## Q&A - PSLE Math

Academic support for Primary 6 and PSLE
Jmaths.SG
YellowBelt
Posts: 19
Joined: Fri Apr 01,

### Re: Q&A - PSLE Math

Find the smallest positive integer, n, such that 42n is a multiple of 180
Reduce 180 and 42 to multiples of prime numbers.
180 = 2 * 2 * 3 * 3 * 5
42 = 2 * 3 * 7

Some numbers must be multipled to 42 in order to become a multiple of 180.
These numbers must include at least one 2, one 3 and one 5, since these are missing from the prime numbers of 42.

The smallest n = 2*3*5 = 30.

Ans: 30

S-H
BrownBelt
Posts: 564
Joined: Mon Feb 08,
Total Likes:1

### Re: Q&A - PSLE Math

Jmaths.SG wrote:
Find the smallest positive integer, n, such that 42n is a multiple of 180
Reduce 180 and 42 to multiples of prime numbers.
180 = 2 * 2 * 3 * 3 * 5
42 = 2 * 3 * 7

Some numbers must be multipled to 42 in order to become a multiple of 180.
These numbers must include at least one 2, one 3 and one 5, since these are missing from the prime numbers of 42.

The smallest n = 2*3*5 = 30.

Ans: 30
Understand, thank you very much!

S-H
BrownBelt
Posts: 564
Joined: Mon Feb 08,
Total Likes:1

### Re: Q&A - PSLE Math

Can someone pls help, thanks!

(a) Solve the following inequality
4x-3
------- less than. 5x+2
7

(b) write down the smallest integer that satisfies this inequality

SAHMom
OrangeBelt
Posts: 60
Joined: Tue Oct 13,
Total Likes:11

### Re: Q&A - PSLE Math

taringal wrote:Hi,

Patrick, Devi and Jiamin each have the same number of beads. How many beads must Devi give to Patrick and Jiamin so that Patrick has 20 beads more than Devi and Jiamin has 4 fewer beads than Patrick?
Hi, hope the below helps...

S-H
BrownBelt
Posts: 564
Joined: Mon Feb 08,
Total Likes:1

### Re: Q&A - PSLE Math

Hi, can someone pls help, thks!

(a) solve the following inequality

4x-3
---------- < 5x + 2
7

(b) write down the smallest integer that satisfies this inequality

ozora
BlueBelt
Posts: 326
Joined: Thu Feb 24,

### Re: Q&A - PSLE Math

Need some guide on the following questions.
Thanks

MathWithModels
YellowBelt
Posts: 11
Joined: Sun Dec 28,

### Re: Q&A - PSLE Math

Question:

Catherine had some money at first. She gave 1/4 of her money to
Anne. As a result, Anne had thrice as much as before. Then, Anne
gave 1/2 of her money to Bernard. Bernard then had 1 1/2 times as
much money as before. Anne still had \$129 left. How much money
did each child have at first?

Anne

We work backwards for Anne. She had \$129 at the end.
This was after giving half her money to Bernard. So, she gave \$129 to Bernard.
Before giving \$129 to Bernard, she had \$129 + \$129 = \$258.
Anne tripled her money after getting 1/4 of Catherine's money.
So Anne had \$258/3 = \$86 at first.

Catherine

Anne tripled her money after getting 1/4 of Catherine's money.
As Anne had \$86 at first, she received 2 x \$86 = \$172 from Catherine.
This was 1/4 of Catherine's money
So Catherine had \$172 x 4 = \$688 at first

Catherine had \$172 x 4 = \$688 at first

Bernard

Bernard received \$129 from Anne, and had 1 1/2 times as much as before.
So the \$129 was half what he had at first.
So Bernard had \$129 x 2 = \$258 at first.

Bernard had \$129 x 2 = \$258 at first

MathWithModels
YellowBelt
Posts: 11
Joined: Sun Dec 28,

### Re: Q&A - PSLE Math

Question:

Eunice, Grace and Joyce collect badges as their hobby. At first,
Eunice gave 2/5 of her badges to Grace. Grace then gave 3/8 of what
she had to Joyce. Then Joyce gave 1/6 of what she had to Eunice. In
have at first?

We work backwards, beginning with each of them having 150 badges in the end.

In the end:

Eunice: 150
Grace: 150
Joyce: 150

Last transfer: Joyce gave 1/6 of what she had to Eunice.

Joyce had 150 after she gave 1/6 of her badges to Eunice.
So 150 is 5/6 of her badges. She had 150 * 6 / 5 = 180 badges before, and gave 30 to Eunice.

Eunice: 150 - 30 = 120 (she had 30 fewer badges before)
Grace: 150 (unchanged)
Joyce: 150 + 30 = 180 (she had 30 more badges before)

Second last transfer: Grace gave 3/8 of what she had to Joyce.

Grace had 150 after she gave 3/8 of her badges to Joyce.
So 150 is 5/8 of her badges. She had 150 * 8 / 5 = 240 badges before, and gave 90 to Joyce.
Before Grace gave some of her badges to Joyce, the girls had

Eunice: 120 (unchanged)
Grace: 150 + 90 = 240 (she had 90 more badges before)
Joyce: 180 - 90 = 90 (she had 90 fewer badges before)

First transfer: Eunice gave 2/5 of her badges to Grace.

She had 120 badges after she gave 2/5 to Grace, so 120 is 3/5 of her badges before the transfer.
She had 120 * 5 / 3 = 200 badges before the transfer, and she gave 80 badges to Grace.

Before Eunice gave some of her badges to Grace, the girls had

Eunice: 120 + 80 = 200 (she had 80 more badges before)
Grace: 240 - 80 = 160 (she had 80 fewer badges before)
Joyce: 90 (unchanged)

ozora
BlueBelt
Posts: 326
Joined: Thu Feb 24,

### Re: Q&A - PSLE Math

MathWithModels wrote:Question:

Catherine had some money at first. She gave 1/4 of her money to
Anne. As a result, Anne had thrice as much as before. Then, Anne
gave 1/2 of her money to Bernard. Bernard then had 1 1/2 times as
much money as before. Anne still had \$129 left. How much money
did each child have at first?

Anne

We work backwards for Anne. She had \$129 at the end.
This was after giving half her money to Bernard. So, she gave \$129 to Bernard.
Before giving \$129 to Bernard, she had \$129 + \$129 = \$258.
Anne tripled her money after getting 1/4 of Catherine's money.
So Anne had \$258/3 = \$86 at first.

Catherine

Anne tripled her money after getting 1/4 of Catherine's money.
As Anne had \$86 at first, she received 2 x \$86 = \$172 from Catherine.
This was 1/4 of Catherine's money
So Catherine had \$172 x 4 = \$688 at first

Catherine had \$172 x 4 = \$688 at first

Bernard

Bernard received \$129 from Anne, and had 1 1/2 times as much as before.
So the \$129 was half what he had at first.
So Bernard had \$129 x 2 = \$258 at first.

Bernard had \$129 x 2 = \$258 at first
thanks
c

peggy
BlueBelt
Posts: 207
Joined: Wed Nov 26,