Directions: In the following questions, two equations numbered are given in variables x and y. You have to solve both the equations and find out the relationship between x and y. Then give answer accordingly-
- I. 3x2 + 10x – 8 = 0,
II. 2y2 – 13y + 6 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option E
Solution:
3x2 + 10x – 8 = 0
3x2 + 12x – 2x – 8 = 0
Gives x = -2, 2/3
2y2 – 13y + 6 = 0
2y2 – 12y – y + 6 = 0
Gives y = 1/2, 6
- I. 16x2 + 8x – 15 = 0,
II. 4y2 + 29y + 30 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option C
Solution:
16x2 + 8x – 15 = 0
16x2 + 20x – 12x – 15 = 0
Gives x = -5/4, 3/4
4y2 + 29y + 30 = 0
4y2 + 24y + 5y + 30 = 0
Gives y = -6, -5/4
- I. 3x2 – 25x + 52 = 0,
II. 15y2 – 38y – 40 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option A
Solution:
3x2 – 25x + 52 = 0
3x2 – 12x – 13x + 52 = 0
Gives x = 4, 13/3
15y2 – 38y – 40 = 0
15y2 + 12y – 50y – 40 = 0
Gives y = -4/5, 10/3
- I. 12x2 – 5x – 3 = 0,
II. 4y2 – 11y + 6 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option D
Solution:
12x2 – 5x – 3 = 0
12x2 + 4x – 9x – 3 = 0
Gives x = -1/3, 3/4
4y2 – 11y + 6 = 0
4y2 – 8y – 3y + 6 = 0
Gives y= 3/4, 2
- I. 3x2 + 7x – 6 = 0,
II. 6y2 – y – 2 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option E
Solution:
Explanation:
3x2 + 7x – 6 = 0
3x2 + 9x – 2x – 6 = 0
Gives x = -3, 2/3
6y2 – y – 2 = 0
6y2 + 3y – 4y – 2 = 0
Gives y = -1/2, 2/3
Put on number line
-3… -1/2 … 2/3
- I. 4x2 + 15x + 9 = 0,
II. 4y2 – 13y – 12 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option D
Solution:
4x2 + 15x + 9 = 0
4x2 + 12x + 3x + 9 = 0
Gives x = -3, -3/4
4y2 – 13y – 12 = 0
4y2 – 16y + 3y – 12 = 0
Gives y = -3/4, 4
- I. 2x2 – (6 + √3)x + 3√3 = 0,
II. 3y2 – (9 + √3)y + 3√3 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
View Answer
Option E
Solution:
2x2 – 6x – √3x + 3√3 = 0
2x (x- 3) – √3 (x – 3) = 0,
So x = 3, √3/2 (0.7)
3y2 – 9y – √3y + 3√3 = 0
3y (y – 3) – √3 (y – 3) = 0
So x = 3, √3/3 (0.6)
- I. 2x2 – (2 + 2√5)x + 2√5 = 0
II. 4y2 – (6 + 2√2)y + 3√2 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option E
Solution:
2x2 – 2x – 2√5x + 2√5 = 0
2x (x – 1) – 2√5 (x – 1) = 0
So x = 1, √5 (2.2)
4y2 – 6y – 2√2y + 3√2 = 0
2y (2y – 3) – √2 (2y – 3) = 0
So, y = 3/2, 1/√2 (0.7)
- I. 2x2 – 15√3x + 84 = 0,
II. 3y2 – 2y – 8 = 0
A) If x > y
B) If x < y
C) If x ≥ y
D) If x ≤ y
E) If x = y or relation cannot be established
View Answer
Option A
Solution:
2x2 – 15√3x + 84 = 0
Now multiply 2 and 84 = 168
we have √3 in equation, so divide, 168/3 = 56
Now make factors so as by multiply you get 56, and by addition or subtraction you get –15
we have factors (-8) and (-7)
So 2x2 – 15√3x + 84 = 0
gives
2x2 – 8√3x – 7√3x + 84 = 0
2x (x – 4√3) – 7√3 (x – 4√3x) = 0
So x = 3.5√3, 4√3
3y2 – 2y – 8 = 0
3y2 – 6y + 4y – 8 = 0
So y = -4/3, 1
Plot on number line
-4/3……1….. 3.5√3….. 4√3
- I. 16x2 + 20x + 6 = 0
II. 10y2 + 38y + 24 = 0
A) x > y
B) x < y
C) x ≥ y
D) x ≤ y
E) x = y or relationship cannot be determined
View Answer
Option A
Solution:
Divide both equations by 2
8x2 + 10x + 3 = 0
8x2 + 4x + 6x + 3 = 0
Gives x = -1/2, -3/4
5y2 + 19y + 12 = 0
5y2 + 15y + 4y + 12 = 0
Gives y = -4, -4/5
ty mam 🙂
@Shubhra_AspirantsZone:disqus mam…… in 10th ques ans shldn’t be A???
yes A
ty mam 🙂
10?? th how
check solution.
ok mam
10th?????????????????????//
check solution
I. 2×2 – (2 + 2√5)x + 2√5 = 0
II. 4y2 – (6 + 2√2)y + 3√2 = 0
MAAM AISE QUESTIONS KO SIMPLIFY KARNA PADEGA USKE BAAD HI SOLVE KARNA HOGA
YA KOI SHORT TRICK HAI SOLVE KARNE KA?
U know the formula of discriminant method?
NO MAAM BATA DJIYE
SHAYAD PATA HO PAR NAAM SUNKE YAAD NHI AARHA
MAAM Q-9 TYPE K AUR QSNS HAI?
Check here
http://aspirantszone.com/quantitative-aptitude-quadratic-equations-set-8-new-pattern/
I ll add more
OKAY MAAM
THANK YOU VERY MUCH:)
ty mam:))
nice ques mam
thanku
Thank you mam……….
tyy mam:)
ty..
ty mam…. 7 n 8 are easy to solve but one has to noe about roots till 10
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