Quadratic Equations — Exercise 5.5

Find two consecutive natural numbers such that the sum of their squares is 61.

Answer

Let the required numbers be = x , x + 1

Given, sum of squares of the numbers = 61

$\Rightarrow x^2 + (x + 1)^2 = 61 \\[1em] \Rightarrow x^2 + (x^2 + 1 + 2x) = 61 \\[1em] \Rightarrow x^2 + x^2 + 1 + 2x = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 - 61 = 0 \\[1em] \Rightarrow 2x^2 + 2x - 60 = 0 \\[1em] \Rightarrow 2(x^2 + x - 30) = 0 \\[1em] \Rightarrow x^2 + x - 30 = 0 \\[1em] \Rightarrow x^2 + 6x - 5x - 30 = 0 \\[1em] \Rightarrow x(x + 6) - 5(x + 6) = 0 \\[1em] \Rightarrow (x + 6)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 6 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -6 \text { or } x = 5$

Since the numbers are natural number so x ≠ -6.

∴ x = 5 , x + 1 = 6.

Hence, the required natural numbers are 5 , 6.

Find two consecutive integers such that the sum of their squares is 61.

Answer

Let the required numbers be = x , x + 1

Given, sum of squares of the numbers = 61

$\Rightarrow x^2 + (x + 1)^2 = 61 \\[1em] \Rightarrow x^2 + (x^2 + 1 + 2x) = 61 \\[1em] \Rightarrow x^2 + x^2 + 1 + 2x = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 = 61 \\[1em] \Rightarrow 2x^2 + 2x + 1 - 61 = 0 \\[1em] \Rightarrow 2x^2 + 2x - 60 = 0 \\[1em] \Rightarrow 2(x^2 + x - 30) = 0 \\[1em] \Rightarrow x^2 + x - 30 = 0 \\[1em] \Rightarrow x^2 + 6x - 5x - 30 = 0 \\[1em] \Rightarrow x(x + 6) - 5(x + 6) = 0 \\[1em] \Rightarrow (x + 6)(x - 5) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 6 = 0 \text{ or } x - 5 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -6 \text { or } x = 5$

∴ When x = -6 , x + 1 = -5 and when x = 5 , x + 1 = 6.

Hence, required integers are 5,6 or -6,-5

If the product of two positive consecutive even integers is 288, find the integers.

Answer

Let the required two positive consecutive even integers be x , x + 2

Given, product of two consecutive even integers = 288

$\Rightarrow x(x + 2) = 288 \\[1em] \Rightarrow x^2 + 2x = 288 \\[1em] \Rightarrow x^2 + 2x - 288 = 0 \\[1em] \Rightarrow x^2 + 18x - 16x - 288 = 0 \\[1em] \Rightarrow x(x + 18) - 16(x + 18) = 0 \\[1em] \Rightarrow (x - 16)(x + 18) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 16 = 0 \text{ or } x + 18 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 16 \text{ or } x = -18$

Since the numbers are natural number so x ≠ -18.

∴ x = 16 , x + 2 = 18.

Hence, required integers are 16, 18.

If the product of two consecutive even integers is 224 , find the integers.

Answer

Let the required two consecutive even integers be x , x + 2

Given, product of two consecutive even integers = 224

$\Rightarrow x(x + 2) = 224 \\[1em] \Rightarrow x^2 + 2x = 224 \\[1em] \Rightarrow x^2 + 2x - 224 = 0 \\[1em] \Rightarrow x^2 + 16x - 14x - 224 = 0 \\[1em] \Rightarrow x(x + 16) - 14(x + 16) = 0 \\[1em] \Rightarrow (x + 16)(x - 14) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 16 = 0 \text{ or } x - 14 = 0. \\[1em] \Rightarrow x = -16 \text{ or } x = 14 \\[1em]$

∴ When x = 14 , x + 2 = 16 and when x = -16 , x + 2 = -14.

Hence, required integers are 14 , 16 or -16, -14 .

Find two consecutive even natural numbers such that the sum of their squares is 340.

Answer

Let the required two consecutive even integers be x , x + 2

Given, sum of squares of two consecutive even natural numbers = 340

$\Rightarrow x^2 + (x + 2)^2 = 340 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x = 340 \\[1em] \Rightarrow 2x^2 + 4x + 4 -340 = 0 \\[1em] \Rightarrow 2x^2 + 4x - 336 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 168) = 0 \\[1em] \Rightarrow x^2 + 2x - 168 = 0 \\[1em] \Rightarrow x^2 + 14x - 12x - 168 = 0 \\[1em] \Rightarrow x(x + 14) - 12(x + 14) = 0 \\[1em] \Rightarrow (x - 12)(x + 14) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x - 12 = 0 \text{ or } x + 14 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = 12 \text{ or } x = -14$

Since the numbers are natural number so x ≠ -14

∴ x = 12 , x + 2 = 14

Hence, required natural numbers are 12 , 14 .

Find two consecutive odd integers such that sum of their squares is 394.

Answer

Let the required two consecutive odd integers be x , x + 2

Given, sum of squares of two consecutive odd integers = 394

$\Rightarrow x^2 + (x + 2)^2 = 394 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x = 394 \\[1em] \Rightarrow 2x^2 + 4x + 4 -394 = 0 \\[1em] \Rightarrow 2x^2 + 4x - 390 = 0 \\[1em] \Rightarrow 2(x^2 + 2x - 195) = 0 \\[1em] \Rightarrow x^2 + 2x - 195 = 0 \\[1em] \Rightarrow x^2 + 15x - 13x - 195 = 0 \\[1em] \Rightarrow x(x + 15) - 13(x + 15) = 0 \\[1em] \Rightarrow (x + 15)(x - 13) = 0 \text{ (Factorising left side) } \\[1em] \Rightarrow x + 15 = 0 \text{ or } x - 13 = 0 \text{ (Zero-product rule) } \\[1em] \Rightarrow x = -15 \text{ or } x = 13$

∴ When x = -15 , x + 2 = -13 and when x = 13 , x + 2 = 15.

Hence required integers are -15 , -13 or 13, 15 .

The sum of two numbers is 9 and the sum of their squares is 41. Find the numbers.

Answer

Let the first number be x.

Since the sum of two numbers is 9, so other number is 9 - x.

Given, the sum of squares of numbers = 41

⇒ x² + (9 - x)² = 41

⇒ x² + x² + 81 - 18x = 41

⇒ 2x² - 18x + 81 - 41 = 0

⇒ 2x² - 18x + 40 = 0

⇒ 2(x² - 9x + 20) = 0

⇒ x² - 9x + 20 = 0

⇒ x² - 4x - 5x + 20 = 0

⇒ x(x - 4) - 5(x - 4) = 0

⇒ (x - 4)(x - 5) = 0

⇒ x - 4 = 0 or x - 5 = 0

⇒ x = 4 or x = 5

∴ x = 5, 9 - x = 4

∴ x = 4, 9 - x = 5

Hence, the numbers are 4 and 5.

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

Answer

Let first number be x.

Since the difference of two numbers is 7, so other number is x + 7.

Given, the products of numbers = 450

⇒ x(x + 7) = 450

⇒ x² + 7x = 450

⇒ x² + 7x - 450 = 0

⇒ x² + 25x - 18x - 450 = 0

⇒ x(x + 25) - 18(x + 25) = 0

⇒ (x + 25)(x - 18) = 0

⇒ x + 25 = 0 or x - 18 = 0

⇒ x = -25 or x = 18.

Since, the numbers are natural numbers,

∴ x ≠ -25

∴ x = 18, x + 7 = 25

Hence, the numbers are 18 and 25.

Five times a certain whole number is equal to three less than twice the square of the number. Find the number.

Answer

Let the number be x

Given, 5 times the number = 3 less than twice the square of the number

$\Rightarrow 5x = 2x^2 - 3 \\[0.5em] \Rightarrow 5x - 2x^2 + 3 = 0 \\[0.5em] \Rightarrow 2x^2 - 5x - 3 = 0 \text{ (on multiplying the equation by -1) } \\[0.5em] \Rightarrow 2x^2 - 6x + x - 3 = 0 \\[0.5em] \Rightarrow 2x(x - 3) + 1(x - 3) = 0 \\[0.5em] \Rightarrow (2x + 1)(x - 3) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow 2x + 1 = 0 \text{ or } x - 3 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = -\dfrac{1}{2} \text{ or } x = 3 \\[0.5em]$

Since the number is a whole number x ≠ $-\dfrac{1}{2}$

∴ x = 3

Hence, the required whole number is 3.

Sum of two natural numbers is 8 and the difference of their reciprocal is $\dfrac{2}{15}.$ Find the numbers.

Answer

Let the first number be x

Since, the sum of two numbers is 8, so other number is 8 - x.

Given, the difference of reciprocal of numbers = $\dfrac{2}{15}$

$\Rightarrow \dfrac{1}{x} - \dfrac{1}{8 - x} = \dfrac{2}{15} \\[0.5em] \Rightarrow \dfrac{8 - x - x}{x(8 - x)} = \dfrac{2}{15} \text{ (On taking L.C.M.) } \\[0.5em] \Rightarrow 15(8 - 2x) = 2x(8 - x) \text{ (On cross multiplication ) } \\[0.5em] \Rightarrow 120 - 30x = 16x - 2x^2 \\[0.5em] \Rightarrow 120 - 30x - 16x + 2x^2 = 0 \\[0.5em] \Rightarrow 2x^2 - 46x + 120 = 0 \\[0.5em] \Rightarrow 2(x^2 - 23x + 60) = 0 \\[0.5em] \Rightarrow x^2 - 20x - 3x + 60 = 0 \\[0.5em] \Rightarrow x(x - 20) - 3(x - 20) = 0 \\[0.5em] \Rightarrow (x - 3)(x - 20) = 0 \text{ (Factorising left side) } \\[0.5em] \Rightarrow x - 3 = 0 \text{ (or) } x - 20 = 0 \text{ (Zero-product rule) } \\[0.5em] \Rightarrow x = 3 \text{ or } x = 20.$

If x = 20 , 8 - x = -12 , Since both are natural numbers hence x ≠ 20.

∴ x = 3 , 8 - x = 5

Hence, the required natural numbers are 3 , 5.

The difference of the squares of two numbers is 45. The square of the smaller number is 4 times the larger number. Determine the numbers.

Answer

Let the larger number be x .

Given, the square of the smaller number is 4 times the larger number.

Hence, square of smaller number = 4x

Given, the difference of squares of two numbers is = 45

$\Rightarrow x^2 - 4x = 45 \\[0.5em] \Rightarrow x^2 - 4x - 45 = 0 \\[0.5em] \Rightarrow x^2 - 9x + 5x - 45 = 0 \\[0.5em] \Rightarrow x(x - 9) + 5(x - 9) = 0 \\[0.5em] \Rightarrow (x + 5)(x - 9) = 0 \\[0.5em] \Rightarrow x + 5 = 0 \text{ or } x - 9 = 0 \\[0.5em] \Rightarrow x = -5 \text{ or } x = 9$

If x = -5 , $\sqrt{4x} = \sqrt{-20}$ , which is not valid as there is no real value of square root of a negative value.

∴ x = 9 , $\sqrt{4x} = \sqrt{36} = 6 \text{ or } -6$.

Hence, the two numbers are 9, 6 or 9, -6.

There are three consecutive positive integers such that the sum of the square of the first and the product of the other two is 154. What are the integers ?

Answer

Let the numbers be x , x + 1 , x + 2.

Given, sum of the square of the first and the product of the other two is = 154

$\Rightarrow x^2 + (x + 1)(x + 2) = 154 \\[1em] \Rightarrow x^2 + (x^2 + 2x + x + 2) = 154 \\[1em] \Rightarrow 2x^2 + 3x + 2 - 154 = 0 \\[1em] \Rightarrow 2x^2 + 3x - 152 = 0 \\[1em] \Rightarrow 2x^2 + 19x - 16x - 152 = 0 \\[1em] \Rightarrow x(2x + 19) - 8(2x + 19) = 0 \\[1em] \Rightarrow (x - 8)(2x + 19) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } 2x + 19 = 0 \\[1em] x = 8 \text{ or } x = -\dfrac{19}{2}$

Since the integers are positive hence , x ≠ $-\dfrac{19}{2}$

∴ x = 8 , x + 1 = 9 , x + 2 = 10.

Hence , the required numbers are 8, 9, 10.

Find three succcessive even natural numbers, the sum of whose squares is 308.

Answer

Let the required numbers be x, x + 2, x + 4.

Given, the sum of squares of these numbers = 308

$\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 308 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 308 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 308 \\[1em] \Rightarrow 3x^2 + 12x - 288 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 96) = 0 \\[1em] \Rightarrow x^2 + 4x - 96 = 0 \\[1em] \Rightarrow x^2 + 12x - 8x - 96 = 0 \\[1em] \Rightarrow x(x + 12) - 8(x + 12) = 0 \\[1em] \Rightarrow (x - 8)(x + 12) = 0 \\[1em] \Rightarrow x - 8 = 0 \text{ or } x + 12 = 0 \\[1em] x = 8 \text{ or } x = -12$

Since the numbers are natural hence , x ≠ -12

∴ x = 8 , x + 2 = 10 , x + 4 = 12.

Hence , the required numbers are 8, 10, 12.

Find three consecutive odd integers , the sum of whose squares is 83.

Answer

Let the required numbers be x , x + 2 , x + 4.

Given, the sum of squares of these numbers = 308

$\Rightarrow x^2 + (x + 2)^2 + (x + 4)^2 = 83 \\[1em] \Rightarrow x^2 + x^2 + 4 + 4x + x^2 + 16 + 8x = 83 \\[1em] \Rightarrow 3x^2 + 12x + 20 = 83 \\[1em] \Rightarrow 3x^2 + 12x - 63 = 0 \\[1em] \Rightarrow 3(x^2 + 4x - 21) = 0 \\[1em] \Rightarrow x^2 + 4x - 21 = 0 \\[1em] \Rightarrow x^2 + 7x - 3x - 21 = 0 \\[1em] \Rightarrow x(x + 7) - 3(x + 7) = 0 \\[1em] \Rightarrow (x + 7)(x - 3) = 0 \\[1em] \Rightarrow x + 7 = 0 \text{ or } x - 3 = 0 \\[1em] x = -7 \text{ or } x = 3$

∴ When x = -7 , x + 2 = -5 , x + 4 = -3 and when x = 3 , x + 2 = 5 , x + 4 = 7.

Hence , the required numbers are -7, -5, -3 and 3, 5, 7.

In a certain positive fraction , the denominator is greater than the numerator by 3. If 1 is subtracted from both the numerator and denominator , the fraction is decreased by $\dfrac{1}{14}$. Find the fraction.

Answer

Let the numerator of the fraction be x

Given, the denominator is greater than numerator by 3 hence, denominator = x + 3

Fraction = $\dfrac{x}{x + 3}$

Given, if 1 is subtracted from both the numerator and denominator , the fraction is decreased by $\dfrac{1}{14}$

$\Rightarrow \dfrac{x}{x + 3} - \dfrac{x - 1}{x + 3 - 1} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x}{x + 3} - \dfrac{x - 1}{x + 2} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x(x + 2) - (x - 1)(x + 3)}{(x + 3)(x + 2)} = \dfrac{1}{14} \text{ (On taking L.C.M) } \\[1em] \Rightarrow \dfrac{x^2 + 2x - (x^2 + 3x - x - 3)}{(x^2 + 2x + 3x + 6)} = \dfrac{1}{14} \\[1em] \Rightarrow \dfrac{x^2 - x^2 + 2x - 2x + 3}{x^2 + 5x + 6} = \dfrac{1}{14} \\[1em] \Rightarrow 3 \times 14 = x^2 + 5x + 6 \text{ (Cross multiplying) } \\[1em] \Rightarrow x^2 + 5x + 6 = 42 \\[1em] \Rightarrow x^2 + 5x + 6 - 42 = 0 \\[1em] \Rightarrow x^2 + 5x - 36 = 0 \\[1em] \Rightarrow x^2 + 9x - 4x - 36 = 0 \\[1em] \Rightarrow x(x + 9) - 4(x + 9) = 0 \\[1em] \Rightarrow (x - 4)(x + 9) = 0 \\[1em] x = 4 \text{ or } x = -9 \\[1em]$

If x = -9 , Fraction = $\dfrac{x}{x + 3} = \dfrac{9}{6}$ In this case numerator > denominator which is incorrect according to the question hence, x ≠ -9.

∴ x = 4 , Fraction = $\dfrac{x}{x + 3} = \dfrac{4}{7}$

Hence, the fraction is $\dfrac{4}{7}$.

The sum of numerator and denominator of a certain positive fraction is 8 . If 2 is added to both the numerator and denominator, the fraction is increased by $\dfrac{4}{35}.$ Find the fraction.

Answer

Let the denominator be = x so, numerator = 8 - x.

Fraction = $\dfrac{8 - x}{x}$

Given , if 2 is added to both the numerator and denominator, the fraction is increased by $\dfrac{4}{35}.$

$\Rightarrow \dfrac{8 - x + 2}{x + 2} - \dfrac{8 - x}{x} = \dfrac{4}{35} \\[1em] \Rightarrow \dfrac{x(10 - x) - (x + 2)(8 - x)}{x(x + 2)} = \dfrac{4}{35} \text{ (On taking L.C.M) }\\[1em] \Rightarrow \dfrac{10x - x^2 - (8x - x^2 + 16 - 2x)}{x^2 + 2x} = \dfrac{4}{35} \\[1em] \Rightarrow 35(-x^2 + x^2 - 8x + 10x + 2x - 16) = 4(x^2 + 2x) \\[1em] \Rightarrow 35(4x - 16) = 4x^2 + 8x \\[1em] \Rightarrow 140x - 560 = 4x^2 + 8x \\[1em] \Rightarrow 4x^2 + 8x - 140x + 560 = 0 \\[1em] \Rightarrow 4x^2 - 132x + 560 = 0 \\[1em] \Rightarrow 4(x^2 - 33x + 140) = 0 \\[1em] \Rightarrow x^2 - 33x + 140 = 0 \\[1em] \Rightarrow x^2 - 28x - 5x + 140 = 0 \\[1em] \Rightarrow x(x - 28) - 5(x - 28) = 0 \\[1em] \Rightarrow (x - 28)(x - 5) = 0 \\[1em] \Rightarrow x - 28 = 0 \text{ or } x - 5 = 0 \\[1em] x = 28 \text{ or } x = 5$

If x = 28 , 8 - x = -20 which will make fraction = $-\dfrac{20}{28}$ negative hence, x ≠ 28

∴ x = 5 , 8 - x = 3 ,fraction = $\dfrac{3}{5}$

Hence, the fraction is $\dfrac{3}{5}$.

A two digit number contains the Larger digit at ten's place. The product of the digits is 27 and the difference between two digits is 6 . Find the number.

Answer

Let the unit's digit be x,
so, ten's digit be = x + 6.

Number = 10(x + 6) + x = 10x + 60 + x = 11x + 60

Given, product of digits is 27

$\Rightarrow x(x + 6) = 27 \\[1em] \Rightarrow x^2 + 6x = 27 \\[1em] \Rightarrow x^2 + 6x - 27 = 0 \\[1em] \Rightarrow x^2 + 9x - 3x - 27 = 0 \\[1em] \Rightarrow x(x + 9) - 3(x + 9) = 0 \\[1em] \Rightarrow (x - 3)(x + 9) = 0 \\[1em] \Rightarrow x - 3 = 0 \text{ or } x + 9 = 0 \\[1em] \Rightarrow x = 3 \text{ or } x = -9$

When x = -9,
Number = 11x + 60
= 11(-9) + 60
= -99 + 60
= -39
In this case the ten's digit is smaller than unit's digit hence x ≠ -9

When x = 3,
Number = 11(x) + 60
= 11(3) + 60
= 33 + 60
= 93

Hence, the required number is 93 .

A two digit positive number is such that the product of its digit is 6. If 9 is added to the number , the digits interchange their place . Find the number.

Answer

Let the digit at unit's place be x .

Since, the product of digits is 6 , it's ten's digit = $\dfrac{6}{x}$

$\therefore \text{ Number } = 10 \times \dfrac{6}{x} + x \\[0.5em] = \dfrac{60}{x} + x \\[0.5em] = \dfrac{60 + x^2}{x} \\[0.5em] = \dfrac{x^2 + 60}{x} \\[0.5em]$

Given, if 9 is added to the number , the digits interchange their place

On interchanging the digits, number becomes = $10 \times x + \dfrac{6}{x}$

Acccording to given,

$\Rightarrow \dfrac{x^2 + 60}{x} + 9 = 10 \times x + \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2 + 9x + 60}{x} = 10x + \dfrac{6}{x} \\[1em] \Rightarrow \dfrac{x^2 + 9x + 60}{x} = \dfrac{10x^2 + 6}{x} \\[1em] \Rightarrow x^2 + 9x + 60 =10x^2 + 6 \\[1em] \Rightarrow 10x^2 - x^2 - 9x + 6 - 60 = 0 \\[1em] \Rightarrow 9x^2 - 9x - 54 = 0 \\[1em] \Rightarrow 9(x^2 - x - 6) = 0 \\[1em] \Rightarrow x^2 - 3x + 2x - 6 = 0 \\[1em] \Rightarrow x(x - 3) + 2(x - 3) = 0 \\[1em] \Rightarrow (x + 2)(x - 3) = 0 \\[1em] \Rightarrow x + 2 = 0 \text{ or } x - 3 = 0 \\[1em] \Rightarrow x = -2 \text{ or } x = 3$

Since the number is positive hence x ≠ -2.

If x = 3 ,

$\text{ Number } = \dfrac{x^2 + 60}{x} \\[1em] = \dfrac{3^2 + 60}{3} \\[1em] = \dfrac{69}{3} \\[1em] = 23$

Hence, the required number is 23.

A rectangle of area 105 cm² has its length equal to x cm. Write down its breadth in terms of x. Given that the perimeter is 44 cm, write down an equation in x and solve it to determine the dimensions of rectangle.

Answer

Length of rectangle = x

Since the area of rectangle = 105 cm², breadth = $\dfrac{105}{x}$ cm.

∴ Perimeter = 2(length + breadth)

$= 2(x + \dfrac{105}{x}) \\[0.5em] = 2(\dfrac{x^2 + 105}{x}) \\[0.5em]$

Given, Perimeter = 44 cm

$\Rightarrow 2(\dfrac{x^2 + 105}{x}) = 44 \\[0.5em] \Rightarrow \dfrac{x^2 + 105}{x} = 22 \\[0.5em] \Rightarrow x^2 + 105 = 22x \\[0.5em] \Rightarrow x^2 - 22x + 105 = 0 \\[0.5em] \Rightarrow x^2 - 15x - 7x + 105 = 0 \\[0.5em] \Rightarrow x(x - 15) - 7(x - 15) = 0 \\[0.5em] \Rightarrow (x - 7)(x - 15) = 0 \\[0.5em] \Rightarrow x - 7 = 0 \text{ or } x - 15 = 0 \\[0.5em] \Rightarrow x = 7 \text{ or } x = 15$

Breadth = $\dfrac{105}{x}$ cm
Equation in x for perimeter, 2(x + $\dfrac{105}{x}$) = 44
Length = 7 cm , Breadth = 15 cm

A rectangular garden 10 m by 16 m is to be surrounded by a concrete walk of uniform width. Given that the area of the walk is 120 square metres , assuming the width of the walk to be x, form an equation in x and solve it to find the value of x.

Answer

Given,

Length of rectangular garden = 10 m

Breadth of rectangular garden = 16 m

Width of walk = x

So, length of garden and walk combined = (10 + x + x) m

Breadth of garden and walk combined = (16 + x + x) m

∴ Area of garden and walk combined = Length $\times$ Breadth = (10 + 2x)(16 + 2x) m²

Given, area of walk = 120m²

Area of walk = Area of combined - Area of garden

$\Rightarrow (10 + 2x)(16 + 2x) - 10 \times 16 = 120 \\[0.5em] \Rightarrow 160 + 20x + 32x + 4x^2 - 160 = 120 \\[0.5em] \Rightarrow 4x^2 + 52x = 120 \\[0.5em] \Rightarrow 4x^2 + 52x - 120 = 0 \\[0.5em] \Rightarrow 4(x^2 + 13x - 30) = 0 \\[0.5em] \Rightarrow x^2 + 13x - 30 = 0 \\[0.5em] \Rightarrow x^2 + 15x - 2x - 30 = 0 \\[0.5em] \Rightarrow x(x + 15) - 2(x + 15) = 0 \\[0.5em] \Rightarrow (x - 2)(x + 15) = 0 \\[0.5em] \Rightarrow x - 2 \text{ or } x + 15 = 0 \\[0.5em] x = 2 \text{ or } x = -15$

Since, width cannot be negative hence x ≠ -15.

The equation in x = (10 + 2x)(16 + 2x) - 10 x 16 = 120.
Value of x = 2m.

The length of a rectangle exceeds its breadth by 5m. If the breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m². Find its dimensions.

Answer

Let breadth of rectangle be x meters

Since , length of rectangle exceeds breadth by 5 meters so, length = (x + 5) meters

Given, if breadth were doubled and the length reduced by 9m, the area of the rectangle would have increased by 140 m

∴ Area of new rectangle = Length $\times$ Breadth = (x + 5 - 9)(2x)

$\Rightarrow (x + 5 - 9)(2x) - x(x + 5) = 140 \\[0.5em] \Rightarrow (x - 4)(2x) - (x^2 + 5x) = 140 \\[0.5em] \Rightarrow 2x^2 - 8x - x^2 - 5x = 140 \\[0.5em] \Rightarrow x^2 - 13x - 140 = 0 \\[0.5em] \Rightarrow x^2 - 20x + 7x - 140 = 0 \\[0.5em] \Rightarrow x(x - 20) + 7(x - 20) = 0 \\[0.5em] \Rightarrow (x + 7)(x - 20) = 0 \\[0.5em] \Rightarrow x + 7 = 0 \text{ or } x - 20 = 0 \\[0.5em] x = -7 \text{ or } x = 20$

Since, breadth cannot be negative hence x ≠ -7

∴ x = 20 and x + 5 = 25

Length of rectangle = 25 m , Breadth of rectangle = 20 m.

The perimeter of a rectangular plot is 180 m and its area is 1800 m². Take the length of the plot as x meters. Use the perimeter 180 m to write the value of the breadth in terms of x . Use the values of length, breadth and the area to write an equation in x . Solve the equation to calculate the length and breadth of the plot.

Answer

Length of rectangular plot = x meters

Perimeter = 2(length + breadth)

$\Rightarrow 2(x + \text{Breadth}) = 180 \\[0.5em] \Rightarrow x + \text{Breadth} = 90 \\[0.5em] \Rightarrow \text{Breadth} = 90 - x$

Area of the rectangle = Length $\times$ Breadth

Given, Area of rectangle = 1800 m²

$\Rightarrow x(90 - x) = 1800 \\[0.5em] \Rightarrow 90x - x^2 = 1800 \\[0.5em] \Rightarrow x^2 - 90x + 1800 = 0 \\[0.5em] \Rightarrow x^2 - 30x - 60x + 1800 = 0 \\[0.5em] \Rightarrow x(x - 30) - 60(x - 30) = 0 \\[0.5em] \Rightarrow (x - 30)(x - 60) = 0 \\[0.5em] \Rightarrow x - 30 = 0 \text{ or } x - 60 = 0 \\[0.5em] \Rightarrow x = 30 \text{ or } x = 60$

Breadth = (90 - x) meters
Equation in x : x(90 - x) = 1800
Length of rectangle = 60 m , Breadth of rectangle = 30 m

The lengths of parallel sides of a trapezium are (x + 9) cm and (2x - 3) cm , and the distance between them is (x + 4) cm . If its area is 540 cm² , find x .

Answer

Given ,

Length of first parallel side = (x + 9) cm

Length of second parallel side = (2x - 3) cm

Distance between parallel side = (x + 4) cm

Area of trapezium = 540 cm²

Area of trapezium is given by,

$=\dfrac{1}{2} \times \text{ (sum of parallel sides) } \times \text{ (distance between them) } \\[1em] \Rightarrow \dfrac{1}{2} \times (x + 9 + 2x - 3) \times (x + 4) = 540 \\[1em] \Rightarrow \dfrac{1}{2} \times (3x + 6)(x + 4) = 540 \\[1em] \Rightarrow \dfrac{1}{2} \times (3x^2 + 12x + 6x + 24) = 540 \\[1em] \Rightarrow \dfrac{1}{2} \times (3x^2 + 18x + 24) = 540 \\[1em] \Rightarrow 3x^2 + 18x + 24 = 540 \times 2 \text{ (On cross multiplication) } \\[1em] \Rightarrow 3x^2 + 18x + 24 = 1080 \\[1em] \Rightarrow 3x^2 + 18x + 24 - 1080 = 0 \\[1em] \Rightarrow 3x^2 + 18x - 1056 = 0 \\[1em] \Rightarrow 3(x^2 + 6x - 352) = 0 \\[1em] \Rightarrow x^2 + 6x - 352 = 0 \\[1em] \Rightarrow x^2 + 22x - 16x - 352 = 0 \\[1em] \Rightarrow x(x + 22) - 16(x + 22) = 0 \\[1em] \Rightarrow (x - 16)(x + 22) = 0 \\[1em] \Rightarrow x - 16 = 0 \text{ or } x + 22 = 0 \\[1em] x = 16 \text{ or } x = -22$

If x = -22 , Length = x + 9 = -22 + 9 = -13 , Breadth = (2x - 3) = -44 - 3 = -47
Since length and breadth cannot be negative hence , x ≠ -22

∴ x = 16

The value of x is 16.

If the perimeter of a rectangular plot is 68 m and length of its diagonal is 26 m , find its area.

Answer

Taking length = l and breadth = b

Perimeter of rectangle = 2(l + b)

Length of diagonal of a rectangle = $\sqrt{l^2 + b^2}$

Given,

Perimeter = 68 m

$\Rightarrow 2(l + b) = 68 \\[0.5em] \Rightarrow l + b = 34 \\[0.5em] \Rightarrow l = 34 - b \qquad \text{ Equation (a)}$

Given,

Diagonal of a rectangle = 26 m

$\Rightarrow \sqrt{l^2 + b^2} = 26$

On squaring both sides,

$\Rightarrow l^2 + b^2 = 26^2$

Putting values of l from equation a,

$\Rightarrow (34 - b)^2 + b^2 = 676 \\[1em] \Rightarrow 1156 + b^2 - 68b + b^2 = 676 \\[1em] \Rightarrow 2b^2 - 68b + 1156 - 676 = 0 \\[1em] \Rightarrow 2b^2 - 68b + 480 = 0 \\[1em] \Rightarrow 2(b^2 - 34b + 240) = 0 \\[1em] \Rightarrow b^2 - 34b + 240 = 0 \\[1em] \Rightarrow b^2 - 24b - 10b + 240 = 0 \\[1em] \Rightarrow b(b - 24) - 10(b - 24) = 0 \\[1em] \Rightarrow (b - 24)(b - 10) = 0 \\[1em] \Rightarrow b - 24 = 0 \text{ or } b - 10 = 0 \\[1em] b = 24 \text{ or } b = 10$

∴ If b = 24 ,l = 34 - b = 34 - 24 = 10

If b = 10 , l = 34 - b = 34 - 10 = 24

Area of rectangle = Length $\times$ Breadth = 24 $\times$ 10 = 240 m²

Hence, the area of rectangle is 240 m².

If the sum of two smaller sides of a right-angled triangle is 17 cm and the perimeter is 30 cm, then find the area of the triangle.

Answer

Let one of the two smaller sides be x cm, then the other side is (17 - x) cm.

Length of hypotenuse = perimeter - sum of other two sides = 30cm - 17cm = 13cm

In right angled triangle

Perpendicular² + Base² = Hypotenuse²

∴ x² + (17 - x)² = 13²

$\Rightarrow x^2 + 289 + x^2 - 34x = 169 \\[1em] \Rightarrow 2x^2 - 34x + 289 - 169 = 0 \\[1em] \Rightarrow 2x^2 - 34x + 120 = 0 \\[1em] \Rightarrow 2(x^2 - 17x + 60) = 0 \\[1em] \Rightarrow x^2 - 17x + 60 = 0 \\[1em] \Rightarrow x^2 - 12x - 5x + 60 = 0 \\[1em] \Rightarrow x(x - 12) - 5(x - 12) = 0 \\[1em] \Rightarrow (x - 5)(x - 12) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x - 12 = 0 \\[1em] \Rightarrow x = 5 \text{ or } x = 12$

If x = 5 , 17 - x = 12 and if x = 12 , 17 - x = 5.

Hence, two small sides are 5 , 12.

Area of right angled triangle = $\dfrac{1}{2} \times \text{base} \times \text{height}$

∴ $\dfrac{1}{2} \times 5\text{cm} \times 12\text{cm} = 30 cm^2$

Hence the area of triangle is 30cm².

The hypotenuse of a grassy land in the shape of a right triangle is 1 metre more than twice the shortest side. If the third side is 7 metres more than the shortest side, find the sides of the grassy land.

Answer

Let the shortest side be x metres

So, hypotenuse = (2x + 1) metres and third side = (x + 7) metres

Hypotenuse² = Perpendicular² + Base²

∴ (2x + 1)² = x² + (x + 7)²

$\Rightarrow 4x^2 + 1 + 4x = x^2 + x^2 + 49 + 14x \\[1em] \Rightarrow 4x^2 + 1 + 4x = 2x^2 + 49 + 14x \\[1em] \Rightarrow 4x^2 - 2x^2 + 1 - 49 + 4x - 14x = 0 \\[1em] \Rightarrow 2x^2 - 48 - 10x = 0 \\[1em] \Rightarrow 2(x^2 - 24 - 5x) = 0 \\[1em] \Rightarrow x^2 - 5x - 24 = 0 \\[1em] \Rightarrow x^2 - 8x + 3x - 24 = 0 \\[1em] \Rightarrow x(x - 8) + 3(x - 8) = 0 \\[1em] \Rightarrow (x + 3)(x - 8) = 0 \\[1em] \Rightarrow x + 3 = 0 \text{ or } x - 8 = 0 \\[1em] x = -3 \text{ or } x = 8$

Since no side of a triangle can be negative hence, x ≠ -3

If x = 8 , (2x + 1) = 17 , (x + 7) = 15

Hence, the hypotenuse of the triangle is 17 metres while the shorter sides are 8 metres and 15 metres.

Mohini wishes to fit three rods together in the shape of a right triangle. If the hypotenuse is 2 cm longer than the base and 4 cm longer than the shortest side, find the lengths of the rods.

Answer

Let the length of hypotenuse be x cm,

So, longer side = (x - 2) cm and shortest side = (x - 4) cm

Hypotenuse² = Sum of the squares of other two sides

∴ x² = (x - 2)² + (x - 4)²

$\Rightarrow x^2 = x^2 + 4 - 4x + x^2 + 16 - 8x \\[1em] \Rightarrow x^2 = 2x^2 - 12x + 20 \\[1em] \Rightarrow 2x^2 - x^2 - 12x + 20 = 0 \\[1em] \Rightarrow x^2 - 12x + 20 = 0 \\[1em] \Rightarrow x^2 - 10x - 2x + 20 = 0 \\[1em] \Rightarrow x(x - 10) - 2(x - 10) = 0 \\[1em] \Rightarrow (x - 2)(x - 10) = 0 \\[1em] \Rightarrow x - 2 = 0 \text{ or } x - 10 = 0 \\[1em] x = 2 \text{ or } x = 10$

x ≠ 2 as that will make shortest side negative (x - 4) = -2.

If x = 10 , x - 2 = 8 , x - 4 = 6.

Hence, the dimensions of triangle are Hypotenuse = 10 cm , Longer side = 8 cm , Shortest side = 6 cm.

In a P.T. display, 480 students are arranged in rows and columns. If there are 4 more students in each row than the number of rows, find the number of students in each row.

Answer

Let the number of rows be x

So, the number of students in each row = x + 4

Given, total students = 480

$\Rightarrow x(x + 4) = 480 \\[1em] \Rightarrow x^2 + 4x = 480 \\[1em] \Rightarrow x^2 + 4x - 480 = 0 \\[1em] \Rightarrow x^2 + 24x - 20x - 480 = 0 \\[1em] \Rightarrow x(x + 24) - 20(x + 24) = 0 \\[1em] \Rightarrow (x - 20)(x + 24) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 24 = 0 \\[1em] x = 20 \text{ or } x = -24.$

Since number of rows cannot be negative hence, x ≠ -24.

If x = 20 , x + 4 = 24.

Hence the number of students in each row are 24.

In an auditorium , the number of rows was equal to number of seats in each row. If the number of rows is doubled and the number of seats in each row is reduced by 5, then the total number of seats is increased by 375. How many rows were there ?

Answer

Let number of rows be = $x$ = number of seats in each row

Hence, total number of seats = $x \times x = x^2$

Given, if the number of rows is doubled and the number of seats in each row is reduced by 5, then the total number of seats is increased by 375

$\therefore 2x \times (x - 5) - x^2 = 375$

$\Rightarrow 2x(x - 5) - x^2 = 375 \\[1em] \Rightarrow 2x^2 - 10x - x^2 = 375 \\[1em] \Rightarrow x^2 - 10x - 375 = 0 \\[1em] \Rightarrow x^2 - 25x + 15x - 375 = 0 \\[1em] \Rightarrow x(x - 25) + 15(x - 25) = 0 \\[1em] \Rightarrow (x - 25)(x + 15) = 0 \\[1em] \Rightarrow x - 25 = 0 \text{ or } x + 15 = 0 \\[1em] \Rightarrow x = 25 \text{ or } x = -15$

Since number of rows cannot be negative hence, x ≠ -15.

Hence the number rows were 25.

At an annual function of a school, each student gives gift to every other student. If the number of gifts is 1980, find the number of students.

Answer

Let the number of students be x

If each student gives gift to every other student so each student gives gift to (x - 1) students

So, x students gives gifts to total = x(x - 1) students.

According to given,

$\Rightarrow x(x - 1) = 1980 \\[1em] \Rightarrow x^2 - x = 1980 \\[1em] \Rightarrow x^2 - x - 1980 = 0 \\[1em] \Rightarrow x^2 - 45x + 44x - 1980 \\[1em] \Rightarrow x(x - 45) + 44(x - 45) \\[1em] \Rightarrow (x - 45)(x + 44) \\[1em] \Rightarrow x - 45 = 0 \text{ or } x + 44 = 0 \\[1em] x = 45 \text{ or } x = -44$

Since number of students cannot be negative hence, x ≠ -44.

Hence, the number of students are 45.

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

Answer

Uniform speed of bus = x km/h

Due to heavy rain the speed reduces to = (x - 10) km/h

Given, due to decrease in speed it takes two hours longer to cover the distance

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

$\therefore \dfrac{240}{x - 10} - \dfrac{240}{x} = 2 \\[1em] \Rightarrow \dfrac{240x - 240(x - 10)}{x(x - 10)} = 2 \text{ (On taking L.C.M.) } \\[1em] \Rightarrow 240x - 240x + 2400 = 2x(x - 10) \\[1em] \Rightarrow 2400 = 2x^2 - 20x \\[1em] \Rightarrow 2x^2 - 20x - 2400 = 0 \\[1em] \Rightarrow 2(x^2 - 10x - 1200) = 0 \\[1em] \Rightarrow x^2 - 10x - 1200 = 0 \\[1em] \Rightarrow x^2 - 40x + 30x - 1200 = 0 \\[1em] \Rightarrow x(x - 40) + 30(x - 40) = 0 \\[1em] \Rightarrow (x + 30)(x - 40) = 0 \\[1em] \Rightarrow x + 30 = 0 \text{ or } x - 40 = 0 \\[1em] x = -30 \text{ or } x = 40$

Since, speed cannot be negative hence, x ≠ -30.

The equation in x is $\rightarrow \dfrac{240}{x - 10} - \dfrac{240}{x} = 2$
Hence, the value of uniform speed is 40 km/h.

The speed of an express train is x km/h and the speed of an ordinary train is 12 km /h less than that of the express train. If the ordinary train takes one hour longer than the express train to cover a distance of 240 km , find the speed of the express train.

Answer

Speed of an express train is x km/h

So, the speed of ordinary train is (x - 12) km/h

Given, ordinary train takes one hour longer than express train to cover 240 km

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

$\therefore \dfrac{240}{x - 12} - \dfrac{240}{x} = 1 \\[1em] \Rightarrow \dfrac{240x - 240(x - 12)}{x(x - 12)} = 1 \\[1em] \Rightarrow 240x - 240x + 2880 = x(x - 12) \\[1em] \Rightarrow 2880 = x^2 - 12x \\[1em] \Rightarrow x^2 - 12x - 2880 = 0 \\[1em] \Rightarrow x^2 - 60x + 48x - 2880 = 0 \\[1em] \Rightarrow x(x - 60) + 48(x - 60) = 0 \\[1em] \Rightarrow (x - 60)(x + 48) = 0 \\[1em] \Rightarrow x - 60 = 0 \text{ or } x + 48 = 0 \\[1em] \Rightarrow x = 60 \text{ or } x = -48$

Since speed of train cannot be negative hence, x ≠ -48

Hence, the speed of express train is 60 km/h.

A car covers a distance of 400 km at a certain speed . Had the speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less . Find the original speed of the car.

Answer

Let the speed of car be x km/h

Given, if speed been 12 km/h more, the time taken for the journey would have been 1 hour 40 minutes less

1 hour 40 minutes = $\dfrac{100}{60}$ hours

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

$\therefore \dfrac{400}{x} - \dfrac{400}{x + 12} = \dfrac{100}{60} \\[1em] \Rightarrow 100(\dfrac{4}{x} - \dfrac{4}{x + 12}) = \dfrac{100}{60} \\[1em] \Rightarrow \dfrac{4}{x} - \dfrac{4}{x + 12} = \dfrac{1}{60} \\[1em] \Rightarrow \dfrac{4(x + 12) - 4x}{x(x + 12)} = \dfrac{1}{60} \\[1em] \Rightarrow 60(4x + 48 - 4x) = x(x + 12) \\[1em] \Rightarrow 60 \times 48 = x^2 + 12x \\[1em] \Rightarrow x^2 + 12x - 2880 = 0 \\[1em] \Rightarrow x^2 + 60x - 48x - 2880 = 0 \\[1em] \Rightarrow x(x + 60) - 48(x + 60) = 0 \\[1em] \Rightarrow (x - 48)(x + 60) = 0 \\[1em] \Rightarrow x - 48 = 0 \text{ or } x + 60 = 0 \\[1em] x = 48 \text{ or } x = -60$

Since speed of train cannot be negative hence, x ≠ -60

Hence, the speed of express train is 48 km/h.

An aeroplane covered a distance of 400 km at an average speed of x km/h. On the return journey, the speed was increased by 40 km/h . Write down an expression for the time taken for :

(i) the onward journey

(ii) the return journey

If the return journey took 30 minutes less than the onward journey , write down an equation in x and find its value.

Answer

(i) Distance covered by plane = 400 km

Average speed of plane = x km/h

Time = $\dfrac{\text{ Distance}}{\text{Speed}} = \dfrac{400}{x} \text{ hrs}$

(ii) Distance covered by plane = 400 km

Average speed of plane = (x + 40) km/h

Time = $\dfrac{\text{ Distance}}{\text{Speed}} = \dfrac{400}{x + 40} \text{ hrs}$

According to question,

$\Rightarrow \dfrac{400}{x} - \dfrac{400}{x + 40} = \dfrac{30}{60} \\[1em] \Rightarrow \dfrac{400(x + 40) - 400x}{x(x + 40)} = \dfrac{1}{2} \\[1em] \Rightarrow \dfrac{400x - 400x + 16000}{x(x + 40)} = \dfrac{1}{2} \\[1em] \Rightarrow 16000 \times 2 = x(x + 40) \\[1em] \Rightarrow 32000 = x^2 + 40x \\[1em] \Rightarrow x^2 + 40x - 32000 = 0 \\[1em] \Rightarrow x^2 + 200x - 160x - 32000 = 0 \\[1em] \Rightarrow x(x + 200) - 160(x + 200) = 0 \\[1em] \Rightarrow (x - 160)(x + 200) = 0 \\[1em] \Rightarrow x - 160 = 0 \text{ or } x + 200 = 0 \\[1em] x = 160 \text{ or } x = -200$

Since speed of aeroplane cannot be negative hence, x ≠ -200

∴ x = 160

Equation in x : $\dfrac{400}{x} - \dfrac{400}{x + 40} = \dfrac{1}{2}$
Hence, the speed of aerolane is 160 km/h.

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/h, and the train travels at a speed which is 16 km/h faster than the car. Calculate :

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

(ii) 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) Speed of car = x km/h

Distance between point A and B by road = 216 km

Time taken = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{216}{x}$ hours

(ii) Speed of train = (x + 16) km/h

Distance between point A and B by rail = 208 km

Time taken = $\dfrac{\text{Distance}}{\text{Speed}} = \dfrac{208}{x + 16}$ hours

(iii) Given,

Train takes 2 hours less than car to reach town B

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

Since speed of car cannot be negative hence x ≠ -48

∴ x = 36

Equation : $\dfrac{216}{x} - \dfrac{208}{x + 16} = 2$

(iv) Speed of train = (x + 16) = (36 + 16) = 52 km/h

Hence, speed of train is 52 km/h.

An aeroplane flying with a wind of 30 km/h takes 40 minutes less to fly 3600 km , then what it would have taken to fly against the same wind. Find the plane's speed of flying in still air.

Answer

Let the speed of plane in still air be x km/h

Speed of wind = 30 km/h

∴ Speed of plane in wind = (x + 30) km/h and Speed of plane against wind = (x - 30) km/h.

40 minutes = $\dfrac{40}{60} \text{ hours } = \dfrac{2}{3} \text{ hours }$

According to question,

$\Rightarrow \dfrac{3600}{x - 30} - \dfrac{3600}{x + 30} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3600(x + 30) - 3600(x - 30)}{(x - 30)(x + 30)} = \dfrac{2}{3} \\[1em] \Rightarrow \dfrac{3600x + 108000 - 3600x + 108000}{x^2 - 30x + 30x - 900} = \dfrac{2}{3} \\[1em] \Rightarrow 216000 \times 3 = 2(x^2 - 900) \text{ (On cross multiplying) }\\[1em] \Rightarrow 648000 = 2(x^2 - 900) \\[1em] \Rightarrow 324000 = x^2 - 900 \text{ (Dividing the complete equation by 2) } \\[1em] \Rightarrow x^2 - 900 - 324000 = 0 \\[1em] \Rightarrow x^2 - 324900 = 0 \\[1em] \Rightarrow x^2 - (570)^2 = 0 \\[1em] \Rightarrow (x - 570)(x + 570) = 0 \\[1em] \Rightarrow x - 570 = 0 \text{ or } x + 570 = 0 \\[1em] x = 570 \text{ or } x = -570 \\[1em]$

Since speed of aeroplane cannot be negative hence, x ≠ -570.

The speed of aeroplane in still air is 570 km/h.

A school bus transported an excursion party to a picnic spot 150 km away. While returning , it was raining and the bus had to reduce its speed by 5 km/h , and it took one hour longer to make the return trip. Find the time taken to return.

Answer

Let the speed of bus while reaching picnic spot be x km/h

Since, the speed of bus decrease by 5 km/h due to rain hence, speed of bus in rain = (x - 5) km/h

∴ Time taken to reach picnic spot = $\dfrac{150}{x}$ and Time taken to reach back to school = $\dfrac{150}{x - 5}$

According to given,

$\Rightarrow \dfrac{150}{x - 5} - \dfrac{150}{x} = 1 \\[1em] \Rightarrow \dfrac{150x - 150(x - 5)}{x(x - 5)} = 1 \\[1em] \Rightarrow \dfrac{150x - 150x + 750}{x^2 - 5x} = 1 \\[1em] \Rightarrow 750 = x^2 - 5x \text{ (On cross multiplying) } \\[1em] \Rightarrow x^2 - 5x - 750 = 0 \\[1em] \Rightarrow x^2 - 30x + 25x - 750 = 0 \\[1em] \Rightarrow x(x - 30) + 25(x - 30) = 0 \\[1em] \Rightarrow (x + 25)(x - 30) = 0 \\[1em] \Rightarrow x + 25 = 0 \text{ or } x - 30 = 0 \\[1em] x = -25 \text{ or } x = 30$

Since speed of bus cannot be negative hence, x ≠ -25.

∴ x = 30

If x = 30 , x - 5 = 25.

Time taken to return = $\dfrac{150}{x - 5}$ hours = 6 hours.

Hence, time taken on return trip is 6 hours.

A boat can cover 10 km up the stream and 5 km down the stream in 6 hours . If the speed of the stream is 1.5 km/h, find the speed of the boat in still water.

Answer

Let the speed of boat in still water be x km/h

Speed of stream = 1.5 km/h

∴ Speed of boat upstream = (x - 1.5) km/h and Speed of boat downstream = (x + 1.5) km/h

According to given,

$\Rightarrow \dfrac{10}{x - 1.5} + \dfrac{5}{x + 1.5} = 6 \\[1em] \Rightarrow \dfrac{10(x + 1.5) + 5(x - 1.5)}{(x + 1.5)(x - 1.5)} = 6 \\[1em] \Rightarrow \dfrac{10x + 15 + 5x - 7.5}{x^2 - 1.5x + 1.5x - 2.25} = 6 \\[1em] \Rightarrow \dfrac{15x + 7.5}{x^2 - 2.25} = 6 \\[1em] \Rightarrow 15x + 7.5 = 6(x^2 - 2.25) \\[1em] \Rightarrow 15x + 7.5 = 6x^2 - 13.50 \\[1em] \Rightarrow 6x^2 - 13.5 - 7.5 - 15x = 0 \\[1em] \Rightarrow 6x^2 - 15x - 21 = 0 \\[1em] \Rightarrow 3(2x^2 - 5x - 7) = 0 \\[1em] \Rightarrow 2x^2 - 5x - 7 = 0 \\[1em] \Rightarrow 2x^2 - 7x + 2x - 7 = 0 \\[1em] \Rightarrow x(2x - 7) + 1(2x - 7) = 0 \\[1em] \Rightarrow (x + 1)(2x - 7) = 0 \\[1em] \Rightarrow x = -1 \text{ or } 2x - 7 = 0 \\[1em] \Rightarrow x + 1 = 0 \text{ or } x = \dfrac{7}{2} \\[1em] \Rightarrow x = -1 \text{ or } x = 3.5$

Since speed of bus cannot be negative hence, x ≠ -1

Hence speed of boat in still water is 3.5 km/h.

Two pipes running together can fill a tank in $11\dfrac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank, find the time in which each pipe would fill the tank.

Answer

The tank is filled by the two pipes together in $11\dfrac{1}{9}$ minutes i.e. in $\dfrac{100}{9}$ minutes,

∴ the part of tank filled in one minute = $\dfrac{9}{100}$

Let the time taken by two pipes to fill tank separately be x minutes and (x + 5) minutes.

∴ the part of tank filled by the first pipe in one minute = $\dfrac{1}{x}$ and

the part of tank filled by the second pipe in one minute = $\dfrac{1}{x + 5}$

According to the question

$\dfrac{1}{x} + \dfrac{1}{x + 5} = \dfrac{9}{100} \\[1em] \Rightarrow \dfrac{x + 5 + x}{x(x + 5)} = \dfrac{9}{100} \\[1em] \Rightarrow \dfrac{2x + 5}{x^2 + 5x} = \dfrac{9}{100} \\[1em] \Rightarrow 100(2x + 5) = 9(x^2 + 5x) \\[1em] \Rightarrow 200x + 500 = 9x^2 + 45x \\[1em] \Rightarrow 9x^2 + 45x - 200x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 155x - 500 = 0 \\[1em] \Rightarrow 9x^2 - 180x + 25x - 500 = 0 \\[1em] \Rightarrow 9x(x - 20) + 25(x - 20) = 0 \\[1em] \Rightarrow (9x + 25)(x - 20) = 0 \\[1em] \Rightarrow x = -\dfrac{25}{9} \text{ or } x = 20$

Since, time cannot be negative hence, x ≠ $-\dfrac{25}{9}$

∴ x = 20 , x + 5 = 25.

Time taken by each pipe is 20 minutes and 25 minutes.

₹480 is divided equally among 'x' children. If the number of children were 20 more, then each would have got ₹12 less . Find 'x'.

Answer

Number of children = x

Money recieved by each = $\dfrac{480}{x}$

If number of children were 20 more then money received by each child = $\dfrac{480}{x + 20}$

Given, if there are 20 more children then each will get ₹12 less

$\therefore \dfrac{480}{x} - \dfrac{480}{x + 20} = 12 \\[1em] \Rightarrow \dfrac{40}{x} - \dfrac{40}{x + 20} = 1 \text{ (Dividing the equation by 12) } \\[1em] \Rightarrow \dfrac{40(x + 20) - 40x}{x(x + 20)} = 1 \\[1em] \Rightarrow \dfrac{40x + 800 - 40x}{x^2 + 20x} = 1 \\[1em] \Rightarrow 800 = x^2 + 20x \\[1em] \Rightarrow x^2 + 20x - 800 = 0 \\[1em] \Rightarrow x^2 + 40x - 20x - 800 = 0 \\[1em] \Rightarrow x(x + 40) - 20(x + 40) = 0 \\[1em] \Rightarrow (x - 20)(x + 40) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 40 = 0 \\[1em] x = 20 \text{ or } x = -40$