# RRB PO Mains Quantitative Aptitude 2021 (Day-03)

Dear Aspirants, Our IBPS Guide team is providing new series of Quantitative Aptitude Questions for RRB PO 2021 Mains 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.

### Start Quiz

Directions (01-04): In each of the following questions, two equations are given. You have to solve both the equations to find the relation between x and y.

1) I: x2 + 11x – 126 = 0

II: y2 – 19y + 84 = 0

A. x < y

B. x > y

C. x ≤ y

D. x ≥ y

E. Relationship between x and y cannot be determined

2) I)
x2 – 33x + 272 = 0

II) y2– 28y + 195 = 0

A.x > y

B.x ≥ y

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

D.x < y

E.x ≤ y

3) I)
x2 – 14x – 207 = 0

II) y2+ 25y + 144 = 0

A.x > y

B.x ≥ y

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

D.x < y

E.x ≤ y

4) I)
x2 – 31x + 238 = 0

II) y2– 25y + 154 = 0

A.x > y

B.x ≥ y

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

D.x < y

E.x ≤ y

Number series

Directions (5-7): The questions below are based on the given Series-I. The series-I satisfy a certain pattern, follow the same pattern in Series-II and answer the questions given below.

6) I) 6, 10, 1, 26, -23

II) 89 …… 60. If 60 is the nthterm find the value of n

A.5

B.6

C.7

D.9

E.8

6) I) 128, 66, 68, 104, 210

II) 72 ….. 126. If 126 is the nthterm, find the value of n

A.5

B.3

C.8

D.4

E.6

7) I) 1, 10, 21, 34, 49

II) 33—-81.If 81 is the nth term, find the value of n

A.2

B.3

C.4

D.5

E.6

Approximation

Directions (8-9): What approximate value should come in the place of (?) in the following questions.

8) √4224 * √15 + ? * 9.99 = 39.91 * 8.91

A.10

B.13

C.15

D.18

E.8

9) 14.93 * 7.89 + 12.07 * 4.901 + 23.98 * 5.96 = ?2

A.25

B.10

C.18

D.13

E.22

Data sufficiency

Directions (10): Following questions contain two statements as statement I and statement II. You have to determine which statement/s is/are necessary to answer the question and give answer as,

10) What is the marked price of the article?

Statement I: Cost price of the article is Rs.3600 and the seller give two successive discounts of Rs.800 and Rs.400 respectively.

Statement II: The shopkeeper gave two successive discounts 40% and 20% respectively on marked price of the article and the selling price of the article is 2400.

A.The data in statement I alone is sufficient to answer the question, while the data in statement II alone is not sufficient to answer the question

B.The data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question

C.The data either in statement I alone or in statement II alone is sufficient to answer the question

D.The data given in both statements I and II together are not sufficient to answer the question

E.The data given in both statements I and II together are necessary to answer the question

From I =>x2 + 11x – 126 = 0

=>x2 + 18x -7x – 126 = 0

=>x(x + 18) -7(x + 18) = 0

=> (x + 18) (x – 7) = 0

=> x = -18, 7

From II =>y2 – 19y + 84 = 0

=> y2 – 12y – 7y + 84 = 0

=>y(y – 12) – 7(y – 12) = 0

=> (y – 12) (y – 7) = 0

=> y = 12, 7

Hence, x ≤ y

x2 – 33x + 272 = 0

x2 – 16x – 17x + 272 = 0

x(x – 16) – 17(x – 16) = 0

(x – 16)(x – 17) = 0

x = 16, 17

y2 – 28y + 195 = 0

y2 – 13y – 15y + 195 = 0

y(y – 13) – 15(y – 13) = 0

(y – 13)(y – 15) = 0

y = 13, 15

x > y

x2 – 14x – 207 = 0

x2 – 23x + 9x – 207 = 0

x(x – 23) + 9(x – 23) = 0

(x + 9)(x – 23) = 0

x = 23, -9

y2 + 25y + 144 = 0

y2 + 9y + 16y + 144 = 0

y(y + 9) + 16(y + 9) = 0

(y + 9)(y + 16) = 0

y = -9, -16

x ≥ y

x2 – 31x + 238 = 0

x2 – 17x – 14x + 238 = 0

x(x – 17) – 14(x – 17) = 0

(x – 17)(x – 14) = 0

x = 14, 17

y2 – 25y + 154 = 0

y2 – 14y – 11y + 154 = 0

y(y – 14) – 11(y – 14) = 0

(y – 14)(y – 11) = 0

y = 14, 11

x ≥ y

Series I,

6 + 22 = 10

10 – 32 = 1

1 + 52 = 26

26 – 72 = -23

Series II,

89 is the 1st term

89 + 22 = 93

93 – 32 = 84

84 + 52 = 109

109 – 72 = 60

60 is 5th term.

Series –I,

128 * 0.5 + 2 = 66

66 * 1 + 2 = 68

68 * 1.5 + 2 = 104

104 * 2 + 2 = 210

Series –II,

72 is the 1st term

72 * 0.5 + 2 = 38

38 * 1 + 2 = 40

40 * 1.5 + 2 = 62

62 * 2 + 2 = 126

126 is 5th term

Series I,

1 + 9 = 10

10 + 11 = 21

21 + 13 = 34

34 + 15 = 49

Series II,

33 is 1st term

33 + 9 = 42

42 + 11 = 53

53 + 13 = 66

66 + 15 = 81

81 is 5th term.

√4224 * √15 + ? * 9.99 = 39.91 * 8.91

260 + ? * 10 = 360

? = 10

14.93 * 7.89 + 12.07 * 4.901 + 23.98 * 5.96 = ?2

120 + 60 + 144 = ?2

? = 18

From statement I,

CP = Rs.3600

SP is not given.

So, Statement I alone is not sufficient to answer the question.

From statement II,

MP * 60/100 * 80/100 = 2400

MP = 5000

So, Statement II alone is sufficient to answer the question. 