PSLE Maths question. How to solve?

[Cooliechang]

From Yahoo Sg:

 

Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim's sweets to chocolates became 1:7 and the ratio of Ken's sweets to chocolates became 1:4. How many sweets did Ken buy?"

 

Ans: 68.

 

Any kind souls to show me the steps to get the answer.

 

I tried but failed.

 

Haiz... really worry for my son next time.

[Sleepyman]

From Yahoo Sg:

 

Jim bought some chocolates and gave half of it to Ken. Ken bought some sweets and gave half of it to Jim. Jim ate 12 sweets and Ken ate 18 chocolates. The ratio of Jim

[Hiphiphoray]

Im oredi worried like hell for my girl going P1 next year. Primary school maths is not as easy as before.

 

So much so we parents cant teach them the method from our time.

 

And sorry, i cant help you. But my wife can.....but she has a 6.40am flight to chicago to catch in a few hours time. Sleeping now.

Edited October 10, 2009 by Hiphiphoray

[Bulls_is_back]

After the exchange both Jim and Ken had the same number of C (no. of chocolates) and S (no. of sweets)

 

From Jim

(S-12)/C=1/7

7(S-12)=C

 

From Ken

S/(C-18)=1/4

4S=(C-18)

C=4S+18

 

7(S-12)=4S+18

7S-84=4S+18

3S=102

S=34

 

Hence at the beginning there were 2S=68 sweets.

[Scenic98]

no need to look at p6 math questions. some of the p3 math questions already siong enuff. if a bunch of adults need to come to a forum and ask around, what do u think a kid can do under exam condition?

 

so let's say a kid can solve the problem and the adult can't, does that mean s/he is smarter than the adult?

[1fast1]

If you can do it with algebra, it's a snap.

 

The giving half to each other thing is just a convoluted way of saying they start out with the same number of sweets and chocolates. Let's say J and K have c chocolates and s sweets each after the exchange.

 

J eats 12 sweets and is therefore left with s-12 sweets and c chocolates.

K eats 18 chocolates and is therefore left with s sweets and c - 18 chocolates.

 

We're given the ratios, putting that into equations we get:

 

(s - 12)/c = 1/7

and

s/(c - 18) = 1/4

 

Solve the simultaneous equations and we get s = 34.

 

Since the orig. no. of sweets K bought is twice that, answer is 68.

 

What I worry about is how the heck do I do it without algebra? I think that's the new (and very stupid) requirement nowadays, if I understood my friend correctly.

[1fast1]

I recently looked at this book of actual primary school exam questions in math. Gosh, things have really moved on from when I was a kid. Now the kids have to answer questions like deducing the nets of a solid shape and summing up arithmetic progressions like 1+3+5+...+19. Presumably, the latter problem is not supposed to be done with canned formulae but with flashes of insight, like Gauss did it as a kid (well, he summed all nos. from 1 to 100).

[Shull]

If you can do it with algebra, it's a snap.

 

The giving half to each other thing is just a convoluted way of saying they start out with the same number of sweets and chocolates. Let's say J and K have c chocolates and s sweets each after the exchange.

 

J eats 12 sweets and is therefore left with s-12 sweets and c chocolates.

K eats 18 chocolates and is therefore left with s sweets and c - 18 chocolates.

 

We're given the ratios, putting that into equations we get:

 

(s - 12)/c = 1/7

and

s/(c - 18) = 1/4

 

Solve the simultaneous equations and we get s = 34.

 

Since the orig. no. of sweets K bought is twice that, answer is 68.

 

What I worry about is how the heck do I do it without algebra? I think that's the new (and very stupid) requirement nowadays, if I understood my friend correctly.

 

need to use dunno what bars etc...did it before but kinda forgot it once i mastered algebra..

 

[Gnolard]

my gosh!! these are PSLE???.....

[Ronnchee]

Simply Chimmmmm

[Huggable]

Both Jim & Ken have same number of sweets & chocolates.

 

Let's say #sweets Jim have = S

Let's say #chocolates Ken have = C

 

Jim ate 12 sweets, and ratio became 1:7. So,

7(S-12) = C -> Equation (1)

 

Ken ate 18 chocolates, and ratio became 1:4. So,

4S = C-18 -> Equation (2)

 

Solving (1) C = 7S-84

Solving (2) C = 4S+18

 

Putting both together,

 

7S-84 = 4S+18

7S-4S = 18+84

3S = 102

S = 34

 

Number of sweets Jim have is 34. Since Jim gave Ken half of what he bought, total number of sweets is 2x34 = 68.

[Spiderman302]

After the exchange both Jim and Ken had the same number of C (no. of chocolates) and S (no. of sweets)

 

From Jim

(S-12)/C=1/7

7(S-12)=C

 

From Ken

S/(C-18)=1/4

4S=(C-18)

C=4S+18

 

7(S-12)=4S+18

7S-84=4S+18

3S=102

S=34

 

Hence at the beginning there were 2S=68 sweets.

 

Pri sch cant use algebra to slove. Illegal.

 

[Babyt]

just call ken and jim on hp and ask them how many they brought can liao mah LOL

 

come to think of it...what kinds of jobs require to use such algebra/equations etc?

 

do we apply them in our daily life?

[Shull]

just call ken and jim on hp and ask them how many they brought can liao mah LOL

 

come to think of it...what kinds of jobs require to use such algebra/equations etc?

 

do we apply them in our daily life?

 

It's not really applicable to daily life..but these type of question are like introduction to prepare them for algebra when these kids reach sec school..

[Alfisti168]

I really feel that those people up there in SEAB managing these PSLE papers contain brains containing Ris Low-material.

 

Really want everyone to BOOMZ their exam then they happy.

 

Although it is true to a certain extent that candidates today are smarter nowadays due to tuitions, enrichment classes in schools and lots of remedials, is there a need to put such a high yardstick to band the students according to the grades that they deserve?

Edited October 10, 2009 by Alfisti168

[Alfisti168]

Who is the actress in your avatar? She's chio.

Edited October 10, 2009 by Alfisti168

[Bystander50]

If you can do it with algebra, it's a snap.

 

The giving half to each other thing is just a convoluted way of saying they start out with the same number of sweets and chocolates. Let's say J and K have c chocolates and s sweets each after the exchange.

 

J eats 12 sweets and is therefore left with s-12 sweets and c chocolates.

K eats 18 chocolates and is therefore left with s sweets and c - 18 chocolates.

 

We're given the ratios, putting that into equations we get:

 

(s - 12)/c = 1/7

and

s/(c - 18) = 1/4

 

Solve the simultaneous equations and we get s = 34.

 

Since the orig. no. of sweets K bought is twice that, answer is 68.

 

What I worry about is how the heck do I do it without algebra? I think that's the new (and very stupid) requirement nowadays, if I understood my friend correctly.

 

I don't think simultaneous equations are taught at this level.

If it's true, might as well ge those P6 kids to go direct for `O' level maths.

 

Looking at the type of questions for the past year, I wonder if they're testing your understanding of English with Maths. Maths questions should be straight forward and not beating around the bush confusing everyone. Seems like GCE `O' Levels Maths papers are easier.

 

 

[Babyt]

we learnt plenty of mathematics equations/formulas in sch but sadly ... we dont really use it in most jobs.