Q&A - P5 Math
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Thank you smartmummy.
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Jane paid $75 for 25 similar pencils and 12 similar files. Mary paid $23.30 for 6 such pencils and 5 such files. Find the total cost of each pencil and each file.
tks -
swordtail:
Hi,Jane paid $75 for 25 similar pencils and 12 similar files. Mary paid $23.30 for 6 such pencils and 5 such files. Find the total cost of each pencil and each file.
tks
For primary, one way is to ‘Make Supposition’.
Suppose Jane buys same number of files
$75 x 5 = $375 for 125 pens and 60 files
$23.3 x 12 = $279.6 for 72 pens and 60 files
125 pens – 72 pens = 53 pens
$375 – $279.6 = $95.4
53 pens will cost $95.4
1 pen will cost $1.8
You can continue with “Suppose Jane buys same number of pens” to find the cost of 1 file or make use of one of the conditions above. I’ll use the 2nd condition here,
$23.3 for 6 pens and 5 files
6 pens cost $10.8
5 files cost ($23.3 - $10.8) = $12.5
1 file cost $2.5
Total cost of 1 pen and 1 file = $1.8 + $2.5 = $4.3
Regards -
Hi, Please help the below 2 questions. Thks.
1) Mrs Lim baked some cookies for sale. She sold 90 cookies in the morning. In the afternoon, she sold 4/9 of the remainder. Then she had 1/3 of the total number of cookies left. How many cookies did Mrs Lim bake at first?
2) Alicia and Josh had some cards at first. In the first round, Alicia lost 1/5 of her cards to Josh. In the second round, Josh lost 1/3 of his cards to Alicia. In the end, both of them had 80 cards each. How many cards did Alicia have at first? -
Musicstar:
Hi,Hi, Please help the below 2 questions. Thks.
1) Mrs Lim baked some cookies for sale. She sold 90 cookies in the morning. In the afternoon, she sold 4/9 of the remainder. Then she had 1/3 of the total number of cookies left. How many cookies did Mrs Lim bake at first?
This can be solved by MD.
At first she has (90 + 9U) cookies.
Morning, she sold --> 90
Afternoon, she sold --> 4U
So, she is left with --> 5U
Since,
1/3 --> 5U
3/3 --> 15U
15U - 9U --> 90
U --> 15
Therefore, at first, she baked 15U --> 225 cookies
Regards -
Musicstar:
Hi,Hi, Please help the below 2 questions. Thks.
2) Alicia and Josh had some cards at first. In the first round, Alicia lost 1/5 of her cards to Josh. In the second round, Josh lost 1/3 of his cards to Alicia. In the end, both of them had 80 cards each. How many cards did Alicia have at first?
This is a \"work backwards\" problem.
(a) At last or After the 2nd round,
A : J : Total
80 : 80 : 160
(b) Before the 2nd round or After the 1st round,
From (a), we know
2/3 of Josh --> 80
3/3 of Josh --> 120
Then Alicia is (160 - 120) = 40
A : J : Total
40 : 120 : 160
(c) Before the 1st round or At first
From (b), we know
4/5 of Alicia --> 40
5/5 of Alicia --> 50
Then Josh is (160 - 50) = 110
A : J : Total
50 : 110 : 160
Therefore, at first, Alicia had 50 cards
Regards -
May spent all her money on 5 shirts and 3 blouses. each blouse cost $7 less than each shirt. if she bought 3 shirts and 2 blouses instead, she would have $56 left. what was the cost of a blouse?
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chloecube:
May spent all her money on 5 shirts and 3 blouses. each blouse cost $7 less than each shirt. if she bought 3 shirts and 2 blouses instead, she would have $56 left. what was the cost of a blouse?
1 Blouse --> 1u
1 Shirt -----> 1u + 7
5 shirts and 3 blouses --> 5*(1u + 7) + 3*(1u) = 8u + 35
3 shirts and 2 blouses --> 3*(1u + 7) + 2*(1u) = 5u + 21
(8u + 35) - (5u + 21) --> 56
3u + 14 --> 56
3u --> 42
1u --> 14
the cost of a blouse was $14 -
jieheng:
thank youchloecube:
May spent all her money on 5 shirts and 3 blouses. each blouse cost $7 less than each shirt. if she bought 3 shirts and 2 blouses instead, she would have $56 left. what was the cost of a blouse?
1 Blouse --> 1u
1 Shirt -----> 1u + 7
5 shirts and 3 blouses --> 5*(1u + 7) + 3*(1u) = 8u + 35
3 shirts and 2 blouses --> 3*(1u + 7) + 2*(1u) = 5u + 21
(8u + 35) - (5u + 21) --> 56
3u + 14 --> 56
3u --> 42
1u --> 14
the cost of a blouse was $14 -
Hi, I have come across this P5 question & will like to know if there is any other ways of solving it instead of using model/graphical…Thks in advance:
6A & 6B have the same number of pupils each. The ratio of the number of boys in 6A to the number of boys in 6B is 1:3. The ratio of the number of girls in 6A to the number of girls in 6B is 4:1. What is the ratio of the number of girls to the number of boys in 6B?