O-Level Additional Math
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Can ignore this, I already got the workings and answer. Thank you.
chrisu:
Please help A Sec.2 Maths question of Proportion.
Q) The scale of map P is 1:p and the scale of map Q is 1:q. If the same area is represented as 32sq.cm. on map P and 50sq.cm. on map Q, calculate the ratio p:q.
Kindly show workings as well.
Thank you. -
Sec1 maths Question:
Two race cars, Car X and Car Y, are at the starting point of a 2km track at the same time. Car X and Car Y make one lap every 60s and 80s respectively.
How long, in minutes, will it take for the faster car to be 5 laps ahead of the slower car?
Thanks all Gurus in advance. -
insanePaPa:
5 laps = 5 x 2km = 10kmSec1 maths Question:
Two race cars, Car X and Car Y, are at the starting point of a 2km track at the same time. Car X and Car Y make one lap every 60s and 80s respectively.
How long, in minutes, will it take for the faster car to be 5 laps ahead of the slower car?
Thanks all Gurus in advance.
Car A's speed = 2/60 = 1/30 kps
Car B's speed = 2/80 = 1/40 kps
Let T be time taken for Car A to be 10km ahead of Car B. Hence
(1/30)*T - (1/40)*T = 10
=> T(1/30 - 1/40) = 10
=> T(4-3/120) = 10
=> T(1/120) = 10
=> T = 1200s = 20mins -
ChiefKiasu:
Thanks a million chief.
5 laps = 5 x 2km = 10kminsanePaPa:
Sec1 maths Question:
Two race cars, Car X and Car Y, are at the starting point of a 2km track at the same time. Car X and Car Y make one lap every 60s and 80s respectively.
How long, in minutes, will it take for the faster car to be 5 laps ahead of the slower car?
Thanks all Gurus in advance.
Car A's speed = 2/60 = 1/30 kps
Car B's speed = 2/80 = 1/40 kps
Let T be time taken for Car A to be 10km ahead of Car B. Hence
(1/30)*T - (1/40)*T = 10
=> T(1/30 - 1/40) = 10
=> T(4-3/120) = 10
=> T(1/120) = 10
=> T = 1200s = 20mins -
chrisu:
1/p = sqrt(x/32) => p = sqrt(32/x)Can ignore this, I already got the workings and answer. Thank you.
chrisu:
Please help A Sec.2 Maths question of Proportion.
Q) The scale of map P is 1:p and the scale of map Q is 1:q. If the same area is represented as 32sq.cm. on map P and 50sq.cm. on map Q, calculate the ratio p:q.
Kindly show workings as well.
Thank you.
1/q = sqrt(x/50) => q = sqrt(50/x)
=> p/q = sqrt(32/50) = sqrt(16/25)
=> p:q = sqrt(16) : sqrt(25) = 4 : 5 -
YumYum:
The hard part to the question is trying to understand what the heck it is saying.Hi, need help with this sec1 Qn:
The key to a code is given by three whole numbers. The key is transmitted from the headquarters by 3 agents, who were given the product of 2 of the 3 numbers. Suppose that the numbers transmitted are 432, 540 and 720. What is the key code?
thanks.
The code is A, B, C
Agent X has A*B = 432
Agent Y has B*C = 540
Agent Z has A*C = 720
X/Y => A/C = 432/540 => A = C*432/540
Since A*C =720
=> C*C*432/540 = 720
=> C*C = 720 * 540/432 = 900
=> C = 30
A = 720/C = 720/30 = 24
B = 432/A = 432/24 = 18
So the 3 numbers are 18, 24 and 30. -
YumYum:
Sorry... don't understand what you typed. Can you scan the question and post it as an image?Hi, 1 more Qn:
Given that S =. 1
___________________________________________________
1/21 + 1/22 + 1/23 + 1/24 + 1/25 + 1/26 +...1/30
Find the largest whole number smaller than S.
Thanks -
Dear chief,
Thank you for your help.
For the 2nd Qn, never mind, it was covered in school.
Thanks ī -
Sec1 Qn:
A class has between 30 to 40 students. Each boy in the class brings 15 chocolates for a class party. The chocolates are shared equally among 20 girls and a teacher with no leftovers.
i) How many students are there in the class?
ii) How many chocolates does their teacher receive?
:thankyou: -
insanePaPa:
Hi insanePapa,Sec1 Qn:
A class has between 30 to 40 students. Each boy in the class brings 15 chocolates for a class party. The chocolates are shared equally among 20 girls and a teacher with no leftovers.
i) How many students are there in the class?
ii) How many chocolates does their teacher receive?
:thankyou:
Since there are between 30-40 students in the class, there should be between 10-20 boys (after deducting 20 girls).
If each boy brings 15 chocolates for the party, and the total chocolates are shared among 20 girls + 1 teacher = 21 people, we must find the number of boys such that when it is multiplied by 15 chocolates, the total number of chocolates can be divided equally among 21 people.
15 = 3 x 5
21= 3 x 7
Hence, possible values for the number of boys will be 7, 14, 21, etc.
Since we know that the value should be between 10-20, the number of boys should be 14.
i) Number of students in the class = 20 girls + 14 boys = 34.
ii) Each of the 20 girls & 1 teacher will receive 15 x 14 / 21 = 10 chocolates. Hence, answer is 10.
Cheers!
Jtutor
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