# SBI Clerk Prelims 2020 Quantitative Aptitude Questions (Day-03)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for SBI Clerk Prelims 2020 so the aspirants can practice it on a daily basis. These questions are framed by our skilled experts after understanding your needs thoroughly. Aspirants can practice these new series questions daily to familiarize with the exact exam pattern and make your preparation effective.

[WpProQuiz 7607]

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Directions (1 – 10): Solve the given equations and find out the values of x and y. Mark the answer

a) If x > y

b) If x < y

c) If x â‰¥ y

d) If x â‰¤ y

e) If x = y, or relation cannot be established

Note: âˆš2= 1.414Â Â Â  âˆš3= 1.732Â Â Â  âˆš5=2.236Â Â Â Â  âˆš7=2.645Â Â  âˆš11=3.316Â Â  âˆš13=3.6Â Â  âˆš17=4.123

1) I) 17x2 +14x -216=0

II) 15y2 -219y +720=0

2) I) 13x2 â€“ 31x â€“ 210=0

II) 7y2 â€“ 4y -96=0

3) I) 11x2 – 51âˆš3x +162=0

II) 3y2 -44âˆš5y +765=0

4) I) 7x2-(6âˆš5+21)x +18âˆš5=0

II) 3y2+(7âˆš17-15âˆš11)y -35âˆš187=0

5) I) 3x2-32âˆš3x+252=0

II) y2-14âˆš5y -255=0

6) I) x2 -20âˆš7x +525=0

II) y2-8âˆš11y- 1155=0

7) I) 2x2 -49x + 272=0

II) 11y2 – 10y -21=0

8) I) 7x2 -42x -385=0

II) 3y2-8y-315=0

9) I) 3x2 -6x -240=0

II) 2y2 -51y â€“ 2835=0

10) I) 2x2-(14+3âˆš7)x +21âˆš7=0

II) 3y2 + (9âˆš7-2âˆš11)y -6âˆš77=0

Directions (1-5) :

17x2 +14x -216=0

17x2+68x -54x -215=0

17x(x+4) -54(x+4)

(17x-54)(x+4)=0

x= 54/17 & -4

=3.18 & -4

15y2 -219y +720=0

15y2â€“ 75y â€“ 144y +720=0

15y(y-5)-144(y-5)=0

(15y-144)(y-5)= 0

y= 144/15 & 5

=9.6 & 5

Thus y>x

13x2 â€“ 31x â€“ 210=0

13x2 -70x + 39x -210=0

x(13x -70) + 3(13x -70)=0

(x+3)(13x-70)= 0

x=-3 & 70/13

=-3 & 5.38

7y2 â€“ 4y -96=0

7y2 -28y +24y -96 =0

7y(y-4) +24(y-4)=0

(7y+24)(y-4)=0

=-3.43 & 4

No relation.

11x2 – 51âˆš3x +162=0 11x2 -33âˆš3x -18âˆš3x+162=0

11x(x-3âˆš3)-183(x-3âˆš3)=0

(11x-18âˆš3)(x-3âˆš3)= 0

x=18âˆš3/11Â  &Â  3âˆš3

=2.83 & 5.196

3y2 -44âˆš5y +765=0

3y2â€“ 27âˆš5y â€“ 17âˆš5y +765=0

3y(y-9âˆš5) â€“ 17âˆš5(y-95)=0

(3y-17âˆš5)(y-9âˆš5)=0

y=17âˆš5/3 & 9âˆš5

=12.67 & 20.124

y>x

7x2-(6âˆš5+21)x +18âˆš5=0

7x2 – 6âˆš5x -21x +18âˆš5=0

x(7x-6âˆš5)-3(7-6âˆš5)=0

(x-3)(7x-6âˆš5)=0

x=3 & 6âˆš5/7

=3 & 1.59

3y2+(7âˆš17-15âˆš11)y -35âˆš187=0

3y2 + 7âˆš17y -15âˆš11y-35âˆš187=0

y(3y+7âˆš17)-5âˆš11(3y+7âˆš17)=0

(y-5âˆš11)(3y+7âˆš17)=0

y=5âˆš11 & -7âˆš17/3

= 16.58 & -9.62

No relation

3x2-32âˆš3x+252=0

3x2– 18âˆš3x -14âˆš3x +252=0

3x(x-6âˆš3)- 14âˆš3(x-6âˆš3)= 0

(3x-14âˆš3)(x-6âˆš3)=0

x= 14âˆš3/3 & 6âˆš3

=8.0826 & 10.392

y2-14âˆš5y -255=0

y2-17âˆš5y + 3âˆš5y -255=0

y(y-17âˆš5) +3âˆš5(y-17âˆš5)=0

(y-17âˆš5)(y+3âˆš5)=0

y= 17âˆš5 & -3âˆš5

= 38.012 & -6.708

No relation.

Shortcut to solving any quadratic equation is to first multiply the constant number with the coefficient of the variable with the highest power and then divide with the numerical value within the under root.; then factorize them to get the middle term and then change the signs to the opposite, then multiply with the under root and figureÂ  divide by the coefficient of the variable with the highest power.

Thus for x2 -20âˆš7x +525=0, we get 525 upon multiplying 1 (coefficient of x2) with 525. Now dividing 525 with 7 we get 75, which we then factorize to get -20. Thus the factors will be -15 & -5. Changing the signs we get 15 & 5. Multiplying with âˆš7 and then dividing with the coefficient of x2, we get the final values of x = 39.675 & 13.225

Similarly for y the factors are 7âˆš11 & 5âˆš11

y=23.212 & 16.58

No relation

2x2 – 49x + 272=0

Factors = -32 & -17

X=16 & 8.5

11y2 – 10y -21=0

Factors = -21 & 11

y = 21/11 & -1

=1.909 & -1

x>y

7x2 -42x -385=0

Factors are= -77 & 35

x=11 & -5

3y2-8y-315=0

Factors are = -35 & 27

y= 11.67 & -9

No relation.

3x2 -6x -240=0

After multiplying 240 & 3 we get the factors = -30 & 24

X= -10 & 8

2y2 -51y â€“ 2835=0

After multiplying 2 & 2835 we get the factors = -105 & 54

Y= 52.5 & -27

No relation

2x2-(14+3âˆš7)x +21âˆš7=03y2 + (9âˆš7-2âˆš11)y -6âˆš77=0

2x2 â€“ 14x -3âˆš7x + 21âˆš7=0

2x(x-7)-3âˆš7(x-7)=0

(2x-3âˆš7)(x-7)=0

X= 3âˆš7/2 & 7

=7.935 & 7

3y(y+3âˆš7) -2âˆš11(y+3âˆš7)=0

3y2 + 9âˆš7y – 2âˆš11y -6âˆš77=0

(3y-2âˆš11)(y+3âˆš7)=0

y=2.21 & 7.935

No relation

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