Answer :

Let, the time taken by smaller tap to fill the tank completely = x

hours

The volume of the tank filled by a smaller tap in 1 hour =

The time taken by larger tap to fill the tank completely = x - 2

hours

The volume of the tank filled by a larger tap in 1 hour

Time taken by the tank to fill hours

hours

The volume of the tank filled by a smaller tap in hours

The volume of the tank filled by a larger tap in hours

⇒ 15(x - 2) + 15x = 8x(x-2)

⇒15(x – 2 + x) = 8x(x – 2)

⇒ 15(2x - 2) = 8x^{2} - 16

^{2}- 16

⇒8x^{2} – 46x + 30 = 0

⇒2(4x^{2} – 23x + 15) = 0

⇒4x^{2} – 23x + 15 = 0

⇒4x^{2} – 20x - 3x + 15 = 0

⇒ 4x(x -5) - 3(x - 5) = 0

⇒ (4x-3) (x -5) = 0

At x = 5

Time taken by smaller tap to fill the tank completely = 5 hours

Time taken by larger tap to fill tank completely = 5 – 2 = 3 hours

At

Time taken by smaller tap to fill the tank completely hours

Time taken by larger tap to fill the tank completely

hours

Time is not possible to be negative.

So, is not possible.

Therefore the time for smaller tap is 5 hrs and for larger tap is 3 hrs.**OR**

Let, speed of the stream = x km/h

And the speed of the boat = y km/h

So, the speed of the upstream = (y – x) km/h

And the speed of the downstream = (y + x) km/h

…(1)

…(2)

Let,

The equations will be:

30u + 44v - 10 = 0

40u + 55v - 13 = 0

By cross multiplication method we know,

Here a_{1} = 30 , b_{1} = 44 , c_{1} = -10

a_{2}= 40, b_{2} = 55, c_{2} = -13

So,

Now,

⇒ 10 = 2(y - x)

⇒ 10 = 2y - 2x

⇒ 5 = y - x .... (3)

⇒ 5 + x = y

And

⇒ y + x = 11 ..... (4)

Put the value of y in eq (4),

⇒ 5 + x + x = 11

⇒ 5 + 2x = 11

⇒2x = 6

⇒ x = 3

Put the value of x in (3) to get,

⇒ 5 = y - 3

⇒ y = 8

So, speed of the stream = 3 km/h

And the speed of the boat = 8 km/h

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