Q&A - PSLE Math

[nnchia]

Hi,


Need help to solve the following question:-

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.
Ai Tong 2011 CA1 Answer - $112.20 (Is this right?)

Thanks in advance for the help!

[MathIzzzFun]

nnchia:

Hi,


Need help to solve the following question:-

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.
Ai Tong 2011 CA1 Answer - $112.20 (Is this right?)

Thanks in advance for the help!

$112.20 is correct.

cheers.

[MathIzzzFun]

Beikiasu:

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


3) A contractor needed to cover an entire hall with tiles. On the first day, he laid 319 tiles. He completed tiling the rest of the hall in 7 days using an equal number of tiles each day. At the end of the 4th day, he was able to tile 8/15 of the hall. How many tiles did the contractor use to cover the entire hall? CHIJ, 2011, Question 15, 4 mark, Ans:1740 tiles.

Thanks

using equal fractions/numerators method:

after laying 319 tiles on 1st day, remaining tiles laid in 7 days

4/7 remaining tiles = 7/15 total number of tiles

28/49 remaining tiles = 28/60 total number of tiles

total number of tiles --> 60 units
remaining tiles --> 49 units
11 units --> 319 tiles
total number of tiles --> 60/11 x 319 = 1740

cheers.

[bookwormkids]

Need help on the following questions.


1)
Log A and Log B are of different lengths. A carpenter wants to saw them into equal number of shorter pieces. The length of each piece from Log A need not be the same as that of each piece from Log B. If he saw Log A into 1.2m pieces and log B into 0.9m pieces, he has no leftover from log A but there will be 0.6m of log B left. If he saw Log A into 1.6m pieces and log B into 1.25m pieces, he has no leftover from log A but there will be 0.6m of log B left. Find the length of log A.

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?

Thanks.

[tianzhu]

nnchia:

Hi,


Need help to solve the following question:-

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.
Ai Tong 2011 CA1 Answer - $112.20 (Is this right?)

Thanks in advance for the help!

Hi

One possible approach is to work in term of the number of coins. You may use Units and Parts.

At first

twenty-cent coins: fifty-cent coins ----- 4u:5u

16 number of fifty-cent coins ----- $8 -------- 40 number of twenty-cent coins

Change

twenty-cent coins: fifty-cent coins ----- 40:-16

In the end

twenty-cent coins: fifty-cent coins ----- 8p:7p

Make the parts the same

4u + 40 ------ 8p
(x7)
28u + 280 ------ 56p

5u – 16 ------ 7p
(x8)
40u – 128 ------ 56p

28u + 280 ------- 40u – 128
12u ------- 408
1u ------ 34

Number of twenty-cent coins ------- 4*34 ------136
Value of twenty-cent coins ------- 27.2

Number of fifty-cent coins ------- 5*34 ------170
Value of fifty-cent coins ------- 85

Total value ------- 85 + 27.2 ------ 112.20

Best wishes

[nnchia]

Hi Tianzhu, thanks a million!

[tianzhu]

bookwormkids:

Need help on the following questions.


1)
Log A and Log B are of different lengths. A carpenter wants to saw them into equal number of shorter pieces. The length of each piece from Log A need not be the same as that of each piece from Log B. If he saw Log A into 1.2m pieces and log B into 0.9m pieces, he has no leftover from log A but there will be 0.6m of log B left. If he saw Log A into 1.6m pieces and log B into 1.25m pieces, he has no leftover from log A but there will be 0.6m of log B left. Find the length of log A.

Hi

Please confirm from from your source.

There should be a difference in length.

Best wishes

[MathIzzzFun]

bookwormkids:

Need help on the following questions.


1)
Log A and Log B are of different lengths. A carpenter wants to saw them into equal number of shorter pieces. The length of each piece from Log A need not be the same as that of each piece from Log B. If he saw Log A into 1.2m pieces and log B into 0.9m pieces, he has no leftover from log A but there will be 0.6m of log B left. If he saw Log A into 1.6m pieces and log B into 1.25m pieces, he has no leftover from log A but there will be 0.6m of log B left. Find the length of log A.



Thanks.

Original question discussed at pg 801


http://www.kiasuparents.com/kiasu/forum/viewtopic.php?f=69&t=280&p=844766#p844766


cheers.

[tianzhu]

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

Scenario 1

When she adds in 15 white counters, 65% of the counters in the box are black.

White:Black ------ 7:13

Scenario 2

If she adds in another 40 black counters, 75% of the counters in the box are black.

Make the units for white counters for both scenarios the same.

White:Black ------ 1:3 -----7:21

21 – 13 ----- 8

8 units ---- 40

1 unit ----- 5

7 units ----- 35

35 – 15 ----- 20 (white counters@first)

Best wishes

[bookwormkids]

tianzhu:

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

Scenario 1

When she adds in 15 white counters, 65% of the counters in the box are black.

White:Black ------ 7:13

Scenario 2

If she adds in another 40 black counters, 75% of the counters in the box are black.

Make the units for white counters for both scenarios the same.

White:Black ------ 1:3 -----7:21

21 – 13 ----- 8

8 units ---- 40

1 unit ----- 5

7 units ----- 35

35 – 15 ----- 20 (white counters@first)

Best wishes

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.