Q&A - PSLE Math

Academic support for Primary 6 and PSLE
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Ice watch
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Re: Q&A - PSLE Math

Post by Ice watch » Mon Mar 21, 2016 1:03 am

forestcadee wrote:Hi, can someone help to solve the following 2 questions on circles? Thank you.

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:thankyou:
(1/2)*(3.14*12*12) = 226.08 cm^2 ==> 226cm^2

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Last edited by Ice watch on Mon Mar 21, 2016 2:40 am, edited 1 time in total.

Ice watch
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Re: Q&A - PSLE Math

Post by Ice watch » Mon Mar 21, 2016 1:12 am

forestcadee wrote:Hi, can someone help to solve the following 2 questions on circles? Thank you.
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:thankyou:
(1/3)*(3.14*10*10) = 104.67 cm^2 ==> 105cm^2

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forestcadee
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Re: Q&A - PSLE Math

Post by forestcadee » Mon Mar 21, 2016 12:23 pm

Thanks, Ice watch and your detailed drawings.

:smile:

angel
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Re: Q&A - PSLE Math

Post by angel » Mon Mar 21, 2016 4:07 pm

1190 people watched a movie. 2/5 of the male audience and 1/7 of the female audience were children. There were an equal number of men and women. How many audience were women?


(I always get stuck with questions like these. I worked on same denominator, then I'm stuck as I am confused its total 35 units or 70. ;))

Ender
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Re: Q&A - PSLE Math

Post by Ender » Mon Mar 21, 2016 4:54 pm

Let total male be 5M and female be 7F

Since there's equal number of Men and Women, therefore 3M = 6F
Hence we can make them common unit. Men is 3 parts, Womem is 6 parts. LCM would be 6

We let Men be 6u, and women be 6u.

Finding total male
3/5 => 6u
5/5 => 10u

Finding total female
6/7 => 6u
7/7 => 7u

Total audience = 17u = 1190
u= 70

Women = 6u = 6(70) = 420


Nebbermind
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Re: Q&A - PSLE Math

Post by Nebbermind » Mon Mar 21, 2016 4:55 pm

angel wrote:1190 people watched a movie. 2/5 of the male audience and 1/7 of the female audience were children. There were an equal number of men and women. How many audience were women?


(I always get stuck with questions like these. I worked on same denominator, then I'm stuck as I am confused its total 35 units or 70. ;))
Male: 2U children, 3U men
Female: 1P children. 6P women

Since 3U = 6P -> U = 2P,

Total = 10P + 7P = 1190
Find 6P

questionable_i
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Re: Q&A - PSLE Math

Post by questionable_i » Mon Mar 21, 2016 5:47 pm

Ice watch wrote:
questionable_i wrote:Can anyone help me with these questions?

1. Mrs Lim will make a profit which is 19% of the import price if she sells the necklace at $4165. What will be her profit if she sells the necklace at $4270 and the import price of the necklace remains constant?

2. Students are donating money to a charity fund that helps less privileged kids. The number of $2, $5 and $10-notes donated are in the ratio of 9 : 2 : 5.
4/9 of the $2-notes and 10 $5-notes are taken out. The remaining notes are worth $300. How many $10-notes are there?

:thankyou:
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:thankyou: Ice watch

Bunnyng
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Re: Q&A - PSLE Math

Post by Bunnyng » Mon Mar 21, 2016 8:48 pm

Can anyone help? Thank you!

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. (Ans $112.20)

PiggyLalala
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Re: Q&A - PSLE Math

Post by PiggyLalala » Mon Mar 21, 2016 10:15 pm

angel wrote:1190 people watched a movie. 2/5 of the male audience and 1/7 of the female audience were children. There were an equal number of men and women. How many audience were women?


(I always get stuck with questions like these. I worked on same denominator, then I'm stuck as I am confused its total 35 units or 70. ;))
I think u shld not make the denominator the same because the given condition is number of men = number of women. Not number of male audience = number of female audience. Male audience here refers to the number of male adults (man) and male children. Likewise for the female adults.

The correct solution is posted by nebbermind and ender.

Ice watch
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Re: Q&A - PSLE Math

Post by Ice watch » Tue Mar 22, 2016 5:04 pm

Bunnyng wrote:Can anyone help? Thank you!

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. (Ans $112.20)
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