Problems on Quadratic Equations — Multiple Choice Questions

If the sum of two natural numbers is 27 and their product is 182, then the smaller number is :

1. 13

2. 14

3. 16

4. 18

Answer

Let two natural numbers be x and y.

Given,

Sum of numbers = 27

⇒ x + y = 27

⇒ y = 27 - x     .........(1)

Given,

Product of numbers is 182.

⇒ xy = 182     .........(2)

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

⇒ x(27 - x) = 182

⇒ 27x - x² = 182

⇒ x² - 27x + 182 = 0

⇒ x² - 13x - 14x + 182 = 0

⇒ x(x - 13) - 14(x - 13) = 0

⇒ (x - 14)(x - 13) = 0

⇒ (x - 14) = 0 or (x - 13) = 0     [Using zero-product rule]

⇒ x = 14 or x = 13.

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

If x = 14, y = 27 − 14 = 13

If x = 13, y = 27 − 13 = 14.

The smallest number among two numbers is 13.

Hence, option 1 is the correct option.

The sum of the squares of two consecutive odd natural numbers is 74. The greater number is :

1. 5

2. 7

3. 9

4. none of these

Answer

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

Given,

Sum of the squares of two consecutive odd natural numbers is 74.

⇒ x² + (x + 2)² = 74

⇒ x² + x² + 4 + 4x = 74

⇒ 2x² + 4x + 4 - 74 = 0

⇒ 2x² + 4x - 70 = 0

⇒ 2x² + 14x - 10x - 70 = 0

⇒ 2x(x + 7) - 10(x + 7) = 0

⇒ (2x - 10)(x + 7) = 0

⇒ (2x - 10) = 0 or (x + 7) = 0     [Using zero-product rule]

⇒ 2x = 10 or x = -7

⇒ x = $\dfrac{10}{2}$ or x = -7

⇒ x = 5 or x = -7

Since the number required is natural number, thus x ≠-7.

x + 2 = 5 + 2 = 7.

The greater number among the two numbers is 7.

Hence, option 2 is the correct option.

Two natural numbers differ by 2 and the sum of their squares is 202. The sum of the numbers is :

1. 14

2. 16

3. 18

4. 20

Answer

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

Given,

Sum of the squares of numbers is 202.

⇒ x² + (x + 2)² = 202

⇒ x² + x² + 4 + 4x = 202

⇒ 2x² + 4x + 4 - 202 = 0

⇒ 2x² + 4x - 198 = 0

⇒ 2x² + 22x - 18x - 198 = 0

⇒ 2x(x + 11) - 18(x + 11) = 0

⇒ (2x - 18)(x + 11) = 0

⇒ (2x - 18) = 0 or (x + 11) = 0     [Using zero-product rule]

⇒ 2x = 18 or x = -11

⇒ x = $\dfrac{18}{2}$ or x = -11

⇒ x = 9 or x = -11.

Since, the number required is natural number, thus x ≠ -11,

⇒ x + 2 = 9 + 2 = 11

Sum of the two numbers = 9 + 11 = 20.

Hence, option 4 is the correct option.

₹ 40 is distributed between two friends such that the product of their shares is 364. The difference of their shares is:

1. ₹ 8

2. ₹ 10

3. ₹ 12

4. ₹ 14

Answer

Given,

The total amount of money = ₹ 40.

Let the shares of two friends be ₹ x and ₹ y respectively.

⇒ x + y = 40

⇒ y = 40 - x     .........(1)

Given,

The product of two parts of amount distributed is 364.

⇒ xy = 364     .........(2)

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

⇒ x(40 - x) = 364

⇒ 40x - x² = 364

⇒ x² - 40x + 364 = 0

⇒ x² - 26x - 14x + 364 = 0

⇒ x(x - 26) - 14(x - 26) = 0

⇒ (x - 14)(x - 26) = 0

⇒ (x - 14) = 0 or (x - 26) = 0     [Using zero-product rule]

⇒ x = 14 or x = 26

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

Case 1: If x = 14, y = 40 − 14 = 26

Case 2: If x = 26, y = 40 − 26 = 14.

The difference between the two parts of amount is, ₹ 26 - ₹ 14 = ₹ 12.

Hence, option 3 is the correct option.

The length of a rectangle is 4 cm more than its breadth. If the area of the rectangle is 96 cm², then the perimeter of the rectangle is :

1. 36 cm

2. 40 cm

3. 44 cm

4. 48 cm

Answer

Let the breadth and length of a rectangle be x cm and (x + 4) cm.

Given,

Area of rectangle = 96 cm².

⇒ x(x + 4) = 96

⇒ x² + 4x = 96

⇒ x² + 4x - 96 = 0

⇒ x² - 8x + 12x - 96 = 0

⇒ x(x - 8) + 12(x - 8) = 0

⇒ (x + 12)(x - 8) = 0

⇒ (x + 12) = 0 or (x - 8) = 0     [Using zero-product rule]

⇒ x = -12 or x = 8

Since length and breadth cannot be negative, thus Breadth = x = 8 cm.

Length = x + 4 = 8 + 4 = 12 cm

Perimeter of the rectangle = 2(l + b)

= 2(12 + 8)

= 2(20)

= 40 cm.

Hence, option 2 is the correct option.

If four times the area of a square is 484 cm², then perimeter of the square is :

1. 32 cm

2. 48 cm

3. 40 cm

4. 44 cm

Answer

Let the area of square be x cm²,

Given,

Four times area of square = 484 cm²

⇒ 4x = 484

⇒ x = $\dfrac{484}{4}$

⇒ x = 121 cm²

Area of Square = 121 cm²

Let side of square be a cm.

⇒ a² = 121

⇒ a = $\sqrt{121}$

⇒ a = ± 11

Since, length cannot be negative, thus a = 11 cm.

Perimeter of Square = 4 × a

⇒ 4 × 11

⇒ 44 cm.

Hence, option 4 is the correct option.

Sum of the squares of the two consecutive positive integers is 365. The sum of the numbers is :

1. 27

2. 31

3. 25

4. 29

Answer

Let two consecutive positive integers be x and x + 1.

Given,

The sum of squares of the two consecutive positive integers = 365.

⇒ x² + (x + 1)² = 365

⇒ x² + x² + 2x + 1 = 365

⇒ 2x² + 2x + 1 - 365 = 0

⇒ 2x² + 2x - 364 = 0

⇒ 2(x² + x - 182) = 0

⇒ x² + x - 182 = 0

⇒ x² + 14x - 13x - 182 = 0

⇒ x(x + 14) - 13(x + 14) = 0

⇒ (x - 13)(x + 14) = 0

⇒ (x - 13) = 0 or (x + 14) = 0     [Using zero -product rule]

⇒ x = 13 or x = -14.

Since, they are consecutive positive integers, x ≠ -14.

⇒ x + 1 = 13 + 1 = 14.

The sum of numbers is = 13 + 14 = 27.

Hence, option 1 is the correct option.

The altitude of a right triangle is 17 cm less than its base. If the hypotenuse is 25 cm, then the perimeter of the triangle is :

1. 48 cm

2. 56 cm

3. 54 cm

4. 64 cm

Answer

Let the base and height of right triangle be x cm and (x - 17) cm respectively.

By pythagoras theorem,

⇒ Base² + Height² = Hypotenuse²

⇒ x² + (x - 17) ² = (25)²

⇒ x² + x² + (17)² - 2 × x × 17 = 625

⇒ x² + x² + 289 - 34x = 625

⇒ 2x² - 34x + 289 - 625 = 0

⇒ 2x² - 34x - 336 = 0

⇒ 2(x² - 17x - 168) = 0

⇒ x² - 17x - 168 = 0

⇒ x² - 24x + 7x - 168 = 0

⇒ x(x - 24) + 7(x - 24) = 0

⇒ (x + 7)(x - 24) = 0

⇒ (x + 7) = 0 or (x - 24) = 0     [Using zero -product rule]

⇒ x = -7 or x = 24.

Since, the length of triangle cannot be negative, x ≠ -7.

x - 17 = 24 - 17 = 7.

The perimeter of right triangle is = 7 + 24 + 25 = 56 cm.

Hence, option 2 is the correct option.

The cost of an article is ₹ 3 more than twice the total number of articles. If the cost of all the articles is ₹ 189, then the number of articles is :

1. 7

2. 9

3. 11

4. 13

Answer

Let the total number of articles be x and the cost of each article be y.

Given,

The cost of an article is ₹ 3 more than twice the total number of articles.

⇒ y = 2x + 3     .........(1)

Given,

Total cost of all articles = ₹ 189.

⇒ xy = 189     .........(2)

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

⇒ x(2x + 3) = 189

⇒ 2x² + 3x = 189

⇒ 2x² + 3x - 189 = 0

⇒ 2x² - 18x + 21x - 189 = 0

⇒ 2x(x - 9) - 21(x - 9) = 0

⇒ (2x - 21)(x - 9) = 0

⇒ (2x - 21) = 0 or (x - 9) = 0     [Using zero -product rule]

⇒ 2x = 21 or x = 9

⇒ x = $\dfrac{21}{2}$ or x = 9

Since, the number of articles cannot be in fraction, thus x ≠ $\dfrac{21}{2}$.

Hence, option 2 is the correct option.

The diagonal of a rectangular field is 60 m more than the shorter side. If the longer side is 30 m more than the shorter side, then the sides are :

1. 60 m, 90 m

2. 80 m, 110 m

3. 90 m, 120 m

4. 110 m, 140 m

Answer

Let the shorter side of rectangular field be x meters.

Given,

The diagonal of rectangular field is 60 m more than shorter side.

Diagonal = (x + 60) meters

Given,

The longer side of rectangle is 30 m more than shorter side, Let the longer side be z,

Longer side = (x + 30) meters

By pythagoras theorem,

In a rectangular field,

⇒ Hypotenuse² = Shorter side² + Longer side²

⇒ (x + 60)² = x² + (x + 30)²

⇒ [x² + (60)² + 2 × x × 60] = x² + [x² + (30)² + 2 × x × 30]

⇒ x² + 3600 + 120x = x² + x² + 900 + 60x

⇒ x² + 3600 + 120x = 2x² + 900 + 60x

⇒ 2x² + 900 + 60x - x² - 3600 - 120x = 0

⇒ 2x² - x² + 60x - 120x - 3600 + 900 = 0

⇒ x² - 60x - 2700 = 0

⇒ x² - 90x + 30x - 2700 = 0

⇒ x(x - 90) + 30(x - 90) = 0

⇒ (x + 30)(x - 90) = 0

⇒ (x + 30) = 0 or (x - 90) = 0     [Using zero -product rule]

⇒ x = -30 or x = 90

Since length of rectangle cannot be negative x ≠ -30

The longer side of rectangle is,

x + 30 = 90 + 30 = 120 meters.

Thus, sides are 90 m and 120 m.

Hence, option 3 is the correct option.

Neha’s father is 28 years older than her. The product of their ages (in years) 4 years ago was 245. If present age of Neha is x years, then the algebraic representation of this information in the form of quadratic equation is:

1. x² − 20x − 341 = 0

2. x² + 20x − 341 = 0

3. x² + 20x + 341 = 0

4. x² − 20x + 341 = 0

Answer

Let Neha's present age be x and the age of her father be y.

Given,

Neha's father is 28 years older than her.

y = 28 + x     .........(1)

Given,

Product of their ages 4 years ago was 245.

⇒ (x - 4)(y - 4) = 245     .........(2)

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

⇒ (x - 4)(28 + x - 4) = 245

⇒ (x - 4)(x + 24) = 245

⇒ x² + 24x - 4x - 96 = 245

⇒ x² + 20x - 96 - 245 = 0

⇒ x² + 20x - 341 = 0.

Hence, option 2 is the correct option.

A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. If the initial speed of the train is x km/hr, then representation of this information algebraically is :

1. x² − 8x − 1280 = 0

2. x² + 8x + 1280 = 0

3. x² − 8x + 1280 = 0

4. x² + 8x − 1280 = 0

Answer

By formula,

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

Initial speed of train = x km/hr

Time taken to cover 480 km = $\dfrac{480}{x}$ hrs

Reduced speed of train = (x - 8) km/hr

Time taken to cover 480 km = $\dfrac{480}{x - 8}$ hrs

Given,

On reducing speed the time taken is 3 hours more.

$\Rightarrow \dfrac{480}{x - 8} - \dfrac{480}{x} = 3 \\[1em] \Rightarrow \dfrac{480x - 480(x - 8)}{x(x - 8)} = 3 \\[1em] \Rightarrow \dfrac{480x - 480x + 3840}{x^2 - 8x} = 3 \\[1em] \Rightarrow 3840 = 3(x^2 - 8x) \\[1em] \Rightarrow 3840 = 3x^2 - 24x \\[1em] \Rightarrow 3x^2 - 24x - 3840 = 0 \\[1em] \Rightarrow 3(x^2 - 8x - 1280) = 0 \\[1em] \Rightarrow x^2 - 8x - 1280 = 0.$

Hence, option 1 is the correct option.

Two cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then :

The number of litres of diesel used by car X is :

1. $\dfrac{72}{x - 3}$ litres

2. $\dfrac{72}{x + 3}$ litres

3. $\dfrac{72}{x}$ litres

4. $\dfrac{12}{x}$ litres

Answer

Given,

Distance covered by cars = 72 km

Given,

Car X travels x km using 1 litre of diesel.

Therefore, to travel 1 km car X uses $\dfrac{1}{x}$ litres of diesel.

To travel 72 km car X will use $\dfrac{72}{x}$ litres of diesel.

Hence, option 3 is the correct option.

Two cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then :

The number of litres of diesel used by car Y is:

1. $\dfrac{72}{x - 3}$ litres

2. $\dfrac{72}{x + 3}$ litres

3. $\dfrac{72}{x}$ litres

4. $\dfrac{12}{x + 3}$ litres

Answer

Given,

Distance covered by cars = 72 km

Given,

Car Y travels x + 3 km using 1 litre of diesel.

Therefore, to travel 1 km car Y uses $\dfrac{1}{x + 3}$ litres of diesel.

To travel 72 km car Y will use $\dfrac{72}{x + 3}$ litres of diesel.

Hence, option 2 is the correct option.

Two cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then :

If car X used 4 litres of diesel more than car Y in the journey, then :

1. $\dfrac{72}{x - 3} - \dfrac{12}{x} = 4$

2. $\dfrac{72}{x + 3} - \dfrac{12}{x} = 4$

3. $\dfrac{72}{x} - \dfrac{72}{x + 3} = 4$

4. $\dfrac{72}{x - 3} - \dfrac{72}{x + 3} = 4$

Answer

Diesel used by car X = $\dfrac{72}{x}$ litres

Diesel used by car Y = $\dfrac{72}{x + 3}$ litres

Given,

Car X used 4 litres of diesel more than car Y in the journey.

$\therefore \dfrac{72}{x} - \dfrac{72}{x + 3} = 4$

Hence, option 3 is the correct option.

Two cars X and Y use 1 litre of diesel to travel x km and (x + 3) km respectively. If both the cars covered a distance of 72 km, then :

The amount of diesel used by car X is:

1. 6 litres

2. 12 litres

3. 18 litres

4. 24 litres

Answer

Solving,

$\Rightarrow \dfrac{72}{x} - \dfrac{72}{x + 3} = 4 \\[1em] \Rightarrow \dfrac{72(x + 3) - 72x}{x(x + 3)} = 4 \\[1em] \Rightarrow \dfrac{72x + 216 - 72x}{x^2 + 3x} = 4 \\[1em] \Rightarrow 216 = 4(x^2 + 3x) \\[1em] \Rightarrow \dfrac{216}{4} = (x^2 + 3x) \\[1em] \Rightarrow 54 = (x^2 + 3x) \\[1em] \Rightarrow x^2 + 3x - 54 = 0 \\[1em] \Rightarrow x^2 + 9x - 6x - 54 = 0 \\[1em] \Rightarrow x(x + 9) - 6(x + 9) = 0 \\[1em] \Rightarrow (x - 6)(x + 9) = 0 \\[1em] \Rightarrow (x - 6) = 0 \text{ or } (x + 9) = 0 \text{....[Using zero-product rule]}\\[1em] \Rightarrow x = 6 \text{ or } x = -9.$

Distance covered cannot be negative.

Thus, x = 6.

Diesel used by car X = $\dfrac{72}{x}$

= $\dfrac{72}{6}$

= 12 litres.

Hence, option 2 is the correct option.