Problems on Quadratic Equations

Let the numbers be x and y.

Given,

Sum of numbers is 50.

⇒ x + y = 50

⇒ y = 50 - x     .........(1)

Given,

Product of numbers is 525.

⇒ xy = 525     .........(2)

Substituting value of y from equation (1) in equation(2), we get:

⇒ x(50 - x) = 525

⇒ 50x - x² = 525

⇒ x² - 50x + 525 = 0

⇒ x² - 35x - 15x + 525 = 0

⇒ x(x - 35) - 15(x - 35) = 0

⇒ (x - 15)(x - 35) = 0

⇒ (x - 15) = 0 or (x - 35) = 0     [Using zero-product rule]

⇒ x = 15 or x = 35.

Substituting value of x in equation (1), we get :

Case 1 :

If x = 15, y = 50 − 15 = 35.

Case 2:

If x = 35, y = 50 − 35 = 15.

Hence, the two natural numbers are 15 and 35.

Let the numbers be x and y.

Given,

Difference of two numbers is 7.

⇒ x - y = 7

⇒ x = 7 + y     .........(1)

Given,

Product of numbers is 450.

⇒ xy = 450     .........(2)

Substituting value of x from equation(1) in equation(2), we get :

⇒ (7 + y)y = 450

⇒ 7y + y² = 450

⇒ y² + 7y - 450 = 0

⇒ y² - 18y + 25y - 450 = 0

⇒ y(y - 18) + 25(y - 18) = 0

⇒ (y + 25)(y - 18) = 0

⇒ (y + 25) = 0 or (y - 18) = 0     [Using zero-product rule]

⇒ y = -25 or y = 18

y ≠ -25 [since they are natural numbers]

Substituting value of y = 18 in equation (1), we get :

⇒ x = 7 + 18 = 25

⇒ x = 25 and y = 18.

Hence, the two natural numbers are 18 and 25.

Let two consecutive even numbers be x and x + 2.

Given,

Product of numbers is 224.

⇒ x(x + 2) = 224

⇒ x² + 2x - 224 = 0

⇒ x² + 16x - 14x - 224 = 0

⇒ x(x + 16) - 14(x + 16) = 0

⇒ (x - 14)(x + 16) = 0

⇒ (x - 14) = 0 or (x + 16) = 0     [Using zero -product rule]

⇒ x = -16 or x = 14

Since we need positive even integers, x = 14

⇒ x + 2 = 14 + 2 = 16.

Hence, two consecutive positive even integers are 14 and 16.

Let two consecutive natural numbers be x and x + 1.

Given,

The sum of squares of the numbers is 145.

⇒ x² + (x + 1)² = 145

⇒ x² + x² + 2x + 1 = 145

⇒ 2x² + 2x + 1 - 145 = 0

⇒ 2x² + 2x - 144 = 0

⇒ 2(x² + x - 72) = 0

⇒ x² + x - 72 = 0

⇒ x² + 9x - 8x - 72 = 0

⇒ x(x + 9) - 8(x + 9) = 0

⇒ (x - 8)(x + 9) = 0

⇒ (x - 8) = 0 or (x + 9) = 0     [Using zero-product rule]

⇒ x = 8 or x = -9.

Since, they are consecutive natural numbers, x ≠ -9.

Thus, x = 8 and x + 1 = 9.

Hence, two consecutive natural numbers 8 and 9.

Let the numbers be x and y.

Given,

Sum of numbers = 12.

⇒ x + y = 12

⇒ x = 12 - y     .........(1)

Given,

Sum of their squares is 74.

⇒ x² + y² = 74     .........(2)

Substituting value of x from equation (1) in equation (2), we get :

⇒ (12 - y)² + y² = 74

⇒ (12)² + y² - 2 × 12 × y + y² = 74

⇒ y² + y² - 24y + 144 = 74

⇒ 2y² - 24y + 144 - 74 = 0

⇒ 2y² - 24y + 70 = 0

⇒ 2y² - 10y - 14y + 70 = 0

⇒ 2y(y - 5) - 14(y - 5) = 0

⇒ (2y - 14) (y - 5) = 0

⇒ (2y - 14) = 0 or (y - 5) = 0     [Using zero-product rule]

⇒ 2y = 14 or y = 5

⇒ y = $\frac{14}{2}$ or y = 5

⇒ y = 7 or y = 5

Substituting value of y in equation (1), we get :

Case 1: If y = 7,

x = 12 - 7 = 5.

Case 2: If y = 5,

x = 12 - 5 = 7.

Hence, the two natural numbers are 5 and 7.

Given,

Let two consecutive multiples of 3 be 3n and 3n + 3.

Given,

The product is 270.

⇒ 3n(3n + 3) = 270

⇒ n(3n + 3) = $\frac{270}{3}$

⇒ 3n² + 3n = 90

⇒ 3n² + 3n - 90 = 0

⇒ 3(n² + n - 30) = 0

⇒ n² + n - 30 = 0

⇒ n² + 6n - 5n - 30 = 0

⇒ n(n + 6) - 5(n + 6) = 0

⇒ (n - 5)(n + 6) = 0

⇒ (n - 5) = 0 or (n + 6) = 0     [Using zero-product rule]

⇒ n = 5 or n = -6

Case 1: n = 5

First multiple is 3n = 3(5) = 15

Second multiple is 3n + 3 = 3(5) + 3 = 15 + 3 = 18

Case 2: n = -6

First multiple is 3n = 3(-6) = -18

Second multiple is 3n + 3 = 3(-6) + 3 = -18 + 3 = -15

Hence, two consecutive multiples of 3 are 15 and 18 or -18 and -15.

Let the numbers be x and its reciprocal be $\frac{1}{x}$.

Given,

Sum of natural number and it's reciprocal = $\frac{65}{8}$

⇒ x + $\frac{1}{x} = \frac{65}{8}$

⇒ $\frac{x^2 + 1}{x} = \frac{65}{8}$

⇒ 8 × (x² + 1) = 65x

⇒ 8x² + 8 - 65x = 0

⇒ 8x² - 64x -x + 8 = 0

⇒ 8x(x - 8) - 1(x - 8) = 0

⇒ (8x - 1)(x - 8)= 0

⇒ (8x - 1) = 0 or (x - 8) = 0     [Using zero-product rule]

⇒ 8x = 1 or x = 8

⇒ x = $\frac{1}{8}$ or x = 8.

Hence, the natural number is 8.

Let two numbers be x and y.

Given,

Sum of two numbers = 2

⇒ x + y = 2

⇒ x = 2 - y .......(1)

Sum of their reciprocals is 2.25.

⇒ $\frac{1}{x} + \frac{1}{y} = 2.25$

⇒ $\frac{x + y}{xy} = 2.25$ ........(2)

Substituting value of x from equation (1) in (2), we get :

$$\begin{aligned}\Rightarrow \frac{2 - y + y}{(2 - y)y} = 2.25 \\[1em] \Rightarrow \frac{2}{2y - y^2} = \frac{225}{100} \\[1em] \Rightarrow 200 = 225(2y - y^2) \\[1em] \Rightarrow 200 = 450y - 225y^2 \\[1em] \Rightarrow 225y^2 - 450y + 200 = 0 \\[1em] \Rightarrow 25(9y^2 - 18y + 8) = 0 \\[1em] \Rightarrow 9y^2 - 18y + 8 = 0 \\[1em] \Rightarrow 9y^2 - 12y - 6y + 8 = 0 \\[1em] \Rightarrow 3y(3y - 4) - 2(3y - 4) = 0 \\[1em] \Rightarrow (3y - 2)(3y - 4) = 0 \\[1em] \Rightarrow (3y - 2)= 0 \text{ or } (3y - 4) = 0 \text{ Using zero product rule } \\[1em] \Rightarrow 3y = 2 \text{ or } 3y = 4 \\[1em] \Rightarrow y = \frac{2}{3} \text{ or } y = \frac{4}{3}. \\[1em]\end{aligned}$$

Hence, the two numbers are $\frac{2}{3},\frac{4}{3}$.

Let the three consecutive odd numbers be x, x + 2, x + 4.

Given,

The sum of the squares of three consecutive odd numbers is 2531.

⇒ x² + (x + 2)² + (x + 4)² = 2531

⇒ x² + x² + (2)² + 2 × 2 × x + x² + (4)² + 2 × 4 × x = 2531

⇒ x² + x² + 4 + 4x + x² + 16 + 8x = 2531

⇒ 3x² + 12x + 20 = 2531

⇒ 3x² + 12x + 20 - 2531 = 0

⇒ 3x² + 12x - 2511 = 0

⇒ 3x² + 93x - 81x - 2511 = 0

⇒ 3x(x + 31) - 81(x + 31) = 0

⇒ (3x - 81)(x + 31) = 0

⇒ (3x - 81) = 0 or (x + 31) = 0     [Using zero-product rule]

⇒ 3x = 81 or x = -31

⇒ x = $\frac{81}{3}$ or x = -31

⇒ x = 27 or x = -31:

Case 1: If x = 27,

x + 2 = 27 + 2 = 29

x + 4 = 27 + 4 = 31

Case 2: If x = -31,

x + 2 = -31 + 2 = -29

x + 4 = -31 + 4 = -27

Hence, the three consecutive odd numbers are 27, 29, 31 or -31, -29, -27.

Given,

Let two consecutive natural numbers multiple of 3 be 3n and 3n + 3.

Given,

The product is 810.

⇒ 3n(3n + 3) = 810

⇒ n(3n + 3) = $\frac{810}{3}$

⇒ 3n² + 3n = 270

⇒ 3n² + 3n - 270 = 0

⇒ 3(n² + n - 90) = 0

⇒ n² + n - 90 = 0

⇒ n² + 10n - 9n - 90 = 0

⇒ n(n + 10) - 9(n + 10) = 0

⇒ (n - 9)(n + 10) = 0

⇒ (n - 9) = 0 or (n + 10) = 0     [Using zero-product rule]

⇒ n = 9 or n = -10.

n = 9 [Since the numbers are natural numbers]

First multiple is 3n = 3(9) = 27

Second multiple is 3n + 3 = 3(9) + 3 = 27 + 3 = 30.

Hence, two consecutive natural multiples of 3 are 27 and 30.

Let the ten's and unit's digits of required number be x and y respectively.

Number = 10x + y

Given,

Product of digits = 18.

⇒ xy = 18

⇒ x = $\frac{18}{y}$     .........(1)

Given,

If 27 is added to the number, the digits interchange the place.

⇒ 27 + 10x + y = 10y + x

⇒ 27 + 10x + y - 10y - x = 0

⇒ 9x - 9y + 27 = 0

⇒ 9(x - y + 3) = 0

⇒ x - y + 3 = 0

⇒ x - y = -3     .........(2)

Substituting the value of x from equation (1) in equation (2),

⇒ $\frac{18}{y}$ - y = -3

⇒ $\frac{18 - y^2}{y}$ = -3

⇒ 18 - y² = -3y

⇒ y² - 3y - 18 = 0

⇒ y² - 6y + 3y - 18 = 0

⇒ y(y - 6) + 3(y - 6) = 0

⇒ (y + 3)(y - 6) = 0

⇒ (y + 3) = 0 or (y - 6) = 0     [Using zero-product rule]

⇒ y = -3 or y = 6.

Since, digits must be positive y ≠-3.

Substituting value of y = 6 in equation (1), we get :

x = $\frac{18}{6}$ = 3.

The number is : 10 × 3 + 6 = 36.

Hence, the required number is 36.

Let the ten's and unit's digits of required number be x and y respectively.

Number : 10x + y

Given,

Product of digits = 40

⇒ xy = 40     .........(1)

Given,

Difference between digits = 3

⇒ x - y = 3

⇒ x = y + 3     .........(2)

Substituting the value of x from equation (2) in equation (1),

⇒ (y + 3)y = 40

⇒ y² + 3y - 40 = 0

⇒ y² + 8y - 5y - 40 = 0

⇒ y(y + 8) - 5(y + 8) = 0

⇒ (y - 5)(y + 8) = 0

⇒ (y - 5) = 0 or (y + 8) = 0     [Using zero-product rule]

⇒ y = 5 or y = -8

Since, digits should be positive, thus y ≠ -8.

⇒ x = y + 3 = 5 + 3 = 8.

The number = 10x + y = 10(8) + 5 = 85.

Hence, the required number is 85.

Let the numerator of fraction be x and denominator be y.

The fraction is $\frac{x}{y}$.

Given,

Sum of numerator and denominator of the fraction is 10.

⇒ x + y = 10

⇒ y = 10 - x     .........(1)

Given,

If 1 is subtracted from both the numerator and denominator, the fraction is decreased by $\frac{2}{21}$.

$$\begin{aligned}\Rightarrow \frac{x}{y} - \frac{x - 1}{y - 1} = \frac{2}{21} \\[1em] \Rightarrow \frac{x(y - 1) - y(x - 1)}{y(y - 1)} = \frac{2}{21} \\[1em] \Rightarrow \frac{xy - x - xy + y }{y(y - 1)} = \frac{2}{21} \\[1em] \Rightarrow \frac{y - x}{y(y - 1)} = \frac{2}{21} \text{ ......(2)}\end{aligned}$$

Substituting the value of y from equation (1) in equation (2),

$$\begin{aligned}\Rightarrow \frac{y - x}{y(y - 1)} = \frac{2}{21} \\[1em] \Rightarrow \frac{(10 - x) - x}{(10 - x)(10 - x - 1)} = \frac{2}{21} \\[1em] \Rightarrow \frac{10 - 2x}{(10 - x)(9 - x)} = \frac{2}{21} \\[1em] \Rightarrow \frac{10 - 2x}{90 - 10x - 9x + x^2} = \frac{2}{21} \\[1em] \Rightarrow \frac{10 - 2x}{90 - 19x + x^2} = \frac{2}{21} \\[1em] \Rightarrow 21(10 - 2x)= 2(90 - 19x + x^2) \\[1em] \Rightarrow 210 - 42x = 180 - 38x + 2x^2 \\[1em] \Rightarrow 2x^2 - 38x + 180 - 210 + 42x = 0 \\[1em] \Rightarrow 2x^2 + 4x - 30 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 15) = 0 \\[1em] \Rightarrow x^2 + 2x - 15 = 0 \\[1em] \Rightarrow x^2 + 5x - 3x - 15 = 0 \\[1em] \Rightarrow x(x + 5) - 3(x + 5) = 0 \\[1em] \Rightarrow (x - 3)(x + 5) = 0 \\[1em] \Rightarrow (x - 3) = 0 \text{ or }(x + 5) = 0 \text{ [Using zero - product rule] } \\[1em] \Rightarrow x = 3 \text{ or } x = -5 \\[1em]\end{aligned}$$

Since fraction is positive.

Thus, x = 3.

Substituting value of x in equation (1), we get :

⇒ y = 10 - 3

⇒ y = 7.

The fraction is $\frac{3}{7}$.

Hence, the required fraction = $\frac{3}{7}$.

Let the present age of man be x and his son be y.

Given,

Two years ago, man’s age was three times the square of son’s age.

⇒ x - 2 = 3(y - 2)²     .........(1)

Given,

In three years time, man’s age will be four times son’s age.

⇒ x + 3 = 4(y + 3)

⇒ x + 3 = 4y + 12

⇒ x = 4y + 12 - 3

⇒ x = 4y + 9     .........(2)

Substituting value of x from equation (2) in equation (1), we get :

⇒ 4y + 9 - 2 = 3(y - 2)²

⇒ 4y + 9 - 2 = 3(y² - 4y + 4 )

⇒ 4y + 7 = 3y² - 12y + 12

⇒ 3y² - 12y + 12 - 4y - 7 = 0

⇒ 3y² - 16y + 5 = 0

⇒ 3y² - 15y - y + 5 = 0

⇒ 3y(y - 5) - 1(y - 5) = 0

⇒ (3y - 1)(y - 5) = 0

⇒ (3y - 1) = 0 or (y - 5) = 0     [Using zero - product rule]

⇒ y = $\frac{1}{3}$ or y = 5

Since, $\frac{1}{3}$ is not whole number, thus y = 5.

Substituting value of y in equation (2), we get :

⇒ x = 4(5) + 9 = 20 + 9 = 29.

Hence, age of man = 29 years and age of his son = 5 years.

Given,

Total distance = 12 km

In first case :

The speed for the first half = x km/hr.

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time₁ = $\frac{6}{x}$

In second case :

Speed for second half = (x + 2) km/hr

Time₂ = $\frac{6}{x + 2}$

Time taken to cover the whole distance = 2 hrs 30 min.

$$\begin{aligned}\Rightarrow 2.5 = \frac{6}{x} + \frac{6}{x + 2} \\[1em] \Rightarrow 2.5 = \frac{6(x + 2) + 6x}{x(x + 2)} \\[1em] \Rightarrow 2.5 = \frac{6x + 12 + 6x}{x^2 + 2x} \\[1em] \Rightarrow 2.5(x^2 + 2x) = 12x + 12 \\[1em] \Rightarrow 2.5x^2 + 5x = 12x + 12 \\[1em] \Rightarrow 2.5x^2 - 12x - 12 + 5x = 0 \\[1em] \Rightarrow 2.5x^2 - 7x - 12 = 0\end{aligned}$$

Multiplying by 2 on both sides of equation,

$$\begin{aligned}\Rightarrow 2(2.5x^2 - 7x - 12) = 0 \\[1em] \Rightarrow 5x^2 - 14x - 24 = 0 \\[1em] \Rightarrow 5x^2 - 20x + 6x - 24 = 0 \\[1em] \Rightarrow 5x(x - 4) + 6(x - 4) = 0 \\[1em] \Rightarrow (5x + 6)(x - 4) = 0 \\[1em] \Rightarrow (5x + 6) = 0 \text{ or } (x - 4) = 0 \text{ ....[Using zero-product rule]}\\[1em] \Rightarrow x = \frac{-6}{5} \text{ or } x = 4 .\end{aligned}$$

(Since speed can’t be negative).

x = 4km/hr

Hence, x = 4km/hr.

Let the original speed of car be x km/hr.

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken to cover 72 km at original speed = $\frac{72}{x}$ hrs

New Speed = x + 10 km/hr

Time taken to cover 72 km at new Speed = $\frac{72}{x + 10}$ hrs

Given,

The time taken to cover the distance is reduced by 36 minutes or $\frac{36}{60}$ hrs.

$$\begin{aligned}\Rightarrow \frac{72}{x} - \frac{72}{x + 10} = \frac{36}{60} \\[1em] \Rightarrow 72 \times \Big[\frac{1}{x} - \frac{1}{x + 10}\Big] = \frac{3}{5} \\[1em] \Rightarrow \frac{1}{x} - \frac{1}{x + 10} = \frac{3}{5} \times \frac{1}{72} \\[1em] \Rightarrow \frac{(x + 10) - x}{x(x + 10)} = \frac{1}{5} \times \frac{1}{24} \\[1em] \Rightarrow \frac{10}{x^2 + 10x} = \frac{1}{120} \\[1em] \Rightarrow 120 \times 10 = x^2 + 10x \\[1em] \Rightarrow 1200= x^2 + 10x \\[1em] \Rightarrow x^2 + 10x - 1200 = 0 \\[1em] \Rightarrow x^2 + 40x - 30x - 1200 = 0 \\[1em] \Rightarrow x(x + 40) - 30(x + 40) = 0 \\[1em] \Rightarrow (x - 30)(x + 40) = 0 \\[1em] \Rightarrow (x - 30) = 0 \text{ or } (x + 40) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = 30 \text{ or } x = -40.\end{aligned}$$

(Since speed can’t be negative).

Hence, the original speed of the car = 30 km/hr.

By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken by train to cover distance of 600 km at x km/hr = $\frac{600}{x}$hrs

Time taken by train to cover distance of 600 km at (x + 20) km/hr = $\frac{600}{x + 20}$hrs

Given,

On increasing the speed, the time taken is reduced by 5 hours.

$$\begin{aligned}\Rightarrow \frac{600}{x} - \frac{600}{x + 20} = 5 \\[1em] \Rightarrow \frac{600(x + 20) - 600x}{x(x + 20)} = 5 \\[1em] \Rightarrow \frac{600x + 12000 - 600x}{x^2 + 20x} = 5 \\[1em] \Rightarrow 12000 = 5(x^2 + 20x) \\[1em] \Rightarrow 12000 = 5x^2 + 100x \\[1em] \Rightarrow 0 = 5x^2 + 100x - 12000 \\[1em] \Rightarrow 5x^2 + 100x - 12000 = 0 \\[1em] \Rightarrow 5(x^2 + 20x - 2400) = 0 \\[1em] \Rightarrow x^2 + 20x - 2400 = 0 \\[1em] \Rightarrow x^2 + 60x - 40x - 2400 = 0 \\[1em] \Rightarrow x(x + 60) - 40(x + 60) = 0 \\[1em] \Rightarrow (x - 40)(x + 60) = 0 \\[1em] \Rightarrow (x - 40) = 0 \text{ or } (x + 60) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = 40 \text{ or } x = -60\end{aligned}$$

Since speed cannot be negative: x = 40km/hr

Hence, obtained equation is x² + 20x - 2400 = 0 and x = 40 km/hr.

By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken by train at normal speed to cover distance of 780 km = $\frac{780}{x}$ hrs

Time taken by train at reduced speed to cover distance of 780 km = $\frac{780}{x - 5}$hrs

Given,

On decreasing the speed, the time taken by train increases by 1 hour.

$$\begin{aligned}\Rightarrow \frac{780}{x - 5} - \frac{780}{x} = 1 \\[1em] \Rightarrow \frac{780x - 780(x - 5)}{x(x - 5)} = 1 \\[1em] \Rightarrow \frac{780x - 780x + 3900}{x^2 - 5x} = 1 \\[1em] \Rightarrow 3900 = x^2 - 5x \\[1em] \Rightarrow x^2 - 5x - 3900 = 0 \\[1em] \Rightarrow x^2 - 65x + 60x - 3900 = 0 \\[1em] \Rightarrow x(x − 65) + 60(x − 65) = 0 \\[1em] \Rightarrow (x + 60)(x − 65) = 0 \\[1em] \Rightarrow (x + 60) = 0 \text{ or } (x − 65) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = -60 \text{ or } x = 65\end{aligned}$$

Since speed must be positive.

Thus, x = 65 km/hr.

Hence, obtained equation is x² - 5x - 3900 = 0 and the speed of the train = 65 km/hr.

(i) By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Speed of car = x km/hr

Distance between A and B by road = 216 km

Time taken by car = $\frac{216}{x}$ hrs

Hence, time taken by the car to reach town B from A is $\frac{216}{x}$ hrs.

(ii) By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Speed of train = (x + 16) km/hr

Distance between A and B by rail = 208 km

Time = $\frac{208}{x + 16}$ hrs

Hence, time taken by the train to reach town B from A is $\frac{208}{x + 16}$ hrs.

(iii) Given,

Train takes 2 hours less than the car to reach town B.

$$\begin{aligned}\Rightarrow \frac{216}{x} - \frac{208}{x + 16} = 2 \\[1em] \Rightarrow \frac{216(x + 16) - 208x}{x(x + 16)} = 2 \\[1em] \Rightarrow \frac{216x + 3456 - 208x}{x^2 + 16x} = 2 \\[1em] \Rightarrow 8x + 3456 = 2(x^2 + 16x) \\[1em] \Rightarrow 8x + 3456 = 2x^2 + 32x \\[1em] \Rightarrow 2x^2 + 32x - 8x - 3456 = 0 \\[1em] \Rightarrow 2x^2 + 24x - 3456 = 0 \\[1em] \Rightarrow 2(x^2 + 12x - 1728) = 0 \\[1em] \Rightarrow x^2 + 12x - 1728 = 0 \\[1em] \Rightarrow x^2 + 48x - 36x - 1728 = 0 \\[1em] \Rightarrow x(x + 48) - 36(x + 48) = 0 \\[1em] \Rightarrow (x - 36)(x + 48) = 0 \\[1em] \Rightarrow (x - 36) = 0 \text{ or } (x + 48) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = 36 \text{ or } x = -48\end{aligned}$$

Since, speed cannot be negative.

∴ x = 36 km/hr

Hence, obtained equation is x² + 12x - 1728 = 0 and the speed of the train = 36 km/hr.

(iv) Speed of the train = x + 16

= 36 + 16 = 52 km/hr.

Hence, the speed of the train = 52 km/hr.

(i) Given,

Car A travels x km for every litre of petrol.

Thus, car A travels 1 km in $\frac{1}{x}$ litres of petrol.

Thus, car A travels 400 km in $\frac{400}{x}$ litres of petrol.

Car B travels x + 5 km for every litre of petrol.

Thus, car B travels 1 km in $\frac{1}{x + 5}$ litres of petrol.

Thus, car B travels 400 km in $\frac{400}{x + 5}$ litres of petrol

Hence, number of litres used by car A and car B in covering a distance of 400 km = $\frac{400}{x}$ and $\frac{400}{x + 5}$ respectively.

(ii) Given,

Car A uses 4 lires more to cover 400 km.

$$\begin{aligned}\Rightarrow \frac{400}{x} - \frac{400}{x + 5} = 4 \\[1em] \Rightarrow \frac{400(x + 5) - 400x}{x(x + 5)} = 4 \\[1em] \Rightarrow \frac{400x + 2000 - 400x}{x^2 + 5x} = 4 \\[1em] \Rightarrow \frac{2000}{x^2 + 5x} = 4 \\[1em] \Rightarrow 2000 = 4(x^2 + 5x) \\[1em] \Rightarrow 2000 = 4x^2 + 20x \\[1em] \Rightarrow 0 = 4x^2 + 20x - 2000 \\[1em] \Rightarrow 4(x^2 + 5x - 500) = 0 \\[1em] \Rightarrow x^2 + 5x - 500 = 0 \\[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \\[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \\[1em] \Rightarrow (x - 20)(x + 25) = 0 \\[1em] \Rightarrow (x - 20) = 0 \text{ or }(x + 25) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = 20 \text{ or } x = -25.\end{aligned}$$

Since, distance cannot be negative.

Thus, x = 20.

Petrol used by car B = $\frac{400}{x + 5} = \frac{400}{20 + 5} = \frac{400}{25}$ = 16 litres.

Hence, obtained equation is $\frac{400}{x} - \frac{400}{x + 5} = 4$ and the petrol used by car B = 16 litres.

(i) Given,

Speed of a boat in still water = x km/hr

Speed of stream = 3 km/hr

Upstream speed = (x − 3) km/hr.

Hence, the speed of the boat upstream = (x - 3) km/hr.

(ii) Given,

Speed of a boat in still water = x km/hr

Speed of stream = 3 km/hr

Downstream speed = (x + 3) km/hr.

Hence, the speed of the boat downstream = (x + 3) km/hr.

(iii) By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Given,

The boat goes 15 km upstream and 22 km downstream in 5 hours.

$\therefore\frac{15}{x - 3} + \frac{22}{x + 3} = 5$

Hence, the required equation : $\frac{15}{x - 3} + \frac{22}{x + 3} = 5$.

(iv) Solving,

$$\begin{aligned}\Rightarrow \frac{15}{x - 3} + \frac{22}{x + 3} = 5 \\[1em] \Rightarrow \frac{15(x + 3) + 22(x - 3)}{(x + 3)(x - 3)} = 5 \\[1em] \Rightarrow \frac{15x + 45 + 22x - 66}{x^2 - 3^2} = 5 \\[1em] \Rightarrow \frac{37x - 21}{x^2 - 9} = 5 \\[1em] \Rightarrow 37x - 21 = 5(x^2 - 9) \\[1em] \Rightarrow 37x - 21 = 5x^2 - 45 \\[1em] \Rightarrow 0 = 5x^2 - 45 - 37x + 21 \\[1em] \Rightarrow 5x^2 - 37x - 24 = 0 \\[1em] \Rightarrow 5x^2 - 40x + 3x - 24 = 0 \\[1em] \Rightarrow 5x(x - 8) + 3(x - 8) = 0 \\[1em] \Rightarrow (5x + 3)(x - 8) = 0 \\[1em] \Rightarrow (5x + 3) = 0 \text{ or }(x - 8) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = -\frac{3}{5} \text{ or } x = 8.\end{aligned}$$

Since, speed must be positive,

∴ x = 8.

Hence, x = 8 km/hr.

Given,

Capacity of tank = 2400 m³

Let filling capacity of pump = x m³/min,

Time required to fill the tank = $\frac{2400}{x}$

Emptying capacity = (x + 10) m³/min

Time required to empty the tank = $\frac{2400}{x + 10}$

Given,

The pump needs 8 minutes lesser to empty the tank than it needs to fill it.

$$\begin{aligned}\Rightarrow \frac{2400}{x} - \frac{2400}{x + 10} = 8 \\[1em] \Rightarrow \frac{2400(x + 10) - 2400x}{x(x + 10)} = 8 \\[1em] \Rightarrow \frac{2400x + 24000 - 2400x}{x^2 + 10x} = 8 \\[1em] \Rightarrow \frac{24000}{x^2 + 10x} = 8 \\[1em] \Rightarrow 24000 = 8(x^2 + 10x) \\[1em] \Rightarrow 8x^2 + 80x - 24000 = 0 \\[1em] \Rightarrow 8(x^2 + 10x - 3000) = 0 \\[1em] \Rightarrow x^2 + 10x - 3000 = 0 \\[1em] \Rightarrow x^2 + 60x - 50x - 3000 = 0 \\[1em] \Rightarrow x(x + 60) - 50(x + 60) = 0 \\[1em] \Rightarrow (x - 50)(x + 60) = 0 \\[1em] \Rightarrow (x - 50) = 0 \text{ or }(x + 60) = 0 \\[1em] \Rightarrow x = 50 \text{ or } x = -60.\end{aligned}$$

Since, rate must be positive.

∴ x = 50.

Hence, filling capacity of pump is 50 m³/min.

Let the shorter side of right angled triangle = x.

Given,

Hypotenuse of triangle = 20 m

Difference between the other two sides of right angled triangle = 4 m

Then the longer side of right angled triangle (aprat from hypotenuse) = (x + 4) m.

By pythagoras theorem,

⇒ x² + (x + 4)² = 20²

⇒ x² + (x² + 8x + 16) = 400

⇒ x² + x² + 8x + 16 - 400 = 0

⇒ 2x² + 8x - 384 = 0

⇒ 2(x² + 4x - 192) = 0

⇒ x² + 4x - 192 = 0

⇒ x² + 16x - 12x - 192 = 0

⇒ x(x + 16) - 12(x + 16) = 0

⇒ (x - 12)(x + 16) = 0

⇒ (x - 12) = 0 or (x + 16) = 0     [Using zero-product rule]

⇒ x = 12 or x = -16.

Since, length cant be negative.

∴ x = 12 m

Shorter side of right angled triangle = 12 m

Longer side of right angled triangle = x + 4 = 12 + 4 = 16 m.

Hence, length of two sides of right angled triangle is 12 m and 16 m.

(i) Given,

Length of three sides of right angled triangle are (2x - 1) m, (4x) m, (4x + 1) m.

The largest side is hypotenuse = (4x + 1) m

By pythagoras theorem,

⇒ (2x - 1)² + (4x)² = (4x + 1)²

⇒ 4x² - 4x + 1 + 16x² = 16x² + 8x + 1

⇒ 4x² - 4x + 1 + 16x² - 16x² - 8x - 1 = 0

⇒ 4x² - 12x = 0

⇒ 4x(x - 3) = 0

⇒ 4x = 0 or x - 3 = 0

⇒ x = 0 or x = 3

Since, x > 0.

∴ x = 3.

Hence, value of x = 3.

(ii) Calculating the sides of triangle,

Length of first side = 2x - 1 = 2(3) - 1

= 6 - 1

= 5 m

Length of second side = 4x

= 4(3)

= 12 m

Length of hypotenuse = 4x + 1

= 4(3) + 1

= 13 m

By formula,

Area of triangle = $\frac{1}{2} \times$ base × height

= $\frac{1}{2}$ × 5 × 12

= 30 m².

Hence, area of triangle = 30 m².

(i) Given,

Two squares have sides x cm and (x + 5) cm.

Sum of areas of squares = 697 cm²

∴ x² + (x + 5)² = 697

⇒ x² + x² + 10x + 25 = 697

⇒ 2x² + 10x + 25 - 697 = 0

⇒ 2x² + 10x - 672 = 0

⇒ 2(x² + 5x - 336) = 0

⇒ x² + 5x - 336 = 0.

Hence, the required equation is x² + 5x - 336 = 0.

(ii) Solving,

⇒ x² + 5x - 336 = 0

⇒ x² + 21x - 16x - 336 = 0

⇒ x(x + 21) - 16(x + 21) = 0

⇒ (x - 16)(x + 21) = 0

⇒ x - 16 = 0 or x + 21 = 0      [Using zero-product rule]

⇒ x = 16 or x = -21.

Since, length cannot be negative.

Thus, x = 16

Length of side of first square = x = 16 cm

Length of side of second square = x + 5 = 16 + 5 = 21 cm

Hence, length of sides of squares are 16 cm and 21 cm.

Let length of altitude be x meters.

Given,

Area of right triangle = 96 m²

Base of triangle is 3 times altitude.

Length of base = 3x meters

By formula,

⇒ Area of triangle = $\frac{1}{2}$ × base × altitude

⇒ 96 = $\frac{1}{2}$ × 3x × x

⇒ 96 × 2 = 3x²

⇒ 192 = 3x²

⇒ x² = $\frac{192}{3}$

⇒ x² = 64

⇒ x = $\sqrt{64}$

⇒ x = 8 meters.

Length of altitude of triangle = 8 m

Length of base of triangle = 3 × 8 = 24 m.

Hence, length of base = 24 m.

Given,

Length of parallel sides of trapezium = (x + 8) cm and (2x + 3) cm.

The distance between the parallel lines (height) = (x + 4) cm.

Area of trapezium = 590 cm²

By formula,

Area of trapezium = $\frac{1}{2}$ × (sum of parallel sides) × height

⇒ 590 = $\frac{1}{2}$ (x + 8 + 2x + 3)(x + 4)

⇒ 590 × 2 = (3x + 11)(x + 4)

⇒ 1180 = 3x² + 12x + 11x + 44

⇒ 1180 = 3x² + 23x + 44

⇒ 0 = 3x² + 23x + 44 - 1180

⇒ 3x² + 23x - 1136 = 0

⇒ 3x² - 48x + 71x - 1136 = 0

⇒ 3x(x - 16) + 71(x - 16) = 0

⇒ (3x + 71)(x - 16) = 0

⇒ (x - 16) = 0 or (3x + 71) = 0     [Using zero-product rule]

⇒ x = 16 or x = $\frac{-71}{3}$

Since, length cannot be negative.

Thus, x = 16

Hence, value of x = 16.

Given,

The ratio between the length and the breadth of a rectangular field is 3 : 2.

Let length of rectangle be 3x and breadth of rectangle be 2x.

Given,

When the length is increased by 5, new length = 3x + 5.

Area of rectangle with increased length = 2600 m²

By formula,

Area of rectangle = length × breadth

⇒ 2600 = (3x + 5)(2x)

⇒ 2600 = 6x² + 10x

⇒ 6x² + 10x - 2600 = 0

⇒ 2(3x² + 5x - 1300) = 0

⇒ 3x² + 5x - 1300 = 0

⇒ 3x² + 65x - 60x - 1300 = 0

⇒ x(3x + 65) - 20(3x + 65) = 0

⇒ (x - 20)(3x + 65) = 0

⇒ (x - 20) = 0 or (3x + 65) = 0     [Using zero-product rule]

⇒ x = 20 or x = $-\frac{65}{3}$

Since, length cannot be negative.

Thus, x = 20.

Breadth of rectangle = 2x = 2 × 20 = 40 m.

Hence, breadth of rectangle = 40 m.

(i) Given,

The length of rectangular plot is x metres

Perimeter of the rectangular plot = 114 m

By formula,

⇒ Perimeter of rectangle = 2(Length + Breadth)

⇒ 114 = 2(x + breadth)

⇒ $\frac{114}{2}$ = x + breadth

⇒ Breadth = (57 - x) meters.

Hence, breadth of rectangle = 57 - x.

(ii) Given,

Area of rectangle = 810 m²

By formula,

Area of rectangle = Length × Breadth

⇒ 810 = x(57 - x)

⇒ 810 = 57x - x²

⇒ x² - 57x + 810 = 0

Hence, obtained equation is x² - 57x + 810 = 0.

(iii) Solving,

⇒ x² - 57x + 810 = 0

⇒ x² - 27x - 30x + 810 = 0

⇒ x(x - 27) - 30(x - 27) = 0

⇒ (x - 30)(x - 27) = 0

⇒ (x - 30) = 0 or (x - 27) = 0     [Using zero-product rule]

⇒ x = 30 or x = 27.

Case 1:

If, x = 30 m then breadth = 57 - x = 57 - 30 = 27 m.

Case 2:

If, x = 27 m then breadth = 57 - x = 57 - 27 = 30 m.

Hence, the dimensions of the rectangular plot are 27 m and 30 m.

Given,

Cost price of given article = ₹ x

Gain % after selling article = x%

Selling price = ₹ 56

By formula,

⇒ Selling price = $\frac{(100 + x)}{100}$ × cost price

⇒ 56 = $\frac{(100 + x)}{100}$ × x

⇒ 56 × 100 = x(100 + x)

⇒ 5600 = 100x + x²

⇒ x² + 100x - 5600 = 0

⇒ x² + 140x - 40x - 5600 = 0

⇒ x(x + 140) - 40(x + 140) = 0

⇒ (x - 40)(x + 140) = 0

⇒ (x - 40) = 0 or (x + 140) = 0     [Using zero-product rule]

⇒ x = 40 or x = -140

Since, cost price and gain% cannot be negative.

Thus, x = 40.

Hence, value of x = 40.

(i) Given,

In first case :

₹ 6400 were divided equally among x persons

Money each person receives = $\frac{6400}{x}$

In second case:

If ₹ 6400 is divided among x + 14 persons, then each gets = $\frac{6400}{x + 14}$

In second case each person receives ₹ 28 less.

$$\begin{aligned}\Rightarrow \frac{6400}{x} - \frac{6400}{x + 14} = 28 \\[1em] \Rightarrow \frac{6400(x + 14) - 6400x}{x(x + 14)} = 28 \\[1em] \Rightarrow \frac{6400x + 14 \times 6400 - 6400x}{x^2 + 14x} = 28 \\[1em] \Rightarrow 14 \times 6400 = 28(x^2 + 14x) \\[1em] \Rightarrow \frac{14 \times 6400}{28} = x^2 + 14x \\[1em] \Rightarrow 3200 = x^2 + 14x \\[1em] \Rightarrow x^2 + 14x - 3200 = 0 \\[1em] \Rightarrow x^2 + 64x - 50x - 3200 = 0 \\[1em] \Rightarrow x(x + 64) - 50(x + 64) = 0 \\[1em] \Rightarrow (x - 50)(x + 64) = 0 \\[1em] \Rightarrow (x - 50) = 0 \text{ or } (x + 64) = 0 \text{.....[Using zero-product rule]} \\[1em] \Rightarrow x = 50 \text{ or } x = -64.\end{aligned}$$

Since, the number of persons cannot be negative.

Thus, x = 50.

Hence, value of x = 50.

(ii) Given,

Let original number of children be x.

In first case:

₹ 7,500 is divided among x children.

Money each child receives = ₹ $\frac{7500}{x}$

In second case:

If ₹ 7500 is divided among (x - 20) children, then each gets = ₹ $\frac{7500}{x - 20}$

In second case, each student will receive ₹ 100 more.

$$\begin{aligned}\Rightarrow \frac{7500}{x - 20} - \frac{7500}{x} = 100 \\[1em] \Rightarrow \frac{7500x - 7500(x - 20)}{x(x - 20)} = 100 \\[1em] \Rightarrow \frac{7500x - 7500x + 7500 \times 20}{x^2 - 20x} = 100 \\[1em] \Rightarrow 7500 \times 20 = 100(x^2 - 20x) \\[1em] \Rightarrow \frac{7500 \times 20}{100} = x^2 - 20x \\[1em] \Rightarrow 1500 = x^2 - 20x \\[1em] \Rightarrow x^2 - 20x - 1500 = 0 \\[1em] \Rightarrow x^2 + 30x - 50x - 1500 = 0 \\[1em] \Rightarrow x(x + 30) - 50(x + 30) = 0 \\[1em] \Rightarrow (x - 50)(x + 30) = 0 \\[1em] \Rightarrow (x - 50) = 0 \text{ or } (x + 30) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = 50 \text{ or } x = -30.\end{aligned}$$

Since, the number of children cannot be negative.

Thus, x = 50.

Hence, the original number of children = 50.

(i) Given,

Cost of x books = ₹ 720

The cost of one book = ₹ $\frac{720}{x}$

Hence, cost of one book = ₹ $\frac{720}{x}$.

(ii) If cost of per book is ₹5 less,

Then new cost per book = $\frac{720}{x} - 5$

According to question,

$$\begin{aligned}\Rightarrow \frac{720}{\frac{720}{x} - 5} = x + 2 \\[1em] \Rightarrow \frac{720}{\frac{720 - 5x}{x}} = x + 2 \\[1em] \Rightarrow \frac{720x}{720 - 5x} = x + 2 \\[1em] \Rightarrow 720x = (x + 2)(720x - 5x) \\[1em] \Rightarrow 720x = 720x - 5x^2 + 1440 - 10x \\[1em] \Rightarrow 720x - 720x + 5x^2 - 1440 + 10x = 0 \\[1em] \Rightarrow 5x^2 + 10x - 1440 = 0 \\[1em] \Rightarrow 5(x^2 + 2x - 288 )= 0 \\[1em] \Rightarrow x^2 + 2x - 288 = 0 \\[1em] \Rightarrow x^2 + 18x - 16x - 288 = 0 \\[1em] \Rightarrow x(x + 18) - 16(x + 18) = 0 \\[1em] \Rightarrow (x - 16)(x + 18) = 0 \\[1em] \Rightarrow (x - 16) = 0 \text{ or } (x + 18) = 0 \text{ ....[Using zero-product rule]} \\[1em] \Rightarrow x = 16 \text{ or } x = -18\end{aligned}$$

Since, number of books cannot be negative.

∴ x = 16

Hence, equation obtained is x² + 2x - 288 = 0 and the value of x = 16.

(i) Given,

Cost of x apples = ₹ 1,200

Cost of one apple = ₹ $\frac{1200}{x}$

Hence, cost of each apple in terms of x = ₹ $\frac{1200}{x}$.

(ii) Given,

Fruit-seller sells each apple at ₹ 3 more than the cost price of each.

Selling price of each apple = ₹ $\frac{1200}{x}$ + 3

Selling price of (x − 10) apples = ₹ $(x - 10)\Big(\frac{1200}{x} + 3 \Big)$

Hence, total selling price = ₹ $(x - 10)\Big(\frac{1200}{x} + 3 \Big)$.

(iii) Cost price = ₹ 1,200

Profit = ₹ 60

Selling price = Cost price + Profit = 1200 + 60 = ₹ 1,260

$$\begin{aligned}\Rightarrow 1260 = (x - 10)\Big(\frac{1200}{x} + 3 \Big) \\[1em] \Rightarrow 1260 = (x - 10)\Big(\frac{1200 + 3x}{x} \Big) \\[1em] \Rightarrow 1260x = (x - 10)(1200 + 3x) \\[1em] \Rightarrow 1260x = 1200x + 3x^2 - 12000 - 30x \\[1em] \Rightarrow 0 = 1200x - 1260x - 30x + 3x^2 - 12000 \\[1em] \Rightarrow 3x^2 - 90x - 12000 = 0 \\[1em] \Rightarrow 3(x^2 - 30x - 4000) = 0 \\[1em] \Rightarrow x^2 - 30x - 4000 = 0 \\[1em] \Rightarrow x^2 - 80x + 50x - 4000 = 0 \\[1em] \Rightarrow x(x - 80) + 50(x - 80) = 0 \\[1em] \Rightarrow (x + 50) = 0 \text{ or } (x - 80) = 0 \text{.....[Using zero-product rule]} \\[1em] \Rightarrow x = -50 \text{ or } x = 80.\end{aligned}$$

Since,

Number of apples cannot be negative.

Thus, x = 80

Hence, obtained equation x² - 30x - 4000 = 0 and the value x = 80.

Let the original number of students be x.

In first case:

Given,

Budget of food = ₹ 2,400

Cost per person = ₹ $\frac{2400}{x}$

In second case :

Given,

8 students failed to join, number of students who went = x - 8

Cost per person = ₹ $\frac{2400}{x - 8}$

When 8 students failed to join the party, the cost of the food for each member increased by ₹ 50.

$$\begin{aligned}\therefore \frac{2400}{x - 8} - \frac{2400}{x} = 50 \\[1em] \Rightarrow \frac{2400x - 2400(x - 8)}{x(x - 8)} = 50 \\[1em] \Rightarrow \frac{2400x - 2400x + 19200}{x^2 - 8x} = 50 \\[1em] \Rightarrow 19200 = 50(x^2 - 8x) \\[1em] \Rightarrow \frac{19200}{50} = (x^2 - 8x) \\[1em] \Rightarrow 384 = x^2 - 8x \\[1em] \Rightarrow x^2 - 8x - 384 = 0 \\[1em] \Rightarrow x^2 - 24x + 16x - 384 = 0 \\[1em] \Rightarrow x(x - 24) + 16(x - 24) = 0 \\[1em] \Rightarrow (x + 16)(x - 24) = 0 \\[1em] \Rightarrow (x + 16) = 0 \text{ or } (x - 24) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = -16 \text{ or } x = 24\end{aligned}$$

Since, number of students cannot be negative.

Thus, x = 24.

Hence, number of students went to picnic = 24.

By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Distance to be covered by bus = 240 km

Time taken by bus at original speed = $\frac{240}{x}$

Time taken by bus at reduced speed = $\frac{240}{x - 10}$

Given,

On decreasing the speed, it takes two hours longer to cover distance.

$$\begin{aligned}\Rightarrow \frac{240}{x - 10} - \frac{240}{x} = 2 \\[1em] \Rightarrow \frac{240x - 240(x - 10)}{x(x - 10)} = 2 \\[1em] \Rightarrow \frac{240x - 240x + 2400}{x^2 - 10x} = 2 \\[1em] \Rightarrow 2400 = 2(x^2 - 10x) \\[1em] \Rightarrow \frac{2400}{2} = (x^2 - 10x) \\[1em] \Rightarrow 1200 = x^2 - 10x \\[1em] \Rightarrow 0 = x^2 - 10x - 1200 \\[1em] \Rightarrow x^2 - 10x - 1200 = 0 \\[1em] \Rightarrow x^2 - 40x + 30x - 1200 = 0 \\[1em] \Rightarrow x(x - 40) + 30(x - 40) = 0 \\[1em] \Rightarrow (x + 30)(x - 40) = 0 \\[1em] \Rightarrow (x + 30) = 0 \text{ or } (x - 40) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = -30 \text{ or } x = 40.\end{aligned}$$

x = 40 [speed must be positive]

Hence, the equation formed is x² - 10x - 1200 = 0 and the speed of bus = 40 km/hr.

Original speed of man be x km/hr.

Given,

Distance to be traveled by man = 100 km

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken by man at speed x km/hr = $\frac{100}{x}$

Time taken by man at increased speed (x + 5) = $\frac{100}{x + 5}$

Given, with increased speed the man takes 1 hour less.

$$\begin{aligned}\Rightarrow \frac{100}{x} - \frac{100}{x + 5} = 1 \\[1em] \Rightarrow \frac{100(x + 5) - 100x}{x(x + 5)} = 1 \\[1em] \Rightarrow \frac{100x + 500 - 100x}{x^2 + 5x} = 1 \\[1em] \Rightarrow 500 = (x^2 + 5x) \\[1em] \Rightarrow 0 = x^2 + 5x - 500 \\[1em] \Rightarrow x^2 + 5x - 500 = 0 \\[1em] \Rightarrow x^2 + 25x - 20x - 500 = 0 \\[1em] \Rightarrow x(x + 25) - 20(x + 25) = 0 \\[1em] \Rightarrow (x + 25)(x - 20) = 0 \\[1em] \Rightarrow (x + 25) = 0 \text{ or } (x - 20) = 0 \text{....[Using zero-product rule]} \\[1em] \Rightarrow x = -25 \text{ or } x = 20.\end{aligned}$$

Since, speed cannot be negative.

Thus, x = 20.

Hence, the original speed of man = 20 km/hr.

Let two natural numbers be x and y.

Given,

Sum of numbers = 27

⇒ x + y = 27

⇒ y = 27 - x     .........(1)

Given,

Product of numbers is 182.

⇒ xy = 182     .........(2)

Substituting value of y from equation (1) in equation (2), we get :

⇒ x(27 - x) = 182

⇒ 27x - x² = 182

⇒ x² - 27x + 182 = 0

⇒ x² - 13x - 14x + 182 = 0

⇒ x(x - 13) - 14(x - 13) = 0

⇒ (x - 14)(x - 13) = 0

⇒ (x - 14) = 0 or (x - 13) = 0     [Using zero-product rule]

⇒ x = 14 or x = 13.

Substituting value of x in equation (1), we get :

If x = 14, y = 27 − 14 = 13

If x = 13, y = 27 − 13 = 14.

The smallest number among two numbers is 13.

Hence, option 1 is the correct option.

Let the two consecutive odd natural numbers be x and x + 2.

Given,

Sum of the squares of two consecutive odd natural numbers is 74.

⇒ x² + (x + 2)² = 74

⇒ x² + x² + 4 + 4x = 74

⇒ 2x² + 4x + 4 - 74 = 0

⇒ 2x² + 4x - 70 = 0

⇒ 2x² + 14x - 10x - 70 = 0

⇒ 2x(x + 7) - 10(x + 7) = 0

⇒ (2x - 10)(x + 7) = 0

⇒ (2x - 10) = 0 or (x + 7) = 0     [Using zero-product rule]

⇒ 2x = 10 or x = -7

⇒ x = $\frac{10}{2}$ or x = -7

⇒ x = 5 or x = -7

Since the number required is natural number, thus x ≠-7.

x + 2 = 5 + 2 = 7.

The greater number among the two numbers is 7.

Hence, option 2 is the correct option.

Let the two natural numbers be x and x + 2.

Given,

Sum of the squares of numbers is 202.

⇒ x² + (x + 2)² = 202

⇒ x² + x² + 4 + 4x = 202

⇒ 2x² + 4x + 4 - 202 = 0

⇒ 2x² + 4x - 198 = 0

⇒ 2x² + 22x - 18x - 198 = 0

⇒ 2x(x + 11) - 18(x + 11) = 0

⇒ (2x - 18)(x + 11) = 0

⇒ (2x - 18) = 0 or (x + 11) = 0     [Using zero-product rule]

⇒ 2x = 18 or x = -11

⇒ x = $\frac{18}{2}$ or x = -11

⇒ x = 9 or x = -11.

Since, the number required is natural number, thus x ≠ -11,

⇒ x + 2 = 9 + 2 = 11

Sum of the two numbers = 9 + 11 = 20.

Hence, option 4 is the correct option.

Given,

The total amount of money = ₹ 40.

Let the shares of two friends be ₹ x and ₹ y respectively.

⇒ x + y = 40

⇒ y = 40 - x     .........(1)

Given,

The product of two parts of amount distributed is 364.

⇒ xy = 364     .........(2)

Substituting value of y from equation (1) in equation (2), we get :

⇒ x(40 - x) = 364

⇒ 40x - x² = 364

⇒ x² - 40x + 364 = 0

⇒ x² - 26x - 14x + 364 = 0

⇒ x(x - 26) - 14(x - 26) = 0

⇒ (x - 14)(x - 26) = 0

⇒ (x - 14) = 0 or (x - 26) = 0     [Using zero-product rule]

⇒ x = 14 or x = 26

Substituting value of x in equation (1), we get:

Case 1: If x = 14, y = 40 − 14 = 26

Case 2: If x = 26, y = 40 − 26 = 14.

The difference between the two parts of amount is, ₹ 26 - ₹ 14 = ₹ 12.

Hence, option 3 is the correct option.

Let the breadth and length of a rectangle be x cm and (x + 4) cm.

Given,

Area of rectangle = 96 cm².

⇒ x(x + 4) = 96

⇒ x² + 4x = 96

⇒ x² + 4x - 96 = 0

⇒ x² - 8x + 12x - 96 = 0

⇒ x(x - 8) + 12(x - 8) = 0

⇒ (x + 12)(x - 8) = 0

⇒ (x + 12) = 0 or (x - 8) = 0     [Using zero-product rule]

⇒ x = -12 or x = 8

Since length and breadth cannot be negative, thus Breadth = x = 8 cm.

Length = x + 4 = 8 + 4 = 12 cm

Perimeter of the rectangle = 2(l + b)

= 2(12 + 8)

= 2(20)

= 40 cm.

Hence, option 2 is the correct option.

Let the area of square be x cm²,

Given,

Four times area of square = 484 cm²

⇒ 4x = 484

⇒ x = $\frac{484}{4}$

⇒ x = 121 cm²

Area of Square = 121 cm²

Let side of square be a cm.

⇒ a² = 121

⇒ a = $\sqrt{121}$

⇒ a = ± 11

Since, length cannot be negative, thus a = 11 cm.

Perimeter of Square = 4 × a

⇒ 4 × 11

⇒ 44 cm.

Hence, option 4 is the correct option.

Let two consecutive positive integers be x and x + 1.

Given,

The sum of squares of the two consecutive positive integers = 365.

⇒ x² + (x + 1)² = 365

⇒ x² + x² + 2x + 1 = 365

⇒ 2x² + 2x + 1 - 365 = 0

⇒ 2x² + 2x - 364 = 0

⇒ 2(x² + x - 182) = 0

⇒ x² + x - 182 = 0

⇒ x² + 14x - 13x - 182 = 0

⇒ x(x + 14) - 13(x + 14) = 0

⇒ (x - 13)(x + 14) = 0

⇒ (x - 13) = 0 or (x + 14) = 0     [Using zero -product rule]

⇒ x = 13 or x = -14.

Since, they are consecutive positive integers, x ≠ -14.

⇒ x + 1 = 13 + 1 = 14.

The sum of numbers is = 13 + 14 = 27.

Hence, option 1 is the correct option.

Let the base and height of right triangle be x cm and (x - 17) cm respectively.

By pythagoras theorem,

⇒ Base² + Height² = Hypotenuse²

⇒ x² + (x - 17) ² = (25)²

⇒ x² + x² + (17)² - 2 × x × 17 = 625

⇒ x² + x² + 289 - 34x = 625

⇒ 2x² - 34x + 289 - 625 = 0

⇒ 2x² - 34x - 336 = 0

⇒ 2(x² - 17x - 168) = 0

⇒ x² - 17x - 168 = 0

⇒ x² - 24x + 7x - 168 = 0

⇒ x(x - 24) + 7(x - 24) = 0

⇒ (x + 7)(x - 24) = 0

⇒ (x + 7) = 0 or (x - 24) = 0     [Using zero -product rule]

⇒ x = -7 or x = 24.

Since, the length of triangle cannot be negative, x ≠ -7.

x - 17 = 24 - 17 = 7.

The perimeter of right triangle is = 7 + 24 + 25 = 56 cm.

Hence, option 2 is the correct option.

Let the total number of articles be x and the cost of each article be y.

Given,

The cost of an article is ₹ 3 more than twice the total number of articles.

⇒ y = 2x + 3     .........(1)

Given,

Total cost of all articles = ₹ 189.

⇒ xy = 189     .........(2)

Substituting value of y from equation (1) in equation (2), we get :

⇒ x(2x + 3) = 189

⇒ 2x² + 3x = 189

⇒ 2x² + 3x - 189 = 0

⇒ 2x² - 18x + 21x - 189 = 0

⇒ 2x(x - 9) - 21(x - 9) = 0

⇒ (2x - 21)(x - 9) = 0

⇒ (2x - 21) = 0 or (x - 9) = 0     [Using zero -product rule]

⇒ 2x = 21 or x = 9

⇒ x = $\frac{21}{2}$ or x = 9

Since, the number of articles cannot be in fraction, thus x ≠ $\frac{21}{2}$.

Hence, option 2 is the correct option.

Let the shorter side of rectangular field be x meters.

Given,

The diagonal of rectangular field is 60 m more than shorter side.

Diagonal = (x + 60) meters

Given,

The longer side of rectangle is 30 m more than shorter side, Let the longer side be z,

Longer side = (x + 30) meters

By pythagoras theorem,

In a rectangular field,

⇒ Hypotenuse² = Shorter side² + Longer side²

⇒ (x + 60)² = x² + (x + 30)²

⇒ [x² + (60)² + 2 × x × 60] = x² + [x² + (30)² + 2 × x × 30]

⇒ x² + 3600 + 120x = x² + x² + 900 + 60x

⇒ x² + 3600 + 120x = 2x² + 900 + 60x

⇒ 2x² + 900 + 60x - x² - 3600 - 120x = 0

⇒ 2x² - x² + 60x - 120x - 3600 + 900 = 0

⇒ x² - 60x - 2700 = 0

⇒ x² - 90x + 30x - 2700 = 0

⇒ x(x - 90) + 30(x - 90) = 0

⇒ (x + 30)(x - 90) = 0

⇒ (x + 30) = 0 or (x - 90) = 0     [Using zero -product rule]

⇒ x = -30 or x = 90

Since length of rectangle cannot be negative x ≠ -30

The longer side of rectangle is,

x + 30 = 90 + 30 = 120 meters.

Thus, sides are 90 m and 120 m.

Hence, option 3 is the correct option.

Let Neha's present age be x and the age of her father be y.

Given,

Neha's father is 28 years older than her.

y = 28 + x     .........(1)

Given,

Product of their ages 4 years ago was 245.

⇒ (x - 4)(y - 4) = 245     .........(2)

Substituting value of y from equation (1) in equation (2), we get :

⇒ (x - 4)(28 + x - 4) = 245

⇒ (x - 4)(x + 24) = 245

⇒ x² + 24x - 4x - 96 = 245

⇒ x² + 20x - 96 - 245 = 0

⇒ x² + 20x - 341 = 0.

Hence, option 2 is the correct option.

By formula,

Time = $\frac{\text{Distance}}{\text{Speed}}$

Initial speed of train = x km/hr

Time taken to cover 480 km = $\frac{480}{x}$ hrs

Reduced speed of train = (x - 8) km/hr

Time taken to cover 480 km = $\frac{480}{x - 8}$ hrs

Given,

On reducing speed the time taken is 3 hours more.

$$\begin{aligned}\Rightarrow \frac{480}{x - 8} - \frac{480}{x} = 3 \\[1em] \Rightarrow \frac{480x - 480(x - 8)}{x(x - 8)} = 3 \\[1em] \Rightarrow \frac{480x - 480x + 3840}{x^2 - 8x} = 3 \\[1em] \Rightarrow 3840 = 3(x^2 - 8x) \\[1em] \Rightarrow 3840 = 3x^2 - 24x \\[1em] \Rightarrow 3x^2 - 24x - 3840 = 0 \\[1em] \Rightarrow 3(x^2 - 8x - 1280) = 0 \\[1em] \Rightarrow x^2 - 8x - 1280 = 0.\end{aligned}$$

Hence, option 1 is the correct option.

Given,

Distance covered by cars = 72 km

Given,

Car X travels x km using 1 litre of diesel.

Therefore, to travel 1 km car X uses $\frac{1}{x}$ litres of diesel.

To travel 72 km car X will use $\frac{72}{x}$ litres of diesel.

Hence, option 3 is the correct option.

Given,

Distance covered by cars = 72 km

Given,

Car Y travels x + 3 km using 1 litre of diesel.

Therefore, to travel 1 km car Y uses $\frac{1}{x + 3}$ litres of diesel.

To travel 72 km car Y will use $\frac{72}{x + 3}$ litres of diesel.

Hence, option 2 is the correct option.

Diesel used by car X = $\frac{72}{x}$ litres

Diesel used by car Y = $\frac{72}{x + 3}$ litres

Given,

Car X used 4 litres of diesel more than car Y in the journey.

$\therefore \frac{72}{x} - \frac{72}{x + 3} = 4$

Hence, option 3 is the correct option.

Solving,

$$\begin{aligned}\Rightarrow \frac{72}{x} - \frac{72}{x + 3} = 4 \\[1em] \Rightarrow \frac{72(x + 3) - 72x}{x(x + 3)} = 4 \\[1em] \Rightarrow \frac{72x + 216 - 72x}{x^2 + 3x} = 4 \\[1em] \Rightarrow 216 = 4(x^2 + 3x) \\[1em] \Rightarrow \frac{216}{4} = (x^2 + 3x) \\[1em] \Rightarrow 54 = (x^2 + 3x) \\[1em] \Rightarrow x^2 + 3x - 54 = 0 \\[1em] \Rightarrow x^2 + 9x - 6x - 54 = 0 \\[1em] \Rightarrow x(x + 9) - 6(x + 9) = 0 \\[1em] \Rightarrow (x - 6)(x + 9) = 0 \\[1em] \Rightarrow (x - 6) = 0 \text{ or } (x + 9) = 0 \text{....[Using zero-product rule]}\\[1em] \Rightarrow x = 6 \text{ or } x = -9.\end{aligned}$$

Distance covered cannot be negative.

Thus, x = 6.

Diesel used by car X = $\frac{72}{x}$

= $\frac{72}{6}$

= 12 litres.

Hence, option 2 is the correct option.

Given,

M.P. of toy = ₹ x

G.S.T. charged = x%

Price including G.S.T. = ₹ 24

∴ 24 = x + $\frac{x}{100}$ × x

⇒ 24 = x + $\frac{x^2}{100}$

⇒ 24 = $\frac{100x + x^2}{100}$

⇒ 2400 = 100x + x²

⇒ x² + 100x - 2400 = 0

⇒ x² + 120x - 20x - 2400 = 0

⇒ x(x + 120) − 20(x + 120) = 0

⇒ (x − 20)(x + 120) = 0

⇒ x − 20 = 0 or x + 120 = 0

⇒ x = 20 or x = −120

Since, M.P. cannot be negative.

∴ M.P. = ₹ 20 and G.S.T. = 20%.

Hence, the marked price of toy = ₹ 20 and G.S.T. = 20%.

1. Let x be the number of students planned a picnic.

Given,

Budget of hiring a bus = ₹ 1440

After 8 students refused, number of students those who went = x - 8

Share for hiring the bus per student who went for the picnic = $\frac{1440}{x - 8}$

Hence, option (4) is the correct option.

2. In first case:

Initially share per student for hiring the bus = $\frac{1440}{x}$

Share per student after 8 students refused to go to the picnic = $\frac{1440}{x - 8}$

According to question,

Cost per student increases by ₹ 30.

$$\begin{aligned}\Rightarrow \frac{1440}{x - 8} - \frac{1440}{x} = 30 \\[1em] \Rightarrow \frac{1440x - 1440(x - 8)}{x(x - 8)} = 30 \\[1em] \Rightarrow \frac{1440x - 1440x + 1440 \times 8}{x^2 - 8x} = 30 \\[1em] \Rightarrow 1440 \times 8 = 30(x^2 - 8x) \\[1em] \Rightarrow \frac{1440 \times 8}{30} = x^2 - 8x \\[1em] \Rightarrow 384 = x^2 - 8x \\[1em] \Rightarrow 0 = x^2 - 8x - 384 \\[1em] \Rightarrow x^2 - 8x - 384 = 0\end{aligned}$$

Hence, option (1) is the correct option.

3. Solving,

⇒ x² - 8x - 384 = 0

⇒ x² - 24x + 16x - 384 = 0

⇒ x(x - 24) + 16(x - 24) = 0

⇒ (x + 16)(x - 24) = 0

⇒ (x + 16) = 0 or (x - 24) = 0     [Using zero-product rule]

⇒ x = -16 or x = 24

⇒ x = 24 [Number of students cannot be negative]

Number of students who went for picnic = x - 8 = 24 - 8 = 16.

Hence, option (2) is the correct option.

4. Share for hiring the bus per student who planned picnic = $\frac{1440}{x} = \frac{1440}{24}$ = ₹ 60.

Number of students who did not go to picnic = 8

Share of eight persons = 60 × 8 = ₹ 480.

Hence, option (4) is the correct option.

5. Share of each student (initially) = ₹ 60.

Hence, option (3) is the correct option.

1. Given,

Original speed of the bus = x km/hr

Next given distance = 90

Speed for next distance = (x + 10) km/hr

Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken for next distance = $\frac{90}{(x + 10)}$.

Hence, option (3) is the correct option.

2. Time taken by bus to travel 75 km = $\frac{75}{x}$

Time taken by bus to travel 90 km = $\frac{90}{x + 10}$

Given,

Total time taken to complete the journey = 3 hours

$$\begin{aligned}\Rightarrow \frac{75}{x} + \frac{90}{x + 10} = 3 \\[1em] \Rightarrow \frac{75(x + 10) + 90x}{x(x + 10)} = 3 \\[1em] \Rightarrow \frac{75x + 750 + 90x}{x^2 + 10x} = 3 \\[1em] \Rightarrow 750 + 165x = 3(x^2 + 10x) \\[1em] \Rightarrow 0 = 3x^2 + 30x - 165x - 750 \\[1em] \Rightarrow 3x^2 - 135x - 750 = 0 \\[1em] \Rightarrow 3(x^2 - 45x - 250) = 0 \\[1em] \Rightarrow x^2 - 45x - 250 = 0.\end{aligned}$$

Hence, option (2) is the correct option.

3. Solving,

⇒ x² - 45x - 250 = 0

⇒ x² + 5x - 50x - 250 = 0

⇒ x(x + 5) - 50(x + 5) = 0

⇒ (x + 5)(x - 50) = 0

⇒ (x + 5) = 0 or (x - 50) = 0     [Using zero-product rule]

⇒ x = -5 or x = 50

⇒ x = 50 [As speed cannot be negative]

Speed = 50 km/hr

Hence, option (1) is the correct option.

4. The new speed of bus = x + 10 = 50 + 10 = 60 km/hr.

Hence, option (3) is the correct option.

5. Given,

Distance = 510 km

New speed = 60 km/hr.

Time = $\frac{\text{Distance}}{\text{Speed}}$

= $\frac{510}{60} = 8\frac{1}{2}$ hrs.

Hence, option (2) is the correct option.

1. Let time taken by smaller tap to fill tank = x hours

Part of tank filled by smaller tap in one hour = $\frac{1}{x}$

Time taken by larger tap to fill tank = (x - 2) hours

Part of tank filled by larger tap in one hour $\frac{1}{x - 2}$

Part of tank filled by larger tap in two hour $\frac{2}{x - 2}$

Hence, option (4) is the correct option.

2. Given,

Together smaller and larger tap take $1\frac{7}{8} = \frac{15}{8}$ hours to fill tank.

Rate at which the taps fill tank in one hour = $\frac{1}{\frac{15}{8}} = \frac{8}{15}$ tank/hr

From question 1 we have,

Part of tank filled by smaller tap in one hour = $\frac{1}{x}$

Part of tank filled by larger tap in one hour = $\frac{1}{x - 2}$

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \\[1em] \Rightarrow \frac{x - 2 + x}{x^2 - 2x}= \frac{8}{15} \\[1em] \Rightarrow 15(2x - 2) = 8(x^2 - 2x) \\[1em] \Rightarrow 30x - 30 = 8x^2 - 16x \\[1em] \Rightarrow 0 = 8x^2 - 16x - 30x + 30 \\[1em] \Rightarrow 8x^2 - 46x + 30 = 0 \\[1em] \Rightarrow 2(4x^2 - 23x + 15) = 0 \\[1em] \Rightarrow 4x^2 - 23x + 15 = 0\end{aligned}$$

Hence, option (1) is the correct option.

3. Solving equation from question 2,

⇒ 4x² - 23x + 15 = 0

⇒ 4x² - 20x - 3x + 15 = 0

⇒ 4x(x - 5) - 3(x - 5) = 0

⇒ (4x - 3)(x - 5) = 0

⇒ (4x - 3) = 0 or (x - 5) = 0     [Using zero-product rule]

⇒ x = $\frac{3}{4}$ or x = 5

Case 1 : x = $\frac{3}{4}$

Larger tap take (x - 2) hours = $\frac{3}{4} - 2 = \frac{3 - 8}{4} = \frac{-5}{4}$

= -1.25 hours, which is not possible.

Case 2 : x = 5

The time taken by the smaller tap is x = 5 hours.

The time taken by the larger tap is x − 2 = 5 - 2 = 3 hours.

Hence, option (2) is the correct option.

4. The smaller tap fills the tank in x = 5 hours.

Part of tank filled by smaller tap in one hour = $\frac{1}{5}$

Part of tank filled by smaller tap in 3 hours = $\frac{3}{5}$

Hence, option (3) is the correct option.

5. The larger tap fills the tank in 3 hours.

Part of tank filled by larger tap in one hour $\frac{1}{3}$

Part of tank filled by larger tap in $1\frac{1}{4} = \frac{5}{4}$ hours

= $\frac{\frac{5}{4}}{3}$

= $\frac{5}{12}$

Hence, option (1) is the correct option.

1. Given,

The speed of the motorboat in still water is 24 km/hr

The speed of the stream be x km/hr

The speed of the motorboat in going upstream = boat speed in still water − stream speed = (24 - x) km/hr

Hence, option (2) is the correct option.

2. The speed of the motorboat in going downstream = boat speed in still water + stream speed = (24 + x) km/hr

Distance to be covered by motorboat = 32km

Time = $\frac{\text{Distance}}{\text{Speed}}$

In first case:

Time taken by motorboat in going upstream = $\frac{32}{24 - x}$

In second case:

Time taken by motorboat in going downstream = $\frac{32}{24 + x}$

The difference in time taken between first case and second case = 1 hour more

$$\begin{aligned}\Rightarrow \frac{32}{24 - x} - \frac{32}{24 + x} = 1 \\[1em] \Rightarrow \frac{32(24 + x) - 32(24 - x)}{(24 - x)(24 + x)}= 1 \\[1em] \Rightarrow \frac{768 + 32x - 768 + 32x}{24^2 - x^2}= 1 \\[1em] \Rightarrow \frac{64x}{576 - x^2}= 1 \\[1em] \Rightarrow 64x = 576 - x^2 \\[1em] \Rightarrow x^2 + 64x - 576 = 0.\end{aligned}$$

Hence, option (3) is the correct option.

3. Solving equation from question 2,

⇒ x² + 64x - 576 = 0

⇒ x² - 8x + 72x - 576 = 0

⇒ x(x - 8) + 72(x - 8) = 0

⇒ (x + 72)(x - 8) = 0

⇒ (x + 72) = 0 or (x - 8) = 0     [Using zero-product rule]

⇒ x = -72 or x = 8

⇒ x = 8km/hr [speed of the stream cannot be negative]

Speed of the motorboat downstream = 24 + x = 24 + 8 = 32 km/hr.

Hence, option (4) is the correct option.

4. The downstream speed of motorboat is 32 km/hr (from Question 3).

The time taken by the motorboat to go 272 km downstream = $\frac{272}{32} = \frac{17}{2}$ = 8.5 hours

Hence, option (1) is the correct option.

5. Speed of the motorboat upstream = 16 km/hr

The time taken by the motorboat to go 80 km upstream = $\frac{80}{16}$ = 5 hours

The time taken by the motorboat to return 80 km downstreamstream = $\frac{80}{32}$ = 2.5 hours

Total time = 5 + 2.5 = 7.5 hours

Hence, option (3) is the correct option.