Q&A - PSLE Math

[mathnoobs]

Barley Tower:

A coin box contained only twenty-cent and fifty-cent coins in the ratio of 4:5. When 16 fifty-cent coins were taken out and replaced by some twenty-cent coins, the number of fifty-cent coins left in the box was 7/8 of the twenty-cent coins. The total value of all the coins remained the same. Find the sum of money in the coin box.


Hi nnchia, you might want to consider this method without using algebra which some children may find too abstract and may not understand.


16 50-cent coins = $ 8.00 which is 40 20-cents coins.\t\t
\t\t
Now:\t\t
Since the Ratio of 20 cents coins to 50-cents coins is 8:7\t\t

20-cents coins: \t8 Units + 40\t
50-cents coins: \t7 Units + 35 \t
\t\t
At First:\t\t
To work backward to the At-First situation, deduct 40 20-cents coins and add back 16 50-cents coints\t\t

20-cents coins: \t8 Units\t
50-cents coins: \t7 Units + 35 + 16\t
\t\t
We are also given that At First, the ratio of 20-cent coins to 50-cents coins is 4:5
\t\t
Hence, if there are 8 units of 20-cent coins, there must be 10 units of 50-cents coins.
\t\t
20-cents coins: \t8 Units\t
50-cents coins: \t10 Units\t
\t\t
Compare the 50-cents coins:

10 Units = 7 Units + 51\t\t
3 units = 51 coins\t\t
1 unit = 17 coins\t\t
\t\t
At first, there are 17 X 8 = 136 20-cents coins and 17 X 10 = 170 50-cents coins.
\t\t
Total value of the coins is $ 27.20 + $ 85.00 = $ 112.20.

I'm quite confused by this:
20-cents coins: \t8 Units + 40\t
50-cents coins: \t7 Units + 35 \t
is that the Before value or the After value ? and if so, how was it derived ? don't know where the 35 comes from.
This method not so easy to understand

[MathIzzzFun]

mathnoobs:

Barley Tower:

A coin box contained only twenty-cent and fifty-cent coins in the ratio of 4:5. When 16 fifty-cent coins were taken out and replaced by some twenty-cent coins, the number of fifty-cent coins left in the box was 7/8 of the twenty-cent coins. The total value of all the coins remained the same. Find the sum of money in the coin box.


Hi nnchia, you might want to consider this method without using algebra which some children may find too abstract and may not understand.


16 50-cent coins = $ 8.00 which is 40 20-cents coins.\t\t
\t\t
Now:\t\t
Since the Ratio of 20 cents coins to 50-cents coins is 8:7\t\t

20-cents coins: \t8 Units + 40\t
50-cents coins: \t7 Units + 35 \t
\t\t
At First:\t\t
To work backward to the At-First situation, deduct 40 20-cents coins and add back 16 50-cents coints\t\t

20-cents coins: \t8 Units\t
50-cents coins: \t7 Units + 35 + 16\t
\t\t
We are also given that At First, the ratio of 20-cent coins to 50-cents coins is 4:5
\t\t
Hence, if there are 8 units of 20-cent coins, there must be 10 units of 50-cents coins.
\t\t
20-cents coins: \t8 Units\t
50-cents coins: \t10 Units\t
\t\t
Compare the 50-cents coins:

10 Units = 7 Units + 51\t\t
3 units = 51 coins\t\t
1 unit = 17 coins\t\t
\t\t
At first, there are 17 X 8 = 136 20-cents coins and 17 X 10 = 170 50-cents coins.
\t\t
Total value of the coins is $ 27.20 + $ 85.00 = $ 112.20.

I'm quite confused by this:
20-cents coins: \t8 Units + 40\t
50-cents coins: \t7 Units + 35 \t
is that the Before value or the After value ? and if so, how was it derived ? don't know where the 35 comes from.
This method not so easy to understand

16 x $0.50 = 40 x $0.20
--> 16 fifty-cents removed
--> 40 twenty-cents added

At first,
number of 20c : 50c = 4: 5 = 8 units : 10 units

In the end,
number of 20c = 8 units + 40 = 8 (1 unit +5)

and the ratio of 20c:50c
= 8: 7
= 8 (1 unit+5) : 7 (1 unit+5)
= 8 units + 40 : 7 units + 35

Comparing the number of 50c coins, we will get:
number at first - number removed = number in the end
10 units - 16 = 7 units + 35
1 unit --> 17

Total value of coins
= 8 x 17 x $0.20 + 10 x 17 x $0.50
= $112.20

cheers.

[iCreative Math]

Never easy to explain Math solution via writing without verbal explanation.


The question mentioned that after taking away 16 50-cents coins, and adding the same value in 20-cents coins (which is 40 20-cents coins), the number of 50-cents coins left in the box is 7/8 of the number of 20-cents coins.

Hence, we deduce that the ratio of 20-cents coins to 50-cents coins is 8:7.

Therefore,

20-cents coins : 50-cents coins is 8 units : 7 units

or (8-units + 40) : (7 units + 35)

We introduce the 40 because we know 40 coins was added.

The number of 20-cents coins At First is therefore 8 units.

If we add 40 coins to the ratio for 20-cents coins, we need to add 35 to the number of 50-cents coins to maintain the ratio of 8:7.

I hope this further explanation is clear.

[mathnoobs]

thanks MathIzzFun and ICreativeMath, I understand now. Appreciate you spending time on the explanations.

[iCreative Math]

most welcome, mathnoonbs … keeping sending your questions … glad to share my methods particularly for challenging questions … i pity the 12 years old … P6 questions are so varied and hence more challenging … they need simplified structured models to help them re-produce the solutions under examination conditions.

[Oracle]

Hi, could someone help me with these questions please. Daughter’s tearing her hair out over them:


1. Mohan drove from Town A towards Town B at 9.50am. Half an hour later, Nora left Town B for Town A. Both of them reached their respective destinations at 10.50am. If Nora travelled at 20km/h faster than Mohan, find the speed of Nora.

2. Abel and Cain cycled from City Mall to East Mall at 11 km/h and 8 km/h respectively. Abel left City Mall an hour later than Cain. He arrived at East Mall 30 minutes earlier than Cain. How long did Abel take?

3. Ann left Town A for Town B. At the same time, Ben left Town B for Town A. Ann drove at an average speed of 72 km/hr while Ben drove at 56 km/h. Ann and Ben passed each other at a point 24 km from the mid point of Towns A and B. What is the distance between A and B?

I have 2 more but my eyelids are too heavy now to go on.

Thank you very much.

[speedmaths.012624com]

Oracle:

Hi, could someone help me with these questions please. Daughter's tearing her hair out over them:


1. Mohan drove from Town A towards Town B at 9.50am. Half an hour later, Nora left Town B for Town A. Both of them reached their respective destinations at 10.50am. If Nora travelled at 20km/h faster than Mohan, find the speed of Nora.

Hi Oracle,

Please see table below.
Can you follow?

Hope this helps.

speedmaths.com

http://i46.tinypic.com/xc5b12.jpg\">

.

[MathIzzzFun]

Oracle:

Hi, could someone help me with these questions please. Daughter's tearing her hair out over them:


2. Abel and Cain cycled from City Mall to East Mall at 11 km/h and 8 km/h respectively. Abel left City Mall an hour later than Cain. He arrived at East Mall 30 minutes earlier than Cain. How long did Abel take?


I have 2 more but my eyelids are too heavy now to go on.

Thank you very much.

Abel left 1 hour later than Cain and arrived 30 minutes earlier than Cain at East Mall, so Abel took 1.5h less than Cain to cycle from City Mall to East mall.

Abel's speed : Cain's speed = 11 : 8

Ratio of Abel's time: Cain's time (to cycle from City Mall to East mall)
= 8u : 11u

Since Abel took 1.5h less than Cain,
3u --> 1.5h
1u --> 0.5h

Total time taken by Abel to cycle from City Mall to East mall = 8 x 0.5 = 4h

cheers.

[MathIzzzFun]

Oracle:

Hi, could someone help me with these questions please. Daughter's tearing her hair out over them:


3. Ann left Town A for Town B. At the same time, Ben left Town B for Town A. Ann drove at an average speed of 72 km/hr while Ben drove at 56 km/h. Ann and Ben passed each other at a point 24 km from the mid point of Towns A and B. What is the distance between A and B?

I have 2 more but my eyelids are too heavy now to go on.

Thank you very much.

Ann's speed : Ben's speed = 72 : 56 --> 9 : 7

When Ann and Ben passed each other,
Ann has travelled a distance of 9 units from Town A, and
Ben has travelled a distance of 7 units from Town B

Total distance between Town A and Town B --> 9 units + 7 units = 16 units

mid point between town A and town B
= 8 units from Town A
= 8 units from Town B

So, 1 unit --> 24 km
Distance between A and B = 16 x 24 km = 384 km

cheers.

[Oracle]

Thanks a lot MathIzzzFun and SpeedMaths (still up at 1:54am!!! Phew …)

Will show them to my daughter tonight and see if she is able to relate.

The other two questions:

Q4. Jeremy walks to school at an average speed of 4 km/h everyday. One day, after walking a distance of 1 km, he realised that his watch was slow. So he ran at a constant speed and managed to reach school just on time. Later, he calculated that if he had run at this speed right from the start of the journey, he would have reached school 5 minutes earlier. What was his running speed?

Q5. Mr Tan took 6 hours to travel from Town A to Town B. Mr Lim took 4 hours to travel from Town B to Town A. If both of them started travelling at 10 am, at what time did they pass each other?

Thank you.