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 T1 be the time taken to cover the distance 30 km in upstream

Let T2 be the time taken to cover the distance 30 km in downstream

Time  =  Distance/Speed

T1  =  30/(15 + x)

T2 =  30/(15 - x)

T1 + T2  =  4 hours 30 minutes

T1 + T2  =  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 - x2)  =  9/2

900 ⋅ 2  =  9 (225 - x²)

By dividing the equation by 9, we get

200  =  (225 - x2)

225 - x2  =  200

225 – 200  =  x2

x =  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  =  x2 -----(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  =  0x  =  1 x-7  =  0x  =  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 cm2. 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 cm2

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)/(x2 – 6x)  =  1/4

4(2x - 6)  =  x2 – 6x

8x – 24  =  x– 6x

x2 – 14x + 24  =  0

(x – 2) (x – 12)  =  0

 x-2  =  0x  =  2 x-12  =  0x  =  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  =  502

[2x+10]² + (2x)2  =  502

[4x+ 100 + 2(2x)(10)] + 4x2  =  2500

8x2 + 40x + 100 – 2500 = 0

8x2 + 40x – 2400 = 0

By dividing the equation by 8, we get

x2 + 5x – 300  =  0

(x + 20) (x -15)  =  0

 x+20  =  0x  =  -20 x-15  =  0x  =  15

Therefore speed of the second train is 15 km/hr.

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