Tutor MathsGuru: Ask me for your burning Maths questions!

[corvette]

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[corneyAmber]

CJM:

Hi all,


need ur help on this question : -

Mr Lim sold a total of 96 papayas and watermelons for $560.
He sold each papaya for $3 and each watermelon for $7.
How many watermelons did Mr Lim sell ?

P = 96-W
3(96-W) + 7W = 560
4W = 560-288 = 272
W = 68

Mr Lim sold 68 watermelons.

[tianzhu]

little_twin_stars:

Hi there...


There is this question:

At 9.20 am, Terence left Town X for Town Y at a constant speed. 1 hour larer, Gary started travelling from Town X on the same road at a constant speed. Gary overtook Terence at 12.20 pm. The speed which Gary was travelling at was 40km/h faster than Terence. Find the speed that Terence was travelling at.

The solution given is as follows:

Time (Terence): Time (Gray) -> 3 : 2
Speed (Terence = D/3
Speed (Gary) = D/2
D/2-D/3 = 1/6 D
1/6 D = 40 km
D = 40km x 6
=240 km/h
Average Speed (T) = 240/3
= 80 km/h


However, I worked it out another way. I would like to know if it is also correct.

Time taked for Gary to overtake Terence -> 2 h
The extra distance that Gary could cover in the 2 h -> 40 x 2
= 80 km
Speed of Terence -> D/T = 80/1
= 80 km/h

Thanks in advance

Hi little_twin_stars

It’s good to have a line diagram when you’re doing speed questions.

You’re right.

Best wishes

[Vanilla Cake]

ksi:

CJM:

Hi all,


need ur help on this question : -

Mr Lim sold a total of 96 papayas and watermelons for $560.
He sold each papaya for $3 and each watermelon for $7.
How many watermelons did Mr Lim sell ?

P = 96-W
3(96-W) + 7W = 560
4W = 560-288 = 272
W = 68

Mr Lim sold 68 watermelons.

Hi CJM,
You may wish to refer to this http://p4mathsquestions.blogspot.com/2009/08/maths-decimals-solutions_21.html and apply the same concept to solve your question.

Selling price of 96 payayas -> 96x$3 = $288
Total amount of money collected for selling 96 papayas and watermelons = $560
Difference (total)= $(560-288) = $272
Difference (1 watermelon and 1 payaya) = $(7-3) = $4
1 payaya + $4 -> 1 watermelon
$272÷4= 68 (This step is enough and follow by the answer statement)
68 payayas + $272 -> 68 watermelons (Additional step to aid your understanding and is not necessary in actual workings.)
96-68 = 28 payayas (Additional step if the question asks for the number of payayas sold.)

Number of watermelons sold by Mr Lim = 68.

Check
Amount collected for selling 68 watermelons @ $7/each -> 68x$7=$476
Amount collected for selling 28 payayas @ $3/each -> 28x$3 = $84
Total amount collected -> $(476+84) = $560

I think this should be a P4 question and there are more examples over http://p4mathsquestions.blogspot.com/.Good luck.
Thanks

PS : Sorry for the above long-winded explanation.

VC's mum

[Vanilla Cake]

little_twin_stars:

At 9.20 am, Terence left Town X for Town Y at a constant speed. 1 hour larer, Gary started travelling from Town X on the same road at a constant speed. Gary overtook Terence at 12.20 pm. The speed which Gary was travelling at was 40km/h faster than Terence. Find the speed that Terence was travelling at.


The solution given is as follows:

Time (Terence): Time (Gray) -> 3 : 2
Speed (Terence = D/3
Speed (Gary) = D/2
D/2-D/3 = 1/6 D
1/6 D = 40 km
D = 40km x 6
=240 km/h
Average Speed (T) = 240/3
= 80 km/h

However, I worked it out another way. I would like to know if it is also correct.

Time taked for Gary to overtake Terence -> 2 h
The extra distance that Gary could cover in the 2 h -> 40 x 2
= 80 km
Speed of Terence -> D/T = 80/1
= 80 km/h

Another way to solve:
Constant distance -> Ratio of constant/average/uniform speed is inversely proportional to ratio of time.

From 12.20 pm to 9.20 am -> 3 hours
Ratio of time
Terence : Gary
3 : 2

Ratio of constant speed
Terence : Gary
2 : 3

Difference in constant speed ->3u-2u = u
u-> 40 km/h
Constant speed of Terence -> 2u -> 2x40km/h = 80 km/h.

VC's mum

[Dharma]

beauty queen:

There are more pupils in School A than School B. 30% of the pupils in School A is 45 more than 40% of the pupils in School B, If 10% of the pupils in School A leaves to join School B, there will be 200 more pupils in School A than School B.

(a) How many pupils are there in School B?

(b) How many percent less pupils are there in School B than School A? Leave your answer as fraction in the simplest form

please help

3/10 A = 4/10 B + 45
A = 4/3 B + 150

School A: 40u + 150
School B : 30u

If10% of pupils in School A join School B

[(40u + 150) – (4u + 15)] – [ 30u + 4u +15] = 200
36u + 135 – 34u – 15 = 200
2u = 80
1u = 40
a)\tNo. of pupils in School B = 30u = 30 x 40 = 1200
b)\t(10u + 150)/(40u + 150) x 100%
550/1750 x 100% = 31 3/7%

[little_twin_stars]

Thanks for the help

[Yu Xuan]

beauty queen:

There are more pupils in School A than School B. 30% of the pupils in School A is 45 more than 40% of the pupils in School B, If 10% of the pupils in School A leaves to join School B, there will be 200 more pupils in School A than School B.

(a) How many pupils are there in School B?

(b) How many percent less pupils are there in School B than School A? Leave your answer as fraction in the simplest form

please help

This is my method of solving;

3/10 A --> 4/10 B + 45
1/10 A --> 2/15 B + 15 [ this is 10% of A]
9/10 A --> 1 1/5 B + 135 [ this is 90% of A]

a)Hence if 10% of School A joins School B;
1 1/5B + 135 - (1B + 2/15B + 15) = 200
1 1/5B + 135 - 1 2/15B - 15 = 200
1/15B = 80
1B = 80 / 1/15 = 1200

b) 1/10A --> 2/15 (1200) + 15 = 175
10 / 10 A --> 1A --> 175 * 10 = 1750
1750 - 1200 = 550
550 / 1750 * 100% = 31 3/7 %

[Herbie]

Hi i hv one qn.

Mrs lim had twice as many ten dollar notes as two dollar notes at first. She bought a digital camera for $288 and pdid for it with thrice as many two dollar notes. Then she realised tat she had 4 times as many ten- dollar notes left.
how many pieces of two dollar notes did mrs lim have at firsT? How much money had mrs lim left after buying the digital camera?Tx

[Yu Xuan]

Herbie:

Hi i hv one qn.

Mrs lim had twice as many ten dollar notes as two dollar notes at first. She bought a digital camera for $288 and pdid for it with thrice as many two dollar notes. Then she realised tat she had 4 times as many ten- dollar notes left.
how many pieces of two dollar notes did mrs lim have at firsT? How much money had mrs lim left after buying the digital camera?Tx

D.Camera;
3u * 2 = 6u
1u * 10 = 10u
6u + 10 u --> 288
1u --> 288/16 = 18
$2--> 3 * 18 = 54
$10 --> 1 * 18 = 18

Initially
$10 --> 2p
$ 2 --> 1p

After spending on D.Camera;
$10 --> 2p - 18
$2 --> 1p - 54

2p - 18 = 4 (1p - 54)
2 p = 198
1 p = 99 ($2 at first)

b) $10 --> (2*99 - 18 ) * $10 = $1800
$2 --> (99-54)*$2 = $90
Total --> $1800 + $90 = $1890

Are these the answers?