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IBPS PO Mains: Quantitative Aptitude Test

Published on Monday, October 19, 2015
aptitude

Do it in 5 minutes
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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