# Numerical Ability for SBI Clerk Prelims Exam 2018

**Dear Students,**

**Numerical Ability for SBI Clerk Prelims 2018**

Numerical Ability Section has given heebie-jeebies to the aspirants when they appear for a banking examination. As the level of every other section is only getting complex and convoluted, there is no doubt that this section, too, makes your blood run cold. The questions asked in this section are calculative and very time-consuming. But once dealt with proper strategy, speed and accuracy, this section can get you the maximum marks in the examination. Following is the with the best of latest pattern questions.

**Numerical Ability quiz**to help you practice**Directions (Q1-5): In these questions, two equations numbered I and II are given. You have to solve both the equations and give**answer :

**(a) if x < y**

**(b) if x > y**

**(c) if x ≤ y**

**(d) if x ≥ y**

**(e) if x = y or relationship between x**and y cannot be determined

**Q1. I. x² – 9x + 18 = 0**

**II. 5y² – 22y + 24 = 0**

**S1. Ans.(b)**

Sol.

I. x² – 9x + 18 = 0

x² – 6x – 3x + 18 = 0

x(x – 6) –3 (x -6) = 0

(x – 3) (x – 6)= 0

x= 3, 6

II. 5y² – 22y + 24 = 0

5y² – 10y – 12y + 24 = 0

5y(y- 2) –12 (y – 2) = 0

(y – 2) (5y – 12) = 0

y=2,12/5

∴ x > y

**Q2. I. 6x² + 11x + 5 = 0**

**II. 2y² + 5y + 3 = 0**

**S2. Ans.(d)**

Sol.

I. 6x² + 11x + 5 = 0

6x² + 6x + 5x + 5 = 0

6x (x + 1) + 5 (x + 1) = 0

(x+ 1) (6x +5)= 0

x=-1,-5/6

II. 2y² + 5y + 3 = 0

2y² + 2y + 3y + 3 = 0

2y(y+ 1) + 3 (y+ 1) = 0

(y+ 1) (2y + 3) = 0

y= -1,-3/2

∴x≥y

**Q3. I. x² + 10x + 24 = 0**

**II. y² – √625=0**

**S3. Ans.(e)**

Sol.

I. x² + 10x + 24 = 0

x² +6x + 4x + 24 = 0

x(x + 6) +4 (x+ 6)= 0

(x + 4) (x + 6) = 0

x = –4, –6

II. y² – √625=0

y²=√625

y^2=25;y= ±5

∴ Relationship between x and y cannot be determined

**Q4. I. 10x² + 11x + 1 = 0**

**II. 15y² + 8y + 1 = 0**

**S4. Ans.(e)**

Sol.

I. 10x² +11x + 1= 0

10x² + 10x+ x+ 1 = 0

10x (x + 1) + 1 (x + 1) = 0

(x + 1) (10x + 1)= 0

x= -1,-1/10

II. 15y² + 8y + 1 = 0

15y² +5y + 3y + 1 = 0

5y (3y + 1)+ 1 (3y+ 1) = 0

(3y + 1) (5y + 1)= 0

y=-1/3,-1/5

∴ Relationship between x and y cannot be determined

**Q5. I.15x² – 11x + 2 = 0**

**II. 10y² – 9y + 2 = 0**

**S5. Ans.(c)**

Sol.

I. 15x² – 11x + 2 = 0

15x² – 5x – 6x + 2 = 0

5x(3x – 1) – 2 (3x – 1)= 0

(3x – 1) (5x – 2)= 0

x=1/3,2/5

II. 10y² – 9y + 2 = 0

10y² – 5y – 4y + 2 = 0

5y(2y – 1) –2 (2y – 1) = 0

(2y – 1) (5y - 2)=0

y=1/2,2/5

∴ x ≤ y

**Directions (Q.6-10): In the following**questions two equations numbered I and II are given. You have to solve both the equations and give answer

**(a) if x < y**

**(b) if x > y**

**(c) if x ≤ y**

**(d) if x ≥ y**

**(e) if x = y or relationship between x**and y cannot be determined

**Q6. I. √x-√6/√x=0**

**II. y^3-6^(3/2)=0**

**Q7. I. 3x-2y=10**

**II. 5x-6y=6**

**Q8. I. x^2+x-12=0**

**II. y^2-5y+6=0**

**Q9. I. x^2+9x+18=0**

**II. y^2-13y+40=0**

**Q10. I. √(x+6)=√121-√36**

**II. y^2+112=473**

**Directions (Q11-15) : In each of the these questions, two equation (I) and (II) are given. You have to solve both the equations and give answer**

**(a) if x < y**

**(b) if x > y**

**(c) if x ≤ y**

**(d) if x ≥ y**

**(e) if x = y or no relationship can be established between x and y.**