Problem 1 :
The speed of a boat in still water is 15 km/hr. It goes 3 km upstream and return downstream to the original point in 4 hrs 30 minutes. Find the speed of the stream.
Solution :
Let x km/hr be the speed of water
Speed of boat is 15 km/hr
So speed is upstream = (15+x) km/hr
Speed in downstream = (15–x) km/hr
Let T_{1} be the time taken to cover the distance 30 km in upstream
Let T_{2} be the time taken to cover the distance 30 km in downstream
Time = Distance/Speed
T_{1} = 30/(15 + x)
T_{2} = 30/(15 - x)
T_{1} + T_{2} = 4 hours 30 minutes
T_{1} + T_{2} = 4 1/2
[30/(15 + x) + 30/(15 - x)] = 9/2
30[1/(15 + x) + 1/(15 - x)] = 9/2
30 [(15 – x + 15 + x)/(225 - x²)] = 9/2
30 (30)/(225 - x^{2}) = 9/2
900 ⋅ 2 = 9 (225 - x²)
By dividing the equation by 9, we get
200 = (225 - x^{2})
225 - x^{2 }= 200
225 – 200 = x^{2}
x^{2 } = 25
x = √25
x = ± 5
Speed must be positive.
So,
speed of water = 5 km/hr.
Problem 2 :
One year ago a man was 8 times as old as his son. Now his age is equal to the square of his son’s age. Find their present age.
Solution :
Let x be the present age of son and y be the age of father.
So (x-1) be the age of son one year ago.
(y-1) be the age of father one year ago.
By using the given information
y = x^{2 }-----(1)
y – 1 = 8(x -1)
y = 8x – 7^{ }-----(2)
(1) = (2)
x² = 8x – 7
x²- 8x + 7 = 0
(x -1)(x – 7) = 0
x-1 = 0 x = 1 |
x-7 = 0 x = 7 |
So, present age of son is 7 years.
Problem 3 :
A chess board contains 64 equal squares and the area of each square is 6.25 cm^{2}. A border around the board is 2 cm wide. Find the length of the side of chess board.
Solution :
Let x be the side length of the square board
Area of one square in the chess board = 6.25 cm^{2}
Area of 64 squares = 64 (62.5)
(x – 4)^{2} = 400
(x – 4) = √400
x-4 = 20
x = 24 cm
Therefore side length of square shaped chess board is 24 cm.
Problem 4 :
A takes less than the time taken by B to finish the piece of work. If both A and B together can finish the work in 4 days. Find the time that B would take to finish the work by himself.
Solution :
Let x be the tune taken by B to finish the work.
So (x – 6) be the time taken by to finish the work.
Work done by A in one day = 1/x
Work done by B in one day = 1/(x - 6)
Number of days taken by both to finish the work = 1/4
1/x + 1/(x-6) = 1/4
(2x - 6)/(x^{2} – 6x) = 1/4
4(2x - 6) = x^{2} – 6x
8x – 24 = x^{2 }– 6x
x^{2} – 14x + 24 = 0
(x – 2) (x – 12) = 0
x-2 = 0 x = 2 |
x-12 = 0 x = 12 |
So, B is taking 12 days to complete the work.
Problem 5 :
Two trains leave railway stations at the same time. The first train travels due west and the second train due north. The first travels 5 km/hr faster than the second train. If after two hours, they are 50 km apart, find the average speed of each train.
Solution :
Let x km/hr be the speed of second train.
So sped of first train will be (x + 5) km/hr
Distance covered by the first train in 2 hours = 2(x + 5)
Distance covered by the second train in 2 hours = 2x
By using Pythagoras theorem,
[2(x+5)]^{2} + (2x)^{2} = 50^{2}
[2x+10]² + (2x)^{2} = 50^{2}
[4x^{2 }+ 100 + 2(2x)(10)] + 4x^{2} = 2500
8x^{2} + 40x + 100 – 2500 = 0
8x^{2} + 40x – 2400 = 0
By dividing the equation by 8, we get
x^{2} + 5x – 300 = 0
(x + 20) (x -15) = 0
x+20 = 0 x = -20 |
x-15 = 0 x = 15 |
Therefore speed of the second train is 15 km/hr.
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