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): Read the following information carefully and answer the given questions.
Shivani goes to a shopping mall from her home at 7.30 AM in her boat (downstream) and the distance between her home and shopping mall is 120 km. She reached the shopping mall at _____ (A) with 10 km/hr stream’s speed. If she increases her boat’s speed by 50%, she would reach 1 hour earlier.
She spends 2 hours in the shopping mall and buys an article at 20% discount from the labeled price. She found the same article in another shop with same labeled price but at 25% discount, if she had bought it in that shop, she would be able to save Rs.50. Then the labeled price of an article is _____ (B)
Then she returns her home (Upstream) in the same boat at _____am/pm(C). The boat’s speed is increased by 175%, since the speed of the stream is increased by 50%.
Her husband bought a mixture containing milk and water in which 33 (1/3) % is water. From that, Shivani uses 15 litres of mixture to make Khoa and again she asks her husband to buy 2 litres of milk. She adds 3 litres of water in the mixture. If the ratio of milk and water in the final mixture is 16: 9, then the initial quantity of milk is _____ (D).
Shivani got her salary credited in her account. If she invests equal sum in two different schemes A and B which offers Simple interest and Compound interest for two years respectively. In scheme A, three-fifth of the sum is invested at the rate of 24% per annum and the remaining sum is invested at 18% per annum. In scheme B, three-eighth of the sum is invested at 20% per annum, three-fifth of the remaining sum is invested at 10% per annum and the remaining sum is invested at ____ (E) % per annum. At the end of two years, interest received in scheme A is Rs. 1905 more than the interest received in scheme B.
1) What should come in the place of (A)?
a) 12.30 pm
b) 11.00 am
c) 11.30 am
d) 1.30 pm
e) None of these
2) What should come in the place of (C)?
a) 11.30 am
b) 4.30 pm
c) 12.30 pm
d) 5.30 pm
e) 6.30 pm
3) If Shivani invests the initial sum at the rate of 36% per annum compounded half yearly for one year, then she received the interest of Rs. 15696. What should come in the place of (E)?
a) 25%
b) 20%
c) 15%
d) 18%
e) 22%
4) What should come in the place of (D)?
a) 60 litres
b) 30 litres
c) 45 litres
d) 40 litres
e) 20 litres
5) What should come in the place of (B)?
a) Rs.1200
b) Rs.1050
c) Rs.2000
d) Rs.1500
e) Rs.1000
6) Income of Raju and Suresh are in the ratio 27 : 22 and ratio of their expenditure is 5:4 respectively. Raju saves Rs._____ and Suresh save Rs._____. Mahesh spends 50% of his income for rent, food and medicine. Amount he spent for medicine is Rs.4000 which is two-ninth of the remaining income. Income of Raju is 75% of Mahesh’s income.
Which of the following satisfies the two blanks given in the questions?
- Rs.2000, Rs.2000
- Rs.3000, Rs.2800
- Rs.7000, Rs.6000
a) Only I
b) Only II
c) Only III
d) Both I and II
e) All I, II and III
7) A _____ litres mixture contains milk and water in the ratio 7:5 respectively. The milkman sold 48 litres of the mixture and added 12 litres of milk and 10 litres of water to the remaining mixture. The milkman again sold ____ litres of the mixture and added six litres of milk in the remaining mixture. The final mixture contains milk to water in the ratio is 5: 3.
Which of the following satisfies the two blanks given in the questions?
- 96 litres, 28 litres
- 216 litres, 38 litres
- 144 litres, 59 litres
a) Only I
b) Only II
c) Only III
d) Only I and II
e) Only II and III
8) Train A started from station P towards station Q. At the same time, train B started from station R towards station Q. All the three stations are in a straight line such that station Q is between station P and station R. Station Q is equidistant from station P and station R. Distance between station P and station Q is 440 Km. If the speed of train A and train B is ______ Km/hr and ______ Km/hr respectively, the distance between both the trains after five hours is 580 km.
Which of the following satisfies the two blanks given in the questions?
- 28 km/hr, 32 km/hr
- 40 km/hr, 20 km/hr
- 25 km/hr, 35 km/hr
a) None
b) Only II
c) Only III
d) Only I and II
e) All I, II and III
9) Tina bought two articles at same price. She sold one at a profit of 20% and other at a loss of 10%. From that amount she bought one refrigerator and marked its price _____% more than cost price and allowed a discount of 10% on the Marked price. If the cost price of each article was Rs.____, and the overall percent profit earned by Tina is 18.125%.
Which of the following satisfies the two blanks given in the questions?
- 20%, Rs.4000
- 25%, Rs.4800
- 25%, Rs.6000
a) Only I
b) Only II
c) Only III
d) Only I and II
e) Only II and III
10) A starts a company and after 3 months B also joins the company. The initial investment of A and B is in the ratio of 3:2, respectively. A and B receives Rs. _____ and Rs. _____ respectively as profit after completion of one year of the company.
Which of the following satisfies the two blanks given in the questions?
- Rs.2698, Rs.1349
- Rs.1920, Rs.960
- Rs.2576, Rs.1293
a) Only I
b) Only II
c) Only III
d) Only I and II
e) Only II and III
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Answers :
Directions (1-5):
From paragraph 1,
Distance between Shivani’s home and shopping = 120 km
Let us take boat’s speed be x km/hr
Speed of the stream = 10 km/hr
Let the time taken to reach shopping mall in downstream be t hours
120/ (x+10) = t —– (1)
If she increased her boat’s speed by 50%, she would able to reach 1 hour earlier
Boat’s new speed = x * 150/100 = 3x/2 km/hr
Then,
120/(3x/2 + 10) = t-1 —- (2)
Substitute t’s value in equation (2), we get
120/(3x/2 + 10) = [120/ (x+10)] -1
120/(x+10) – 120/(3x/2 + 10) = 1
120 [(3x/2 + 10) – (x +10)]/[(x+10)*(3x/2 +10)] = 1
120 [(3x/2 + 10) – (x +10)] = [(x+10)*(3x/2 +10)]
120*[(3x + 20 – 2x – 20)/2] = 3x2/2 + 50x/2 + 100
120x/2 = 3x2/2 + 50x/2 + 200/2
120x = 3x2 + 50x + 200
= > 3x2 – 70x + 200 = 0
= > 3x2 – 60x – 10x + 200 = 0
= > 3x (x – 20) – 10 (x – 20) =0
= > (3x – 10) (x – 20) =0
= > x = 20 and 10/3 (Eliminate (10/3) because boat’s speed always greater than the stream’s speed)
Substitute the value of x in equation (1), we get
120/ (20 + 10) = t
120/30 = t
4 = t
Shivani reach the shopping mall at 7.30 + 4 hrs = 11.30 am
From paragraph 2,
Let us take the labeled price of an article be Rs. x
Selling price in shop 1, = x * 80/100
Selling price in shop 2, = x * 75/100
If she had bought it in that shop, she would be able to save Rs.50
(x*80/100) – (x*75/100) = 50
5x/100 = 50
= > x = (50/5)*100
= > x = Rs.1000
Then the labeled price of an article is Rs.1000
From paragraph 3,
She spends 2 hours in the shopping mall, and then she started to return 11.30 + 2 hrs = 1.30 pm.
Boat’s new speed = 20 * 275/100 = 55 km/hr
Stream’s new speed = 10 * 150/100 = 15 km/hr
Upstream speed = 55 – 15 = 40 km/hr
Time to returns her home,
= > 120/40
= > 3 hours
Shivani returns her home = 1.30 + 3 hrs = 4.30 pm
From paragraph 4,
Ratio of Milk to water = 66 (2/3): 33 (1/3)
= 200/3: 100/3
= 2: 1
From that, Shivani uses 15 litres of mixture to make Khoa and again she asks her husband to buy 2 litres of milk. She adds 3 litres of water in the mixture. If the ratio of milk and water in the final mixture is 16: 9
From the above statement we conclude,
(2x-10+2)/(x-5+3) =16/9
(2x-8)*9 = (x-2)*16
= > 18x – 72 = 16x – 32
= > 2x = 72 – 32 = 40
= > x = 20 litres
Initial quantity of milk = 2 * 20 = 40 litres
From paragraph 5,
In scheme A,
Let us take the equal sum be Rs.x
Interest received in scheme A,
= > [(3x/5 * 24*2)/100] + [(2x/5 *18*2)/100]
= > 144x/500 + 72x/500
= > 216x/500
In scheme B,
Three –eighth of the sum = 3x/8
Remaining = x – 3x/8 = 5x/8
Three-fifth of the remaining = 5x/8 * 3/5 = 3x/8
Remaining = x – 3x/8 – 3x/8
= (8x-3x-3x)/8
= 2x/8 = x/4
Interest received in scheme B,
= [3x/8 *(1+20/100)2 + 3x/8*(1+10/100)2 + x/4 *(1+R/100)2] – x
= [3x/8 * 36/25 + 3x/8 * 121/100 + x/4* (1+R/100)2]–x
= 27x/50 + 363x/800 –x + x/4 *(1+R/100)2
= (432x+363x-800x)/800 + x/4 *(1+R/100)2
= x/4 *(1+R/100)2 – 5x/800
Difference of interest received in scheme A and B,
216x/500 – x/4 *(1+R/100)2 + 5x/800 = 1905 —- (3)
1) Answer: c)
Shivani reach the shopping mall at 7.30 + 4 hrs = 11.30 am
2) Answer: b)
Shivani returns her home = 1.30 + 3 hrs = 4.30 pm
3) Answer: a)
Interest for two years when compounded half yearly, r = 36/2 = 18%
According to the question,
= > x*(1 + R/100)2 – x = 15696
= > x*[(1 + 18/100)2 – 1] = 15696
= > x*((118/100)*(118/100)) – 1) = 15696
= > x*[(3481/2500) – 1] = 15696
= > x = 15696 * (2500/981)
= > x = Rs. 40000
If we substitute the value of x in equation (3), we get
216x/500 – (x/4) *(1+R/100)2 + 5x/800 = 1905
(216 * 40000)/500 – (40000/4) *(1 + R/100)2 + (5*40000)/800= 1905
17280 – 10000 *(1 + R/100)2 + 250 = 1905
17280 + 250 -1905 = 10000 *(1 + R/100)2
15625/10000 = (1 + R/100)2
125/100 = 1 + R/100
125 = 100 +R
R = 25%
4) Answer: d)
Initial quantity of milk = 2 * 20 = 40 litres
5) Answer: e)
Then the labeled price of an article is Rs.1000
6) Answer: e)
Let us take income of Mahesh be x
Amount spent for medicine = 2/9 * (x-50/100*x) = 4000
= > 2/9 * (50x/100) = 4000
= > 2/9 * x/2 = 4000
= > x/9 = 4000
= > x = 36000
Income of Raju = 36000 * 75/100
= 27000
Income of Suresh = 27000/27 * 22
= 22000
From I:
According to the question,
= > (27000 – 2000)/(22000-2000)
= > 25000/20000
= > 5/4
This is satisfies the given condition.
From II:
According to the question,
= > (27000 – 3000)/(22000-2800)
= > 24000/19200
= > 5/4
This is satisfies the given condition.
From III:
According to the question,
= > (27000 – 7000)/(22000-6000)
= > 20000/16000
= > 5/4
This is satisfies the given condition.
7) Answer: a)
From I:
Quantity of milk in the initial mixture = (7/12) x 96 = 56 litres
Quantity of water in the initial mixture = (5/12) x 96 = 40 litres
After selling 48 litres and adding 12 litres milk and 10 litres water to the remaining mixture:
Quantity of Milk in 48 litres = 48 * (7/12) = 28 litres
Quantity of Water in 48 litres = 48 * (5/12) = 20 litres
Quantity of milk after the replacement = 56 – 28 + 12
= 28 + 12 = 40 litres
Quantity of water after replacement = 40 – 20 + 10
= 20 + 10 = 30 litres
Total mixture = 70 litres
Milk: water = 40: 30 = 4:3
After selling 28 litres of the mixture and adding six litres of milk to the remaining mixture:
Quantity of milk in 28 litres = 28 * (4/7) = 16 litres
Quantity of water in 28 litres = 28 * (3/7) = 12 litres
Quantity of milk after replacement= 40 – 16 + 6
= 24 + 6 = 30 litres
Quantity of water after replacement= 30 – 12 = 18 litres
Ratio of milk and water in the final mixture = 30/18 = 5/3
This is satisfies the given condition.
From II:
Quantity of milk in the initial mixture = (7/12) x 216 = 126 litres
Quantity of water in the initial mixture = (5/12) x 216 = 90 litres
After selling 48 litres and adding 12 litres milk and 10 litres water to the remaining mixture:
Quantity of Milk in 48 litres = 48 * (7/12) = 28 litres
Quantity of Water in 48 litres = 48 * (5/12) = 20 litres
Quantity of milk after the replacement = 126 – 28 + 12
= 98 + 12
= 110 litres
Quantity of water after replacement = 90 – 20 + 10
= 70 + 10
= 80 litres
Total mixture = 190 litres
Milk: water = 110: 80 = 11: 8
After selling 38 litres of the mixture and adding six litres of milk to the remaining mixture:
Quantity of milk in 38 litres = 38 * (11/19) = 22 litres
Quantity of water in 38 litres = 38 * (8/19) = 16 litres
Quantity of milk after replacement= 110 – 22 + 6
= 88 + 6 = 94 litres
Quantity of water after replacement= 80 – 16 = 64 litres
Ratio of milk and water in the final mixture = 94/64 = 47/32
This is not satisfies the given condition.
From III:
Quantity of milk in the initial mixture = (7/12) x 144 = 84 litres
Quantity of water in the initial mixture = (5/12) x 144 = 60 litres
After selling 48 litres and adding 12 litres milk and 10 litres water to the remaining mixture:
Quantity of Milk in 48 litres = 48 * (7/12) = 28 litres
Quantity of Water in 48 litres = 48 * (5/12) = 20 litres
Quantity of milk after the replacement = 84 – 28 + 12
= 56 + 12 = 68 litres
Quantity of water after replacement = 60 – 20 + 10
= 40 + 10 = 50 litres
Total mixture = 118 litres
Milk: water = 68: 50 = 34: 25
After selling 59 litres of the mixture and adding six litres of milk to the remaining mixture:
Quantity of milk in 59 litres = 59 * 34/59 = 34 litres
Quantity of water in 59 litres = 59 * 25/59 = 25 litres
Quantity of milk after replacement= 68 – 34 + 6
= 34 + 6 = 40 litres
Quantity of water after replacement= 50 – 25 = 25 litres
Ratio of milk and water in the final mixture = 40/25 = 8/5
This is not satisfies the given condition.
8) Answer: e)
From I:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train A in 5 hours = 28 x 5 = 140 Km
Distance travelled by train B in 5 hours = 32 x 5 = 160 Km
Required distance = 880 – 140 – 160
= 880 – 300
= 580 Km
This is satisfies the given condition.
From II:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train A in 5 hours = 40 x 5 = 200 Km
Distance travelled by train B in 5 hours = 20 x 5 = 100 Km
Required distance = 880 – 200 – 100
= 880 – 300
= 580 Km
This is satisfies the given condition.
From III:
Distance between station P and station R = 440 x 2 = 880 Km
Distance travelled by train A in 5 hours = 25 x 5 = 125 Km
Distance travelled by train B in 5 hours = 35 x 5 = 175 Km
Required distance = 880 – 125 – 175
= 880 – 300
= 580 Km
This is satisfies the given condition.
9) Answer: e)
From I:
Cost price of the two article for Tina = 4000*2 = Rs. 8000
Selling price of two article = 4000 x 120/100 + 4000 x 90/100
= 4800 + 3600
= Rs.8400 = cost price of the refrigerator
Marked price of the refrigerator = 8400 x 120/100 = Rs.10800
Selling price of the refrigerator = 10800 x 90/100 = Rs.9072
% profit = [(9072 – 8000)/8000] x 100
= 1072/8000 x 100
= 13.4%
This is not satisfies the given condition.
From II:
Cost price of the two article for Tina = 4800 x 2 = rs.9600
Selling price of two article = 4800 x 120/100 + 4800 x 90/100
= 5760 + 4320
= Rs.10080 = cost price of the refrigerator
Marked price of the refrigerator = 10080 x 125/100 = Rs.12600
Selling price of the refrigerator = 12600 x 90/100 = Rs.11340
% profit = [(11340 – 9600)/9600] x 100
= (1740/9600) x 100
= 18.125%
This is satisfies the given condition.
From III:
Cost price of the two article for Tina = 6000 x 2 = Rs.12000
Selling price of two article = 6000 x 120/100 + 6000 x 90/100
= 7200 + 5400
= Rs.12600 = cost price of the refrigerator
Marked price of the refrigerator = 12600 x 125/100 = Rs.15750
Selling price of the refrigerator = 15750 x 90/100 = Rs.14175
% profit = (14175 – 12000)/12000 x 100
= 2175/12000 x 100
= 18.125%
This is satisfies the given condition.
10) Answer: d)
The ratio of profit share of A and B is
A: B = 3 × 12: 2 × (12 – 3) = 2: 1
From I:
The profit ratio of A and B = 2698: 1349
= 2: 1
This is satisfies the given condition.
From II:
The profit ratio of A and B = 1920: 960
= 2: 1
This is satisfies the given condition.
From III:
The profit ratio of A and B = 2576: 1293
This is not satisfies the given condition.
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