Explore topic-wise InterviewSolutions in .

Given: The sum of a number and its positive square root is 6/25. 

To find: the number.Solution:Let the number be ‘a’.

⇒ a + √a = 6/25 

⇒ √a = (6/25) – a 

Squaring both sides

⇒ a = 36/625 + a² – 12a/25 

⇒ a² – 37a/25 + 36/625 = 0

factorise by splitting the middle term. 

⇒ a² – a/25 – 36a/25 + 36/625 = 0 

⇒ a(a – 1/25) – (36/25) × (a – 1/25) = 0 

⇒ (a – 36/25)(a – 1/25) = 0 

⇒ a = 36/25 , 1/25 

But only 1/25 is possible as its sum with its positive root is 6/25.

Hence the number is 1/25.

(b) 23

Suppose the number of students in each row = x 

Then, number of students in each column = x + 3

⇒ x(x + 3) = 460 ⇒ x² + 3x - 460 = 0

⇒ (x + 23)(x - 20) = 0 ⇒ x = - 23, 20

Rejecting negative value, number of students in each row = 20 

∴ No. of students in each column = 23.

(b) 24

Let the tens' digit be x. Then, ones' digit = \(\frac8x\) 

Original number = 10x + \(\frac8x\) = \(\frac{10x^2+8}{x}\)

Number after reversing

= \(10\times\frac8x+x=\frac{80}x+x=\frac{80+x^2}{x}\)

Given, \(\frac{10x^2+8}{x}\) + 18 = \(\frac{80 +x^2}{x}\)

⇒ \(\frac{10x^2 + 8 + 18x}{x}\) = \(\frac{80 +x^2}{x}\) 

⇒ 9x² + 18x - 72 = 0

⇒ 9x² + 36x -18x - 72 = 0

⇒ 9x(x + 4) - 18 (x + 4) = 0

⇒ (9x - 18)(x + 4) = 0

⇒ 9x -18 = 0 or x + 4 = 0

⇒ x = 2, - 4

Rejecting the negative value, x = 2

∴ Original number = 10 x 2 + \(\frac82\) = 20 + 4 = 24.

(d) \(\frac1{25}\)

Let the number be x.

Then x + \(\sqrt{x}= \frac{6}{25}\)

⇒ 25x + 25\(\sqrt{x} = 6\)

Let \(\sqrt{x}\) = y. Then, the equation becomes

25y² + 25y - 6 = 0

⇒ 25y² + 30y - 5y - 6 = 0

⇒ 5y(5y + 6) - 1(5y + 6) = 0

⇒ (5y + 6)(5y - 1) = 0 ⇒ y = \(-\frac65,\frac15\)

Rejecting the negative value , \(\sqrt{x}\) = \(\frac15\) ⇒ x = \(\frac1{25}.\)

(c) 24 years

Let the daughter's age be x years. Then, the mother's age = (\(x\) + 21) years 

According to the question,

(\(x\) + 21) \(-\frac1{12}\) x \(x\) x (\(x\) + 21) = 18

⇒ 12(x + 21 ) - (x² + 21x) = 216

⇒ 12x + 252 - x² - 21x = 216

⇒ x² + 9x - 36 = 0

⇒ (x + 12 )(x - 3) = 0

⇒ x = -12 , 3

 Rejecting negative value, daughter's age = 3 years

∴ Mother's age = 24 years

Let Mr. Kasam make x number of pots on daily basis. 

Production cost of each pot = Rs. (10x + 40) 

According to the given condition, 

x(10x + 40) = 600 

∴ 10x² + 40x = 600 

∴ 10x² + 40x- 600 = 0 

∴ x² + 4x – 60 = 0 …[Dividing both sides by 10]

∴ x² + 10x – 6x – 60 = 0 

∴ x(x + 10) – 6(x + 10) = 0 

∴ (x + 10) (x – 6) = 0 

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get

∴ x + 10 = 0 or x – 6 = 0 

∴ x = – 10 or x = 6 

But, number of pots cannot be negative. 

∴ x = 6 

∴ Production cost of each pot = 7(10 x + 40)

= Rs. [(10×6)+ 40] 

= Rs.(60 + 40) = Rs. 100 

Production cost of one pot is Rs.100 and the number of pots Mr. Kasam makes per day is 6.

Let the two consecutive positive integers be x and (x + 1)

According to the given condition,

x(x + 1) = 306

⇒ x² + x - 306 = 0

⇒ x² + 18x - 17x - 306 = 0

⇒ x(x + 18) - 17(x + 18) = 0

⇒ (x + 18)(x - 17) = 0

⇒ x + 18 = 0 or x - 17 = 0

⇒ x = - 18  or x = 17

∴ x = 17  (x is a positive integers)

When x = 17,

x + 1 = 17 + 1 = 18

Hence, the required integers are 17 and 18.

Given that x denotes the number of toys produced in a day.

So, the cost of production of each toy = (55 – x)

And, the total cost of production is the product of number of toys produced in a day and cost of production of each toy i.e, x (55 – x)

From the question, it’s given that

The total cost of production on that particular day is Rs.750

So,

⇒ x (55 – x) = 750

⇒ 55x – x² = 750

⇒ x² – 55x + 750 = 0

Thus, the required quadratic equation is x² – 55x + 750 = 0.

Let the number of toys produced on that day be ‘x’. 

∴ The cost of production (in rupees) of each toy that day = 55 – x 

∴ The total cost of production (in rupees) that day = x(55 – x) 

∴ x(55 – x) = – 750 

55x – x²= 750 

-x² + 55x – 750 = 0 

x² – 55x + 750 = 0 

∴ The number of toys produced that day satisfies the quadratic equation x² – 55x + 750 = 0. 

Now, we have to find out the value of ’x’ 

x² – 55x + 750 = 0 

x² – 30x – 25x + 750 = 0 

x(x – 30) – 25(x – 30) = 0 

(x – 30) (x – 25) = 0 

If x – 30 = 0, then x = 30 

If x – 25 = 0, then x = 25 

∴ x = 30 OR x = 25.

Number of marbles John has is x. 

Given, John and Jivanti together have 45 marbles. 

Number of marbles which Jivanti has = 45 – x 

Now, both of them lost 5 marbles each, and the product of the number of marbles they now have is 128. 

So John will have x - 5 marbles and Jivanti will have 45 - x - 5 = 40 - x marbles. 

⇒ (x – 5)(40 – x) = 128 

⇒ 40x – 200 + 5x – x² = 128 

⇒ 40x – 200 + 5x – x² - 128 = 0 

⇒ 45x – 328 – x² = 0 

⇒ x² – 45x + 328 = 0

By Pythagoras theorem : 

Hypotenuse² = perpendicular² + base² 

Given, height of a right TRIANGLE IS 7 CM LESS THAN ITS BASE and the hypotenuse is 13 cm. 

Let the base be ‘x’ ⇒ 132 = (x – 7)² + x² 

⇒ 169 = x² – 14x + 49 + x² 

⇒ x² – 7x – 60 = 0

Number of toys produced that day is ‘x’. 

Cost of production of each toy (in rupees) was found to be 55 minus the number of articles produced in a day. 

∴ Cost of production of each toy = 55 – x 

Given, total cost of production = Rs. 750 

⇒ x × (55 – x) = 750 

⇒ - x² + 55x – 750 = 0 

⇒ x² – 55x + 750 = 0

Time = distance/speed 

Given, while boarding an aeroplane, a passenger got hurt and the plane started late by 30 minutes to reach the destination, 1500 km away in time, the pilot increased the speed by 100km/hr.

Let the usual speed be ‘a’.

\(\frac{1500}{a}-\frac{1500}{a\,+\,100}=\frac{30}{60}\)

⇒ 2 × 1500 × (a + 100 – a) = a² + 100a 

⇒ a² + 100a – 300000 = 0 

⇒ a² + 600a – 500a – 300000 = 0 

⇒ a(a + 600) – 500(a + 600) = 0 

⇒ (a + 750)(a – 500) = 0 

⇒ a = 500 km/hr

 \(x^2 -(\sqrt{2}+i)x + \sqrt{2}i = 0\)

Given \(x^2 -(\sqrt{2}+i)x + \sqrt{2}i = 0\)

⇒ \(x^2 -(\sqrt{2}x+ix) + \sqrt{2}i = 0\)

⇒ \(x^2 -\sqrt{2}x-ix + \sqrt{2}i = 0\) 

⇒ \(x(x-\sqrt{2})-i(x-\sqrt{2})=0\)

⇒\((x-\sqrt{2})(x-i)=0\) 

⇒  \(x-\sqrt{2}=0\,or\, x=i=0\)

∴ x = \(\sqrt{2}\,or\,i\)

Thus, the roots of the given equation are \(\sqrt{2}\) and i.

Given as x² – (3√2 + 2i)x + 6√2i = 0

x² – (3√2x + 2ix) + 6√2i = 0

x² – 3√2x – 2ix + 6√2i = 0

x(x – 3√2) – 2i(x – 3√2) = 0

(x – 3√2) (x – 2i) = 0

(x – 3√2) = 0 or (x – 2i) = 0

x = 3√2 or x = 2i

∴ The roots of the given equation are 3√2, 2i

(i) (x + 1)² = 2(x – 3) 

x² + 2x + 1 = 2x – 6 

x² + 2x – 2x +1 + 6 = 0 

x² + 0x + 7 = 0 

This is in the form of ax² + bx + c = 0. 

∴ This is a quadratic equation. 

(ii) x² – 2x = (-2) (3 – x) 

x² – 2x = -6 + 2 

x² – 2x – 2x + 6 = 0 

x² – 4x + 6 = 0 

This is in the form of ax² + bx + c = 0. 

∴ This is a quadratic equation. 

(iii) (x – 2) (x + 1) = (x – 1) (x + 3) 

x² + x – 2x – 2 = x² + 3x – x – 3 

x² – x² = x² + 2x – 3 

x² – x² – x – 2x – 2 + 3 = 0 

-x – 2x – 2 + 3 = 0 

-3x + 1 = 0 

3x – 1 = 0 

This is not in the form of ax² + bx + c = 0 

∴ Given equation is not quadratic equation.

(iv) (x – 3) (2x + 1) = x (x + 5) 

2x² + x – 6x – 3 = x² + 5x 

2x² – 5x – 3 = x² + 5x 

2x² – x² – 5x – 5x – 3 = 0 

x² – 10x – 3 = 0 

This is in the form of ax² + bx + c = 0. 

∴ This is a quadratic equation. 

(v) (2x – 1) (x – 3) = (x + 5) (x – 1) 

2x² – 6x – x + 3 = x² – x + 5x – 5 

2x² – 7x + 3 = x² + 4x – 5 

2x² – x² – 7x – 4x + 3 + 5 = 0 

x² – 11x + 8 = 0 

This is in the form of ax² + bx + c = 0. 

∴ This is a quadratic equation. 

(vi) x² + 3x + 1 = {x – 2}² 

x² + 3x + 1 = x² – 4x + 4 

x² + 3x + 1 – x² + 4x – 4 = 0 

7x – 3 = 0 

This is not in the form of ax² + bx + c = 0 

∴ Given equation is not quadratic equation. 

(vii) (x + 2)³ = 2x (x² – 1) 

x³ + 8 + 3(x)(2)(x+2) = 2x³ – 2x 

x³ + 8 = 6x(x + 2) = 2x³ – 2x 

x³ + 8 + 6x² + 12x = 2x³ – 2x 

x³ – 2x³ + 6x² + 12x + 2x + 8 = 0 

-x³ + 6x² + 14x + 8 = 0 

x³ – 6x² – 14x – 8 = 0 

This is not in the form of ax² + bx + c = 0 

∴ Given equation is not quadratic equation. 

(viii) x³ – 4x² – x + 1 = (x – 2)³ 

x³ – 4x² – x + 1 = x³ – 8 – 3(x)(2)(x – 2) 

x³ – 4x² – x + 1 = x² – 8 – 6x (x – 2) 

x³ – 4x² – x + 1 = x³ – 8 – 6x² + 12x 

x³ – x³ – 4x² + 6x² – x – 12x + 1 + 8 = 0 

+2x² – 13x + 9 = 0 

This is in the form of ax² + bx + c = 0. 

∴ This is a quadratic equation.

Let the present age of Kishor be x.

∴ Present age of Vivek = (x + 5) years According to the given condition,

     \(\frac{1}{x}\) + \(\frac{1}{x + 5}\) = \(\frac{1}{6}\)

∴ \(\frac{x+5+x}{x(x+5)}\) = \(\frac{1}{6}\)

∴ \(\frac{2x +5}{x(x+5)}\) = \(\frac{1}{6}\)

∴ 6(2x + 5) = x(x + 5)

∴ 12x + 30 = x² + 5x 

∴ x² + 5x – 12x – 30 = 0 

∴ x² – 7x – 30 = 0 

∴ x² – 10x + 3x – 30 = 0 

∴ x(x – 10) + 3(x – 10) = 0 

∴ (x – 10)(x + 3) = 0

By using the property, if the product of two numbers is zero, then at least one of them is zero, we get 

∴ x – 10 = 0 or x + 3 = 0

∴ x = 10 or x = – 3 

But, age cannot be negative

∴ x = 10 andx + 5 = 10 + 5 = 15 

∴ Present ages of Kishor and Vivek are 10 years and 15 years respectively.

Let the breadth of rectangular plot (b) be ’x’ m. 

Then the length of th plot is one more than twice its breadth, 

∴ Length (l)= 2x + 1 m. 

But Length × Breadth = Area of rectangle l × b = A 

∴ x × (2x + 1) = 528 sq.m. 

2x² + x = 528 

∴ 2x² + x – 528 = 0 is the required equation. 

Now, we have to find out the value of ‘x’ : 

2x² + x – 528 = 0 

2x² – 32x + 33x – 528 = 0 

2x(x – 16) + 33(x – 16) = 0 

(x – 16) (2x + 33) = 0 

If x – 16 = 0, then x = 16 

If 2x + 33 = 0, then x = -33/2 

∴ Breadth (b) = 16 m. 

Length (l) = (2x + 1) = 2(16) + 1 = 32 + 1 = 33m 

∴ Length (l) = 33 m 

Breadth (b) = 16 m.

Let one positive integer be x’. 

The Next integer is (x + 1) 

Their product is 306. 

∴ x (x + 1) = 306 

x² + x = 306 

∴ x² + x – 306= 0. 

This is required equation. 

Now, we have to solve for positive integer. 

x² + x – 306 = 0 

x² + 18x – 17x – 306 = 0 

x(x + 18) – 17(x + 18) = 0 

(x + 18) (x – 17) = 0 

If x + 18 = 0, then x = -18 

If x – 17 = 0, then x = 17 

∴ x = 18, OR x = 17.

Let the present age of Rohan be ‘x’, then His mother’s age will be (x + 26) 

After 3 years, Age of Rohan is (x + 3). 

After 3 years his mother’s age will be 

= (x + 26 + 3) = (x + 29) 

Then product of their ages is 360. 

∴ (x + 3) (x + 29) = 360 

x² + 29x + 3x + 87 = 360 

x² + 32x + 87 = 360 

x² + 32x + 87 – 360 = 0 

x² + 32x – 273 = 0. 

This is required equation. 

Now, we have to solve for the value of ‘x’: 

x² + 32x – 273 = 0 

x² + 39x – 7x – 273 = 0 

x(x + 39) – 7(x + 39) = 0 

(x + 39) (x – 6) = 0 

If x + 39 = 0, then x = -39 

If x – 6 = 0, then x = 7 

Present age of Rohan’s mother 

= x + 26 

= 7 + 26 = 33 years.

Let the present age of Rohan be x years. 

Then age of Rohan’s mother = x + 26 

After 3 years: 

Age of Rohan would be = x + 3 

Rohan’s mother’s age would be = (x + 26) + 3 = x + 29 

By problem (x + 3) (x + 29) = 360 

⇒ x(x + 29) + 3(x + 29) = 360 

⇒ x² + 29x + 3x + 87 = 360 

⇒ x² + 32x + 87 – 360 = 0 

⇒ x² + 32x – 273 = 0 

⇒ x² + 39x – 7x – 273 = 0 

⇒ x (x + 39) – 7 (x + 39) = 0 

⇒ (x – 7) (x + 39) = 0 

⇒ x = 7 or x = -39 ‘x’ being age cannot be negative. 

∴ x = Present age of Rohan = 7 years.

Let the breadth of the rectangular plot be x m. 

Then its length (by problem) = 2x + 1. 

Area = l . b = (2x + 1) . x = 2x² + x 

But area = 528 m² (∵ given) 

∴ 2x² + x = 528 

⇒ 2x² + x – 528 = 0 where x is the breadth of the rectangle.

Let the two integers be x and x+1, x taken as the smaller integer.

From the question, the product of these two integers is 306

So,

x(x + 1) = 306

⇒ x² + x – 306 = 0

Thus, the required quadratic equation is x² + x – 306 = 0

Let the consecutive integers be x and x + 1. 

Their product = x(x + 1) = x² + x 

By problem x² + x = 306 

⇒ x² + x – 306 = 0 

where x is the smaller integer.

Quadratic equation has equal roots then d = b² – 4ac = 0

Here a = 1, b = k and c = 4

So b² – 4ac = 0

⇒ k² – 4 × 1 × 4 = 0

⇒ k² – 16 = 0

⇒ k = \(\pm4\)

The given equation is x² + 6x + 9 = 0

Putting x = -3 in the given equation, we get

LHS = (-3)² + 6 x (-3) + 9 = 9 - 18 + 9 = 0 = RHS

∴ x = -3 is a solution of the given equation.

Let the number of articles produced be x. 

Then the cost of each article = 2x + 3 

Total cost of the articles produced = x [2x + 3] = 2x² + 3x 

By problem 2x² + 3x = 90 

⇒ 2x² + 3x – 90 = 0 

⇒ 2x² + 15x – 12x – 90 = 0 

⇒ x (2x + 15) – 6 (2x + 15) = 0

⇒ (2x + 15) (x – 6) = 0 

⇒ 2x + 15 = 0 (or) x – 6 = 0 

⇒ x = \(\frac{-15}{2}\) or x = 6 

But x can’t be negative.

∴ x = 6 

2x + 3 = 2 × 6 + 3 = 15 

∴ Number of articles produced = 6 Cost of each article = Rs. 15.

Perimeter of a rectangle = 2(l + b) 

Area of the rectangle = l × b 

Given, perimeter of a rectangular field is 82 m and its area is 400 m² 

Let the breadth be ‘a’ m and length be ‘b’ m 

⇒ 2(a + b) = 82 

⇒ b = 41 – a 

Also, a × b = 400 

⇒ a × (41 – a) = 400 

⇒ a² – 41a + 400 = 0 

⇒ a² – 25a – 16a + 400 = 0 

⇒ a(a – 25) – 16(a – 25) = 0 

⇒ (a – 16)(a – 25) = 0 

⇒ a = 16, 25 

Assuming breadth to smaller, thus breadth = 16m

Let the length of the rectangle = x 

Given perimeter = 2(1 + b) = 28 

⇒ (1 + b) = 28/2 = 14 

Breadth of the rectangle = 14 – x 

Area = length . breadth = x (14 – x) 

= 14x – x²

By problem, 14x – x² = 40. 

⇒ x² – 14x + 40 = 0 

⇒ x² – 10x – 4x + 40 = 0 

⇒ x(x – 10) – 4(x – 10) = 0 

⇒ (x – 10) (x – 4) = 0 

⇒ x – 10 = 0 (or) x – 4 = 0 

⇒ x = 10 (or) 4 

∴ Length = 10 m or 4 m 

Then breadth = 14 – 10 = 4 m (or) 14 – 4 = 10 m

Let the altitude of the triangle h = x cm 

Then its base ‘b’ = x + 4. 

Area = 1/2 × base × height

= 1/2 (x + 4)(x) 

= \(\frac{x^2+4x}{2}\)

By problem \(\frac{x^2+4x}{2}\) = 48 

⇒ x² + 4x = 2 × 48 

⇒ x² + 4x – 96 = 0 

⇒ x² + 12x – 8x – 96 = 0 

⇒ x(x + 12) – 8(x + 12) = 0 

⇒ (x + 12)(x – 8) = 0 

⇒ x + 12 = 0 (or) x – 8 = 0 

⇒ x = -12 (or) x = 8 

But x can’t be negative. 

∴ x = 8 and x + 4 = 8 + 4 = 12 

Hence altitude = 8 cm and base = 12 cm.

Area of a square = side × side 

Given, squares have sides x cm and (x + 4) cm. 

The sum of their areas is 656 cm²

⇒ x² + (x + 4)² = 656 

⇒ x² + x² + 16 + 8x = 656 

⇒ x² + 4x – 320 = 0 

⇒ x² - 16x + 20x - 320 = 0 

⇒ x(x – 16) + 20(x - 16) = 0 

⇒ (x – 16)(x + 20) = 0 

⇒ x = 16, - 20 

The sides are 16, 20.

Let the altitude of the triangle be x m. 

Therefore, the base will be 3x m.

Area of a triangle = 1/2 x Base x Altitude

∴ 1/2 \(\times\) 3x \(\times\) x = 90 

(∵ Area = 96 sq m)

⇒ x²/2 = 32

⇒ x2 = 64

⇒ x = ± 8

The value of x cannot be negative 

Therefore, the altitude and base of the triangle are 8 m and (3 x 8 = 24m),respectively

Let S₁ and S₂ be the two squares.

And, let x cm be the side square S₁ and (x + 4) cm be the side of the square S₂.

So,

Area of the square S₁ = x² cm²

Area of the square S₂ =(x + 4)² cm²

From the question, we have

Area of the square S₁ + Area of the square S₂ = 656 cm²

⇒ x² cm² + (x + 4)² cm² = 656 cm²

x²+ x² + 16 + 8x – 656 = 0

2x² + 16 + 8x – 656 = 0

2(x² + 4x – 320) = 0

x² +4 x – 320 = 0

x² + 20x – 16x – 320 = 0

x(x + 20) – 16(x + 20) = 0

(x + 20)(x – 16) = 0

Now, either x + 20 = 0 ⇒ x = -20

Or, x – 16 = 0 ⇒ x = 16

As the value of x cannot be negative, we choose the value of x = 16 ⇒ x + 4 = 20

Therefore,

The side of the square S₁= 16 cm

The side of the square S₂ = 20 cm

(x + 3)² – 4(x + 3) – 5 = 0

Let x + 3 = y

Then y² – 4y – 5 = 0

⟹ y² – 5y + y – 5 = 0

⟹ y (y – 5) + 1 (y – 5) = 0

⟹ (y – 5) (y + 1) = 0

If y – 5 = 0 Or y + 1 = 0

Then y = 5 Or y = – 1

⟹ x + 3 = 5 or x + 3 = -1

⟹ x = 2 or x = – 4

x² – (a + b) x + ab = 0

⟹ X² – ax – bx + ab = 0

⟹ x(x – a) – b (x – a) = 0

⟹ x(x – a) (x – b) = 0

Since x – a = 0 Or x – b = 0

Then x = a Or x = b

(2x – 3)² = 49

Taking square root on both sides

2x – 3 = ± 7

When 2x – 3 = 7 ⟹ 2x = 10 ⟹ x = 5

And, when 2x – 3 = – 7 ⟹ 2x = – 4 ⟹ x = – 2

2(x² – 6) = 3 ( x – 4)

⟹ 2x² – 12 = 13x – 12

⟹ 2x² – 3x = 0

⟹ x(2x – 3) = 0

Since x = 0 Or 2x – 3 = 0

Then x = 0 Or x = 3/2