**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.

**“Be not afraid of growing slowly; be afraid only of standing still”**

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**Click here to View Quantitative Aptitude Questions in Hindi**

**1) Rakesh, Trishul, Sonu and Kushal starts running around a circular field having area 1386 Km ^{2} at the same time. Speed of Rakesh, Trishul, Sonu and Kushal are 4 Km/h, 6 Km/h, 12 Km/h and 3 Km/h. After what time all of the four will meet each other at the starting point for the first time?**

a) 66 hours

b) 80 hours

c) 132 hours

d) 264 hours

e) None of these

**2) Average of a set of five consecutive even numbers is 46 and average of another set of five consecutive odd numbers is 87. Sum of the smallest even number and smallest odd number is approximately what percent of the sum of the biggest even number and biggest odd number from the given two sets?**

a) 78%

b) 89%

c) 43%

d) 93%

e) 79%

**3) Difference between compound interest and simple interest on a certain sum at 6% per annum after two years will be Rs.216. If half of the sum is invested on compound interest at 10% per annum for three years and remaining half is invested on simple interest at 10% per annum for 3 years, then what will be the total interest earned?**

a) Rs.15340

b) Rs.20440

c) Rs.14660

d) Rs.18930

e) None of these

**Directions (4 – 5): Study the following information carefully and answer the questions given below:**

A group contains 6 engineers, 5 doctors, 8 dancers, 4 artists, 10 players. Out of them, 50% of the engineers, 40% of the doctors, ½ of the dancers, 1 artist and 60% of the players are females.

**4) A committee of 10 members is to be formed such that the committee contains one female from each profession and rest of the members are males from any profession. Find the number of ways in which this can be done.**

a) 754320

b) 891072

c) 786032

d) 642406

e) None of these

**5) A committee of 6 members is to be formed such that the committee contains equal number of males and females. Find the number of ways in which this can be done.**

a) 180800

b) 280800

c) 480800

d) 380800

e) None of these

**Directions (6 – 10): Each question below contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.**

**6)** **Quantity I: **Mohan, Raju and Shyam entered into a partnership with investment in the ratio 6:3:5 respectively. After one year, Raju doubled his investment. After one more year, Shyam doubled his investment. At the end of three years, they earned a total profit of Rs.159000. Find the share of Raju in the profit.

**Quantity II: **Mina, Vishal and Meera entered into a partnership with investment in the ratio 4:5:3 respectively. After one year, Mina made her investment 1.5 times. At the end of two years, they earned a total profit of Rs.130000. Find the share of Mina in the profit.

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

**7)** **Quantity I: **If m^{2a} x (m^{5}÷ m^{9}) x m^{6} = m^{(a + 5)}, then find the value of a?

**Quantity II: **If n^{6}÷n^{b}x n^{3}÷ n^{9} = 1/n^{4}, then find the value of b?

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

**8) Quantity I: **If (x – 3)^{2} + (y + 5)^{2} + z^{2} = 0, then find xyz?

**Quantity II: **If (a + 6)^{2} + (b – 10)^{2} = 0, then find ab?

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

**9)** **Quantity I:** A and B together can complete a work in 15 days. A and C together can complete the work in 20 days. If B and C together can complete the work in 18 days, find the number of days taken by A to complete half of the work alone.

**Quantity II: **P, Q and R together can complete a work in 10 days. Q and T together can complete the work in 15 days. S and T together can complete the work in 12 days. If S can complete the work in 20 days, then find the number of days taken by P and R together to complete the work?

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

**10)** **Quantity I: **If (x – 1/x) = 3, then find (x^{2} + 1/x^{2})?

**Quantity II: **If y^{3} + y^{2} – 2y = 0, then find the value of y?

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

**Answers:**

**1) Answer: c)**

We know that

Area of a circular field = πr^{2}

=> 1386 = 22/7 x r^{2}

=> r^{2} = 1386 x 7/22

=> r^{2} = 441

=> r = √441

=> r = 21 Km

We know that

Circumference of a circular field = 2πr

Circumference of the circular field = 2 x 22/7 x 21 = 132 Km

Time taken by Rakesh to complete one round = 132/4 = 33 hours

Time taken by Trishul to complete one round = 132/6 = 22 hours

Time taken by Sonu to complete one round = 132/12 = 11 hours

Time taken by Kushal to complete one round = 132/3 = 44 hours

LCM of 33, 22, 11and 44 = 132

Hence, they will meet each other at the starting point after 132 hours

**2) Answer: b)**

We know that

Average of n consecutive even/odd numbers = first number + (n – 1)

Let, smallest even number = a

And smallest odd number = b

46 = a + (5 – 1)

=> a = 46 – 4

=> a = 42

And

87 = b + (5 – 1)

=> b = 87 – 4

=> b = 83

Set of even numbers: 42, 44, 46, 48, 50

Set of odd numbers: 83, 85, 87, 89, 91

Required percentage = [(42 + 83)/(50 + 91)] x 100

= (125/141) x 100

= 88.65% = 89% approx.

**3) Answer: d)**

We know that, for two years

CI – SI = P x (r/100)^{2}

=> 216 = P x (6/100)^{2}

=> P = 216 x 100/6 x 100/6

=> P = Rs. 60000

We know that

CI = P x (1 + r/100)^{t}– P

= 30000 x (1 + 10/100)^{3} – 30000

= 30000 x (1 + 1/10)^{3} – 30000

= 30000 x (11/10)^{3} – 30000

= 30000 x 1331/1000 – 30000

= 39930 – 30000

= Rs. 9930

We know that

SI = (P x r x t)/100

= (30000 x 10 x 3)/100

= Rs. 9000

Required sum = 9930 + 9000 = Rs. 18930

**Directions (4 – 5): **

**Engineers:**

Total = 6

Females = (50/100) x 6 = 3

Males = 6 – 3 = 3

**Doctors:**

Total = 5

Females = (40/100) x 5 = 2

Males = 5 – 2 = 3

**Dancers:**

Total = 8

Females = ½ x 8 = 4

Males = 8 – 4 = 4

**Artists:**

Total = 4

Females = 1

Males = 4 – 1 = 3

**Players:**

Total = 10

Females = (60/100) x 10 = 6

Males = 10 – 6 = 4

Total number of males = 3 + 3 + 4 + 3 + 4 = 17

Total number of females = 3 + 2 + 4 + 1 + 6 = 16

**4) Answer: b)**

Required number of ways = ^{3}c_{1} x ^{2}c_{1} x ^{4}c_{1} x ^{1}c_{1} x ^{6}c_{1} x ^{17}c_{5}

= 3 x 2 x 4 x 1 x 6 x 6188

= 891072

**5) Answer: d)**

Required number of days = ^{17}c_{3} x ^{16}c_{3}

= 680 x 560

= 380800

**Directions (6 – 10):**

**6) Answer: a)**

**Quantity I: **Mohan, Raju and Shyam entered into a partnership with investment in the ratio 6:3:5 respectively. After one year, Raju doubled his investment. After one more year, Shyam doubled his investment. At the end of three years, they earned a total profit of Rs.159000. Find the share of Raju in the profit.

Let, amount invested by Mohan, Raju and Shyam be Rs.6k, Rs.3k, Rs.5k respectively.

Ratio of share in the profit:

Mohan : Raju : Shyam = (6k x 3) : (3k + 6k x 2) : (5k x 2 + 10k)

= 18k : 15k : 20k

= 18:15:20

Share of Raju in the profit = [15/(18 + 15 + 20)] x 159000

= (15/53) x 159000

= Rs. 45000

**Quantity II: **Mina, Vishal and Meera entered into a partnership with investment in the ratio 4:5:3 respectively. After one year, Mina made her investment 1.5 times. At the end of two years, they earned a total profit of Rs.130000. Find the share of Mina in the profit.

Let, amounts invested by Mina, Vishal and Meera be Rs.4k, Rs.5k and Rs.3k respectively.

Mina : Vishal : Meera = (4k + 6k) : (5k x 2) : (3k x 2)

= 10k : 10k : 6k

= 5:5:3

Share of Mina in the profit = [5/(5 + 5 + 3)] x 130000

= (5/13) x 130000

= Rs. 50000

Hence, **Quantity I < Quantity II**

**7) Answer: a)**

**Quantity I:**

m^{2a} x m^{5}÷ m^{9} x m^{6} = m^{(a + 5)}

m^{(2a + 5 – 9 + 6) }= m^{(a + 5)}

=> 2a + 5 – 9 + 6 = a + 5

=> 2a – a = 5 – 2

=> a = 3

**Quantity II: **

n^{6}÷n^{b}x n^{3}÷ n^{9} = 1/n^{4}

n^{(6 – b + 3 – 9)} = n^{-4}

=> 6 – b + 3 – 9 = -4

=> b = 4

Hence, **Quantity I < Quantity II**

**8) Answer: c)**

**Quantity I:** (x – 3)^{2} + (y + 5)^{2} + z^{2} = 0

=> x – 3 = 0, y + 5 = 0, z = 0 [If sum of positive numbers is zero, each number is equal to zero]

=> x = 3, y = – 5, z = 0

Now,

xyz = 3 x (-5) x 0 = 0

**Quantity II:** (a + 6)^{2} + (b – 10)^{2} = 0

=> a + 6 = 0 and b – 10 = 0 [If sum of positive numbers is zero, each number is equal to zero]

=> a = -6, b = 10

Now,

ab = -6 x 10 = -60

**Quantity I > Quantity II**

**9) Answer: c)**

**Quantity I:** A and B together can complete a work in 15 days. A and C together can complete the work in 20 days. If B and C together can complete the work in 18 days, find the number of days taken by A to complete half of the work alone.

1/A + 1/B = 1/15 ——- (i)

1/A + 1/C = 1/20 ——– (ii)

1/B + 1/C = 1/18 ——— (iii)

Equation (i) + Equation (ii) – Equation (iii)

1/A + 1/B + 1/A + 1/C – 1/B – 1/C = 1/15 + 1/20 – 1/18

=> 2/A = (12 + 9 – 10)/180

=> 1/A = 11/180 x ½

=> 1/A = 11/360

Number of days taken by A to complete half of the work = 360/11 x ½ = 180/11 days

**Quantity II: **P, Q and R together can complete a work in 10 days. Q and T together can complete the work in 15 days. S and T together can complete the work in 12 days. If S can complete the work in 20 days, find the number of days taken by P and R together to complete the work.

1/P + 1/Q + 1/R = 1/10 ————- (i)

1/Q + 1/T = 1/15 —————- (ii)

1/S + 1/T = 1/12 ————– (iii)

1/S = 1/20 ———— (iv)

From (iii) and (iv)

1/20 + 1/T = 1/12

=> 1/T = 1/12 – 1/20

=> 1/T = (5 – 3)/60

=> 1/T = 2/60

=> 1/T = 1/30

From (ii)

1/Q + 1/30 = 1/15

=> 1/Q = 1/15 – 1/30

=> 1/Q = (2 – 1)/30

=> 1/Q = 1/30

From (i)

1/P + 1/R + 1/30 = 1/10

=> 1/P + 1/R = 1/10 – 1/30

=> 1/P + 1/R = (3 – 1)/30

=> 1/P + 1/R = 2/30

=> 1/P + 1/R = 1/15

Required number of days = 15

Hence, **Quantity I > Quantity II**

**10) Answer: c)**

**Quantity I: **If (x – 1/x) = 3, then find (x^{2} + 1/x^{2})?

x – 1/x = 3

Squaring both sides

(x – 1/x)^{2} = 3^{2}

=>x^{2} + 1/x^{2} – 2*x*(1/x) = 9

=>x^{2} + 1/x^{2} – 2 + 2 = 9 + 2

=>x^{2} + 1/x^{2} = 11

**Quantity II: **If y^{3} + y^{2} – 2y = 0, then find the value of y?

y^{3} + y^{2} – 2y = 0

=>y (y^{2} + y – 2) = 0

=>y (y^{2} + 2y – y – 2) = 0

=>y [y(y + 2) – 1(y + 2)] = 0

=>y (y – 1) (y + 2) = 0

=> y = 0, 1, -2

Hence, **Quantity I > Quantity II**

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