IBPS PO Mains: Quantitative Aptitude Test

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In the following questions two equations numbered I and II are given .You have to solve both the equations and based on that give answer

Give answer (1) if x>y

Give answer (2) if x≥y

Give answer (3) if x<y

Give answer (4) if x≤y

Giver answer(5) if x=y or the relationship cannot be established .

Q1.    I.   12x² -28x+15=0
          II  4y² -20y+21=0

Q2.    I.  3x² +11x+10=0
          II  2y² +13y+21=0

Q3.    I.  4x² -29x+21=0
          II  3y² -19y+28=0

Q4.    I.   2x² -13x+21=0
          II  5y² -22y+21=0

Q5.    I.   12x² -28x+15=0
          II  4y² -20y+21=0

Q6.    I.   6x² -5x+1=0
          II  12y² -17y+6=0

Q7.    I.   2x² +7x+6=0
          II  2y² +11y+15=0

Q8.    I.   20x² -9x+1=0
          II  12y² -7y+1=0

Q9.    I.   3x² +23x+44=0
          II  3y² +20y+33=0

Answers:

Q1-(4)

Q2-(1)

Q3-(5)

Q4-(4)

Q5-(2)

Q6-(3)

Q7-(1)

Q8-(4)

Q9-(4)

Solutions :

 

Problem 1:  To solve these type of equation based questions ,we have to make factors of both these statements and find the value of x and y to answer these questions.

Statement I

12x² -28x+15=0

12x²-18x-10x+15=0

6x(2x-3)-5(2x-3)

(2x-3)(6x-5)

X=3/2,or 5/6

Statement II

4y²-20y+21=0

4y²-14y-6y+21=0

2y(2y-7)-3(2y-7)=0

(2y-7)(2y-3)=0

Y=7/2,or 3/2

Conclusion

x=y=3/2

7/2>5/6 means y>x

So in conclusion x and y are equal and y is greater than x

y≥x

Answer is (4)

Problem 2:

Solution:

Statement 1 

I.  3x² +11x+10=0

3x²+6x+5x+10=0

3x(x+2)+5(x+2)=0

(x+2)(3x+5)=0

X=-2,x=-5/3

Statement II

  2y² +13y+21=0                  

2y²+6y+7y+21=0

2y(y+3)+7(y+3)=0

(y+3)(2y+7)=0

y=-3,-7/2

We got the value of x and y , now based on that we have to find the conclusion .

As both of value of x i.e -2 and -5/3 are greater than y so conclusion x>y is correct .

Answer –(1)

Problem 3:

Solution : Statement I

 I.  4x² -29x+21=0

4x²-20x-9x+45=0

4x(x-5)-9(x-5)=0

(x-5)(4x-9)=0

X=5,9/4

Statement II

  II  3y² -19y+28=0

3y²-12x-7x+28=0

3y(y-3)-7(x-4)=0

(y-3)(3y-7)=0

y=3,7/3

In this when we compared the values of x and y ,value 5 of x is greater than 3 and 7/3 but value of y =7/3 is greater than 9/4 , so no definite conclusion

Answer (5) No relationship cannot be established.

Problem 4:

Solution: 2x² -13x+21=0

2x²-6x-7x+21=0

2x(x-3)-7(x-3)=0

(2x-7)(x-3)=0

X=7/2,3

 5y² -22y+21=0

 5y²-15y-7y+21=0

5y(y-3)-7(y-3)=0

(y-3)(5y-7)

Y=3,7/5

y is either equal or greater than x so  x≤y answer (4)

Problem 5:

Solution:

I. 12x²+7x+3x+21=0

X(x+7) +3(x+3)=0

(x+7)(x+3)

X=-7,-3

II.  4y²+11y+7y+77=0

y(y+11)+7(y+11)=0

(y+11)(y+7)=0

Y=-11,-7

So x≥y answer (2)

Problem 6:

I. 6x²-3x-2x+1=0

3x(2x-1)-1(2x-1)=0

(2x-1)(3x-1)=0

X=1/2,1/3=0.5,.33

II. 12y²-9y-8y+6=0

3y(4y-3)-2(4y-3)

(4y-3)(3y-2)=0

y=3/4,2/3=.75,.66

y>x

Problem 7:

I. 2x²+4x+3x+6=0

2x(x+2)+3(x+2)=0

(2x+3)(x+2)=0

X=-3/2,-2

II. 2y² +6y+5y+15=0

   2y(y+3)+5(y+3)=0

Y=-3,-5/2    x>y  answer (1)

Problem 8

I. 20x²-5x-4x+1=0

5x(4x-1)-1(4x-1)=0

(4x-1)(5x-1)=0

X=1/4,1/5

II. 12y²-4y-3y+1=0

4y(3y-1)-1(3y-1)=0

(4y-1)(3y-1)=0

Y=1/4,1/3

y≥x

Problem 9:

I.  3x²+12x+11x+44=0

3x(x+4)+11(x+4)=0

(x+4)(3x+11)

X=-4,-11/3

II. 3y²+9y+11y+33=0

3y(y+3)+11(y+3)=0

(y+3)(3y+11)=0

Y=-3,-11/3

y≥x

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