Hi everybody. Can anyone help me to solve these problems?
1) Two types of coffee, P and Q, are blended in the ratio 3:13 by weight to form a standard blend X. A second standard blend Y, is formed by blending type P coffee, type Q Coffee, and type R coffee in the ratio 1:7:12 by weight.
Calculate,
(a) the ratio of the three types of coffee in a third standard blend Z, formed by mixing equal weights of X and Y. (ANS P:Q:R= 19:93:48)
(b) the weight of type P coffee in 800g of Z. (ANS 95g)
2) Peanuts sell for $3 per round. Cashews sell for $6 per round. How many pounds of cashews should be mixed with 12 pounds of peanuts to obtain a mixture that sells for $4.2 per round? (ANS 8 pounds)
3) Steven left Town A and walked towards Town B at a speed of 100m/min. At the same time, Jason and Melvin started from Town B and walked towards Town A at a speed of 80m/min and 75m/min respectively. If Steven met Melvin 6 mins after passing Jason, find the distance, in km, between Town A and Town B. (ANS 37.8km)
Pls help. .
Lower Secondary Mathematics
PSLE marks the graduation of Primary school students and their entry into Secondary schools as teenagers. Discuss all issues about Secondary schooling here.
Lower Secondary Mathematics
by PuffyCakes » Wed Mar 20, 2013 3:23 pm

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Re: Sec1 Math Problem Sums needed
by ChewingPencilLine » Sun Mar 24, 2013 1:04 pm
Question 1(a)
For X, P : Q = 3 : 13, total 3 + 13 = 16 units
For Y, P : Q : R = 1 : 7 : 12, total 1 + 7 + 12 = 20 units
Since blend Z is formed by mixing equal weights of X and Y, the total number of units for X and Y must be the same in order for reasonable comparison (units must be of the same size!)
Lowest common multiple of 16 and 20 = 80
(16 x 5 = 80, 20 x 4 = 80)
Therefore, 5 'sets' of X and 4 'sets' of Y (are blended together to get Z. (Only then will there be equal weight of X and Y blended. Though of course, 10 and 8 sets and other integer multiples are possible too. However, since this is about ratio, eventually, you will still need to simplify; might as well work with the lowest common multiple!)
3 : 13 = 15 : 65
1 : 7 : 12 = 4 : 28 : 48
Add up the P and the Q...
Final answer is 15 + 4 : 65 + 28 : 48 = 19 : 93 : 48
For X, P : Q = 3 : 13, total 3 + 13 = 16 units
For Y, P : Q : R = 1 : 7 : 12, total 1 + 7 + 12 = 20 units
Since blend Z is formed by mixing equal weights of X and Y, the total number of units for X and Y must be the same in order for reasonable comparison (units must be of the same size!)
Lowest common multiple of 16 and 20 = 80
(16 x 5 = 80, 20 x 4 = 80)
Therefore, 5 'sets' of X and 4 'sets' of Y (are blended together to get Z. (Only then will there be equal weight of X and Y blended. Though of course, 10 and 8 sets and other integer multiples are possible too. However, since this is about ratio, eventually, you will still need to simplify; might as well work with the lowest common multiple!)
3 : 13 = 15 : 65
1 : 7 : 12 = 4 : 28 : 48
Add up the P and the Q...
Final answer is 15 + 4 : 65 + 28 : 48 = 19 : 93 : 48

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Re: Sec1 Math Problem Sums needed
by ChewingPencilLine » Sun Mar 24, 2013 1:07 pm
Question 1(b)
Once you get your answer for 1(a), 1(b) is not difficult...
From (a), we know that for blend Z, P : Q : R = 19 : 93 : 48
Total of 19 + 93 + 48 = 160 units
(notice how it is twice of 80, the LCM of 16 and 20)
Therefore, weight of type P coffee in 800g of blend Z is simply
19 / 160 * 800 = 95g
Once you get your answer for 1(a), 1(b) is not difficult...
From (a), we know that for blend Z, P : Q : R = 19 : 93 : 48
Total of 19 + 93 + 48 = 160 units
(notice how it is twice of 80, the LCM of 16 and 20)
Therefore, weight of type P coffee in 800g of blend Z is simply
19 / 160 * 800 = 95g

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Re: Sec1 Math Problem Sums needed
by ChewingPencilLine » Sun Mar 24, 2013 1:12 pm
Question (2)
12 pounds of peanuts would cost 12 * $3 = $36
Let the number of pounds of cashews mixed in be represented by x.
Given the price of the cashews and peanuts, it can be calculated that the total price of the mixture is given by
(x * $6) + $36 = $(6x + 36)  (1)
Given the total weight of the mixture (x + 12) pounds, it can be calculated that the total price of the mixture is given by
$4.2 * (x + 12) = $(4.2x + 50.4)  (2)
This is a set of simultaneous linear equations, which a Secondary 1 student should be able to solve easily:
6x + 36 = 4.2x + 50.4
1.8x = 14.4
x = 8
Recall that x stands for the number of pounds of cashew nuts to be mixed in.
Therefore, the answer is 8 pounds.
12 pounds of peanuts would cost 12 * $3 = $36
Let the number of pounds of cashews mixed in be represented by x.
Given the price of the cashews and peanuts, it can be calculated that the total price of the mixture is given by
(x * $6) + $36 = $(6x + 36)  (1)
Given the total weight of the mixture (x + 12) pounds, it can be calculated that the total price of the mixture is given by
$4.2 * (x + 12) = $(4.2x + 50.4)  (2)
This is a set of simultaneous linear equations, which a Secondary 1 student should be able to solve easily:
6x + 36 = 4.2x + 50.4
1.8x = 14.4
x = 8
Recall that x stands for the number of pounds of cashew nuts to be mixed in.
Therefore, the answer is 8 pounds.
Last edited by ChewingPencilLine on Sun Mar 24, 2013 1:21 pm, edited 2 times in total.

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Re: Sec1 Math Problem Sums needed
by ChewingPencilLine » Sun Mar 24, 2013 1:19 pm
Question 3
Let x be the time at which Steven met Jason in minutes, after the 3 of them started walking from Town A to B or B to A respectively.
in x minutes, Steven would have walked 100x meters.
in x minutes, Jason would have walked 80x meters.
Since Steven and Jason must have met some point in between town A and town B, we know that the distance between Town A and Town B is given by
100x + 80x = 180x  (1)
in x + 6 minutes, Steven would have walked 100 (x+6) = 100x + 600 meters.
in x + 6 minutes, Melvin would have walked 75 (x+6) = 75x + 450 meters.
Since Steven and Melvin must have met some point in between town A and town B, we know that the distance between Town A and Town B is given by
100x + 600 + 75x + 450 = 175x + 1050  (2)
Equating the 2 equations, you get
180x = 175x + 1050
5x = 1050
x = 210
Recall that the total distance between town A and town B is given by 180x (equation 1).
Therefore, 180 * 210 = 37800 meters = 37.8 km
Let x be the time at which Steven met Jason in minutes, after the 3 of them started walking from Town A to B or B to A respectively.
in x minutes, Steven would have walked 100x meters.
in x minutes, Jason would have walked 80x meters.
Since Steven and Jason must have met some point in between town A and town B, we know that the distance between Town A and Town B is given by
100x + 80x = 180x  (1)
in x + 6 minutes, Steven would have walked 100 (x+6) = 100x + 600 meters.
in x + 6 minutes, Melvin would have walked 75 (x+6) = 75x + 450 meters.
Since Steven and Melvin must have met some point in between town A and town B, we know that the distance between Town A and Town B is given by
100x + 600 + 75x + 450 = 175x + 1050  (2)
Equating the 2 equations, you get
180x = 175x + 1050
5x = 1050
x = 210
Recall that the total distance between town A and town B is given by 180x (equation 1).
Therefore, 180 * 210 = 37800 meters = 37.8 km

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Re: Sec1 Math Problem Sums needed
by ChewingPencilLine » Sun Mar 24, 2013 1:24 pm
The key is to read the question and understand what x should be used to stand for. A lot of the times, x is used to stand for the quantity you are looking for. For example, look at question 2. I used x to stand for the amount of cashew nuts to be mixed in, in pounds.
However, this is not always the case. Try to think and see what quantity x should represent to simplify the question. For example, in question 3, it is obvious that the common quantity between scenario 1 (Steven meeting Jason) and scenario 2 (Steven meeting Melvin) is the time at which Steven passed Jason (they tell you that Steven met Melvin 6 mins after passing Jason!).
Hope I have helped some!
However, this is not always the case. Try to think and see what quantity x should represent to simplify the question. For example, in question 3, it is obvious that the common quantity between scenario 1 (Steven meeting Jason) and scenario 2 (Steven meeting Melvin) is the time at which Steven passed Jason (they tell you that Steven met Melvin 6 mins after passing Jason!).
Hope I have helped some!

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Re: Sec1 Math Problem Sums needed
by ChewingPencilLine » Sun Mar 24, 2013 9:08 pm
Haha, no problem. And it is just one soul :p Me!

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How to avoid Careless Mistakes for Maths?
by mathtuition88 » Sun May 05, 2013 10:10 pm
Dear Parents,
Many parents have feedback to me that their child often makes careless mistakes in Maths, at all levels, from Primary, Secondary, to JC Level.
How to avoid Careless Mistakes for Maths?
Please read my article, which includes some suggestions and tips.
http://mathtuition88.wordpress.com/sing ... tionblog/
Many parents have feedback to me that their child often makes careless mistakes in Maths, at all levels, from Primary, Secondary, to JC Level.
How to avoid Careless Mistakes for Maths?
Please read my article, which includes some suggestions and tips.
http://mathtuition88.wordpress.com/sing ... tionblog/

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