Q&A - PSLE Math

[anneshirleygilbert]

Hi MathIzzzFun,


Separate the following 8 numbers 26, 34, 57, 65, 69, 95, 119, 161 into 2 groups, each with 4 numbers so that the product of the 4 numbers in each group is equal.


26 = 2 x 13
34 = 2 x 17
57 = 3 x 19
65 = 5 x 13
69 = 3 x 23
95 = 5 x 19
119 = 7 x 17
161 = 7 x 23

Group 1 –> 26 (2 x 13) , 69 (3 x 23), 95 (5 x19), 119 (7x17)
Group 2 –> 34 (2 x 17) , 57 (3 x 19), 65 (5x 13), 161 (7 x 23)

You’re a whizz at numbers!!! Can you share how you figured this one out? Honestly, I’d be stumped were I faced with such a question! Given the time constraint in an exam, even with a calculator, it’d have taken me ages.

[MathIzzzFun]

anneshirleygilbert:

Hi MathIzzzFun,


Separate the following 8 numbers 26, 34, 57, 65, 69, 95, 119, 161 into 2 groups, each with 4 numbers so that the product of the 4 numbers in each group is equal.


26 = 2 x 13
34 = 2 x 17
57 = 3 x 19
65 = 5 x 13
69 = 3 x 23
95 = 5 x 19
119 = 7 x 17
161 = 7 x 23

Group 1 --> 26 (2 x 13) , 69 (3 x 23), 95 (5 x19), 119 (7x17)
Group 2 --> 34 (2 x 17) , 57 (3 x 19), 65 (5x 13), 161 (7 x 23)

You're a whizz at numbers!!!! Can you share how you figured this one out? Honestly, I'd be stumped were I faced with such a question! Given the time constraint in an exam, even with a calculator, it'd have taken me ages.

Hi

the first 4 numbers gave clues to using factors to solve this problem...
26 - 2 x 13
34 - 2 x 17
57 - 3 x 19
65 - 5 x 13
So, once the factors for the remaining numbers are worked out, what is left is just to group the numbers by factors - either by 2,3,5,7 or 13,15,17,19

cheers.

[tianzhu]

small:

Hi chan09,


I think the answer should be 80 red and blue ribbons in the bottle at first

Hi

I have some red & blue ribbons in a bottle. If I add in 20 red ribbons, 60% of my ribbons are blue. If I add in another 60 blue ribbons, 75% of my ribbons are blue. How many ribbons have I in the bottle ?

To get this answer(80), I gathered that the assumptions are as follows:

Condition1 ------ Add in 10 red beads.
Condition 2 ------ Without removing the first 10 red beads, add in another 30 blue beads.

Now our confusion starts,try to use the same assumptions on this question.

Q1) Mrs Lim bought some bowls and plates. If she buys another 5 bowls, there are 60% as many bowls as plates. If she buys another 9 plates, there will be thrice as many plates as bowls. How many bowls did she buy? (CHS P5 SA2 2009)

The issue here lies in one’s interpretation of the additional word “another”
There are two different answers, depending on the assumptions one makes.

Best wishes

[Nebbermind]

tianzhu:

Q1) Mrs Lim bought some bowls and plates. If she buys another 5 bowls, there are 60% as many bowls as plates. If she buys another 9 plates, there will be thrice as many plates as bowls. How many bowls did she buy? (CHS P5 SA2 2009)

The issue here lies in one’s interpretation of the additional word “another”
There are two different answers, depending on the assumptions one makes.

Best wishes

I did the add-then-add-again...and got a fraction of plates and bowls!! :oops:

[cimman]

Nebbermind:

tianzhu:

Q1) Mrs Lim bought some bowls and plates. If she buys another 5 bowls, there are 60% as many bowls as plates. If she buys another 9 plates, there will be thrice as many plates as bowls. How many bowls did she buy? (CHS P5 SA2 2009)

The issue here lies in one’s interpretation of the additional word “another”
There are two different answers, depending on the assumptions one makes.

Best wishes

I did the add-then-add-again...and got a fraction of plates and bowls!! :oops:

I think there should be some sort of quality control check on exam questions to make sure that there is no room for ambguity. Maths questions are starting to look like legal documents where the emphasis is on language.

[tianzhu]

Nebbermind:

I did the add-then-add-again...and got a fraction of plates and bowls!! :oops:

Hi

I’m using these two questions to illustrate that the assumptions made by small in the NH question is not applicable to the CHS question.

I have some red & blue ribbons in a bottle. If I add in 20 red ribbons, 60% of my ribbons are blue. If I add in another 60 blue ribbons, 75% of my ribbons are blue. How many ribbons have I in the bottle?

This question is from NH CA1 2011
http://www.orlesson.org/orp/11Ma/2011-P6-Math-CA1-NanHua.pdf
The given answer is 180.

Mrs Lim bought some bowls and plates. If she buys another 5 bowls, there are 60% as many bowls as plates. If she buys another 9 plates, there will be thrice as many plates as bowls. How many bowls did she buy? (CHS P5 SA2 2009)

This question is from CHS SA2 2009
http://www.orlesson.org/orp/09Ma/2009-Math-SA2-CHS.pdf
The given answer is 13.

Best wishes

[tianzhu]

Hi


To get an answer of 180 in the NH question and 13 in the CHS question, you’ve to interpret that the “if” in a statement signifies the event didn’t really happen.

For questions on “double if”, students work on the assumption that those scenarios didn’t happen. The two conditions are also independent or exclusive.

Consider this question.

Susan has some vanilla and chocolate sweets in her bag. If she adds 20 vanilla sweets into the bag, the percentage of vanilla sweets in the bag becomes 40%. If she adds 30 chocolate sweets, there will be 70% of chocolate sweets in the bag. How many vanilla and chocolate sweets does she have?

Best wishes

[Nebbermind]

tianzhu:

Hi


To get an answer of 180 in the NH question and 13 in the CHS question, you’ve to interpret that the “if” in a statement signifies the event didn’t really happen.

For questions on “double if”, students work on the assumption that those scenarios didn’t happen. The two conditions are also independent or exclusive.

Consider this question.

Susan has some vanilla and chocolate sweets in her bag. If she adds 20 vanilla sweets into the bag, the percentage of vanilla sweets in the bag becomes 40%. If she adds 30 chocolate sweets, there will be 70% of chocolate sweets in the bag. How many vanilla and chocolate sweets does she have?

Best wishes

OK. let me try to get it right.

For double 'if', ONLY use the 'original qty' as base for both computation. YA?

[Winx5015]

Hi,


Got problem with this question. Ps help!

Mother baked a total of 176 chocolate and vanilla cupcakes. After giving away 4/5 of the chocolate cupcakes and 2/3 of the vanilla cupcakes, she was left with 52 cupcakes altogether. How many chocolate cupcakes did she bake at first?

Thanks!

[tianzhu]

Winx5015:

Hi,


Got problem with this question. Ps help!

Mother baked a total of 176 chocolate and vanilla cupcakes. After giving away 4/5 of the chocolate cupcakes and 2/3 of the vanilla cupcakes, she was left with 52 cupcakes altogether. How many chocolate cupcakes did she bake at first?

Thanks!

Hi

This question touches on simultaneous concept.

You may use Units and Parts, letters of the alphabet or solve it the pictorial way.

Please refer to some of the earlier posts for examples of such solution.

I’ll use the alphabet way.

Chocolate
Gave ----- CCCC
Left ----- C

Vanilla
Gave ----- VV
Left ----- V

She was left with 52 cupcakes altogether.
C + V ----- 52

CCCC + VV ----- 124 (176-52)

CC + (CC+VV) ----- 124
CC + 104 ----- 124
CC ----- 20

C ----- 10

Chocolate ----- CCCCC ------- 50 at first.

Best wishes