Please help! Sec 1 Maths.
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1. Kishore and Ajay can do a piece of work in 12 days, Ajay and Vijay in 15 days and Vijay and Kishore in 20 days. In how many days will they finish it if they work together? 2. Mahesh and Ramesh can finish a piece of work in 12 days, Ramesh and Satish finish in 15 days while Satish and Mahesh can finish it in 20 days. All the three started the work but Ramesh left after two days. After working for four more days, Satish also leaves. In how many more days will Mahesh finish the work? How do I do these questions?
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1 In 1 day, K + A ==> 1/12 work done A + V ==> 1/15 work done V + K ==> 1/20 work done K+ K + A + A + V + V ==> 1/5 work dones If there is 1 of each guy, K + A + V ====> 1/10 work done work will be done in 10 days... edit here due to weird formating in the reply below
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3 ppl work will become so slow meh? Did consider that method but it seems abit wierd that 3 ppl work does it even slower so i didn't post that answer.
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Originally posted by dkcx:
3 ppl work will become so slow meh? Did consider that method but it seems abit wierd that 3 ppl work does it even slower so i didn't post that answer.
because got 3 guys so they slack more, haha... actually, time flows concurrently for all guys, so my method is wrong, haha, thanks for telling me 1 In 1 day, K + A ==> 1/12 work done A + V ==> 1/15 work done V + K ==> 1/20 work done K+ K + A + A + V + V ==> 1/5 work dones If there is 1 of each guy, K + A + V ====> 1/10 work done work will be done in 10 days... -
Not slack bah, 3 ppl more things to talk, talk talk talk 4get to do work :p
Oh, i didn't think of fractions and yeah thats alot more logical. Should be able to solve the 2nd equation using the same method except its abit more complicated with the leaving. Maybe solve for M, R and S using simulataneous equations might work but i don't have time to work it out with exams in 3 days. U can give it a try :)
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lol memories
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why all indian names ?
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18 days.
use the previous working, K+A+V ---> 1/10
A+V ---> 1/15
K ---> 1/10 - 1/15 = 1/30
i.e. Mahesh finish 1/30 work in a day.
3 ppl work together for 2 days => 2/10 done, left 8/10.
Ramesh leave => 1/20 work per day x 4 days => 1/5 work done, left 6/10.
This 6/10 Mahesh will do alone.
work / rate = time to do work
i.e. 6/10 / 1/30 = 18days to finish
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bonus question will be: what kind of work do these indians do? (a mental image comes to mind... oops) =D
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sure this is sec 1 maths? i am sec one n as far as i know i haf never met such a questions..most schools are at the algebra topic for sec ones arent they?
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This is considered algebra.
I think the indians was making pratas?
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The 2 questions under the topic Rate, ratio, and speed (something liddat).
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Okay, shall use this thread instead of starting a new one. Number pattern question.
1st: 24 toothpicks
2nd: 36 toothpicks
3rd: 52 toothpicks
4th: 72 toothpicks
So I know the difference between the numbers are like 12 and 16 and then 20 subsequently by adding 4 to the difference.
How do I come out with a algebraic equation such that when i substitute the number (1, 2, 3, 4, 5) in, I get the number I want?
Like if the question asks for the 109th term of this patten, what would be the answer?
Hah thanks.
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1st ==> 24 + 12(0)
2nd ==>24 + 12(1) + 4(0)
3rd ==> 24 + 12(2) +4(1)
4th ==> 24 + 12(3) + 4(2)
5th ==> 24 + 12(4) + 4(3)
is this right so far?
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Originally posted by skythewood:
1st ==> 24 + 12(0)
2nd ==>24 + 12(1) + 4(0)
3rd ==> 24 + 12(2) +4(1)
4th ==> 24 + 12(3) + 4(2)
5th ==> 24 + 12(4) + 4(3)
is this right so far?
Eh, the 4th one should be 72? But according to your equation it's 68. -
Originally posted by d3sT1nY:
Eh, the 4th one should be 72? But according to your equation it's 68.ok. you adjust the stuff til number 5. than find the pattern and write the forumla.
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Originally posted by d3sT1nY:
Okay, shall use this thread instead of starting a new one. Number pattern question.
1st: 24 toothpicks
2nd: 36 toothpicks
3rd: 52 toothpicks
4th: 72 toothpicks
So I know the difference between the numbers are like 12 and 16 and then 20 subsequently by adding 4 to the difference.
How do I come out with a algebraic equation such that when i substitute the number (1, 2, 3, 4, 5) in, I get the number I want?
Like if the question asks for the 109th term of this patten, what would be the answer?
Hah thanks.
1 ==> 24
2 ==> 24 + 12
3 ==> 24 + 12 + 16
4 ==> 24 + 12 + 16 + 20
5 ==> 24 + 12 + 16 + 20 + 24
....
109 ==> 24 + 12 + 16 + 20 + 24 + ... + 440
(pattern is in the last term, 4 * (n+1) for nth term)
= 24 + 4 * (3 + 4 + 5 + ... + 110)
= 24 + 4 * (108/2) (113) ===> sum up of 3 + 110, or 4 + 109, or 5 + 108, etc, ... pri 6 PSLE question came up similar one before
= 6126 toothpicks
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not sure if this is in syllabus, but 1 + 2 + ... + n = (n/2)(n+1)
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Originally posted by eagle:
1 ==> 24
2 ==> 24 + 12
3 ==> 24 + 12 + 16
4 ==> 24 + 12 + 16 + 20
5 ==> 24 + 12 + 16 + 20 + 24
....
109 ==> 24 + 12 + 16 + 20 + 24 + ... + 440
(pattern is in the last term, 4 * (n+1) for nth term)
= 24 + 4 * (3 + 4 + 5 + ... + 110)
= 24 + 4 * (108/2) (113) ===> sum up of 3 + 110, or 4 + 109, or 5 + 108, etc, ... pri 6 PSLE question came up similar one before
= 6126 toothpicks
hey hey, summation... nice.
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Alright thanks.
Anyway the 108 is because 109-1=108, the 1 is the first term which is 24 right? Ok, right.
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108 is because of 110 - 2