Solving Problems (Based on Quadratic Equations)

Let two natural numbers be x and y.

Given,

Sum = 5

⇒ x + y = 5

⇒ x = 5 - y ........(1)

Sum of reciprocals = $\frac{5}{6}$

⇒ $\frac{1}{x} + \frac{1}{y} = \frac{5}{6}$ ...........(2)

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

$$\begin{aligned}\Rightarrow \frac{1}{5 - y} + \frac{1}{y} = \frac{5}{6} \\[1em] \Rightarrow \frac{y + 5 - y}{y(5 - y)} = \frac{5}{6} \\[1em] \Rightarrow \frac{5}{5y - y^2} = \frac{5}{6} \\[1em] \Rightarrow 5y - y^2 = \frac{6 \times 5}{5} \\[1em] \Rightarrow 5y - y^2 = 6 \\[1em] \Rightarrow y^2 - 5y + 6 = 0 \\[1em] \Rightarrow y^2 - 2y - 3y + 6 = 0 \\[1em] \Rightarrow y(y - 2) - 3(y - 2) = 0 \\[1em] \Rightarrow (y - 3)(y - 2) = 0 \\[1em] \Rightarrow y - 3 = 0 \text{ or } y - 2 = 0 \\[1em] \Rightarrow y = 3 \text{ or } y = 2.\end{aligned}$$

If y = 3,

x = 5 - y = 5 - 3 = 2.

If y = 2,

x = 5 - y = 5 - 2 = 3.

∴ Numbers are 2 and 3.

Hence, Option 3 is the correct option.

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

Given,

Product of two consecutive even whole numbers = 24.

∴ x(x + 2) = 24

⇒ x² + 2x = 24

⇒ x² + 2x - 24 = 0

⇒ x² + 6x - 4x - 24 = 0

⇒ x(x + 6) - 4(x + 6) = 0

⇒ (x - 4)(x + 6) = 0

⇒ x - 4 = 0 or x + 6 = 0

⇒ x = 4 or x = -6.

Since, x is a whole number so x cannot be equal to -6.

x = 4 and x + 2 = 6.

Hence, Option 4 is the correct option.

Let two consecutive integers be x and x + 1.

Given,

Sum of squares of two consecutive integers = 41.

⇒ x² + (x + 1)² = 41

⇒ x² + x² + 1 + 2x = 41

⇒ 2x² + 2x = 41 - 1

⇒ 2x² + 2x = 40

⇒ 2(x² + x) = 40

⇒ x² + x = $\frac{40}{2}$

⇒ x² + x = 20

⇒ x² + x - 20 = 0

⇒ x² + 5x - 4x - 20 = 0

⇒ x(x + 5) - 4(x + 5) = 0

⇒ (x - 4)(x + 5) = 0

⇒ x - 4 = 0 or x + 5 = 0

⇒ x = 4 or x = -5

If x = 4,

⇒ x + 1 = 4 + 1 = 5.

If x = -5,

⇒ x + 1 = -5 + 1 = -4.

Numbers = 4 and 5 or -4 and -5.

Hence, Option 2 is the correct option.

Let number be x.

Given,

Sum of a number and its reciprocal is 5.2

$$\begin{aligned}\Rightarrow x + \frac{1}{x} = 5.2 \\[1em] \Rightarrow \frac{x^2 + 1}{x} = \frac{52}{10} \\[1em] \Rightarrow 10(x^2 + 1) = 52x \\[1em] \Rightarrow 10x^2 + 10 = 52x \\[1em] \Rightarrow 10x^2 - 52x + 10 = 0 \\[1em] \Rightarrow 10x^2 - 50x - 2x + 10 = 0 \\[1em] \Rightarrow 10x(x - 5) - 2(x - 5) = 0 \\[1em] \Rightarrow (10x - 2)(x - 5) = 0 \\[1em] \Rightarrow 10x - 2 = 0 \text{ or } x - 5 = 0 \\[1em] \Rightarrow 10x = 2 \text{ or } x = 5 \\[1em] \Rightarrow x = \frac{2}{10} \text{ or } x = 5 \\[1em] \Rightarrow x = \frac{1}{5} \text{ or } x = 5.\end{aligned}$$

Hence, Option 1 is the correct option.

Let two integers be x and x - 2.

Give,

Sum of squares = 52

⇒ x² + (x - 2)² = 52

⇒ x² + x² + 4 - 4x = 52

⇒ 2x² - 4x + 4 - 52 = 0

⇒ 2x² - 4x - 48 = 0

⇒ 2(x² - 2x - 24) = 0

⇒ x² - 2x - 24 = 0

⇒ x² - 6x + 4x - 24 = 0

⇒ x(x - 6) + 4(x - 6) = 0

⇒ (x + 4)(x - 6) = 0

⇒ x + 4 = 0 or x - 6 = 0

⇒ x = -4 or x = 6.

If x = -4,

x - 2 = -4 - 2 = -6.

If x = 6,

x - 2 = 6 - 2 = 4.

Numbers = -4 and -6 or 6 and 4.

Hence, Option 4 is the correct option.

Let numbers be x and (15 - x).

According to the question,

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{15 - x} = \frac{3}{10} \\[1em] \Rightarrow \frac{15 - x + x}{x(15 - x)} = \frac{3}{10} \\[1em] \Rightarrow \frac{15}{x(15 - x)} = \frac{3}{10} \\[1em] \Rightarrow 150 = 3x(15 - x) \\[1em] \Rightarrow 150 = 45x - 3x^2 \\[1em] \Rightarrow 3x^2 - 45x + 150 = 0 \\[1em] \Rightarrow 3(x^2 - 15x + 50) = 0 \\[1em] \Rightarrow x^2 - 10x - 5x + 50 = 0 \\[1em] \Rightarrow x(x - 10) - 5(x - 10) = 0 \\[1em] \Rightarrow (x - 5)(x - 10) = 0 \\[1em] \Rightarrow x - 5 = 0 \text{ or } x - 10 = 0 \\[1em] \Rightarrow x = 5 \text{ or } x = 10.\end{aligned}$$

Hence, numbers = 5, 10.

Let larger number be x and smaller number be y,

According to first part,

x² + y² = 208 ........(i)

x² = 18y ........(ii)

Substituting value of x² from (ii) in (i) we get,

⇒ 18y + y² = 208

⇒ y² + 18y - 208 = 0

⇒ y² + 26y - 8y - 208 = 0

⇒ y(y + 26) - 8(y + 26) = 0

⇒ (y - 8)(y + 26) = 0

⇒ y - 8 = 0 or y + 26 = 0

⇒ y = 8 or y = -26.

Since, numbers are positive integers,

∴ y ≠ -26.

Substituting value of y = 8 in (ii),

⇒ x² = 18(8) = 144

⇒ x = $\sqrt{144} = \pm 12$.

Since, numbers are positive integers,

∴ x ≠ -12.

Hence, numbers are 12 and 8.

Let two consecutive positive odd numbers be x and (x + 2).

⇒ (x)² + (x + 2)² = 74

⇒ x² + x² + 4 + 4x = 74

⇒ 2x² + 4x + 4 - 74 = 0

⇒ 2x² + 4x - 70 = 0

⇒ 2(x² + 2x - 35) = 0

⇒ x² + 2x - 35 = 0

⇒ x² + 7x - 5x - 35 = 0

⇒ x(x + 7) - 5(x + 7) = 0

⇒ (x + 7)(x - 5) = 0

⇒ (x + 7) = 0 or (x - 5) = 0

⇒ x = -7 or x = 5.

Since, numbers are positive and even,

∴ x ≠ -7.

∴ x = 5 and (x + 2) = 7.

Hence, numbers are 5, 7.

Let two parts be x and (20 - x).

According to question,

3x² - (20 - x) = 10

3x² + x - 20 = 10

3x² + x - 30 = 0

3x² + 10x - 9x - 30 = 0

x(3x + 10) - 3(3x + 10) = 0

(x - 3)(3x + 10) = 0

x - 3 = 0 or (3x + 10) = 0

x = 3 or x = $-\frac{10}{3}$.

20 - x = 20 - 3 = 17.

Hence, numbers are 3 and 17.

Let three consecutive numbers be (x - 1), x and (x + 1).

According to question,

⇒ x² - [(x + 1)² - (x - 1)²] = 60

⇒ x² - [x² + 1 + 2x - (x² + 1 - 2x)] = 60

⇒ x² - [x² + 1 + 2x - x² - 1 + 2x] = 60

⇒ x² - 4x = 60

⇒ x² - 4x - 60 = 0

⇒ x² - 10x + 6x - 60 = 0

⇒ x(x - 10) + 6(x + 10) = 0

⇒ (x + 6)(x - 10) = 0

⇒ x + 6 = 0 or x - 10 = 0

⇒ x = -6 or x = 10.

Since, numbers are natural numbers,

∴ x = 10.

∴ x - 1 = 9 and x + 1 = 11.

Hence, the numbers are 9, 10, 11 and quadratic equation = x² - 4x - 60 = 0.

Let the numbers be (p - 1), p and (p + 1).

According to question,

⇒ 3(p + 1)² = (p - 1)² + p² + 67

⇒ 3(p² + 1 + 2p) = (p² + 1 - 2p) + p² + 67

⇒ 3p² + 3 + 6p = 2p² - 2p + 68

⇒ 3p² - 2p² + 6p + 2p + 3 - 68 = 0

⇒ p² + 8p - 65 = 0

⇒ p² + 13p - 5p - 65 = 0

⇒ p(p + 13) - 5(p + 13) = 0

⇒ (p - 5)(p + 13) = 0

⇒ p - 5 = 0 or p + 13 = 0

⇒ p = 5 or p = -13

Since, numbers are natural numbers,

∴ p ≠ -13.

∴ p - 1 = 4 and p + 1 = 6.

Hence, p = 5.

Let smaller part be x.

According to second part of question,

Larger part = $\sqrt{8x}$

According to first part of question,

x² + $(\sqrt{8x})^2$ = 20

x² + 8x = 20

x² + 8x - 20 = 0

x² + 10x - 2x - 20 = 0

x(x + 10) - 2(x + 10) = 0

(x - 2)(x + 10) = 0

x - 2 = 0 or x + 10 = 0

x = 2 or x = -10.

Since number is positive,

∴ x ≠ -10

Larger part = $\sqrt{8x} = \sqrt{8(2)} = \sqrt{16} = 4.$

Number = Smaller part + Larger part = 2 + 4 = 6.

Hence, number = 6.

It is given that the difference between two natural numbers = 5.

Let one number be x. So, the other number = x + 5

And, the difference of their reciprocals is $\frac{1}{10}$.

$$\begin{aligned}\Rightarrow \frac{1}{x} - \frac{1}{x + 5} = \frac{1}{10}\\[1em] \Rightarrow \frac{x + 5}{x(x + 5)} - \frac{x}{x(x + 5)} = \frac{1}{10}\\[1em] \Rightarrow \frac{x + 5 - x}{x(x + 5)} = \frac{1}{10}\\[1em] \Rightarrow \frac{5}{x(x + 5)} = \frac{1}{10}\\[1em] \Rightarrow 5 \times 10 = x(x + 5)\\[1em] \Rightarrow 50 = x^2 + 5x\\[1em] \Rightarrow x^2 + 5x - 50 = 0\\[1em] \Rightarrow x^2 + 10x - 5x - 50 = 0\\[1em] \Rightarrow x(x + 10) - 5(x + 10) = 0\\[1em] \Rightarrow (x + 10)(x - 5) = 0\\[1em] \Rightarrow (x + 10) = 0 \text{ or }(x - 5) = 0\\[1em] \Rightarrow x = -10 \text{ or }x = 5\end{aligned}$$

It is given that numbers are natural numbers. So, number cannot be -10.

∴ x = 5

Other number = 5 + 5 = 10

Hence, the two numbers are 5 and 10.

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

It is given that the sum of the squares of the numbers = 514.

⇒ x² + (x + 2)² = 514

⇒ x² + x² + 2² + 4x = 514

⇒ 2x² + 4 + 4x - 514 = 0

⇒ 2x² + 4x - 510 = 0

⇒ 2(x² + 2x - 255) = 0

⇒ x² + 2x - 255 = 0

⇒ x² + 17x - 15x - 255 = 0

⇒ x(x + 17) - 15(x + 17) = 0

⇒ (x - 15)(x + 17) = 0

⇒ (x - 15) = 0 or (x + 17) = 0

⇒ x = 15 or x = -17

As x is a natural number, x cannot be equal to -17. So, x = 15.

The other odd natural number = 15 + 2 = 17

Hence, the consecutive odd natural numbers are 15 and 17.

Let the multiplies of 7 be 7x and 7(x + 1).

It is given that the sum of the squares of two consecutive multiplies of 7 is 637.

⇒ (7x)² + [7(x + 1)]² = 637

⇒ 49x² + [7x + 7]² = 637

⇒ 49x² + 49x² + 7² + 98x - 637 = 0

⇒ 98x² + 49 + 98x - 637 = 0

⇒ 98x² + 98x - 588 = 0

⇒ x² + x - 6 = 0

⇒ x² + 3x - 2x - 6 = 0

⇒ x(x + 3) - 2(x + 3) = 0

⇒ (x + 3)(x - 2) = 0

⇒ (x + 3) = 0 or (x - 2) = 0

⇒ x = -3 or x = 2

It is given that x is bigger positive number.

So, the multiplies of 7 is 7 x 2 = 14 and 7(2 + 1) = 7 x 3 = 21

Hence, the smaller angle is 14.

It is given that difference between two positive numbers = 5.

Let the one number be x and other number be 5 + x.

Three times the square of the larger number exceeds twice the square of smaller number by 334.

⇒ 3 $\times$ (5 + x)² - 2 $\times$ x² = 334

⇒ 3 $\times$ (25 + x² + 10x) - 2 $\times$ x² = 334

⇒ 75 + 3x² + 30x - 2x² - 334 = 0

⇒ x² + 30x - 259 = 0

⇒ x² + 37x - 7x - 259 = 0

⇒ x(x + 37) - 7(x + 37) = 0

⇒ (x + 37)(x - 7) = 0

⇒ (x + 37) = 0 or (x - 7) = 0

⇒ x = -37 or x = 7

Larger number = 5 + x = 5 + 7 = 12

Hence, the larger number = 12.

Let the three consecutive positive odd integers be : x, x + 2, x + 4

Given,

The sum of squares of three consecutive positive odd integers is 155.

⇒ x² + (x + 2)² + (x + 4)² = 155

⇒ x² + x² + 4x + 4 + x² + 8x + 16 = 155

⇒ 3x² + 12x + 20 = 155

⇒ 3x² + 12x + 20 - 155 = 0

⇒ 3x² + 12x - 135 = 0

⇒ 3(x² + 4x - 45) = 0

⇒ x² + 4x - 45 = 0

⇒ x² + 9x - 5x - 45 = 0

⇒ x(x + 9) - 5(x + 9) = 0

⇒ (x - 5)(x + 9) = 0

⇒ (x - 5) = 0 or (x + 9) = 0

⇒ x = 5 or x = -9.

Since the integers are positive, x = 5

The three consecutive positive odd integers,

x = 5,

x + 2 = 5 + 2 = 7,

x + 4 = 5 + 4 = 9.

Hence, three consecutive positive odd integers are 5, 7, 9.

Let the three consecutive positive even integers be: x, x + 2, x + 4.

Given,

Three times the square of least integer plus seven times the square of middle integer minus five time the square of the third integer is 328

⇒ 3x² + 7(x + 2)² - 5(x + 4)² = 328

⇒ 3x² + 7(x² + 4x + 4) - 5(x² + 8x + 16) = 328

⇒ 3x² + 7x² + 28x + 28 - 5x² - 40x - 80 = 328

⇒ 3x² + 7x² - 5x² + 28x - 40x - 80 + 28 = 328

⇒ 5x² - 12x - 52 = 328

⇒ 5x² - 12x - 52 - 328 = 0

⇒ 5x² - 12x - 380 = 0

⇒ 5x² - 50x + 38x - 380 = 0

⇒ 5x(x - 10) + 38(x - 10) = 0

⇒ (5x + 38)(x - 10) = 0

⇒ (5x + 38) = 0 or (x - 10) = 0

⇒ x = $-\frac{38}{5}$ or x = 10

Since integers must be positive even numbers, x = 10.

Three consecutive positive even integers are :

x = 10

x + 2 = 10 + 2 = 12

x + 4 = 10 + 4 = 14

Hence, three consecutive positive even integers are 10, 12, 14.

$\frac{5}{3}$ or $\frac{3}{5}$

Reason

It is given that the sum of the numerator and the denominator is 8.

Let the numerator of the fraction be x. So, denominator of the fraction = 8 - x

And, the product of numerator and denominator is 15.

⇒ x(8 - x) = 15

⇒ 8x - x² = 15

⇒ 8x - x² - 15 = 0

⇒ x² - 8x + 15 = 0

⇒ x² - 5x - 3x + 15 = 0

⇒ x(x - 5) - 3(x - 5) = 0

⇒ (x - 5)(x - 3) = 0

⇒ (x - 5) = 0 or (x - 3) = 0

⇒ x = 5 or x = 3

When x = 5, denominator = 8 - 5 = 3

When x = 3, denominator = 8 - 3 = 5

So, the fraction = $\frac{5}{3} \text{ or } \frac{3}{5}$.

Hence, option 2 is the correct option.

45

Reason

Let tens digit be x and units digit be y.

Given,

x + y = 9
xy = 20

From x + y = 9, we have y = 9 - x

Substituting in xy = 20, we get,

⇒ x(9 - x) = 20

⇒ 9x - x² = 20

⇒ 9x - x² - 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

When x = 4, y = 9 - 4 = 5

When x = 5, y = 9 - 5 = 4

The problem states "the unit digit is greater than the tens digit", so we need y > x.

∴ x = 4 and y = 5

∴ The number is 45

Hence, option 1 is the correct option.

6 and 4

Reason

It is given that two whole numbers are in ratio 3:2.

Let the numbers be 3x and 2x.

The sum of their square = 52.

⇒ (3x)² + (2x)² = 52

⇒ 9x² + 4x² = 52

⇒ 13x² = 52

⇒ x² = $\frac{52}{13}$

⇒ x² = 4

⇒ x = $\sqrt{4}$

⇒ x = $\pm$ 2

So, the numbers = 3x = 3 x 2 or 3 x (-2) = 6 or -6

2x = 2 x 2 or 2 x (-2) = 4 or -4

Since, the given numbers are whole numbers. So, the numbers cannot be -6 and -4.

Hence, option 2 is the correct option.

45

Reason

Let the two digit number be 10x + y.

It is given that two digit number is 5 times the sum of its digits

⇒ 10x + y = 5(x + y)

⇒ 10x + y = 5x + 5y

⇒ 10x - 5x = 5y - y

⇒ 5x = 4y

⇒ y = $\frac{5}{4}$x

As x and y are digits, so

x ∈ {1, 2, 3, 4, 5, 6, 7, 8, 9}

and

y ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

Since y must be a single digit integer, x must be a multiple of 4. The only possible values for x from the above set are x = 4 and x = 8.

For x = 4, y = $\frac{5}{4} \times$ 4 = 5

For x = 8, y = $\frac{5}{4} \times$ 8 = 10 which is not a valid digit.

∴ Valid digits are x = 4 and y = 5.

∴ The number = 10 × 4 + 5 = 45

Hence, option 3 is the correct option.

8, 6 and 4

Reason

It is given that three positive numbers are in ratio 4 : 3 : 2.

Let the numbers be 4x, 3x and 2x.

The difference between square of the largest and the smallest number = 48.

⇒ (4x)² - (2x)² = 48

⇒ 16x² - 4x² = 48

⇒ 12x² = 48

⇒ x² = $\frac{48}{12}$

⇒ x² = 4

⇒ x = $\sqrt{4}$

⇒ x = $\pm$ 2

So, the numbers are 4x = 4 x 2 or 4 x (-2) = 8 or -8

3x = 3 x 2 or 3 x (-2) = 6 or -6

2x = 2 x 2 or 2 x (-2) = 4 or -4

The numbers are positive numbers. So, the numbers cannot be -8, -6 and -4.

∴ The numbers are 8, 6 and 4

Hence, option 4 is the correct option.

Let the fraction be $\frac{a}{b}$.

It is given that the numerator of a fraction is 3 less than its denominator.

⇒ a = b - 3

And, one is added to the denominator, the fraction is decreased by $\frac{1}{15}$.

$$\begin{aligned}\Rightarrow \frac{a}{b + 1} = \frac{a}{b} - \frac{1}{15}\\[1em] \Rightarrow \frac{b - 3}{b + 1} = \frac{b - 3}{b} - \frac{1}{15}\\[1em] \Rightarrow \frac{b - 3}{b + 1} - \frac{b - 3}{b} = - \frac{1}{15}\\[1em] \Rightarrow \frac{b(b - 3) - (b + 1)(b - 3)}{b(b + 1)} = - \frac{1}{15}\\[1em] \Rightarrow \frac{(b^2 - 3b) - (b^2 - 3b + b - 3)}{b(b + 1)} = - \frac{1}{15}\\[1em] \Rightarrow \frac{b^2 - 3b - b^2 + 3b - b + 3}{b(b + 1)} = - \frac{1}{15}\\[1em] \Rightarrow \frac{ - b + 3}{b^2 + b} = - \frac{1}{15}\\[1em] \Rightarrow 15(-b + 3) = - 1(b^2 + b)\\[1em] \Rightarrow - 15b + 45 = - b^2 - b\\[1em] \Rightarrow - 15b + 45 + b^2 + b = 0\\[1em] \Rightarrow b^2 - 14b + 45 = 0\\[1em] \Rightarrow b^2 - 9b - 5b + 45 = 0\\[1em] \Rightarrow b(b - 9) - 5(b - 9) = 0\\[1em] \Rightarrow (b - 9)(b - 5) = 0\\[1em] \Rightarrow (b - 9) = 0 \text{ or }(b - 5) = 0\\[1em] \Rightarrow b = 9 \text{ or }b = 5\\[1em]\end{aligned}$$

If b = 9, then a = 9 - 3 = 6

And, if b = 5, then a = 5 - 3 = 2

So, the fraction = $\frac{6}{9} = \frac{2}{3} \text{ or }\frac{2}{5}$

Hence, the fraction = $\frac{2}{5}$.

Let the fraction be $\frac{a}{b}$.

It is given that the denominator of a fraction is 3 more than its numerator.

⇒ b = a + 3

And, the sum of the fraction and its reciprocal is $2\frac{9}{10}$.

$$\begin{aligned}\Rightarrow \frac{a}{b} + \frac{b}{a} = 2\frac{9}{10}\\[1em] \Rightarrow \frac{a}{a + 3} + \frac{a + 3}{a} = \frac{29}{10}\\[1em] \Rightarrow \frac{(a \times a) + [(a + 3)(a + 3)] }{(a + 3)a} = \frac{29}{10}\\[1em] \Rightarrow \frac{a^2 + (a + 3)^2}{(a + 3)a} = \frac{29}{10}\\[1em] \Rightarrow \frac{a^2 + (9 + 6a + a^2)}{(a + 3)a} = \frac{29}{10}\\[1em] \Rightarrow \frac{a^2 + 9 + 6a + a^2}{3a + a^2} = \frac{29}{10}\\[1em] \Rightarrow 10(a^2 + 9 + 6a + a^2) = 29(3a + a^2)\\[1em] \Rightarrow 10(2a^2 + 9 + 6a) = 29(3a + a^2)\\[1em] \Rightarrow 20a^2 + 90 + 60a = 87a + 29a^2\\[1em] \Rightarrow 20a^2 + 90 + 60a - 87a - 29a^2 = 0\\[1em] \Rightarrow -9a^2 - 27a + 90 = 0\\[1em] \Rightarrow a^2 + 3a - 10 = 0\\[1em] \Rightarrow a^2 + 5a - 2a - 10 = 0\\[1em] \Rightarrow a(a + 5) - 2(a + 5) = 0\\[1em] \Rightarrow (a + 5)(a - 2) = 0\\[1em] \Rightarrow (a + 5) = 0 \text{ or }(a - 2) = 0\\[1em] \Rightarrow a = -5 \text{ or } a = 2\\[1em]\end{aligned}$$

If a = -5, then b = 3 + (-5) = -2

And, if a = 2, then b = 3 + 2 = 5

So, the fraction = $\frac{-5}{-2} = \frac{5}{2} \text{ or }\frac{2}{5}$

Since, denominator is 3 more than the numerator,

∴ The fraction is $\frac{2}{5}$

Hence, the fraction = $\frac{2}{5}$.

Let unit's digit be x and ten's digit be y.

According to question,

⇒ xy = 24 ........(i)

⇒ x = 2y + 2 ........(ii)

Substituting value of x from (ii) in (i) we get,

⇒ (2y + 2)y = 24

⇒ 2y² + 2y = 24

⇒ y² + y = 12

⇒ y² + y - 12 = 0

⇒ y² + 4y - 3y - 12 = 0

⇒ y(y + 4) - 3(y + 4) = 0

⇒ (y - 3)(y + 4) = 0

⇒ y - 3 = 0 or y + 4 = 0

⇒ y = 3 or y = -4.

Since digit at ten's place cannot be negative

∴ y ≠ -4.

⇒ x = 2y + 2 = 2(3) + 2 = 8

Number = 10(y) + x = 10(3) + 8 = 38.

Hence, number = 38.

Let tens digit be x and units digit be y.

Given,

$$\begin{aligned}xy = 18 \\[1em] \Rightarrow y = \frac{18}{x}\end{aligned}$$

The number = 10x + y = $10 \times x + \frac{18}{x} = \frac{10x^2 + 18}{x}$

Reversed number = 10y + x = $10 \times \frac{18}{x} + x = \frac{180 + x^2}{x}$

Given,
if 63 is subtracted from the number, the digits are reserved.

$$\begin{aligned}\Rightarrow \frac{10x^2 + 18}{x} - 63 = \frac{180 + x^2}{x}\\[1em] \Rightarrow \frac{10x^2 + 18 - 63x}{x} = \frac{180 + x^2}{x}\\[1em] \Rightarrow 10x^2 + 18 - 63x = 180 + x^2\\[1em] \Rightarrow 10x^2 + 18 - 63x - 180 - x^2 = 0\\[1em] \Rightarrow 9x^2 - 63x - 162 = 0\\[1em] \Rightarrow x^2 - 7x - 18 = 0\\[1em] \Rightarrow x^2 - 9x + 2x - 18 = 0\\[1em] \Rightarrow x(x - 9) + 2(x - 9) = 0\\[1em] \Rightarrow (x - 9)(x + 2) = 0\\[1em] \Rightarrow (x - 9) = 0 \text{ or } (x + 2) = 0\\[1em] \Rightarrow x = 9 \text{ or } x = - 2\end{aligned}$$

As the digit of a number cannot be negative. So x = 9.

y = $\frac{18}{x} = \frac{18}{9}$ = 2

The number = 10x + y = 10 $\times$ 9 + 2 = 90 + 2 = 92

Hence, the number = 92.

It is given that the ratio between two positive numbers is $\frac{1}{5}:\frac{1}{7}$.

Let the two numbers be $\frac{1}{5}x$ and $\frac{1}{7}x$.

The sum of the squares of the numbers = 666.

$$\begin{aligned}\Rightarrow \Big(\frac{1}{5}x\Big)^2 + \Big(\frac{1}{7}x\Big)^2 = 666\\[1em] \Rightarrow \frac{1}{25}x^2 + \frac{1}{49}x^2 = 666\\[1em] \Rightarrow \frac{1 \times 49}{25 \times 49}x^2 + \frac{1 \times 25}{49 \times 25}x^2 = 666\\[1em] \Rightarrow \frac{49}{1,225}x^2 + \frac{25}{1,225}x^2 = 666\\[1em] \Rightarrow \frac{74}{1,225}x^2 = 666\\[1em] \Rightarrow x^2 = \frac{666 \times 1,225}{74}\\[1em] \Rightarrow x^2 = \frac{8,15,850}{74}\\[1em] \Rightarrow x^2 = 11,025\\[1em] \Rightarrow x = \sqrt{11,025}\\[1em] \Rightarrow x = 105\end{aligned}$$

So, the numbers = $\frac{1}{5}x = \frac{1}{5} \times 105$ = 21 and $\frac{1}{7}x = \frac{1}{7} \times 105$ = 15

Hence, the two numbers = 21 and 15.

It is given that the ratio between three positive numbers = $\frac{1}{4} : \frac{1}{3} : \frac{1}{2}$

Let the three number be $\frac{1}{4}x, \frac{1}{3}x \text{ and } \frac{1}{2}x$.

If the square of the middle number is subtracted from the sum of the squares of the other, the result is 725.

$$\begin{aligned}\Rightarrow \Big[\Big(\frac{1}{4}x\Big)^2 + \Big(\frac{1}{2}x\Big)^2\Big] - \Big(\frac{1}{3}x\Big)^2 = 725\\[1em] \Rightarrow \Big[\frac{1}{16}x^2 + \frac{1}{4}x^2\Big] - \frac{1}{9}x^2 = 725\\[1em] \Rightarrow \Big[\frac{1}{16}x^2 + \frac{1 \times 4}{4 \times 4}x^2\Big] - \frac{1}{9}x^2 = 725\\[1em] \Rightarrow \Big[\frac{1}{16}x^2 + \frac{4}{16}x^2\Big] - \frac{1}{9}x^2 = 725\\[1em] \Rightarrow \frac{5}{16}x^2 - \frac{1}{9}x^2 = 725\\[1em] \Rightarrow \frac{5 \times 9}{16 \times 9}x^2 - \frac{1 \times 16}{9 \times 16}x^2 = 725\\[1em] \Rightarrow \frac{45}{144}x^2 - \frac{16}{144}x^2 = 725\\[1em] \Rightarrow \frac{29}{144}x^2 = 725\\[1em] \Rightarrow x^2 = \frac{725 \times 144}{29}\\[1em] \Rightarrow x^2 = \frac{104,400}{29}\\[1em] \Rightarrow x^2 = 3,600\\[1em] \Rightarrow x = \sqrt{3,600}\\[1em] \Rightarrow x = 60\end{aligned}$$

So, the numbers = $\frac{1}{4}x = \frac{1}{4} \times 60 = 15, \frac{1}{3}x = \frac{1}{3} \times 60 = 20 \text{ and } \frac{1}{2}x = \frac{1}{2} \times 60 = 30$

Hence, the number = 15, 20 and 30.

Let the fraction be $\frac{a}{b}$.

It is given that the numerator of a fraction is 3 less than its denominator.

⇒ a = b - 3

And, If 2 is added to both the numerator and denominator, the sum of the new fraction and the original fraction is $1\frac{9}{20}$

$$\begin{aligned}\Rightarrow \frac{a}{b} + \frac{a + 2}{b + 2} = 1\frac{9}{20}\\[1em] \Rightarrow \frac{b - 3}{b} + \frac{(b - 3) + 2}{b + 2} = \frac{29}{20}\\[1em] \Rightarrow \frac{b - 3}{b} + \frac{b - 1}{b + 2} = \frac{29}{20}\\[1em] \Rightarrow \frac{(b - 3) \times (b + 2)}{b \times (b + 2)} + \frac{(b - 1) \times b}{(b + 2) \times b} = \frac{29}{20}\\[1em] \Rightarrow \frac{b^2 - 3b + 2b - 6}{b^2 + 2b} + \frac{b^2 - b}{b^2 + 2b} = \frac{29}{20}\\[1em] \Rightarrow \frac{(b^2 - 3b + 2b - 6) + (b^2 - b)}{b^2 + 2b} = \frac{29}{20}\\[1em] \Rightarrow \frac{b^2 - b - 6 + b^2 - b}{b^2 + 2b} = \frac{29}{20}\\[1em] \Rightarrow \frac{2b^2 - 2b - 6}{b^2 + 2b} = \frac{29}{20}\\[1em] \Rightarrow 20(2b^2 - 2b - 6) = 29(b^2 + 2b)\\[1em] \Rightarrow 40b^2 - 40b - 120 = 29b^2 + 58b\\[1em] \Rightarrow 40b^2 - 40b - 120 - 29b^2 - 58b = 0\\[1em] \Rightarrow 11b^2 - 98b - 120 = 0\\[1em] \Rightarrow 11b^2 - 110b + 12b - 120 = 0\\[1em] \Rightarrow 11b(b - 10) + 12(b - 10) = 0\\[1em] \Rightarrow (b - 10)(11b + 12) = 0\\[1em] \Rightarrow (b - 10) = 0 \text{ or } (11b + 12) = 0\\[1em] \Rightarrow b = 10 \text{ or } b = -\frac{12}{11}\\[1em]\end{aligned}$$

As b cannot be fraction. So, b = 10.

When b = 10, a = b - 3 = 10 - 3 = 7

The fraction = $\frac{a}{b} = \frac{7}{10}$

New fraction = $\frac{a + 2}{b + 2} = \frac{7 + 2}{10 + 2} = \frac{9}{12} = \frac{3}{4}$

Hence, the fraction = $\frac{3}{4}$.

Let the two-digit number be 10x + y.

If the number is 4 times the sum of its digits.

⇒ 10x + y = 4(x + y)

⇒ 10x + y = 4x + 4y

⇒ 10x - 4x = 4y - y

⇒ 6x = 3y

⇒ 2x = y .......... (1)

And, the number is twice the product of its digits.

⇒ 10x + y = 2(xy)

Using equation (1), we get

⇒ 10x + 2x = 2(x $\times$ 2x)

⇒ 12x = 2(2x²)

⇒ 12x = 4x²

⇒ 12 = 4x

⇒ 3 = x

Putting the value of x in equation (1), we get

y = 2x = 2 $\times$ 3 = 6

The number = 10x + y = 10 $\times$ 3 + 6 = 30 + 6 = 36

Hence, the number = 36.

Let one number be x and the other number = (27 - x).

The sum of their reciprocal = $\frac{3}{20}$

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{27 - x} = \frac{3}{20}\\[1em] \Rightarrow \frac{27 - x}{x(27 - x)} + \frac{x}{x(27 - x)} = \frac{3}{20}\\[1em] \Rightarrow \frac{27 - x + x}{x(27 - x)} = \frac{3}{20}\\[1em] \Rightarrow \frac{27}{27x - x^2} = \frac{3}{20}\\[1em] \Rightarrow 27 \times 20 = 3 \times (27x - x^2)\\[1em] \Rightarrow 540 = 81x - 3x^2\\[1em] \Rightarrow 81x - 3x^2 - 540 = 0\\[1em] \Rightarrow x^2 - 27x + 180 = 0\\[1em] \Rightarrow x^2 - 12x - 15x + 180 = 0\\[1em] \Rightarrow x(x - 12) - 15(x - 12) = 0\\[1em] \Rightarrow (x - 12)(x - 15) = 0\\[1em] \Rightarrow (x - 12) = 0 \text{ or } (x - 15) = 0\\[1em] \Rightarrow x = 12 \text{ or } x = 15\\[1em]\end{aligned}$$

If one number = 12, other number = 27 - 12 = 15

If one number = 15, other number = 27 - 15 = 12

Ratio = 12 : 15 = 4 : 5 or 15 : 12 = 5 : 4

Hence, the ratio of two numbers = 5 : 4.

18 days

Reason

Given,

A alone can do the same work in 9 days.

So, one day work of A = $\frac{1}{9}$

Let B alone can do the same work in y days.

So, one day work of B = $\frac{1}{y}$

As together they can do work in 6 days.

∴ In one day, both of them do $\frac{1}{6}$th part of work.

$$\begin{aligned}∴ \frac{1}{9} + \frac{1}{y} = \frac{1}{6}\\[1em] \Rightarrow \frac{1 \times y}{9\times y} + \frac{1\times 9}{y\times 9} = \frac{1}{6}\\[1em] \Rightarrow \frac{y}{9y} + \frac{9}{9y} = \frac{1}{6}\\[1em] \Rightarrow \frac{y + 9}{9y} = \frac{1}{6}\\[1em] \Rightarrow 6(y + 9) = 9y\\[1em] \Rightarrow 6y + 54 = 9y\\[1em] \Rightarrow 9y - 6y = 54\\[1em] \Rightarrow 3y = 54\\[1em] \Rightarrow y = \frac{54}{3}\\[1em] \Rightarrow y = 18\end{aligned}$$

∴ B alone can complete the work in 18 days.

Hence, option 3 is the correct option.

A can do a piece of work in 5 days.

So, in 1 day

A will do $\frac{1}{5}$ th of the work.

B can do a piece of work in (x + 5) days.

So, in 1 day

B will do $\frac{1}{x + 5}$ of the work.

Work done by A and B in 1 day = $\frac{1}{5} + \frac{1}{x + 5}$

Days taken to complete work = $\frac{1}{\text{Work done by A and B in 1 day}}$

$$\begin{aligned}\Rightarrow 3\frac{1}{3} = \frac{1}{\frac{1}{5} + \frac{1}{x + 5}} \\[1em] \Rightarrow \frac{10}{3} = \frac{1}{\frac{x + 5 + 5}{5(x + 5)}} \\[1em] \Rightarrow \frac{10}{3} = \frac{5(x + 5)}{x + 10} \\[1em] \Rightarrow \frac{10}{3} = \frac{5x + 25}{x + 10} \\[1em] \Rightarrow 10(x + 10) = 3(5x + 25) \\[1em] \Rightarrow 10x + 100 = 15x + 75 \\[1em] \Rightarrow 15x - 10x = 100 -75 \\[1em] \Rightarrow 5x = 25 \\[1em] \Rightarrow x = \frac{25}{5} = 5.\end{aligned}$$

Hence, Option 1 is the correct option.

Given,

C.P. = ₹ x

Profit = x %

S.P. = C.P. + Profit

$$\begin{aligned}\Rightarrow 56 = x + \frac{x}{100} \times x \\[1em] \Rightarrow 56 = x + \frac{x^2}{100} \\[1em] \Rightarrow \frac{x^2 + 100x}{100} = 56 \\[1em] \Rightarrow x^2 + 100x = 5600 \\[1em] \Rightarrow x^2 + 100x - 5600 = 0\\[1em] \Rightarrow x^2 + 140x - 40x - 5600 = 0 \\[1em] \Rightarrow x(x + 140) - 40(x + 140) = 0 \\[1em] \Rightarrow (x - 40)(x + 140) = 0 \\[1em] \Rightarrow x - 40 = 0 \text{ or }x + 140 =0 \\[1em] \Rightarrow x = 40 \text{ or } x = -140.\end{aligned}$$

Since, cost cannot be negative.

∴ x ≠ -140.

Hence, Option 3 is the correct option.

6 hours

Reason

Given,

An empty tank is filled by a pipe in 2 hours.

So, one hour work of the pipe = $\frac{1}{2}$

Another pipe empties the full tank in 3 hours.

So, one hour work of other pipe = -$\frac{1}{3}$

When both pipes are opened together

$$\begin{aligned}= \frac{1}{2} + \Big(-\frac{1}{3}\Big)\\[1em] = \frac{1}{2} - \frac{1}{3}\\[1em] = \frac{1 \times 3}{2 \times 3} - \frac{1 \times 2}{3 \times 2}\\[1em] = \frac{3}{6} - \frac{2}{6}\\[1em] = \frac{1}{6}\end{aligned}$$

Total time taken to fill the tank = $\frac{1}{\frac{1}{6}}$ = 6 hours.

Hence, option 3 is the correct option.

A can do work in x days

B can do work in (x + 16) days

In one day, A completes $\frac{1}{x}$ part of work

In one day, B completes $\frac{1}{x + 16}$ part of work

Given, both can do work in 15 days,

$$\begin{aligned}\therefore \frac{1}{x} + \frac{1}{x + 16} = \frac{1}{15} \\[1em] \frac{x + 16 + x}{x(x + 16)} = \frac{1}{15} \\[1em] \frac{2x + 16}{x^2 + 16x} = \frac{1}{15} \\[1em] 15(2x + 16) = x^2 + 16x \\[1em] 30x + 240 = x^2+ 16x \\[1em] x^2 + 16x - 30x - 240 = 0 \\[1em] x^2 - 14x - 240 = 0 \\[1em] x^2 - 24x + 10x - 240 = 0 \\[1em] x(x - 24) + 10(x - 24) = 0 \\[1em] (x + 10)(x - 24) = 0 \\[1em] x + 10 = 0 \text{ or } x - 24 = 0 \\[1em] x = -10 \text{ or } x = 24.\end{aligned}$$

Since, no. of days cannot be negative,

∴ x ≠ -10.

Hence, x = 24 days.

Let second pipe fill in x hours, and first pipe in (x - 3) hours.

In one hour second pipe will fill $\frac{1}{x}$ and first pipe $\frac{1}{x - 3}$.

Given, together pipes can fill the cistern in 6 hours 40 minutes i.e. $\frac{400}{60}$ hours.

Hence, in one hour they will fill $\frac{1}{\frac{400}{60}} = \frac{60}{400}$.

$$\begin{aligned}\therefore \frac{1}{x} + \frac{1}{x - 3} = \frac{60}{400} \\[1em] \Rightarrow \frac{x - 3 + x}{x(x - 3)} = \frac{3}{20} \\[1em] \Rightarrow \frac{2x - 3}{x^2 - 3x} = \frac{3}{20} \\[1em] \Rightarrow 20(2x - 3) = 3(x^2 - 3x) \\[1em] \Rightarrow 40x - 60 = 3x^2 - 9x \\[1em] \Rightarrow 3x^2 - 9x - 40x - 60 = 0 \\[1em] \Rightarrow 3x^2 - 49x - 60 = 0 \\[1em] \Rightarrow 3x^2 - 45x - 4x - 60 = 0 \\[1em] \Rightarrow 3x(x - 15) - 4(x - 15) = 0 \\[1em] \Rightarrow (3x - 4)(x - 15) = 0 \\[1em] \Rightarrow 3x - 4 = 0 \text{ or } x - 15 = 0 \\[1em] \Rightarrow x = \frac{4}{3} \text{ or } x = 15.\end{aligned}$$

Since first pipe takes 3 hours less than second pipe,

∴ x ≠ $\frac{4}{3}$ as in this case time will be negative which is not possible.

∴ x = 15, x - 3 = 12.

Hence, pipes will take 12 hours and 15 hours to fill cistern separately.

C.P. = ₹ x

S.P. = ₹ 52

Profit = S.P. - C.P. = ₹(52 - x)

Profit % = $\frac{\text{Profit}}{\text{C.P.}} \times 100$

Substituting value in above equation we get,

$$\begin{aligned}\Rightarrow x - 10 = \frac{52 - x}{x} \times 100 \\[1em] \Rightarrow x(x - 10) = 100(52 - x) \\[1em] \Rightarrow x^2 - 10x = 5200 - 100x \\[1em] \Rightarrow x^2 - 10x + 100x - 5200 = 0 \\[1em] \Rightarrow x^2 + 90x - 5200 = 0 \\[1em] \Rightarrow x^2 + 130x - 40x - 5200 = 0 \\[1em] \Rightarrow x(x + 130) - 40(x + 130) = 0 \\[1em] \Rightarrow (x - 40)(x + 130) = 0 \\[1em] \Rightarrow (x - 40) = 0 \text{ or } (x + 130) = 0 \\[1em] \Rightarrow x = 40 \text{ or } x = -130.\end{aligned}$$

Since, cost cannot be negative

∴ x ≠ -130.

Hence, C.P. = ₹ 40

Given,

C.P. = ₹ x

S.P. = ₹ 27

Loss = (x - 5) %

Loss = C.P. - S.P.

= ₹ (x - 27)

As we know that Loss % = $\frac{\text{Loss}}{\text{C.P.}} \times 100$

Substituting the values, we get

$$\begin{aligned}\Rightarrow (x - 5) = \frac{(x - 27)}{x} \times 100 \\[1em] \Rightarrow x(x - 5) = (x - 27) \times 100 \\[1em] \Rightarrow x^2 - 5x = 100x - 2700 \\[1em] \Rightarrow x^2 - 5x - 100x + 2700 = 0\\[1em] \Rightarrow x^2 - 105x + 2700 = 0\\[1em] \Rightarrow x^2 - 45x - 60x + 2700 = 0\\[1em] \Rightarrow x(x - 45) - 60(x - 45) = 0\\[1em] \Rightarrow (x - 45)(x - 60) = 0\\[1em] \Rightarrow (x - 45) = 0 \text{ or } (x - 60) = 0\\[1em] \Rightarrow x = 45 \text{ or } x = 60\\[1em]\end{aligned}$$

Hence, the value of C.P. = ₹ 45 or ₹ 60.

Given,

C.P. = ₹ x

S.P. = ₹ 152

Profit = (x + 10) %

Profit = S.P. - C.P.

= ₹ (152 - x)

As we know that Profit % = $\frac{\text{Profit}}{\text{C.P.}} \times 100$

Substituting the values, we get

$$\begin{aligned}\Rightarrow (x + 10) = \frac{(152 - x)}{x} \times 100 \\[1em] \Rightarrow x(x + 10) = (152 - x) \times 100 \\[1em] \Rightarrow x^2 + 10x = 15200 - 100x \\[1em] \Rightarrow x^2 + 10x - 15200 + 100x = 0\\[1em] \Rightarrow x^2 + 110x - 15200 = 0\\[1em] \Rightarrow x^2 + 190x - 80x - 15200 = 0\\[1em] \Rightarrow x(x + 190) - 80(x + 190) = 0\\[1em] \Rightarrow (x + 190)(x - 80) = 0\\[1em] \Rightarrow (x + 190) = 0 \text{ or } (x - 80) = 0\\[1em] \Rightarrow x = -190 \text{ or } x = 80\\[1em]\end{aligned}$$

The cost price of the article cannot be negative.

Hence, the value of C.P. = ₹ 80.

Let the time taken by each pipe be x minutes and y minutes.

It is given that one pipe takes 5 minutes more than the other pipe to fill the empty cistern.

⇒ x = y + 5

So, one minute work of the pipe = $\frac{1}{x} = \frac{1}{y + 5}$

And, one minute work of the other pipe = $\frac{1}{y}$

Both pipes running together can fill the empty cistern in $4\frac{20}{21} = \frac{104}{21}$ minutes.

$$\begin{aligned}\Rightarrow \frac{1}{x} + \frac{1}{y} = \frac{21}{104}\\[1em] \Rightarrow \frac{1}{y + 5} + \frac{1}{y} = \frac{21}{104}\\[1em] \Rightarrow \frac{1 \times y}{y(y + 5)} + \frac{1 \times (y + 5)}{y(y + 5)} = \frac{21}{104}\\[1em] \Rightarrow \frac{y}{y(y + 5)} + \frac{y + 5}{y(y + 5)} = \frac{21}{104}\\[1em] \Rightarrow \frac{y + y + 5}{y(y + 5)} = \frac{21}{104}\\[1em] \Rightarrow \frac{2y + 5}{y^2 + 5y} = \frac{21}{104}\\[1em] \Rightarrow 104(2y + 5) = 21(y^2 + 5y)\\[1em] \Rightarrow 208y + 520 = 21y^2 + 105y\\[1em] \Rightarrow 21y^2 + 105y - 208y - 520 = 0\\[1em] \Rightarrow 21y^2 - 103y - 520 = 0\\[1em] \Rightarrow 21y^2 + 65y - 168y - 105 = 0\\[1em] \Rightarrow y(21y + 65) - 8(21y + 65) = 0\\[1em] \Rightarrow (21y + 65)(y - 8) = 0\\[1em] \Rightarrow (21y + 65) = 0 \text{ or }(y - 8) = 0\\[1em] \Rightarrow y = -\frac{65}{21} \text{ or }y = 8\\[1em]\end{aligned}$$

As minutes cannot be negative.

So, one pipe takes 8 minutes and the other pipe takes 8 + 5 = 13 minutes.

Hence, the time taken by pipes = 8 minutes and 13 minutes.

Let the time taken by the larger diameter pipe to fill the pool alone be L hours and the time taken by the smaller diameter pipe to fill the pool alone be S hours.

From the problem, we know that the larger diameter pipe takes 10 hours less than the smaller diameter pipe. Hence, we can write the relationship as:

⇒ L = S - 10

The rate of the larger pipe is $\frac{1}{\text{L}}$ of the pool per hour.

The rate of the smaller pipe is $\frac{1}{\text{S}}$ of the pool per hour.

It is given that when the larger pipe is used for 4 hours and the smaller pipe is used for 9 hours, half of the pool is filled. So, the total work done in this scenario can be expressed as:

⇒ $4 \times \frac{1}{\text{L}} + 9 \times \frac{1}{\text{S}} = \frac{1}{2}$

Substituting the value of L = S - 10, we get

$$\begin{aligned}\Rightarrow 4 \times \frac{1}{S - 10} + 9 \times \frac{1}{\text{S}} = \frac{1}{2}\\[1em] \Rightarrow \frac{4}{S - 10} + \frac{9}{\text{S}} = \frac{1}{2}\\[1em] \Rightarrow \frac{4S}{S(S - 10)} + \frac{9(S - 10)}{S(S - 10)} = \frac{1}{2}\\[1em] \Rightarrow \frac{4S}{S(S - 10)} + \frac{9S - 90}{S(S - 10)} = \frac{1}{2}\\[1em] \Rightarrow \frac{4S + 9S - 90}{S(S - 10)} = \frac{1}{2}\\[1em] \Rightarrow \frac{13S - 90}{S^2 - 10S} = \frac{1}{2}\\[1em] \Rightarrow 2(13S - 90) = S^2 - 10S\\[1em] \Rightarrow 26S - 180 = S^2 - 10S\\[1em] \Rightarrow S^2 - 10S - 26S + 180 = 0\\[1em] \Rightarrow S^2 - 36S + 180 = 0\\[1em] \Rightarrow S^2 - 30S - 6S + 180 = 0\\[1em] \Rightarrow S(S - 30) - 6(S - 30) = 0\\[1em] \Rightarrow (S - 30)(S - 6) = 0\\[1em] \Rightarrow (S - 30) = 0 \text{ or } (S - 6) = 0\\[1em] \Rightarrow S = 30 \text{ or } S = 6\\[1em]\end{aligned}$$

Substituting the value of S in equation, L = S - 10

⇒ L = 30 - 10 = 20 or 6 - 10 = -4

As time cannot be negative.

Hence, pipe of the larger diameter alone will take 20 hours to fill the pool.

Let present ages of Radha and Meena be x and y years respectively.

Given,

Sum of present ages = 39

⇒ x + y = 39

⇒ x = 39 - y ........(1)

Given,

Two years ago the ages of Radha and Meena were in the ratio 5 : 2.

$$\begin{aligned}\Rightarrow \frac{x - 2}{y - 2} = \frac{5}{2} \\[1em] \Rightarrow \frac{39 - y - 2}{y - 2} = \frac{5}{2} \\[1em] \Rightarrow \frac{37 - y}{y - 2} = \frac{5}{2} \\[1em] \Rightarrow 2(37 - y) = 5(y - 2) \\[1em] \Rightarrow 74 - 2y = 5y - 10 \\[1em] \Rightarrow 5y + 2y = 74 + 10 \\[1em] \Rightarrow 7y = 84 \\[1em] \Rightarrow y = \frac{84}{7} = 12.\end{aligned}$$

x = 39 - y = 39 - 12 = 27.

∴ Radha's ages = 27 years and Meena's age = 12 years.

Two years ago their age will be :

x - 2 = 27 - 2 = 25 years and y - 2 = 12 - 2 = 10 years.

Hence, Option 3 is the correct option.

Let length and breadth of rectangular field be l and b meters respectively.

By formula,

⇒ Perimeter = 2(l + b)

⇒ 28 = 2(l + b)

⇒ l + b = 14

⇒ l = 14 - b ......(1)

By formula,

⇒ Area = lb

⇒ 40 = lb

Substituting value of l from equation (1), we get :

⇒ 40 = b(14 - b)

⇒ 40 = 14b - b²

⇒ b² - 14b + 40 = 0

⇒ b² - 10b - 4b + 40 = 0

⇒ b(b - 10) - 4(b - 10) = 0

⇒ (b - 4)(b - 10) = 0

⇒ b - 4 = 0 or b - 10 = 0

⇒ b = 4 or b = 10.

If b = 4,

l = 14 - b = 14 - 4 = 10.

If b = 10,

l = 14 - b = 14 - 10 = 4.

∴ Sides = 4 m and 10 m.

Hence, Option 2 is the correct option.

Let side of square be x units.

Given,

Area = Perimeter

⇒ (side)² = 4 × side

⇒ x² = 4x

⇒ x² - 4x = 0

⇒ x(x - 4) = 0

⇒ x = 0 or x - 4 = 0

⇒ x = 0 or x = 4.

Since, side cannot be equal to zero.

∴ Side = 4 units

Perimeter = 4 × 4 = 16 units.

Hence, Option 4 is the correct option.

From figure,

EF = 16 - x - x = (16 - 2x) meters

FG = 20 - x - x = (20 - 2x) meters

Area of shaded portion = Area of rectangle ABCD - Area of rectangle EFGH

= AB × BC - EF × FG

= 16 × 20 - (16 - 2x)(20 - 2x)

= 320 m² - (16 - 2x)(20 - 2x) m².

Hence, Option 1 is the correct option.

Speed of boat upstream = Speed of boat - Speed of stream

= 32 km/h - 8 km/h

= 24 km/h

Hence, Option 4 is the correct option.

By formula,

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

For train A :

Time taken = $\frac{400}{x}$ hrs.

For train B :

Time taken = $\frac{400}{x - 5}$ hrs.

Hence, Option 3 is the correct option.

Let the age of sister be x years.

It is given in question that the boy is twice as old as her sister.

⇒ Boy's age = 2x

Four years hence, the product of their ages = 160

⇒ (Sister's age + 4) x (Boy's age + 4) = 160

⇒ (x + 4) $\times$ (2x + 4) = 160

⇒ x $\times$ (2x + 4) + 4 $\times$ (2x + 4) = 160

⇒ 2x² + 4x + 8x + 16 = 160

⇒ 2x² + 12x + 16 - 160 = 0

⇒ 2x² + 12x - 144 = 0

⇒ x² + 6x - 72 = 0

⇒ x² + 12x - 6x - 72 = 0

⇒ x(x + 12) - 6(x + 12) = 0

⇒ (x + 12)(x - 6) = 0

⇒ (x + 12) = 0 or (x - 6) = 0

⇒ x = -12 or x = 6

Since age cannot be negative,

∴ Present age of sister = 6 years

Boy's age = 2 x 6 = 12 years

Thus, the present age of boy = 12 years and sister = 6 years.

Let the age of Geeta be x years and that of Rohit be y years.

It is given that 7 years ago, Rohit's age was five times the square of Geeta's age.

⇒ y - 7 = 5(x - 7)²

⇒ y - 7 = 5(x² + 49 - 14x)

⇒ y - 7 = 5x² + 245 - 70x

⇒ y = 5x² + 245 - 70x + 7

⇒ y = 5x² + 252 - 70x .......... (1)

And, 3 years hence, Geeta's age will be two fifths of Rohit's age.

⇒ (x + 3) = $\frac{2}{5}$ (y + 3)

⇒ 5(x + 3) = 2(y + 3)

⇒ 5x + 15 = 2y + 6

⇒ 5x + 15 - 6 = 2y

⇒ 5x + 9 = 2y .......... (2)

Substituting the value of y in the above equation,

⇒ 5x + 9 = 2(5x² + 252 - 70x)

⇒ 5x + 9 = 10x² + 504 - 140x

⇒ 10x² + 504 - 140x - 5x - 9 = 0

⇒ 10x² - 145x + 495 = 0

⇒ 2x² - 29x + 99 = 0

⇒ 2x² - 18x - 11x + 99 = 0

⇒ 2x(x - 9) - 11(x - 9) = 0

⇒ (x - 9)(2x - 11) = 0

⇒ (x - 9) = 0 or (2x - 11) = 0

⇒ x = 9 or 2x = 11

⇒ x = 9 or x = $\frac{11}{2}$

Since, age cannot be in fraction. So, the age of Geeta = 9 years.

Substituting the value of x in equation (2),

⇒ 5 $\times$ 9 + 9 = 2y

⇒ 45 + 9 = 2y

⇒ 54 = 2y

⇒ y = $\frac{54}{2}$ = 27

Thus, the age of Rohit = 27 years and that of Geeta = 9 years.

Let present age of Joseph be x years.

It is given that the sum of the reciprocals of Joseph's age 3 years ago and five years from now is $\frac{1}{3}$.

$$\begin{aligned}\Rightarrow \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3}\\[1em] \Rightarrow \frac{(x + 5) + (x - 3)}{(x - 3)(x + 5)} = \frac{1}{3}\\[1em] \Rightarrow \frac{2x + 2}{(x - 3)(x + 5)} = \frac{1}{3}\\[1em] \Rightarrow 3(2x + 2) = (x - 3)(x + 5)\\[1em] \Rightarrow 6x + 6 = x(x + 5) - 3(x + 5)\\[1em] \Rightarrow 6x + 6 = x^2 + 5x - 3x - 15\\[1em] \Rightarrow x^2 + 5x - 3x - 15 - 6x - 6 = 0\\[1em] \Rightarrow x^2 - 4x - 21 = 0\\[1em] \Rightarrow x^2 - 7x + 3x - 21 = 0\\[1em] \Rightarrow x(x - 7) + 3(x - 7) = 0\\[1em] \Rightarrow (x - 7)(x + 3) = 0\\[1em] \Rightarrow (x - 7) = 0 \text{ or } (x + 3) = 0\\[1em] \Rightarrow x = 7 \text{ or } x = -3\\[1em]\end{aligned}$$

Since, age cannot be negative.

Hence, the present age of Joseph = 7 years.

Let the shorter side of the rectangular field be x m.

It is given that the diagonal of a rectangular field is 16 m more than the shorter side.

⇒ d = x + 16

And, the longer side is 14 m more than the shorter side.

⇒ l = x + 14

As we know that angles of rectangle are 90°. Using Pythagoras theorem,

⇒ diagonal² = length² + breadth²

⇒ (x + 16)² = (x + 14)² + x²

⇒ x² + 16² + 32x = x² + 14² + 28x + x²

⇒ x² + 256 + 32x = 2x² + 196 + 28x

⇒ 2x² + 196 + 28x - x² - 256 - 32x = 0

⇒ x² - 4x - 60 = 0

⇒ x² - 10x + 6x - 60 = 0

⇒ x(x - 10) + 6(x - 10) = 0

⇒ (x - 10)(x + 6) = 0

⇒ (x - 10) = 0 or (x + 6) = 0

⇒ x = 10 or x = -6

As length of shorter side cannot be negative,

Length = 10 + 14 = 24 m

Hence, length and breath of the rectangle = 24 m and 10 m.

Let the sides of two squares be 'a' m and 'b' m.

As we know that area of square = (side)² and perimeter of square = 4 x side

It is given that sum of the areas of two squares is 260 m².

⇒ a² + b² = 260 .......... (1)

And, the difference of their perimeters is 24 m.

⇒ 4a - 4b = 24

⇒ 4(a - b) = 24

⇒ a - b = $\frac{24}{4}$

⇒ a - b = 6

⇒ a = 6 + b .......... (2)

Substituting the value of a in equation (1), we get

⇒ (6 + b)² + b² = 260

⇒ 36 + b² + 12b + b² = 260

⇒ 36 + 2b² + 12b - 260 = 0

⇒ 2b² + 12b - 224 = 0

⇒ 2(b² + 6b - 112) = 0

⇒ b² + 6b - 112 = 0

⇒ b² + 14b - 8b - 112 = 0

⇒ b(b + 14) - 8(b + 14) = 0

⇒ (b + 14)(b - 8) = 0

⇒ (b + 14) = 0 or (b - 8) = 0

⇒ b = -14 or b = 8

Since, side of square cannot be negative.

Side of one square = 8 m

Side of other square = 6 + 8 = 14 m

Thus, side of squares = 8 m and 14 m.

Let the usual speed of man be x km/hr.

Given that when he walks 1 km/h faster than usual, his speed is (x + 1) km/h, and he covers a distance of 3 km in 15 minutes less time than usual.

Using the formula; Time = $\frac{\text{Distance}}{\text{Speed}}$

The time it takes for him to cover 3 km at his usual speed x km/h is given by:

Usual time = $\frac{3}{x}$ hr

The time it takes for him to cover 3 km at the faster speed x + 1 km/h is given by:

Faster time = $\frac{3}{x + 1}$ hr

The faster time is 15 minutes ($\frac{15}{60} = \frac{1}{4}$ hours) less than the usual time. Therefore, we can write:

$$\begin{aligned}\Rightarrow \frac{3}{x} - \frac{3}{x + 1} = \frac{1}{4}\\[1em] \Rightarrow \frac{3(x + 1) - 3x}{x(x + 1)} = \frac{1}{4}\\[1em] \Rightarrow \frac{3x + 3 - 3x}{x(x + 1)} = \frac{1}{4}\\[1em] \Rightarrow \frac{3}{x(x + 1)} = \frac{1}{4}\\[1em] \Rightarrow 3 \times 4 = x(x + 1)\\[1em] \Rightarrow 12 = x^2 + x\\[1em] \Rightarrow x^2 + x - 12 = 0\\[1em] \Rightarrow x^2 + 4x - 3x - 12 = 0\\[1em] \Rightarrow x(x + 4) - 3(x + 4) = 0\\[1em] \Rightarrow (x + 4)(x - 3) = 0\\[1em] \Rightarrow (x + 4) = 0 \text{ or } (x - 3) = 0\\[1em] \Rightarrow x = -4 \text{ or } x = 3\end{aligned}$$

Since, speed cannot be negative,

∴ Usual speed of man = 3 km/hr

Hence, usual speed of man = 3 km/hr.

Let's assume the bus's original speed is x km/hr.

Using the formula; Time = $\frac{\text{Distance}}{\text{Speed}}$

The time taken to travel 75 km at the original speed x km/hr = $\frac{75}{x}$

The time taken to travel 90 km at a speed of (x + 10) km/hr (since the second speed is 10 km/h more than the original speed) = $\frac{90}{x + 10}$

The total time taken for the journey is 3 hours. Therefore, we can write the equation:

$$\begin{aligned}\Rightarrow \frac{75}{x} + \frac{90}{x + 10} = 3\\[1em] \Rightarrow \frac{75(x + 10) + 90x}{x(x + 10)} = 3\\[1em] \Rightarrow \frac{75x + 750 + 90x}{x(x + 10)} = 3\\[1em] \Rightarrow \frac{165x + 750}{x(x + 10)} = 3\\[1em] \Rightarrow 165x + 750 = 3x(x + 10)\\[1em] \Rightarrow 165x + 750 = 3x^2 + 30x\\[1em] \Rightarrow 3x^2 + 30x - 165x - 750 = 0\\[1em] \Rightarrow 3x^2 - 135x - 750 = 0\\[1em] \Rightarrow 3(x^2 - 45x - 250) = 0\\[1em] \Rightarrow x^2 - 45x - 250 = 0\\[1em] \Rightarrow x^2 - 50x + 5x - 250 = 0\\[1em] \Rightarrow x(x - 50) + 5(x - 50) = 0\\[1em] \Rightarrow (x - 50)(x + 5) = 0\\[1em] \Rightarrow (x - 50) = 0 \text{ or }(x + 5) = 0\\[1em] \Rightarrow x = 50 \text{ or }x = -5\\[1em]\end{aligned}$$

Since, speed of bus cannot be negative.

Hence, speed of bus = 50 km/hr.

(i) We know that,

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

Time taken by ordinary train = $\frac{300}{x}$

Time taken by express train = $\frac{300}{x + 25}$

Hence, time taken by ordinary train = $\frac{300}{x}$ hours and $\frac{300}{x + 25}$ hours.

(ii) According to question,

$$\begin{aligned}\Rightarrow \frac{300}{x} - \frac{300}{x + 25} = 2 \\[1em] \Rightarrow \frac{300(x + 25) - 300x}{x(x + 25)} = 2 \\[1em] \Rightarrow \frac{300x + 7500 - 300x}{x^2 + 25x} = 2 \\[1em] \Rightarrow \frac{7500}{x^2 + 25x} = 2 \\[1em] \Rightarrow 7500 = 2(x^2 + 25x) \\[1em] \Rightarrow 2x^2 + 50x - 7500 = 0 \\[1em] \Rightarrow 2(x^2 + 25x - 3750) = 0 \\[1em] \Rightarrow x^2 + 25x - 3750 = 0 \\[1em] \Rightarrow x^2 + 75x - 50x - 3750 = 0 \\[1em] \Rightarrow x(x + 75) - 50(x + 75) = 0 \\[1em] \Rightarrow (x - 50)(x + 75) = 0 \\[1em] \Rightarrow x - 50 = 0 \text{ or } x + 75 = 0 \\[1em] \Rightarrow x = 50 \text{ or } x = -75.\end{aligned}$$

Since, speed cannot be negative

∴ x = 50 and x + 25 = 75.

Hence, speed of express train = 75 km/hr.

Let speed of goods train be x km/hr

Speed of express train will be (x + 20) km/hr

Time taken to cover 1040 km by,

Goods train = $\frac{1040}{x}$ hours

Express train = $\frac{1040}{x + 20}$ hours

Since express train leaves 2 hours after the goods train and arrives 36 minutes before the goods train,

∴ It takes 2 hours 36 minutes less i.e., (2 x 60) + 36 = 156 minutes

$$\begin{aligned}\therefore \frac{1040}{x} - \frac{1040}{x + 20} = \frac{156}{60} \\[1em] \Rightarrow \frac{1040(x + 20) - 1040x}{x(x + 20)} = \frac{13}{5} \\[1em] \Rightarrow \frac{1040x + 20800 - 1040x}{x^2 + 20x} = \frac{13}{5} \\[1em] \Rightarrow \frac{20800}{x^2 + 20x} = \frac{13}{5} \\[1em] \Rightarrow 20800 \times 5 = 13(x^2 + 20x) \\[1em] \Rightarrow 104000 = 13x^2 + 260x \\[1em] \Rightarrow 13x^2 + 260x - 104000 = 0 \\[1em] \Rightarrow 13(x^2 + 20x - 8000) = 0 \\[1em] \Rightarrow x^2 + 20x - 8000 = 0 \\[1em] \Rightarrow x^2 + 100x - 80x - 8000 = 0 \\[1em] \Rightarrow x(x + 100) - 80(x + 100) = 0 \\[1em] \Rightarrow (x - 80)(x + 100) = 0 \\[1em] \Rightarrow x - 80 = 0 \text{ or } x + 100 = 0 \\[1em] \Rightarrow x = 80 \text{ or } x = -100.\end{aligned}$$

Since speed cannot be negative,

∴ x = 80, (x + 20) = 100.

Hence, speed of goods train = 80 km/hr and express train = 100 km/hr.

7 × 5 = 35, where 7 and 5 are whole numbers and greater than 4.

Hence, Option 2 is the correct option.

Let ten's digit be x and one's digit be y.

Given,

Ten's digit is bigger and difference between digits = 2.

∴ x - y = 2

⇒ x = 2 + y .......(1)

Given,

Product of digits = 24

⇒ xy = 24

⇒ y(2 + y) = 24 .......[From Eq. (1)]

⇒ 2y + y² = 24

⇒ y² + 2y - 24 = 0

⇒ y² + 6y - 4y - 24 = 0

⇒ y(y + 6) - 4(y + 6) = 0

⇒ (y - 4)(y + 6) = 0

⇒ y - 4 = 0 or y + 6 = 0

⇒ y = 4 or y = -6.

Since, digit cannot be negative.

∴ y ≠ -6.

x = 2 + y = 2 + 4 = 6.

Number = 10x + y = 10 × 6 + 4 = 60 + 4 = 64.

Hence, Option 4 is the correct option.

Let x be ten's digit and y be unit's digit.

Given,

Product of digits of number = 24

⇒ xy = 24 .........(1)

Number = 10x + y

Given,

If 18 is added to a two-digit number, its digits are reversed.

⇒ 10x + y + 18 = 10y + x

⇒ 10x - x = 10y - y - 18

⇒ 9x = 9y - 18

⇒ 9x = 9(y - 2)

⇒ x = y - 2 .........(2)

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

⇒ (y - 2)y = 24

⇒ y² - 2y = 24

⇒ y² - 2y - 24 = 0

⇒ y² - 6y + 4y - 24 = 0

⇒ y(y - 6) + 4(y - 6) = 0

⇒ (y + 4)(y - 6) = 0

⇒ y + 4 = 0 or y - 6 = 0

⇒ y = -4 or y = 6.

Digit cannot be negative, so y = 6.

⇒ x = y - 2 = 6 - 2 = 4.

Number = 10x + y = 10(4) + 6 = 40 + 6 = 46.

Hence, Option 1 is the correct option.

none of these

Reason

Let the present ages of Radha and Meena be a and b.

It is given, three years hence, the ages of Radha and Meena will be in the ratio 5:2.

$$\begin{aligned}\Rightarrow \frac{a + 3}{b + 3} = \frac{5}{2}\\[1em] \Rightarrow 2(a + 3) = 5(b + 3)\\[1em] \Rightarrow 2a + 6 = 5b + 15\\[1em] \Rightarrow 2a = 5b + 15 - 6\\[1em] \Rightarrow 2a = 5b + 9\\[1em] \Rightarrow a = \frac{5b + 9}{2} \space .......... \space (1)\end{aligned}$$

The sum of their present ages = 62 years.

⇒ a + b = 62

Substituting the value of a from equation (1), we get

⇒ $\frac{5b + 9}{2}$ + b = 62

⇒ 5b + 9 + 2b = 124

⇒ 7b + 9 = 124

⇒ 7b = 124 - 9

⇒ 7b = 115

⇒ b = $\frac{115}{7}$

Substituting the value of b in equation (1),

$$\begin{aligned}\Rightarrow a = \frac{5 \times \frac{115}{7} + 9}{2}\\[1em] \Rightarrow a = \frac{575 + 63}{14}\\[1em] \Rightarrow a = \frac{638}{14}\\[1em] \Rightarrow a = \frac{319}{7}\\[1em]\end{aligned}$$

Radha's age 2 years ago = a - 2 = $\frac{319}{7} - 2 = \frac{319 - 14}{7} = \frac{305}{7} = 43.57$

Meena's age 2 years ago = b - 2 = $\frac{115}{7} - 2 = \frac{115 - 14}{7} = \frac{101}{7} = 14.42$

Hence, option 4 is the correct option.

Given,

₹ 320 were divided equally among 8 children.

So, each child will get = ₹ $\frac{320}{8}$ = ₹ 40.

If 2 more children join, total children = 10.

So, now each child will get = ₹ $\frac{320}{10}$ = ₹ 32.

Each child will now be getting (₹ 40 - ₹ 32) ₹ 8 less.

Hence, Option 3 is the correct option.

Let unit's digit be x and ten's digit be y.

According to question,

⇒ xy = 24 ........(i)

⇒ x = 2y + 2 ........(ii)

Substituting value of x from (ii) in (i) we get,

⇒ (2y + 2)y = 24

⇒ 2y² + 2y = 24

⇒ y² + y = 12

⇒ y² + y - 12 = 0

⇒ y² + 4y - 3y - 12 = 0

⇒ y(y + 4) - 3(y + 4) = 0

⇒ (y - 3)(y + 4) = 0

⇒ y - 3 = 0 or y + 4 = 0

⇒ y = 3 or y = -4.

Since digit at ten's place cannot be negative

∴ y ≠ -4.

⇒ x = 2y + 2 = 2(3) + 2 = 8

Number = 10(y) + x = 10(3) + 8 = 38.

Hence, number = 38.

Let in x years the product of their ages be 304.

In x years age of sisters will be (11 + x) and (14 + x).

∴ (11 + x)(14 + x) = 304

⇒ 154 + 11x + 14x + x² = 304

⇒ x² + 25x + 154 - 304 = 0

⇒ x² + 25x - 150 = 0

⇒ x² + 30x - 5x - 150 = 0

⇒ x(x + 30) - 5(x + 30) = 0

⇒ (x - 5)(x + 30) = 0

⇒ x - 5 = 0 or x + 30 =0

⇒ x = 5 or x = -30.

Since time cannot be negative,

∴ x ≠ -30.

Hence, in 5 years the product of age of sisters will be 304.

Let one year ago,

son's age = x years

So, man's age = 8x years

Present age of,

son = x + 1

man = (x + 1)²

We can write,

⇒ (x + 1)² - 1 = 8x

⇒ x² + 1 + 2x - 1 = 8x

⇒ x² + 2x - 8x = 0

⇒ x² - 6x = 0

⇒ x(x - 6) = 0

⇒ x = 0 or x - 6 = 0

Since, age cannot be zero,

∴ x = 6

∴ x + 1 = 7 and (x + 1)² = 7² = 49.

Hence, present age of son = 7 years and man = 49 years.

Let present age of,

son = x years

So, man's age = 2x²

After eight years,

son = (x + 8) years

man's age = 3(x + 8) + 4 = 3x + 24 + 4 = (3x + 28) years.

We can write,

⇒ 2x² + 8 = 3x + 28

⇒ 2x² - 3x + 8 - 28 = 0

⇒ 2x² - 3x - 20 = 0

⇒ 2x² - 8x + 5x - 20 = 0

⇒ 2x(x - 4) + 5(x - 4) = 0

⇒ (2x + 5)(x - 4) = 0

⇒ 2x + 5 = 0 or x - 4 = 0

⇒ x = $-\frac{5}{2}$ or x = 4.

Since, x cannot be negative

∴ x ≠ $-\frac{5}{2}$

⇒ 2x² = 2(4)² = 2(16) = 32 years.

Hence, present age of son = 4 years and father = 32 years.

No. of oranges Mr. Mehra receives = x

Total no. of oranges = x + 3

Actual cost of each orange = $\frac{15}{x + 3}$

Cost of one orange for Mr. Mehra = $\frac{15}{x}$

According to question,

$$\begin{aligned}\Rightarrow \frac{15}{x} - \frac{15}{x + 3} = \frac{25}{100} \\[1em] \Rightarrow \frac{15(x + 3) - 15x}{x(x + 3)} = \frac{1}{4} \\[1em] \Rightarrow \frac{15x + 45 - 15x}{x^2 + 3x} = \frac{1}{4} \\[1em] \Rightarrow \frac{45}{x^2 + 3x} = \frac{1}{4} \\[1em] \Rightarrow 180 = x^2 + 3x \\[1em] \Rightarrow x^2 + 3x - 180 = 0 \\[1em] \Rightarrow x^2 + 15x - 12x - 180 = 0 \\[1em] \Rightarrow x(x + 15) - 12(x + 15) = 0 \\[1em] \Rightarrow (x + 15)(x - 12) = 0 \\[1em] \Rightarrow x + 15 = 0 \text{ or } x - 12 = 0 \\[1em] \Rightarrow x = -15 \text{ or } x = 12.\end{aligned}$$

Since, no. of oranges cannot be negative,

∴ x ≠ -15.

Hence, no. of oranges = 12 and equation = x² + 3x - 180.

Let no. of children be x,

Each student receives = ₹ $\frac{250}{x}$

If 25 children are increased then,

No. of children = x + 25

Now each student receives = $\frac{250}{x + 25}$

According to question,

$$\begin{aligned}\Rightarrow \frac{250}{x} - \frac{250}{x + 25} = \frac{50}{100} \\[1em] \Rightarrow \frac{250(x + 25) - 250x}{x(x + 25)} = \frac{1}{2} \\[1em] \Rightarrow \frac{250x + 6250 - 250x}{x^2 + 25x} = \frac{1}{2} \\[1em] \Rightarrow \frac{6250}{x^2 + 25x} = \frac{1}{2} \\[1em] \Rightarrow 12500 = x^2 + 25x \\[1em] \Rightarrow x^2 + 25x - 12500 = 0 \\[1em] \Rightarrow x^2 + 125x - 100x - 12500 = 0 \\[1em] \Rightarrow x(x + 125) - 100(x + 125) = 0 \\[1em] \Rightarrow (x - 100)(x + 125) \\[1em] \Rightarrow x - 100 = 0 \text{ or } x + 125 = 0 \\[1em] \Rightarrow x = 100 \text{ or } x = -125.\end{aligned}$$

Since no. of children cannot be negative,

∴ x = 100

Hence, no. of children = 100.

Initial weekly wage = ₹ 3150,

Let weekly wage of each worker be ₹ x,

No. of employees = $\frac{3150}{x}$

New wage = ₹ (x + 5)

No. of employees after reducing 5 employees = $\frac{3150}{x}$ - 5

Given, new total wage = ₹ 3250

$$\begin{aligned}\Rightarrow (x + 5)\Big(\frac{3150}{x} - 5\Big) = 3250 \\[1em] \Rightarrow (x + 5)\Big(\frac{3150 - 5x}{x}\Big) = 3250 \\[1em] \Rightarrow \frac{(x + 5)(3150 - 5x)}{x} = 3250 \\[1em] \Rightarrow 3150x - 5x^2 + 15750 - 25x = 3250x \\[1em] \Rightarrow 5x^2 + 3250x - 3150x + 25x - 15750 = 0 \\[1em] \Rightarrow 5x^2 + 125x - 15750 = 0 \\[1em] \Rightarrow 5(x^2 + 25x - 3150) = 0 \\[1em] \Rightarrow x^2 + 25x - 3150 = 0 \\[1em] \Rightarrow x^2 + 70x - 45x - 3150 = 0 \\[1em] \Rightarrow x(x + 70) - 45(x + 70) = 0 \\[1em] \Rightarrow (x - 45)(x + 70) = 0 \\[1em] \Rightarrow x - 45 = 0 \text{ or } x + 70 = 0 \\[1em] \Rightarrow x = 45 \text{ or } x = -70.\end{aligned}$$

Since wage cannot be negative,

∴ x ≠ -70.

Hence, weekly wage of worker = ₹ 45 and quadratic equation = x² + 25x - 3150 = 0.

Let no. of articles bought be x,

Cost price of each article = ₹ $\frac{1200}{x}$

S.P of each article = ₹ $\Big(\frac{1200}{x} + 2\Big)$

S.P. of (x - 10) articles = (x - 10)$\Big(\frac{1200}{x} + 2\Big)$

Profit = ₹ 60

$$\begin{aligned}\therefore (x - 10)\Big(\frac{1200}{x} + 2\Big) - 1200 = 60 \\[1em] \Rightarrow 1200 + 2x - \frac{12000}{x} - 20 - 1200 = 60 \\[1em] \Rightarrow 2x - \frac{12000}{x} = 80 \\[1em] \Rightarrow \frac{2x^2 - 12000}{x} = 80 \\[1em] \Rightarrow 2x^2 - 12000 = 80x \\[1em] \Rightarrow 2x^2 - 12000 - 80x = 0 \\[1em] \Rightarrow 2(x^2 - 6000 - 40x = 0) \\[1em] \Rightarrow x^2 - 40x - 6000 = 0 ........(i) \\[1em] \Rightarrow x^2 - 100x + 60x - 6000 = 0 \\[1em] \Rightarrow x(x - 100) + 60(x - 100) = 0 \\[1em] \Rightarrow (x + 60)(x - 100) = 0 \\[1em] \Rightarrow x = -60 \text{ or } x = 100.\end{aligned}$$

Since, no. of articles cannot be negative,

∴ x ≠ -60.

From (i) quadratic equation = x² - 40x - 6000 = 0.

Hence, no. of articles bought = 100 and quadratic equation = x² - 40x - 6000 = 0.

Let no. of articles bought be x.

C.P. of each article = ₹ $\frac{4800}{x}$

Profit earned = ₹ $15 \times \frac{4800}{x}$

We know that,

Profit = S.P. - C.P.

$$\begin{aligned}\Rightarrow \frac{72000}{x} = 100x - 4800 \\[1em] \Rightarrow 72000 = x(100x - 4800) \\[1em] \Rightarrow 72000 = 100(x^2 - 48x) \\[1em] \Rightarrow 720 = x^2 - 48x \\[1em] \Rightarrow x^2 - 48x - 720 = 0 \\[1em] \Rightarrow x^2 - 60x + 12x - 720 = 0 \\[1em] \Rightarrow x(x - 60) + 12(x - 60) = 0 \\[1em] \Rightarrow (x + 12)(x - 60) = 0 \\[1em] \Rightarrow x = -12 \text{ or } x = 60.\end{aligned}$$

Since, no. of articles cannot be negative,

∴ x ≠ -12.

Hence, no. of articles bought = 60.

Let the number be x.

Given,

Sum of a number and its reciprocal is 4.25

$$\begin{aligned}\Rightarrow x + \frac{1}{x} = 4.25 \\[1em] \Rightarrow \frac{x^2 + 1}{x} = \frac{425}{100} \\[1em] \Rightarrow \frac{x^2 + 1}{x} = \frac{17}{4} \\[1em] \Rightarrow 4(x^2 + 1) = 17x \\[1em] \Rightarrow 4x^2 + 4 = 17x \\[1em] \Rightarrow 4x^2 - 17x + 4 = 0 \\[1em] \Rightarrow 4x^2 - 16x - x + 4 = 0 \\[1em] \Rightarrow 4x(x - 4) - 1(x - 4) = 0 \\[1em] \Rightarrow (4x - 1)(x - 4) = 0 \\[1em] \Rightarrow 4x - 1 = 0 \text{ or } x - 4 = 0 \\[1em] \Rightarrow 4x = 1 \text{ or } x = 4 \\[1em] \Rightarrow x = \frac{1}{4} \text{ or } x = 4.\end{aligned}$$

Hence, Option 2 is the correct option.

Let numbers be x and y.

Given,

Sum of whole numbers = 18

⇒ x + y = 18

⇒ x = 18 - y ......(1)

Product = 45

⇒ xy = 45 ........(2)

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

⇒ (18 - y)y = 45

⇒ 18y - y² = 45

⇒ y² - 18y + 45 = 0

⇒ y² - 15y - 3y + 45 = 0

⇒ y(y - 15) - 3(y - 15) = 0

⇒ (y - 3)(y - 15) = 0

⇒ y - 3 = 0 or y - 15 = 0

⇒ y = 3 or y = 15.

If y = 3,

x = 18 - y = 18 - 3 = 15.

If y = 15,

x = 18 - y = 18 - 15 = 3.

Hence, Option 1 is the correct option.

Let width be x meters and length be (x + 3) meters.

Given,

Area = 180 m²

⇒ x(x + 3) = 180

⇒ x² + 3x = 180

⇒ x² + 3x - 180 = 0

⇒ x² + 15x - 12x - 180 = 0

⇒ x(x + 15) - 12(x + 15) = 0

⇒ (x - 12)(x + 15) = 0

⇒ x - 12 = 0 or x + 15 = 0

⇒ x = 12 or x = -15.

Since, width cannot be negative.

∴ x = 12 meters.

Length = x + 3 = 12 + 3 = 15 meters.

Hence, Option 3 is the correct option.

Speed of boat (downstream) = Speed of boat in still water + Speed of stream = 15 + 5 = 20 km/hr.

Speed of boat (upstream) = Speed of boat in still water - Speed of stream = 15 - 5 = 10 km/hr.

By formula,

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

Time taken by boat to go x km downstream = $\frac{x}{20}$

Time taken by boat to go x km upstream = $\frac{x}{10}$

Total time taken = $\Big(\frac{x}{20} + \frac{x}{10}\Big)$ hrs.

Hence, Option 3 is the correct option.

Let time taken by 2nd pipe be x min.

So, in 1 min it will fill $\frac{1}{x}$ of cistern.

So, 1st pipe will take 3 hrs i.e. 180 min less = (x - 180) min

So, in 1 min it will fill $\frac{1}{x - 180}$ of cistern.

Both pipes will fill the cistern together in 2 hrs i.e. 120 mins.

So, in 1 min they will fill = $\frac{1}{120}$ of cistern.

$$\begin{aligned}\therefore \frac{1}{x} + \frac{1}{x - 180} = \frac{1}{120} \\[1em] \Rightarrow \frac{x - 180 + x}{x(x - 180)} = \frac{1}{120} \\[1em] \Rightarrow \frac{2x - 180}{x^2 - 180x} = \frac{1}{120} \\[1em] \Rightarrow 120(2x - 180) = x^2 - 180x \\[1em] \Rightarrow 240x - 21600 = x^2 - 180x \\[1em] \Rightarrow x^2 - 180x - 240x + 21600 = 0 \\[1em] \Rightarrow x^2 - 420x + 21600 = 0 \\[1em] \Rightarrow x^2 - 420x + 21600 = 0 \\[1em] \Rightarrow x^2 - 360x - 60x + 21600 = 0 \\[1em] \Rightarrow x(x - 360) - 60(x - 360) = 0 \\[1em] \Rightarrow (x - 60)(x - 360) = 0 \\[1em] \Rightarrow x - 60 = 0 \text{ or } x - 360 = 0 \\[1em] \Rightarrow x = 60 \text{ or } x = 360.\end{aligned}$$

x cannot be equal to 60 as then x - 180 will be negative, which is not possible as time cannot be negative.

∴ x = 360 mins = 6 hours.

Hence, Option 2 is the correct option.

Both A and R are true and R is correct reason for R.

Reason

Let the age of Rohan be x years.

It is given that the sum of the ages of Rohan and his father is 35 years.

⇒ Father's age = 35 - x

And, the product of their ages is 150.

⇒ x(35 - x) = 150

⇒ 35x - x² = 150

⇒ 35x - x² - 150 = 0

⇒ x² - 35x + 150 = 0

⇒ x² - 30x - 5x + 150 = 0

⇒ x(x - 30) - 5(x - 30) = 0

⇒ (x - 30)(x - 5) = 0

⇒ (x - 30) = 0 or (x - 5) = 0

⇒ x = 30 or x = 5

When son's age is 5 years, father's age = (35 - x) = (35 - 5) = 30 years.

When son's age is 30 years, father's age = (35 - x) = (35 - 30) = 5 years (father's age is less than son's age which is not possible).

So, Assertion (A) and Reason (R) both are true and R is correct reason for A.

Hence, option 3 is the correct option.

A is false, R is true.

Reason

Given,

Speed of boat downstream = 20 km/hr

Speed of boat upstream = 16 km/hr

Let the speed of boat in still water is x km/hr and the speed of stream is y km/hr.

Speed of boat downstream = x + y

⇒ x + y = 20 .......... (1)

Speed of boat upstream = x - y

⇒ x - y = 16 .......... (2)

Adding equation (1) and (2), we get

$$\begin{matrix} & x & + & y & = & 20 \\ & x & - & y & = & 16 \\ & + & + & & & + \\ \hline & 2x & & & = & 20 + 16 \\ \Rightarrow & x & & & = & \frac{20 + 16}{2} \\ \end{matrix}$$

⇒ x = $\frac{36}{2}$ = 18 km/hr

Subtracting equation (2) from (1), we get

$$\begin{matrix} & x & + & y & = & 20 \\ & x & - & y & = & 16 \\ & - & - & & & - \\ \hline & & & 2y & = & 20 - 16 \\ \Rightarrow & & & y & = & \frac{20 - 16}{2} \\ \end{matrix}$$

⇒ y = $\frac{4}{2}$ = 2 km/hr

According to Assertion, the speed of the boat in still water = $\Big(\frac{20 - 18}{2}\Big)$ km/hr = $\frac{2}{2}$ km/hr = 1 km/hr.

So, Assertion (A) is false but Reason (R) is true.

Hence, option 2 is the correct option.

Both the statement are true.

Reason

Let the two consecutive natural numbers each of which is multiple of 3 be 3x and 3(x + 1).

Product = 108

⇒ 3x $\times$ 3(x + 1) = 108

⇒ 3x $\times$ (3x + 3) = 108

⇒ 9x² + 9x = 108

⇒ 9x² + 9x - 108 = 0

⇒ 9(x² + x - 12) = 0

⇒ x² + x - 12 = 0

⇒ x² + 4x - 3x - 12 = 0

⇒ x(x + 4) - 3(x + 4) = 0

⇒ (x + 4)(x - 3) = 0

⇒ (x + 4) = 0 or (x - 3) = 0

⇒ x = -4 or x = 3

Since, number are two natural numbers. So, x = 3.

And, when x = 3, two consecutive numbers = 3 x 3 = 9 and 3 x (3 + 1) = 3 x 4 = 12

So, both statements are true.

Hence, option 1 is the correct option.

Statement 1 is false, and statement 2 is true.

Reason

Given,

C.P. = ₹x

S.P. = ₹24

Loss% = x%

Using the formula, Loss = C.P. - S.P.

⇒ S.P. = C.P. - Loss

⇒ S.P. = C.P. - Loss% of C.P.

Substituting the values, we get

⇒ 24 = x - x% of x

So, statement 2 is true.

Hence, option 4 is the correct option.

(i) Time taken by car = $\frac{\text{Distance}}{\text{Speed}} = \frac{216}{x}$ hours.

Hence, time taken by car = $\frac{216}{x}$ hours.

(ii) Time taken by train = $\frac{\text{Distance}}{\text{Speed}} = \frac{208}{x + 16}$ hours.

Hence, time taken by train = $\frac{208}{x + 16}$ hours.

(iii) According to question,

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

Since speed cannot be negative,

∴ x = 36

Hence, required equation = $\frac{216}{x} - \frac{208}{x + 16} = 2$ and x = 36.

(iv) Speed of train = x + 16 = 36 + 16 = 52.

Hence, speed of train = 52 km/hr.

(i) Total cost = ₹ 600

No. of articles = x

Cost of each article = ₹ $\frac{600}{x}$

Hence, cost of each article = ₹ $\frac{600}{x}$.

(ii) According to question,

$$\begin{aligned}\Rightarrow \frac{600}{\frac{600}{x} + 5} = x - 4 \\[1em] \Rightarrow \frac{600}{x - 4} = \frac{600}{x} + 5 \\[1em] \Rightarrow \frac{600}{x - 4} - \frac{600}{x} = 5 ......(i) \\[1em] \Rightarrow \frac{600x - 600(x - 4)}{x(x - 4)} = 5 \\[1em] \Rightarrow \frac{600x - 600x + 2400}{x^2 - 4x} = 5 \\[1em] \Rightarrow \frac{2400}{x^2 - 4x} = 5 \\[1em] \Rightarrow 2400 = 5(x^2 - 4x) \\[1em] \Rightarrow x^2 - 4x = 480 \\[1em] \Rightarrow x^2 - 4x - 480 = 0 \\[1em] \Rightarrow x^2 - 24x + 20x - 480 = 0 \\[1em] \Rightarrow x(x - 24) + 20(x - 24) = 0 \\[1em] \Rightarrow (x + 20)(x - 24) = 0 \\[1em] \Rightarrow x + 20 = 0 \text{ or } x - 24 = 0 \\[1em] \Rightarrow x = -20, 24.\end{aligned}$$

Since no. of articles cannot be negative,

∴ x ≠ -20.

From (i) we get quadratic equation,

$\frac{600}{x - 4} - \frac{600}{x} = 5.$

Hence, required quadratic equation $\frac{600}{x - 4} - \frac{600}{x} = 5$ and x = 24.

Let no. of people be x.

Each person's bill = ₹ $\frac{4800}{x}$

According to question,

$$\begin{aligned}\Rightarrow \frac{4800}{x + 4} = \frac{4800}{x} - 200 \\[1em] \Rightarrow \frac{4800}{x} - \frac{4800}{x + 4} = 200 \\[1em] \Rightarrow \frac{4800(x + 4) - 4800x}{x(x + 4)} = 200 \\[1em] \Rightarrow \frac{4800x + 19200 - 4800x}{x^2 + 4x} = 200 \\[1em] \Rightarrow 19200 = 200x^2 + 800x \\[1em] \Rightarrow 96 = x^2 + 4x \\[1em] \Rightarrow x^2 + 4x - 96 = 0 \\[1em] \Rightarrow x^2 + 12x - 8x - 96 = 0 \\[1em] \Rightarrow x(x + 12) - 8(x + 12) = 0 \\[1em] \Rightarrow (x + 12)(x - 8) = 0 \\[1em] \Rightarrow x = -12, 8\end{aligned}$$

No. of people cannot be negative,

∴ x = 8.

Hence, no. of people staying overnight = 8.

(i) Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken for onward journey = $\frac{400}{x}$ hours.

(ii) Time = $\frac{\text{Distance}}{\text{Speed}}$

Time taken for return journey = $\frac{400}{x + 40}$ hours.

According to question,

$$\begin{aligned}\Rightarrow \frac{400}{x} - \frac{400}{x + 40} = \frac{30}{60} \\[1em] \Rightarrow \frac{400}{x} - \frac{400}{x + 40} = \frac{1}{2} ......(i) \\[1em] \Rightarrow \frac{400(x + 40) - 400x}{x(x + 40)} = \frac{1}{2} ......(i) \\[1em] \Rightarrow \frac{400x + 16000 - 400x}{x^2 + 40x} = \frac{1}{2} ......(i) \\[1em] \Rightarrow \frac{16000}{x^2 + 40x} = \frac{1}{2} ......(i) \\[1em] \Rightarrow 32000 = x^2 + 40x \\[1em] \Rightarrow x^2 + 40x - 32000 = 0 \\[1em] \Rightarrow x^2 + 200x - 160 - 3200 = 0 \\[1em] \Rightarrow x(x + 200) - 160(x + 200) = 0 \\[1em] \Rightarrow (x - 160)(x + 200) = 0 \\[1em] \Rightarrow x - 160 = 0 \text{ or } x + 200 = 0 \\[1em] \Rightarrow x = 160 \text{ or } x = -200.\end{aligned}$$

Since speed cannot be negative,

∴ x = 160

Hence, required equation = $\frac{400}{x} - \frac{400}{x + 40} = \frac{1}{2}$ and speed = 160 km/hr.

Let usual speed be x km/hr

Time taken = $\frac{1500}{x}$ hours

If speed is increased by 250 km/hr then time taken = $\frac{1500}{x + 250}$ hours

Given, on increasing speed it takes 30 minutes less,

$$\begin{aligned}\therefore \frac{1500}{x} - \frac{1500}{x + 250} = \frac{30}{60} \\[1em] \Rightarrow \frac{1500(x + 250) - 1500x}{x(x + 250)} = \frac{1}{2} \\[1em] \Rightarrow \frac{1500x + 375000 - 1500x}{x^2 + 250x} = \frac{1}{2} \\[1em] \Rightarrow \frac{375000}{x^2 + 250x} = \frac{1}{2} \\[1em] \Rightarrow 750000 = x^2 + 250x \\[1em] \Rightarrow x^2 + 250x - 750000 = 0 \\[1em] \Rightarrow x^2 + 1000x - 750x - 750000 = 0 \\[1em] \Rightarrow x(x + 1000) - 750(x + 1000) = 0 \\[1em] \Rightarrow (x - 750)(x + 1000) = 0 \\[1em] \Rightarrow x - 750 = 0 \text{ or } x + 1000 = 0 \\[1em] \Rightarrow x = 750 \text{ or } x = -1000.\end{aligned}$$

Since, speed cannot be negative

∴ x ≠ -1000.

Hence, usual speed = 750 km/hr.

(i) Let no. of rows be x and no. of seats in each row = x.

Total no. of seats = x²

If row is doubled then rows = 2x
No. of seats in each row are reduced by 10 = (x - 10)

Total no. of seats = 2x(x - 10)

According to question,

⇒ 2x(x - 10) - x² = 300

⇒ 2x² - 20x - x² = 300

⇒ x² - 20x - 300 = 0

⇒ x² - 30x + 10x - 300 = 0

⇒ x(x - 30) + 10(x - 30) = 0

⇒ (x - 30)(x + 10) = 0

⇒ x - 30 = 0 or x + 10 = 0

⇒ x = 30 or x = -10.

Since, no. of rows cannot be negative,

∴ x = 30

Hence, no. of rows in original arrangement = 30.

(ii) No. of seats after rearrangement = 2x(x - 10) = 2(30)(30 - 10) = 2 × 30 × 20 = 1200.

Hence, no. of seats after rearrangement = 1200.

Let numerator be x and denominator = x + 3.

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

New fraction's numerator = x - 1 and denominator = x + 3 - 1 = x + 2

New Fraction = $\frac{x - 1}{x + 2}$

According to question,

$$\begin{aligned}\Rightarrow \frac{x}{x + 3} - \frac{x - 1}{x + 2} = \frac{1}{14} \\[1em] \Rightarrow \frac{x(x + 2) - (x - 1)(x + 3)}{(x + 2)(x + 3)} = \frac{1}{14} \\[1em] \Rightarrow \frac{x^2 + 2x - (x^2 + 3x - x - 3)}{x^2 + 3x + 2x + 6} = \frac{1}{14} \\[1em] \Rightarrow \frac{x^2 + 2x - x^2 - 2x + 3}{x^2 + 5x + 6} = \frac{1}{14} \\[1em] \Rightarrow \frac{3}{x^2 + 5x + 6} = \frac{1}{14} \\[1em] \Rightarrow 42 = x^2 + 5x + 6 \\[1em] \Rightarrow x^2 + 5x + 6 - 42 = 0 \\[1em] \Rightarrow x^2 + 5x - 36 = 0 \\[1em] \Rightarrow x^2 + 9x - 4x - 36 = 0 \\[1em] \Rightarrow x(x + 9) - 4(x + 9) = 0 \\[1em] \Rightarrow (x + 9)(x - 4) = 0 \\[1em] \Rightarrow x + 9 = 0 \text{ or } x - 4 = 0 \\[1em] \Rightarrow x = -9 \text{ or } x = 4.\end{aligned}$$

Since fraction is positive,

∴ x ≠ -9.

Fraction = $\frac{x}{x + 3} = \frac{4}{7}$.

Hence, fraction = $\frac{4}{7}$.

Let ten's digit be x and one's digit be y.

According to question,

⇒ xy = 27 ........(i)

⇒ x - y = 6

⇒ x = 6 + y .......(ii)

Substituting value of x from (ii) in (i) we get,

⇒ (6 + y)y = 27

⇒ y² + 6y = 27

⇒ y² + 6y - 27 = 0

⇒ y² + 9y - 3y - 27 = 0

⇒ y(y + 9) - 3(y + 9) = 0

⇒ (y - 3)(y + 9) = 0

⇒ y - 3 = 0 or y + 9 = 0

⇒ y = 3 or y = - 9.

Since digits cannot be negative,

∴ y ≠ -9

x = 6 + y = 6 + 3 = 9

Number = 10(x) + y = 10(9) + 3 = 93.

Hence, number = 93.

No. of children = x and money = ₹ 480

Each children received = ₹ $\frac{480}{x}$

No. of new children = x + 20

Each children received = ₹ $\frac{480}{x + 20}$

According to question,

$$\begin{aligned}\Rightarrow \frac{480}{x} - \frac{480}{x + 20} = 12 \\[1em] \Rightarrow \frac{480(x + 20) - 480x}{x(x + 20)} = 12 \\[1em] \Rightarrow \frac{480x + 9600 - 480x}{x^2 + 20x} = 12 \\[1em] \Rightarrow \frac{9600}{x^2 + 20x} = 12 \\[1em] \Rightarrow 9600 = 12(x^2 + 20x) \\[1em] \Rightarrow 12x^2 + 240x - 9600 = 0 \\[1em] \Rightarrow 12(x^2 + 20x - 800) = 0 \\[1em] \Rightarrow x^2 + 20x - 800 = 0 \\[1em] \Rightarrow x^2 + 40x - 20x - 800 = 0 \\[1em] \Rightarrow x(x + 40) - 20(x + 40) = 0 \\[1em] \Rightarrow (x - 20)(x + 40) = 0 \\[1em] \Rightarrow x - 20 = 0 \text{ or } x + 40 = 0 \\[1em] \Rightarrow x = 20 \text{ or } x = -40.\end{aligned}$$

Since no. of children cannot be negative,

∴ x ≠ -40.

Hence, x = 20.

Let the length and the breadth of rectangular plot be l m and b m.

Area of rectangle: l × b = 24 .....(1)

Perimeter of rectangle = 20 m

⇒ 2(l + b) = 20

⇒ (l + b) = $\frac{20}{2}$

⇒ (l + b) = 10

⇒ b = 10 - l

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

⇒ l × (10 - l) = 24

⇒ 10l - l² = 24

⇒ l² - 10l + 24 = 0

⇒ l² - 6l - 4l + 24 = 0

⇒ l(l - 6) - 4(l - 6) = 0

⇒ (l - 4)(l - 6) = 0

⇒ (l - 4) = 0 or (l - 6) = 0

⇒ l = 4 or 6.

Length of rectangle = 6 m or 4 m

Hence, the length of the rectangular plot is 6 m or 4 m.

Let ABCD be a rectangular garden.

Let the width of the walk be x m

ABCD is the rectangular garden,

Length of rectangular garden = 16 m

Breadth of rectangular garden = 10 m

From figure,

Length of PQRS = 16 + x + x = (16 + 2x) m

Breadth of PQRS = 10 + x + x = (10 + 2x) m

Area of PQRS = (16 + 2x)(10 + 2x)

Area of rectangular garden ABCD = 16 × 10 = 160 m²

Area of concrete walk = 120

Now,

From figure,

⇒ Area of walk = Area of rectangle PQRS - Area of rectangle ABCD

⇒ (16 + 2x)(10 + 2x) - 160 = 120

⇒ (16 + 2x)(10 + 2x) = 120 + 160

⇒ (16 + 2x)(10 + 2x) = 280

⇒ 160 + 32x + 20x + 4x² = 280

⇒ 4x² + 52x + 160 = 280

⇒ 4x² + 52x + 160 - 280 = 0

⇒ 4x² + 52x - 120 = 0

⇒ 4(x² + 13x - 30) = 0

⇒ x² + 13x - 30 = 0

⇒ x² + 15x - 2x - 30 = 0

⇒ x(x + 15) - 2(x + 15) = 0

⇒ (x - 2)(x + 15) = 0

⇒ (x - 2) = 0 or (x + 15) = 0

⇒ x = 2 or x = -15

Since width cannot be negative,

x = 2 m.

Hence, the width of the concrete walk is 2 m.

(i) Given,

The height of the stone above the ground after t seconds: h (t) = 5t² + 30t + 2

t = 2

Substituting value of t = 2 in h(t):

⇒ h(2) = 5(2)² + 30(2) + 2

= 5(4) + 60 + 2

= 20 + 60 + 2

= 82 m.

Hence, the height after 2 seconds is 82 m.

(ii) The initial height of stone,

Substituting value of t = 0 in h(t):

⇒ h(0) = 5(0)² + 30(0) + 2

= 2 m.

Hence, stone is thrown from a height of 2 m.

(iii) Given,

h(t) = 37

⇒ 5t² + 30t + 2 = 37

⇒ 5t² + 30t + 2 - 37 = 0

⇒ 5t² + 30t - 35 = 0

⇒ 5(t² + 6t - 7) = 0

⇒ t² + 6t - 7 = 0

⇒ t² + 7t - t - 7 = 0

⇒ t(t + 7) - 1(t + 7) = 0

⇒ (t - 1)(t + 7) = 0

⇒ (t - 1) = 0 or (t + 7) = 0

⇒ t = 1 or t = -7

Since time cannot be negative,

t = 1 second.

Hence, stone will be 37 m above the ground after 1 second.

(i) Given,

The speed of Malik's jeep = x km/h.

Speed of Rakesh's jeep = (x - 10) km/h

Distance covered by Rakesh in 5 hours

= 5(x − 10) km

Hence, distance covered by Rakesh in 5 hours = 5(x − 10) km.

(ii) The distance between Gurugram and Ahmedabad = 975 km

Time taken by Malik = $\frac{975}{x}$

Time taken by Rakesh = $\frac{975}{x - 10}$

Given, that Rakesh took 2 hours more than Malik to reach Ahmedabad,:

$\Rightarrow \frac{975}{x - 10} = \frac{975}{x}$ + 2

$\Rightarrow \frac{975}{x - 10} = \frac{975 + 2x}{x}$

⇒ 975x = 975(x − 10) + 2x(x − 10)

⇒ 975x = 975x − 9750 + 2x² − 20x

⇒ 0 = 2x² − 20x − 9750

⇒ 0 = 2(x² − 10x − 4875)

⇒ x² − 10x − 4875 = 0

Hence, quadratic equation is x² − 10x − 4875 = 0.

(iii) From part (ii),

⇒ x² − 75x + 65x − 4875 = 0

⇒ x(x - 75) + 65(x − 75) = 0

⇒ (x − 75)(x + 65) = 0

⇒ (x − 75) = 0 or (x + 65) = 0

⇒ x = 75 or −65

Since speed cannot be negative,

x = 75 km/h

Hence, speed of Malik's jeep = 75 km/h.

(iv) Speed of Rakesh's jeep = x - 10

= 75 - 10 = 65 km/h.

Time taken by Rakesh to reach Ahmedabad =

= $\frac{975}{65}$

= 15 hours.

Hence, Rakesh takes 15 hours to reach Ahmedabad.