Dear Aspirants,
We have introduced "the 17 days study plan to crack IPPB Scale-1 Prelims 2017". This plan is specifically dedicated to Quantitative Aptitude and cover all types of questions asked in Bank PO exams. This will help you to prepare in a more efficient and systematic way. So, follow this post to keep updated with the plan.
We have introduced "the 17 days study plan to crack IPPB Scale-1 Prelims 2017". This plan is specifically dedicated to Quantitative Aptitude and cover all types of questions asked in Bank PO exams. This will help you to prepare in a more efficient and systematic way. So, follow this post to keep updated with the plan.
Directions(1-15): In each question two equations are given, find x and y and give the answer:
A. x > y
B. x < y
C. x ≥ y
D. x ≤ y
E. x = y or relation can not be established.
1. a) 3x² - 22x + 7 = 0
b) y² - 15y + 56 = 0
Sol.
a) 3x² - 21x - x + 7 = 0
3x(x-7) -1(x-7) = 0
(3x-1)(x-7) = 0
x = 7, 1/3.
b) y² -7y - 8y + 56 = 0
y(y-7) - 8(y-7) = 0
(y-8)(y-7) = 0
y = 8, 7.
Thus, y≥x.
2. a) 2x² - 17x + 36 = 0
b) 2y² - 19y + 44 = 0
Sol.
a) 2x² - 8x - 9x + 36 = 0
2x(x-4) - 9(x-4) = 0
(2x-9)(x-4) = 0
x = 9/2, 4.
b) 2y² - 8y - 11y + 44 = 0
2y(y-4) - 11(y-4) = 0
(2y-11)(y-4) = 0
y = 11/2, 4.
Thus, no relation.
3. a) x - √169 = 0
b) y² - 169 = 0
Sol.
a) x = 13
b) y = ±√169 = ±13.
Thus, x≥y.
4. a) 3x² + 20x + 25 = 0
b) 3y² + 14y + 8 = 0
Sol.
a) 3x² + 15x + 5x + 25 = 0
3x(x+5) + 5(x+5) = 0
(3x+5)(x+5) = 0
x = -5/3, -5.
b) 3y² + 12y + 2y + 8 = 0
3y(y+4) + 2(y+4) = 0
(3y+2)(y+4) = 0
y = -2/3, -4.
Thus, no relation.
5. a) 3x² + 5x + 2 = 0
b) 3y² + 18y + 24 = 0
Sol.
a) 3x² + 3x + 2x + 2 = 0
3x(x+1) + 2(x+1) = 0
(3x+2)(x+1) = 0
x = -2/3, -1.
b) 3(y² + 6y + 8) = 0
y² + 6y + 8 = 0
y² + 4y + 2y + 8 = 0
y(y+4) + 2(y+4) = 0
y = -2, -4.
Thus, x>y.
6. a) 6x² + 31x + 35 = 0
b) 2y² + 3y + 1 = 0
Sol.
a) 6x² + 21x + 10x + 35 = 0
3x(2x+7) + 5(2x+7) = 0
(3x+5)(2x+7) = 0
x = -5/3, -7/2.
b) 2y² + 2y + y + 1 = 0
2y(y+1) + 1(y+1) = 0
(2y+1)(y+1) = 0
y = -1/2, -1.
Thus, x<y.
7. a) 2x² + 9x + 10 = 0
b) 4y² + 28y + 45 = 0
Sol.
a) 2x² + 4x + 5x + 10 = 0
2x(x+2) + 5(x+2) = 0
(2x+5)(x+2) = 0
x = -5/2, -2.
b) 4y² +18y + 10y + 45 = 0
2y(2y+9) + 5(2y+9) = 0
(2y+5)(2y+9) = 0
y = -5/2, -9/2.
Thus, x≥y.
8. a) 15x² - 11x - 12 = 0
b) 20y² - 49y + 30 = 0
Sol.
a) 15x² - 20x + 9x - 12 = 0
5x(3x-4) +3(3x-4) = 0
(5x+3)(3x-4) = 0
x = -3/5, 4/3.
b) 20y² -25y - 24y + 30 = 0
5y(4y-5) -6(4y-5) = 0
(5y-6)(4y-5) = 0
y = 6/5, 5/4.
Thus, no relation.
9. a) 2x² - 15 = 7x
b) 17y = -7 - 6y²
Sol.
a) 2x² - 10x + 3x - 15 = 0
2x(x-5) +3(x-5) = 0
(2x+3)(x-5) = 0
x = -3/2, 5.
b) 6y² + 3y + 14y + 7 = 0
3y(2y+1) +7(2y+1) = 0
(3y+7)(2y+1) = 0
y = -7/3, -1/2.
Thus, no relation.
10. a) 3x² - 19x - 14 = 0
b) 2y² +15y + 13 = 0
Sol.
a) 3x² - 21x - 2x - 14 = 0
3x(x-7) -2(x-7) = 0
(3x-2)(x-7)
x = 2/3, 7.
b) 2y² +2y +13y + 13 = 0
2y(y+1) +13(y+1) = 0
(2y+13)(y+1) = 0
y = -13/2, -1.
Thus, x>y.
11. a) (x³ -13x + 12)/(x-1) = 0
b) (y³ + 5y² - 2y - 24)/(y-2) = 0
Sol.
a) (x³ - x - 12x + 12)/(x-1) = 0
[x(x²-1) -12(x-1)]/(x-1) = 0
[(x-1)(x+1)(x-12)]/(x-1) = 0
x² + x - 12 = 0
Solving, x = 3, -4
b) y³ + 5y² - 2y - 24/(y-2) = 0
y³ + 7y² - 2y² -14y + 12y - 24/(y-2) = 0
(y-2)(y² + 7y + 12)/(y-2) = 0
Solving, y = -4, -3.
Thus, x≥y.
12. a) y = 2x + 1
b) 2y = 3x - 1
Sol. Solving the two using substitution method,
x = -3
y = -5
Thus, x>y.
13. a) 9x² - 29x + 22 = 0
b) y² - 7y + 12 = 0
Sol.
a) 9x² - 18x - 11x + 22 = 0
9x(x-2) - 11(x-2) = 0
(9x-11)(x-2) = 0
x = 11/9, 2.
b) y² - 4y - 3y + 12 = 0
y(y-4) - 3(y-4) = 0
(y-4)(y-3) = 0
y = 3, 4.
Thus, x<y.
14. a) 3x² - 4x - 32 = 0
b) 12y² - 109y + 247 = 0
Sol.
a) 3x² - 12x + 8x - 32 = 0
3x(x-4) + 8(x-4) = 0
(3x+8)(x-4) = 0
x = -8/3, 4.
b) 12y² - 52y - 57y + 247 = 0
4y(3y - 13) - 19(y - 13) = 0
(4y-19)(3y-13) = 0
y = 19/4, 13/3.
Thus, x<y.
15. a) 4x + 7y = 42
b) 3x - 11y = -1
Sol. Solving the two equations using the substitution method,
x = 7
y = 2
Thus, x>y.
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