A man increases his walking speed by 1/2 mph, so he can go 30 miles in 2 hrs less time. His rate =?
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- Опубліковано 30 жов 2024
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Mr math man you are so smart I could never figure this out
Using v for his current speed and using time = distance/speed, we have:
30/(v + 0.5) + 2 = 30/v
Multiply both sides by (v + 0.5)
30 + 2v + 1 = 30(v + 0.5)/v
Multiply both sides by v
30v + 2v² + v = 30v + 15
Subtract 30v from both sides and subtract 15 from both sides
2v² + v - 15 = 0
It's a quadratic and the b²-4ac value is a perfect square, so it solves easily with the quadratic formula
v = (-1 ±√(1 + 120))/4
v = (-1 ± 11)/4
v = 2.5, -3
We know we need a positive answer so discard the extraneous solution.
v = 2.5mph
As a check, this would cover 30 miles in 12 hours, and 0.5mph faster is 3 mph which would cover 30 miles in 10 hours.
Good problem.
This problem can be solved using a system of linear equations--and a substitution.
rt = 30
(r + 1/2)(t - 2) = 30
rt -2r + (1/2)t -1 = 30
-(rt = 30)
-rt = -30
-2r +(1/2)t -1 = 0
(1/2)t -2r = 1
Because t=(30/r), we can reduce the equation to one variable...
(1/2)(30/r) -2r = 1
(15/r) -2r = 1
r[(15/r) -2r = 1]
15 - 2r^2 = r
0 = 2r^2 + r - 15
2r^2 + r - 15 =0
Factorization for -30 and +1 are...
(2r - 5)(r + 3) = 0
2r - 5 = 0
2r = 5
r = 5/2
r + 3 = 0
r = -3
Since the rate must be positive, then r = 2-1/2 mph
Let’s say that the man walks 30 miles
in t hours.
Then, the man’s rate of walking is
30/t mph.
When the man increases his rate of walking by 1/2 mph, his new rate of walking is
30/t + 1/2 mph
Then, he covers 30 miles in 2 hours less time and his rate of walking can be also expressed as
30/(t - 2)mph.
30 30 1
-- = - + -
(t-2) t 2
30 = 60 + t
-- ----
(t-2) 2t
Cross multiplying,
60t = (t-2)(60+t)
60t = t^2+60t-2t-120
60t-60t=0=t^2-2t-120
t^2-2t-120 = 0
(t-12)(t+10)= 0
Either t-12 =0
OR
t+10 = 0
If t-12 = 0,
t = 12 hours
If t+10 = 0,
t = -10 hours.
Since time t cannot be a negative number,
t must be equal to
12 hours.
The initial rate of walking of the man was
30/t mph
If t= 12 hours,
30/t = 30/12 = 5/2 mph.
5/2 = 2 1/2 mph.
That's how I did this: calculating on time, not on speed. I think this way is much easier. After finding the walking time it is very simple to establish the speed.
_Answer_ : 2.5 mph
_Calculation_ (from the thumbnail):
Normal rate:
30 miles in T hours = 30/T [mph]
Increased rate:
(30/T + ½) [mph]
which must equal
30 miles in (T-2) hours = 30/(T-2) [mph]
(30/T + ½) = 30/(T-2)
30(T-2) + ½T(T-2) = 30T
30T - 60 + ½T² - T = 30T
½T² - T = 60
T² - 2T = 120
T² - 2T + 1 = 121
(T - 1)² = 11²
(T - 1) = 11 OR (T - 1) = -11
T = 12 OR T = -10
T = -10 is negative, which results in an unrealistic outcome.
T = 12 : Normal rate is 30 miles in 12 hours = 30/12 = 2.5 mph
got 2.5 my brain hurts thanks for the lesson
i'd like to solve it, but i'm in hospital, now, promise to figure it out, later. 🥰🥰🥰🙏🙏
distance 30 miles = speed x time
30 =s.t =
(s+.5)(t-2)
=st+.5t-1-2s .. find s
1+2s=.5t=15/s
2s^2+s-15=0
(2s-5)(s+3)
s>0
s=2.5 answer
check
s_1=2.5 t_1=12
s_2=3 t_2=10
ok
Thank you
unknowns:
M: man's walking rate (mph)
D: distance traveled (miles)
T: time of travel (hours)
I.E. D (miles) =
T (hours) × R (mph)
this analysis:
30 (miles) =
(M+(1/2)) × (T-2) eq.1
30 = MT - 2M +(1/2)T -1 eq.1
30 = M × T
M = 30 - T eq.2
eq.2 => eq.1
30=(M+(1/2)) × (T-2) eq.1
30=(30-T+(1/2))(T-2)
30=(30-T+(0.5))(T-2)
30=(30.5-T)(T-2)
30=30.5T-61-T^2+2T
T^2-2T-30.5T+61+30=0
T^2+30.5T+91=0
[-b+/-sqrt(b^2-4ac)]/2a
a = 1
b = 30.5
c = 91
[-30.5+/-sqrt(30.5^2
-4×1×91)]/2
[-30.5+/-sqrt(930.25-364)]/2
[-30.5+/-sqrt(566.25)]/2
[-30.5+/-(23.796)]/2
sol.1 -27.148
sol.2 -3.352
That did not land right...
I calculated it correctly in my head! But wasn’t sure how to write it down in a mathematical language 😂
What? I needed 15 minutes with paper. How can you solve a not-that-simple quadratic equation in your head?
speed now 30 mls / 2.5 mph = 12 hrs : increased speed 30 mls / 3 mph = 10hrs
10/29/2024 2:45 AM
initial speed = x
increase speed = x + 1/2 mph
time reduced = t -2
distance = 30 mph (I better work with kilometers)
time at initial speed t = 30 / x
reduced time at increased speed t -2 = 30 / (x + 1/2 )
substitute initial time t = 30 /x in --> t -2 = 30 / (x + 1/2)
(30 / x) - 2 = 30 / (x + 1/2) -----> isolate the x variable
(30 / x) - 30 / (x + 1/2) = 2 --> eliminate the fractions by the common denominator
multiply every term by x(x + 1/2) --> lcd
(30 / x) * x(x + 1/2) - 30 / (x + 1/2) * x(x + 1/2) = 2 * x(x + 1/2)
30(x + 1/2) - 30x = 2x(x + 1/2) --> simplify
30x + 15 - 30x = 2x² + x
2x² + x = 15 --> solve by completing the square
2x²/2 + x/2 = 15/2 --> divide every term by 2 in the equation
x² + x/2 + [ ] = 15/2 + [ ] --> take the coefficient of the middle term divide it by 2 and square it ((1/2) / 2)² = 1/16
paste 1/16 in the slots
x² + x/2 + [ 1/16] = 15/2 + [ 1/16]
(x + 1/4)² = 120/16 + 1/16
(x + 1/4)² = 121/16
√(x + 1/4)² = ±√(121/16)
x + 1/4 = 11/4
x = 11/4 - 1/4 = 10/4 = 5/2 = 2 .5 miles -->ANS
x + 1/4 = -11/4
x = -11/4 - 1/4 = -12/4 = -3/1 = -3 miles Not practical
Calculating on time t results in t² - 2t -120 = 0 or (t-12).(t+10) = 0 That looks much easier to me.
That's interesting how to get there, and what I'd have guessed at. But were you looking for his initial rate or his new rate?
Hmm, if the man Walks 1/2mph faster he can do the 30 miles in 2 hours less.
Walking 2,5mph, he does 30 miles in 12 hours.
Walking 2mph, he does 30 miles in 15 hours. I am missing 1 hour, cause 15 hours minus 12 is 3 hours faster, not 2.
But I guess my mind got scrambled....
😂 I get it
30 miles, 2,5 mph = 30 miles in 12 hours
30 miles, 3 mph = 30 miles in 10 hours
🤦♀️
Brain got fried
Alternative route to solve this, not calculating the speed but better/easier calculating the time...
v = 30 / t and v + ½ = 30 / (t-2) so together (30 / t) + ½ = 30 / (t-2)
30 / t - 30 / (t-2) = -½ so we have 2 fractions with different denominators, to equal them:
t . (t-2) = t² - 2t = T so ( 30 . (t-2) - 30 . t ) / T = -½ with 30.t - 60 - 30.t = -60
-60 / T = -½ or T = -60 / -½ = 120 so t² - 2t = 120
And there we have a very nice quadratic equation t² - 2t - 120 = 0 factorised in
(t - 12) . (t + 10) = 0 so t = 12 hour (t = -10 dismissed)
Now we can calculate the speed as v = 30 miles / 12 hour = 10 /4 = 5/2 = 2½ mph
Check: v + ½ = 30 / (12 - 2) = 3 so v = 3 - ½ = 2½ mph
Did you enjoy this alternative?
2.5 mph
It took me two tries :-)
Km’s
Whoa 😂👍👏🙏💪😎🌎