# RRB Clerk Prelims Quantitative Aptitude Questions 2021 (Day-06)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for RRB Clerk Prelims 2021 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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Pie chart

Directions (01-05): Study the following information carefully and answer the questions given below.

The given pie chart shows the number of five different products manufactured in 2018. 1) If the ratio of the number of sold to unsold Watch and TV is 7:8 and 3:2 respectively, then what is the difference between the number of sold watch and TV together and unsold watch and TV together?

A.238

B.243

C.256

D.240

E.None of these

2) What is the difference between the number of Laptop and AC manufactured in 2018?

A.340

B.450

C.330

D.350

E.None of these

3) If the number of non defective to defective mobile, laptop and AC in the ratio of 5:4, 3:1 and 3:2 respectively, then find the sum of the number of defective mobile, laptop and AC together?

A.1570

B.1590

C.1610

D.1620

E.None of these

4) In 2019, the number of manufactured mobile and TV is increased by 20% and 10% respectively, then find the total number of mobile and TV manufactured in 2019.

A.3206

B.3346

C.3118

D.3406

E.None of these

5) The number of defective laptop is 380 which is 76% of the number of defective AC. What is the difference between the number of defective and non defective AC in 2018?

A.800

B.750

C.600

D.700

E.850

Directions (06-10): 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,

6) I) x2 + 12x – 189 = 0

II) y2– 22y + 117 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

7) I)
x2 + x – 552 = 0

II) y2+ 51y + 650 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

8) I)
x2 – 32x + 247 = 0

II) y2– 35y + 304 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

9) I) x2 – 42x – 135 = 0

II) y2+ 42y + 117 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

10) I)
x2 – 5x + 6 = 0

II) y2–8y + 15 = 0

A.x > y

B.x ≥ y

C.x = y or relationship can’t be determined.

D.x < y

E.x ≤ y

Number of sold Watch = 15/100 * 7000 * 7/15 = 490

Number of unsold watch = 15/100 * 7000 * 8/15 = 560

Number of sold TV = 22/100 * 7000 * 3/5 = 924

Number of unsold TV = 22/100 * 7000 * 2/5 = 616

Required difference = (490 + 924) – (560 + 616) = 238

Required ratio = 5/100 * 7000 = 350

Defective mobile = 18/100 * 7000 * 4/9 = 560

Defective Laptop = 20/100 * 7000 * ¼ = 350

Defective AC = 25/100 * 7000 * 2/5 = 700

Required sum = 560 + 350 + 700 = 1610

Number of mobile in 2019 = 18/100 * 7000 * 120/100 = 1512

Number of TV in 2019 = 22/100 * 7000 * 110/100 = 1694

Required total = 1512 + 1694 = 3206

Total number of defective AC = 100/76 * 380 = 500

Total number of AC manufactured in 2018 = 25/100 * 7000 = 1750

Number of non defective AC = 1750 – 500 = 1250

Difference = 1250 – 500 = 750

x2 + 12x – 189 = 0

x2 + 21x – 9x – 189 = 0

x(x + 21) – 9(x + 21) = 0

(x – 9)(x + 21) = 0

x = 9, -21

y2 – 22y + 117 = 0

y2 – 13y – 9y + 117 = 0

y(y – 13) – 9(y – 13) = 0

(y – 9)(y – 13) = 0

y = 9, 13

Hence, x ≤ y

x2 + x – 552 = 0

x2 + 24x – 23x – 552 = 0

x(x + 24) – 23(x + 24) = 0

(x – 23)(x + 24) = 0

x = 23, -24

y2 + 51y + 650 = 0

y2 + 25y + 26y + 650 = 0

y(y + 25) + 26(y + 25) = 0

(y + 26)(y + 25) = 0

y = – 26, -25

Hence, x > y

x2 – 32x + 247 = 0

x2 – 19x – 13x + 247 = 0

x(x – 19) – 13(x – 19) = 0

(x – 13)(x – 19) = 0

x = 13, 19

y2 – 35y + 304 = 0

y2 – 19y – 16y + 304 = 0

y(y – 19) – 16( y – 19) = 0

(y – 19)(y – 16) = 0

y = 19, 16

x2 – 42x – 135 = 0

x2 – 45x + 3x – 135 = 0

x(x – 45) + 3(x – 45) = 0

(x + 3)(x – 45) = 0

x = -3, 45

y2 + 42y + 117 = 0

y2 + 39y + 3y + 117 = 0

y(y + 39) + 3(y + 39) = 0

(y + 39)(y + 3) = 0

y = – 39, -3

Hence, x ≥ y

x2 – 5x + 6 = 0

x2 – 2x – 3x + 6 = 0

x(x – 2) – 3(x – 2) = 0

(x – 3)(x – 2) = 0

x = 3, 2

y2 – 8y + 15 = 0

y2 – 5y – 3y + 15 = 0

y(y – 5) – 3(y – 5) = 0

(y – 3)(y – 5) = 0

y = 3, 5

Hence, x ≤ y 