# SBI PO Quantitative Aptitude Questions 2019 (Day-39) 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-39) High Level New Pattern

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Directions (1 – 5): Study the following information carefully and answer the question given below:

There are six boxes, each contain five different coloured balls. The following table represents probabilities of drawing one ball of different colours from the six boxes. Some data in the following table are missing.

 Box Red Green Blue Yellow White A 1/5 — 1/10 — 3/10 B — 1/3 1/6 1/8 — C 1/5 — 2/5 — 1/5 D — 1/5 1/3 — 1/5 E 1/9 2/9 — 1/9 — F 1/4 7/20 — 3/20 —

1) Box A contains 4 red, 2 blue and 5 yellow balls. Find the respective ratio of the probability of drawing 2 green balls from box A and probability of drawing 2 yellow balls from box A.

a) 3:5

b) 3:10

c) 1:2

d) 4:9

e) None of these

2) Box C contains 3 green, 10 blue and 5 white balls. Probability of drawing three balls from the box such that 2 balls are of red colour and 1 ball is of green colour is approximately what percent of the probability of drawing 2 yellow balls from the box?

a) 351%

b) 391%

c) 289%

d) 271%

e) 411%

3) Box B contains 4 blue balls. Number of red balls in box B is two more than the number of blue balls in that box and number of white balls in box B is half of the number of red balls in that box. Find the probability of drawing five balls from box B such that each ball is of different colour.

a) 37/1881

b) 64/1661

c) 59/1991

d) 72/1771

e) None of these

4) Find the relation between following two quantities:

Quantity I: Box D contains 2 red, 3 green and 2 yellow balls. Find the probability of drawing 2 blue and 2 white balls from box D.

Quantity II: Box E contains 2 red, 4 green and 6 blue balls. Find the probability of drawing 2 yellow and 2 white balls from box E.

a) Quantity I > Quantity II

b) Quantity I < Quantity II

c) Quantity I ≥ Quantity II

d) Quantity I ≤ Quantity II

e) Quantity I = Quantity II or Relation cannot be determined.

5) Find the probability of drawing five balls from box F such that all the balls are of same colour.

Statement I: Box F contains 5 red, 7 green and 3 yellow balls.

Statement II: Number of white balls in box F is one more than the number of blue balls in the box

a) Statement I alone is sufficient to answer the question, but the statement II alone is not sufficient.

b) Statement II alone is sufficient to answer the question, but the statement I alone is not sufficient.

c) Either statement I alone or statement II alone is sufficient to answer the question.

d) Both statements I and II together are needed to answer the question.

e) Both statements I and II together are not sufficient to answer the question.

6) The selling price of two articles A and B is Rs.—— and Rs.3600 respectively and the marked price of article A is 15% above its cost price and article B is 10% above its cost price. If the cost price of article B is 50% of the cost price of article A and the discount amount of article B is Rs.——.

a) Rs.8970, Rs.680

b) Rs.9200, Rs.800

c) Rs.9430, Rs.1000

d) Rs.9545, Rs.960

a) Only B

b) Only A, B

c) Only C

d) Only B, D

e) Only A

7) Rahul travels D km at the speed of x kmph in t hours. If he travels the same distance at the speed of (x + 4) kmph, he will reach in (t – 2) hours.

From the statement given in the above question which of the following can be determined.

a) Value of x

b) Value of D

c) Value of t

d) Cannot be determine

a) Only A and B

b) Only C and B

c) All A, B, and C

d) Only D

e) Only C

8) A, B and C three friends invests their amount on two different schemes S and T and scheme S offer the simple interest and T offers the compound interest. A invested Rs.6000 on scheme S at 12% per annum for 5 years. The interest received by A and B in the ratio of 9:8 and B invest his amount on scheme S at 8% per annum for 5 years. C invested the amount on scheme T which is 20% more than that of the amount invested by B at the rate of 10% per annum for three years.

From the statement given in the above question which of the following can be determined.

a) Amount invested by B

b) Amount invested by C

c) Interest received by C after three years.

d) Difference between the interest received by A and C

a) Only A and B

b) Only C and D

c) All A, B, C and D

d) Only D

e) Cannot be determine

9) If the ratio of the length of the train A to B is 3:2 and the ratio of the speed of the train A to train B is 4:5. Train A crosses train B in 20 seconds while travelling in opposite directions and also train B cross the pole in 14.4 seconds.

From the statement given in the above question which of the following can be determined.

a) Length of train A

b) Length of train B

c) Time taken by train B to cross a pole

d) Speed of train B

a) All A, B, C and D

b) Only B and C

c) Only D and B

d) Only A

e) Cannot be determine

10) The ratio of the present ages Ram to Sunil is 3:4 and the product of the ages of Sunil and Rahul is 864 years. If the ratio of the ages of Ram to Shon is 3:2 and the product of the ages of Shon and Sunil is 648 years.

From the statement given in the above question which of the following can be determined.

a) Age of Ram

b) Age of Rahul

c) If the ratio of the ages of Rahul to Surya is 4:3, then the age of Surya.

d) If the sum of the ages of Shon and Nitish is 30 years, then the age of Nitish.

a) All A, B, C and D

b) Only B and C

c) Only D and B

d) Only A

e) Cannot be determine

In box A:

Red = 4

Blue = 2

Yellow = 5

Let,

Green = a

White = b

=> Total number of balls = 4 + 2 + 5 + a + b = 11 + a + b

4/(11 + a + b) = 1/5

=> 11 + a + b = 20

=> a + b = 20 – 11

=> a + b = 9 ———— (i)

b/(11 + a + b) = 3/10

=> 10b = 33 + 3a + 3b

=> 10b – 3a – 3b = 33

=> 7b – 3a = 33 ———— (ii)

3 x equation (i) + equation (ii)

3a + 3b + 7b – 3a = 27 + 33

=> 10b = 60

=> b = 60/10

=> b = 6

From (i)

a + 6 = 9

=> a = 9 – 6

=> a = 3

Hence,

Green = 3

White = 6

Total number of balls = 11 + 3 + 6 = 20

Probability of drawing 2 green balls from box A = 3c2/20c2

= 3/190

Probability of drawing 2 yellow balls from box A = 5c2/20c2

= 1/19

Required ratio = 3/190 : 1/19

= 3:10

In box C:

Green = 3

Blue = 10

White = 5

Let,

Red = a

Yellow = b

Total = 3 + 10 + 5 + a + b = 18 + a + b

10/(18 + a + b) = 2/5

=> 5/(18 + a + b) = 1/5

=> 25 = 18 + a + b

=> a + b = 25 – 18

=> a + b = 7 ————– (i)

a/(18 + a + b) = 1/5

=> 5a = 18 + a + b

=> 5a – a – b = 18

=> 4a – b = 18 —————– (ii)

Equation (ii) + Equation (i)

4a – b + a + b = 18 + 7

=> 5a = 25

=> a = 25/5

=> a = 5

From (i)

5 + b = 7

=> b = 7 – 5

=> b = 2

Hence,

Red = 5

Yellow = 2

Total = 18 + 5 + 2 = 25

Probability of drawing three balls from the box such that 2 balls are of red colour and 1 ball is of green colour = (5c2 x 3c1)/25c3

= (10 x 3)/2300

= 3/230

Probability of drawing 2 yellow balls from the box = 2c2/25c2

= 1/300

Required percentage = (3/230)/(1/300) x 100

= 391.30%

= 391% approx.

In box B:

Blue = 4

Red = 4 + 2 = 6

White = 6/2 = 3

Let,

Green = a

Yellow = b

Total = 4 + 6 + 3 + a + b = 13 + a + b

4/(13 + a + b) = 1/6

=> 24 = 13 + a + b

=> a + b = 24 – 13

=> a + b = 11 ———- (i)

a/(13 + a + b) = 1/3

=> 3a = 13 + a + b

=> 3a – a – b = 13

=> 2a – b = 13 ————- (ii)

Equation (i) + Equation (ii)

a + b + 2a – b = 11 + 13

=> 3a = 24

=> a = 24/3

=> a = 8

From (i)

8 + b = 11

=> b = 11 – 8

=> b = 3

Hence,

Green = 8

Yellow = 3

Total = 13 + 8 + 3 = 24

Required probability = (6c1 x 8c1 x 4c1 x 3c1 x 3c1)/24c5

= (6 x 8 x 4 x 3 x 3)/42504

= 72/1771

Quantity I:

In box D:

Red = 2

Green = 3

Yellow = 2

Let,

Blue = a

White = b

Total = 2 + 3 + 2 + a + b = 7 + a + b

3/(7 + a + b) = 1/5

=> 15 = 7 + a + b

=> a + b = 15 – 7

=> a + b = 8 ————- (i)

a/(7 + a + b) = 1/3

=> 3a = 7 + a + b

=> 3a – a – b = 7

=> 2a – b = 7 ————— (ii)

Equation (i) + Equation (ii)

a + b + 2a – b = 8 + 7

=> 3a = 15

=> a = 15/3

=> a = 5

From (i)

5 + b = 8

=> b = 8 – 5

=> b = 3

Hence,

Blue = 5

White = 3

Total = 7 + 5 + 3 = 15

Required probability = (5c2 x 3c2)/15c4

= (10 x 3)/1365

= 2/91

Quantity II:

In box E:

Red = 2

Green = 4

Blue = 6

Let,

Yellow = a

White = b

Total = 2 + 4 + 6 + a + b = 12 + a + b

2/(12 + a + b) = 1/9

=> 18 = 12 + a + b

=> a + b = 18 – 12

=> a + b = 6 ———— (i)

a/(12 + a + b) = 1/9

=> 9a = 12 + a + b

=> 9a – a – b = 12

=> 8a – b = 12 ————— (ii)

Equation (i) + Equation (ii)

a + b + 8a – b = 6 + 12

=> 9a = 18

=> a = 18/9

=> a = 2

From (i)

2 + b = 6

=> b = 6 – 2

=> b = 4

Hence,

Yellow = 2

White = 4

Total = 12 + 2 + 4 = 18

Required probability = (2c2 x 4c2)/18c4

= (1 x 6)/3060

= 1/510

Hence, Quantity I > Quantity II

From I:

In box F:

Red = 5

Green = 7

Yellow = 3

Let,

Blue = a

White = b

Total = 5 + 7 + 3 + a + b = 15 + a + b

5/(15 + a + b) = ¼

=> 20 = a + b + 15

=> a + b = 20 – 15

=> a + b = 5 ——- (i)

7/(15 + a + b) = 7/20

=> 15 + a + b = 20

=> a + b = 20 – 15

=> a + b = 5

3/(15 + a + b) = 3/20

=> 15 + a + b = 20

=> a + b = 20 – 15

=> a + b = 5

From II:

White = Blue + 1

From I and II:

b = a + 1

From (i)

a + b = 5

=> a + a + 1 = 5

=> 2a = 5 – 1

=> a = 4/2

=> a = 2

=> b = 2 + 1 = 3

Hence,

Blue = 2

White = 3

Total = 15 + 2 + 3 = 20

Required probability = (5c5 + 7c5)/20c5

= (1 + 21)/15504

= 22/15504

= 11/7752

Hence, both statements I and II together are needed to answer the question.

From option (A)

CP of article A =x

CP of article B =x/2

SP of article A=(115/100)*x=8970

x=7800

CP of article B=7800/2 =Rs.3900

MP of article B=3900*110/100=Rs.4290

Discount amount of article B=4290-3600=Rs.690

This not satisfied the given condition.

From Option (B)

CP of article A =x

CP of article B =x/2

SP of article A=115/100*x=9200

x=8000

CP of article B=Rs.4000

MP of article B=4000*110/100=Rs.4400

Discount amount of article B=4400-3600=Rs.800

This satisfies the given condition.

From Option (C)

CP of article A =x

CP of article B =x/2

SP of article A=115/100*x=9430

x=8200

CP of article B=Rs.4100

MP of article B=4100*110/100=Rs.4510

Discount amount of article B=4510-3600=Rs.910

This not satisfied the given condition.

From Option (D)

CP of article A =x

CP of article B =x/2

SP of article A=115/100*x=9545

x=8300

CP of article B=Rs.4150

MP of article B=4150*110/100=Rs.4565

Discount amount of article B=4565-3600=Rs.965

This not satisfied the given condition.

Distance travelled in t hours at x kmph is,

D=x * t——(1)

Distance travelled in t-2 hours at x+4 kmph is,

D=(x + 4) * (t – 2)

D=xt – 2x + 4t – 8

D=D – 2x + 4t – 8—–(sub (1))

2x – 4t + 8 =0

If speed (x) or time (t) is given we can find other values otherwise we can’t determine any value

The interest received by A is,

SI=6000 * 12 * 5/100=Rs.3600

Interest received by B=3600 * 8/9=3200

Amount invested by Bin SI=x

3200 = X * 8 * 5/100

X=8000

Amount invested by C=8000 * 120/100=9600

The interest received by C is,

CI=9600 * (1 + 10/100)3 – 9600

CI=3177.6

Difference of interest received by A and C=3600 – 3177.6=Rs.422.4

We can find the answers of all the given questions.

Distance travelled by A is,

5x=9y * (5/18) * 20

X=10y

Length of train B is,

2x=5y * (5/18) * 14.4

2x=20y

X=10y

We cannot find the answer of any given questions.

The ratio of ages of Ram, Sunil and Shon is,

Ram: Sunil: Shon=3:4:2

The product of the ages of Shon and Sunil is,

4x * 2x=648

X=9 years

Shon’s age=2 * 9=18 years

Sunil’s age=4 * 9=36 years

Ram’s age=3 * 9=27 years

The product of the ages of Sunil and Rahul is 864 years. So,

Rahul’s age=864/36=24 years

The ratio of the ages of Rahul to Surya is 4:3, then

Surya’s age= (3/4) * 24= 18 years

The sum of the ages of Shon and Nitish is 30 years. So,

Nitish’s age=30 – 12=18 years

We can find answers of all the given questions.

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