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Can you explain step by step how to approach this problem?
Together John and Jane eat 2/3 a pizza in 4 seconds.
How long would it take each of them to eat the whole pizza alone if Jane can eat a pizza 5 seconds faster than John?
2 Answers
- AlanLv 71 month ago
Alternate Method
(4) Jane' Rate + 4* (John's Rate) = 2/3 of a pizza
then, multiply both sides by 3/2
6*Jane's Rate + 6* John's Rate = 1 pizza
so it takes them 6 seconds together
Jane's Rate = 1/Jane'S time to eat one pizza
John's rate = 1/ John's Time to eat one pizza
John's rate = 1/ (Jane's time + 5)
This gives
6/ (1/Jane_time) + 6(1/ (jane's_time+5) = 1
Common denominator
6(jane's Time +5) + 6(Jane's Time) / (jane's Time^2 + 5*Jane's Time) = 1
jane'sTime^2 + 5*Jane's Time = 12*jane's Time + 30
jane's Time^2 -7*Jane's Time - 30 = 0
(Jane's Time -10) (jane's Time +3) =0
Jane's time =10 or -3 seconds
Jane's time = 10 seconds
John's Time = 15 seconds
Checking
(4)(1/10) + 4(1/15) = 4/10 + 4/15 = 12/30 + 8/30 = 20/30 = 2/3
6(1/10) + 6(1/15) = 6/10 + 6/15 = 18/30 + (12/30) = 1
Since my answer check out, the first answer has a mistake
It is right to here :
2(t² - 5t) = 24t - 60
2t² - 10t = 24t - 60 (still right)
2t² - 24t + 60 = 0 (wrong)
should be
2t^2 -34t + 60 = 0 (divide by 2)
t² - 17t + 30 = 0
(t - 15)(t-2) = 0
so t = 15 or t = 2 for John
but t =2 is impossible
so John's time = 15
and
jane's time = 10
- llafferLv 71 month ago
work = rate * time
Work here is the fraction of a pizza eaten in the given time.
When they work together, they add their rates together.
Let r = John's rate of pizza per time
Let R = Jane's rate of pizza per time
separately, it takes Jane 5 fewer seconds to eat a whole pizza than it takes for John to eat. We need a new variable:
Let t = time it takes for John to eat a pizza
Then the time it takes for Jane to eat one is (t - 5).
Now we can create two equations for r and R in terms of t:
w = rt
work here is 1 as it's the entire pizza.
1 = rt and 1 = R(t - 5)
1 / t = r and 1 / (t - 5) = R
When they eat pizza together, we add the rates, so:
1 / t + 1 / (t - 5)
(t - 5) / [t(t - 5)] + t / [t(t - 5)]
(t - 5 + t) / [t(t - 5)]
(2t - 5) / [t(t - 5)]
That's the rate when they work together. We know they can eat 2/3 of a pizza in 4 seconds together, so:
w = rt
2/3 = (2t - 5) / [t(t - 5)] * 4
We now have an equation with one unknown. We can solve for t:
2 = 12(2t - 5) / [t(t - 5)]
2 = (24t - 60) / [t(t - 5)]
2[t(t - 5)] = 24t - 60
2(t² - 5t) = 24t - 60
2t² - 10t = 24t - 60
2t² - 24t + 60 = 0
t² - 12t + 30 = 0
I'll solve this by completing the square. Subtracting both sides by 30, then adding 36 to both sides to complete the square:
t² - 12t = -30
t² - 12t + 36 = -30 + 36
(t - 6)² = 6
t - 6 = ± √6
t = 6 ± √6
there are two possible times. We can now solve these for r and R so we can see which ones make sense (we can't have a negative rate, etc.)
r = 1 / t and R = 1 / (t - 5)
r = 1 / (6 - √6) and R = 1 / (6 - √6 - 5) and r = 1 / (6 + √6) and R = 1 / (6 + √6 - 5)
r = (6 + √6) / [(6 - √6)(6 + √6)] and R = 1 / (1 - √6) and r = (6 - √6) / [(6 - √6)(6 + √6)] and R = 1 / (1 + √6)
The first R here will end up being negative so we can throw out that solution, leaving the last answer to check:
r = (6 - √6) / 30 and R = (1 - √6) / [(1 - √6)(1 + √6)]
r = (6 - √6) / 30 and R = (1 - √6) / (1 - 6)
r = (6 - √6) / 30 and R = (1 - √6) / (-5)
r = (6 - √6) / 30 and R = (√6 - 1) / 5
These are the exact rates. Decimal approximations work out to be :
r = 0.11835 and 0.28990 pizza per second
If we use these rates to find the times, one should be 5 seconds faster than the other:
w = rt
1 = 0.11835t and 1 = 0.28990t
1 / 0.11835 = t and 1 / 0.28990 = t
8.44951 = t and 3.44947 = t
It's not exact due to rounding, but the times are about 5 seconds off.
So again the exact rates in terms of pizza per second for each person is:
r = (6 - √6) / 30 and R = (√6 - 1) / 5