Tutor MathsGuru: Ask me for your burning Maths questions!
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powforhighschool:
Two candles of the same height are lit at the same time. the first is consumed in 4 hours and the second in 3 hours. Assuming that each candle burns at a constant rate, in how many hours after being lit was the first candle twice the height of the second?
Hi
If you are looking for algebra solution, here's one from vanilla cake for a similar type of question.
Candle A and candle B are of the same length. Candle A, which is broader, can burn for 5 h while Candle B, the thinner candle, can burn for 4 h. If both candles are lighted at the same time, how long does it take for Candle A to be twice as long left as Candle B?
Using algebra :
Assume the number of hours burnt be n
1 on the left side is 5/5 and 1 on the right hand side is 4/4
1-n/5=2(1-n/4)
1-n/5=2-n/2
(5-n)/5=(4-n)/2
2(5-n)=5(4-n)
10-2n=20-5n
3n=10
n= 3⅓ hours
It takes 3⅓ hours for Candle A to be twice as long left as Candle B.
Best wishes -
Hifive:
Exactly... Lydia had 4 times as much money as Jenny at last.Hi Dharma
Thanks for the solution. Can you pls explain why you place
$255 – 1p = 4($370 – 2p)
is it to \"balance\" up both sides since Lydia has 4 units left? -
Hi Dharma
Thanks! Have a nice day! -
Hifive:
No worries. You have a nice day too!Hi Dharma
Thanks! Have a nice day! -
Thank you that helped alot.
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1.the average cost of a 1990 car depreciates linearly, so that in 1993, the average cost of the car was $14,500 and in 1995 the average cost of the car was $12,700. Let y represent the year and c represent the aerage cost of the car in a linear function. Determine the anticipated cost of the sports car in 1996.
2. A local radio station is holding a contest to give away cash. THe announcer makes a telephone call, and if the person who answers guesses the correct amount of money to be given away, he or she wins the money. If the person misses, $20 is added to the jackpot. At the beginning of the 5th call, the jackpot was $240. On what call could the winner receive $1060 -
powforhighschool:
Hi.1.the average cost of a 1990 car depreciates linearly, so that in 1993, the average cost of the car was $14,500 and in 1995 the average cost of the car was $12,700. Let y represent the year and c represent the average cost of the car in a linear function. Determine the anticipated cost of the sports car in 1996.
2. A local radio station is holding a contest to give away cash. THe announcer makes a telephone call, and if the person who answers guesses the correct amount of money to be given away, he or she wins the money. If the person misses, $20 is added to the jackpot. At the beginning of the 5th call, the jackpot was $240. On what call could the winner receive $1060
Q1: $11800
Q2: the 45th call -
I got the answer to the 2nd question as the 46th caller, can you explain how you got the 45th caller Muffins?
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powforhighschool:
1) Depreciation per year = (14500 - 12700)/2 = 1800/2 = 9001.the average cost of a 1990 car depreciates linearly, so that in 1993, the average cost of the car was $14,500 and in 1995 the average cost of the car was $12,700. Let y represent the year and c represent the aerage cost of the car in a linear function. Determine the anticipated cost of the sports car in 1996.
2. A local radio station is holding a contest to give away cash. THe announcer makes a telephone call, and if the person who answers guesses the correct amount of money to be given away, he or she wins the money. If the person misses, $20 is added to the jackpot. At the beginning of the 5th call, the jackpot was $240. On what call could the winner receive $1060
Cost of car in 1996 = cost of car in 1995 - depreciation = 12700 - 900 = 11800
2) 1060 = 240 + (n-5)20
820/20 = n-5
n = 46.
On 46th call, the winner could receive $1060. -
This thread has benefitted so many PSLE students and parents. It will be most useful if such a thread continues to support Sec-1 Maths also.
I am sure all those who wrote PSLE this year will be glad to have such a thread next year also, as they pursue Sec-1. Does it already exist in this forum? Or will anyone be kind enough to start such a ‘Sec-1 thread’ soon?
Thanks and regards.
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