Q&A - PSLE Math

[wkong]

CoffeeCat:

wkong:

Hi All,


Would like to seek advice for the following question:

At 9am John was cycling from Point A to Point B. At the same time, Mary was cycling from point B to point A using the same route as John. Cycliing at a speed of 4km/h faster than Mary, they would pass each other 300 meter away from the mid point.

a) What time did they pass each other?
b) If John took another 3 minutes to reach Point B, what time would Mary reach Point A?


Thanks,
wkong

If u draw the \"Distance line\", just a straight line from A to B, with the midpoint, and another point (call it D) , where D is between midpoint and B, you will see that line segment AD represent distance travelled by John while DB represent distance travelled by Mary.
Currently D is 300m away from midpoint. Now we are given the difference in speed, using DST formula, in order to find time, we need difference in distance. So by reflecting the point D about the midpoint, call the new point D', you will find that D'D represents the difference in distance, namely 600m = 0.6km.
Time taken = 0.6 / 4 = 0.15 hour = 9 mins

Part B)
This is an inverse speed ratio question, 1 of the most challenging type of speed question that can appear for p6 school syllabus.
Notice that John travel DB in 3 mins
while Mary travel DB in 9 mins (from part A).
Intuitively Mary travel 3 times as slow as John and hence will always take 3 times as much time as John to cover the same distance.
So if John travel from A to B in 3+ 9 = 27 min, Mary will travel from B to A in 3*27 = 81 mins.

Thanks for the help.

[CoffeeCat]

wkong:

Another Maths question that I need advice:


In the 2 classes 6A and 6B, the total number of girls was 100% more than the total number of boys. The ratio of boys to girls in 6A was 3:4 and the ratio of boys to girls in class 6B was 1:6. If there were 8 more girls in class 6A than class 6B, find the number of pupils in class 6A.


Thanks,
wkong

As usual to solve using ratios, we need to convert the units to be that of the same size. (The problem is A and B are of different sizes, we need to find their size relative to each other)
To accomplish this we can use algebra or guess & check.



Let A and B represent the number of students in 6A & 6B respectively.
Then 3/7 * A + 1/7 * B = boys
4/7 *A + 6/7 * B = girls
Notice girls = 2* boys
So 6/7 * A + 2/7 * B = 4/7 * A + 6/7 *B
2/7 * A = 4/7 * B
A = 2B

Now we can use ratios to finish the question
6A
B : G : Total
3 : 4 : 7
6 : 8 : 14
6B
B : G : Total
1: 6 : 7
(Check that 8+ 6 = 14 is 2 times (6+1) , satisfying that \" the total number of girls was 100% more than the total number of boys.\")
2 units -> 8
6A -> 14 units -> 56

[wkong]

CoffeeCat:

wkong:

Another Maths question that I need advice:


In the 2 classes 6A and 6B, the total number of girls was 100% more than the total number of boys. The ratio of boys to girls in 6A was 3:4 and the ratio of boys to girls in class 6B was 1:6. If there were 8 more girls in class 6A than class 6B, find the number of pupils in class 6A.


Thanks,
wkong

As usual to solve using ratios, we need to convert the units to be that of the same size. (The problem is A and B are of different sizes, we need to find their size relative to each other)
To accomplish this we can use algebra or guess & check.



Let A and B represent the number of students in 6A & 6B respectively.
Then 3/7 * A + 1/7 * B = boys
4/7 *A + 6/7 * B = girls
Notice girls = 2* boys
So 6/7 * A + 2/7 * B = 4/7 * A + 6/7 *B
2/7 * A = 4/7 * B
A = 2B

Now we can use ratios to finish the question
6A
B : G : Total
3 : 4 : 7
6 : 8 : 14
6B
B : G : Total
1: 6 : 7
(Check that 8+ 6 = 14 is 2 times (6+1) , satisfying that \" the total number of girls was 100% more than the total number of boys.\")
2 units -> 8
6A -> 14 units -> 56

Thanks again.

[atutor2001]

raghupavithra:

2/3​ of X is equal to 1/5​ of Y. Find the value of 80% of Y if X is 930. how to do this sum

Hi

Step 1 : Find the ratio of x :y
2/3 of x = 1/5 of y
2/3 of x = 2/10 of y
The ratio of x : y = 3 : 10 (the denominators of the fractions)

Step 2 : Find y
3 unit is 930
1 unit is 930/3 = 310
10 units (y) = 310 x 10 = 3100

Step 3 : Find 80% of y
80/100 x 3100 = 2480 (answer)

[bobsg01]

Answer sheet given was 10:3


Need help to know how it was derived.

Thanks.

http://i56.tinypic.com/255rsjo.jpg\">

[MathIzzzFun]

bobsg01:

Answer sheet given was 10:3


Need help to know how it was derived.

Thanks.

http://i56.tinypic.com/255rsjo.jpg\">

Hi

Hope this helps :lol:

http://www.flickr.com/photos/62167097@N02/5703951396/in/photostream

cheers.

[bobsg01]

Thanks MathIzzzFun.


Think I need to learn some math modeling.

:imcool: :imcool: :imcool:

http://i53.tinypic.com/295qufl.jpg\">

[MathIzzzFun]

bobsg01:

Thanks MathIzzzFun.


Think I need to learn some math modeling.

:imcool: :imcool: :imcool:

http://i53.tinypic.com/295qufl.jpg\">

hi bobsg01,

glad to be of help :lol:

cheers.

[Emelyn]

Hi,


Need help on this :

2010 Ai Tong School Paper 2 Question 15
The ration of the number of $100-notes to $5-notes that Angel had was 7:4. She exchanged 12 pieces of $100-notes for some $5-notes. After which, the ration of the number of $100-notes to $5-notes becamse 1:16. What was the amount of money Angel had at first ?

Many thanks.

[Cheerfuldad]

Hi all,


Please help!

Q 1. While on a Europe tour, Sam read 544 pages of a book at first. He read the remaining pages in 23 days, with the same number of pages each day. During these 23 days, he read 1/13 of the book in 7 days. How many pages of the book did he read in the 7 days?

Q 2. There were 65 more boys than girls at a carnival on Sunday. 60% of the boys and 2/7 of the girls who were at the carnival on Sunday went on Saturday too. The remaining 143 boys and girls went only on Sunday.
a) How many more boys than girls went to the carnival on both days?
b) Find the total number of boys and girls at the carnival on Sunday.

Thank you.