Quantitative Aptitude Quiz for IBPS, SBI and other Bank Exams (Application Problems)

Mentor for Bank Exams
Quantitative Aptitude Quiz for IBPS, SBI and other Bank Exams (Application Problems)
1. Peter wants to attend the drama performance of Mary Jane in the theatre near Peter's home. The theatre is at a distance of 6 km from his home and he travels at a constant speed. He travelled half the way with the specified speed and stopped for 5 minutes to buy flowers. To reach the theatre on time, he had to increase his speed by 6 km/hr for the rest of the way. Next week, he again went to see the drama, but this time he stopped for 10 minutes to buy flowers. By what value must he increase his speed for the remaining half of the distance to reach the theatre as per the schedule?
A) 24kmph
B) 8 kmph
C) 6 kmph
D) 12 kmph
E) Can’t be determined
2. Gary and Harry have some amount of money in the form of Rs. 2 and Rs. 5 coins. The average amount of money Gary and Harry have in form of Rs. 2 coins is Rs. 15, while that in form of Rs. 5 coins is Rs. 50. What is the number of Rs. 2 coins that Gary has if Gary and Harry have money in the ratio 8 : 5?
A) 15
B) 10
C) 5
D) 0
E) Can’t be determined
3. Nathan is a middleman. He buys chairs from factory and sells to public with profit. The difference between the prices at which he bought and at which he sold a chair is Rs. 20. If Nathan had sold at 10% more price, he would have made a profit of 21.58%. Find the price at which Nathan bought a chair from factory.
A) Rs.180
B) Rs.190
C) Rs.200
D) Rs.210
E) Rs.240
4. A sample of 20 litres air is taken in which there are no impurities. It has 78% Nitrogen and 22% Oxygen. In this mixture, some pure oxygen is mixed so that the amount of oxygen and nitrogen becomes equal. How much more pure oxygen should be added so that ratio of Nitrogen and Oxygen becomes 1:4?
A) 25.2 litres
B) 29.8 litres
C) 38.7 litres
D) 41.2 litres
E) 46.8 litres
5. If 25 students can paint a wall in 2 hours. But at the end of every 10 mins, 10 additional students join. In how much time will the whole wall be painted?
A) 40 mins
B) 50 mins
C) 53 mins
D) 60 mins
E) 55 mins
6. The length, breadth and height of an open box is 60 cm, 50 cm and 30 cm respectively. Its thickness is 5 cm. What would be the cost of the box if 1 cm3 of wood costs Rs. 0.08?
A) Rs.3740
B) Rs.3800
C) Rs.3860
D) Rs.3200
E) Rs.3840
7. Every day, Bobby eats three chocolates while Gary eats five candies. The price of seven candies and two chocolates is Rs. 69. Also, price of three chocolates is Rs. 16 more than the price of two candies. How much do Bobby and Gary spend together every day on chocolates and candies?
A) Rs.17
B) Rs.51
C) Rs.65
D) Rs.34
E) Rs.9
8. Jack owned 950 gold coins all of which he distributed amongst his three Friends John, James and Nick. John gave 25 gold coins to his wife. James donated 15 gold coins and Nick made jewellery out of 30 gold coins. The new respective ratio of the coins left with them was 20 : 73 : 83. How many gold coins did James receive from Jack?
A) 380
B) 415
C) 400
D) 350
E) None of these
9. Harry is an elite customer of a bank, while Gary is a normal customer. Bank gives a simple interest of 5% to Harry and 4% to Gary. So, instead of investing in Bank, Gary gave his sum of money of Rs. 34000 to Harry so that he can invest on his behalf for a year. Harry promised a 4.5% interest rate to Gary. Harry also invested Rs. 19000. Harry returned Gary’s money along with promised interest at the end of year. How much money Harry has at the end of year?
A) Rs.2000
B) Rs.19860
C) Rs.20100
D) Rs.20120
E) Rs.20480
10. Stacy and James entered into a partnership and invested equal sums of for a year. The profit that this investment offers is 15% for a year. Due to a financial constraint, James withdrew his share after 7 months. As a clause of partnership, James only received back his original sum and his earned profit went to Stacy. In total, Stacy earned a profit of Rs. 1543.75. How much money each of them invested?
A) Rs.5500
B) Rs.6000
C) Rs.6500
D) Rs.7200
E) Rs.7500
Solutions:
1. A) Let the regular speed of the Peter = x km/hr.
Now, time = distance/speed
regular time taken by Peter to travel the full distance = (6/x) hrs
Case-1:
Distance travelled at regular speed = 6/2= 3 km
time taken to travel this distance = (3/x) hrs       ---(1)
Waiting time = 5 mins = 1/12 hrs
for the remaining 3 kms, the man increases his speed by 6 km/h (i.e. x + 6)
time taken to travel last 3 km = 3/(x + 6)            ---(2)
From the given data, we can clearly deduce that:
Regular time = (time taken to travel first 3 km) + (waiting time) + (time taken to travel last 3 km)
=> 6/x = 3/x + 1/12 + 3/(x + 6) => 3/x – 3/(x + 6) = 1/12 => 1/(x2 + 6x) = 1/216 => x2 + 6x – 216 = 0 (x + 18) (x - 12) = 0
Omitting the negative value of speed,
x = 12 km/h 
Now, in the second case:
Peter travels first 3 kms at its regular speed and waits for 10 mins.
In this case: waiting time = 1/6  hrs
Let y = the speed with which he travels the remaining distance.
Again, he reaches the destination in regular time.
=> 6/x = 3/x + 1/6 + 3/y => 3/x= 1/6 + 3/y
Substituting x = 12,
y = 36 km/h
Increase in usual speed = y – x = 36 – 12 = 24 km/hr
2. E) The average amount of money Gary and Harry have in form of Rs. 2 coins is Rs. 15, while that in form of Rs. 5 coins is Rs. 50.
Total coins of Rs. 2 = (15 × 2)/2 = 15
Total coins of Rs. 5 = (50 × 2)/5 = 20
Suppose Gary has M coins of Rs. 2, and N coins of Rs. 5.
=> {2M + 5N}/{[2(15 – M) + 5(20 – N)]}
10M + 25N = 240 – 16M + 800 – 40N
26M + 65N = 1040
2M + 5N = 80
M = 40 2.5N
We know, M and N must be greater than or equal to 0. M should be less than equal to 15, and N should be less than equal to 20.
In this equation, for M to be integer, N must be even.
For M to be less than 15, N should be at least 10.
Following pairs of M and N will satisfy given conditions, (15, 10), (10, 12), (5, 14), (0, 16).
Hence, number of Rs. 2 coins with Gary cannot be uniquely determined.
3. B) Suppose Nathan bought a chair from factory at Rs. T.
Price at which Nathan sold chair to Public = Rs. (T + 20)
If Nathan had sold at 10% more price, he would have made a profit of 21.58%.
In this case, selling price would become Rs. (T + 20) × (1 + 10/100) = Rs. (1.1T + 22)
We know, Selling Price = Cost price × (1 + (Profit Percentage)/100)
1.1T + 22 = T × (1 + 21.58/100)
1.1T + 22 = 1.2158T
T = 22/0.1158 = 190
Price at which Nathan bought chair from factory is Rs. 190.
4. E) Amount of Nitrogen in 20 litres air = (20 × (78/100)) = 15.6 litres
Amount of Oxygen in 20 litres air = (20 × (22/100)) = 4.4 litres
To make oxygen and nitrogen equal, amount of oxygen to be added will be (15.6 – 4.4) litres, i.e., 11.2 litres.
Total amount of air now = 20 litres + 11.2 litres = 31.2 litres
And it has 15.5 litres of each of nitrogen and oxygen.
Suppose T litres of oxygen are added to mixture, so that ratio of Nitrogen and Oxygen becomes 1:4.
15.6/(15.6 + T) = ¼
62.4 = 15.6 + T
T = 46.8
46.8 litres of oxygen should be added.
5. D) 25 students can paint the wall in 2 hours = 2 × 60 = 120 min.
part of wall painted by 25 students in one minute = 1/120
part of wall painted by 1 student in 1 min = 1/(120 × 25)
For the 1st 10 min, there are only 25 students
part of wall painted by 25 students in 10 min can paint = (25 × 10)/(120 × 25) = 1/12
Total part of the wall painted after 10 min = 1/12
The whole wall be painted in 60 min
6. D) The outer dimensions of the box are as follows:
Length = l1 = 60 cm; Breadth = b1 = 50 cm; Height = h1 = 30 cm
Now, since the box is wooden and has a thickness of 5 cm, only the edges would be made of wood and there’ll be cavity inside the box. The dimensions of the cavity would be as shown in the figure.
The dimensions of cavity are as follows:
Length = l2 = 50 cm; Breadth = B2 = 40 cm; Height = H2 = 25 cm
Now, Volume of wood used for box = volume of box – volume of cavity
volume of wood used for the box = (l1 × b1 × h1) – (l2 × b2 × h2)
volume of wood used for the box = (60 × 50 × 30) (50 × 40 × 25)
volume of wood used for the box = 90000 50000 = 40000 cm3
Cost of wood = rate of wood × volume of wood
Cost of wood = 0.08 × 40000 = Rs. 3200
7. C) Let price of chocolate is M rupees and price of candy is N rupees
Price of 7 candies and two chocolates is Rs. 69  
7N + 2M = 69 -------- (i)
 Difference between prices of three chocolates and two candies is Rs. 16.
3M – 2N = 16 -------- (ii)
Let us solve (i) and (ii), multiply (i) by 2 and (ii) by 7, and then add
2(7N + 2M) + 7(3M – 2N) = 2 × 69 + 7 × 16
25M = 250
M = 10
Substitute this value in equation (ii)
3 × 10 – 2N = 16
2N = 14
N = 7
Price of one chocolate = Rs. 10
Price of one candy = Rs. 7
Bobby eats three chocolates and Gary eats five candies, everyday.
Money Spent every day = Rs (3 × 10 + 5 × 7) = Rs.  65
8. A) Jack have total gold coins = 950
No. of total gold coins left after donating to three friends = 950 – (25 + 15 + 30) = 880
According to question, the new respective ratio of the coins left with them was 20 : 73 : 83
Let us consider coins left with them are 20k, 73k and 83k respectively.
20k + 73k + 83k = 176k
Now, in order to find the ratio of coins for one part
k = 880/176 = 5
So, James’s coins = 73 × 5 = 365
James has donated 15 coins.
So, Total coins that James received = 365 + 15 = 380 coins
James got 380 from his friend Jack.
9. D) Bank gives a simple interest of 5% to Harry and 4% to Gary.
Gary gave his sum of money of Rs. 34000 to Harry so that he can invest on his behalf for a year.  Harry promised a 4.5% interest rate to Gary, but got an interest of 5% from bank. So, Harry got 0.5% profit on Gary’s investment.
 Harry also invested Rs. 19000. On this, he will get 5% interest.
We know, Simple interest = (Principal × Rate × Time)/100
Total interest earned by Harry = (34000 × 0.5 × 1)/100 + (19000 × 5 × 1)/100 = 1120
Total amount of money that Harry will have at end of year = Money Harry initially had + Interest earned = Rs. 19000 + Rs. 1120 = Rs. 20120
10. C) Suppose both Stacy and James invested Rs. T each.
Profit earned by James in 7 months = T × (15/100) × (7/12) = 0.0875T
As a clause of partnership, James only received back his original sum and his earned profit went to Stacy.
Profit earned by in Stacy 12 months = T × (15/100) × 1 = 0.15T
In total, Stacy earned a profit of Rs. 1543.75.
0.0875T + 0.15T = 1543.75
T = 1543.75/0.2375 = 6500
Each of them invested an amount of Rs. 6500.