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Q&A - PSLE Math

Academic support for Primary 6 and PSLE
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Nebbermind
KiasuGrandMaster
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Re: Q&A - PSLE Math

Post by Nebbermind » Wed May 25, 2016 3:41 pm

MathWithModels wrote:Question:

A circular wheel of diameter 35cm makes 100 revolutions in 1 minute.
Calculate the distance covered by the wheel in half an hour. Express your answer in km.


Answer:

Radius of the wheel: 35 / 2 = 17.5 cm
Circumference of the wheel: 2 x Pi x Radius = 2 x 3.14 x 17.5 cm = 109.9 cm
Distance covered by the wheel in 1 minute = 109.9 x 60 = 6594 cm
Distance covered by the wheel in half an hour = 6594 x 30 = 197820 cm = 1.9782 km
it's 100rpm...no?
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cindy28
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Re: Q&A - PSLE Math

Post by cindy28 » Fri May 27, 2016 12:46 pm

need help with the following questions :thankyou:

Q1. Arnold and Sandy were given some money each. If Arnold spent $25 and Sandy spent $75 each day, Arnold would still have $2100 while Sandy would have spent all her money. If Arnold spent $75 and Sandy spent $25 each day, Arnold would still have $900 while Sandy spent all her money. How much money was given to each of them?

Q2. At first Edwin had some $10 and $2 notes in the ratio of 3 : 8. After exchanging 2 $10 notes for $2 notes, the ratio of $10 to $2 becomes 1 : 3. How much did money did Edwin have at first?

Q3. There are 260 students in a school. 30% of them were girls while the rest were boys. When the school reopened, more student were admitted. The new students had boys and girls in the ratio of 5 : 4. After new students were admitted, 40% of the students was girls.
a. How many girls were admitted?
b. How many students were there in the end?
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Ender
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Re: Q&A - PSLE Math

Post by Ender » Fri May 27, 2016 1:52 pm

cindy28 wrote:need help with the following questions :thankyou:

Q1. Arnold and Sandy were given some money each. If Arnold spent $25 and Sandy spent $75 each day, Arnold would still have $2100 while Sandy would have spent all her money. If Arnold spent $75 and Sandy spent $25 each day, Arnold would still have $900 while Sandy spent all her money. How much money was given to each of them?
Scenario 1 :Spending Ratio
A : S
25 : 75

Total Arnold had = 25u + $2100
Total Sandy had = 75u ->She spent all, which is also what she had

Scenario 2 Sepnding Ratio
A : S
75 : 25

Total Arnold had = 75p + $900
Total Sandy had = 25p

25p=75u , Base on Sally's money
75p = 225u

25u + 2100 = 225u + 900 , Base on what Arnold had
200u = 1200
u = 6

Arnold had 25u + 2100 = $2250 #####
Sally had 75u = $450 #####
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Ender
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Re: Q&A - PSLE Math

Post by Ender » Fri May 27, 2016 2:01 pm

cindy28 wrote:need help with the following questions :thankyou:



Q2. At first Edwin had some $10 and $2 notes in the ratio of 3 : 8. After exchanging 2 $10 notes for $2 notes, the ratio of $10 to $2 becomes 1 : 3. How much did money did Edwin have at first?
2 piece of $10 notes will exchange for 10 piece of $2 notes

Code: Select all

$10 : $2
 3u : 8u
 -2   +10
 1p : 3p
3u - 2 = 1p
8u + 10 = 3p

9u - 6 = 3p

8u + 10 = 9u - 6
u = 16

At 1st,
Pieces of $10 notes = 3u = 48
Pieces of $2 notes = 8u = 128

Value of $10 notes = $480
Value of $2 notes = $256

Total he had at first = $480+$256 = $736 ##
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Ender
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Re: Q&A - PSLE Math

Post by Ender » Fri May 27, 2016 3:27 pm

cindy28 wrote: Q3. There are 260 students in a school. 30% of them were girls while the rest were boys. When the school reopened, more student were admitted. The new students had boys and girls in the ratio of 5 : 4. After new students were admitted, 40% of the students was girls.
a. How many girls were admitted?
b. How many students were there in the end?
Before,
girls = 30% of 260 = 78
Boys = 70% of 260 = 182

Code: Select all

Boys : Girls
182     78
+5u     +4u
60p     40P
182 + 5u = 60p --(1)
78 + 4u = 40p --(2)

(2) X 1.5,
117 + 6u = 60p

117 + 6u = 182 + 5u
u = 65

a) Girls admitted = 4u = 260 ###

b) from (2)
78 + 260 = 40p
40p = 338
p=8.45

In the end there are 100p student = 100 (8.45) = 845 ###
Last edited by Ender on Sat May 28, 2016 2:20 pm, edited 1 time in total.
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westbb
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Re: Q&A - PSLE Math

Post by westbb » Fri May 27, 2016 8:06 pm

is it possible to solve this question without using guess n check or quadratic equation?

the area of a rectangle is 84cm2. the length is 5cm more than the width. find the length of the rectangle.

thanks!
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alfretztay
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Re: Q&A - PSLE Math

Post by alfretztay » Fri May 27, 2016 9:04 pm

westbb wrote:is it possible to solve this question without using guess n check or quadratic equation?

the area of a rectangle is 84cm2. the length is 5cm more than the width. find the length of the rectangle.

thanks!
Without using guess and check or quadratic equation, I would solve the above question as follows :

84 = 1 x 84 = 2 x 42 = 3 x 28 = 4 x 21 = 6 x 14 = 7 x 12
Observe that 12 - 7 = 5 (the length is 5cm more than the width)
Thus, length = 12

Ans : 12cm.
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sushi88
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Re: Q&A - PSLE Math

Post by sushi88 » Fri May 27, 2016 11:13 pm

westbb wrote:is it possible to solve this question without using guess n check or quadratic equation?

the area of a rectangle is 84cm2. the length is 5cm more than the width. find the length of the rectangle.

thanks!
L x B = 84
L - B = 5
Factoring 84 with 2 factors within a difference of 5,
84 = 7 x 12 = 7 x (7+5)
Hence length -> 12
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questionable_i
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Re: Q&A - PSLE Math

Post by questionable_i » Sat May 28, 2016 11:50 am

Can anyone help me with these questions? Thank you

Image

Image
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EconsPhDTutor
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Re: Q&A - PSLE Math

Post by EconsPhDTutor » Sat May 28, 2016 12:31 pm

questionable_i wrote:Can anyone help me with these questions? Thank you

Image

Image

(13) The shaded triangle in △ABC has the same base as, but ½ the height of △ABC. Thus, it has ½ the area of △ABC.

Thus ▱ABCD has area 4 × 14 = 56 cm².

(15) Let F be the number of 5-bead packets and S be the number of 7-bead packets.

Then the total number of packets is F + S and the total number of beads is 5F + 7S. The ratio of the former to the latter is (F + S) ÷ (5F + 7S). We are told that this ratio is 5:29.

Cross-multiplying, we have 29 (F + S) = 5 (5F + 7S).

Doing the algebra, 4F = 6S. Rearranging, S:F = 4:6. That is, there are four 7-bead packets for every six 5-bead packets. Hence, 40% or ⅖ of all packets contain 7 beads.

Hope this helps and feel free to ask for any clarifications.

- Dr. Choo Yan Min

www.EconsPhDTutor.com

If you found this answer helpful, then please help me out by spreading word of my services. I’m a new tutor in Singapore so I need some help getting word out! Thank you!
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