Solving Problems (Based on Quadratic Equations) — Exercise 6(B)

The sum of the numerator and the denominator is 8 and their product is 15. Then the fraction is:

1. $\dfrac{5}{3}$ and $\dfrac{3}{5}$

2. $\dfrac{5}{3}$ or $\dfrac{3}{5}$

3. 3 and 5

4. 3 or 5

Answer

$\dfrac{5}{3}$ or $\dfrac{3}{5}$

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 = $\dfrac{5}{3} \text{ or } \dfrac{3}{5}$.

Hence, option 2 is the correct option.

The sum of the digits of a two digit number is 9 and the product of the digits is 20. If the unit digit is greater than the tens digit. The number is

1. 45

2. 54

3. none of these

Answer

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.

Two whole numbers are in ratio 3:2. If the sum of their square is 52, the numbers are:

1. 9 and 6

2. 6 and 4

3. 9 and 4

4. none of these

Answer

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 = $\dfrac{52}{13}$

⇒ 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.

A two digit number is 5 times the sum of its digits. The number is:

1. 63

2. 36

3. 45

4. 54

Answer

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 = $\dfrac{5}{4}$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 = $\dfrac{5}{4} \times$ 4 = 5

For x = 8, y = $\dfrac{5}{4} \times$ 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.

Three positive numbers are in ratio 4 : 3 : 2. If the difference between square of the largest and the smallest number is 48, the numbers are;

1. 48, 36 and 24

2. 24, 18 and 12

3. 8, 6 and 2

4. 8, 6 and 4

Answer

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 = $\dfrac{48}{12}$

⇒ 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.

The numerator of a fraction is 3 less than its denominator. If one is added to the denominator, the fraction is decreased by $\dfrac{1}{15}$. Find the fraction.

Answer

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

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 $\dfrac{1}{15}$.

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

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

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

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

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

The denominator of a fraction is 3 more than its numerator. The sum of the fraction and its reciprocal is $2\dfrac{9}{10}$. Find the fraction.

Answer

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

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 $2\dfrac{9}{10}$.

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

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

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

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

Since, denominator is 3 more than the numerator,

∴ The fraction is $\dfrac{2}{5}$

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

The product of the digits of a two digit number is 24. If it's unit's digits exceeds twice it's ten's digit by 2; find the number.

Answer

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.

A two-digit number is such that the product of its digit is 18. When 63 is subtracted from the number, the digits are reserved. Find the number.

Answer

Let tens digit be x and units digit be y.

Given,

$xy = 18 \\[1em] \Rightarrow y = \dfrac{18}{x}$

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

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

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

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

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

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

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

Hence, the number = 92.

The ratio between two positive numbers is $\dfrac{1}{5}:\dfrac{1}{7}$. If the sum of the squares of the numbers is 666, find the numbers.

Answer

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

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

The sum of the squares of the numbers = 666.

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

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

Hence, the two numbers = 21 and 15.

The ratio between three positive numbers is $\dfrac{1}{4} : \dfrac{1}{3} : \dfrac{1}{2}$. When the square of the middle number is subtracted from the sum of the squares of the other, the result is 725. Find the numbers.

Answer

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

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

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

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

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

Hence, the number = 15, 20 and 30.

The numerator of a fraction is 3 less than its denominator. If 2 is added to both the numerator and denominator, the sum of the new fraction and the original fraction is $1\dfrac{9}{20}$. Find the new fraction.

Answer

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

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 $1\dfrac{9}{20}$

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

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

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

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

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

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

A two digit number is 4 times the sum of its digits and twice the product of its digits. Find the number.

Answer

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.

27 is divided into two parts such that the sum of their reciprocal is $\dfrac{3}{20}$. Find the ratio between the numbers (5 : 4).

Answer

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

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

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

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.