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

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

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Quadratic Equations

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 ≥ y

d) If x ≤ y

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

Note: √2= 1.414    √3= 1.732    √5=2.236     √7=2.645   √11=3.316   √13=3.6   √17=4.123

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

II) 15y2 -219y +720=0

 

2) I) 13x2 – 31x – 210=0

II) 7y2 – 4y -96=0

 

3) I) 11x2 – 51√3x +162=0

II) 3y2 -44√5y +765=0

 

4) I) 7x2-(6√5+21)x +18√5=0

II) 3y2+(7√17-15√11)y -35√187=0

 

5) I) 3x2-32√3x+252=0

II) y2-14√5y -255=0

 

6) I) x2 -20√7x +525=0

II) y2-8√11y- 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 – 2835=0

 

10) I) 2x2-(14+3√7)x +21√7=0

II) 3y2 + (9√7-2√11)y -6√77=0

Answers :

Directions (1-5) :

1) Answer: b)

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

15y2– 75y – 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

2) Answer: e)

13x2 – 31x – 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 – 4y -96=0

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

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

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

y= -24/7 & 4

=-3.43 & 4

No relation.

3) Answer: b)

11x2 – 51√3x +162=0 11x2 -33√3x -18√3x+162=0

11x(x-3√3)-183(x-3√3)=0

(11x-18√3)(x-3√3)= 0

x=18√3/11  &  3√3

=2.83 & 5.196

3y2 -44√5y +765=0

3y2– 27√5y – 17√5y +765=0

3y(y-9√5) – 17√5(y-95)=0

(3y-17√5)(y-9√5)=0

y=17√5/3 & 9√5

=12.67 & 20.124

y>x

4) Answer: e)

7x2-(6√5+21)x +18√5=0

7x2 – 6√5x -21x +18√5=0

x(7x-6√5)-3(7-6√5)=0

(x-3)(7x-6√5)=0

x=3 & 6√5/7

=3 & 1.59

3y2+(7√17-15√11)y -35√187=0

3y2 + 7√17y -15√11y-35√187=0

y(3y+7√17)-5√11(3y+7√17)=0

(y-5√11)(3y+7√17)=0

y=5√11 & -7√17/3

= 16.58 & -9.62

No relation

5) Answer: e)

3x2-32√3x+252=0

3x2– 18√3x -14√3x +252=0

3x(x-6√3)- 14√3(x-6√3)= 0

(3x-14√3)(x-6√3)=0

x= 14√3/3 & 6√3

=8.0826 & 10.392

y2-14√5y -255=0

y2-17√5y + 3√5y -255=0

y(y-17√5) +3√5(y-17√5)=0

(y-17√5)(y+3√5)=0

y= 17√5 & -3√5

= 38.012 & -6.708

No relation.

6) Answer: e)

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 figure  divide by the coefficient of the variable with the highest power.

Thus for x2 -20√7x +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 √7 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 7√11 & 5√11

y=23.212 & 16.58

No relation

7) Answer: a)

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

8) Answer: e)

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.

9) Answer: e)

3x2 -6x -240=0

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

X= -10 & 8

2y2 -51y – 2835=0

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

Y= 52.5 & -27

No relation

10) Answer: e)

2x2-(14+3√7)x +21√7=03y2 + (9√7-2√11)y -6√77=0

2x2 – 14x -3√7x + 21√7=0

2x(x-7)-3√7(x-7)=0

(2x-3√7)(x-7)=0

X= 3√7/2 & 7

=7.935 & 7

3y(y+3√7) -2√11(y+3√7)=0

3y2 + 9√7y – 2√11y -6√77=0

(3y-2√11)(y+3√7)=0

y=2.21 & 7.935

No relation

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