Q&A - P5 Math

[MathIzzzFun]

YumYum:

Hi, can anyone pls help with these 2questions?


1. I am thinking of a number. If the decimal point of this number is shifted to the right once, the sum of of 5 times of the new number and the initial number is 216.75. Find the number I have in mind.

2. Jane has some one-dollar and fifty cent coins. The number of one-dollar coins is thrice the number of fifty cent coins. If the total value of the coins is $171.50, how many fifty-cent coins doesjane have?

Thanks

Hi

Q1.
original number --> 1 unit
...if decimal point shifted to the right once --> new number is 10 times original number.
so, 216.75 --> 51 units, 1 unit --> 4.25
the number is 4.25

Q2.
for every group of (3x $1 coins + 1x50-cent coin) --> $3.50
$171.50 / $3.5 = 49 groups of 3 x $1 coins & 1x 50-cent coin

Number of 50-cent coins = 49

cheers.

[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.

[YumYum]

MathIzzzFun:

YumYum:

Hi, can anyone pls help with these 2questions?


1. I am thinking of a number. If the decimal point of this number is shifted to the right once, the sum of of 5 times of the new number and the initial number is 216.75. Find the number I have in mind.

2. Jane has some one-dollar and fifty cent coins. The number of one-dollar coins is thrice the number of fifty cent coins. If the total value of the coins is $171.50, how many fifty-cent coins doesjane have?

Thanks

Hi

Q1.
original number --> 1 unit
...if decimal point shifted to the right once --> new number is 10 times original number.
so, 216.75 --> 51 units, 1 unit --> 4.25
the number is 4.25

Q2.
for every group of (3x $1 coins + 1x50-cent coin) --> $3.50
$171.50 / $3.5 = 49 groups of 3 x $1 coins & 1x 50-cent coin

Number of 50-cent coins = 49

cheers.

Hi MathIzzzFun,

Thank you for your help

[concerned1970]

MathIzzzfun, Tysm for yr detailed explanation.

[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

[happyheart]

Can someone help to explain this sort of questions using ratio?I am lost by the third sentence. Ans is 94 stickers.



Q) 4 children, Amy, Ben, Candace and Don, each have some stickers.
The number of stickers that Amy has 1/2 of the total number of stickers that Ben, Candace and Don have.
The number of stickers that Ben has 1/4 of the total number of stickers that Amy, Candace and Don have.
The number of stickers that Candace has is 2/3 of the total number of stickers that Amy, Ben and Don have.
If Don has 47 stickers, how many stickers should Candace give to Ben so that Amy and Ben have the same number of stickers?

Thanks in advance!

[tianzhu]

happyheart:

Can someone help to explain this sort of questions using ratio?I am lost by the third sentence. Ans is 94 stickers.



Q) 4 children, Amy, Ben, Candace and Don, each have some stickers.
The number of stickers that Amy has 1/2 of the total number of stickers that Ben, Candace and Don have.
The number of stickers that Ben has 1/4 of the total number of stickers that Amy, Candace and Don have.
The number of stickers that Candace has is 2/3 of the total number of stickers that Amy, Ben and Don have.
If Don has 47 stickers, how many stickers should Candace give to Ben so that Amy and Ben have the same number of stickers?

Thanks in advance!

Hi

A:(B+C+D) ----- 1:2 (3) ----- 5:10 (15)

B:((A+C+D) ------ 1:4 (5) ----- 3:12 (15)

C:(A+B+D) ----- 2:3 (5) ----- 6:9 (15)

Make the total number of units the same.

A common multiple of 3 and 5 is 15.

A ----- 5 units

B ----- 3 units

C ----- 6 units

D ----- 1 unit

1 unit ----- 47

C must give 2 units to B so that A and B have the same number of stickers.

2 units ----- 2*47 ----- 94

Best wishes

[happyheart]

tianzhu:

happyheart:

Can someone help to explain this sort of questions using ratio?I am lost by the third sentence. Ans is 94 stickers.



Q) 4 children, Amy, Ben, Candace and Don, each have some stickers.
The number of stickers that Amy has 1/2 of the total number of stickers that Ben, Candace and Don have.
The number of stickers that Ben has 1/4 of the total number of stickers that Amy, Candace and Don have.
The number of stickers that Candace has is 2/3 of the total number of stickers that Amy, Ben and Don have.
If Don has 47 stickers, how many stickers should Candace give to Ben so that Amy and Ben have the same number of stickers?

Thanks in advance!

Hi

A:(B+C+D) ----- 1:2 (3) ----- 5:10 (15)

B:((A+C+D) ------ 1:4 (5) ----- 3:12 (15)

C:(A+B+D) ----- 2:3 (5) ----- 6:9 (15)

Make the total number of units the same.

A common multiple of 3 and 5 is 15.

A ----- 5 units

B ----- 3 units

C ----- 6 units

D ----- 1 unit

1 unit ----- 47

C must give 2 units to B so that A and B have the same number of stickers.

2 units ----- 2*47 ----- 94

Best wishes

Tianzhu, Thank you as always!

[chokos]

http://i45.tinypic.com/9t3ltf.jpg\">

[MathIzzzFun]

chokos:

http://i45.tinypic.com/9t3ltf.jpg\">

Hi

The point H is not on the figure, I suppose it is the intersection of line DG and line CF.

Area of triangle DFG = Area of triangle CFG = 1/2 of area of rectangle
= 1/2 x 7 x 17 = 59.5 cm2

Area of triangle CGH
= Area of triangle CFG - Area of triangle FGH
= 59.5 cm2 - 51 cm2
= 8.5 cm2

cheers.