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General Chat :: Re: What to do after exams?

sixonechampion — Fri, 08 Jun 2018 17:36:53 +0100

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Question, Feedback and Comments :: observer

peterworrallania — Wed, 08 Nov 2017 14:37:00 +0100

interesting!

Primary 6 Matters :: Re: An interesting question.

echeewh — Thu, 21 Sep 2017 11:55:15 +0100

Hey <drstevewu> ,

Following pls find my worked solution.

Given that Ben (B ) used all his fifty-cent coins to buy stationery, B would have left with only 17 twenty-cent coins.

17 × 20 = 340

And this is $9.10 less than Carl (C), we have ...

C = 910 + 340 = 1250

As C had only fifty-cent coins, we have ...

Qty of fifty-cent coins = 1250 ÷ 50 = 25

=====

Trust this helps.

Do let me know again if there is further clarification.

Cheers,
Edward

Primary 6 Matters :: An interesting question.

drstevewu — Tue, 19 Sep 2017 18:51:27 +0100

Question: Three boys Alan, Ben And Carl had the same number of coins. Alan and Ben each had a mix of twenty-cents and fifty-cents coins. Alan had 7 twenty-cents coins while Ben had 17 twenty-cent coins. Carl had only fifty-cent coins. Ben used all his fifty-cent coins to buy stationery. He then had $9.10 less than Carl. How many fifty-cents coins did Carl have? (the answer is 25 fifty-cents coins. I thought it was 24)

Hope to see your working and your explanation if your answer is 25 fifty-cent coins.

Thank you very much.

Primary 6 Matters :: Re: Heuristic question?

echeewh — Wed, 30 Aug 2017 17:25:15 +0100

Hey Steve,

Following pls find my explanation and worked solution.

A : (1/3) boys
B + C : (2/3) boys

Given that the number of boys in Class C is equal to the number of girls in Class B, we have ...

C (boys) = B (girls)
C (girls) = B (boys)

Hence,
C (boys) + B (boys) can fill up one class given that each class contains equal number of pupils.

And, we also know ...

C (boys) + B (boys) --> (2/3) boys

Therefore, (2/3) boys will occupy 1 class (for e.g. B ), while the other remaining class (C) will be occupied by all girls.

In other words, the (2/3) boys can form one class; while the girls (from B and C) will form another class.

Given that (1/3) boys are in A and (2/3) boys can form 1 class,

(1/3) boys will occupy half (1/2) of class A while the other half of class A will be occupied by girls.

Since all the boys can occupy 1 1/2 classes, the remaining 1 1/2 classes will be occupied by the girls

(1 1/2) ÷ 3 = (1/2)

Half (1/2) of the classes are girls

=======

Trust this helps.

Do let me know again if there is further clarification.

Cheers,
Edward

Primary 6 Matters :: Heuristic question?

drstevewu — Thu, 17 Aug 2017 22:34:43 +0100

Hi Edward, please help me look at the following question:

There are three classes that have an equal number of pupils. 331/3% (thirty-three and one-third percentage) of the boys in the classes are in Class A. The number of boys in Class C is equal to the number of girls in Class B. What fraction of pupils in the classes are girls? (Ans: 1/2)

Thank you for your time.

Cheers,
Steve

Primary 6 Matters :: Re: Need help in this question.

drstevewu — Sun, 06 Aug 2017 15:53:22 +0100

Thank you Edward,

I got what you meant already. Had try to explain to the kid about the Usual Price and Cost Price and Selling Price but he seem confused about the whole thing.

Cheers,
Steve Wu

Primary 6 Matters :: Re: Need help in this question.

echeewh — Tue, 01 Aug 2017 22:59:57 +0100

Hey <drstevewu>

Pls note the Usual Price ( UP ) may not necessarily be the Cost Price ( CP ).

Hence, if you can calculate the UP of the camera, you should be able to find the answer to its CP in this case.

Thank you.

Best regards,
Edward

Primary 6 Matters :: Need help in this question.

drstevewu — Tue, 01 Aug 2017 16:55:12 +0100

Hi there, i am thinking if this question make sense? Need help from you guys.

Question: If Sam sells his camera at a discount of 10% of the usual price, he will make a loss of $250. If he sells it at a discount of 30%, he will incur a loss of $650. Find the cost price of his camera. (Answer: $2050)

Thank you for your time. :)