Tutor MathsGuru: Ask me for your burning Maths questions!

[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?

[Brenda10]

Muffins:

Brenda10:

[quote=\"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!!! :)[/quote]Hi Muffins,

Thank you for thinking highly upon her. :oops:


In fact, DD is just average. She has done more than 50 ratio/percentage questions during the school holiday and may be the numerous practices really help. The questions in this forum really provide her the good opportunity on trial and error. The more important thing is that she could also learn from all of you as well.

[Vanilla Cake]

ruyu:

dont understand :? care to explain?

Thks for pointing out, I have amended the workings.
Average speed should be 3 km/h

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

[Dharma]

Vanilla Cake:

ruyu:

dont understand :? care to explain?

Thks for pointing out, I have amended the workings.
Average speed should be 3 km/h

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

Time ratio: Uphill : Downhill = 3 : 1
Ave speed = ((2 x 3) + (6 x 1))km/(3 +1)h = 12km/4hr = 3km/h

[YLH88]

[quote]Vanilla Cake wrote:

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

Dharma wrote :
Time ratio: Uphill : Downhill = 3 : 1
Ave speed = ((2 x 3) + (6 x 1))km/(3 +1)h = 12km/4hr = 3km/h[/quote]Hi VC, Dharma,

Why do you change the time as 12/2 and 12/6 ?

For speed questions, when do we use the inverse proportion method ? any standard method we can follow ? Like the question on NanHua P6 SA1 2009 Paper 1 Qn 14, VC uses the rate method. My DS find it very confusing and he is very confused when to use the inverse method and when to use other method for all the speed questions. Can help to advise ? :?

Thank you!
[quote]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 ?

Vanilla Cake wrote :
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)
[/quote]

[Dharma]

YLH88:

[quote]Hi VC, Dharma,


Why do you change the time as 12/2 and 12/6 ?

For speed questions, when do we use the inverse proportion method ? any standard method we can follow ? Like the question on NanHua P6 SA1 2009 Paper 1 Qn 14, VC uses the rate method. My DS find it very confusing and he is very confused when to use the inverse method and when to use other method for all the speed questions. Can help to advise ? :?

Thank you!
[quote]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 ?

Vanilla Cake wrote :
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)

[/quote][/quote]Hi YLH88,

For using ratios, just remember
Distance = Speed x Time
Don’t worry too much about inverse method (there isn’t such a thing) ….just understand the concept (i.e. relationship between distance, speed and time)

For the above qn, we know the distance is the same after Adrian and Bob travelled for 3 hours. So, the speed ratios and time ratios of Bob and Adrian are inversely proportional to each other.

For constant distance (to complete journey after 3 hours)
Time ratio => Bob : Adrian = (6-3) : (9-3) = 3 : 6 = 1 : 2
Speed ratio => Bob : Adrian = 2 : 1

Now, total distance travelled by
Bob : 2u x 6 = 12u
Adrian : 1u x 9 = 9u
Distance ratio => Bob: Adrian = 4 : 3

[Dharma]

YLH88:

[quote]Vanilla Cake wrote:

LCM of 2 and 6 is 12

Time taken to climb uphill @ 2 km/h = 12/2 = 6h
Time taken to go downhill @ 6 km/h = 12/6 = 2h
Total time taken = 8h
Total distance covered = 12x2 = 24 km
Average speed for the entire trip = 24km/8h = 3 km/h

Dharma wrote :
Time ratio: Uphill : Downhill = 3 : 1
Ave speed = ((2 x 3) + (6 x 1))km/(3 +1)h = 12km/4hr = 3km/h

Hi VC, Dharma,

Why do you change the time as 12/2 and 12/6 ?

[/quote][/quote]

Hi YLH88,

Hope it is clearer now.

For this qn, the distance uphill/downhill is constant.

So, speed ratio=> Uphill : Downhill = 1 : 3
Time ratio => Uphill : Downhill = 3 :1

Total Distance
= Distance Uphill + Distance Downhill (#Distance Uphill = Distance downhill)
= (3u)hrs x 2km/h + (1u)hr x 6km/h
= (12u) km

Total time
= (3u + 1u)hrs
= (4u)hrs

Average Speed = Total Distance  Total time = (12u/4u) km/h = 3 km/h