Melissa drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 10 hours. When Melissa drove home, there was no traffic and the trip only took 7 hours. If her average rate was 18 miles per hour faster on the trip home, how far away does Melissa live from the mountains? Do not do any rounding.
Okay let's say that the mountains are at a distance say x away from home...
In other words she had to travel x units distance... is that fine?
yes
Now, she took 10 hours for going from home to mountains.. So speed = distance/time That means.. speed = x/10
Now, when she returned, she had to travel x miles again... this time she took 7 hours.. So, speed will be x/7
Any doubt till this step?
nope
Great!
Now, the question says... "If her average rate was 18 miles per hour faster on the trip home" That means, \(\frac{x}{7} = \frac{x}{10} + 18\)
Now, can you solve this equation?
hmmm how do I solve that
First let's simplify this a bit..
\(\frac{x}{7} = \frac{x}{10} + 18\) First let's multiply 10 on both sides, so that we can get rid of that 10 in denominator
\(10\frac{x}{7} = x + 18*10\) Now, we multiply on both sides by 7 to get rid of that 7 in denominator..
We get, \(7*10*x = 7*x + 7*18*10\)
I will simplify this: \(70x = 7x + 1260\)
Any problem till this step?
nope
Great! Now, let's bring 7x to the left hand side, we get: \(70x - 7x = 1260\) \(\implies 63x = 1260\) \(\implies x = \frac{1260}{63}\)
so 20?
Perfect!!
So, she lives 20 miles away from mountains...
It was wrong :(
Wait... let me check ..
Damn! I did a calculation mistake... The equation was: \(\frac{x}{7} = \frac{x}{10} + 18\) \(\implies 10x = 7x + 1260\) \(\implies 3x = 1260\) \(\implies x = \frac{1260}{3} = 420\)
So, the answer should have been 420 miles...
that was right. lol
Can you help me with 2-3 more? :o
Yeah sure
Ok it's Two trains leave the station at the same time, one heading east and the other west. The eastbound train travels 14 miles per hour slower than the westbound train. If the two trains are 750 miles apart after 5 hours, what is the rate of the eastbound train? Do not do any rounding.
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Now, both trains are travelling in opposite directions so their distance will increase with time...
mhm
Let's assume the speed of eastward train = v1 And the speed of westward train = v2
Now, it is given that: "The eastbound train travels 14 miles per hour slower than the westbound train." Thus, v1 = v2 - 14
We also know that: Distance = Speed * Time Thus, distance traveled by Eastward train: Distance1 = v1 * 5 Similarly, distance traveled by Westward train: Distance2 = v2*5
mhhmmm
Now, total distance between trains after 5 hours will be: Distance1 + Distance2 = 750 => v1*5 + v2*5 = 750 We also know v1 = v2 - 14 Put this value of v1 in first equation
ok
So, we get: (v2 - 14)*5 + v2 * 5 = 750 => 5* v2 - 5 * 14 + v2 * 5 = 750 => 10*v2 - 70 = 750 => 10 * v2 = 750 + 70 So, what will be the value of v2 ?
82?
Good :)
Now, v1 = v2 - 14 What will be v1?
ummm 14?
No no ... see you know v2 = 82.. just subtract 14 from it
OOHH
68
Great!
so its 68? :o
Yeps! So the speed of eastward train is 68 miles/hour
I believe I have 2 more problems left. :D
next one is, A car passes a landmark on a highway traveling at a constant rate of 45 kilometers per hour. Two hours later, a second car passes the same landmark traveling in the same direction at 70 kilometers per hour. How much time after the second car passes the landmark will it overtake the first car?
See... this time both cars are travelling in same direction so distance between them will decrease
i see
Now, second car will overtake first car when distance between them becomes zero
mhm
Now, distance traveled by first car = d1 = speed * time = 45 * t
I see
Now, second car traveled the same landmark 2 hours after the first car
Thus, during this time, first car traveled an extra distance of 45*2 = 90 km
Any doubt?
nopp
Great!
Now, in same time (t hours), car 2 will travel a distance = 70*t
To overtake the first car, this distance must be equal to 90 + 45*t .. so we get: 70*t = 90 + 45*t
Any doubt?
no :p
OKay ... So we can solve this equation to find out t...
so um 75t = 135? right?
No no ... see... 70t = 90 + 45t => 70t - 45t = 90 => 25t = 90 Calculate t from here
um 3.6?
Very good :D
hopefully this is the last one.. Ivanna drove to the mountains last weekend. There was heavy traffic on the way there, and the trip took 8 hours. When Ivanna drove home, there was no traffic and the trip only took 6 hours. If her average rate was 16 miles per hour faster on the trip home, how far away does Ivanna live from the mountains?
It's exactly like the one we did before the last one...
First, assume the distance between home and mountains = x Then, you know that speed = distance/ time So, on her way to mountains, her speed will be x/8 = v1 And on her way back to home, her speed will be x/6 = v2
"her average rate was 16 miles per hour faster on the trip home" This means that v2 = 16 + v2 => x/6 = 16 + x/8
mhmm
Now can you solve this equation?
I multiply the 16 and 8 right?
Exactly!
No no not by 16 and 8 but by 6 and 8
Notice that denominators are 6 and 8...
ohh yes
Can you tell me what will you obtain?
would it be 3?
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