Q&A - PSLE Math

[Easy-going]

Hi

Please help me with this question. Thank you very much in advance.
Question 14 (Rosyth CA 1 2012 Paper Two)
Jack, Meili and Raj shared some 50-cent coins.
They each had an average of 28 coins.
The total amount of money that Jack and Meili had was $7 more than the total amount of Jack and Raj had.
If Raj’s amount was 4 times of Jack’s amount, how much more money did Meili have than Jack?

[PapayaDad]

Cannot get round number leh…

My method:
Average = 28 coins
Total = 28 x 3 = 84 coins = $42

J+M=J+R+7 so M=R+7
R=4J

J=u
R=4u
M=4u+7
Total = 9u+7 = 42
u=3.8888?

[Chalupa]

(I need help for this question !) Mr. Lee left town at 6.45am and drove at a uniform speed of 60km/h to Town Q.Mrs Chen left town P at 7:15am and drove towards Town Q along the same expressway as Mr Lee.She passed him at 10:15am and continued driving at the same speed until she reached Town Q.The distance between Town P to Town Q was 560km.How far was Mr Lee from Town Q when Mrs Chen reached Town Q ?

Thanks in advance.

[PapayaDad]

Mr.Lee left town P at 6.45am? did u miss the P?


Also the correct thread for these is here :

http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&start=8950

[PapayaDad]

Mr Lee start 6.45 ; speed 60kmh

At 10.15 (3.5h later) ; Mr Lee travel 60+60+60+30=210km already

Mrs Chen took 3h (7.15 to 10.15) to travel 210km (catch up Mr Lee)
Mrs Chen Speen is 70kmh.
Mrs Chen need to travel another 350km to reach Q ie another 5h

Another 5h, Lee travel 300km only…still 50km away…

[Musicstar]

Hi,


Can someone enlighten me the below question with two types of method to solve the question but why different answer? Please advise the mistake. I suppose no matter which methods should also get the same answer. Thank you.

A box contained some twenty-coins and some fifty-cent coins in the ratio 3:4. When 26 fifty-cent coins were taken out and replaced by the same number of twenty-cent coins, the ratio became 4:1. Find the sum of money in the box in the box at first.


Working No 1
Since the number of coins did not change, we make the total the same

Before
T : F
3 : 4 (Total = 7u)
15 : 20 (Total = 35u)

After
T : F
4 : 1 (Total = 5u)
28 : 7 (Total = 35u)

13u = 26
1u = 2

15u of 20 cent = 15 x 2 x 0.2 = $6
20u of 50 cent = 20 x 2 x 0.5 = $20

Total = $26

Working No 2
Using units and parts
20cents : 50cents
3 units : 4 units
+65 -26 x 4
4 parts 1 part

3u +65 = 4p
16u -104 = 4p
3u + 65 = 16u - 104
16u - 3u = 65 + 104
13u = 169
u = 13
3u = 13 x 3 = 39
39 x $0.20 = $7.80
4u = 13 x 4 = 52
52 x $0.50 = $26
$7.80 + $26. = $33.80

[Oldschool]

Musicstar:

Hi,


Can someone enlighten me the below question with two types of method to solve the question but why different answer? Please advise the mistake. I suppose no matter which methods should also get the same answer. Thank you.

A box contained some twenty-coins and some fifty-cent coins in the ratio 3:4. When 26 fifty-cent coins were taken out and replaced by the same number of twenty-cent coins, the ratio became 4:1. Find the sum of money in the box in the box at first.


Working No 1
Since the number of coins did not change, we make the total the same

Before
T : F
3 : 4 (Total = 7u)
15 : 20 (Total = 35u)

After
T : F
4 : 1 (Total = 5u)
28 : 7 (Total = 35u)

13u = 26
1u = 2

15u of 20 cent = 15 x 2 x 0.2 = $6
20u of 50 cent = 20 x 2 x 0.5 = $20

Total = $26

Working No 2
Using units and parts
20cents : 50cents
3 units : 4 units
+65 -26 x 4
4 parts 1 part

3u +65 = 4p
16u -104 = 4p
3u + 65 = 16u - 104
16u - 3u = 65 + 104
13u = 169
u = 13
3u = 13 x 3 = 39
39 x $0.20 = $7.80
4u = 13 x 4 = 52
52 x $0.50 = $26
$7.80 + $26. = $33.80

Hi,

This is because in (b), instead of replaced with same number of coins, it is substituted with same amount of coins.

i.e. 65x20cents= 26x50cents in amount


But the question said \"same number\" though.

Regards

[Musicstar]

Oldschool,


Thank you for pointing out to me. I am really careless and did not read the question properly. The two words are really important. If don't have these two words \"the same\" which mean I need to change to the number of coin to the amount.

Oldschool:

Musicstar:

Hi,

Can someone enlighten me the below question with two types of method to solve the question but why different answer? Please advise the mistake. I suppose no matter which methods should also get the same answer. Thank you.

A box contained some twenty-coins and some fifty-cent coins in the ratio 3:4. When 26 fifty-cent coins were taken out and replaced by the same number of twenty-cent coins, the ratio became 4:1. Find the sum of money in the box in the box at first.


Working No 1
Since the number of coins did not change, we make the total the same

Before
T : F
3 : 4 (Total = 7u)
15 : 20 (Total = 35u)

After
T : F
4 : 1 (Total = 5u)
28 : 7 (Total = 35u)

13u = 26
1u = 2

15u of 20 cent = 15 x 2 x 0.2 = $6
20u of 50 cent = 20 x 2 x 0.5 = $20

Total = $26

Working No 2
Using units and parts
20cents : 50cents
3 units : 4 units
+65 -26 x 4
4 parts 1 part

3u +65 = 4p
16u -104 = 4p
3u + 65 = 16u - 104
16u - 3u = 65 + 104
13u = 169
u = 13
3u = 13 x 3 = 39
39 x $0.20 = $7.80
4u = 13 x 4 = 52
52 x $0.50 = $26
$7.80 + $26. = $33.80

Hi,

This is because in (b), instead of replaced with same number of coins, it is substituted with same amount of coins.

i.e. 65x20cents= 26x50cents in amount


But the question said \"same number\" though.

Regards

[rocklee]

Anyone can help?


There were a total of 20800 toys in Factory A and Factory B. After 3/4 of toys in Factory A and 3/5 in Factory B were sold, there were 1040 more toys in B than A. How many toys were there in each factory at first?

[CloudeeDaz]

rocklee:

Anyone can help?


There were a total of 20800 toys in Factory A and Factory B. After 3/4 of toys in Factory A and 3/5 in Factory B were sold, there were 1040 more toys in B than A. How many toys were there in each factory at first?

hihi
there is a solution http://www.kiasuparents.com/kiasu/forum/viewtopic.php?p=775568#p775568/