SBI Clerk Prelims 2021 Quantitative Aptitude Questions (Day-05)

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.

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Quadratic equations

Directions (01-05): Following question contains two equations as I and II. You have to solve both equations and determine the relationship between them and give answer as,

1) I) x2 + 17x – 84 = 0

II)y2– 22y + 21 = 0

A.x > y

B.x ≥ y

C.x = y or relationship cannot be determined.

D.x < y

E.x ≤ y


2) I)
 2x2 – 30x + 108 = 0

II) y2– 11y + 30 = 0

A.x > y

B.x ≥ y

C.x = y or relationship cannot be determined.

D.x < y

E.x ≤ y


3) I)
 x2 – 27x + 182 = 0

II)y2– 12y – 189 = 0

A.x = y or relationship cannot be determined.

B.x ≥ y

C.x > y

D.x < y

E.x ≤ y


4) I)
 x2 + 17x + 72 = 0

II)y2– 2y – 80 = 0

A.x > y

B.x ≥ y

C.x = y or relationship cannot be determined.

D.x < y

E.x ≤ y


5) I) 
x2 – 13x – 14 = 0

II)y2+ 8y + 12 = 0

A.x < y

B.x ≥ y

C.x = y or relationship cannot be determined.

D.x > y

E.x ≤ y


6) A car travels from A to B at the speed of 60 kmph then reached B at 6 am. If the car decreased the speed by 10 kmph, then reached B at 7 am, then what is the distance between A and B?

A.150 km

B.180 km

C.240 km

D.210 km

E.None of these


7) 20 boys can finish a work in 15 days and 15 girls can finish the same work in 12 days. If 40 boys and 20 girls work together, in how many days required to finish the work?

A.4 1/11 days

B.3 1/11 days

C.6 1/11 days

D.5 1/11 days

E.None of these


8)
If the ratio of the length to breadth of a rectangle is 8: 3 and the perimeter of a square is 56 cm. If the radius of a circle is half of the side of the square and the circumference of the circle is equal to the perimeter of the rectangle, then what is the area of the rectangle?

A.48 cm2

B.72 cm2

C.88 cm2

D.96 cm2

E.None of these


9) If the marked price of the article is 25% more than the cost price of the article and the shopkeeper offers a discount of 15% while he gets the profit of Rs.250, then find the cost price of the article?

A.Rs.2000

B.Rs.2800

C.Rs.3600

D.Rs.4000

E.None of these


10) P, Q and R invested in a business in the ratio of 2: 5: 4. If Q invested for a period whose numerical value is 120% of Q’s investment ratio but P and R invested for one year. If profit of Q at the end of the year is Rs.25450, then what is the share of profit of R?

A.Rs.25670

B.Rs.34560

C.Rs.40720

D.Rs.78960

E.None of these


Answers :

Directions (1-5) :

1) Answer: C

x2 + 17x – 84 = 0

x2 + 21x – 4x – 84 = 0

x(x + 21) – 4(x + 21) = 0

(x – 4)(x + 21) = 0

x = 4, -21

y2 – 22y + 21 = 0

y2 – 21y – y + 21 = 0

y(y – 21) – 1(y – 21) = 0

(y – 1)(y – 21) = 0

y = 1, 21

Relationship between x and y cannot be established.


2) Answer: B

2x2 – 30x + 108 = 0

÷2 = > x2 – 15x + 54 = 0

x2 – 9x – 6x + 54 = 0

(x – 9) (x – 6) = 0

x = 9, 6

y2 – 11y + 30 = 0

y2 – 6y – 5y + 30 = 0

y(y – 6) – 5(y – 6) = 0

(y – 5)(y – 6) = 0

y = 5, 6

x ≥ y


3) Answer: A

x2 – 27x + 182 = 0

x2 – 13x – 14x + 182 = 0

x(x – 13) – 14(x – 13) = 0

(x – 14)(x – 13) = 0

x = 14, 13

y2 – 12y – 189 = 0

y2 – 21y + 9y – 189 = 0

y(y – 21) + 9(y – 21) = 0

(y + 9)(y – 21) = 0

y = 21, -9

Relationship between x and y cannot be established.


4) Answer: E

x2 + 17x + 72 = 0

x2 + 9x + 8x + 72 = 0

x(x + 9) + 8(x + 9) = 0

(x + 8)(x + 9) = 0

x = -8, -9

y2 – 2y – 80 = 0

y2 – 10y + 8y – 80 = 0

y(y – 10) + 8(y – 10) = 0

(y + 8)(y – 10) = 0

y = -8, 10

x ≤ y


5) Answer: D

x2 – 13x – 14 = 0

x2 – 14x + x – 14 = 0

x(x – 14) + 1(x – 14) = 0

(x + 1)(x – 14) = 0

x = -1, 14

y2 + 8y + 12 = 0

y2 + 6y + 2y + 12 = 0

y(y + 6) + 2(y + 6) = 0

(y + 2)(y + 6) = 0

y = -2, -6

x > y


6) Answer: E

Distance = Speed * Time

Time = t

Distance = x

60 * t = (60 – 10) * (t + 1)

60t = 50t + 50

t = 5

Distance = 60 * 5 = 300 km


7) Answer: A

20 boys complete the work = 1/15

40 boys complete the work = 2/15 = 1/7.5

15 girls complete the work = 1/12

20 girls complete the work = 20/(15 * 12) = 1/9 days

20 girls and 40 boys together can complete the work = 1/7.5 + 1/9

= 16.5/67.5

Required time = 67.5/16.5 = 4 1/11 days


8) Answer: D

Side of the square = 56/4 = 14 cm

Radius of the circle = 14/2 = 7 cm

Circumference of the circle = 2πr = 2 * 22/7 * 7 = 44 cm

Perimeter of the rectangle = 2 * (l + b) = 44 cm

2 * (8x + 3x) = 44

11x = 22

x = 2 cm

Length of the rectangle = 8 * 2 = 16 cm

Breadth of the rectangle = 3 * 2= 6 cm

Area of the rectangle = 16 * 6 = 96 cm2


9) Answer: D

CP = x

MP = x * 125/100 = 5x/4

SP = 5x/4 * 85/100 = 17x/16

17x/16 – x = 250

x/16 = 250

x = 4000


10) Answer: C

Profit will be shared in ratio = (2 * 12): (5 * 120/100 * 5): (4 * 12)

= 24 : 30 : 48

= 4: 5: 8

R’s profit = 25450/5 * 8 = Rs.40720

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