Tutor MathsGuru: Ask me for your burning Maths questions!

[Vanilla Cake]

wkong:

The figure is not drawn to scale. It shows a rectangle divided into 6 parts. Each part has a different area.

a) Find the total area of P and Q.
b) Find the perimeter of the figure.

http://www.postimage.org/

Use guess and check,
Area of P = 8 cmx 6 cm = 48 cm²
Area of Q = 5 cm x 4 cm = 20 cm²
(a) Total area of P and Q = (48+20) cm² = 68 cm²

Length of rectangle = (5+6+4) cm = 15 cm
Breadth of rectangle = (8+5) cm = 13 cm
(15x2)+(13x2) = 30+26 = 56 cm
(b) Perimeter of the figure = 56 cm

[youngtay]

kancheongmum Thanks for your help.


Best wishes

[wkong]

Thanks again VC for your help

[Vanilla Cake]

YLH88:

Hi all,

Need help in the below question :

This is an interesting question but nobody has forward his/her solution yet.I am not please with my algebra method and let's hope that others will forward with better methods for sharing and learning.Could you pls advise the source of this question?

VC's mum

YLH88:

Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
Both of them were running in the clockwise direction.
Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
How long did Sam take to run past Harry for the first time ?
(ans : 30 minutes)

http://postimage.org/image/1z70q6hqc/

LCM of 12 and 15 is 60.
Work out 60m as the length (circumference) of the circular track even though 60m is not a realistic number.
Sam's average speed = 60m÷12 = 5 m/min
Harry's average speed = 60m÷15 = 4 m/min

Note that both of them started running at the same time from the opposite ends of the diameter of a circular track means the difference in their distance = ½x60m = 30 m.
For them to meet for the first time, Sam must run more than Harry by 30 m in the same amount of time.

Assume the time taken to be T,
5xT= (4xT)+30
5T=4T+30
T=30

Sam took 30 min to run past Harry for the first time.

[YLH88]

Vanilla Cake:

YLH88:

Hi all,

Need help in the below question :

This is an interesting question but nobody has forward his/her solution yet.I am not please with my algebra method and let's hope that others will forward with better methods for sharing and learning.Could you pls advise the source of this question?

VC's mum

YLH88:

Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
Both of them were running in the clockwise direction.
Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
How long did Sam take to run past Harry for the first time ?
(ans : 30 minutes)

http://postimage.org/image/1z70q6hqc/

LCM of 12 and 15 is 60.
Work out 60m as the length (circumference) of the circular track even though 60m is not a realistic number.
Sam's average speed = 60m÷12 = 5 m/min
Harry's average speed = 60m÷15 = 4 m/min

Note that both of them started running at the same time from the opposite ends of the diameter of a circular track means the difference in their distance = ½x60m = 30 m.
For them to meet for the first time, Sam must run more than Harry by 30 m in the same amount of time.

Assume the time taken to be T,
5xT= (4xT)+30
5T=4T+30
T=30

Sam took 30 min to run past Harry for the first time.

Hi VC's mum,

Thank you for the solution. But I find it a bit difficult to explain to my DS on this.

This question is taken from the A star booklet on Rate and Speed.

[kancheongmum]

YLH88:

Hi all,


Need help in the below question :

Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
Both of them were running in the clockwise direction.
Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
How long did Sam take to run past Harry for the first time ?
(ans : 30 minutes)

http://postimage.org/image/1z70q6hqc/

Thank you!

Hi YLH88

This is catching up question without the distance given. I too had a hard time explaining to my DS. Let me try and see it helps.

Sam took 12 min to complete one round
Harry took 15 min to complete one round
So Sam is for sure faster than Harry
We name Sam starting point S and Harry starting point H
We imagine both Sam and Harry start from point S. Harry start running first and when he reached point H, then Sam start running. So Sam has half the circular track to catch up.

Time ratio --- Sam: Harry ---- 12 : 15 ----4:5
Same distance, constant speed, time ratio is inversely proportional to distance ratio and speed ratio
distance ratio --- 5:4
5u - 4u = 1u ---- 1/2 of the track
When Sam past Harry the 1st time Sam would have run

1/2 x 5 = 5/2 or 2 1/2 of the track
He would take --- 5/2 x 12 min = 30 mins

You can also use Harry to get the answer
When Sam past Harry the lst time Harry would have run

1/2 x 4 = 2 round
He would take --- 2 x 15 min = 30 mins

hope this help

[iFruit]

YLH88:

Hi all,


Need help in the below question :

Sam and Harry started running at the same time from the opposite ends of the diameter of a circular track.
Both of them were running in the clockwise direction.
Sam took 12 minutes to run round the track once while Harry took 15 minutes to do the same.
How long did Sam take to run past Harry for the first time ?
(ans : 30 minutes)

http://postimage.org/image/1z70q6hqc/

Thank you!

Sam took 12 mins to run round the track.

So distance run by Sam in 1 min = 1/12 of track.

distance run by Harry in 1min = 1/15 of track.

Distance caught up by Sam in 1 min = (1/12 - 1/15) of track = 1/60 of track

Time taken by sam to cover the distance between them, which is 1/2 the track or 30/60 of track = 30mins.

[zyberk]

Hi all,


Appreciate if anyone can help with the steps to achieve the answer to this math number problem. Thanks

| Y |
| 16 | 5 |
| 28 | 10 | 3 |
| 50 | 20 | 8 | 3 |

The answer given is Y=23

Thanks again.

[CJM]

Dave had 59 toy cars and toy soldiers. After giving away 1/3 of his toy cars and 4 of his toy soldiers, he had an equal number of toy cars and toy soldiers. How many toy soldiers did he have at first ?

[small]

CJM:

Dave had 59 toy cars and toy soldiers. After giving away 1/3 of his toy cars and 4 of his toy soldiers, he had an equal number of toy cars and toy soldiers. How many toy soldiers did he have at first ?

Hi CJM,

Here is the solution, hope it helps.

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