Tutor MathsGuru: Ask me for your burning Maths questions!

[Muffins]

Hi Vanilla Cake, Tang has addressed this question in the PSLE-Math thread. The answer is: A – 25 : 75 = 1 : 3


B – 75 : 25 = 3 : 1

A + 3 B –> 250 + 600 = 850

3 A + B –> 200 + 150 = 350

4 A + 4 B –> 850 + 350 = 1200
1 A + 1 B –> 1200 / 4 = 300

2 B –> 850 - 300 = 550
B –> 550 / 2 = 275

2 A –> 350 - 300 = 50
A –> 50 / 2 = 25

250 - 25 = 225 red

200 - 25 x 3 = 125 blue (or 275 - 150 = 125 blue)

So move 225 red marbles and 125 blue marbles from Container A to Container B.

[CoffeeCat]

i think vanilla is hoping for mathsguru to come up with an alternative solution based on models.

[Muffins]

Oh... :imsorry:

[Vanilla Cake]

CoffeeCat:

i think vanilla is hoping for mathsguru to come up with an alternative solution based on models.

Yes, CoffeeCat. You are right, I am looking for alternative solution based on models from Mathsguru besides the solution using simultaneous equations as posted by Tang.

Muffins:

Oh... :imsorry:

It's ok. . You are cute!

firebird:

1) Victor wants to travel from Town X to Town Y. If he drives at an average speed of 60km/h, he would reach Town Y at 10.30am. If he drives at an average speed of 80km/h, he would reach Town Y at 9.30am.

(a) Find the distance between Town X and Town Y .
(b) At what average speed should he drive if he wants to reach Town Y at 9.30 am?

\"At what average speed should he drive if he wants to reach Town Y at 9.30 am?\", is there a typo error?

(a) Distance between Town X and Town Y is 240 km
Difference in speed = (80-60) km/h = 20 km/h.
60÷20=3
3 x 80 = 240 km

OR

use algebra, let number of hours taken to travel at 80 km/h be n.
80n = 60(n+1)
80n = 60n + 60
20n = 60
n = 3

80x3 = 240 km

OR

Wait for Mathguru's solutions to be posted.

2) A piggy bank contained some fifty-cents coins and twenty-cents coins in the ratio of 5:2. When 36 fifty-cents coins were taken out and replaced by the same number of twenty-cents coins, the ratio of the number of fifty-cents coins to twenty-cents coins became 1:4.

a) What was the value of the twenty-cents coins at first?
b) How much money was in the piggy bank at first?

(a)
Before
50¢ coins : 20¢ coins
5 : 2
25 : 10

After
50¢ coins : 20¢ coins
1 : 4
7 : 28

Ratio: Make the total (before)->25+10=35 and total (after)->7+28=35 be equal.

18u=36
10u =20

20x 20 cents = 400 cents = $4.00
At first, value of twenty-cents coins was $4.00

(b)
25u = 50
50 x 50 cents = 2500 cents = $25

$(25+4) = $29
At first, amount of money in the piggy bank was $29

OR

Wait for Mathguru's solutions to be posted.

[underthesea]

Can someone help me with this maths question:


There are 2 boxes. Box and Box Y, both of which contain pens and erasers only. The respective numbers of pens and erasers in Box X are in the ratio 1:3. The respective numbers of pens and erasers in Box Y are in the ratio 5:6. The number of items in Box X is thrice the number of items in Box Y.

(a) Find the ratio of the number of pens in Box Y to the number of pens in Box X.

(b) Knowing that there are 100 more erasers in Box Y than in Box X, find the number of erasers in Box Y.

anything wrong with the question?

TIA

[YLH88]

Hi Mathsguru,


Need your help with the below questions :

1) Doran had 8/11 of the number of game cards Thomas had. After Thomas lost 18 cards to Doran, they both had the same number of cards.

a) How many more game cards did Thomas have than Doran at first ?
b) As the game continued, Doran won more cards from Thomas. How many more cards did Thomas lose to Doran such that Doran has 3 times as many cards as him ?

I’m able to get the answer for (a), which is 36. But can’t find for (b)

Thank you!

[CoffeeCat]

YLH88:

Hi Mathsguru,


Need your help with the below questions :

1) Doran had 8/11 of the number of game cards Thomas had. After Thomas lost 18 cards to Doran, they both had the same number of cards.

a) How many more game cards did Thomas have than Doran at first ?
b) As the game continued, Doran won more cards from Thomas. How many more cards did Thomas lose to Doran such that Doran has 3 times as many cards as him ?

I'm able to get the answer for (a), which is 36. But can't find for (b)

Thank you!

draw the new model for part (b) and you know the total number of cards

[Vanilla Cake]

underthesea:

Can someone help me with this maths question:


There are 2 boxes. Box and Box Y, both of which contain pens and erasers only. The respective numbers of pens and erasers in Box X are in the ratio 1:3. The respective numbers of pens and erasers in Box Y are in the ratio 5:6. The number of items in Box X is thrice the number of items in Box Y.

(a) Find the ratio of the number of pens in Box Y to the number of pens in Box X.
(b) Knowing that there are 100 more erasers in Box Y than in Box X, find the number of erasers in Box Y.

anything wrong with the question?TIA

Sorry, underthesea. This is my view as I am also waiting for Mathsguru to provide solution to my Maths problem too.Where is this question from? There is a mistake in this question stating that there are 100 more erasers in Box Y than in Box X.

I can only solve part (a) and let's wait for Mathsguru to solve part (b)

(a)
Box X
P : E
1 : 3
33 : 99

Box Y
P : E
5 : 6
20 : 24

Ratio: Total for Box X is 33+99=132 and Total for Box Y is 20+24=44
132÷44=3
fits the statement-\"The number of items in Box X is thrice the number of items in Box Y\".

Ratio of the number of pens in Box Y to the number of pens in Box X is 20:33

(b)
From workings in part (a), the number of erasers in Box X is more than the number of erasers in Box Y. This contradicts the statement given in part (b) of the question ie \"Knowing that there are 100 more erasers in Box Y than in Box X.....\"

Pls wait for Mathsguru to provide her solutions.

[CoffeeCat]

underthesea:

Can someone help me with this maths question:


There are 2 boxes. Box and Box Y, both of which contain pens and erasers only. The respective numbers of pens and erasers in Box X are in the ratio 1:3. The respective numbers of pens and erasers in Box Y are in the ratio 5:6. The number of items in Box X is thrice the number of items in Box Y.

(a) Find the ratio of the number of pens in Box Y to the number of pens in Box X.

(b) Knowing that there are 100 more erasers in Box Y than in Box X, find the number of erasers in Box Y.

anything wrong with the question?

TIA

yea i think sth's wrong too

[Vanilla Cake]

YLH88:

Hi Mathsguru,

Need your help with the below questions :

1) Doran had 8/11 of the number of game cards Thomas had. After Thomas lost 18 cards to Doran, they both had the same number of cards.

a) How many more game cards did Thomas have than Doran at first ?
b) As the game continued, Doran won more cards from Thomas. How many more cards did Thomas lose to Doran such that Doran has 3 times as many cards as him ?

I'm able to get the answer for (a), which is 36. But can't find for (b)
Thank you!

Hi YLH88,
While waiting for Mathsguru's model solution, pls see whether mine using ratio is of any use or not?

(a)
D : T
8 : 11
16 : 22

D : T
19 : 19

(16+22)÷2=19

3u=18
u=6
6u=36

At first, Thomas had 36 more game cards than Doran.

(b)
D : T
19:19
38: 38

D : T
3 : 1
57:19

6u=18 [from part (a) ie 6u=36, reduce half of 36 = 18 since 19:19 has been doubled to 38:38 in part (b)]
u=3
19u = 57
Thomas lost 57 more cards to Doran.