# SBI PO Quantitative Aptitude Questions 2019 (Day-43) High Level New Pattern

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Dear Aspirants, the most awaited notification of SBI PO – 2019 has been released. 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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SBI PO Quantitative Aptitude Questions 2019 (Day-43) High Level New Pattern

Directions (1-6):Study the following information carefully and answer the given questions.

Arjun goes to park and runs for 60 minutes without stoppage. For starting 15 minutes he runs at a uniform speed of 5 km/hr and after that he runs at a uniform speed of 9km/hr for remaining time. He runs total (A) km in the Park. After that he comes to his house and get ready for Library which is 75 km away from his house. He reaches Library in 1.5 hours at 9:00 a.m.

In Library he gives some questions to his Friends Jinny and Mitu at (B).Jinny can solve those questions in 12 hours. While efficiency of Jinny and Mitu is in the ratio 5: 4.Jinny and Mitu together completes 60% of that work at 2:30 p.m. Arjun and Mitu together can solve same questions in 6 hours. Arjun is (C)% more efficient than Jinny. After solving questions, he comes back to home in upstream (Speed of stream is 5km/hr and his speed in still water and distance between his house and library are same as earlier). He takes (D) hours to reach home.

When he reaches home, one of his friends Manjeer come at his house. Two start to play a game in which 2 coins are used by each person. (E)is the number of possible outcomes for Arjun. In a game, both two throw their coins, in which if head comes, it is marked as 2 and when tail comes, it is marked as 3  and each one of them get a certain sum and the result contains the same outcomes in both the times (like (2,2)). Winner is the one who gets highest number as the sum of the square of the number comes in coins. (F)Should be the outcomes of the coins of Manjeer if Manjeer is winner of the game.

1) What value will come at the place of ‘A’?

a) 4.25 km

b) 3.75 km

c) 8 km

d) 5.25 km

e) None of these

2) What value will come at the place of ‘B’?

a) 10:45 a.m.

b) None of these

c) 11 a.m.

d) 10:30 a.m.

e) 10 a.m.

3) What value will come at the place of ‘C’?

a) 44(4/9)%

b) 20%

c) 25%

d) 80%

e) 50%

4) What value will come at the place of ‘D’?

a) 2 hours

b) 1.5 hours

c) 1 (7/8) hours

d) 1 (2/3) hours

e) 1 (4/11) hours

5) What value will come at the place of ‘E’?

a) 64

b) 216

c) 36

d) 8

e) 4

6) What value will come at the place of ‘F’?

a) None of these

b) Cannot be determined

c) 3 and 3

d) 2 and 3

e) 2 and 2

7) A consignment should deliver on time to customer for which Naman start his journey with uniform speed, after 2 hours customer call Naman to deliver his order 1 hour before the decided time. Naman increased his speed by 50% to deliver it 1 hour before the decided time. Find the total time taken by Naman to deliver the consignment?

a) 4 hours

b) 5 hours

c) 6 hours

d) 7 hours

e) 3 hours

8) Cost price of a Nokia mobile is 40% more than cost price of a Lenovo mobile. Shopkeeper marked up Nokia mobile at 25% above its cost price and Lenovo mobile at 50% above its cost price. If shopkeeper allowed discount of 20% on Lenovo mobile and 25% on Nokia mobile and the difference between selling price of Nokia  mobile and Lenovo mobile was Rs 675 then find sum of marked price of one Nokia mobile and one Lenovo mobile?

a) Rs 20000

b) Rs 20500

c) Rs 19000

d) Rs 18500

e) Rs 19500

9) The speed of the truck is 150 % more than that of the speed of the bus and the speed of the bus is 25 % less than that of speed of the car. If the car can take to travel certain distance in 9 hours, then find the difference between the time taken by the truck and that of car to cover the same distance?

a) 3 hours 40 mins

b) 4 hours 12mins

c) 5 hours 28mins

d) 2 hours 50 mins

e) None of these

10) Ashok, Karthi and Mahesh entered into a partnership to construct a building by investing in the ratio of 5 : 4 : 7. After one year, Ashok invested Rs. 15000 more and after another one more year, Mahesh invested Rs. 12000 more. At the end of 3 years, their profits are in the ratio of 30 : 20 : 37. Find the initial investment of Karthi?

a) Rs. 40000

b) Rs. 35000

c) Rs. 20000

d) Rs. 25000

e) None of these

Directions (1 – 6):

Total Distance (in Km), he runs in the park = (15/60)*5 + (45/60)*9 = 8km = A

Efficiency of Jinny, Mitu and Arjun is:

J     M      A

5     4

100           100+C

100   80    100+C

According to the question:

(M+A)*6 =J*12

(180+C)*6 = 100*12

C=20

C is 20% (As percentage increase is asked)

Total work= Jinny efficiency*12hours = 100*12=1200

60% of 1200= 720

720 work is completed by Jinny and Mituin = 720/180= 4hrs

So, B=2:30-4= 10:30am

Now, Time takes to reach home from library (D)= Distance/speed = 75km/(still speed-stream speed)

Still speed= 75km/1.5hrs =50kmph

So, D= 75/(50-5) = 75/45 = 5/3hrs = 1 (2/3) hrs

E= Total Number of outcomes = 4[(T,T), (H,H), (T,H), (H,T)]

According to the question:

Result can be [(2,2), (3,3)]

Square of the sum of outcomes can highest only in the case is (3,3), As, 3^2+3^2 =18. While 2^2 +2^2 =8. So, 2 and 2, (F) should be the outcomes of the coins of Manjeer, if Manjeer is winner of the game.

Total Distance (in Km), he runs in the park = (15/60)*5 + (45/60)*9 = 8 km = A

Efficiency of Jinny, Mitu and Arjun is:

J       M       A

5       4

100            100+C

100    80     100+C

According to the question:

(M+A)*6 =J*12

(180+C)*6 = 100*12

C=20

C is 20% (As percentage increase is asked)

Total work= Jinny efficiency*12 hours = 100*12=1200

60% of 1200= 720

720 work is completed by Jinny and Mituin = 720/180= 4 hrs

So, B = 2:30 – 4 = 10:30 am

Efficiency of Jinny, Mitu and Arjun is:

J       M       A

5       4

100            100+C

100    80     100+C

According to the question:

(M+A)*6 =J*12

(180+C)*6 = 100*12

C=20

C is 20% (As percentage increase is asked)

Now, Time takes to reach home from library (D) = Distance/speed = 75km/(still speed-stream speed)

Still speed= 75km/1.5hrs = 50 kmph

So, D= 75/(50-5) = 75/45 = 5/3 hrs = 1 (2/3) hrs

E= Total Number of outcomes = 4 [(T,T), (H,H), (T,H), (H,T)]

According to the question:

Result can be [(2, 2), (3, 3)]

Square of the sum of outcomes can highest only in the case is (3, 3), As, 3^2+3^2 =18. While 2^2 +2^2 = 8. So, 2 and 2, (F) should be the outcomes of the coins of Manjeer, if Manjeer is winner of the game.

Let, total time be= 2hrs+x

As, the speed is increased by 50%,

Initial      Final

Speed 2               3

Time   3               2   diff= 1hrs, So, value of time, x= 2hrs

Total Time= 2+2 = 4hrs

Let CP of Lenovo mobile is 100x

CP of Nokia mobile is (100x + 40%*100x) = 140x

MRP of Lenovo = (100x + 50%*100) = 150x

Discount on Lenovo given is 20% So,

SP of Lenovo = (150x – 20%*150x)

SP = 120x …………. (i)

MRP of Nokia = (140x + 25%*100) = 175x

And discount given is 25% So,

SP = (175x – 25%*175x)

SP = 131.25x …………….. (ii)

ATQ,

Difference between selling price of Nokia mobile and Lenovo mobile was Rs 67.5

Subtracting Equ i from ii,

131.25x – 120x = 675

11.25x = 675

x = 60

Sum of marked price of one Nokia mobile and one Lenovo mobile

= 675/11.25 *325

=Rs. 19500

The speed of the truck = 150 % more than that of the speed of the bus

The ratio of speed of the truck to that of the speed of the bus

= >250 : 100 = 5 : 2

The speed of the bus = 25 % less than that of speed of the car

The ratio of speed of the bus to that of speed of the car = 75 : 100 = 3 : 4

The ratio of speed of the truck, bus and car = 15 : 6 : 8

The ratio of time taken by the truck, bus and car to cover the certain distance

= > (1/15) : (1/6) : (1/8) = 8 : 20 : 15

15’s = 9 hours

1’s = 9/15 = 3/5 hours

The difference between the time taken by the truck and that of car to cover the same distance

= > (15 – 8)’s = 7’s = (3/5)*7 = 21/5 hours = 4 hours 12mins

The share of Ashok, Karthi and Mahesh

= > [5x*1 + (5x + 15000)*2] : [4x*3] : [7x*2 + (7x + 12000)*1] = 30 : 20 : 37

= > [5x + 10x + 30000] : 12x : [14x + 7x + 12000] = 30 : 20 : 37

= > [15x + 30000] : 12x : [21x + 12000] = 30 : 20 : 37

According to the question,

= > 12x/(21x + 12000) = 20/37

= > 111x = 105x + 60000

= > 6x = 60000

= > x = 10000

Initial investment of Karthi = 4x = Rs. 40000

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