Tutor MathsGuru: Ask me for your burning Maths questions!

[Vanilla Cake]

Q1

Teams X and Y work separately on two different projects.
On sunny days, team X can complete the work in 12 days while team Y needs 15 days.
On rainy days, team X's efficiency decreases by 50% while team Y's efficiency decreases by 25%.
Given that the two teams started and ended the the projects at the same time, how many rainy days are there?

Q2
2004 students arrange themselves in a row.
In the first round of counting, they number themselves
1,2,3,1,2,3,1,2,3,........ from left to right.
In the second round of counting, they number themselves
1,2,3,4,5,1,2,3,4,5,1,2,3,4,5........ from right to left.
Find the number of students whose sum of numbers in the first and second rounds of counting is 5.

Q3
Tom walks up a staircase.
Each time he can either take one step or two steps.
How many ways are there for Tom to walk up a ten-step staircase?

Q4
Two points A and B are 1100 m apart.
Alice and Ben leave point A at the same time and travel to and fro along a straight road between A and B at uniform speeds. Alice and Ben travel at 60 m/min and 160 m/min respectively. They both stop after 40 minutes.
(i) At which meeting are they nearest to point B?
(ii) Find the nearest distance in metre.

Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

Sorry, I could not find the given answers for the above questions. Your effort and time to provide worked solutions for them are appreciated.

[Muffins]

iFruit:

Muffins:

Hi guys, do you have a quick way of doing this???


Thanks!

Grandma sent Johnny some money for his birthday. Johnny spent all of it in five stores. In each store, he spent $1.00 more than half of what he had when he came in. How much money did he get from grandma?

Can be done backwards way..

Amount he had when he came into 5th store = $2 (since 1 dollar more than half is all he has)
Amount he had when he came into 4th store = (2+1)2 = 6 (as he spent $1 more than half in 4th store leaving $1 less than half for 5th store)

Amount he had when he came into 3th store = (6+1)2 = 14

Amount he had when he came into 2th store = (14+1)2 = 30

Amount he had when he came into 1th store = (30+1)2 = 62


Money he got from grandma = $62

HTH

Thanks iFruit, I had gotten it via this method as well, but just wanted to find whether there was a quicker way of doing this

[iFruit]

Vanilla Cake:

Q1

Teams X and Y work separately on two different projects.
On sunny days, team X can complete the work in 12 days while team Y needs 15 days.
On rainy days, team X's efficiency decreases by 50% while team Y's efficiency decreases by 25%.
Given that the two teams started and ended the the projects at the same time, how many rainy days are there?

Amount of work done by Team X on sunny day = 1/12
Amount of work done by Team X on rainy day = 1/2 x 1/12 = 1/24

Amount of work done by Team X on sunny day = 1/15
Amount of work done by Team X on rainy day = 3/4 x 1/15 = 1/20

Suppose it took s sunny days and r rainy days to finish project.

Then

s/12 + r/24 = s/15 + r/20

(2s+r)/24 = (4s+3r)/60 -----> (2s+r)/2 = (4s+3r)/5 --> 10s + 5r = 8s + 6r ---> 2s = r

we also know s/12 + r/24 = 1 (total work)

so s/12 + 2s/24 = 1 --> s= 6

so rainy days = 12

[iFruit]

Vanilla Cake:

Q1



Q2
2004 students arrange themselves in a row.
In the first round of counting, they number themselves
1,2,3,1,2,3,1,2,3,........ from left to right.
In the second round of counting, they number themselves
1,2,3,4,5,1,2,3,4,5,1,2,3,4,5........ from right to left.
Find the number of students whose sum of numbers in the first and second rounds of counting is 5.

because 2004 is divisible by 3, there are 2004/3 = 668 groups three students each.

if you take (5x3) = 15 groups of students ( starting from right most side) the arrangement from left to right and right to left will be as below, with students in bold get a sum count of 5

123123123123123----> left to right
543215432154321----> right to left

because 668 = 133x5 + 3, we will have 133 groups of 15 students in above manner and there will be three students left out with
123
321

so the number of students with sum count of 5 = 133 x 3 = 399

[Vanilla Cake]

Hi iFruit,


Thank you very much for your quick response and helpful solutions.
The questions are from http://www.hci.sg/aphelion/apmops/2007/pdf/English/2004%20English%20IR.pdf but no answer keys are given.

[iFruit]

Vanilla Cake:

Q1


Q3
Tom walks up a staircase.
Each time he can either take one step or two steps.
How many ways are there for Tom to walk up a ten-step staircase?

This is a fibonacci series. It is explained in the math hub olympiad challenge thread.

so the number ways for n steps taken will be in this form.

1 2 3 5 8 13 21 34 55 89 144...

so for 10 steps = 89 ways

[Maths Hub]

Vanilla Cake:

Q2
2004 students arrange themselves in a row.
In the first round of counting, they number themselves
1,2,3,1,2,3,1,2,3,........ from left to right.
In the second round of counting, they number themselves
1,2,3,4,5,1,2,3,4,5,1,2,3,4,5........ from right to left.
Find the number of students whose sum of numbers in the first and second rounds of counting is 5.



Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

Sorry, I could not find the given answers for the above questions. Your effort and time to provide worked solutions for them are appreciated.

The answer is 402.

First, 2004 is divisible by 3, but not by 5 (So we will count from the right)

123123123123123
543215432154321

The last 15 digits yield 3 columns with sum count of 5.
Because 2004 = 133 x 15 + 9, we will have 133 groups of 15 students in which every group has 3 students with sum count of 5.

The remaining 9 students also gives us 3 students with sum count of 5:
123123123
432154321

Hence, the number of students with sum count of 5 = 134 x 3 = 402

[Maths Hub]

Vanilla Cake:

Q4
Two points A and B are 1100 m apart.
Alice and Ben leave point A at the same time and travel to and fro along a straight road between A and B at uniform speeds. Alice and Ben travel at 60 m/min and 160 m/min respectively. They both stop after 40 minutes.
(i) At which meeting are they nearest to point B?
(ii) Find the nearest distance in metre.

Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

(i) 2nd meeting
(ii) 100m from point B

Since both Alice and Ben started at point A, the first time they will meet each other again will be after both of them have traveled 2200m altogether.
eg.
A-------------------->|<-------------(continue)
B--------------------------------------(make a u-turn)

Hence the total length traveled by both of them is 2 x length = 2 x 1100 = 2200m.

The total speed of the two is 60m/min + 160m/min = 220m/min

They will meet each other the first time in 2200/220 = 10 mins

The next time they meet each other (the distance have to be 4x1100m = 4400m, 6 x 1100m = 6600m and 8x1100m= 8800m)

We stop at 8800m because that's the total distance both of them can travel in 40 mins.

The meeting points are listed below:
Alice|Ben|Total
600m|1600m|2200m
1200m|3200m|4400m
1800m|4800m|6600m
2400m|6400m|8800m

Hence, from the table, we can see that at the second meeting, they are nearest from B (100m away).

[Maths Hub]

iFruit:

Vanilla Cake:

Q1


Q3
Tom walks up a staircase.
Each time he can either take one step or two steps.
How many ways are there for Tom to walk up a ten-step staircase?

This is a fibonacci series. It is explained in the math hub olympiad challenge thread.

so the number ways for n steps taken will be in this form.

1 2 3 5 8 13 21 34 55 89 144...

so for 10 steps = 89 ways

If Tom can only take 1 step or 2 steps at a time,
No of ways to climb up a 1-step staircase: 1 way
No of ways to climb up a 2-steps staircase: 1 + 1 = 2 ways
3-steps staircase: 1+2 = 3 ways
4-steps staircase: 2+3 = 5 ways
5-steps staircase: 3+5 = 8 ways
6-steps staircase: 5+8 = 13 ways
7-steps staircase: 8+13 = 21 ways
8-steps staircase: 13+21 = 34 ways
9-steps staircase: 21+34 = 55 ways
10-steps staircase: 34+55 = 89 ways

This is because the boy has only 2 choices at first: either to take 1 or 2 steps case. From 10 steps, if he chooses 1 step, then the no of ways immediately reduced to 9-steps, if he chooses 2 steps, then the no of ways reduced to 8-steps' case. Hope this helps.

This is in fact the fibonacci series, as mentioned by iFruit.

All Parents/Students, you can find more Maths Olympiad Questions at our thread:
http://www.kiasuparents.com/kiasu/forum/viewtopic.php?t=14953&postdays=0&postorder=asc&start=0

[iFruit]

Maths Hub:

Vanilla Cake:

Q2
2004 students arrange themselves in a row.
In the first round of counting, they number themselves
1,2,3,1,2,3,1,2,3,........ from left to right.
In the second round of counting, they number themselves
1,2,3,4,5,1,2,3,4,5,1,2,3,4,5........ from right to left.
Find the number of students whose sum of numbers in the first and second rounds of counting is 5.



Source: Asia Pacific Mathematical Olympiad for Primary Schools 2004.

Sorry, I could not find the given answers for the above questions. Your effort and time to provide worked solutions for them are appreciated.

The answer is 402.

First, 2004 is divisible by 3, but not by 5 (So we will count from the right)

123123123123123
543215432154321

The last 15 digits yield 3 columns with sum count of 5.
Because 2004 = 133 x 15 + 9, we will have 133 groups of 15 students in which every group has 3 students with sum count of 5.

The remaining 9 students also gives us 3 students with sum count of 5:
123123123
432154321

Hence, the number of students with sum count of 5 = 134 x 3 = 402

ah..right