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

The product of two whole numbers, each greater than 4, is 35; the numbers are :

1. -7 and -5

2. 7 and 5

3. 7 and -5

4. -7 and 5

Answer

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

Hence, Option 2 is the correct option.

The difference between digits of a two-digit number is 2 and the product of digits is 24. If tens digit is bigger, the number is :

1. 24

2. 26

3. 46

4. 64

Answer

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.

If 18 is added to a two-digit number, its digits are reversed. If the product of the digits of the number is 24, the number is :

1. 46

2. 64

3. 56

4. 48

Answer

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.

Three years hence, the ages of Radha and Meena will be in the ratio 5 : 2. If the sum of their present ages is 62 years, their ages, 2 years ago, were:

1. 43 years and 19 years

2. 41 years and 17 years

3. 40 years and 16 years

4. none of these

Answer

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.

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

The sum of their present ages = 62 years.

⇒ a + b = 62

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

⇒ $\dfrac{5b + 9}{2}$ + b = 62

⇒ 5b + 9 + 2b = 124

⇒ 7b + 9 = 124

⇒ 7b = 124 - 9

⇒ 7b = 115

⇒ b = $\dfrac{115}{7}$

Substituting the value of b in equation (1),

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

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

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

Hence, option 4 is the correct option.

₹ 320 were divided equally among 8 children. If 2 more children join them, each will get, how many rupees more or less ?

1. ₹ 40 more

2. ₹ 32 more

3. ₹ 8 less

4. ₹ 8 more

Answer

Given,

₹ 320 were divided equally among 8 children.

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

If 2 more children join, total children = 10.

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

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

Hence, Option 3 is the correct option.

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.

The ages of two sisters are 11 years and 14 years. In how many years time will the product of their ages be 304 ?

Answer

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.

One year ago, a man was 8 times as old as his son. Now his age is equal to square of his son's age. Find their present ages.

Answer

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.

The age of a father is twice the square of the age of his son. Eight years hence, the age of the father will be 4 years more than three times the age of the son. Find their present ages.

Answer

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 = $-\dfrac{5}{2}$ or x = 4.

Since, x cannot be negative

∴ x ≠ $-\dfrac{5}{2}$

⇒ 2x² = 2(4)² = 2(16) = 32 years.

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

Mr. Mehra sends his servant to the market to buy oranges worth ₹ 15. The servant having eaten three oranges on the way, Mr. Mehra pays 25 paise per orange more than the market price. Taking x to be the number of oranges which Mr. Mehra receives, form a quadratic equation in x. Hence, find the value of x.

Answer

No. of oranges Mr. Mehra receives = x

Total no. of oranges = x + 3

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

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

According to question,

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

Since, no. of oranges cannot be negative,

∴ x ≠ -15.

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

₹ 250 is divided equally among a certain number of children. If there were 25 children more, each would have received 50 paise less. Find the number of children.

Answer

Let no. of children be x,

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

If 25 children are increased then,

No. of children = x + 25

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

According to question,

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

Since no. of children cannot be negative,

∴ x = 100

Hence, no. of children = 100.

An employer finds that if he increases the weekly wages of each worker by ₹ 5 and employs five workers less, he increases his weekly wage bill from ₹ 3150 to ₹ 3250. Taking the original weekly wage of each worker as ₹ x; obtain an equation in x and then solve it to find the weekly wages of each worker.

Answer

Initial weekly wage = ₹ 3150,

Let weekly wage of each worker be ₹ x,

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

New wage = ₹ (x + 5)

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

Given, new total wage = ₹ 3250

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

Since wage cannot be negative,

∴ x ≠ -70.

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

A trader bought a number of articles for ₹ 1200. Ten were damaged and he sold each of the remaining articles at ₹ 2 more than what he paid for it, thus getting a profit of ₹ 60 on the whole transaction.

Taking the number of articles he bought as x, form an equation in x and solve it.

Answer

Let no. of articles bought be x,

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

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

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

Profit = ₹ 60

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

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.

The total cost price of a certain number of identical articles is ₹ 4800. By selling the articles at ₹ 100 each, a profit equal to the cost price of 15 articles is made. Find the number of articles bought.

Answer

Let no. of articles bought be x.

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

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

We know that,

Profit = S.P. - C.P.

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

Since, no. of articles cannot be negative,

∴ x ≠ -12.

Hence, no. of articles bought = 60.