The year-end holiday starts tomorrow!


Here is the next installment for the Maths Olympiad Challenge! Looking forward to your participation! :rahrah:

If this appears as a bonus question in an examination, which will you choose? Option A or Option B?

A) You get 1 bonus point. (Only if more than 50% choose option A)
B) You get 3 bonus points. (Only if less than 50% choose option B)

Looking forward to your participation! :rahrah:

wanting05:

Maths Hub:

[quote=\"wanting05\"]Two different tanks, A and B, were filled with water. If tank A leaked 10 ml of water each hour and tank B leaked 5ml of water each hour, tank A would still have 300ml of water left when tank B became empty. If tank A leaked 5ml of water and tank B leaked 10ml of water each hour, tank A would have 750ml of water left when tank B was empty. How much water was there in tank A at first?


Thanks for the help!

Hi wanting05,

Try model method:
First case:
A [][]+300ml
B []

Second case:
A [] +750ml
B [][]

In both cases, B finished leaking so we must make B the same. So cut the model in first case and we get the following:

First case:
A [][][][]+300ml
B [][]

Second case:
A [] +750ml
B [][]

From here, we deduce the difference in 3 units --> 750 - 300 = 450
1 unit-> 450/3 = 150
150+750 = 900ml

Hope that helps.

Are u using 5ml per unit as a model gauge?[/quote]Hi wanting05,

It does not matter whether it is 5ml and 10ml OR 1ml and 2ml. We use ratio to solve. The question can even use 0.1 ml and 0.2 ml and the answer is the same. Hope this answer your doubts.=)

wanting05:

Two different tanks, A and B, were filled with water. If tank A leaked 10 ml of water each hour and tank B leaked 5ml of water each hour, tank A would still have 300ml of water left when tank B became empty. If tank A leaked 5ml of water and tank B leaked 10ml of water each hour, tank A would have 750ml of water left when tank B was empty. How much water was there in tank A at first?


Thanks for the help!

Hi wanting05,

Try model method:
First case:
A [][]+300ml
B []

Second case:
A [] +750ml
B [][]

In both cases, B finished leaking so we must make B the same. So cut the model in first case and we get the following:

First case:
A [][][][]+300ml
B [][]

Second case:
A [] +750ml
B [][]

From here, we deduce the difference in 3 units --> 750 - 300 = 450
1 unit-> 450/3 = 150
150+750 = 900ml

Hope that helps.

Michaelia0816:

Pls help,


Mr Lee and Mr Chew started travelling in opposite directions from the same point by car. Mr Lee travelled at an average speed of x km/h, while Mr Chew travelled at an average speed which is 4km/h slower than Mr Lee. After 5 hours, they were 880km apart.
(i) Write down an equation, in terms of x, to represent the information given above.
(ii) Find the average speed at which Mr Chew travelled.

Hi Michaelia0816,

i) 5x + 5(x-4) = 880
ii) 5x + 5x -20 = 880
10 x = 900
x = 90
x-4 = 86
ans: 86km/h

Hope that helps.

wanting05:

Please help me. Thanks in advance!

Some children sat for a selection test. There were 15 more boys than girls in the group. 3/4 of the girls and 4/5 of the boys passed the test. There were 2 more girls than boys who failed the test. How many children sat for the selection test?

Let total boys be 5 parts (4 parts pass, 1 part fail) and girls be 4 units (3 units pass, 1 unit fail).
Looking at difference in total boys and girls: 5p-15-->4u

Looking at difference in failures: 1p+2-->1u
Make the 1 u into 4u:
4p+8-->4u

Combine both 5p-15-->4u and 4p+8-->4u
we get 5p-15 --> 4p+8
1 p --> 23
5 p--> 23 X 5 = 115
4 u--> 115 -15 = 100
Total--> 115+100=215

Hope that helps.

Michaelia0816:

I got 2 questions and pls explain the solutions.

Questions:
1)
Solve 3(2x+1/-5) more or equal to -9.
Hence, find the largest possible value of x if x is a composite number.
PS. Pls note something in this equation: (2x+1) divide -5 and pls do not be mistaken as 2x+1/-5.
2)
I post it on twitter and the link is below:

t.co/xNlg6oLChI

PS. Pls help me with part (b) and (c) only. Do ignore my writings, very ugly lol.

Thank you and that all.

Hi Michaelia0816,

This is secondary school work right?

For question 1:
3(2x+1/-5) >= -9
(2x+1/-5) >= -3
2x+1 <= 15 (we flip the sign as we multiply by a negative number, you should have covered the concept in school)
2x<=14
x<=7
So the biggest composite number less than 7 is 6. (7 is a prime)

for 2)
Your a) should be (8-2)X180 divide by 8 = 135
The one you have is the total sum of angles, not each angle.
b) Each angle in equilateral triangle is 60.
360 - 135 - 60 = 165
c) (n-2)X180/n = 165
180n - 360 = 165n
15n = 360
n = 24

Hope that helps.

Mary Joy:

Hello pls help me with this qn from Nanyang prelims 2013


http://i39.tinypic.com/29205tt.jpg\">

Hi Mary Joy,

First we look at number of coins for A.
31 - 3 - 3 - 7 = 18
18 / 3 = 6
So A has 6 20-cents worth $1.20.

Now we minus off the extra values,
12.20 - 2.60 - 0.90 - 1.20 = $7.50
This $7.50 comprises the 31-7-3-6=15 coins

Using assumption method, assume all are 50-cents:
0.50 X 15 = $7.50 so all the $7.50 must be only 50-cents coins.

We know that B has 6+3= 9 coins and this is made up of the 6 50-cents and 3 coins making up the 90-cents but we are not interested in the composition of 90-cents here.
Therefore, C has 9 + 7 = 16 and this is made up of the 9 50-cents coins and 7 coins making up the 260-cents but again, we are not interested in the composition of 260 cents here.

So answer is 9X0.50+2.60= $7.10.

Hope that helps.

Mary Joy:

Hi pls help me with this qn from Nanyang prelims 2013


http://i44.tinypic.com/etxfl3.jpg\">

Hi Mary Joy,

We get rid of the volume of water from the excess heights.

4960 - 8X120 - 4X50 = 3800 cm^3

Now we find the combined surface area then we can find the height of L.
3800 / (120+50+80) = 15.2 cm

Volume of water in L in the end -> 15.2 X 80 = 1216 cm^3
Volume of water in J in the end -> (15.2+8) X 120 = 2784 cm^3
2784 - 1216 = 1568 cm^3

Hope that helps.

thehybrid:

hi, could someone help me with this question?

A bus set off from the airport at 10.15 am and took 3 hours to reach the hotel.
A car set off from the airport at 10.30 am and took 2 hours to reach the hotel.
At what time did the car catch up with the bus?

Hi thehybrid,

At 10.30 am,

Bus is already 15/180 = 1/12 of the journey ahead.
Catch up distance-> 1/12 of distance
Speed of bus->1/3 of distance per hour
Speed of car->1/2 of distance per hour
Catch up speed -> 1/2-1/3 = 1/6 per hour

Catch up time-> 1/12 divide by 1/6 = 1/2 h
So 10.30 am --> +1/2 h --> 11.00 am

Hope that helps.

Mary Joy:

Hello everyone,pls help me with this Nanyang prelims 2013


http://i44.tinypic.com/sym7gg.jpg\">

We can divide the shaded area into 3 parts: one small 2cmX2cm square (at bottom left corner) , two rectangles with breadth 2 cm (Length of rectangle is length of unshaded square).

Area of the two rectangles will be 44 cm^2 - 2X2 cm^2= 40 cm^2
Area of 1 small rectangle will be 40 cm^2 / 2 = 20 cm^2
Length of 1 small rectangle = 20 cm^2 / 2 cm = 10 cm

So the length of the unshaded square is 10 cm.

Hope that helps.