Q&A - PSLE Math

[MathAche]

Is there a simpler method to solve this question (extract from My Pals are Here, Maths Homework, Primary 6A)


Question
-----------
Keith and Shawn each have some money. If Keith spends $38, the ratio of the amount of money Keith has to the amount of money Shawn has will become 2:5. If Shawn spends $38, the ratio will become 8:13. How much money does each boy have?

Answer
--------
Eqn 1: (K-38)/S = 2/5 -> 2S = 5K-190
Eqn 2: K/(S-38) = 8/13 -> 13K = 8S-304

Substitute eqn 1 into eqn 2 to get
13K = 4 (5K-190) - 304
K = 152

Solve eqn 1 using K=152

S = ((5x152) -190)/2 = 285

Keith has $152 and Shawn has $285

[MathIzzzFun]

MathAche:

Would somebody please check this (extract from My Pals are Here, Maths Homework - P6)


Question
-----------
The ratio of Jame's age to Andy's age is 2:7. In 8 years time, the ratio will become 2:5. Find Jame's present age.

Answer
---------

Eqn 1: J/A = 2/7 -> 7J=2A

Eqn 2: (J+8) / (A+8) = 2/5 -> 5J + 40 = 2A + 16

Substitute eqn 1 into eqn 2

5J + 40 = 7J +16
7J-5J = 40-16
2J = 24
J = 12
James present age is 12 years old.

James : Andy
2 : 7
12 : 42

Andy is 42 years old.

After 8 years
James : Andy
2 : 5
20 : 50 (or 12+8 : 42+8)

Hi

Age difference will remain the same.

Present-
James : Andy --> 2 : 7 --> 6u : 21u
8 years' time-
James : Andy --> 2 : 5 --> 10u : 25u
So, 4u --> 8, 1u--> 2
Present age :
James --> 6u = 12 years old
Andy --> 21u --> 42 years old.

cheers.

[MathIzzzFun]

MathAche:

Is there a simpler method to solve this question (extract from My Pals are Here, Maths Homework, Primary 6A)


Question
-----------
Keith and Shawn each have some money. If Keith spends $38, the ratio of the amount of money Keith has to the amount of money Shawn has will become 2:5. If Shawn spends $38, the ratio will become 8:13. How much money does each boy have?

Answer
--------
Eqn 1: (K-38)/S = 2/5 -> 2S = 5K-190
Eqn 2: K/(S-38) = 8/13 -> 13K = 8S-304

Substitute eqn 1 into eqn 2 to get
13K = 4 (5K-190) - 304
K = 152

Solve eqn 1 using K=152

S = ((5x152) -190)/2 = 285

Keith has $152 and Shawn has $285

Hi

Total amount Keith and Shawn had left in both cases are the same.

Keith spends $ 38:
Keith : Shawn --> 2:5 --> 6u : 15u (total 21u)
Shawn spends $38:
Keith : shawn --> 8u : 13u (total 21u)
so 2u --> $38, 1u--> $19
Keith's money --> 8u --> 8 x $ 19 = $ 152
Shawn's money --> 15u --> 15 x $ 19 = $ 285

cheers.

[small]

MathIzzzFun:

MathAche:

Would somebody please check this (extract from My Pals are Here, Maths Homework - P6)


Question
-----------
The ratio of Jame's age to Andy's age is 2:7. In 8 years time, the ratio will become 2:5. Find Jame's present age.

Answer
---------

Eqn 1: J/A = 2/7 -> 7J=2A

Eqn 2: (J+8) / (A+8) = 2/5 -> 5J + 40 = 2A + 16

Substitute eqn 1 into eqn 2

5J + 40 = 7J +16
7J-5J = 40-16
2J = 24
J = 12
James present age is 12 years old.

James : Andy
2 : 7
12 : 42

Andy is 42 years old.

After 8 years
James : Andy
2 : 5
20 : 50 (or 12+8 : 42+8)

Hi

Age difference will remain the same.

Present-
James : Andy --> 2 : 7 --> 6u : 21u
8 years' time-
James : Andy --> 2 : 5 --> 10u : 15u
So, 4u --> 8, 1u--> 2
Present age :
James --> 6u = 12 years old
Andy --> 21u --> 42 years old.

cheers.

MathIzzzFun.

typo error in your solution,

It should be 25u instead of 15u

cheers.

[MathIzzzFun]

small:

MathIzzzFun:

[quote=\"MathAche\"]Would somebody please check this (extract from My Pals are Here, Maths Homework - P6)


Question
-----------
The ratio of Jame's age to Andy's age is 2:7. In 8 years time, the ratio will become 2:5. Find Jame's present age.

Answer
---------

Eqn 1: J/A = 2/7 -> 7J=2A

Eqn 2: (J+8) / (A+8) = 2/5 -> 5J + 40 = 2A + 16

Substitute eqn 1 into eqn 2

5J + 40 = 7J +16
7J-5J = 40-16
2J = 24
J = 12
James present age is 12 years old.

James : Andy
2 : 7
12 : 42

Andy is 42 years old.

After 8 years
James : Andy
2 : 5
20 : 50 (or 12+8 : 42+8)

Hi

Age difference will remain the same.

Present-
James : Andy --> 2 : 7 --> 6u : 21u
8 years' time-
James : Andy --> 2 : 5 --> 10u : 25u
So, 4u --> 8, 1u--> 2
Present age :
James --> 6u = 12 years old
Andy --> 21u --> 42 years old.

cheers.

MathIzzzFun.

typo error in your solution,

It should be 25u instead of 15u

cheers.[/quote]thanks .. partially into dream land...

cheers.

[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.