Equations and Inequations Questions for IBPS Clerk
Equations and Inequations Previous Year Questions for IBPS Clerk 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 Clerk main 2016)
1. 3x2 + 11x + 6 = 0
3y2 + 10y + 8 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established
2. 2x2 – 13x + 20 = 0, 2y2 – 7y + 6 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established
3. y2 – x2 = 32, y – x = 2
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established
4. x2 – 15x + 28 = 0
4y2 – 23y + 30 = 0
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established
5. (x – 4) (x – 3) = (x – 6) (x – 5) (y – 9) (y – 3) = (y – 4) (y – 3)
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or relation cannot be established
Equations and Inequations Previous Year Questions for IBPS Clerk 2016 Answers
1. (e)
3x2 + 11x + 6 = 0
3x2 + 9x + 2x+6 = 0
x = -0.66, -3
3y2 + 10y + 8 = 0
3y2 + 6y + 4y + 8 = 0
y = -1.33, -2
Put on number line
-3, -2, -1.33, -0.66
2. (a)
2x 2 – 13x + 20 = 0
2x2 – 8x – 5x + 20 = 0
x = 2.5, 4
2y2 – 7y + 6 = 0
2y2 – 3y – 4y + 6 = 0
y = 1.5, 2
Put on number line
1.5, 2, 2.5, 4
3. (b)
y 2 – x2 = 32
(y + x)(y – x) = 32
From equation (2) y – x = 2
(y + x)2 = 32
y + x = 16
y – x = 2
2y = 18 ⇒ y = 9
9 – x = 2
x = 7
4. (e)
2x2 -15x + 28 = 0
2x2 -8x -7x + 28 = 0
x = 4, 3.5
4y2 – 23y + 30 = 0
4y2 – 15y – 8y + 3 = 0
y =3.75, 2
Put on number line
2, 3.5, 3.75, 4
5. (a)
(x – 4) (x – 3) = (x – 6) (x – 5)
4x = 18
x = 4.5
(y – 9) (y – 3) = (y – 4) (y – 3)
y = 3
Equations and Inequations Previous Year Questions for IBPS Clerk 2015 Exam
(a) if x > y
(b) if x < y
(c) if x ≥ y
(d) if x ≤ y
(e) if x = y or relation cannot be established between ‘x’ and ‘y’. (IBPS Clerk 2015)
1. x2 – 1 = 0, y2 + 4y + 3 = 0
2. x2 – 10x + 24=0, y2 – 14y + 48 = 0
3. 2x2 – 13x + 20=0, 2y2 -7y + 6 = 0
4. [latex]\frac { 15 } { \sqrt { x } } + \frac { 9 } { \sqrt { x } } = \frac { 11 } { \sqrt { x } } , \left( \frac { \sqrt { y } } { 4 } \right) + \left( \frac { 5 \sqrt { y } } { 12 } \right) = \left( \frac { 1 } { \sqrt { y } } \right)[/latex]
5. x4 – 227 = 398, y2 + 321 = 346
Equations and Inequations Previous Year Questions for IBPS Clerk 2015 Answers
1. (c)
x2 = 1
x = ± 1
y2 + 4y + 3 = 0
y2 + y + 3y + 3 = 0
y = -1, -3
2. (d)
x2 – 10x + 24 = 0
x2 – 6x – 4x + 24 0
x = 4, 6
y2 – 14y + 48 = 0
y2 – 6y – 8y + 48 = 0
y = 6, 8
3. (a)
2x2 – 13x + 20 = 0
2x2 – 8x – 5x + 20 = 0
x = 2.5, 4
2y2 – 7y + 6 = 0
2y2 – 3y – 4y + 6 = 0
y = 1.5, 2
4. (a)
[latex]( 15 / \sqrt { x } ) + ( 9 / \sqrt { x } ) = 11 \sqrt { x }[/latex]
[latex]24 / \sqrt { x } = 11 \sqrt { x }[/latex]
[latex]\mathrm { x } = 24 / 11 = 2.18[/latex]
[latex]( \sqrt { y } / 4 ) + ( 5 \sqrt { y } / 12 ) = ( 1 / \sqrt { y } )[/latex]
[latex]( 8 \sqrt { y } / 12 ) = ( 1 / \sqrt { y } )[/latex]
y = 1.5
5. (e) x4 – 227 = 398
x4 = 625
Take square root on both sides
x2 = 25
x = 5, -5
y2 + 321 = 346
y2 = 25
y2 = 25
y = ± 5
Equations and Inequations Previous Year Questions for IBPS Clerk 2013 Exam
1. 12 yr ago the ratio between the ages of A and B was 3 : 4 respectively. The present age of A is 3 3/4 times of C’s present age. If C’s present age is 10 yr, then what is B’s present age? (in years) (IBPS Clerk 2013)
(a) 48
(b) 46
(c) 60
(d) 54
(e) 36
2. A and B are two numbers. 6 times of square of B is 540 more than the square of A. If the respective ratio between A and B is 3 : 2, what is the value of B? (IBPS Clerk 2013)
(a) 10
(b) 12
(c) 16
(d) 8
(e) 14
Equations and Inequations Previous Year Questions for IBPS Clerk 2013 Answers
1. (d)
[latex]\frac { \mathrm { A } + 12 } { \mathrm { B } + 12 } = \frac { 3 } { 4 }[/latex]
[latex]\mathrm { A } = \frac { 15 } { 4 } \mathrm { C }[/latex]
[latex]\mathrm { A } = \frac { 15 } { 4 } \times 10 = 37.5[/latex]
[latex]\frac { 37.5 + 12 } { B + 12 } = \frac { 3 } { 4 }[/latex]
B = 54
2. (b)
6B2 = A2 + 540
[latex]\frac { \mathrm { A } } { \mathrm { B } } = \frac { 3 } { 2 }[/latex]
[latex]\mathrm { A } = \frac { 3 \mathrm { B } } { 2 }[/latex]
[latex]6 \mathrm { B } ^ { 2 } = \frac { 9 \mathrm { B } ^ { 2 } } { 4 } + 540[/latex]
3.75B2 = 540
[latex]\mathrm { B } = \sqrt { 144 } = 12[/latex]
Equations and Inequations Previous Year Questions for IBPS Clerk 2012 Exam
1. The cost of 5 pens and 8 pencils is ₹ 31. What would be the cost of 15 pens and 24 pencils? (IBPS Clerk 2012)
(a) ₹ 93
(b) ₹ 99
(c) ₹ 96
(d) Cannot be determined
(e) None of these
2. ‘A’, ‘B’ and ‘C’ are three consecutive even integers such that four times ‘A’ is equal to three times ‘C’. What is the value of B? (IBPS Clerk 2012)
(a) 12
(b) 10
(c) 6
(d) 14
(e) None of these
3. The sum of the squares of two odd numbers is 11570. The square of the smaller number is 5329. What is the other number? (IBPS Clerk 2012)
(a) 73
(b) 75
(c) 78
(d) 79
(e) None of these
Equations and Inequations Previous Year Questions for IBPS Clerk 2012 Answers
1. (a)
Cost of 5 pens + 8 pencils = ₹ 31
On multiplying by 3
15 pens + 24 pencils = 3 × 31 = ₹ 93
2. (d)
Let A = x,
B = x + 2,
C = x + 4
∴ According to the Question
4x = 3(x + 4)
⇒ 4x – 3x + 12 ⇒ x = 12
∴ B = x + 2 = 12 + 2 = 14
3. (d)
(larger number)2
= 11570 – 5329 = 6241
∴ Larger number [latex]= \sqrt { 6241 } = 79[/latex]
Equations and Inequations Previous Year Questions for IBPS Clerk 2011 Exam
1. Mala’s monthly income is two-third of Kajal’s monthly income. Kajal’s annual income is ₹ 4,32,000, What is Mala’s annual income? (In some cases monthly income and in some cases annual income is used) (IBPS Clerk 2011)
(a) ₹ 2,92,000
(b) ₹ 2,63,500
(c) ₹ 2,48,200
(d) ₹ 2,88,000
(e) None of these
2. At present Sheetal is three times Surabhi’s age. After seven years Sheetal will be twice Surabhi’s age then. How many times will Sheetal’s age be in another fourteen years time with respect to Surabhi’s age then? (IBPS Clerk 2011)
(a) 1
(b) 3
(c) 1
(d) 1.5
(e) None of these
3. On teacher’s day sweets were to be equally distributed amongst 540 children. But on that particular day 135 children remained absent, hence each child got 2 sweets extra. How many sweets was each child originally supposed to get? (IBPS Clerk 2011)
(a) 4
(b) 8
(c) 10
(d) 6
(e) Cannot be determined
4. At present, Tarun is twice the age of Vishal and half the age of Tanvi. After Four years, Tarun will be 1.5 times Vishal age and Tanvi will be 2.5 times Vishal age, what is Tanvi present age? (IBPS Clerk 2011)
(a) 12 years
(b) 8 years
(c) 20 years
(d) 16 years
(e) None of these
5. At present, Aruna’s age is 1.5 times the age of Pujan and twice the age of Somy. After six years, Aruna will be 1.4 times the age of Pujan and Somy will be 0.8 times the age of Pujan then, what is the present age of Somy? (IBPS Clerk 2011)
(a) 30 years
(b) 36 years
(c) 24 years
(d) 18 years
(e) None of these
6. At present, Anil’s age is 1.5 times the age of Purvi. Eight years hence, the ratio of the ages of Anil and Purvi will be 25 : 18. What is Purvi’s present age? (IBPS Clerk 2011)
(a) 50 years
(b) 28 years
(c) 42 years
(d) 36 years
(e) None of these
7. At present Kavita is twice Sarita’s age. Eight years hence, the respective ratio between Kavita’s and Sarita’s ages then will be 22: 13. What is Kavita’s present age? (IBPS Clerk 2011)
(a) 26 years
(b) 18 years
(c) 42 years
(d) 36 years
(e) None of these
Equations and Inequations Previous Year Questions for IBPS Clerk 2011 Answers
1. (d)
Kajal’s annual income = ₹ 4,32,000
Kajal’s Monthly income = [latex] \frac { 4,32,000 } { 12 }[/latex] = ₹ 36,000
Mala’s Monthly income = [latex]\frac { 2 } { 3 } \times 36,000[/latex] =₹ 24,000
Mala’s annual income = ₹ 24,000 × 12 = ₹ 2,88,000
2. (d)
Surabhi’s present age = x year
Sheetal’s present age = 3x year
After seven years,
3x + 7 = 2(x + 7)
⇒ 3x + 7 = 2x +14
⇒ x = 7
∴ Surabji’s present age = 7 year
Sheetal’s present age = 21 years
After another 14 years i.e. after 21 year
Surabhi’s age = 28 years
Sheetal’s age = 42 years
∴ Required answer = [latex]\frac { 42 } { 28 } = 1.5[/latex]
3. (d)
Let total numbers of Sweets be x
Each child got [latex]\left( \frac { x } { 540 } \right)[/latex] Sweets
On a Particular day no. of children = 540 – 135 = 405
On that day each child got [latex]\left( \frac { x } { 405 } \right)[/latex] Sweets
According to question
[latex]\frac { x } { 405 } = \frac { x } { 540 } + 2[/latex]
[latex]x \left( \frac { 1 } { 405 } – \frac { 1 } { 540 } \right) = 2[/latex]
[latex]x \left( \frac { 135 } { 405 \times 540 } \right) = 2[/latex]
[latex]x = \frac { 2 \times 405 \times 540 } { 135 } = 3240[/latex]
Each child originally get [latex]\left( \frac { 3240 } { 540 } \right) = 6[/latex] Sweets
4. (d)
Let Tanvi’s age be x years
∴ Trun’s age [latex]= \frac { x } { 2 }[/latex]
∴ Vishal’s age is [latex]\frac { x } { 4 }[/latex] years
After four years,
[latex]( x + 4 ) = \left( \frac { x } { 4 } + 4 \right) 2.5[/latex]
or, [latex]x + 4 = \frac { 2.5 x } { 4 } + 10[/latex]
or, 4x + 16 = 2.5x + 40
or, 1.5x = 24
or, [latex]x = \frac { 24 } { 1.5 } = 16[/latex]
5. (d)
Let Punjan’s present age be x years
Aruna’s present age = 1.5x years
Somy’s present age [latex]= \frac { 1.5 x } { 2 }[/latex] years
After six years,
Punjan’s age = (x + 6) years
Aruna’s age = (1.5x + 6) years
Now, 1.4(x + 6) = (1.5x + 6)
or, 1.4 + 8.4 = 1.5x + 6
or, 0.7x – 24
or, x = 24 years
∴ Somy’s present age [latex]= \frac { 1.5 \times 24 } { 2 } = 18[/latex] years
6. (b)
Let Purvi’s present age be x years
Anil’s present age = 1.5x years.
Age after eight eight years [latex]= \frac { 1.5 x + 8 } { x + 8 } = \frac { 25 } { 18 }[/latex]
or, 27x + 144 = 25x + 200 or, 2x = 56
∴ x = 28
7. (d)
Let sarita’s age = x year, then Kavita’s age = 2x year
According to question after eight years, [latex]\frac { 2 x + 8 } { x + 8 } = \frac { 22 } { 13 }[/latex]
⇒ 26x + 104 = 24x + 176
⇒ 26x – 22x = 176 – 104
⇒ 4x = 72 ∴ [latex]x = \frac { 72 } { 4 } = 18[/latex]
∴ Kavita’s present age = 2x = 2 × 18 = 36 years
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