SBI PO 2020 Quantitative Aptitude Questions (Day-02)

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SBI PO 2020 Quantitative Aptitude Questions (Day-02)

Click Here to View Quantitative Aptitude Video Solution (Q. 1-5)

Click Here to View Quantitative Aptitude Video Solution (Q. 6-10)

Inequality

Directions (1 – 10): Each question contains a statement followed by Quantity I and Quantity II. Read the contents clearly and answer your questions accordingly.

a) If Quantity I > Quantity II

b) If Quantity I < Quantity II

c) If Quantity II ≥ Quantity I

d) If Quantity II ≤ Quantity I

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

1) Quantity 1: X, Y, Z are batsmen of a cricket team. The ratio of X: Y & Y: Z is 5:4. If the total runs scored by the three of them be 244 find out X’s score.

Quantity 2: A train takes 25 seconds to pass a platform. It takes 15 seconds to pass a man running in the opposite direction with speed 18 km/hr. Find the length of the platform if the speed of the train is 72 km/hr.

2) Quantity 1: 2 pipes X & Y can fill a tank in 20 & 25 minutes. Z can empty it in 30 minutes. X & Y are opened for 10 minutes. How much time will Z take to empty the tank?

Quantity 2: X, Y, Z can together fill a tank in 40 minutes. X & Y can together fill in 60 minutes. In what time does Z fill the entire tank?

3) Quantity 1: A shopkeeper sold an article at 15% profit. If he had sold it at 10% profit, he would have earned Rs 70 less. Find the cost price.

Quantity 2: Marked price of an article is Rs 2000. The shopkeeper gives 10% off & earns 25% profit. Find the cp.

4) Quantity 1: Difference between CI & SI on a certain sum of money for 2 years is Rs 450 at 3% rate of interest. Find the difference between CI at 4% for 3 years and SI at 4% for 4 years on the same amount.

Quantity 2: A man deposits Rs 7000 at 10% SI for 2 years. He withdrew ‘X’ after 2 years and deposited rest at CI in bank at 10% CI to get Rs 1050 interest after 2 years. Find X.

5) A’s efficiency is 66 2/3% of B

Quantity 1:  A can do 3/5 work in’ x’ days. In how many days can B finish the work?

Quantity 2: B can do 2/5 work in ‘y’ days. In how many days can A finish the work?

6)Quantity 1: There is a race between 4 athletes. Each cover 1 m, 1.1 m, 1.05 m, 1.15 m in one step. Find the minimum distance which all can cover in complete number of steps.

Quantity 2: A can eat 5.5 sweets, B can eat 6.75 sweets, C can eat 4.8 sweets and D can eat 5.25 sweets in one bite. Find the minimum number of sweets required all of them can finish all the sweets in complete number of bites.

7) Quantity 1: Find the least number of 4 digits divisible by 6,9,12 and also a perfect cube.

Quantity 2: Find the least square divisible by 36 & 49

8)  Quantity 1: Find the number of pairs of numbers who have 13 as their H.C.F. and 170 as their L.C.M.

Quantity 2: Find the number of pair of numbers which have 17 as their H.C.F. and 459 as their L.C.M.

9) Quantity 1: Find the remainder when 1010101010101 is divided by 999.

Quantity 2: find the remainder when 1100110011010 is divided by 999

10) Quantity 1: 1234 is written 27 times consecutively. Find the remainder when divided by 4

Quantity 2: Find the remainder when the same number is divided by 8.

Answers:

Directions (1-10):

1) Answer: b)

Quantity 1: Ratio of X:Y:Z = 5:4::5:4= 25:20:16

X scored = (244*25)/61= 100

Quantity 2: Relative speed of the train= 90 km/hr or 25 m/s

Thus length of the train= 25*15 m= 375 m

Thus length of the platform= (25*20) – 375 = 125m

Q2>Q1

2) Answer: b)

Quantity 1:

Total work done together in 1 minute= 15+12 – 10= 17

X & Y in 1 minute can do= 27 work

Thus in 10 minutes they can do= 270 work

Thus Z can empty 270 in = 270 / 10= 27 minutes.

Quantity 2:

X+Y+Z= 3 ; X+Y=2

Thus X= 1

Thus X can complete the work in =120/1 minutes = 120 minutes.

Q2>Q1

3) Answer: b)

Quantity 1:

Profit percentage is calculated on cp. Thus 5% = 70.

Thus cp= Rs 70 *20= Rs 1400

Quantity 2:

After giving 10% discount Selling price= Rs 1800

Thus cp= Rs 1800/ 1.25= Rs 1440

Q2>Q1

4) Answer: a)

Quantity 1:

PR2/10000= 450

Or P= (450* 10000)/9 =5,00,000

CI at 4% for 3 years = 5,00,000 *{ [1+(1/25)3] – 1 }

= 5,00,000 * {[(26)3 – (25)3]/ 253}= 62,432

SI at 4% for 4 years = (500000*4*4)/100= 80000

Thus difference = 17,568

Quantity 2:

At the end of 2 years he gets =Rs 7000 + Rs (7000*10*2)/100= Rs 8400

Cumulative interest rate at the end of 2 years= 21%

21%= Rs 1050

100% = 5000

Thus X= Rs (7000-5000)= Rs 2000

Q1>Q2

5) Answer: e)

Quantity 1:

A can finish the work in 5x/3 days.

B can finish the work in (5x/3) * (3/2) = 5x/2 days

Quantity 2:

B can complete the work in 5y/2 days.

A can finish the work in (5y/2) * (3/2)= 15y/4 days

As no relation between ‘x’ and ‘y’ is given, no relationship can be established.

Q1 < Q2

6) Answer: b)

Here we have to take the L.C.M. of all the numbers in both the cases and find out which one is greater in magnitude.

Quantity 1: L.C.M. of 1, 1.1, 1.05 and 1.15 = 531300

Quantity 2: L.C.M. 5.5, 6.75, 4.8 and 5.25 = 1663200

Q1 < Q2

7) Answer: b)

Quantity 1:

Least number of 4 digits which is a cube is 103 or 1000. But that is not divisible by any of the numbers given. Neither is the next cube 1331, cube of 11. 1728 which is the cube of 12 is divisible by all the three.

Hence Q1= 1728

Quantity 2:

L.C.M. of 36 & 49 is 1764, and it is also a perfect square. 422= 1764

Hence Q2 = 1764

Thus Q2 > Q1

8) Answer: b)

Quantity 1:

H.C.F. must be present in L.C.M. also. But in this case LC.M. is not divisible by H.C.F.

Thus no such pairs exist.

Q1 = 0

Quantity 2:

Product of the two numbers = Product of H.C.F. & L.C.M.

Let the numbers be denoted by 17x and 17y (as 17 is HCF we take 17 common, such that x & y are co-prime)

17x * 17 y = 17 * 459

xy = 27

Thus the possible prime factor pairs are = (1, 27) (3,9)

Thus 2 such pairs are possible.

Q2 = 2

Q2 > Q1

9) Answer: b)

For mod 9 = > add the numbers from right end

1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 + 0 + 1 = 7

For mod 99 = > group into 2 numbers from right end and add

1 + 01 + 01 + 01 + 01 + 01 + 01 = 7

For Q1:

For mod 999, group into 3 numbers from right end and add

1 + 010 + 101 + 010 + 101 = 223

For Q2:

For mod 999, group into 3 numbers from right end and add

1 + 100+ 110 + 011 + 010 = 232

Q2 > Q1

10) Answer: e)

To find the divisibility by 4 the last 2 digits must be 0 or divisible by 4.

Thus Q1= 34/4 or remainder = 2

For Q2 = the last 3 digits are taken: 234/8 or remainder = 2.

Thus Q1 = Q2

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