Q&A - PSLE Math

[TheSolver]

There were fewer than 50 exercise books left at a school bookshop on Monday.
On Tuesday, 1/7 of the exercise books were sold.
On Wednesday, one book fewer than 4/9 of the remaining exercise books were sold.
On Thursday, 2 books more than 3/7 of the remaining exercise books were sold.
How many exercise books did the bookshop have on Monday?

First, you have to find the lowest common multiple of the two denominators of the fractions, which is 7 and 9. So, the LCM is 21. Now, list out the multiples of 21 below 50, which will give you only 21 and 42. Now, you try both numbers.
21:
21*6/7=18
18*5/9=10
10+1=11
11*4/7=6 2/7
6 2/7-2=4 2/7
So, that means that you can't use 21.

42:
42*6/7=36
36*5/9=20
20+1=21
21*4/7=12
12-2=10

Ans:42

[leo_mums]

There were fewer than 50 exercise books left at a school bookshop on Monday.
On Tuesday, 1/7 of the exercise books were sold.
On Wednesday, one book fewer than 4/9 of the remaining exercise books were sold.
On Thursday, 2 books more than 3/7 of the remaining exercise books were sold.
How many exercise books did the bookshop have on Monday?

First, you have to find the lowest common multiple of the two denominators of the fractions, which is 7 and 9. So, the LCM is 21. Now, list out the multiples of 21 below 50, which will give you only 21 and 42. Now, you try both numbers.
21:
21*6/7=18
18*5/9=10
10+1=11
11*4/7=6 2/7
6 2/7-2=4 2/7
So, that means that you can't use 21.

42:
42*6/7=36
36*5/9=20
20+1=21
21*4/7=12
12-2=10

Ans:42

TheSolver!

[belnanna]

Hi all,

THis question has been asked in the forum before..However, there wasnt a detailed explanation to this question... Would some kind soul please help to solve this ..THanks in advance!

If Kenny gives 18 marbles to Andrew, he will have thrice as many marbles as Andrew. If Andrew gives 12 marbles to kenny, he will have 1/9 of the number of marbles that kenny has. How many marbles does Kenny have at first?

[belnanna]

Hi all,

Not too sure how to solve these 2 questions.. Please help.. THanks so much in advance.

1: Mrs Low bought a total of 20 hens and geese for $126.Each goose cost $2 more than a hen. How much did she pay for each goose if she bought 6 fewer hens than geese?

2: Andrew , Annie and Ravin opened a bank account each in January 2008.Every month, Andrew, Annie and Ravin deposited $20,$30 and $45 into their bank accounts respectively. IF Annie's account was $900 after a few months,
a) How much longer did Andrew take than Annie to top up his savings to $900?
b)How much would Ravin have when Andrew had topped up to $900?

[Ender]

Mr Ali and Mr Raju started driving to town A at the same time.
Mr Raju took 6 hours to reach town a while Mr Ali took 9 hours to reach town A.
After 3 hours, both men were left with the same distance to complete.
Find the ratio of the total distance travelled by Mr Raju to that travelled by Mr Ali to reach town A if the both of them drove at a constant speed.

After 3 hrs, Mr Rahu left 3 hrs to go, i.e he traveled half of his distance. Let the total Distance for Mr Raju be 2R.

After 3 hrs, Ali have traveled 1/3 of his distance, since he still have 6 hours to go.Let total distance for Ali be 3A.

The remaining distance for Mr Raju is 1R and Ali is 2A and they are equal.
1R = 2A

Total distance for Mr Raju = 2R = 4A ...... (Since 1R=2A)
Total distance for Ali = 3A

Distance Raju : Ali -> 4 : 3

[EC72]

Mr Ali and Mr Raju started driving to town A at the same time.
Mr Raju took 6 hours to reach town a while Mr Ali took 9 hours to reach town A.
After 3 hours, both men were left with the same distance to complete.
Find the ratio of the total distance travelled by Mr Raju to that travelled by Mr Ali to reach town A if the both of them drove at a constant speed.

After 3 hrs, Mr Rahu left 3 hrs to go, i.e he traveled half of his distance. Let the total Distance for Mr Raju be 2R.

After 3 hrs, Ali have traveled 1/3 of his distance, since he still have 6 hours to go.Let total distance for Ali be 3A.

The remaining distance for Mr Raju is 1R and Ali is 2A and they are equal.
1R = 2A

Total distance for Mr Raju = 2R = 4A ...... (Since 1R=2A)
Total distance for Ali = 3A

Distance Raju : Ali -> 4 : 3

Ender!

[leo_mums]

There were fewer than 50 exercise books left at a school bookshop on Monday.
On Tuesday, 1/7 of the exercise books were sold.
On Wednesday, one book fewer than 4/9 of the remaining exercise books were sold.
On Thursday, 2 books more than 3/7 of the remaining exercise books were sold.
How many exercise books did the bookshop have on Monday?

First, you have to find the lowest common multiple of the two denominators of the fractions, which is 7 and 9. So, the LCM is 21. Now, list out the multiples of 21 below 50, which will give you only 21 and 42. Now, you try both numbers.
21:
21*6/7=18
18*5/9=10
10+1=11
11*4/7=6 2/7
6 2/7-2=4 2/7
So, that means that you can't use 21.

42:
42*6/7=36
36*5/9=20
20+1=21
21*4/7=12
12-2=10

Ans:42

TheSolver, how is it the LCM for 7 and 9 is 21?

[berryspice]

Hi,

Need help to solve this question.
Thanks in advance.

[Ender]

Hi,

Need help to solve this question.
Thanks in advance.

http://www.kiasuparents.com/kiasu/forum ... 3#p1540853

[Ender]

A cash register contained some fifty-cent coins, some twenty-cent coins and some ten-cent coins in the ratio 4:6:3.
80% of the fifty-cent coins were taken out and replaced with the same number of ten-cent coins.
Then 90 twenty-cent coins were taken out and replaced with the same number of ten-cent coins.
In the end, the ratio of the number of fifty-cent coins to the number of twenty-cent coins to the number of ten-cent coins became 4:25:36.

a) Find the total number of fifty-cent and twenty-cent coins taken out of the cash register.
b) Find the percentage decrease in the total value of the coins in the cash register in the end. Round off your answer to 2 decimal places.

50c : 20c : 10c -> 4:6:3 -> 20 : 30 : 15 -> total = 65

After 80% of 50c removed, replace with same number of 10c. (80% of 20u = 16u)
50c : 20c : 10c -> 4u : 30 : 31u

90 coins of 20c removed replaced with 90 coins of 10c.
50c : 20c : 10c -> 4u : 30u-90 : 31u+90 -> 4u : 25u : 36u

30u-90= 25u or 31u + 90 = 36u
5u = 90
u= 18

a) Number of 50c taken out = 16u = 16(18) = 288
Total number of 50c and 20c = 288+90 = 378 coins.

b) Original value of 50c = 20(18) X 0.5 = $180
Original value of 20c = 30(18) X 0.2 = $108
Original value of 10c = 15(18) X 0.1 = $27
Original total = 180+108+27 = $315

Final value of 50c = 4(18) X 0.5= $36
Final value of 20c = 25(18) X 0.2 = $90
Final value of 10c = 36(18) X 0.1 = $64.8
Final total = 36+90+64.8 = 190.8

Decrease in value = 315- 190.8 = 124.2
Percentage decrease = (124.2/315) X 100 = 39.43%