Practice Quantitative Aptitude – Application Sums (Day14):
One Time Offer @ Rs.345 Yearly Platinum Pack with 1 Year Validity
 Free IBPS Clerk Mains Mock Test 2019 – Take Test Now
 Free LIC Assistant Mains Mock Test 2019 – Take Test Now
 Free IBPS SO Mock Test 2019 – Take Test Now
Dear Readers, Important Practice Quantitative Aptitude – Application Sums for IBPS Exams 2017 was given here with Solutions. Aspirants those who are preparing for the Bank Examination and other Competitive Examination can use this material.
Quizsummary
0 of 15 questions completed
Questions:
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
Information
Application Sum Day14
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Results
0 of 15 questions answered correctly
Your time:
Time has elapsed
You have reached 0 of 0 points, (0)
Average score 

Your score 

Categories
 Not categorized 0%
Pos.  Name  Entered on  Points  Result 

Table is loading  
No data available  
 1
 2
 3
 4
 5
 6
 7
 8
 9
 10
 11
 12
 13
 14
 15
 Answered
 Review

Question 1 of 15
1. Question
A’s present age is 3/4 of B’s and B’s present age is 11/17 of C’s present age. The sum of their present ages is 145 years. What is the difference between the ages of A and C?
Correct
Correct Answer is : c
Solution :
B= 11C/17, A=3B/4=33C/68
So, 33C/68+ 11C/17+C=145
Therefore, age of C=68 years, A=33 years
CA=6833=35 years
Hence, difference between the ages of A and C=35 years
Incorrect
Correct Answer is : c
Solution :
B= 11C/17, A=3B/4=33C/68
So, 33C/68+ 11C/17+C=145
Therefore, age of C=68 years, A=33 years
CA=6833=35 years
Hence, difference between the ages of A and C=35 years

Question 2 of 15
2. Question
A gives 58% of his salary to B. From this amount, B spends 28% on his food items and 8% on repair of furniture and the remaining Rs 24128 puts in the saving account. What is the salary of A?
Correct
Correct Answer is: c
Solution:
Let salary of A=X
B gets=58X/100
He spends 28%+8%=36%, i.e. left with 64% of (58X/100)
So (64/100)*(58/100)*X=24128
X=24128*100*100/(64*58)=Rs 65000
Hence, salary of A=Rs 65000
Incorrect
Correct Answer is: c
Solution:
Let salary of A=X
B gets=58X/100
He spends 28%+8%=36%, i.e. left with 64% of (58X/100)
So (64/100)*(58/100)*X=24128
X=24128*100*100/(64*58)=Rs 65000
Hence, salary of A=Rs 65000

Question 3 of 15
3. Question
If the numerator of a fraction is increased by 2 and denominator is increased by 3. the fraction becomes 7/9; and if numerator as well as denominator is decreased by 1 the fraction becomes 4/5. What is the original fraction?
Correct
Correct Answer is: c
Solution:
Let the numerator and denominator be x and y respectively.
Then, (x+2)/(y+3)= 7/9
=> 9(x + 2) = 7(y + 3)
=> 9x – 7y = 3 …. (i)
(x1)/(y1)= 4/5
=> 5(x 1) = 4(y1)
=> 5x – 4y = 1 …. (ii)
Solving (i) and (ii), we get
X = 5, Y = 6
Required fraction = x/y=5/6
Incorrect
Correct Answer is: c
Solution:
Let the numerator and denominator be x and y respectively.
Then, (x+2)/(y+3)= 7/9
=> 9(x + 2) = 7(y + 3)
=> 9x – 7y = 3 …. (i)
(x1)/(y1)= 4/5
=> 5(x 1) = 4(y1)
=> 5x – 4y = 1 …. (ii)
Solving (i) and (ii), we get
X = 5, Y = 6
Required fraction = x/y=5/6

Question 4 of 15
4. Question
Adam Goldberg and Barry Goldberg start walking around a circular park of radius 105 meters. They start at the same point. Adam Goldberg goes clockwise at 15 m/s while Barry Goldberg goes anticlockwise at 12m/s. How many times will they cross each other at the starting point if they walk for 1 hour and 50 minutes?
Correct
Correct Answer is: b
Solution:
Given, radius of the park = 105 m.
Circumference of the park = 2 * (22/7) * 105 = 660 m.
=> Time taken by Adam Goldberg to complete 1 round = 660/15 = 44 s.
=> Time taken by Barry Goldberg to complete 1 round = 660/12 = 55 s.
They will meet at the starting point at a time that is the LCM of the time taken by each of them to complete one full round.
They will meet at the starting point after every LCM of 44 s and 55 s = 220 s.
They walked for a total of 1 hour 50 minutes = 110 minutes = 110 * 60 = 6600 s.
=> Number of times they will cross each other = 6600/220 = 30
Incorrect
Correct Answer is: b
Solution:
Given, radius of the park = 105 m.
Circumference of the park = 2 * (22/7) * 105 = 660 m.
=> Time taken by Adam Goldberg to complete 1 round = 660/15 = 44 s.
=> Time taken by Barry Goldberg to complete 1 round = 660/12 = 55 s.
They will meet at the starting point at a time that is the LCM of the time taken by each of them to complete one full round.
They will meet at the starting point after every LCM of 44 s and 55 s = 220 s.
They walked for a total of 1 hour 50 minutes = 110 minutes = 110 * 60 = 6600 s.
=> Number of times they will cross each other = 6600/220 = 30

Question 5 of 15
5. Question
Sam invested Rs. 20,000 @ 15% per annum for one year. If the interest is compounded half yearly, then what will be the amount received by Sam at the end of the year?
Correct
Correct Answer is : b
Solution :
P = Rs 20000; R = 15% p.a. = 7.5 % per halfyear;
T = 1 year = 2 halfyears
If an amount P is compounded at interest rate of r for T years then the total amount which we
Will get = P (1 + r/100)^{T}
Amount = Rs.[20000 X (1 +7.5/100) X (1 +7.5/100)]
= Rs.(20000 X 43/40 X 43/40)=Rs 23112.5
Incorrect
Correct Answer is : b
Solution :
P = Rs 20000; R = 15% p.a. = 7.5 % per halfyear;
T = 1 year = 2 halfyears
If an amount P is compounded at interest rate of r for T years then the total amount which we
Will get = P (1 + r/100)^{T}
Amount = Rs.[20000 X (1 +7.5/100) X (1 +7.5/100)]
= Rs.(20000 X 43/40 X 43/40)=Rs 23112.5

Question 6 of 15
6. Question
A shopkeeper bought 150 calculators at the rate of Rs. 250 per calculator. He spent Rs. 2500 on transportation and packing. If the marked price of calculator is Rs.320 per calculator and the shopkeeper gives a discount of 5 % on the marked price then what will be the percentage profit gained by the shopkeeper?
Correct
Correct Answer is : a
Solution :
Cost Price (CP) of 150 calculators = 150 X 250 = Rs 37500
Since he spent Rs. 2500 on transportation and packing so,
Total cost Price = 37500 + 2500 = Rs 40000
Marked price of 150 calculators = 150 X 320 = Rs 48000
Since shopkeeper gives 5% discount so selling price will be 95% (1005) of the marked price.
So, total Selling price after discount = 0.95 X 48000
= Rs. 45600
As we know that profit % = (Selling price – Cost Pricey)/(cost price) X 100
Percentage profit = (45,600 – 40,000)/40,000 X 100=14%
Incorrect
Correct Answer is : a
Solution :
Cost Price (CP) of 150 calculators = 150 X 250 = Rs 37500
Since he spent Rs. 2500 on transportation and packing so,
Total cost Price = 37500 + 2500 = Rs 40000
Marked price of 150 calculators = 150 X 320 = Rs 48000
Since shopkeeper gives 5% discount so selling price will be 95% (1005) of the marked price.
So, total Selling price after discount = 0.95 X 48000
= Rs. 45600
As we know that profit % = (Selling price – Cost Pricey)/(cost price) X 100
Percentage profit = (45,600 – 40,000)/40,000 X 100=14%

Question 7 of 15
7. Question
A tank can be filled by three pipes A, B and C in 8 hours. Pipes were opened for 2 hours and then pipe C was closed. A and B filled the remaining tank in 12 hours. Find the time taken by pipe C alone to completely fill the tank.
Correct
Correct Answer is: c
Solution :
Tank filled in 2 hours = 2/8
= 1/4
Tank left = 3/4
In 12 hours, 3/4th tank is filled by A+B
In 1 hour, 3/(4*12) = 1/16 tank is filled
Tank filled by C in 1 hour = 1/8 – 1/16
= (21)/16
= 1/16
C fills in 16 hours.
Incorrect
Correct Answer is: c
Solution :
Tank filled in 2 hours = 2/8
= 1/4
Tank left = 3/4
In 12 hours, 3/4th tank is filled by A+B
In 1 hour, 3/(4*12) = 1/16 tank is filled
Tank filled by C in 1 hour = 1/8 – 1/16
= (21)/16
= 1/16
C fills in 16 hours.

Question 8 of 15
8. Question
Ram gives David a head start of 10 meter in a 100 meter race and still beats him by 10 meters. If Ram is able to complete the 200 meters race in 20 seconds then how much time David will take to complete the 100 meters race?
Correct
Correct Answer is c
Solution :
Ram gives David a head start of 10 meters means when David has covered the distance of 10 meters then Ram starts running. Still he beats him by 10 meters which means when Ram is at 100 meters mark David is at 80 meters mark only.
Means the time in which Ram is able to complete 100 meters in the same time David is able to complete only 80 m.
Hence the ratio of speeds of Ram and David = 100:80 = 5:4
Since Ram is able to run 200 meter in 20 seconds hence
his speed = (200/20) = 10 m/sec
Therefore the speed of the David = (10 X 4)/5=8 m/sec
Hence the time in which David will complete the 100 meters race = 100/8 seconds
= 12.5 seconds
Incorrect
Correct Answer is c
Solution :
Ram gives David a head start of 10 meters means when David has covered the distance of 10 meters then Ram starts running. Still he beats him by 10 meters which means when Ram is at 100 meters mark David is at 80 meters mark only.
Means the time in which Ram is able to complete 100 meters in the same time David is able to complete only 80 m.
Hence the ratio of speeds of Ram and David = 100:80 = 5:4
Since Ram is able to run 200 meter in 20 seconds hence
his speed = (200/20) = 10 m/sec
Therefore the speed of the David = (10 X 4)/5=8 m/sec
Hence the time in which David will complete the 100 meters race = 100/8 seconds
= 12.5 seconds

Question 9 of 15
9. Question
There are 10 tickets to the theater, four of which are for the seats in the front row. 5 tickets are selected at random. What is the probability that three of them are for the front row?
Correct
Correct Answer is: b
Solution :
Number of ways of choosing 5 tickets out of 10 = ^{10}C_{5} = 252
Number of ways of choosing 3 front row tickets = ^{4}C_{3} X ^{6}C_{2 }= 4 X 15 = 60
Therefore required probability = 60/252= 15/63 =5/21
Incorrect
Correct Answer is: b
Solution :
Number of ways of choosing 5 tickets out of 10 = ^{10}C_{5} = 252
Number of ways of choosing 3 front row tickets = ^{4}C_{3} X ^{6}C_{2 }= 4 X 15 = 60
Therefore required probability = 60/252= 15/63 =5/21

Question 10 of 15
10. Question
Water is flowing through a rectangular tunnel having a cross section of 0.8 m X 0.65 m. If the water is flowing at a speed of 12 km/hr then what is the approximate volume of water in liters which is flowing through the tunnel in a second?
Correct
Correct Answer is: b
Solution :
Speed of the water = 12 km/hr = 12 X 5/18=10/3 m/sec
Area of the rectangular cross section = 0.8 X 0.65 = 0.52 m^{2}
Volume of the water which flows through the tunnel in 1 sec
= 0.52 X 10/3 = 26/15 m^{3}
As we know that;
1 m^{3} = 1000 liter
Therefore required volume = 26/15 X 1000 = 1733.33 liter
=1734 liter (Approx)
Incorrect
Correct Answer is: b
Solution :
Speed of the water = 12 km/hr = 12 X 5/18=10/3 m/sec
Area of the rectangular cross section = 0.8 X 0.65 = 0.52 m^{2}
Volume of the water which flows through the tunnel in 1 sec
= 0.52 X 10/3 = 26/15 m^{3}
As we know that;
1 m^{3} = 1000 liter
Therefore required volume = 26/15 X 1000 = 1733.33 liter
=1734 liter (Approx)

Question 11 of 15
11. Question
A and B always work on alternate days, but none of them work on Saturday and Sunday. Working alone A and B can finish the work in 20 days and 30 days respectively. If the work was started by A on Monday then on which day work will be finished?
Correct
Correct Answer is : e
Solution :
Work done in week 1 = 1/20 + 1/30 + 1/20 + 1/30 + 1/20 = 13/60
Work done in week 2 = 1/30 + 1/20 + 1/30 + 1/20 + 1/30 = 12/60
Work done in 2 weeks = 13/60 + 12/60 = 25/60 = 5/12
Work done in 4 weeks = 5/12 x 2 = 5/6
Work left = 1/6 which will get completed in the 5th week.
Now A will start working on the Monday of the filth week as it was B who was working on Friday in the fourth week.
Work done on Monday = 1/20 Work done on Tuesday = 1/30
Work done on Wednesday = 1/20
Work done on Thursday = 1/30
Work done in the four days = 1/20 + 1/30 + 1/20 + 1/30 = 1/6
Hence the work will get completed on Thursday.
Incorrect
Correct Answer is : e
Solution :
Work done in week 1 = 1/20 + 1/30 + 1/20 + 1/30 + 1/20 = 13/60
Work done in week 2 = 1/30 + 1/20 + 1/30 + 1/20 + 1/30 = 12/60
Work done in 2 weeks = 13/60 + 12/60 = 25/60 = 5/12
Work done in 4 weeks = 5/12 x 2 = 5/6
Work left = 1/6 which will get completed in the 5th week.
Now A will start working on the Monday of the filth week as it was B who was working on Friday in the fourth week.
Work done on Monday = 1/20 Work done on Tuesday = 1/30
Work done on Wednesday = 1/20
Work done on Thursday = 1/30
Work done in the four days = 1/20 + 1/30 + 1/20 + 1/30 = 1/6
Hence the work will get completed on Thursday.

Question 12 of 15
12. Question
Kalpesh, Chitresh and Prabhakar enter into a business of telecommunication. Kalpesh invested some amount in the starting. After 1 year, Chitresh invested double the amount invested by Kalpesh and after 1 year 4 months. Prabhakar invested thrice the amount invested by Kalpesh. They earn a profit of Rs 90000 at the end of 2 years. What is Chitresh’s share in the profit?
Correct
Correct Answer is: c
Solution:
Let the amount invested by Kalpesh is P
So, Chitresh’s investment will be 2P and Prabhakar’s investment will be 3P
Ratio of investment by 3 of them in 2 years = P * 24: 2P *12:8 * 3P = 1:1:1
Therefore, Chitresh’s share in profit = 1/3 * 90000 = Rs 30000
Incorrect
Correct Answer is: c
Solution:
Let the amount invested by Kalpesh is P
So, Chitresh’s investment will be 2P and Prabhakar’s investment will be 3P
Ratio of investment by 3 of them in 2 years = P * 24: 2P *12:8 * 3P = 1:1:1
Therefore, Chitresh’s share in profit = 1/3 * 90000 = Rs 30000

Question 13 of 15
13. Question
The average monthly expenditure of a family for the first five months is Rs 5600, Rs 5500 for the next 3 months and for the next 4 months it is Rs 5900. If the family saves Rs 6900 during the year, find the average monthly income.
Correct
Correct Answer is: e
Solution:
Income = Expenditure + savings
Total expenditure = 5600*5 + 5500*3 + 5900*4
= 28000 + 16500 + 23600
= 68100
Savings = 6900 given
Total income = 68100 + 6900 = 75000
Average income = 75000/12 = 6250
Incorrect
Correct Answer is: e
Solution:
Income = Expenditure + savings
Total expenditure = 5600*5 + 5500*3 + 5900*4
= 28000 + 16500 + 23600
= 68100
Savings = 6900 given
Total income = 68100 + 6900 = 75000
Average income = 75000/12 = 6250

Question 14 of 15
14. Question
Amit and Manju are best friends. Both secured a job during the campus placements. Ratio of their salary is in the ratio of 7:8. After getting their joining locations, Amit got a job in his home town whereas Manju got a job outside her hometown due to which Manju’s expenditure becomes more than Amit’s expenditure. Ratio of their expenditure is in the ratio of 6:7. If both of them saved 8000, then, find the salary of Manju?
Correct
Correct Answer is: e
Solution :
Let the Amit and Manju’s salary be 7k and 8k.
Expenditure be 6e and 7e
According to the question,
7k6e=8000– (1)
8k7e=8000– (2)
Using 1 and 2
e=8000, k=8000
Manju’s salary=8k=8*8000=64000
Incorrect
Correct Answer is: e
Solution :
Let the Amit and Manju’s salary be 7k and 8k.
Expenditure be 6e and 7e
According to the question,
7k6e=8000– (1)
8k7e=8000– (2)
Using 1 and 2
e=8000, k=8000
Manju’s salary=8k=8*8000=64000

Question 15 of 15
15. Question
In an Olympics game, if each of the 1000 participants is given an identity card, and the cards were numbered with consecutive natural numbers starting from 1 to 1000, then how many of the cards will have numbers such that the digit 6 does not occur more than once?
Correct
Correct Answer is : d
Solution :
There is only one two digit number that has more than one 6 and that is 66.
In the three digit number 6×6, the place ‘x’ can be filled in 10 ways; out of them one possibility is 666.
In the three digit number x66, the place ‘x’ can be filled in 9 ways; out of them one possibility is 666.
In the three digit number 66x, the place ‘x’ can be filled in 10 ways; out of them one possibility is 666.
In the above calculations, we have included the number 666 three times instead of only once.
The total number of numbers that have more than one 6, are 1 + 10 + 9 + 10 – 2 = 28.
Hence, the required number is 1000 – 28 = 972.
Hence, the correct answer is option 4.
Incorrect
Correct Answer is : d
Solution :
There is only one two digit number that has more than one 6 and that is 66.
In the three digit number 6×6, the place ‘x’ can be filled in 10 ways; out of them one possibility is 666.
In the three digit number x66, the place ‘x’ can be filled in 9 ways; out of them one possibility is 666.
In the three digit number 66x, the place ‘x’ can be filled in 10 ways; out of them one possibility is 666.
In the above calculations, we have included the number 666 three times instead of only once.
The total number of numbers that have more than one 6, are 1 + 10 + 9 + 10 – 2 = 28.
Hence, the required number is 1000 – 28 = 972.
Hence, the correct answer is option 4.