Q&A - PSLE Math

[MrsKiasu]

[quote="questionable_i"]Can anyone help me? Thank you in advance.



Q2)

W : B
Meg 7 : 8 , say total is "X" : 7/15X+8/15X-4/15X=132, X=180
given away is 4/15 X = 48 (Ans for a)

Lea 2/3Y:1/3Y , say total is "Y" and white is 2times of black

After receiving 48 black bands from Meg...

Lea 5/9(New Y ie Y+48) : 4/9 (New Y)

Lea : workings on black bands : 1/3Y + 48 = 4/9 (new Y ie Y+48)
1/3Y+48=4/9(Y+48)
(Original) Y=240
(New) Y after receiving 48 from Meg=240+48=288 (Ans for b )

[MrsKiasu]

Can anyone help me? Thank you in advance.


Q3)

Ans a) 120%=$2400 (Shop A)
therefore 100% = $2000 (shop B)

Ans b) Say Selling Price for A = x% A, SP for B=x%B and discount = 100%-x%

x%B=x%A-$320
$2000(x%)=$2400(x%)-$320
$400(x%)=$320
x%=80%

and the discount will be 100%-80%=20%

[sushi88]

Can anyone help me? Thank you in advance.

Q1)


Q2)


Q3)

Q1)

Difference in amount between A and F = $5
Difference in amount in savings = 72-32 = $40
40/5 = 8 weeks
A saved weekly 72/8 = $9 and spent $15
(a)Hence total money A has per week is 9+15=$24
(b) F saved $32/8 = $4 per week.
44/4 = 11 weeks
So F has to save (11-8)= 3 more weeks to buy the watch.

Q2)

------------------------white-------black
M original--------------7u---------8u
M give 4u B to L-------7u---------8u-4u=4u
L original --------------2p----------p
after rec 4u from M---2p----------p+4u
L New ratio-------------5-----------4

7u+4u->132
11u-> 132
u->12
(a)So M gave 4 x 12 = 48 to L

2p/(p+48) -> 5/4
5p+240 -> 8p
3p->240

(b)Hence L has 3p+48 = 240+48 = 288 beads in the end.

Q3)

u = price sold by Shop B
1.2u -> 2400
u -> 2000
(a)Hence shop B is selling the same laptop at $2000

If x is the % discount by both shops,
(1-x)(2400) = 2000 (1-x) + 320
2400-2400x = 2000-2000x + 320
400x = 80
x = 0.2
(b)Hence discount is 20%

[wilburlim]

1760 children took part in a run last year. This year, the number of boys increased by 20% while the number of girls decreased by 20%. As a result, the number of girls became twice as many as the number of boys this year.
(a) How many boys took part in the run last year?
(b) What is the percentage decrease in the number of children who took part in the run this year?

[wilburlim]

Someone please help me. Thanks!

A lorry left Town A at 8 a.m. and travelled towards Town B at an average speed of 55 km/h. At 11a.m., a car left Town A for Town B, travelling at an average speed of 85 km/h along the same road. Both the lorry and the car travelled at a uniform speed.
(a) At what time did the car pass the lorry?
(b) How far was the lorry from Town A when the car passed the lorry?

[Ender]

1760 children took part in a run last year. This year, the number of boys increased by 20% while the number of girls decreased by 20%. As a result, the number of girls became twice as many as the number of boys this year.
(a) How many boys took part in the run last year?
(b) What is the percentage decrease in the number of children who took part in the run this year?

Start with boys =5u last year. 5u selected because it's easier to work with a 20% increase.

This year the boys = 1.2 X 5u = 6u , i.e 20% increasse from 5u
This year the girls= 12u , i.e. twice of boys this year.

Girls this year is 80% of girls last year
80% => 12u
100% => 12u/0.8 = 15u <-last year girls

5u + 15u = 1760
20u =1760
u = 88

a) Last year boys = 5u = 5(88) = 440#

b) Total participants for last year = 20u
Total paticipants for this year = 6u+12u = 18u
Reduce by 2u
% reduction = 2u/20u X 100 = 10% ###

[Ender]

Someone please help me. Thanks!

A lorry left Town A at 8 a.m. and travelled towards Town B at an average speed of 55 km/h. At 11a.m., a car left Town A for Town B, travelling at an average speed of 85 km/h along the same road. Both the lorry and the car travelled at a uniform speed.
(a) At what time did the car pass the lorry?
(b) How far was the lorry from Town A when the car passed the lorry?

Lorry had a 3 hr head start, so at 11am the lorry is at a distance of 55 X 3 = 165km

a) Car is faster by 85 - 55 = 30km/h
At 11am, the car has to close the gap of 165km, at 30km/h
Time take to close the gap = 165/30 = 5.5 hrs
11am plus 5.5 hr => 4:30 pm #

b)Lorry would have travelled the same distance as the car travelled, so use the car information.
distance when car passed the lorry= 85 X 5.5 = 467.5km #

[wilburlim]

1760 children took part in a run last year. This year, the number of boys increased by 20% while the number of girls decreased by 20%. As a result, the number of girls became twice as many as the number of boys this year.
(a) How many boys took part in the run last year?
(b) What is the percentage decrease in the number of children who took part in the run this year?

Start with boys =5u last year. 5u selected because it's easier to work with a 20% increase.

This year the boys = 1.2 X 5u = 6u , i.e 20% increasse from 5u
This year the girls= 12u , i.e. twice of boys this year.

Girls this year is 80% of girls last year
80% => 12u
100% => 12u/0.8 = 15u <-last year girls

5u + 15u = 1760
20u =1760
u = 88

a) Last year boys = 5u = 5(88) = 440#

b) Total participants for last year = 20u
Total paticipants for this year = 6u+12u = 18u
Reduce by 2u
% reduction = 2u/20u X 100 = 10% ###

Hi, thanks for the reply! But is there a easier way to solve this? I don't understand the 5u for boys at first.

[questionable_i]

Can anyone help me with these questions? in advance.

Q1)


Q2)


Q3)


Q4)


Q5)


Q6)

[Ender]

1760 children took part in a run last year. This year, the number of boys increased by 20% while the number of girls decreased by 20%. As a result, the number of girls became twice as many as the number of boys this year.
(a) How many boys took part in the run last year?
(b) What is the percentage decrease in the number of children who took part in the run this year?

Start with boys =5u last year. 5u selected because it's easier to work with a 20% increase.

This year the boys = 1.2 X 5u = 6u , i.e 20% increasse from 5u
This year the girls= 12u , i.e. twice of boys this year.

Girls this year is 80% of girls last year
80% => 12u
100% => 12u/0.8 = 15u <-last year girls

5u + 15u = 1760
20u =1760
u = 88

a) Last year boys = 5u = 5(88) = 440#

b) Total participants for last year = 20u
Total paticipants for this year = 6u+12u = 18u
Reduce by 2u
% reduction = 2u/20u X 100 = 10% ###

Hi, thanks for the reply! But is there a easier way to solve this? I don't understand the 5u for boys at first.

5u is selected because 20% of 5u is just 1u.. Easier to work.
We can also select boys to be 10u at first if you want,, which will also work well with a 20% increase.
e.g Let last year boys be 10u,
This year boys will be 12u, i.e 20% increase,
This year girls will be 24u , i.e. twice of this year boys

80% =>24u
100%=> 24u/0.8 = 30u Last year girls

30u+10u = 1760
40u=1760
u = 44

a) 10u = 440 #
b) reduction by 40u - 36u = 4u
% reduction = 4u/40u X 100 = 10% ###