Problems on Quadratic Equations — Exercise 6

Find two natural numbers whose sum is 50 and product is 525.

Answer

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.

The difference of two natural numbers is 7 and their product is 450. Find the numbers.

Answer

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.

Find two consecutive positive even integers whose product is 224.

Answer

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.

Find two consecutive natural numbers, the sum of whose squares is 145.

Answer

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.

The sum of two natural numbers is 12 and the sum of their squares is 74. Find the numbers.

Answer

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 = $\dfrac{14}{2}$ 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.

Find two consecutive multiples of 3 whose product is 270.

Answer

Given,

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

Given,

The product is 270.

⇒ 3n(3n + 3) = 270

⇒ n(3n + 3) = $\dfrac{270}{3}$

⇒ 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.

The sum of a natural number and its reciprocal is $\dfrac{65}{8}$. Find the natural number.

Answer

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

Given,

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

⇒ x + $\dfrac{1}{x} = \dfrac{65}{8}$

⇒ $\dfrac{x^2 + 1}{x} = \dfrac{65}{8}$

⇒ 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 = $\dfrac{1}{8}$ or x = 8.

Hence, the natural number is 8.

The sum of two numbers is 2 and the sum of their reciprocals is 2.25. Find the numbers.

Answer

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.

⇒ $\dfrac{1}{x} + \dfrac{1}{y} = 2.25$

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

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

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

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

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

Answer

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 = $\dfrac{81}{3}$ 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.

The product of two consecutive natural numbers which are multiples of 3 is equal to 810. Find the two numbers.

Answer

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) = $\dfrac{810}{3}$

⇒ 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.

A number consists of two digits whose product is 18. If 27 is added to the number, the digits interchange their places. Find the number.

Answer

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 = $\dfrac{18}{y}$     .........(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),

⇒ $\dfrac{18}{y}$ - y = -3

⇒ $\dfrac{18 - y^2}{y}$ = -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 = $\dfrac{18}{6}$ = 3.

The number is : 10 × 3 + 6 = 36.

Hence, the required number is 36.

A two-digit number contains the smaller of the two digits in the unit place. The product of the digits is 40 and the difference between the digits is 3. Find the number.

Answer

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.

The sum of the numerator and denominator of a certain fraction is 10. If 1 is subtracted from both the numerator and denominator, the fraction is decreased by $\dfrac{2}{21}$. Find the fraction.

Answer

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

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

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 $\dfrac{2}{21}$.

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

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

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

Since fraction is positive.

Thus, x = 3.

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

⇒ y = 10 - 3

⇒ y = 7.

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

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

Two years ago, a man’s age was three times the square of his son’s age. In three years time, his age will be four times his son’s age. Find their present ages.

Answer

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 = $\dfrac{1}{3}$ or y = 5

Since, $\dfrac{1}{3}$ 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.

Ashish goes to his friends house which is 12 km away from his house. He covers half of the distance at a speed of x km per hour and the remaining at (x + 2) km per hour. If he takes 2 hrs 30 min. to cover the whole distance, find the value of x.

Answer

Given,

Total distance = 12 km

In first case :

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

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

Time₁ = $\dfrac{6}{x}$

In second case :

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

Time₂ = $\dfrac{6}{x + 2}$

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

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

Multiplying by 2 on both sides of equation,

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

(Since speed can’t be negative).

x = 4km/hr

Hence, x = 4km/hr.

By increasing the speed of a car by 10 km/hr, the time of journey of 72 km is reduced by 36 minutes. Find the original speed of the car.

Answer

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

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

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

New Speed = x + 10 km/hr

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

Given,

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

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

(Since speed can’t be negative).

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

A train covers a distance of 600 km at x km/hr. Had the speed been (x + 20) km/hr, the time taken to cover the same distance would have been reduced by 5 hours. Write down an equation in x and solve it to evaluate x.

Answer

By formula,

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

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

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

Given,

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

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

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

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

A train covers a distance of 780 km at x km/hr. Had the speed been (x − 5) km/hr, the time taken to cover the same distance would have been increased by 1 hour. Write down an equation in x and solve it to evaluate x.

Answer

By formula,

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

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

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

Given,

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

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

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.

The distance by road between two towns A and B, is 216 km, and by rail it is 208 km. A car travels at a speed of x km/hr and the train travels at a speed which is 16 km/hr faster than the car.

(i) Write down the time taken by the car to reach town B from A, in terms of x.

(ii) Write down the time taken by the train to reach town B from A, in terms of x.

(iii) If the train takes 2 hours less than the car to reach town B, obtain an equation in x and solve it.

(iv) Hence, find the speed of the train.

Answer

(i) By formula,

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

Speed of car = x km/hr

Distance between A and B by road = 216 km

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

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

(ii) By formula,

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

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

Distance between A and B by rail = 208 km

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

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

(iii) Given,

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

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

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.

Car A travels x km for every litre of petrol, while car B travels (x + 5) km for every litre of petrol.

(i) Write down the number of litres used by car A and car B in covering a distance of 400 km.

(ii) If car A used 4 litres of petrol more than car B in covering 400 km, write an equation in x and solve it to determine the number of litres of petrol used by car B for the journey.

Answer

(i) Given,

Car A travels x km for every litre of petrol.

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

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

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

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

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

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

(ii) Given,

Car A uses 4 lires more to cover 400 km.

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

Since, distance cannot be negative.

Thus, x = 20.

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

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

The speed of a boat in still water is x km/hr and the speed of the stream is 3 km/hr.

(i) Write the speed of the boat upstream, in terms of x.

(ii) Write the speed of the boat downstream, in terms of x.

(iii) If the boat goes 15 km upstream and 22 km downstream in 5 hours, write an equation in x to represent the statement.

(iv) Solve the equation to evaluate x.

Answer

(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 = $\dfrac{\text{Distance}}{\text{Speed}}$

Given,

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

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

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

(iv) Solving,

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

Since, speed must be positive,

∴ x = 8.

Hence, x = 8 km/hr.

A booster pump can be used for filling as well as for emptying a tank. The capacity of the tank is 2400 m³. The emptying capacity of the tank is 10 m³ per minute higher than its filling capacity and the pump needs 8 minutes lesser to empty the tank than it needs to fill it. What is the filling capacity of the pump?

Answer

Given,

Capacity of tank = 2400 m³

Let filling capacity of pump = x m³/min,

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

Emptying capacity = (x + 10) m³/min

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

Given,

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

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

Since, rate must be positive.

∴ x = 50.

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

The hypotenuse of a right triangle is 20 m. If the difference between the lengths of other sides be 4 m, find the other sides.

Answer

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.

The lengths of the sides of a right triangle are (2x − 1) m, (4x) m and (4x + 1) m, where x > 0. Find :

(i) the value of x,

(ii) the area of the triangle.

Answer

(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 = $\dfrac{1}{2} \times$ base × height

= $\dfrac{1}{2}$ × 5 × 12

= 30 m².

Hence, area of triangle = 30 m².

Two squares have sides x cm and (x + 5) cm. The sum of their areas is 697 sq. cm.

(i) Express this as an algebraic equation in x.

(ii) Solve this equation to find the sides of the squares.

Answer

(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.

The area of a right-angled triangle is 96 m². If its base is three times its altitude, find the base.

Answer

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 = $\dfrac{1}{2}$ × base × altitude

⇒ 96 = $\dfrac{1}{2}$ × 3x × x

⇒ 96 × 2 = 3x²

⇒ 192 = 3x²

⇒ x² = $\dfrac{192}{3}$

⇒ 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.

The lengths of the parallel sides of a trapezium are (x + 8) cm and (2x + 3) cm, and the distance between them is (x + 4) cm. If its area is 590 cm², find the value of x.

Answer

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 = $\dfrac{1}{2}$ × (sum of parallel sides) × height

⇒ 590 = $\dfrac{1}{2}$ (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 = $\dfrac{-71}{3}$

Since, length cannot be negative.

Thus, x = 16

Hence, value of x = 16.

The ratio between the length and the breadth of a rectangular field is 3 : 2. If only the length is increased by 5 metres, the new area of the field will be 2600 sq. metres. What is the breadth of the rectangular field ?

Answer

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 = $-\dfrac{65}{3}$

Since, length cannot be negative.

Thus, x = 20.

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

Hence, breadth of rectangle = 40 m.

The perimeter of a rectangular plot of land is 114 metres and its area is 810 square metres.

(i) Take the length of plot as x metres. Use the perimeter 114 m to write the value of the breadth in terms of x.

(ii) Use the values of length, breadth and area to write an equation in x.

(iii) Solve the equation to find the length and breadth of the plot.

Answer

(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)

⇒ $\dfrac{114}{2}$ = 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.

A man buys an article for ₹ x and sells it for ₹ 56 at a gain of x%. Find the value of x.

Answer

Given,

Cost price of given article = ₹ x

Gain % after selling article = x%

Selling price = ₹ 56

By formula,

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

⇒ 56 = $\dfrac{(100 + x)}{100}$ × 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) ₹ 6,400 were divided equally among x persons. Had this money been divided equally among (x + 14) persons, each would have got ₹ 28 less. Find the value of x.

(ii) ₹ 7,500 were divided equally among a certain number of children. Had there been 20 less children, each would have received ₹ 100 more. Find the original number of children.

Answer

(i) Given,

In first case :

₹ 6400 were divided equally among x persons

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

In second case:

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

In second case each person receives ₹ 28 less.

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

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 = ₹ $\dfrac{7500}{x}$

In second case:

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

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

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

Since, the number of children cannot be negative.

Thus, x = 50.

Hence, the original number of children = 50.

A shopkeeper buys x books for ₹ 720.

(i) Write the cost of 1 book in terms of x.

(ii) If the cost per book be ₹ 5 less, the number of books that could be bought for ₹ 720 would be 2 more.

Write down the equation in x and solve it to find x.

Answer

(i) Given,

Cost of x books = ₹ 720

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

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

(ii) If cost of per book is ₹5 less,

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

According to question,

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

Since, number of books cannot be negative.

∴ x = 16

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

A fruit-seller bought x apples for ₹ 1,200.

(i) Write the cost price of each apple in terms of x.

(ii) If 10 of the apples were rotten and he sold each of the rest at ₹ 3 more than the cost price of each, write the selling price of (x − 10) apples.

(iii) If he made a profit of ₹ 60 in this transaction, form an equation in x and solve it to evaluate x.

Answer

(i) Given,

Cost of x apples = ₹ 1,200

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

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

(ii) Given,

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

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

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

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

(iii) Cost price = ₹ 1,200

Profit = ₹ 60

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

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

Since,

Number of apples cannot be negative.

Thus, x = 80

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

Some students planned a picnic. The budget for the food was ₹ 2,400. As 8 of them failed to join the party, the cost of the food for each member increased by ₹ 50. Find how many students went to the picnic.

Answer

Let the original number of students be x.

In first case:

Given,

Budget of food = ₹ 2,400

Cost per person = ₹ $\dfrac{2400}{x}$

In second case :

Given,

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

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

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

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

Since, number of students cannot be negative.

Thus, x = 24.

Hence, number of students went to picnic = 24.

A bus covers a distance of 240 km at a uniform speed. Due to heavy rain, its speed gets reduced by 10 km/hr and as such it takes two hours longer to cover the total distance. Assuming the uniform speed to be x km/hr, form an equation and solve it to evaluate x.

Answer

By formula,

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

Distance to be covered by bus = 240 km

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

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

Given,

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

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

x = 40 [speed must be positive]

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

A man covers a distance of 100 km, traveling with a uniform speed of x km/hr. Had the speed been 5 km/hr more, it would have taken 1 hour less. Find x, the original speed.

Answer

Original speed of man be x km/hr.

Given,

Distance to be traveled by man = 100 km

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

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

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

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

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

Since, speed cannot be negative.

Thus, x = 20.

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