Explore topic-wise InterviewSolutions in .

I. a² + 18a + 56 = 0

⇒ a² + 14a + 4a + 56 = 0

⇒ a(a + 14) + 4(a + 14) = 0

⇒ (a + 14) (a + 4) = 0

Then, a = (-4) or a = (-14)

II. 39b² - 624b = 0

⇒ 39b(b - 16)0

⇒ (b) (b - 16) = 0

Then, b = (0) or b = 16

So, when a = (-4), a < b for b = (16) and a < b for b = 0

And when a = (-14), a < b for b = (16) and a < b for b = 0

2X² + 9X + 10 = 0

⇒ 2x² + 4x + 5X + 10 = 0

⇒ 2x(x + 2) + 5(x + 2) = 0

⇒ (2x + 5)(x + 2) = 0

x = -5/2, -2

3y² + 20y + 33 = 0

⇒ 3y² + 9Y + 11y + 33 = 0

⇒ 3y(y + 3) + 11(y + 3) = 0

⇒ (3y + 11)(y + 3) = 0

y = -11/3, -3

Thus, x = -5/2, -2 and y = -11/3, -3

⇒ Let the original number of CHILDREN be x

⇒ 2400 / (x – 100) – 2400 / x = 2

⇒ (2400x - 2400x + 240000)/x² - 100X = 2

⇒ 240000 = 2x² – 200x

⇒ 120000 = x² – 100x

⇒ 400(400 – 100) = x(x – 100)

∴ x = 400 

I. 3x² + 32x + 85 = 0

⇒ 3x² + 15x + 17x + 85 = 0

⇒ 3x(x + 5) + 17(x + 5) = 0

⇒ (x + 5) (3x + 17)

∴ x = -5, -17/3

II. 2y² + 19y + 45 = 0

⇒ 2y² + 10Y + 9y + 45 = 0

⇒ 2y(y + 5) + 9(y + 5) = 0

⇒ (y + 5) (2y + 9) = 0

∴ y = -5, -9/2

When x = -5, for y = -5, x = y and for y = -9/2, x < y

When x = -17/3, for y = -5, x < y and for y = -9/2, x < y

∴ x ≤ y

I. 8a² - 6a + 1 = 0

⇒ 8a² - 4A - 2A + 1 = 0

⇒ 4a(2a - 1) - 1(2a - 1) = 0

⇒ (4a - 1) (2a - 1) = 0

Then, a = 1/4 or a = 1/2

II. 4b² - 9b + 2 = 0

⇒ 4b² - 8b - b + 2 = 0

⇒ 4b(b - 2) - 1(b - 2) = 0

⇒ (4b - 1) (b - 2) = 0

Then, b = (1/4) or b = 2

So, when a = (1/4), a = b for b = (1/4) and a < b for b = 2

And when a = (1/2), a > b for b = (1/4) and a < b for b = 2

I. x² – 18X – 115 = 0

⇒ x² – 23x + 5x – 115 = 0

⇒ x(x – 23) + 5(x – 23) = 0

⇒ (x – 23)(x + 5) = 0

Then, x = + 23 or x = - 5

II. y² + 22y + 112 = 0

⇒ y² + 14y + 8y + 112 = 0

⇒ y(y + 14) + 8(y + 14) = 0

⇒ (y + 14)(y + 8) = 0

Then, y = - 14 or y = - 8

So, when x = + 23, x > y for y = - 14 and x > y for y = - 8

And when x = - 5, x > y for y = - 14 and x > y for y = - 8

I. 10x² + 37X + 7 = 0

⇒ 10x² + 35X + 2x + 7 = 0

⇒ 5x(2x + 7) + 1(2x + 7) = 0

⇒ (2x + 7)(5x + 1) = 0

Then, x = - 7/2 or x = - 1/5

II. 5Y² – 14y – 3 = 0

⇒ 5y² – 15y + y – 3 = 0

⇒ 5y(y – 3) + 1(y – 3) = 0

⇒ (y – 3)(5y + 1) = 0

Then, y = + 3 or y = - 1/5

So, when x = - 7/2, x < y for y = + 3 and x < y for y = - 1/5

And when x = - 1/5, x < y for y = + 3 and x = y for y = - 1/5

I. 4x² + 4√3x + 3 = 0

⇒ 4x² + 2√3x + 2√3x + 3 = 0

⇒ 2x(2x + √3) + √3(2x + √3) = 0

⇒ (2x + √3)² = 0

∴ x =-√3/2

II. 35y² + 48y - 27 = 0

⇒ 35y² + 63y - 15Y - 27 = 0

⇒ 7y(5Y + 9) - 3(5y + 9) = 0

⇒ (5y + 9)(7y - 3) = 0

∴ y =-9/5 or y = 3/7

If x = -√3/2, x > y for y = -9/5

If x = -√3/2, x < y for y = 3/7

∴ This shows that relationship cannot be ESTABLISHED between x and y

I. X² + 75X + 1406 = 0

⇒ x² + 37x + 38x + 1406 = 0

⇒ x (x + 37) + 38(x + 37) = 0

⇒ (x + 37) (x + 38) = 0

⇒ x = -37 or -38

II. y² + 76y + 1444 = 0

⇒ y² + 38y + 38y + 38² = 0

⇒ (y + 38)² = 0

⇒ y + 38 = 0

⇒ y = -38

When x = -37, x > y for y = -38

And when x = -38, x = y for y = -38

I. 16a² - 52a + 22 = 0

⇒ 16a² - 8a - 44a + 22 = 0

⇒ 8a(2A - 1) - 22(2a - 1) = 0

⇒ (8a - 22) (2a - 1) = 0

Then, a = 22/8 or a = 1/2

II. 30b² + 56b + 10 = 0

⇒ 30b² + 6B + 50b + 10 = 0

⇒ 6b(5b + 1) + 10(5b + 1) = 0

⇒ (6b + 10) (5b + 1) = 0

Then, b = (-10/6) or b = (-1/5)

So, when a = (22/8), a > b for b = (-10/6) and a > b for b = (-1/5)

And when a = (1/2), a > b for b = (-10/6) and a > b for b = (-1/5)

Let the NUMERATOR and denominator be X and y.

$(\Rightarrow \frac{{x + 8}}{y} - \frac{x}{y} = \frac{4}{9})$

⇒ 8 × 9 = 4y

⇒ y = 18

∴ Denominator of a FRACTION is 18.

I. x² – 12x + 35 = 0

⇒ x² – 5X – 7x + 35 = 0

⇒ x(x – 5) – 7(x – 5) = 0

⇒ (x – 7) (x – 5) = 0

⇒ x = 7 OR x = 5

II. y² – 16Y + 63 = 0

⇒ y² – 7y – 9y + 63 = 0

⇒ y (y – 7) – 9(y – 7) = 0

⇒ (y – 9) (y – 7) = 0

⇒ y = 9 OR y = 7

So, when x = + 7, x = y when y = 7 and x < y when y = 9.

And when x = 5, x < y when y = 7 and x < y when y = 9.

∴ x ≤ y. 

I. x² – 5X – 24 = 0

⇒ x² – 8x + 3x – 24 = 0

⇒ x(x – 8) + 3(x – 8) = 0

⇒ (x – 8)(x + 3) = 0

Then, x = + 8 or x = - 3 

II. y² – 8y – 33 = 0

⇒ y² – 11y + 3y – 33 = 0

⇒ y(y – 11) + 3(y – 11) = 0

⇒ (y – 11)(y + 3) = 0

Then, y = + 11 or y = - 3 

So, when x = + 8, x < y for y = + 11 and x > y for y = - 3

And when x = - 3, x < y for y = + 11 and x = y for y = - 3

∴ So, we can observe that no clear RELATIONSHIP cannot be determined between x and y.

LET the QUANTITY of guava purchased be ‘g’ and kiwi be ‘k’

ACCORDING to the question,

⇒ g + k = 5----(1)

⇒ 60g + 20k = 180----(2)

Solving (1) and (2), we get

Multiplying equation(1) by 20

⇒ 20g + 20k = 100----(3)

⇒ 60g + 20k = 180----(4)

Substracting equation(3) from equation(4)

⇒ 40g = 80

I. 2x² – 7x + 5 = 0

⇒ 2x² – 5X – 2x + 5 = 0

⇒ x(2x – 5) – 1(2x – 5) = 0

⇒ (2x – 5)(x – 1) = 0

Then, x = + 5/2 or x = + 1

II. 2Y² – 3Y + 1 = 0

⇒ 2y² – 2y – y + 1 = 0

⇒ 2y(y – 1) – 1(y – 1) = 0

→ (y – 1)(2y – 1) = 0

Then, y = + 1 or y = + ½

So, when x = + 5/2, x > y for y = + 1 and x > y for y = + ½

And when x = + 1, x = y for y = + 1 and x > y for y = + ½

Let’s assume that present age of FATHER is X and age of son is Y

∴ Sum of their present ages = X + Y = 38

Four YEARS ago the age of Father was = X – 4

Four years ago the age of son was = Y – 4

? Four years ago the product of their ages was 3 times the father’s age at that time

∴ (X – 4) × (Y – 4) = 3 (X – 4)

⇒ Y – 4 = 3

⇒ Y = 7

∴ present age of son is = 7 YRS

I. x² + 22x – 104 = 0

⇒ x² + 26x – 4X – 104 = 0

⇒ x(x + 26) – 4(x + 26) = 0

⇒ (x + 26)(x – 4) = 0

Then, x = - 26 or x = + 4

II. y² + 31y – 140 = 0

⇒ y² + 35Y – 4y – 140 = 0

⇒ y(y + 35) – 4(y + 35) = 0

⇒ (y + 35)(y – 4) = 0

Then, y = - 35 or y = + 4

So, when x = - 26, x > y for y = - 35 and x < y for y = + 4

And when x = + 4, x > y for y = - 35 and x = y for y = + 4

Let the unit digit of number be X and tenth digit be y

Then,

Number = 10y + x

Number OBTAINED by reversing its digits = 10x + y

Number obtained by interchanging the digits of a two digit number is less than the original number by 54

∴ (10y + x) – (10x + y) = 54

⇒ 9y – 9x = 54

⇒ y – x = 6----EQ (1)

Sum of the digits of the number = 8

∴ y + x = 8----eq (2)

By adding eq (1) and eq (2), we get

2y = 14

⇒ y = 7

By putting this VALUE of y in eq(2), we get

7 + x = 8

⇒ x = 1

∴ Original two digit number = (10 × 7) + 1 = 71

We will SOLVE both the equations separately.

Equation I:

12x² – x – 1 = 0

⇒ 12x² – 4X + 3x – 1 = 0

⇒ 4x(3x – 1) + 1(3x – 1) = 0

⇒ (4x + 1) (3x – 1) = 0

⇒ x = -(1/4) = - 0.25

Or, x = 1/3 = 0.33

Equation II:

20y² – 41y + 20 = 0

⇒ 20y² – 25Y – 16y + 20 = 0

⇒ 5y (4y – 5) – 4(4y – 5) = 0

⇒ (5y – 4) (4y – 5) = 0

⇒ y = 4/5 = 0.8 or y = 5/4 = 1.25

Comparing the VALUES of x and y we get,

I. $(\frac{{17}}{{\sqrt X }} = \sqrt x - \frac{7}{{\sqrt x }} = 0)$

⇒ $(\frac{{17}}{{\sqrt x }})$ + $(\frac{7}{{\sqrt x }} = \sqrt x)$

⇒ $(\frac{{17\; + \;7}}{{\sqrt x }})$ = √x

⇒ 24 = x

II. y³ – $(\frac{{\left( {3\; \times \;8} \right)\frac{9}{2}}}{{y\sqrt y }})$ = 0

⇒ y^9/2 = 24^9/2

⇒ y = 24

When x = 24, x = y for y = 24

x² + 91 = 20x

(x - 7) (x - 13)

Solving we get, x = 7, 13

10y² - 29y + 21 = 0

( 5X - 7 ) (2X - 3)

Solving we get, y = 7/5, 3/2

So x is GREATER than y

I. X² + 19X – 120 = 0

⇒ x² + 24x – 5x – 120 = 0

⇒ x(x + 24) – 5(x + 24) = 0

⇒ (x + 24)(x – 5) = 0

Then, x = - 24 or x = + 5

II. y² – 14y + 45 = 0

⇒ y² – 9Y – 5y + 45 = 0

⇒ y(y – 9) – 5(y – 9) = 0

⇒ (y – 9)(y – 5) = 0

Then, y = + 9 or y = + 5

So, when x = - 24, x < y for y = + 9 and x < y for y = + 5

And when x = + 5, x < y for y = + 9 and x = y for y = + 5

I. ?X = 3

⇒ x = 3³

∴ x = 27

II. Y² – 729 = 0

⇒ Y^{2 =} 729

⇒ y = 27 or y = -27

When x = 27, x = y for y = 27 and x > y for y = -27

∴ x ≥ y

I. a² - 61A + 424 = 0

⇒ a² - 53a - 8a + 424 = 0

⇒ a(a - 53) - 8(a - 53) = 0

⇒ (a - 8) (a - 53) = 0

Then, a = 53 or a = 8

II. b² + 54B + 53 = 0

⇒ b² + 53b + b + 53 = 0

⇒ b(b + 53) + 1(b + 53) = 0

⇒ (b + 1) (b + 53) = 0

Then, b = (-53) or b = (-1)

So, when a = (53), a > b for b = (-53) and a > b for b = (-1)

And when a = 8, a > b for b = (-53) and a > b for b = (-1)

I. a² - 8A + 12 = 0

⇒ a² - 6a - 2A + 12 = 0

⇒ a(a - 6) - 2(a - 6) = 0

⇒ (a - 2) (a - 6) = 0

Then, a = 2 or a = 6

II. 8b² + 10b + 3 = 0

⇒ 8b² + 6b + 4B + 3 = 0

⇒ 2B(4b + 3) + 1(4b + 3) = 0

⇒ (2b + 1) (4b + 3) = 0

Then, b = (-1/2) or b = (-3/4)

So, when a = 2, a > b for b = (-1/2) and a > b for b = (-3/4)

And when a = 6, a > b for b = (-1/2) and a > b for b = (-3/4)

I. x² - 324 = 0

⇒ x² = 324

⇒ x = 18 or x = - 18

II. y² - 36Y + 323 = 0

⇒ y² -17y - 19y + 323 = 0

⇒ y(y - 17) -19(y - 17) = 0

⇒ (y - 17) (y - 19) = 0

⇒ y = 17 or y = 19

When x = 18, x > y for y = 17 and x < y for y = 19

And when x = -18, x < y for y = 17 and x < y for y = 19

Let ‘x’ be the first number.

Other TWO numbers are (x + 3) & (x + 6)

GIVEN that,

x + x + 3 + x + 6 = 633

⇒ 3x = 624, x = 624/3 = 208

⇒ Third number = (x + 6) = 208 + 6 = 214

$(4{x^2} - \LEFT( {8 + \SQRT {15} } \right)x + 2\sqrt {15} {\rm{}} = {\rm{}}0)$

⇒ 4x(x – 2) – √15(x – 2) = 0

⇒ (x – 2) (4x - √15) = 0

⇒ x = 2 or x = √15/4 ≈ 0.97

$(12{y^2} - \left( {18 + 2\sqrt {15} } \right)y + 3\sqrt {15} = 0)$

⇒ 6y(2y – 3) – √15(2y – 3) = 0

⇒ (2y – 3) (6y - √15) = 0

⇒ y = 3/2 or y = √15/6 ≈ 0.645

STATEMENT I:

5x – 35/x = 18

⇒ 5x² – 18x – 35 = 0

5x² – 25x + 7x – 35 = 0

5x (x – 5) + 7 (x – 5) = 0

x = 5 or x = - 7/5

Statement II:

25y² + 110y + 121 = 0

25y² + 55y + 55y + 121 = 0

5Y (5y + 11) + 11 (5y + 11) = 0

y = - 11/5

∴ y < x

I. x² – 15x – 154 = 0

⇒ x² – 22X + 7x – 154 = 0

⇒ x(x – 22) + 7(x – 22) = 0

⇒ (x – 22)(x + 7) = 0

Then, x = + 22 or x = - 7

II. y² + 38Y + 217 = 0

⇒ y² + 31y + 7y + 217 = 0

⇒ y(y + 31) + 7(y + 31) = 0

⇒ (y + 31)(y + 7) = 0

Then, y = - 31 or y = - 7

So, when x = + 22, x > y for y = - 31 and x > y for y = - 7

And when x = - 7, x > y for y = - 31 and x = y for y = - 7

Let the total number of students be X.

Then, according to question:

Number of students passed in Mathematics = 0.62X

Number of students passed in ENGLISH = 0.52X

Number of students passed in both Mathematics and English = 108

Number of students fail in both Mathematics and English = 0.22X

We know that, Number of students passed + Number of students FAILED = Total number of students

Now, number of students passed = 0.62X + 0.52X – 108 (Because 108 factor got COUNTED twice)

∴ 0.62X + 0.52X – 108 + 0.22X = X

⇒ 1.36X – 108 = X

I. x² + 9x - 22 = 0

⇒ x² + 11x - 2x - 22 = 0

⇒ x(x + 11) - 2(x + 11) = 0

⇒ (x + 11) (x - 2) = 0

⇒ x = -11 or x = 2

II. y² + 25y + 156 = 0

⇒ y² + 12y + 13Y + 156 = 0

⇒ y(y + 12) + 13(y + 12) = 0

⇒ (y + 12) (y + 13) = 0

⇒ y = -12 or y = -13

When x = -11, x > y for y = -12 and x > y for y = -13

And when x = 2, x > y for y = -12 and x > y for y = -13

∴ x > y

5X² – 18x + 9 = 0

⇒ 5x² – 15x – 3x + 9 = 0

⇒ (5x – 3)(X - 3) = 0

⇒ x = 3/5 or x = 3

3Y² + 5y – 2 = 0

⇒ 3y² + 6y – y -2 = 0

⇒ (3y - 1)(y + 2) = 0

⇒ y = 1/3 or -2

∴ x = 3/5, 3 and y = 1/3, -2

∴ comparing these VALUES of x and y, we get x > y.

I. a² + 10a + 21 = 0

⇒ a² + 7a + 3A + 21 = 0

⇒ a(a + 7) + 3(a + 7) = 0

⇒ (a + 3) (a + 7) = 0

Then, a = (-3) or a = (-7)

II. 28b² + 15B + 2 = 0

⇒ 28b² + 7b + 8B + 2 = 0

⇒ 7b(4b + 1) + 2(4b + 1) = 0

⇒ (7b + 2) (4b + 1) = 0

Then, b = (-2/7) or b = (-1/4)

So, when a = (-3), a < b for b = (-2/7) and a < b for b = (-1/4)

And when a = (-7), a < b for b = (-2/7) and a < b for b = (-1/4)

I. a² - 36A + 323 = 0

⇒ a² - 19a - 17a + 323 = 0

⇒ a(a – 19) - 17(a - 19) = 0

⇒ (a - 19)(a - 17) = 0

Then, a = 19 or a = 17

II. b² - 33b + 272 = 0

⇒ b² - 17b – 16b + 272 = 0

⇒ b(b - 17) - 16(b - 17) = 0

⇒ (b - 17)(b - 16) = 0

Then, b = 17 or b = 16

So, when a = 19, a > b for b = 17 and a > b for b = 16

And when a = 17, a = b for b = 17 and a > b for b = 16

I. a² - 46a - 147 = 0

⇒ a² - 49a + 3A - 147 = 0

⇒ a(a - 49) + 3(a - 49) = 0

⇒ (a - 49) (a + 3) = 0

Then, a = (-3) or a = 49

II. B² + 27b + 180 = 0

⇒ b² + 15b + 12B + 180 = 0

⇒ b(b + 15) + 12(b + 15) = 0

⇒ (b + 12) (b + 15) = 0

Then, b = (-12) or b = (-15)

So, when a = (-3), a > b for b = (-12) and a > b for b = (-15)

And when a = 49, a > b for b = (-12) and a > b for b = (-15)

A is as much YOUNGER than B as he is older than C.

So, the age of A < the age of B

And the age of A > the age of C

The summation of the age of B and C = 40 years

From the above data, we cannot determine the difference in age between B and A.

I. a² - 25a – 26A + 650 = 0

⇒ a² - 25a - 26a + 650 = 0

⇒ a(a – 25) - 26(a - 25) = 0

⇒ (a - 25)(a - 26) = 0

Then, a = 25 or a = 26

II. b² – 53b + 702 = 0

⇒ b² - 27b – 26B + 702 = 0

⇒ b(b - 27) - 26(b - 27) = 0

⇒ (b - 27)(b - 26) = 0

Then, b = 27 or b = 26

So, when a = 25, a < b for b = 27 and a < b for b = 26

And when a = 26,a < b for b = 27 and a = b for b = 26

LET the AGES of father and son be x and y

The SUM of ages of a father and son is 45 years.

∴ x + y = 45----(1)

Five years ago, the product of their ages was FOUR times the father’s age at that time.

Five years ago their ages would be (x - 5) and (y - 5)

∴ (x - 5) (y - 5) = 4(x - 5)

⇒ y - 5 = 4

⇒ y = 9

Put the value in equation (1)

⇒ x + 9 = 45

⇒ x = 45 – 9 = 36 years

∴ The present age of father will be = 36 years

I. 10x² – 7x + 1 = 0

⇒ 10x² – 5X – 2x + 1 = 0

⇒ 5x(2x – 1) – 1(2x – 1) = 0

⇒ (5x – 1)(2x – 1) = 0

Then, x = (1/5) or x = (1/2)

II. 35y² – 12y + 1 = 0

⇒ 35y² – 7y – 5y + 1 = 0

⇒ 7y(5y – 1) – 1(5y – 1) = 0

⇒ (7y – 1)(5y – 1) = 0

Then, y = (1/7) or y = (1/5)

So, when x = (1/5), x = y for y = (1/5) and x > y for y = (1/7)

And when x = (1/2), x > y for y = (1/5) and x > y for y = (1/7)

x² – 19X + 84 = 0

⇒ x² – 12x – 7X + 84 = 0

⇒ x(x – 12) – 7(x – 12) = 0

⇒ (x – 12) (x – 7) = 0

⇒ x = 12 or x = 7

y² – 3Y – 108 = 0

⇒ y² – 12y + 9y – 108 = 0

⇒ y(y – 12) + 9(y – 12) = 0

⇒ (y – 12) (y + 9) = 0

⇒ y = 12 or y = -9

Quantity A:

For the first woman:

Amount for 4 hrs = RS.2000

Amount for each extra hr = Rs 100

⇒ Amount to be paid for 6 hr = 2000 + 2 × 100 = 2200

For the second woman:

Amount for 6 hr = Rs 800

Amount for each extra hr = Rs. 400

⇒ Amount to be paid for 8 hr = 800 + 400 × 2 = 1600

⇒ Total amount to be paid by lady = 2200 + 1600 = Rs. 3800

Quantity B:

Considering confusion,

For the first woman:

Amount for 4 hrs = Rs.2000

Amount for each extra hr = Rs 100

⇒ Amount to be paid for 8 hr = 2000 + 4 × 100 = 2400

For the second woman:

Amount for 6 hr = Rs 800

Amount for each extra hr = Rs. 400

Amount to be paid for 6 hr = 800

⇒ Total amount to be paid by lady = 2400 + 800 = Rs. 3200

∴ Quantity A > Quantity B

We will solve both the equations separately.

Equation I:

8x² + 6X - 5 = 0

⇒ 8x² + 10x – 4x – 5 = 0

⇒ 4x (2x – 1) + 5(2x – 1) = 0

⇒ (4x + 5) (2x – 1) = 0

⇒ x = - 5/4 = - 1.25 or, x = ½ = 0.5

Equation II:

12y² – 22Y + 8 = 0

⇒ 12y² – 16y – 6Y + 8 = 0

⇒ 6y (2y – 1) – 8(2y – 1) = 0

⇒ (6y – 8) (2y – 1) = 0

⇒ y = 8/6 = 1.33 or, y = ½ = 0.5

Comparing the values of x and y, we GET,