# High Level New Pattern Quantitative Aptitude Questions For SBI PO 2019 (Day-02)

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SBI PO 2019 Notification is about to come and it is the most awaited exam among the aspirants. We all know that new pattern questions are introducing every year in the SBI PO exam. Further, the questions are getting tougher and beyond the level of the candidate’s expectations.

Our IBPS Guide is providing High-Level New Pattern Quantitative Aptitude Questions for SBI PO 2019 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 high-level questions daily to familiarize with the exact exam pattern. We wish that your rigorous preparation leads you to a successful target of becoming SBI PO.

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New Pattern Quantitative Aptitude Questions For SBI PO (Day-02)

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Direction (1 – 5): Read the following information carefully and answer the given questions:

Sachin scored 60% runs in Match1 where as score of Virat in the same match is 60 which is 40% of Sachin in Match1. Sachin scored 50% runs in Match2 which is 60 more than the runs of Virat in the same match. The total Score of Match2 is 360. The ratio between the runs of Virat and Sachin in Match3 is 2 : 3, where as the difference in their runs in Match3 is 48. The total runs of Match3 are 80% of the total score of Match5. Sachin scored 78 runs in Match4 which is 18 more than runs of Virat in the same Match. The total score of Match4 is equal to the total score of Match1. The total runs of Sachin and Virat in Match5 is 180 and the Sachin scored 60 less than runs that of Virat in the same Match.

1)

Quantity I: What is average score of Sachin in all the 5 matches together?

Quantity II: What is the average score of Virat in all the 5 matches together?

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

2) Find the total score in Match6?

Statement I: Sachin’s score in Match6 is 50% of the runs in his score in Match2 and the Virat score in Match6 is 80% of Sachin in the same match.

Statement II: Sachin scored 40% runs in Match6.

a) Only I

b) Only II

c) Either I or II sufficient

d) All I and II necessary to the answer the question

e) The question can’t be answered even with all I and II

3)

Quantity I: Virat score in Match5 is what percent of the total score in Match3?

Quantity II: Sachin score in Match3 is what percent of the total score in the same match?

a) Quantity I > Quantity II

b) Quantity I ≥ Quantity II

c) Quantity II > Quantity I

d) Quantity II ≥ Quantity I

e) Quantity I = Quantity II or Relation cannot be established

4) Find the difference between the total score of Virat in all the given five Matches together and the sum of the total score in Match4 and Match2 together?

a) 148

b) 152

c) 154

d) 147

e) None of these

5) What is the average total score of the all the five matches together?

a) 280

b) 260

c) 270

d) 250

e) None of these

Directions (6-10): Each question contains a statement followed by Quantity I, quantity II and Quantity III.

6)

Quantity I: 4, 4, 7, 19, 71, 361, 2161. Find the wrong number

Quantity II: 6, 8, 11, 17, 23, 34. Find the wrong number

Quantity III: 10, 18, 27, 90, 116, 332, 381. Find the wrong number

Which of the following should be placed in the blank spaces of the expression “Quantity I__ Quantity II___Quantity III” from left to right with respect to the above statements?

a) >, >

b) <, <

c) >, <

d) <, >

e) =, >

7)

Quantity I: x2 – 30x + 221 = 0

Quantity II: y2 – 36y + 323 = 0

Quantity III: z = √289

Which of the following should be placed in the blank spaces of the expression “Quantity I__ Quantity II___Quantity III” from left to right with respect to the above statements?

a) ≥, >

b) ≤, <

c) =, ≤

d) ≤, ≥

e) ≥, =

8)

Quantity I: Find the cost price of the item, if the selling price of the article is Rs 2500 and he got 25% profit after selling the item.

Quantity II: Rs.2500

Quantity III: Find the profit earned by selling an item, if the difference between the cost price and the selling price is 124% of 75% of 500.

Which of the following should be placed in the blank spaces of the expression “Quantity I__ Quantity II___Quantity III” from left to right with respect to the above statements?

a) <, >

b) >, >

c) ≤, ≤

d) =, ≥

e) ≥, =

9)

Quantity I: Value of (230– 229)/2

Quantity II: Value of 228

Quantity III: Value of (256 * 1024)

Which of the following should be placed in the blank spaces of the expression “Quantity I__ Quantity II___Quantity III” from left to right with respect to the above statements?

a) >, >

b) =, <

c) =, =

d) <, >

e) =, >

10) A sells an item to B at 30% profit. B sells it to C at 40% profit and C sells it to D at Rs 350 profit. The difference between the cost price of D and the cost price of A is Rs 1252. How much did B pay to A for the item?

Quantity I: Cost price of D

Quantity II: Cost price of B

Quantity III: Cost price of C

Which of the following should be placed in the blank spaces of the expression “Quantity I __ Quantity II ___ Quantity III” from left to right with respect to the above statements?

a) <, >

b) >, <

c) =, <

d) =, >

e) =, =

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Direction (1 – 5)

Match1:

Sachin score in Match1=60%

Virat score in Match1=60

Sachin score in match1=60*(100/40) =150

Total score in Match1= (150/60)*100=250

Match2:

Total score in match2=360

Sachin score in match2=360*50/100=180

Virat score in match2=180-60=120

Match3:

Sachin score in match3 = (3/1)*48 = 144

Virat score in Match3 = (2/1)*48 = 96

Total score in match3 = 144 + 96 = 240

Match4:

Sachin score in Match4=78

Virat score in match4=78-18=60

Total score in match4=250

Match5:

Total Score in Match5 = 240*(100/80) = 300

Sachin score in Match5 = x – 60

Virat score in Match5 = x

X + x – 60 =180

2x=240

X=120

Sachin score in Match5= 60

Virat score in Match5=120

From quantity I,

Average = (150+180+144+78+60)/5 = 122.4

From quantity II,

Average = (60+120+96+60+120)/5 = 91.2

From Statement I,

Sachin score in Match6=180*(50/100) = 90

Virat score in Match6=180*(80/100) = 144

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

From Statement II,

Sachin scored 40% runs in Match6.

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

From statement I, II

Total score in Match6 = 90*(100/40) = 225

From quantity I,

Required percentage = (120/240)*100 = 50%

From quantity II,

Required percentage = (144/240)*100 = 60%

Virat’s score= (60+120+96+60+120) = 456

Sum of Match4 and Match2=360+250 = 610

Difference = 610 – 456 =154

Average= (250+360+240+250+300)/5=280

Direction (6-10) :

Quantity I:

4 * 1 – 0= 4

4 * 2 – 1 = 7

7 * 3 – 2 = 19

19 * 4 – 3 =73

73 * 5 – 4 = 361

361 * 6 – 5 = 2161

Wrong number = 71

Quantity II:

8 – 6 = 2

11 – 8 = 3

16 – 11 = 5

23 – 16 = 7

34 – 23 = 11

Wrong number = 17

Quantity III:

10 + 23 = 18

18 + 32= 27

27 + 43= 91

91 + 52= 116

116 + 63= 332

332 + 72= 381

Wrong number = 90

Quantity I > Quantity II < Quantity III

Quantity I:

x2 – 30x + 221 = 0

= > (x – 17) (x – 13) = 0

= > x = 17, 13

Quantity II:

y2 – 36y + 323 = 0

= > (y – 19) (y – 17) =0

= > y = 19, 17

Quantity III:

z = √289

= > z = 17

Quantity I Quantity II Quantity III

Quantity I:

Let Rs x be the cost price of an item.

Then according to the question, we have

X + (25x/100) = 2500

X = (100/125)*2500

X = Rs 2000

Quantity II: Rs.2500

Quantity III:

Required profit=124% of 75% 500 = (124/100)*(75/100)*500 = Rs 465

Quantity I < Quantity > Quantity III

Quantity I:

(230– 229)/2 = {229(2-1)}/2 = 228

Quantity II:

228

Quantity III:

(256 * 1024)

(16*16 * 16 * 16 * 4)

(24 * 24 *24 *24 *22)

2(4+4+4+4+2)

218

Quantity I = Quantity II > Quantity III

Let the cost price of the item be Rs x,

Then CP of B=130x/100

CP of C=130x/100 * 140/100

=182x/100

Then CP of D=182x/100+350

Hence 182x/100+350-x=1252

82x/100=1252 – 350 = 902

x=Rs 1100

Quantity I:

Cost price of D = (1100 * 130/100 * 140/100) + 350

= 2352

Quantity II:

Cost price of B = 1100 * 130/100

= 1430

Quantity III:

Cost price of C = 1430 * 140/100

= 2002

Quantity I > Quantity II < Quantity III

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