Q&A - P5 Math

[MathIzzzFun]

simple88:

Hi,


Can someone help me with the following question? Besides using the guess and check method, is there any other method that is less time consuming?

Mr Gopal purchased a certain number of shirts. He bought 11 small-sized shirts at $24 each and the large-sized shirts at $27 each. If the average cost of all the shirts was $25.35, how many large-sized shirts did he buy?

Thank you in advance.

hi

one simple way is to use the area model ...

http://i49.tinypic.com/724ysl.png\">

cheers.

[simple88]

Dear MathIzzzFun,


Thank you so much for your help.

[SAHMwith2boys]

MathIzzzFun:

SAHMwith2boys:

Hi, need solution for the math question below:


The average thickness of 3 identical books is 5mm thicker than the average thickness of 2 magazines. Find the thickness of a book if the average thickness of the books and magazines is 12mm.

Thanks.

Hi
Total thickness = 5 x 12mm = 60mm
Thickness of each book = (60 mm + 5mm x 2)/5 = 14 mm

http://www.flickr.com/photos/62167097@N02/7463411044/in/photostream

cheers.

Thanks for the solution.

[SAHMwith2boys]

Xiao Hu:

SAHMwith2boys:

Hi, need solution for the math question below:


The average thickness of 3 identical books is 5mm thicker than the average thickness of 2 magazines. Find the thickness of a book if the average thickness of the books and magazines is 12mm.

Thanks.

Hi,
B= average thickness of the 3 books.
M= average thickness of the 2 mags.
B-M=5 ---(1)

Average thickness of 3 books and 2 mags:
(3B+2M)/5=12
3B+2M=60 ---(2)

Solve by eliminating M:
(1)*2: 2B-2M=10 ---(3)
(2)+(3): 5B = 70
B=14mm

Hope it helps.
May I know where is this question from?

Rgds,
Xiaohu

Thanks for your solution. The question was from an assessment book that DS was working on.

[happyheart]

Need help on this questions. Kindly illustrate with model. Thank you.


Q) A restaurant had the same number of apples, pears and oranges at first. After 38 pears , some oranges and apples were used, there were 90 fruits left. The number of oranges is 1/2 the number of apples. The number of pears left is 15 fewer than the number of apples left. How many oranges were left?

[MathIzzzFun]

happyheart:

Need help on this questions. Kindly illustrate with model. Thank you.


Q) A restaurant had the same number of apples, pears and oranges at first. After 38 pears , some oranges and apples were used, there were 90 fruits left. The number of oranges is 1/2 the number of apples. The number of pears left is 15 fewer than the number of apples left. How many oranges were left?

Hi

http://i46.tinypic.com/29w21bl.png\">

cheers.

[ftwm]

Hi need help.


Ali has some singaore stamps, malaysia stamps & australia stamps in two albums. There were 765 stamps altogether. He moved 0.25 of the australia stamps, 0.6 of the malaysia stamps and 45 singapore stamps from the first album to the second album. He now found that there were 2 1/4times as many australia stamps in the first album. How many stamps did ali move from the first album to the second album?

[ftwm]

Another question, pls help:


0.25 of the coins in a box were 20¢ coins 0.5 of the remaining coins were 10¢ coins and the rest were 50¢ coins. The total value of the coins in the box was $101.20.
(a) Find the value of 50¢ coins in the box in dollars.
(b) If all the 20¢ coins were exchanged for 10¢ coins, express the number of 10¢ coins as a fraction of the total number of coins in the box? Give your answer as a decimal (Can you discover a shortcut?)

[ftwm]

Pls help.


There were some red, blue and green jelly beans in a bag. The number of red jelly beans was 2 1/3 times the number of green jelly beans. The number of blue jelly beans was 0.5 times the total number of the red and green jelly beans. When 200 jelly beans were added into the box, the number of green beans become 1.5 times the original number of green jelly bean, the number of red jelly beans was increased by 1/7 and the number of blue beans was increased by 50. Find the total numberof jelly beans in the bag in the end.

Thank you.

[MathIzzzFun]

ftwm:

Pls help.


There were some red, blue and green jelly beans in a bag. The number of red jelly beans was 2 1/3 times the number of green jelly beans. The number of blue jelly beans was 0.5 times the total number of the red and green jelly beans. When 200 jelly beans were added into the box, the number of green beans become 1.5 times the original number of green jelly bean, the number of red jelly beans was increased by 1/7 and the number of blue beans was increased by 50. Find the total numberof jelly beans in the bag in the end.

Thank you.

Hi

for questions involving decimals, change the decimals to fractions first, and the questions will not be that daunting.

Q3. There were some red, blue and green jelly beans in a bag. The number of red jelly beans was 2 1/3 (=7/3)times the number of green jelly beans. The number of blue jelly beans was 1/2 times the total number of the red and green jelly beans. When 200 jelly beans were added into the box, the number of green beans become 1 1/2 (=3/2) times the original number of green jelly bean, the number of red jelly beans was increased by 1/7 and the number of blue beans was increased by 50. Find the total numberof jelly beans in the bag in the end.

At first,
Red : Green : Blue = 7 : 3 : 5 --> 14u : 6u : 10u
in the end,
Red : Green : Blue --> 16u : 9u : 10u + 50 -- total = 35u + 50

added --> 2u + 3u + 50 = 200
1u --> 30
Total --> 35 x 30 + 50 = 1100

cheers.