SOLUTION: 1. A father is 34 years old and his son is 12 years old. In how many years will the father be twice as old as his son.
2. Sam is 18 and Bill is 24. How many years ago was Bill thr
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Question 996415: 1. A father is 34 years old and his son is 12 years old. In how many years will the father be twice as old as his son.
2. Sam is 18 and Bill is 24. How many years ago was Bill three times as old as her Sam?
3. John has four times as many marbles as Peter. If he gives Peter 8 marble he will have twice as many marble as Peter. How many did each have originally.
4. The sum of two numbers is 17, and three times the smaller is 3 more than the bigger, Find the numbers.
5. If each stroke of a pump removes 20% of the air in the cylinder, what percentage of the air will be left after the third stroke?
Found 2 solutions by lwsshak3, ikleyn:
Answer by lwsshak3(11628) (Show Source): You can put this solution on YOUR website!
1. A father is 34 years old and his son is 12 years old. In how many years will the father be twice as old as his son.
let x=number of years father will be twice as old as his son.
34+x=2(12+x)
34+x=24+2x
x=10
number of years father will be twice as old as his son=10
******
2. Sam is 18 and Bill is 24. How many years ago was Bill three times as old as her Sam?
let x= How many years ago was Bill three times as old as her Sam.
24-x=3(18-x)
24-x=54-3x
2x=30
x=15
How many years ago was Bill three times as old as her Sam. 15
*****
3. John has four times as many marbles as Peter. If he gives Peter 8 marble he will have twice as many marble as Peter. How many did each have originally.
let x=number of marbles Peter had originally
4x= number of marblesJohn had originally
4x-8=2(x+8)
4x-8=2x+16
2x=24
x=12
number of marbles Peter had originally=12
4x= number of marblesJohn had originally=48
*****
4. The sum of two numbers is 17, and three times the smaller is 3 more than the bigger, Find the numbers.
let x=larger number
17-x=smaller number
3(17-x)=3+x
51-3x=3+x
4x=48
x=12
larger number=12
smaller number=5
*****
5. If each stroke of a pump removes 20% of the air in the cylinder, what percentage of the air will be left after the third stroke? 40%
Answer by ikleyn(52781) (Show Source): You can put this solution on YOUR website!
.
I'd like to contribute to the question 5)
"If each stroke of a pump removes 20% of the air in the cylinder, what percentage of the air will be left after the third stroke?"
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More correct formulation is:
"If each stroke of a pump removes 20% of the current air mass in the cylinder, what percentage of the air will be left after the third stroke?"
The answer in the previous response is not correct. If to follow that logic then what percentage of the air will be left after the sixth stroke? Minus 20% ??
The correct analysis is like this:
1) After the first stroke 20% of the initial mass M goes out; and 80%, or 0.8 of the initial mass, 0.8*M, remains.
2) After the second stroke 20% of the remained mass (i.e. 0.2*(0.8*M) = 0.16*M) goes out; and 0.8*M - 0.16*M = 0.64*M remains.
3) After the third stroke 20% of the remained mass (i.e. 0.2*(0.64*M) = 0.128*M) goes out; and 0.64*M - 0.128*M = 0.512*M remains.
Thus the answer is: After the third stroke 0.512 = 51.2% of the initial air mass remains.
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