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22 Jul 2024, 01:30
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C
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
59% (02:22) correct
41%
(02:26)
wrong
based on 215
sessions
History
Date
Time
Result
Not Attempted Yet
The distance between City A and City B is 120 miles. A freight train leaves City A to City B and after some time a passenger train leaves City B to City A. If the trains met at exactly halfway between City A and City B, find the speed of the passenger train.
(1) The passenger train leaves City B exactly 30 minutes after the freight train leaves City A.
(2) The passenger train is 6 mph faster than the freight train.
(1) The passenger train leaves City B exactly 30 minutes after the freight train leaves City A.
(2) The passenger train is 6 mph faster than the freight train.
22 Jul 2024, 07:37
Kudos
Bookmarks
Givne - The distance between City A and City B is 120 miles. A freight train leaves City A to City B and after some time a passenger train leaves City B to City A.
To find - If the trains met at exactly halfway between City A and City B, find the speed of the passenger train.
City A ---------------------<120>------------------------ City B
Frieght train speed = a mph and passenger train = b mph
train a traveled for t hrs while train b traveled for (t-t')
Where t' is that some time after which a passenger train leaves City B to City A.
\(at = b(t-t') = 60\) (as distance = speed * time)
1st - The passenger train leaves City B exactly 30 minutes after the freight train leaves City A.
\(t' = \frac{1}{2} Hr\)
Not sufficient.
2nd - The passenger train is 6 mph faster than the freight train.
b = a + 6
Not sufficient.
lets combine both
now at = 60 so
\(t = \frac{60}{a}\) and \(b(t-t') = 60\)
and \(t' = \frac{1}{2} Hr\)
\(b(\frac{60}{a} - \frac{1}{2}) = 60\)
Now solving further,
\(b(120-a) = 120a\)
\(120b - ab = 120a\)
\(120(b-a) = ab\)
but \(b-a=6\) and \(b=a+6\)
\(120*6 = a(a+6)\)
\(720 = a(a+6)\)
but \(720 = 24*30\)
\(a=24 mph\)
and \(b=a+6=30 mph\)
Answer is C.
To find - If the trains met at exactly halfway between City A and City B, find the speed of the passenger train.
City A ---------------------<120>------------------------ City B
Frieght train speed = a mph and passenger train = b mph
train a traveled for t hrs while train b traveled for (t-t')
Where t' is that some time after which a passenger train leaves City B to City A.
\(at = b(t-t') = 60\) (as distance = speed * time)
1st - The passenger train leaves City B exactly 30 minutes after the freight train leaves City A.
\(t' = \frac{1}{2} Hr\)
Not sufficient.
2nd - The passenger train is 6 mph faster than the freight train.
b = a + 6
Not sufficient.
lets combine both
now at = 60 so
\(t = \frac{60}{a}\) and \(b(t-t') = 60\)
and \(t' = \frac{1}{2} Hr\)
\(b(\frac{60}{a} - \frac{1}{2}) = 60\)
Now solving further,
\(b(120-a) = 120a\)
\(120b - ab = 120a\)
\(120(b-a) = ab\)
but \(b-a=6\) and \(b=a+6\)
\(120*6 = a(a+6)\)
\(720 = a(a+6)\)
but \(720 = 24*30\)
\(a=24 mph\)
and \(b=a+6=30 mph\)
Answer is C.
14 Mar 2026, 08:50
Kudos
Bookmarks
Statement 1
(x+1/2)y = 60
xz = 60
Too many equations.
INSUFFICIENT
Statement 2:
we have a, a+6 as speeds
a(t1)=60
(a+6)t2 = 60
We dont know when either leave.
INSUFFICIENT
Combining the 2:
(x+1/2)(y+6)=60
xy=60
Two equations two variables. Solvable
SUFFICIENT.
===========================
Common mistakes people make(I've especially done many times):
The equations once combined you might have the tendency to see oh, both are 60 lets equate quickly and then u realise you have just 1 equation with x,y still remaining! I mean you could still substitute 60 somewhere but in time pressure you will forget about the existence of 60 when u quickly equate LHS of 1 and 2. Instead better to expand equation 1 first and then wherever one finds xy substitute those terms with 60 and it becomes solvable.
(x+1/2)y = 60
xz = 60
Too many equations.
INSUFFICIENT
Statement 2:
we have a, a+6 as speeds
a(t1)=60
(a+6)t2 = 60
We dont know when either leave.
INSUFFICIENT
Combining the 2:
(x+1/2)(y+6)=60
xy=60
Two equations two variables. Solvable
SUFFICIENT.
===========================
Common mistakes people make(I've especially done many times):
The equations once combined you might have the tendency to see oh, both are 60 lets equate quickly and then u realise you have just 1 equation with x,y still remaining! I mean you could still substitute 60 somewhere but in time pressure you will forget about the existence of 60 when u quickly equate LHS of 1 and 2. Instead better to expand equation 1 first and then wherever one finds xy substitute those terms with 60 and it becomes solvable.
Bunuel
