Sec 2 Arithmetics :)

aYexa SG

1.Find the HCF and LCM of 2^3 x 3^2 x 5 and 2^2 x 3^3 x 7

2. Three bells ring at intervals of 6 mins, 15mins and 24mins respectively.
If they ring together at 8.00am, at what time will they next ring together again?

3. Johnathan and Michael decided to get the same desktop computer priced at $2300.

(a) Michael paid 25% of the selling price as a deposit and paid the remaining balence in 12 equal monthly installments. Find the amount Michael needed to pay each month.

(b) Johnathan paid for the computer in cash and was given a cash rebate discount. In the end, he paid $2012.50 for the computer.

(i) Calculate the percentage discount he received.
(ii) Given that the store still makes a profit of 15% from the sale, find the original cost price of the computer.

4. London time is 7hours behind Singapore time. A non-stop flight from London to Singapore is scheduled to take 13hours and 29mins. If the flight is to arrive in Singapore on saturday at 1718 Singapore time, find the departure day and time of the flight in London time.

wee_ws SG

Hi,

For Q2, consider LCM of 6, 15 and 24.

Do consult your school textbook to understand the approach to solve the others, which are typical.

Thanks!

Cheers,
Wen Shih

dkcx SG

1) As written, its clearly shown that the HCF is 3^2 or 9 base on what is given.
LCM should be 2^3 X 3^3 X 5 X 7 since 2^2 and 3^2 are already covered by 2^3 and 3^3.

3a) 2300 X 75% = a
       a / 12 = answer
bi)   (2300 - 2012.50)/2300 X 100% = ans
bii) 2012.50 = 115%
        100% = ans

4) Departure in Singapore time = 17 18 - 13 hrs 29mins = b
     Departure time in London = b - 7 = ans

nightzip SG

1.Find the HCF and LCM of 2^3 x 3^2 x 5 and 2^2 x 3^3 x 7

This one got method, for HCF, just look at both sets of factors....get all the lowest amount of similar factors.

E.g., 2^3 and 2^2, you take 2^2, then 3^2 and 3^3, you take 3^2

And since 5 is only available on one number, and not the other, while 7 is available for the other number but not the first, they cannot be taken.

That leaves you with 2^2 x 3^2 (This will be the HCF).

For LCM, method is to look at both sets of factors, then get all the highest amount of similar factors PLUS the dissimilar/different factors.

E.g. 2^3 and 2^2, take 2^3

3^3 and 3^2, take 3^3

Then take also both 5 & 7.

So, LCM = 2^3 x 3^4 x 5 x 7.

Remember, DON'T go and multiply everything out and then go and refactorise again. Not the right way since it is already factorised for you. =)

nightzip SG

2. Three bells ring at intervals of 6 mins, 15mins and 24mins respectively.
If they ring together at 8.00am, at what time will they next ring together again?

This is an LCM question. Somehow MOE didnt have the ability to devise HCF problems sums yet I supposed, so if it doubt, it is good to hazard using the LCM.

Why this is an LCM question?

Cuz, just remember that if your problem sum is some group of things/people/vehicles etc starting together at the same time, then asking you to find the next time that they will meet altogether again, this is equivalent to LCM because you need the lowest/fastest time where the next time they will meet again, the the "highest" or next "highest" time, but the IMMEDIATE next time they will meet (common).

Thus, 6,15,24 <== find LCM.

Remember what you are doing is in terms of minutes. So, you will need to add the extra minutes in "time mode", i.e. now is 8:00am, then you need to add loh.

If e.g. LCM is 138min, then it becomes 2hrs and 18 mins more, so the next time they meet will be 10:18am.

However, sometimes they will ask you "when will they meet again for the 2ND time or 3RD time?"

Ah...then you just add every 2hrs 18mins again and again depending on the question.

Sometimes, they will also ask "How many times will the 3 bells ring concurrently/in unison from 12am to 12pm?"

Then you will have to do 12hrs divide by 2hrs 18mins see how many times the 3 bells can ring in unison. =)