Tutor MathsGuru: Ask me for your burning Maths questions!

[Vanilla Cake]

super star:

There were 500 beads in 4 jars-PQR&S.If the number of beads inP was halved,the number of beads in Q was increased by 18,the number of beads in R was tripled &the number of beads in S was decreasede by 24, the four jars would have the same number of beads .what is the difference in the number of beads between jars P &Q?

Working backwards
In the end
P: u
Q: u
R: u
S: u

At first
P: 2u
Q: u-18
R: u/3
S: u+24

2u+u-18+u/3+u+24=500
4u+u/3=494
12u+u=1452
13u=1482
u=114

Difference ->2u-( u-18 )=2u-u+18=u+18=114+18=132
Difference in the number of beads between jars P &Q = 132

[Vanilla Cake]

super star:

A box contain some 50 cent coins,some 20 cent coins & some 10 cent coins in the ratio of 3:2:1 respectively.3/4 of the 50 cent coins were taken & replaced by the same number of 10 cent coins.Then180 twenty cent coins were taken out & replaced by same number of ten cent coins.In the end,the ratio of the 50 centcoins ,twenty cent coins & ten cent coins become 3:4:17 respectively.

(a)what is the total number of coins taken out of the box?
(b)what is the total value of coins taken out of the box?

Before
50¢ -> 3u
20¢ -> 2u
10¢ -> u

After
50¢ -> 0.75u (3p)
20¢ -> 2u-180 (4p)
10¢ -> 3.25u+180 (17p)

0.75u -> 3p
u -> 4p

2u-180=u
u=180
2.25u = 2.25x180=405
405+180=585
(a) Total number of coins taken out of the box = 585

405x$0.50+180x$0.20
$202.50+$36=$238.50

(b) Total value of coins taken out of the box = $238.50

Did this in a rush (otherwise I will be late for school), pls help to check the calculations.

[Dharma]

YLH88:

Hi all,


I need help with the below questions :? :

1) ABCD is a rectangle. X is a point on AB. XB is 3/4 of AB.
The area of AXCD is 90 cm2. Find the area of XBC.

http://postimage.org/image/1c1ogtk3o/


Thank you very much!

http://postimage.org/image/28adtcfgk/

[Dharma]

Vanilla Cake:

super star:

A box contain some 50 cent coins,some 20 cent coins & some 10 cent coins in the ratio of 3:2:1 respectively.3/4 of the 50 cent coins were taken & replaced by the same number of 10 cent coins.Then180 twenty cent coins were taken out & replaced by same number of ten cent coins.In the end,the ratio of the 50 centcoins ,twenty cent coins & ten cent coins become 3:4:17 respectively.

(a)what is the total number of coins taken out of the box?
(b)what is the total value of coins taken out of the box?

Before
50¢ -> 3u
20¢ -> 2u
10¢ -> u

After
50¢ -> 0.75u (3p)
20¢ -> 2u-180 (4p)
10¢ -> 3.25u+180 (17p)

0.75u -> 3p
u -> 4p

2u-180=u
u=180
2.25u = 2.25x180=405
405+180=585
(a) Total number of coins taken out of the box = 585

405x$0.50+180x$0.20
$202.50+$36=$238.50

(b) Total value of coins taken out of the box = $238.50

Did this in a rush (otherwise I will be late for school), pls help to check the calculations.

50 cents : 60u – 45u = 15u
20 cents : 40u - 180
10 cents : 20u + 45u + 180 = 65u + 180

3p = 15u
4p = 20u

40u – 180 = 20u
20u = 180
1u = 9

a)\tTotal no. of coins taken out = 45u + 180 = 45(9) + 180 = 585
b)\tValue of coins taken out = 405(50) + 180(20) = = 23850 cents = $238.50

[YLH88]

Hi VC,


Thank you for the link and solution. Yes, these are from Catholic High Prelim 2 and they are really challenging ....

Hi Dharma,

I'm not able to open up your attachment for Qn 1

[laikiasu]

Hi need some advise.


My boy is in P1. Generally, his calcutions are correct but I find that he is a tat too slow to find the answers.
Is there any way to increase the speed?
Is there any courses or tuition centers that can help to increase his number sense?
More for P2 since cannot do much now, nearing to exam time.

[Dharma]

YLH88:

Hi VC,


Thank you for the link and solution. Yes, these are from Catholic High Prelim 2 and they are really challenging ....

Hi Dharma,

I'm not able to open up your attachment for Qn 1

AX = 1u
XB = 3u
For simplicity; I assume the height AD = BC = 1cm

Unshaded area = 1u + ½(3u) = 90cm2
2.5u = 90cm2
1u = 36cm2

Shaded area XBC = ½ x 3u = 1.5u = 54cm2

[YLH88]

Dharma:

YLH88:

Hi VC,


Thank you for the link and solution. Yes, these are from Catholic High Prelim 2 and they are really challenging ....

Hi Dharma,

I'm not able to open up your attachment for Qn 1

AX = 1u
XB = 3u
For simplicity; I assume the height AD = BC = 1cm

Unshaded area = 1u + ½(3u) = 90cm2
2.5u = 90cm2
1u = 36cm2

Shaded area XBC = ½ x 3u = 1.5u = 54cm2

Hi Dharma,

Thank you!

[Almighty]

kancheongmum:

Vanilla Cake:

[quote=\"Ler\"]Maha Bohi SA2 math 2010, Paper 2

Many thanks for working out most of the solutions. I hv problem understanding but still trying to work out.

Quick one as I need to go to school,my mum will continue for the rest of your problem sums.

Ler:

Question 7 - In the diagram show above, O is the centre of the circle and OAB is a right angle triange. Given that AB is 10cm and the shaded parts of the circle add up to 133 m square. Find the area of the circle.

Pls post the diagram otherwise how can others be expected to solve without reading the diagram? I notice that you have changed your question as the original question stated \"AB is 10m\" and the diagram showed length of AB labeled as 10m. The original question also stated 133 m² and this is where the confusion comes.So, I change all metres and metres square to cm and cm² to make life simple.

Square of length of AO+Square of length of OB=10cmx10cm (http://en.wikipedia.org/wiki/Pythagorean_theorem)
but length of AO=length of OB (radii of circle)
Assume length of AO be n, so
n²+n²=100
2n²=100
n²=50

Area of triangle AOB = 1/2xABxOB=1/2xn²=1/2x50=25 cm²
Since area of the shaded part of the circle was given as 133 cm², area of the circle = (25+133) cm² = 158 cm².

Need to go now and mum will continue the rest.

Hi Vanilla Cake and Ler

This Maha Bodhi prelim 2010 paper is very challenging and have a few very interesting questions. Vanilla Cake I don't think Q7 should be solved this way. Pythagoras' Theorem is not taught in P6 yet.

Area of the triangle = 1/2 x base x height
if the radius is given you can use that to work out the answer. But not given in this case. So look at the triangle in another way. You will note that it is 1/4 of the square inscribe in the circle.

base = 10m height = 10m/2 = 5 m

area of the circle = 133m3 + (1/2 x 10x5) =158m3

I always tell my ds, don't think too complicated when doing the questions. They will not give you anything out of the syllabus. Make use of the basic concept taught.[/quote]Hi,
Can anyone give me the link for
1) Maha Bodhi prelim 2010 paper qt : 7
2) Nanyang 2010 P6 prelim paper?
TIA

[Herbie]

Hi! I hv one qn.

4800 children registered to take part in a fitness test. 3/5 of the children were boxs. On the actual day of the fitness test, some boys did not turn up. Hence the % of the boys who actually took the test was 40% of the total no of children present tat day. At the end of the fitness test the no of girls who pasSed was 20% more than the no of boys who passed. 340 children the testHow many boys turned up for test