Q&A - P5 Math
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Need some help. TIA!
1. 5/8 of Ann’s stamps is equal to 2/5 of Tony’s stamps. If Tony gives 18 stamps to Ann, they will then have the same number of stamps. How many stamps do they have altogether?
2. Mary spent $270 on a necklace. She spent 1/7 of her money on a dress. She then had 3/8 of her money left. How much money did she have at first?
3. Yesterday, Ron had 72 more stamps than Lisa. Today, Ron gave Lisa 2/5 of his stamps. As a result, he has 1/2 as many stamps as Lisa now. How many stamps did Ron have yesterday? -
let me try…
1. Ann - 16u
Tony - 25u
total 16+25=41u
41/2=20.5
25-20.5=4.5u
4.5u - 18 stamps
1u - 4 stamps
4x41 =164 -
![](\"http://i50.tinypic.com/5l6g7k.jpg\")
http://i50.tinypic.com/5l6g7k.jpg\">
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![](\"http://i49.tinypic.com/2zinaco.jpg\")
Pl help :? -
Jamesbond:
Pl help :?
Shaded part is 5/8 of the square.
Therefore answer is 240 x (5/8) = 150 cm square.
Explanation:
We know that AP = PB and AQ = QD.
Therefore, it is clear that the square can be divided into 4 equal small squares, which can then be cut up into 8 small triangles.
It should be obvious from the drawing that half of APXQ is shaded where X is the center point of the square.
Also, it should be obvious that the other 2 shaded triangles FBC and QDC can be pieced together to form a rectangle that is half of the entire square.
Therefore, out of the 8 small triangles, we have 6 shaded triangles. -
[quote]2. Mary spent $270 on a necklace. She spent 1/7 of her money on a dress. She then had 3/8 of her money left. How much money did she have at first?[/quote]
It might be good to draw the model for this question (if you have trouble visualizing) although I am not going to provide one in the following solution.
After spending $270, Mary has some money left. Let us call that the remainder.
She then spent 1/7 of the remainder on a dress.
Thereafter, she is left with 6/7 of the remainder.
Therefore, 6/7 of remainder = 3/8 of her total money
For reasonable comparison, you need the unit size to be the same!
As such, find the lowest common multiple of 6 and 3. Then make the numerator the same (same amount represented by the same number of units -> units are of the same size), i.e.
6/7 of remainder = 6/16 of her total money
As such, her total money can be represented using 16 units. The remainder is then represented using 7 units.
The money spent on the necklace is hence represented by
16 - 7 = 9 units
9 units = $270
1 unit = $270 / 9 = $30
16 units = $30 x 16 = $480 -
ChewingPencilLine:
[quote]2. Mary spent $270 on a necklace. She spent 1/7 of her money on a dress. She then had 3/8 of her money left. How much money did she have at first?
It might be good to draw the model for this question (if you have trouble visualizing) although I am not going to provide one in the following solution.
After spending $270, Mary has some money left. Let us call that the remainder.
She then spent 1/7 of the remainder on a dress.
Thereafter, she is left with 6/7 of the remainder.
Therefore, 6/7 of remainder = 3/8 of her total money
For reasonable comparison, you need the unit size to be the same!
As such, find the lowest common multiple of 6 and 3. Then make the numerator the same (same amount represented by the same number of units -> units are of the same size), i.e.
6/7 of remainder = 6/16 of her total money
As such, her total money can be represented using 16 units. The remainder is then represented using 7 units.
The money spent on the necklace is hence represented by
16 - 7 = 9 units
9 units = $270
1 unit = $270 / 9 = $30
16 units = $30 x 16 = $480[/quote]Hi, Sorry to interrupt..She only spent 1/7 of the money on the dress and not the remainder. So the steps are
1/7=8/56
3/8=21/56
21/56+8/56=29/56
56-29=27
270/27=10
56*10=560
Correct me if I am wrong. -
ChewingPencilLine:
Got it...Thank you..Jamesbond:
Pl help :?
Shaded part is 5/8 of the square.
Therefore answer is 240 x (5/8) = 150 cm square.
Explanation:
We know that AP = PB and AQ = QD.
Therefore, it is clear that the square can be divided into 4 equal small squares, which can then be cut up into 8 small triangles.
It should be obvious from the drawing that half of APXQ is shaded where X is the center point of the square.
Also, it should be obvious that the other 2 shaded triangles FBC and QDC can be pieced together to form a rectangle that is half of the entire square.
Therefore, out of the 8 small triangles, we have 6 shaded triangles. -
Anna, Bee Bee and Cari shared the cost of a gift for their friend. 2/5 of Cari’s share was equal to 1/2 of Bee Bee’s and Anna’s share. 1/3 of Bee Bee share was equal to 1/3 of Cari and Anna share. If Anna paid $40 less than Cari, how much did the gift cost
Pls help -
Jamesbond:
Hi, Sorry to interrupt..She only spent 1/7 of the money on the dress and not the remainder. So the steps areChewingPencilLine:
[quote]2. Mary spent $270 on a necklace. She spent 1/7 of her money on a dress. She then had 3/8 of her money left. How much money did she have at first?
It might be good to draw the model for this question (if you have trouble visualizing) although I am not going to provide one in the following solution.
After spending $270, Mary has some money left. Let us call that the remainder.
She then spent 1/7 of the remainder on a dress.
Thereafter, she is left with 6/7 of the remainder.
Therefore, 6/7 of remainder = 3/8 of her total money
For reasonable comparison, you need the unit size to be the same!
As such, find the lowest common multiple of 6 and 3. Then make the numerator the same (same amount represented by the same number of units -> units are of the same size), i.e.
6/7 of remainder = 6/16 of her total money
As such, her total money can be represented using 16 units. The remainder is then represented using 7 units.
The money spent on the necklace is hence represented by
16 - 7 = 9 units
9 units = $270
1 unit = $270 / 9 = $30
16 units = $30 x 16 = $480
1/7=8/56
3/8=21/56
21/56+8/56=29/56
56-29=27
270/27=10
56*10=560
Correct me if I am wrong.[/quote]Oh my! I am so sorry! Misread the question, haha! I did not check if your workings are right (in a rush right now) but thank you very much for pointing out my mistake!!
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