Q&A - PSLE Math

[tianzhu]

Beikiasu:

Hi, can someone help me to solve these problem sum?

1) A rectangular tank A, 180 cm by 60 cm by 80 cm, is 5/8 filled with water. When some water from tank A is pumped into another rectangular tank B, tank A is left with 15% of its water.
(a)How many litres of water is pumped into tank B? Ans:459 litres
(b)Given that tank B has a square base of sides 50 cm, what is the height of water in tank B? Ans:183.6 cm.SCG, 2011,SA1, Question 13, 4 mark.

Hi

A rectangular tank A, 180 cm by 60 cm by 80 cm, is 5/8 filled with water.

5 units ----- 540 litres (180*80*60*5/8) (convert from cm3 to litres)

15% of 540 litres ------ 81 litres

Amount of water transferred to tank B ------ 540 – 81 ------ 459 litres

Height of water in tank B ------ 459000/50*50 ------ 183.6cm

Best wishes

[cimman]

bookwormkids:

tianzhu:

[quote=\"bookwormkids\"]
2)
Catherine has a box containing some black and white counters. When she adds in 15 white counters, 65% of the counters in the box are black. If she adds in another 40 black counters, 75% of the counters in the box are black. How many white counters are there in the box at first?

Hi Tianzhu, my son would like to know:

When 40 black counters are added, the number of white counters remain unchanged. So when 15 white counters are added,did the number of black counters remain the same?

Thanks for your time.

[/quote]this is a query on problem interpretation. In the context of Singapore Maths Problem sums (regardless of Maths topic), whenever something is not mentioned, it has a value of 0. eg. Tom and Mary each brought $10 to school. During recess time, Tom spent $5 on food and $2 on stationary. How much did Mary have in the end ?

Notice that the problem sum gave information on how much Tom spent but no information on how Mary spent her money. If nothing was mentioned about Mary, we must assume Mary spent nothing.
Of course, in the real world, if you boss never mentioned anything about your bonus for the year, we just say the bonus is unknown and not assume there's no bonus. The language of Maths problem sum does not reflect how we normally express ourselves. Because of this, we must learn how Maths problem sums are typically structured (from a language viewpoint) and what needs to be inferred.

This problem belongs to a category of problems I call \"Double If\" problems. The word \"when\" can be substituted for \"if\".
In most Double If questions, the 2 If scenarios are not linked, except at the initial conditions (Before values). ie. if something happened to the white counters in the first scenario, then in the second scenario, we start off with a clean slate, ie. assume nothing happened to the white counters. Nothing is carried over from the 1st If scenario to the 2nd If scenario.

This is an unusual question because it carried information from the 1st If scenario to the 2nd If scenario. The clue to this lies in the keyword \"another\", \"...adds in another 40 black counters\". If the word \"another\" is used, it means there is a precedent, or a prior condition. The next question to ask is, what does the precedent refers to ? It refers to conditions in the 1st If scenario.

Which leads to the next question: What are the conditions for the 1st If scenario ? - no information was given about any activities on the black counters, and 15 white counters were added. This means that we must assume that 0 Black counters were added or removed, and 15 white counters were added.

In the 2nd If scenario, because of the word \"another\", we must provide the initial condition of 0 change in Black counters and 15 white counters added. On top of this, we have to add in 40 black counters.
http://i49.tinypic.com/iwjdjo.png\">

Contrast this to how the problem is normally worded:
Catherine has a box containing some black and white counters. When she adds in 15 white counters, 65% of the counters in the box are black. If she adds in 40 black counters, 75% of the counters in the box are black. How many white counters are there in the box at first?

in this case, because there is no keyword \"another\", the 2nd scenario is decoupled from the 1st scenario. We start off the 2nd IF scenario with a clean slate. 40 Black counters are added, and no White counters are added or removed. Notice the zero values in the boxes. There is no prior condition.
http://i47.tinypic.com/jqtijt.png\">

I call this concept, the \"Zero Assumption\" concept - if nothing is mentioned, assume 0 Change value. Problem interpretation concepts are an important component in my workshops.

[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.

[marstolive]

Hi MathIzzzFun,

Thanks for all the help!

[marstolive]

cimman:

marstolive:

Two teams had to complete a certain job in 30days. 6 days after they started the job, one team was transferred to another zone while the other continued working alone and completed the remaining part of the job in 40 days.

In how many days could each team,working separately, complete the whole job.

if you don't mind working in algebra, here is a solution. Don't worry, there's no simultaneous equations here. The alphabets are there for a structured presentation with minimal words. Black fonts indicates values transferred directly from the problem sum, Red fonts indicates derived values.
http://i50.tinypic.com/314frme.png\">

:thankyou:

[marstolive]

cimman:

marstolive:

Two teams had to complete a certain job in 30days. 6 days after they started the job, one team was transferred to another zone while the other continued working alone and completed the remaining part of the job in 40 days.

In how many days could each team,working separately, complete the whole job.

if you don't mind working in algebra, here is a solution. Don't worry, there's no simultaneous equations here. The alphabets are there for a structured presentation with minimal words. Black fonts indicates values transferred directly from the problem sum, Red fonts indicates derived values.
http://i50.tinypic.com/314frme.png\">

:thankyou:

[Mary Joy]

Pls. help me to solve this;


At noon a van driver left Town C for Town B travelling at a constant speed. Two hours later a motorist also left Town C for Town D.The motorist overtook the van driver at 5 pm.The speed of the motorist was 28km / hr faster than that of the van driver.
A) Find the speed of the van driver.
B) Find the distance between Town C & Town D if the motorist was 80 km away from Town D at 5 pm.

[MathIzzzFun]

Mary Joy:

Pls. help me to solve this;


At noon a van driver left Town C for Town D travelling at a constant speed. Two hours later a motorist also left Town C for Town D.The motorist overtook the van driver at 5 pm.The speed of the motorist was 28km / hr faster than that of the van driver.
A) Find the speed of the van driver.
B) Find the distance between Town C & Town D if the motorist was 80 km away from Town D at 5 pm.

using time ratio/speed ratio:

Point X --> where motorist catch up with van

Ratio of time taken to travel from Town C to Point X
van : motorist --> 5h : 3h = 5:3

Ratio of speed of van : motorist --> 3 : 5 = 3 units : 5 units

2 units --> 28 km/h, van's speed --> 3/2 x 28 km/h = 42 km/h

Distance from Town C to Town D
= distance from Town C to point X + Distance from point X to Town D
= 5 x 42 + 80
= 290 km


cheers.

[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.