## Tuesday, June 26, 2018

### Quadratic Equations Comparision With Solution Asked in SBI Clerk Prelims 2018(23rd June & 24th June)

Solved Quadratic Equations Asked in SBI Clerk
Dear students, we are sharing answer and detailed explanation of Quadratic Equations(Numerical Ability) Asked in SBI Clerk Prelims 2018 Exam held on 23rd June and 24th June.

Quadratic Equations Asked in SBI Clerk Prelims 2018(23 June):
Question1) In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly:
Give Answer:
(a)x>y
(b)x<y
(c)x=y or relation can’t be established between x & y
(d)x≥y
(e)x≤y
I.x2 – 5x + 6 = 0
II.y2 – 9y + 20= 0
Answer & Explanation:
Solve I.x2 – 5x + 6 = 0
Or,x2 – 5x + 6 = 0
Or,x2 – 3x-2x + 6 = 0
Or,x(x – 3)-2(x – 3) = 0
Or,(x-2)(x – 3) = 0
X=2 or 3
Solve II.y2 – 9y + 20= 0
Or, y2 – 5y– 4y + 20= 0
Or,y(y – 5)-4(y – 5) = 0
Or,(y-5)(x – 4) = 0
y=5 or 4
Now, Compare x and y:
2<5 means x<y
2<4 means x<y
3<5 means x<y
3<4 means x<y
Therefore x<y is the correct answer.
Question 2) In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly:
Give Answer:
(a)x>y
(b)x<y
(c)x=y or relation can’t be established between x & y
(d)x≥y
(e)x≤y
I.6x2 – 11x + 4 =0
II.50y2 – 25y + 3 =0
Answer & Explanation:
Solve I.6x2 – 11x + 4 =0
Or, 6x2 – 8x-3x + 4 =0
Or, 2x(3x –4) -1(3x – 4) =0
Or, (2x -1)(3x – 4) =0
Or, x=1/2=0.5 or x=4/3=1.33
Solve II.50y2 – 25y + 3 =0
Or, 50y2 – 15y– 10y + 3 =0
Or, 5y(10y – 3)– 1(10y – 3) =0
Or, (5y– 1)(10y – 3) =0
Or,y=1/5=0.2 or y=3/10=0.3
Now, Compare x and y:
0.5>0.2 means x>y
0.5>0.3 means x>y
1.33>0.2 means x>y
1.33>0.3 means x>y
Therefore x>y is the correct answer.
Question 3) In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly:
Give Answer:
(a)x>y
(b)x<y
(c)x=y or relation can’t be established between x & y
(d)x≥y
(e)x≤y
I.x2 – 4x – 12 =0
II.y2 – 5y – 14 =0
Answer & Explanation:
Solve I.x2 – 4x – 12 =0
Or, x2 – 6x+ 2x – 12 =0
Or, x(x – 6) + 2(x – 6) =0
Or, (x + 2)(x – 6) =0
Or,x=-2 or 6
Solve II.y2 – 5y – 14 =0
Or, y2 – 7y+ 2y – 14 =0
Or, y(y – 7)+ 2(y –7) =0
Or, (y+ 2)(y –7) =0
Or,y=-2 or 7
Now, Compare x and y:
-2>-2 means x=y
-2<7 means x<y
6>-2 means x>y
6<7 means x<y
Therefore No relationship can be established because >,< & = never occur together.
Question 4) In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly:
Give Answer:
(a)x>y
(b)x<y
(c)x=y or relation can’t be established between x & y
(d)x≥y
(e)x≤y
I.3x2 + 8x + 4=0
II.6y2 + 7y + 2 =0
Answer & Explanation:
Solve I.3x2 + 8x + 4=0
Or, 3x2 + 6x+ 2x + 4=0
Or, 3x(x + 2) + 2(x + 2)=0
Or, (3x + 2) + (x + 2)=0
Or, x=-2/3=-0.66 or x=-2
Solve II.6y2 + 7y + 2 =0
Or, 6y2 + 4y + 3y + 2 =0
Or, 2y(3y + 2) +1( 3y + 2) =0
Or, (2y+1)( 3y + 2) =0
Or,y=-1/2=-0.5 or y=-2/3=-0.66
Now, Compare x and y:
-0.66>-0.5 means x<y
-0.66=-0.66 means x=y
-2>-0.5 means x<y
-2<-0.66 means x<y
Therefore x≤y  is the correct answer.