ken. The ratio of Yvonne's age to Kenny's age is 3:2. In how many years' time will the ratio of Yvonne's age to Kenny's age be 9:7.

Thanks..

3u – 2u = 1u = 4
Yvonne’s age now = 3 x 4 = 12
Kenny’s age now = 2 x 4 = 8

9p – 7p = 2p = 4
1p = 2

Yvonne’s age = 9 x 2 = 18

No of years taken for the ratio of Yvonne’s age to Kenny’s age be 9 : 7
= 18 – 12
= 6