Let, the time taken by large tap to fill the tank = x hours So, the time taken by smaller tap to fill the tank = x + 10 hours Therefore, in 1 hour, tank filled by larger tap = 1/x+10 According to question, 1/x + 1/x + 10 = 1/(9 (3/8)) ⇒ x + 10 + x/x(x + 10) = 8/75 ⇒ 75(2x + 10) = 8x(x + 10) ⇒ 150x +Read more
Let, the time taken by large tap to fill the tank = x hours
So, the time taken by smaller tap to fill the tank = x + 10 hours
Therefore, in 1 hour,
tank filled by larger tap = 1/x+10
According to question,
1/x + 1/x + 10 = 1/(9 (3/8))
⇒ x + 10 + x/x(x + 10) = 8/75
⇒ 75(2x + 10) = 8x(x + 10)
⇒ 150x +750 = 8x² + 80x
⇒ 8x² – 70x – 750 = 0
⇒ 4x² – 35x – 375 = 0
⇒ 4x² – 60x + 25x – 375 = 0
⇒ 4x(x – 15) + 25 (x – 15) = 0
⇒ (x – 15) (x + 25) = 0
⇒ (x – 15) = 0 or (x + 25) = 0
Either x = 15 or x = – 25
But, x ≠ – 25, as x is the time taken to fill the tank which can’t be negative. So, x
Hence, the time taken by larger top to fill the tank = 15 hours
and the time by smaller tap to fill the tank = 15 + 10 = 25 hours.
2x² – 3x + 5 = 0 The given equations is of the form ax² + bx + c = 0, in which a = 2, b = - 3, c = 5. Therefore, D = b² - 4ac = (-3)² - 4 × 2 × 5 = 9 - 40 = - 31 < 0 Hence, there is no real roots for this quadratic equation.
2x² – 3x + 5 = 0
The given equations is of the form ax² + bx + c = 0, in which a = 2, b = – 3, c = 5.
Therefore, D = b² – 4ac = (-3)² – 4 × 2 × 5 = 9 – 40 = – 31 < 0
Hence, there is no real roots for this quadratic equation.
Let, the breadth of mango grove = x m Therefore, the lenght = 2x m So, the area = x × 2x = 2x² According to question, 2x² = 800 ⇒ x² = 400 ⇒ x = ±20 Since, the breadth of the mango grove can't be negative, so the breadth = 20 m Hence, the lenght of the mango grove = 2 × 20 = 40 m.
Let, the breadth of mango grove = x m
Therefore, the lenght = 2x m
So, the area = x × 2x = 2x²
According to question, 2x² = 800
⇒ x² = 400
⇒ x = ±20
Since, the breadth of the mango grove can’t be negative, so the breadth = 20 m
Hence, the lenght of the mango grove = 2 × 20 = 40 m.
Let, the age of the first friend = x years So, the age of the other friend = 20 - x years Four years ago: Age of the first friend = x - 4 years Age of the second friend = 20 - x - 4 = 16 - x years According to question, (x - 4) (16 - x) = 48 ⇒ 16x - x² - 64 + 4x = 48 ⇒ x² - 20x + 112 = 0 For the quaRead more
Let, the age of the first friend = x years
So, the age of the other friend = 20 – x years
Four years ago:
Age of the first friend = x – 4 years
Age of the second friend = 20 – x – 4 = 16 – x years
According to question,
(x – 4) (16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
For the quadratic equation x² – 20x + 112 = 0, we have a = 1, b = – 20, c = 112.
Therefore,
D = b² – 4ac = (-20)² – 4 × 1 × 112 = 400 – 448 = – 48 < 0
So, there is no real roots of this quadratic equation.
Hence, this situation is not possible.
Is it possible to design a rectangular park of perimeter 80 m and area 400 m2? If so, find its length and breadth.
Let the lenght of the park = x Perimeter = 80 m Therefore, the breadth = 40 - x m [As Perimeter = 2 (Lenght + Breadth] According to question, Area = x (40 - x) = 400 ⇒ 40x - x² = 400 ⇒ x² - 40x + 400 = 0 ⇒ x² - 20x - 20x + 400 = 0 ⇒ x(x - 20) - 20(x - 20) = 0 ⇒ (x - 20) (x - 20) = 0 ⇒ (x - 12)² = 0Read more
Let the lenght of the park = x
See lessPerimeter = 80 m
Therefore, the breadth = 40 – x m [As Perimeter = 2 (Lenght + Breadth]
According to question,
Area = x (40 – x) = 400
⇒ 40x – x² = 400
⇒ x² – 40x + 400 = 0
⇒ x² – 20x – 20x + 400 = 0
⇒ x(x – 20) – 20(x – 20) = 0
⇒ (x – 20) (x – 20) = 0
⇒ (x – 12)² = 0 or (x – 13) = 0
⇒ (x – 20) = 0
⇒ x = 20
Hence the lenght = 20 m and breadth of the park = 40 – 20 = 20 m.
Two water taps together can fill a tank in 9(3/8) hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.
Let, the time taken by large tap to fill the tank = x hours So, the time taken by smaller tap to fill the tank = x + 10 hours Therefore, in 1 hour, tank filled by larger tap = 1/x+10 According to question, 1/x + 1/x + 10 = 1/(9 (3/8)) ⇒ x + 10 + x/x(x + 10) = 8/75 ⇒ 75(2x + 10) = 8x(x + 10) ⇒ 150x +Read more
Let, the time taken by large tap to fill the tank = x hours
So, the time taken by smaller tap to fill the tank = x + 10 hours
Therefore, in 1 hour,
tank filled by larger tap = 1/x+10
According to question,
1/x + 1/x + 10 = 1/(9 (3/8))
⇒ x + 10 + x/x(x + 10) = 8/75
See less⇒ 75(2x + 10) = 8x(x + 10)
⇒ 150x +750 = 8x² + 80x
⇒ 8x² – 70x – 750 = 0
⇒ 4x² – 35x – 375 = 0
⇒ 4x² – 60x + 25x – 375 = 0
⇒ 4x(x – 15) + 25 (x – 15) = 0
⇒ (x – 15) (x + 25) = 0
⇒ (x – 15) = 0 or (x + 25) = 0
Either x = 15 or x = – 25
But, x ≠ – 25, as x is the time taken to fill the tank which can’t be negative. So, x
Hence, the time taken by larger top to fill the tank = 15 hours
and the time by smaller tap to fill the tank = 15 + 10 = 25 hours.
Find the nature of the roots of the following quadratic equations. If the real roots exist, find them: 2 x² – 3x + 5 = 0
2x² – 3x + 5 = 0 The given equations is of the form ax² + bx + c = 0, in which a = 2, b = - 3, c = 5. Therefore, D = b² - 4ac = (-3)² - 4 × 2 × 5 = 9 - 40 = - 31 < 0 Hence, there is no real roots for this quadratic equation.
2x² – 3x + 5 = 0
See lessThe given equations is of the form ax² + bx + c = 0, in which a = 2, b = – 3, c = 5.
Therefore, D = b² – 4ac = (-3)² – 4 × 2 × 5 = 9 – 40 = – 31 < 0
Hence, there is no real roots for this quadratic equation.
Find the values of k for each of the following quadratic equations, so that they have two equal roots. 2 x² + kx + 3 = 0
2x² + kx + 3 = 0 For the quadratic equation 2x² + kx + 3 = 0 we have a = 2, b = k, c = 3. Therefore b² - 4ac = (k)² - 4 × 2 × 3 = k² - 24 For two equal roots, we have k² - 24 = 0 ⇒ k² = 24 ⇒ k = ±√24 ⇒ k = ±2√6.
2x² + kx + 3 = 0
See lessFor the quadratic equation 2x² + kx + 3 = 0 we have a = 2, b = k, c = 3.
Therefore
b² – 4ac = (k)² – 4 × 2 × 3 = k² – 24
For two equal roots, we have k² – 24 = 0
⇒ k² = 24
⇒ k = ±√24
⇒ k = ±2√6.
Is it possible to design a rectangular mango grove whose length is twice its breadth, and the area is 800 m²? If so, find its length and breadth.
Let, the breadth of mango grove = x m Therefore, the lenght = 2x m So, the area = x × 2x = 2x² According to question, 2x² = 800 ⇒ x² = 400 ⇒ x = ±20 Since, the breadth of the mango grove can't be negative, so the breadth = 20 m Hence, the lenght of the mango grove = 2 × 20 = 40 m.
Let, the breadth of mango grove = x m
See lessTherefore, the lenght = 2x m
So, the area = x × 2x = 2x²
According to question, 2x² = 800
⇒ x² = 400
⇒ x = ±20
Since, the breadth of the mango grove can’t be negative, so the breadth = 20 m
Hence, the lenght of the mango grove = 2 × 20 = 40 m.
Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is 20 years. Four years ago, the product of their ages in years was 48.
Let, the age of the first friend = x years So, the age of the other friend = 20 - x years Four years ago: Age of the first friend = x - 4 years Age of the second friend = 20 - x - 4 = 16 - x years According to question, (x - 4) (16 - x) = 48 ⇒ 16x - x² - 64 + 4x = 48 ⇒ x² - 20x + 112 = 0 For the quaRead more
Let, the age of the first friend = x years
See lessSo, the age of the other friend = 20 – x years
Four years ago:
Age of the first friend = x – 4 years
Age of the second friend = 20 – x – 4 = 16 – x years
According to question,
(x – 4) (16 – x) = 48
⇒ 16x – x² – 64 + 4x = 48
⇒ x² – 20x + 112 = 0
For the quadratic equation x² – 20x + 112 = 0, we have a = 1, b = – 20, c = 112.
Therefore,
D = b² – 4ac = (-20)² – 4 × 1 × 112 = 400 – 448 = – 48 < 0
So, there is no real roots of this quadratic equation.
Hence, this situation is not possible.