Q&A - P3 Math
-
The cost of 7 erasers and 3 rulers is $5.05.
The cost of 2 erasers and 6 rulers is $3.95
Find the cost of 2 erasers and 2 rules.
7 erasers + 3 rulers = 5.05 --------(1) x 2
14 erasers + 6 rulers = 10.1 --------(2)
2 erasers + 6 rulers = 3.95 ---------(3)
(2)-(3),
12 erasers = 6.15
2 erasers = 1.025
Then 6 rulers = 3.95-1.025 = 2.925
2 rulers = 0.975
So 2 erasers + 2 rulers = 1.025 + 0.975 = $2
or (1) + (2)
9 erasers + 9 rulers = $9
2 erasers + 2 rulers = $2 (proportionately) -
ksi:
Thanks ksi. But I do not how to explain this simultaneous equation to a P3The cost of 7 erasers and 3 rulers is $5.05.
The cost of 2 erasers and 6 rulers is $3.95
Find the cost of 2 erasers and 2 rules.
7 erasers + 3 rulers = 5.05 --------(1) x 2
14 erasers + 6 rulers = 10.1 --------(2)
2 erasers + 6 rulers = 3.95 ---------(3)
(2)-(3),
12 erasers = 6.15
2 erasers = 1.025
Then 6 rulers = 3.95-1.025 = 2.925
2 rulers = 0.975
So 2 erasers + 2 rulers = 1.025 + 0.975 = $2
or (1) + (2)
9 erasers + 9 rulers = $9
2 erasers + 2 rulers = $2 (proportionately)
My initial solution is :
7 Erasers + 3 Rulers = $5.05
2 Erasers + 6 Rulers = $ 3.95
Add them together we have
9 Erasers + 9 Rulers = $5.05 + 3.95 = $9
Each set of 1 Eraser 1 Ruler = $1
Therefore 2 sets of Eraser & Ruler = $2.
I'm thinking since the question ask for the price of 2 sets, the working out of each price is not necessary.
Anyone to shed some lights? Thanks in advance. -
Actually the second method i gave is the same as yours and it is not necessary to find the cost for 1, you can directly give the answer for 2 of each item.
Also in method 1, I was not thinking of simultaneous equation, I was using common sense but I guess it looked like simultaneous equation. :lol: -
Devnahouse:
http://www.postimage.org/image.php?v=TsKNGfAHello,
Can help me solve this one.
The cost of 7 erasers and 3 rulers is $5.05.
The cost of 2 erasers and 6 rulers is $3.95
Find the cost of 2 erasers and 2 rules.
Many thanks. -
Dharma, Thanks for taking time to explain in model form. Thanks ksi for your further clarifications. Appreciated.
-
Easiest method to solve this one?
A snail climbed 1cm up a wall in Day 1,
2cm in Day 2, 3cm in Day 3 and so on..
Each day, it climbed 1cm more than it did
the previous day. How many days did the
snail take to climb 21cm? -
buds:
Hi BudsEasiest method to solve this one?
A snail climbed 1cm up a wall in Day 1,
2cm in Day 2, 3cm in Day 3 and so on..
Each day, it climbed 1cm more than it did
the previous day. How many days did the
snail take to climb 21cm?
Easiest way is by listing
Day 1 => 1
Day 2 => 1 + 2 = 3
Day 3 => 1 + 2 + 3 = 6
Day 4 => 1 + 2 + 3 + 4 = 10
Day 5 => 1 + 2 + 3 + 4 + 5 = 15
Day 6 => 1 + 2 + 3 + 4 + 5 + 6 = 21
The snail took 6 days to climb 21cm
But if we have a large number, eg number of days taken by the snail to climb 210cm; using listing method will be very time consuming.
We need to establish a pattern
This is a question a series of consecutive numbers starting from 1
1 + 2 + 3 + ….+ N = N x (N + 1)/2
So, to find no. of days for snail to climb 210cm
N x (N +1 ) / 2 = 210
N x ( N + 1 ) = 420
We know 20 x 21 = 420
N = 20
So, in 20 days the snail will be able to climb 210cm -
Hi Dharma.
Sorry to be a buzzkill, but the actual question was 21 cm, not 210 cm… -
Muffins:
Hi Dharma.
Sorry to be a buzzkill, but the actual question was 21 cm, not 210 cm....
Hi Muffins,
To climb 21cm
N x (N + 1) / 2 = 21
N x (N + 1) = 42
6 x 7 = 42
N = 6 days
By using 210cm I was trying to show that it will be easier to solve the problem by finding a pattern. The listing method will
be just too long. -
Can you help to explain why is it wrong.
I am a 3 digit number.
My hundreds and tens digit are the same.
My ones digit is 4 more than my tens digit.
What number can I be?
My son writes 448, but the teacher marked it wrong and said the correct ans is 559. I thought the ans can be either 559, 448, 337, 226 or 115
Thanks
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