# SBI PO Quantitative Aptitude Questions 2019 (Day-5) High Level New Pattern

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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Directions (1 – 5): Study the following information carefully and answer the questions give below:

Total amount = 225000 Total amount = 300000 1) If Meena invested her amount on simple interest at 6% per annum for 4 years and her amount on compound interest at 5% per annum for two years, then find the difference between compound interest and simple interest earned by her.

a) Rs.5120

b) Rs.4980

c) Rs.4650

d) Rs.5730

e) None of these

2) Total amount of Reena on simple interest at 8% per annum after 6 years will be approximately what percent of total amount of Reena on compound interest at 10% per annum after 3 years?

a) 172%

b) 178%

c) 169%

d) 162%

e) 148%

3) If Tina invests 20% more amount on compound interest, then find the sum of her total amount on compound interest at 6% per annum after three years and total amount on simple interest at 5% per annum after 4 years?

a) Rs.123536.65

b) Rs.139591.44

c) Rs.145231.36

d) Rs.165832.24

e) None of these

4) Find the respective ratio of simple interest and compound interest earned by Sona at 4% per annum after 3 years?

a) 1107:5133

b) 2152:3093

c) 2625: 7804

d) 1123: 5321

e) None of these

5) Total amount of Leena on simple interest at 6% per annum after 4 years is approximately what percent more/less than the total amount of Leena on compound interest on 5% per annum after 2 years?

a) 69% more

b) 55% more

c) 69% less

d) 55% less

e) 32% more

Directions (6 – 10): In each of the following questions, a question is followed by three statements I, II and III. Read all the statements to find the answer to given question and then answer accordingly that which statement/s can give the answer alone/together.

6) Find the difference between compound interest on that sum at 5% per annum after 2 years and simple interest on that sum at 6% per annum after 3 years.

1. Difference between compound interest and simple interest on that sum at 4% per annum after two years is Rs.86.4.
2. Simple interest on that sum at 8% per annum after six years is Rs.25920.
3. The sum amounts to Rs.60674.4 on compound interest at 6% per annum after 2 years.

a) All I, II and III

b) Any two of the three

c) Only I and III

d) Any one of the three

e) Even I, II and III together are not sufficient.

7) Diya, Maya and Sunita entered into a partnership for two years with investment in the ratio 5 : 9 : a respectively. At the end of two years, they earned a total profit of Rs. 141000. Find the share Sunita in the profit.

1. If Sunita had invested her amount on simple interest at 7% per annum for five years, she would have earned an interest of Rs. 24500.
2. After one year, Diya doubled her investment.
3. Had Diya invested her amount on simple interest at 5% per annum for four years, she would have earned an interest of Rs. 10000.

a) Only I and II

b) All I, II and III

c) Only I and III

d) Any one of the three

e) Even I, II and III together are not sufficient.

8) Anita with double her efficiency and Kiran, both of them can complete a piece of work 12 days. Find the time taken by Anita and Simran to complete the work.

1. Kiran, Madhav and Simran together can complete the work in 8 days.
2. Madhav and Kiran together can complete the work in 15 days.
3. Madhav and Simran together can complete the work in 10 days.

a) Only II and III

b) All I, II and III

c) Only I and III

d) Any one of the three

e) Even I, II and III together are not sufficient.

9) Ratio of the ages of Vikash and Rana before four years was 14:13 respectively. Find the present average age of Vikash, Atul and Vinay.

1. Ratio of the present ages of Vikash and Atul is 4:3 respectively. After four years, ratio of their ages will be 9:7 respectively.
2. Average of the present ages of Rana and Vinay is 33 years.
3. Ratio of the present ages of Atul and Vinay is 2:3 respectively.

a) Only II and III

b) All I, II and III

c) Only I and either II or III

d) Any one of the three

e) Even I, II and III together are not sufficient.

10) Find the time taken by train A to cover a distance of 462 Km?

1. Train A can cross a platform of length 520 m in 72 seconds.
2. Train A can cross another train of length 460 m coming from opposite direction with the speed of 36 Km/h in 36 seconds.
3. Length of train A is less than the length of train B by 140 m.

a) Only II and III

b) All I, II and III

c) Only II and either I or III

d) Any one of the three

e) Even I, II and III together are not sufficient

Direction (1-5) :

Amount invested by Meena on simple interest = 22/100 x 225000 = Rs. 49500

Amount invested by Meena on compound interest = 20/100 x 300000 = Rs. 60000

We know that,

SI = (P x r x t)/100

= (49500 x 6 x 4)/100

= Rs. 11880

We know that

CI = P x (1 + r/100)t – P

= 60000 x (1 + 5/100)2 – 60000

= 60000 x 105/100 x 105/100 – 60000

= 66150 – 60000

= Rs. 6150

Required difference = 11880 – 6150 = Rs. 5730

Amount invested by Reena on simple interest = 32/100 x 225000 = Rs.72000

Amount invested by Reena on compound interest = 15/100 x 300000 = Rs.45000

We know that amount on simple interest = (P x r x t)/100 + P

Total amount of Reena on simple interest = (72000 x 8 x 6)/100 + 72000

= 34560 + 72000

= Rs.106560

We know that amount on compound interest = p x (1 + r/100)t

Total amount of Reena on compound interest

= 45000 x 110/100 x 110/100 x 110/100

= Rs.59895

Required percentage = (106560/59895) x 100 = 177.91% = 178% approx.

Amount invested by Tina on compound interest = 25/100 x 300000 = Rs.75000

Increased amount invested by Tina on compound interest

= 75000 x 120/100 = Rs.90000

We know that amount on compound interest = p x (1 + r/100)t

Total amount of Tina on compound interest

= 90000 x 106/100 x 106/100 x 106/100

= Rs.107191.44

Amount invested by Tina on simple interest = 12/100 x 225000 = Rs.27000

We know that amount on simple interest = (P x r x t)/100 + P

= (27000 x 5 x 4)/100 + 27000

= 5400 + 27000

= Rs.32400

Required sum = 107191.44 + 32400 = Rs.139591.44

Amount invested by Sona on simple interest = 14/100 x 225000 = Rs.31500

We know that

SI = (P x r x t)/100

= (31500 x 4 x 3)/100

= Rs. 3780

Amount invested by Sona on compound interest = 30/100 x 300000 = Rs.90000

We know that

CI = P x (1 + r/100)t – P

= 90000 x (1 + 4/100)3 – 90000

= 90000 x 104/100 x 104/100 x 104/100 – 90000

= 101237.76 – 90000

= Rs.11237.76

Required ratio = 3780: 11237.76 = 2625: 7804

Amount invested by Leena on simple interest = 20/100 x 225000 = Rs.45000

Amount invested by Leena on compound interest = 10/100 x 300000 = Rs.30000

We know that amount on simple interest = (P x r x t)/100 + P

= (45000 x 6 x 4)/100 + 45000

= 10800 + 45000

= Rs.55800

We know that amount on compound interest = p x (1 + r/100)t

= 30000 x (1 + 5/100)2

= 30000 x 105/100 x 105/100

= Rs. 33075

Required percentage = [(55800 – 33075)/33075] x 100

= (22725/33075) x 100

= 68.707% more

= 69% more

Direction (6 – 10) :

From I:

We know that, for two year

CI – SI = P x (r/100)2

=> 86.4 = P x (4/100)2

=> 86.4 = P x (1/25)2

=> P = 86.4 x 625

=> P = Rs.54000

We know that

CI = P x (1 + r/100)2 – P

= 54000 x (1 + 5/100)2 – 54000

= 54000 x 105/100 x 105/100 – 54000

= 59535 – 54000

= Rs.5535

We know that

SI = (P x r x t)/100

= (54000 x 6 x 3)/100

= Rs.9720

Required difference = 9720 – 5535 = Rs.4185

From II:

We know that

SI = (P x r x t)/100

=>25920 = (P x 8 x 6)/100

=> P = 2592000/48

=> P = Rs.54000

We know that

CI = P x (1 + r/100)2 – P

= 54000 x (1 + 5/100)2 – 54000

= 54000 x 105/100 x 105/100 – 54000

= 59535 – 54000

= Rs.5535

We know that

SI = (P x r x t)/100

= (54000 x 6 x 3)/100

= Rs.9720

Required difference = 9720 – 5535 = Rs.4185

From III:

We know that

Amount on CI = P x (1 + r/100)t

=> 60674.4 = P x (1 + 6/100)2

=> 60674.4 = P x 106/100 x 106/100

=> P = 60674.4 x 100/106 x 100/106

=> P = Rs.54000

We know that

CI = P x (1 + r/100)2 – P

= 54000 x (1 + 5/100)2 – 54000

= 54000 x 105/100 x 105/100 – 54000

= 59535 – 54000

= Rs.5535

We know that

SI = (P x r x t)/100

= (54000 x 6 x 3)/100

= Rs.9720

Required difference = 9720 – 5535 = Rs.4185

Hence, any one of the three statements is sufficient.

From I:

Let, amount invested by Sunita on simple interest be Rs. P

We know that

SI = (P x r x t)/100

=> 24500 = (P x 7 x 5)/100

=> P = 2450000/35

=> P = Rs. 70000

From II:

After one year, Diya doubled her investment.

From III:

Let, amount invested by Diya be Rs. K

We know that

SI = (P x r x t)/100

=> 10000 = (K x 5 x 4)/100

=> K = 1000000/20

=> K = Rs. 50000

From I, II and III:

Let, amount invested by Sunita on simple interest be Rs. P

We know that,

SI = (P x r x t)/100

=> 24500 = (P x 7 x 5)/100

=> P = 2450000/35

=> P = Rs. 70000

Let, amount invested by Diya be Rs. K

We know that,

SI = (P x r x t)/100

=> 10000 = (K x 5 x 4)/100

=> K = 1000000/20

=> K = Rs. 50000

Amount invested by Maya = 9/5 x 50000 = Rs.90000

Ratio of share in the profit:

Diya : Maya : Sunita = (50000 + 100000) : (90000 x 2) : (70000 x 2)

= 150000 : 180000 : 140000

= 15 : 18 : 14

Share of Sunita in the profit = (14/47) x 141000 = Rs. 42000

Hence, All I, II and III together are sufficient.

2/Anita + 1/Kiran = 1/12 ——— (a)

From I:

1/Kiran + 1/Madhav + 1/Simran = 1/8

From II:

From III:

From I, II and III:

1/Kiran + 1/Madhav + 1/Simran = 1/8 —— (i)

1/Madhav + 1/Kiran = 1/15 ——– (ii)

1/Madhav + 1/Simran = 1/10 ——- (iii)

Equations (ii) + (iii) – (i)

2/Madhav + 1/Kiran + 1/Simran – 1/Kiran – 1/Madhav – 1/Simran = 1/15 + 1/10 – 1/8

=>1/Madhav = (8 + 12 – 15)/120

From (ii)

1/24 + 1/Kiran = 1/15

=> 1/Kiran = 1/15 – 1/24

=> 1/Kiran = (8 – 5)/120

=> 1/Kiran = 3/120

=> 1/Kiran = 1/40

From (a)

2/Anita + 1/40 = 1/12

=> 2/Anita = 1/12 – 1/40

=> 2/Anita = (10 – 3)/120

=> 2/Anita = 7/120

=> 1/Anita = 7/240

From (iii)

1/24 + 1/Simran = 1/10

=> 1/Simran = 1/10 – 1/24

=> 1/Simran = (12 – 5)/120

=> 1/Simran = 7/120

Let required number of days = n

n x (7/240 + 7/120) = 1

=> n x (7 + 14)/240 = 1

=> n = 240/21

=> n = 80/7 days

Hence, All I, II and III together are sufficient.

4 years ago, Vikash: Rana = 14:13

From I:

Let, the present ages of Vikash and Atul be 4k years and 3k years respectively

(4k + 4)/(3k + 4) = 9/7

=> 28k + 28 = 27k + 36

=> 28k – 27k = 36 – 28

=> k = 8

Present age of Vikash = 4k = 4 x 8 = 32 years

Present age of Atul = 3k = 3 x 8 = 24 years

From II:

Rana + Vinay = 2 x 33 = 66

From III:

Atul : Vinay = 2:3

From I and II:

Let, the present ages of Vikash and Atul be 4k years and 3k years respectively

(4k + 4)/(3k + 4) = 9/7

=> 28k + 28 = 27k + 36

=> 28k – 27k = 36 – 28

=> k = 8

Present age of Vikash = 4k = 4 x 8 = 32 years

Present age of Atul = 3k = 3 x 8 = 24 years

Rana + Vinay = 2 x 33 = 66

Rana = 13/14 x (32 – 4) + 4 = 13/14 x 28 + 4 = 30 years

Vinay = 66 – 30 = 36 years

Required average = (32 + 24 + 36)/3 = 92/3 years

From I and III:

Let, the present ages of Vikash and Atul be 4k years and 3k years respectively

(4k + 4)/(3k + 4) = 9/7

=> 28k + 28 = 27k + 36

=> 28k – 27k = 36 – 28

=> k = 8

Present age of Vikash = 4k = 4 x 8 = 32 years

Present age of Atul = 3k = 3 x 8 = 24 years

And

Atul : Vinay = 2:3

Vinay = 3/2 x 24 = 36 years

Required average = (32 + 24 + 36)/3 = 92/3 years.

Hence, Only I and either II or III are sufficient.

Let, length of train A = l metres

And speed of train A = s Km/h

From I:

(l + 520) = s x 5/18 x 72

=> l + 520 = 20s

From II:

(l + 460) = (s + 36) x 5/18 x 36

=> l + 460 = (s + 36) x 10

From III:

l = length of train B – 140

From I and II:

(l + 520) = s x 5/18 x 72

=> l + 520 = 20s

=> l = 20s – 520 —— (i)

And

(l + 460) = (s + 36) x 5/18 x 36

=> l + 460 = (s + 36) x 10 ——- (ii)

From (i) and (ii),

20s – 520 + 460 = 10s + 360

=> 20s – 10s = 360 + 60

=> 10s = 420

=> s = 42 km/h

Required time = 462/42 = 11 hours

From II and III:

(l + 460) = (s + 36) x 5/18 x 36

=> l + 460 = (s + 36) x 10 —– (i)

And

l = 460 – 140 = 320 m

Putting this value in equation (i)

320 + 460 = 10s + 360

=> 10s = 780 – 360

=> 10s = 420

=> s = 42 Km/h

Required time = 462/42 = 11 hours

Hence, Only II and either I or III are sufficient.

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 Topic Daily Publishing Time Daily News Papers & Editorials 8.00 AM Current Affairs Quiz 9.00 AM Current Affairs Quiz (Hindi) 9.30 AM IBPS SO/NIACL AO Prelims – Reasoning 10.00 AM IBPS SO/NIACL AO Prelims – Reasoning (Hindi) 10.30 AM IBPS SO/NIACL AO Prelims – Quantitative Aptitude 11.00 AM IBPS SO/NIACL AO Prelims – Quantitative Aptitude (Hindi) 11.30 AM Vocabulary (Based on The Hindu) 12.00 PM IBPS SO/NIACL AO Prelims – English Language 1.00 PM SSC Practice Questions (Reasoning/Quantitative aptitude) 2.00 PM IBPS Clerk – GK Questions 3.00 PM SSC Practice Questions (English/General Knowledge) 4.00 PM Daily Current Affairs Updates 5.00 PM SBI PO/IBPS Clerk Mains – Reasoning 6.00 PM SBI PO/IBPS Clerk Mains – Quantitative Aptitude 7.00 PM SBI PO/IBPS Clerk Mains – English Language 8.00 PM 