SDT Practice SET – 4
Speed, Distance & Time: Level 4
1.
A truck travels at 60 mph. A car traveling at 75 mph leaves the same place 1 hour later. How many hours will the car travel before catching the truck?
Solution (C):
Truck head start = 60 miles.
Relative gain = 15 mph.
Time for car = \( 60 / 15 = 4 \) hours.
2.
A person covers half of his journey at 40 mph and the remaining half at 60 mph. His average speed for the whole journey is:
Solution (A):
Harmonic mean formula applies for equal distances: \( \frac{2xy}{x+y} = \frac{2(40)(60)}{100} = 48 \) mph.
3.
Two trains are 300 miles apart. Train A moves at 40 mph towards B. Train B moves at 60 mph towards A. How far from A’s starting point do they meet?
Solution (D):
Combined speed = 100 mph.
Time = 300 / 100 = 3 hours.
Distance A = \( 40 \times 3 = 120 \) miles.
4.
A man can row 6 mph in still water. If the river current is 2 mph, how long does it take him to row 8 miles upstream and back?
Solution (E):
Upstream rate = \( 6 – 2 = 4 \) mph. Time = 8/4 = 2 hrs.
Downstream rate = \( 6 + 2 = 8 \) mph. Time = 8/8 = 1 hr.
Total = \( 2 + 1 = 3 \) hours.
5.
A car travels \(d\) miles at \(r\) mph and returns at \(r/2\) mph. The total time is \(t\). Which equation relates these?
Solution (B):
Time 1 = \( d/r \).
Time 2 = \( d/(r/2) = 2d/r \).
Total \( t = d/r + 2d/r = 3d/r \).
6.
If a runner increases his speed by 25%, by what percent does his time to cover the same distance decrease?
Solution (C):
New Speed = \( 1.25S = \frac{5}{4}S \).
New Time = \( D / (\frac{5}{4}S) = \frac{4}{5} (D/S) = 0.8 T \).
Time decreased by \( 1 – 0.8 = 0.2 = 20\% \).
7.
A train passes a pole in 10 seconds and a platform of length 200m in 20 seconds. What is the length of the train?
Solution (A):
Speed = \(L/10\).
Speed = \((L+200)/20\).
\(L/10 = (L+200)/20 \implies 2L = L + 200 \implies L = 200\).
8.
A man walks at 3 mph and runs at 6 mph. He covers a total of 12 miles in 3 hours. How many miles did he run?
Solution (D):
Let run distance = \(r\), walk = \(12-r\).
\( \frac{12-r}{3} + \frac{r}{6} = 3 \).
Multiply by 6: \( 2(12-r) + r = 18 \).
\( 24 – 2r + r = 18 \implies 24 – r = 18 \implies r = 6 \).
9.
Two cars start from X and Y, 100 miles apart, towards each other. They meet after 1 hour. If Car A is 20 mph faster than Car B, find speed of Car B.
Solution (B):
Relative speed = \(100/1 = 100\) mph.
\(B + (B+20) = 100 \implies 2B = 80 \implies B = 40\).
10.
A bus leaves city P at 8 AM. Another bus leaves city Q at 9 AM. They meet at 11 AM. If the distance PQ is 300 km and the first bus is 20 km/h slower than the second, what is the speed of the first bus?
Solution (C):
Time A = 3 hrs (8 to 11). Time B = 2 hrs (9 to 11).
Let Speed A = \(x\), Speed B = \(x+20\).
\(3x + 2(x+20) = 300\).
\(3x + 2x + 40 = 300 \implies 5x = 260 \implies x = 52\).
Score: 0 / 0