Q&A - PSLE Math

[Ender]

need help with the following questions



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 ##

[Ender]

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 ###

[westbb]

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!

[alfretztay]

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.

[sushi88]

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

[questionable_i]

Can anyone help me with these questions? Thank you

[EconsPhDTutor]

Can anyone help me with these questions? Thank you

(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!

[sushi88]

Can anyone help me with these questions? Thank you

height of shaded triangle -> h
height of big triangle -> 2h
Area of big triangle -> 1/2(b)(2h) =bh
Area of shaded triangle -> 1/2b(h) = 14
Area of unshaded part between the big triangle and shaded = bh-1/2bh =1/2bh = 14
Hence area of total unshaded part -> 14 x 2 = 28 cm2

Ans: (2)

P : B -> 5 : 29
29 = 5 x 5 + 4 = 5 x 4 + 9 = 5 x 3 + 14
Packets of 7 -> 2
Fraction is 2/5

Ans: (2)

[lynn01sg]

pls help:

Saj spent 50 minutes on his project. He spent e minutes more than Thor. Thor spent 10 minutes less than Udell. How much time did Udell spend on his project? Leave your answer in minutes.

[alfretztay]

pls help:

Saj spent 50 minutes on his project. He spent e minutes more than Thor. Thor spent 10 minutes less than Udell. How much time did Udell spend on his project? Leave your answer in minutes.

Saj ——- 50
Thor ——- (50 – e)
Udell ——- (50 – e) + 10 = (60 – e)

Ans : (60 – e)min.