## Equations and Inequations Questions for IBPS PO

### Equations and Inequations Previous Year Questions for IBPS PO 2017 Exam

**Two equations I and II are given below in each question. You have to solve these equations and give answer. (IBPS PO Prelim 2017 Exam)**

(a) If x < y

(b) If x > y

(c) If x ≤ y

(d) If x ≥ y

(e) If x = y or no relation can be established

**1.**

I.

II.

**2.**

I. x^{2} – 3481 = 0

II.

**3.****
**I. x

^{2}– 5x – 14 = 0

II. y

^{2}+ 7y + 10 = 0

**4.****
**I. 5x

^{2}+2x – 3 = 0

II. 2y

^{2}+ 7y + 6 = 0

**5.****
**I. (17)

^{2}+ 144 ÷ 18 = x

II. (26)

^{2}– 18 × 21 = y

**Equations and Inequations Questions for SBI PO 2017 Answers**

**1.**(a)

I.

⇒ x = (18)^{3}

II.

**2. **(e)

I. x = ± 59

II. 3y^{2} = 60

⇒ y = ± √20

∴ No relation exists.

**3.**(d)

I. x^{2} – 7x +2x – 14 = 0

x(x – 7) + 2 (x – 7) = 0

x = 7, – 2

II. y^{2} + 5y +2y + 10 = 0

y = -2, -5

x ≥ y

**4. **(b)

I. 5x^{2} + 5x – 3x – 3 = 0

5x (x + 1) – 3(x + 1) = 0

II. 2y^{2} + 4y + 3y + 6 = 0

2y(y + 2) + 3(y + 2) = 0

**5.**(a)

I.

II. y = 298

### Equations and Inequations Previous Year Questions for IBPS PO 2016 Exam

**In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer (IBPS PO Pre 2016)**

**1.**

x² + 30x + 221 = 0

y^{2} – 53y + 196 = 0

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**2.**

2x^{2} – 9x + 10 = 0

y^{2}– 18y + 72 = 0

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**3.**

x (35 – x) = 124

y(2y + 3) = 90

(a) x>y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**4.
**

(y + 2)(27 – y) = 210

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**5.**

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**In the following questions two equations numbered I and II are given. You have to solve both the equations and give answer. (IBPS PO Mains 2016) **

**6.**

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**7.**

5x + 2y = 31

3x + 7y = 36

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**8.**

7x + 6y + 4z = 122

4x + 5y + 3z = 88

9x + 2y + z = 78

(a) x < y = z

(b) x ≤ y < z

(c) x < y > z

(d) x = y > z

(e) x = y = z or relation cannot be established

**9.**

4x + 2y =8.5,

2x + 4y = 9.5

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established

**10.**

I. 56 x^{2} + 37x + 6 = 0

II. 66y^{2}– 13y – 4 = 0

(a) if x ≥ y

(b) if x > y

(c) if x < y

(d) if x ≤ y

(e) if the relation between x and y cannot be established

**Equations and Inequations Previous Year Questions for IBPS PO 2016 Answers**

**1.** (b)

(x + 13) (x + 17) = 0

x = – 13, – 17

y^{2} – 53y + 196 = 0

y = 49, 4

**2.** (b)

2x^{2} – 9x + 10 = 0

x = 2.5, 2

y^{2} – 18y + 72 = 0

y = 12, 6

**3.** (e)

x(35 – x) = 124

x = 31, 4

y(2y + 3) = 90

y = -7.5, 6

**4.** (b)

x^{2} – 4x -21 = 0

x = 7, -3

(y + 2)(27 – y) = 210

y^{2} – 25x + 156 = 0 ⇒ y = 12, 13

**5.** (a)

6x + 8 = 0

x = -1.33

9y + 16 = 0

y = – 1.77

**6.** (e)

x = 1, 2

y = –3, 7

**7.** (a)

5x + 2y = 31 —— (1)

3x + 7y = 36 —— (2)

By solving eqn(1) and (2)

x = 5 ; y = 3

**8.** (a)

7x + 6y + 4z = 122 —— (1)

4x + 5y + 3z = 88 —— (2)

9x + 2y + z = 78 —— (3)

From (1) and (2) ⇒ 5x – 2y =4 —— (a)

From (2) and (3) ⇒ 23x + y = 146 —— (b)

From (a) and (b) ⇒ x = 6, y = 8. Put values in eqn (3)

⇒ z = 8

**9.** (b)

4x + 2y = 8.5 —— (I)

2x + 4y = 9.5 —— (II)

By solving eqn (I and II)

x = 1.25, y = 1.75

Put on number line

1.25 1.75

**10.** (c)

### Equations and Inequations Previous Year Questions for IBPS PO 2015 Exam

**In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answe**r

(a) x > y

(b) x < y

(c) x ≥ y

(d) x ≤ y

(e) x = y or relation cannot be established between ‘x’ and ‘y’ **(IBPS PO Prelim 2015)**

**1.**

I. 8x + y =10

II. 4x + 2y = 13

**2.**

I. (x + 3) (y + 2) = 12

II. 2xy + 4x + 5y = 11

**3.**

I. (3x – 2)/y = (3x + 6)/(y + 16)

II. (x + 2)/(y + 4) = (x + 5)/(Y + 10)

**4.**

I. x^{2} + 20x + 4 = 50 – 25x

II. y^{2} – 10y – 24 = 0

**5.**

I. (x^{2} – 10x + 16)/(x^{2} – 12x + 24) = 2/3

II. y^{2} – y – 20 = 0

**6.**

The number obtained by interchanging the two digits of a two digit number is lesser than the original number by 54. If the sum of the two digits of the number is 12, then what is the original number?** (IBPS PO Prelim 2015**)

(a) 28

(b) 39

(c) 82

(d) Can’t say

(e) None of these

**7.**

At present Geeta is eight times her daughter’s age. Eight years from now, the ratio of the ages of Geeta and her daughter will be 10 : 3 respectively. What is Geeta’s present age ?** (IBPS PO Prelim 2015)**

(a) 32 years

(b) 40 years

(c) 36 years

(d) Can’t say

(e) None of these

**8.**

The ages of Samir and Tanuj are in the ratio of 8 : 15 years respectively. After 9 years the ratio oftheir ages will be 11 : 18. What is the difference in years between their ages? **(IBPS PO Main 2015)**

(a) 24 years

(b) 20 years

(c) 33 years

(d) 21 years

(e) None of these

**For the two given equations I and II.**

Give answer (a) if p is greater than q.

Give answer (b) if p is smaller than q.

Give answer (c) if p is equal q.

Give answer (d) if p is either equal to or greater than **(IBPS PO Main 2015)**

**9.**

I. p + 5p + 6 = 0

II. q^{2} + 3q + 2 = 0

**10.**

I. p^{2} = 4

II. q^{2} + 4q = – 4

**11.**

I. p^{2} + p = 56

II. q^{2}– 17q+ 72 = 0

**12.**

I. 3p + 2q – 58 = 0

II. 4q + 4p = 92

**13.**

I. 3p^{2} + 17p + 10 = 0

II. 10q^{2} + 9q + 2 = 0

**14.**

The denominators of two fractions are 5 and 7 respectively. q. Give answer (e) if p is either equal or smaller than q. The sum of these fractions is . On interchanging the numerators, their sum becomes . The fractions are **(IBPS PO Main 2015)**

(a)

(b)

(c)

(d)

(e) None of these

**In each of the following questions two equations are given. You have to solve them and**

Give answer (a) if p < q;

Give answer (b) if p > q;

Give answer (c) if p £ q;

Give answer (d) if p 3 q;

Give answer (e) if p = q; **(IBPS PO Prelim 2015)**

**15.**

I. p^{2} – 7p = – 12

II. q^{2} – 3q + 2 = 0

**16.**

I. 12p^{2} – 7q = -1

II. 6q^{2} – 7q + 2 = 0

**17.**

I. p^{2} + 12p + 35 = 0

II. 2q^{2} + 22q + 56 = 0

**18.**

I. p^{2} – 8p +15 = 0

II. q^{2} – 5q = – 6

**19.**

I. 2p^{2} + 20p + 50 = 0

II. q^{2} = 25

**Equations and Inequations Previous Year Questions for IBPS PO 2015 Answers**

**1.** (b)

From both equation

x = 7/12, y = 16/3

y > x

**2.** (e)

xy + 3y + 2x + 6 =12

2xy + 6y + 4x = 12 —— (i)

2xy + 5y + 4x = 11 —— (ii)

From equation (i) and (ii)

y = 1

From equation (i)

x = 1

x = y

**3.** (b)

(3x – 2)/y = (3x + 6)/(y + 16)

48x – 8y = 32 —— (i)

(x + 2)/(y + 4) = (x + 5)/(y + 10)

y = 2x —— (ii)

x = 1, y = 2

y > x

**4.** (b)

From the given equation

x = 1, – 46& y = – 4, 6

y > x

**5.** (a)

From 1st equation

x^{2} – 6x = 0

x = 0,6

From 2nd equation

(y + 4) (y – 5)

y = – 4, 5

x > y

**6.** (e)

Let the number be xy

(10x + y) – (10y + x) = 54

x – y = 6 And x + y = 12

Solving the equations we get x = 9 and y = 3

So the number is 93.

**7.** (a)

Let the age of Geeta’s daughter be x. Then Geeta’s age is 8x.

(8x + 8)/(x + 8) = 10/3

x = 4

Geeta’s present age = 8x = 32 years.

**8.** (d)

Let present ages of Samir and Tanuj are 8x and 15x years respectively.

Difference between their ages = 15x – 8x = 7x

Ratio of ages after 9 years,

⇒ 144x + 162 = 165x + 99

⇒ 21x = 63 ⇒ x = 3

**9.** (e)

I. p^{2} + 3p + 2p + 6 = 0

p (p + 3) + 2(p + 3) = 0

(p + 3)(p + 2) = 0

p = 2 or – 3

II. q^{2} + q + 2q + 2 = 0

q (q + 1) + 2 (q + 1) = 0

(q +1) + (q + 2) = 0

q = – 1 or – 2

Obviously p < q

**10.** (d)

I. p = ± 2

II.q^{2} + 2q + 2q + 4 = 0

q (q + 2) + 2 (q + 2) = 0

(q + 2) + (q + 2) = 0

q = – 2

Obviously p > q

**11.** (b)

I. p^{2} + p – 56 = 0

p^{2} + 8p – 7p – 56 = 0

p (p + 8) -7 (p + 8) = 0

(p + 8) (p – 7) = 0

p = 7 or – 8

II. q^{2} – 8q – 9q + 72 = 0

q (q – 8) – 9 (q – 8) = 0

(q – 8) (q – 9) = 0

q = 8 or 9

Obviously p < q 12.

**12. **(a)

We have,

3p + 2q = 58 —— (i)

4p + 4q = 92 ⇒ 2p + 2q = 46 —— (ii)

By (i), (ii) we get p =12

From (i), 3 × 12 + 2q = 58

⇒ 2q = 58 – 36 = 22

⇒ q = 11

Hence, p > q

**13.** (b)

I. ⇒ 3p^{2} + 15p + 2p +10 = 0

⇒ 3p (p + 5) + 2(p + 5) = 0

⇒ (p + 5) (3p + 2) = 0

p – – 5 or -j

II. ⇒10q^{2} + 5q + 4q + 2 = 0

⇒ 5q (2q + 1) + 2(2q + 1) = 0

⇒ (2q + 1) (5q + 2) = 0

Hence, p < q

**14.** (b)

**15.** (b)

(I) p^{2} – 7p = – 12

p^{2} – 4p +12 = 0

⇒ p^{2} – 4p – 3p + 12 = 0

⇒ p (p – 4) -3 (p -4) = 0

⇒ (p – 4) (p – 3) = 0

⇒ p = 3 or 4

(II) q^{2} – 3q + 2 = 0

q^{2} – 2q – 9 + 2 = 0

⇒ q (q – 2) -1 (q – 2) = 0

⇒ (q – 2) (q – 1) = 0

⇒ q =1 or 2

Obviously p > q

**16.** (a)

(I) 12p^{2} – 7p = – 1

⇒ 12p^{2} – 7p + 1 = 0

⇒ 12p^{2} – 4p – 3p + 1 = 0

⇒ 4p (3p -1) -1 (3p-1) = 0

⇒ (3p -1) (4p – 1) = 0

(II) 6q^{2} – 7q + 2 = 0

⇒ 6q^{2} – 4q – 3q + 2 = 0

⇒ 2q (3q – 2) -1 (3q – 2) = 0

⇒ (3q – 2) (2q – 1) = 0

Obviously, p £ q

**17.** (c)

(I) p^{2} + 12p + 35 = 0

⇒ p^{2} + 7p + 5p + 35 = 0

⇒ p (p + 7) + 5 (p + 7) = 0

⇒ (p + 7) (p + 5) = 0

⇒ p = – 5 or – 7

(II) 2q^{2} + 22q +56 = 0

⇒ 2q^{2} + 14q + 8q + 56 = 0

⇒ 2q (q + 7) + 8 (q + 7) = 0

⇒ (q + 7) (2q + 8) = 0

⇒ q = – 7 or – 4

Obviously, p £ q

**18.** (d)

(I) p^{2} – 8p +15 = 0

⇒ p^{2} – 3p – 5p + 15 = 0

⇒ p (p – 3) – 5 (p – 3) = 0

⇒ (p – 3) (p – 5) = 0

⇒ p = 3 or 5

(II) q^{2} – 5q = – 6

⇒ q^{2} – 5q + 6 = 0

⇒ q^{2} – 3q – 2q + 6 = 0

⇒ q (q – 3) – 2 (q – 3) = 0

⇒ q (q – 3) (q – 2) = 0

⇒ q = 3 or 2

Obviously, p 3 q.

**19.** (c)

(I) 2p^{2} + 20p +50 = 0

⇒ p^{2} + 10p + 25 = 0

⇒ (p + 5)2 = 0

⇒ p + 5 = 0

⇒ p = -5

(II) q^{2} = 25

⇒ q = ±5

Obviously, p £ q,

### Equations and Inequations Previous Year Questions for IBPS PO 2014 Exam

**In each of the following questions two equations are given. You have to solve them and give answer accordingly.**

(a) If x > y

(b) If x < y

(c) If x = y

(d) If x ≥ y

(e) If x ≤ y** (IBPS PO/MT 2014)**

**1.**

I. 2x^{2} + 5x +1 = x^{2} + 2x – 1

II. 2y^{2} – 8y +1 = -1

**2.**

I.

II. 3y^{2} – 10y + 8 = y^{2} + 2y – 10

**3.**

I. 4x^{2} – 20x +19 = 4 x -1

II. 2y^{2} = 26y + 84

**4.**

I. y^{2} + y – 1 = 4 – 2y – y^{2}

II.

**5.**

I. 6x^{2} + 13x = 12 – x

II.

**Equations and Inequations Previous Year Questions for IBPS PO 2014 Answers**

**1.** (b)

I. 2x^{2} + 5x +1 = x^{2} + 2x -1

x^{2} + 3x + 2 = 0

x^{2} + 2x + x + 2 = 0

x (x + 2) + 1 (x – 2) = 0

(x + 2) (x + 1) = 0 x = -2 , -1

II. 2y^{2} – 8y + 1 = -1

2y^{2} – 8y + 2 = 0

y^{2} – 4y + 1 = 0

Hence, y > x

**2.** (b)

I. x^{2}+ 2x – 1 = 2

x^{2} + 2x – 3 =0

x + 3x – x – 3 = 0

x(x + 3)-1(x + 3) = 0

(x+3)(x- 1) = 0

x = -3, 1

II. 2y^{2} – 12y +18 = 0

y^{2} – 6y + 9 = 0

(y – 3)^{2} = 0

y = 3, 3

Hence, y > x

**3.** (b)

I. 4x^{2} – 24x+20 = 0

x^{2} – 6x + 5 = 0

x^{2} – 5x – x + 5 = 0

x (x – 5) -1 (x – 5) = 0

(x – 5) (x – 1) = 0

x = 5 , 1

II. y^{2} – 13y + 42 = 0

y^{2} – 7y – 6y + 42 = 0

y (y – 7) – 6 (y – 7) = 0

(y – 7) (y – 6) = 0

y = 7, 6

Hence, y > x.

**4.** (a)

2y^{2} + 3y – 5 = 0

2y^{2} + 5y – 2y – 5 = 0

y (2y + 5) -1 (2y + 5) = 0

(2y + 5) (y – 1) = 0

II. x^{2} – 3x = 2x – 6

x^{2} -5x + 6 = 0

x^{2} – 3x – 2x + 6 = 0

x (x – 3) -2 (x – 3) = 0

(x – 3) (x – 2) = 0

x = 3 , 2

Hence, x > y

**5.** (e)

6x^{2} + 14x = 12

3x^{2} + 7x – 6 = 0

(x + 3) (3x – 2) = 0

II.

12y^{2} – 17y + 6 = 0

12y^{2} – 8y – 9y + 6 = 0

4y (3y – 2) – 3 (3y – 2) = 0

(3y – 2) (4y – 3) = 0

Hence, x < y

### Equations and Inequations Previous Year Questions for IBPS PO 2013 Exam

**In each of the following questions two equations are given. Solve these equations and give answer:**

(a) if x ≥ y, i.e., x is greater than or equal to y.

(b) if x > y, i.e., x is greater than y.

(c) if x ≤ y, i.e., x is less than or equal to y.

(d) if x < y, i.e., x is less than y.

(e) x = y or no relation can be established between x and y **(IBPS SO 2013)**

**1.**

I. x^{2} + 5x + 6 = 0

II. y^{2} + 7y + 12 = 0

**2.**

I. x^{2} + 20 = 9x

II. y^{2} + 42 = 13y

**3.**

I. 2x + 3y = 14

II. 4x + 2y = 16

**4.**

I.

II.

**5.**

I. x^{2} + 4x + 4 = 0

II. y^{2} – 8y + 16 = 0

**6.**

If the positions of the digits of a two-digit number are interchanged, the number obtained is smaller than the original number by 27. If the digits of the number are in the ratio of 1 : 2, what is the original number? **(IBPS PO/MT 2013)**

(a) 36

(b) 63

(c) 48

(d) Cannot be determined

(e) None of these

**7.**

One of the angles of a quadrilateral is thrice the smaller angle of a parallelogram. The respective ratio between the adjacent angles of the parallelogram is 4:5. Remaining three angles of the quadrilateral are in ratio 4 : 11: 9 respectively. What is the sum of the largest and the smallest angles of the quadrilateral? (IBPS PO/MT 2013)

(a) 255°

(b) 260°

(c) 265°

(d) 270°

(e) None of these

**In the following questions,**

(a) if x > y

(b) if x ≥ y

(c) if x < y

(d) if x ≤ y

(e) if x = y or the relationship cannot be established

**8.**

I. 12x^{2} +11x + 12 = 10x^{2} + 22x

II. 13y^{2} – 18y + 3 = 9y^{2} – 10y

**9.**

I.

II. y^{3} + 9.68 + 5.64 = 16.95

**10.**

I.

II.

**11.**

I.

II.

**12.**

I.

II.

**Equations and Inequations Previous Year Questions for IBPS PO 2013 Answers**

**1.** (a)

x^{2} + 5x + 6 = 0

x^{2} + 2x + 3x + 6 = 0

⇒ x(x + 2) + 3(x + 2) = 0

⇒ (x + 3)(x + 2) = 0

⇒ x = -3 or – 2

II. y^{2} + 7y +12 = 0

⇒ y^{2} + 4y + 3y +12 = 0

⇒ y(y + 4) + 3(y + 4) = 0

⇒ (y + 3)(y + 4) = 0

⇒ y= -3 or – 4

On comparing the value of eqn. (i) and equ. (ii)

x > y

**2.** (d)

I. x^{2} – 9x + 20 = 0

⇒ x^{2} – 5x – 4x + 20 = 0

⇒ x(x – 5) – 4 (x – 5) = 0 = (x – 4) (x – 5) = 0

x = 4or 5

II. y^{2} -13y + 42 = 0

⇒ y^{2} – 7y – 6y + 42 = 0

⇒ y(y – 7) – 6(y – 7) = 0

⇒ (y – 6)(y – 7) =0

⇒ y =6 or 7

Here, y > x

**3.** (d)

2x + 3 y =14 —— (I)

4x+ 2y =16 —— (II)

By equation (I) x 2 – equation II.

4 x + 6y – 4 x – 2 y = 28 – 16

⇒ 4 y =12 ⇒ y = 3

From equation I,

2x + 3 x 3 = 14

Here, y > x

**4.** (e)

I.

II.

No relation can be established between x and y.

**5.** (d)

I. x^{2} + 4x +4 = 0

(x + 2)^{2} = 0

⇒ x = – 2

II. y^{2} – 8 y +16 = 0

⇒( y – 4 )^{2} = 0

⇒ y = 4

Here, y > x

**6.** (b)

Let one’s digit = x

ten’s digit = 2x

Number =10 (2x) + x = 21 x

After interchange the digit number = 12x

∴ 21x – 12x=27

9x = 27

x = 3

∴ one’s digit = 3

Let’s digit = 2 × 3 = 6

Number = 10 × 6 + 3 = 63

**7.** (b)

Let the adjacent angles of the parallelogram be 4x and 5x.

∴ 4x + 5x = 180 or 9x = 180 ∴ x=20

One angle of quadrilateral = 3 × 80° = 240°

Again, sum of angles of quadrilateral

4y + 11y + 9y + 240° = 360°

24y = 120° ∴ y = 5

Hence, the sum of the largest and the smallest angles of the quadrilateral = 4 × 5 + 240 = 260°

**8.** (b)

I. 12x^{2} + 11x + 12 = 10x^{2} + 22x

2x^{2} – 11x + 12 = 0

2x^{2} – 8x – 3x + 12 = 0

(x – 4) (2x – 3) = 0

x = 4, x = 3/2

II. 13y^{2} – 18y + 3 = 9y^{2} – 10y

4y^{2} – 8y + 3 = 0

4y^{2} – 6y – 2y + 3 = 0

(2y – 3) (2y – 1) = 0

∴ x ≥ y

**9.** (c)

II. y^{3} + 9.68 + 5.64 = 16.95

⇒ y^{3} = 16.95 – 15.32

**10.** (a)

I. 35x + 70 = 0

II.

⇒ 3y + 7 = 0 ⇒ 3y = – 7

**11.** (a)

I.

II.

**12.** (d)

I.

II.

∴ x ≤ y

### Equations and Inequations Previous Year Questions for IBPS PO 2012 Exam

**1.**

Rachita enters a shop to buy ice-creams, cookies and pastries. She has to buy atleast 9 units of each. She buys more cookies than ice-creams and more pastries than cookies. She picks up a total of 32 items. How many cookies does she buy ?** (IBPS PO/MT 2012)**

(a) Either 12 or 13

(b) Either 11 or 12

(c) Either 10 or 11

(d) Either 9 or 11

(e) Either 9 or10

**2.**

The fare of a bus is ₹ X for the first five kilometers and ₹ 13/- per kilometer thereafter. If a passenger pays ₹ 2402/ for a journey of 187 kilometers, what is the value of X ? **(IBPS PO/MT 2012)**

(a) ₹ 29/-

(b) ₹ 39/-

(c) ₹ 36/-

(d) ₹ 31/-

(e) None of these

**Equations and Inequations Previous Year Questions for IBPS PO 2012 Answers**

**1.** (c)

By options

(a) Either 12 or 13

then ice-creams should not be given atleast 9. This can be rejected.

(b) Either 11 or 12

Ice-cream should be atleast 9. By this combination ice cream gets less than 9.

(c) Either 10 or 11

By giving cookies 10 or 11, we get all the possible condition fulfilled.

(d) and (e), the ice-cream distribution can be more than cookies which violates our condition.

∴ option (c) is the write answer.

**2**. (c)

₹ [(x for first 5 km) + 13 x remaining kms] = Total pay

₹ x + ₹ 13 × 182 = ₹ 2402

x + 2366 = 2402

x = ₹ 36

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