Q&A - PSLE Math

[MathIzzzFun]

Michaelia0816:

MathIzzFun, this is the Q I needed help with,


A train 210m long was travelling at 90km/h and it overtook a train travelling 72km per hour in 55s. What was the length of the second train?

Hi

need to assume that time taken to overtake is the time when train A catches up with train B (ie its front end just reaches rear of second train) to the time train A completely pulls away from train B ie rear of A just passes front end of B.

http://i50.tinypic.com/1pgzu9.png\">

cheers.

[Michaelia0816]

Thx!

[Michaelia0816]

MathIzzzFun:

Michaelia0816:

MathIzzFun, this is the Q I needed help with,


A train 210m long was travelling at 90km/h and it overtook a train travelling 72km per hour in 55s. What was the length of the second train?

Hi

need to assume that time taken to overtake is the time when train A catches up with train B (ie its front end just reaches rear of second train) to the time train A completely pulls away from train B ie rear of A just passes front end of B.

http://i50.tinypic.com/1pgzu9.png\">

cheers.

But I got something unsure yet,
Why got 1000 and 60 x 60 ?
How do you get it?

[MathIzzzFun]

Michaelia0816:

MathIzzzFun:

[quote=\"Michaelia0816\"]MathIzzFun, this is the Q I needed help with,


A train 210m long was travelling at 90km/h and it overtook a train travelling 72km per hour in 55s. What was the length of the second train?

Hi

need to assume that time taken to overtake is the time when train A catches up with train B (ie its front end just reaches rear of second train) to the time train A completely pulls away from train B ie rear of A just passes front end of B.

http://i50.tinypic.com/1pgzu9.png\">

cheers.

But I got something unsure yet,
Why got 1000 and 60 x 60 ?
How do you get it?[/quote]Train A is faster --> 90-72 = 18 km/h = 18000 m/h = 18000 / (60x60) m/s

cheers.

[pixiedust]

Appreciate help with this question (4marks) :

(I am continually challenged by circles type of questions!!! Any tips for me ?)

http://i1197.photobucket.com/albums/aa424/pixie_dust8/mq5.jpg\">

[dicoyote]

MathIzzzFun:

dicoyote:

Hi,


Need help, what method can be use for this question? guess and check?

A logistics company charges $15 for every 2 vases delivered on time and $10 for every 3 vases delivered late.
In June, the company had to deliver 500 vases.
The company collected $3675 for the delivery.
How many vases were delivered late?

Hi

one way is to \"convert\" the question to a typical \"chicken & cow\" question.
3 vases delivered late --> $10 -- 30 units (3 units/$1), 1 vase delivered late --> 10 units
2 vases delivered on time --> $ 15 --- 45 units, 1 vase delivered on time --> 22.5 units

$3675 --> 11025 units
22.5 units - 10 units = 12.5 units

Number of vases delivered late = (500 x 22.5 - 11025) ÷ 12.5 = 18

cheers.

Hi MathIzzzFun,

Thanks for the solution, though pardon me for question, I understand why the uses of \"x3\" for the values, though need a little more help on the explanation on this equations, am a bit lost, thanks

\"(500 x 22.5 - 11025) ÷ 12.5\"

[tianzhu]

pixiedust:

Appreciate help with this question (4marks) :

(I am continually challenged by circles type of questions!!! Any tips for me ?)

http://i1197.photobucket.com/albums/aa424/pixie_dust8/mq5.jpg\">

Hi

Perimeter of shaded parts ------ Perimeter of quadrant (r4) + Perimeter of semicircle (r3)

14.28 + 15.42 ----- 29.7

For part (b)

Let the unshaded area be labeled as B. It is shared by the quadrant and the semi circle.

Area of quadrant (r4) ------- A (shaded area) + B (unshaded area)

Area of semicircle (r3) ------- B (unshaded area) + C (shaded area)

Difference in shaded area ----- area of semi circle (r3) – area of quadrant (r4)

14.13 - 12.56 ----- 1.57

I’ll leave it to you to work out the details.

Best wishes

[MathIzzzFun]

dicoyote:

MathIzzzFun:

[quote=\"dicoyote\"]Hi,


Need help, what method can be use for this question? guess and check?

A logistics company charges $15 for every 2 vases delivered on time and $10 for every 3 vases delivered late.
In June, the company had to deliver 500 vases.
The company collected $3675 for the delivery.
How many vases were delivered late?

Hi

one way is to \"convert\" the question to a typical \"chicken & cow\" question.
3 vases delivered late --> $10 -- 30 units (3 units/$1), 1 vase delivered late --> 10 units
2 vases delivered on time --> $ 15 --- 45 units, 1 vase delivered on time --> 22.5 units

$3675 --> 11025 units
22.5 units - 10 units = 12.5 units

Number of vases delivered late = (500 x 22.5 - 11025) ÷ 12.5 = 18

cheers.

Hi MathIzzzFun,

Thanks for the solution, though pardon me for question, I understand why the uses of \"x3\" for the values, though need a little more help on the explanation on this equations, am a bit lost, thanks

\"(500 x 22.5 - 11025) ÷ 12.5\"[/quote]3 vases delivered late --> $10 -- 30 units (3 units/$1), 1 vase delivered late --> 10 units
2 vases delivered on time --> $ 15 --- 45 units, 1 vase delivered on time --> 22.5 units
$3675 --> 11025 units

question now becomes :
Total number of vases delivered = 500 (**think: total number of animals)
1 vase delivered on time --> 22.5 units (**think: 1 cow --> 4 legs)
1 vase delivered late --> 10 units (** think: 1 chicken --> 2 legs)
22.5 units - 10 units = 12.5 units (**think: 4-2 = 2 legs -- difference in number of legs)
(difference in amount collected)
Total amount collected --> 11025 units (**think: total number of legs)

All vases delivered on time --> 22.5 units x 500 = 11250 units (**think: all are cows)
11250 - 11025 = 225 units, extra amount collected if all delivered on time (think: extra legs if all are cows)
Number of vases delivered late = 225 units / 12.5 units = 18

using assumption method:
for every 6 vases delivered time, amount collected = 6/2 x $15 = $45
for every 6 vases delivered late, amount collected = 6/3 x $ 10 = $20

If all delivered on time, amount collected = 500/2 x $ 15 = $3750
$3750 - $3675 = $75
Number of vases delivered late = $75 / ($45-$20) x 6 = 18

cheers.

[dicoyote]

Hi MathIzzzFun,


Thanks for the solution, though pardon me for question, I understand why the uses of \"x3\" for the values, though need a little more help on the explanation on this equations, am a bit lost, thanks

\"(500 x 22.5 - 11025) ÷ 12.5\"[/quote]

3 vases delivered late --> $10 -- 30 units (3 units/$1), 1 vase delivered late --> 10 units
2 vases delivered on time --> $ 15 --- 45 units, 1 vase delivered on time --> 22.5 units
$3675 --> 11025 units

question now becomes :
Total number of vases delivered = 500 (**think: total number of animals)
1 vase delivered on time --> 22.5 units (**think: 1 cow --> 4 legs)
1 vase delivered late --> 10 units (** think: 1 chicken --> 2 legs)
22.5 units - 10 units = 12.5 units (**think: 4-2 = 2 legs -- difference in number of legs)
(difference in amount collected)
Total amount collected --> 11025 units (**think: total number of legs)

All vases delivered on time --> 22.5 units x 500 = 11250 units (**think: all are cows)
11250 - 11025 = 225 units, extra amount collected if all delivered on time (think: extra legs if all are cows)
Number of vases delivered late = 225 units / 12.5 units = 18

using assumption method:
for every 6 vases delivered time, amount collected = 6/2 x $15 = $45
for every 6 vases delivered late, amount collected = 6/3 x $ 10 = $20

If all delivered on time, amount collected = 500/2 x $ 15 = $3750
$3750 - $3675 = $75
Number of vases delivered late = $75 / ($45-$20) x 6 = 18

cheers.[/quote]

Hi MathIzzzFun,

Thank you so much, am amazed at the details you have provided, not only that, even applying side by side with the \"chicken & cow\" logic, which make understanding more easier

Just to clarify my understanding, the \"x3\" uses across for the value is to find a similarity basis for the equation, am I correct? :scratchhead:

[charsen]

How to solve this question ?

The sum of 4 numbers is 80.If you add 3 to the first number, subtract 3 from the second number,multiply the third number by 3 d divide the fourth number by 3,the result will be equal.What is the difference between the largest and the smallest of the original numbers.