Tutor MathsGuru: Ask me for your burning Maths questions!

[Vanilla Cake]

YLH88:

Hi all,

Sorry, to clarify, the below questions are all for P6 level, the P1 and P2 is referring to whether Paper 1 or Paper 2. I have made the amendments for the below. Ya, I too wonder where is Mathsguru .....

Hi YLH88,
Mathsguru's last post in this forum was on http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=6373&postdays=0&postorder=asc&start=1290.Hope she will come back ASAP to provide Maths solutions for us.While waiting for her solution, pls see whether my amateurish solutions can help you or not.

YLH88:

1) Nan Hua P6 SA1 2009 Paper 1 Qn 14
Adrian and Bob started driving at the same time. Bob took 6 hours to complete his journey while Adrian took 9 hours to complete his. After 3 hours, the two men were left with the same distance to complete. What is the ratio of the distance of Bob's journey to the distance of Adrian's journey ?

a) 1:2
b) 1:3
c) 2:3
d) 4:3

Distance travelled by Bob in 1 hour = 1/6 of the journey
Distance travelled by Adrian in 1 hour = 1/9 of the journey

Distance travelled in 3 hours for Bob and Adrian:
Bob = 1/6x3 = 1/2 of the journey
Adrian = 1/9x3 = 1/3 of the journey

Remaining distance to be completed for Bob and Adrian:
Bob = 1/2 = 2/4 of the journey
Adrian = 2/3 of the journey

You have to make both numerators to be same as \"two men were left with the same distance to complete.\"

Ratio of the distance of Bob's journey to the distance of Adrian's journey = 4 : 3
Answer option is (d)
(Pls see notes below, answer option should be 4)

I believe the answer options given in the actual Maths exam script is in numerical option (1,2,3,4) and not alphabetical option (a,b,c,d). Pls clarify.

YLH88:

3) AiTong P6 SA1 2009 Paper 2 Qn 15
Tommy and Wilson jog on a hexagon track (6 equal sides) at 8 am. Tommy starts at point C while Wilson starts at point F. Both jog in the direction as shown by the arrows. Tommy's speed is 1.5 times Wilson's.
a) At which point will they meet ?
b) If they meet at 8.20 am, find Tommy's speed ?

http://www.postimage.org/image.php?v=aVD5NPS

Pls refer to my solution dated http://www.kiasuparents.com/kiasu/forum/viewtopic.php?p=178599#178599 and CoffeeCat's solution dated Thu May 06, 2010 11:00 p.m.Both solutions are self-explanatory and hope this helps while waiting for Mathsguru's solutions.

Thanks.

[Brenda10]

YLH88:

Hi Brenda10,


I understand how you get 4 for the shaded part. But for the unshaded part, why use only 30 ? how about A and B ?

Sorry still a bit confused .. :?

Hi YLH88

May be explain this way, figure means outline.

Although there were 3 circles, A, B & C. However the two small circles are insided the big circle C. the figure refers to the outline of the shape.

Thank you and hope you can understand.

[Vanilla Cake]

YLH88:

2) Nan Hua SA1 2009 P1 Qn 15

In the figure, not drawn to scale, the ratio of the area of Circle A to Circle B to Circle C is 3 : 4 : 10. If 1/3 of B is shaded, what is the ratio of the shaded part to the unshaded part of the figure ?

a) 1:17
b) 2:13
c) 3: 17
d) 4:13

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

Ratio of area
A : B : C
3 : 4 : 10

Since 1/3 of B is shaded, look for the LCM of 3 and 4 which is 12
So, multiply for B ->4x3 = 12.Do the same for A and C.

A : B : C
9 : 12 : 30

1/3 of B is shaded = shaded part of the figure ie 1/3x12 = 4
See the picture to visualize. Got it so far?

Whole figure is C itself ie 30.
Remaining part of the figure (after deducting the shaded portion)= Unshaded part of the figure
-> 30-4 = 26

Finally, ratio of the shaded part to the unshaded part of the figure = 4 :26
= 2: 13

Answer option is (b)

[YLH88]

Hi Brenda10,


Thank you so much!! Now I understand liao ..

Hi VC,

Thank you so much for the solution!! You are right, the options given in the questions are in 1, 2, 3, 4 Just that I wrote in (a, b, c, d) to easy differentiate between the question and options You are so well-informed of where the answers to all the related questions are !!
[quote]I believe the answer options given in the actual Maths exam script is in numerical option (1,2,3,4) and not alphabetical option (a,b,c,d). Pls clarify.[/quote]

[Herbie]

Hi, I have one qn which needs help?




Paya Lebar MGS, Paper 2, Q15


5x5 = 25
5x5x5=125
5x5x5x5=625
5x5x5x5x5=3125

(a) What is the sum of the last three digits in 5 to the power of 15?
(b) What is the sum of the last four digits in 5 to the power of 210?

Can someone provide us with the solution?

Many thanks!

[Maths Monster]

delete

[CoffeeCat]

Herbie:

Hi, I have one qn which needs help?




Paya Lebar MGS, Paper 2, Q15


5x5 = 25
5x5x5=125
5x5x5x5=625
5x5x5x5x5=3125

(a) What is the sum of the last three digits in 5 to the power of 15?
(b) What is the sum of the last four digits in 5 to the power of 210?

Notice that the last 3 digits of 3125 is 125, which corresponds to the 125 from two lines before. The cycle will repeat itself 125, 625, 125,625...
As for the pattern of the last 4 digits more calculations is needed
just remember to work with the last 4 digits and discard the rest and u will find the pattern.

[corneyAmber]

Herbie:

Hi, I have one qn which needs help?

Paya Lebar MGS, Paper 2, Q15
5x5 = 25
5x5x5=125
5x5x5x5=625
5x5x5x5x5=3125

(a) What is the sum of the last three digits in 5 to the power of 15?
(b) What is the sum of the last four digits in 5 to the power of 210?

Can someone provide us with the solution?

Many thanks!

(a) odd number power => it ends with 125
so power of 15 is odd number => it will end with 125,
Hence sum of last 3 digits = 1+2+5 = 8

I need to mull over how to present (b).

[Muffins]

Brenda10:

Muffins:

Hi Brenda10, your daughter just managed to summarise my workings into a few steps! :shock: How old is your DD???

Thank you Muffins.

She is 11 years old.

:shock: :shock: Very young and VERY BRIGHT!!!

[ruyu]

Vanilla Cake:

YLH88:

6) Mark can climb uphill at an average speed of 2km/h and go downhill at an average speed of 6km/h. Going first uphill and then downhill, without stopping and using the same route, what will his average speed be for the entire trip ?

Assume the distance of the route be unit.
Total time taken = unit/2 + unit/6 = 4units/6 = 2units/3
Distance = unit
Average speed = unit÷2 units/3 = unitx3/2units = 3/2 = 1.5 km/h

This is similar to NMOS 2009 Prelim round Q3. Where is this question taken from?

dont understand :? care to explain?