IBPS Clerk Mains Quantitative Aptitude Question Paper 2021
Every year the Institute of banking personnel selection IBPS clerk exam is conducted at the Institute of banking personnel selection. To clear this exam one should be thorough with the concept and solve an ample amount of questions daily to maintain speed and concept. Solving the previous year’s question plays a vital role in enhancing concept building along with enhancing speed.
Here we will discuss IBPS Clerk Mains 2021 Quantitative Aptitude Question Paper.
Directions ( 1-6) :
Read the following Pie-charts carefully and answer the following questions. The following pie chart shows the percentage of the total population (both Males and Females) in five societies
The following pie charts show the percentage of Males in all the five societies.
Note :
I) The difference between the total population in society A and society B is 21.
II) The number of males in society B is 33.33% more than the number of females in society B.
Solution (1-6):
Let, Total population in all society together = P
According to question,
P×(18/100) – P×(15/100) = 21
=> 3P/100 = 21
=> P = 700
Total Population = 700
From chart 1 :
Total population in society B=700 × (15/100) =105
Let, Number of females in society B = F
Number of males in society B = F×(133⅓/100) =4F/3
Now, (F + 4F/3) = 105
=> 7F/3 = 105
=> F = 45
Females in society B = 45
Males in society B = (105 – 45) = 60
From chart 2 :
Let, total number of males in all society together= M
Males in society B = M×(15/100) = 3M/20
3M/20 = 60
=> M = 400
Total number of males in five societies= 400.
Society A:
Total Population in A = (18/100)×700 = 126
Males population = (20/100)×400 = 80
Females population = (126-80) = 46
Society B :
Total population in B = (15/100)×700 = 105
Male population = (15/100)×400 = 60
Female population = (105-60) = 45
Society C :
Total population in C = (22/100)×700 = 154
Male population = (15/100)×400 = 60
Female population = (154-60) = 94
Society D :
Total population in D = (20/100)×700 = 140
Male population = (25/100)×400 = 100
Female population= (140-100) = 40
Society E :
Total population in E = (25/100)×700 = 175
Male population = (25/100)×400 = 100
Female population = (175-100) = 75
| Society | No. of males | No. of females |
|---|---|---|
| A | 80 | 46 |
| B | 60 | 45 |
| C | 60 | 94 |
| D | 100 | 40 |
| E | 100 | 75 |
Que 1. What is the ratio between the number of males in societies D and E together to the number of females in societies A and C together?
A) 7 : 10
B) 10 : 7
C) 6 : 1
D) 5 : 3
E) 11 : 5
Answer : B
Explanation :
Number of males in society D and E together = (100 + 100) = 200
Number of females in society A and C together = (46 + 94) = 140
∴ The required ratio = 200 : 140 = 10 : 7
Que 2. The number of males in B and C is what percentage of the number of females in the same societies?
A) 86.33%
B) 92.58%
C) 48.76%
D) 110.58%
E) 88.66%
Answer : A
Explanation :
Males in society B and C together = (60 + 60) = 120
Females in society B and C together = (45 + 94) = 139
∴ Required percentage=(120/139) × 100 = 86.33%.
Que 3. What is the difference between the average number of males in societies A, B and D together to the average number of females in societies C, D and E together?
A) 15.66
B) 10.33
C) 12.66
D) 11.33
E) 18.66
Answer : B
Explanation :
Average number of males in society A, B and D together = (80 + 60 + 100)/3 = 80
Average number of females in C,D and E together = (94 + 40 + 75)/3 = 69.67
∴ Required difference = (80 – 69.67) = 10.33
Que 4. If 60% of males in society C and 40% of females in society B went to a function, then what is the number of people that attended the function?
A) 45
B) 30
C) 54
D) 28
E) 66
Answer : C
Explanation :
60% of 60 = 60 × (60/100) = 36
40% of 45 = 45 × (40/100) = 18
∴ Number of people attended the function = (36 + 18) = 54
Que 5. What is the ratio between the number of females in B and E together and the number of females in the rest of the societies?
A) 2 : 3
B) 3 : 2
C) 1 : 2
D) 5 : 3
E) 2 : 5
Answer : A
Explanation :
∴ Required ratio = (45 + 75) : (46 + 94 + 40) = 2 : 3.
Que 6. The number of females in society B and E together is what percentage of the number of males in society A and E together?
A) 33.33%
B) 48%
C) 66.66%
D) 82.5%
E) 58.6%
Answer: C
Explanation:
Females in society B and E together = (45 + 75) = 120
Males in society A and E together = (80 + 100) = 180
∴Required percentage =(120/180) × 100 = 66.66%
Directions ( 7 – 11) :
What approximately will come in the place of the question mark ‘?’ in the following question?
Que 7. [(1105.05/13) – 4.8% of 250 – 7.8% of (35×10)] =?% × 300
A) 15
B) 18
C) 32
D) 12
E) 10
Answer : A
Explanation :
[(1105.05/13) – 4.8% of 250 – 7.8% of (35×10)] = ?% × 300
=> [85 – 12 – 7.8% of 350] = (?/100) × 300
=> [85 – 12 – 27.3] = 3 × ?
=> 45/3 = ? (Taking approx value)
∴? = 15
∴ The required value=15
Que 8. 53.003 × 9.95 × (257.81 ÷ 6.01) – (45×5) + 152.21 – 25.31 = ?
A) 22465
B) 22692
C) 15234
D) 21123
E) 28234
Answer: B
Explanation :
53.003 × 9.95 × (257.81 ÷ 6.01) – (45 × 5) + 152.21 – 25.31 = ?
=> 53 × 10 × (258/6) – 225 + 152 – 25 = ?
=> 53 × 10 × 43 – 225 + 152 – 25 = ?
=> 22790 – 98 = ?
∴ ? = 22692
∴ The required value=22692
Que 9. 555.05 ÷ 18.5 + 10⁵ ÷ 10³ + 11⁵ ÷ 11³ = ?
A) 130
B) 250
C) 251
D) 127
E) 156
Answer : C
Explanation :
555.05 ÷ 18.5 + 10⁵ ÷ 10³ + 11⁵ ÷ 11³ = ?
=> 555/18.5 + 10⁵/10³ + 11⁵/11³ = ?
=> 30 + 100 + 121 = ?
∴ ? = 251
∴ The required value = 251
Que 10. 144.062 ÷ 12.02 + 11.99 ÷ 3.99 × 16 ÷ 7.99 = ?
A) 25
B) 18
C) 19
D) 32
E) 10
Answer : B
Explanation :
144.062 ÷ 12.02 + 11.99 ÷ 3.99 × 16 ÷ 7.99 = ?
=> 144/12 + 12/4 × 16/8 = ?
=> 12 + 3 × 2 = ?
∴ ? = 18
∴ The required value=18
Que 11. [29.99% of (2/5) + 50.01% of (2/5) + 30.1% of (2/5)] = ?
A) 11/15
B) 22/25
C) 11/25
D) 6/5
E) 17
Answer : C
Explanation :
[29.99% of (2/5) + 50.01% of (2/5) + 30.1% of (2/5)] = ?
=> [(30/100) × (2/5) + (50/100) × (2/5) + (30/100) × (2/5) = ?
=> [3/25 + 1/5 + (3/25)] = ?
=> (3 + 5 +3)/25 = ?
∴ ? = 11/25
∴ The required value = 11/25
Que 12. The ratio of monthly income and savings of A is 5 : 2. The annual expenditure of A is Rs.108000. If the monthly income of A is increased by 20% and the monthly saving of A is increased by 10%. Then find by what percentage expenditure of A is increased?
A) 20/3%
B) 50/3%
C) 70/3%
D) 80/3%
E) 40/3%
Answer : D
Explanation :
Income Saving Expenditure
5x 2x (5x – 2x) = 3x
3x = 108000
=> x = 36000
Income = 5 × 36000 = 180000
Income after 20% increased = 180000 × (120/100) = 216000
Savings = 2 × 36000 = 72000
Savings after 10% increased = 72000 × (110/100) = 79200
Now, Expenditure =(216000 – 79200) = 136800
Expenditure increased = (136800 – 108000) = 28800
∴ Required percentage increased = (28800/108000)×100 = 80/3%
Que 13. A, B and C have a certain number of chocolates in the ratio of 3 : 2 : 5. If A takes 10 chocolates from B and 20 chocolates from C, the ratio of the chocolates of B and C becomes 1 : 3. What is the total number of chocolates they have initially?
A) 80
B) 90
C) 100
D) 120
E) 110
Answer : C
Explanation :
Let, the number of chocolates that A, B and C have be 3x, 2x and 5x
According to question,
Now, A’s chocolates = (2x + 10 + 10)
B’s chocolates = (2x – 10)
C’s chocolates = (5x – 20)
(2x – 10)/(5x – 20) = 1/3
=> 6x – 30 = 5x – 20
=> x = 10
∴ Initially total number of chocolates among them =(3 × 10 + 2 × 10 + 5 × 10) = 100.
Que 14. Find the value of K if one root of equation 4x² – 3Kx + 2 = 0 is 2 .
A) -2
B) -3
C) 3
D) 6
E) 2
Answer : C
Explanation :
4x² – 3kx + 2 = 0 (x = 2 ,is a root of the equation)
=> 4×2² – 3k × 2 + 2 = 0.
=> 16 – 6k + 2 = 0
=> 6k = 18
=> k = 3
∴ The value of k = 3
Que 15. In an office, the average height of 48 employees is 120 feet. If ‘n’ employees were included whose average height was 108 feet and now the average height of total employees is 116 feet. Find the total number of employees in an office.
A) 72
B) 70
C) 62
D) 54
E) None of these
Answer : A
Explanation :
Total height of 48 employees = 48 × 120 feet.
Total height of n employees = n × 108 feet.
Total height of (n + 48) employees = (n + 48) × 116
Now,
(n + 48) × 116 = (120 × 48) + (n × 108)
=> 116n + (116 × 48) = (120 × 48) + 108n
=> 8n = 48 × (120 – 116)
=> n = (48×4)/8
=> n = 24
∴ The total number of employees = 24 + 48 = 72
Que 16. The average age of 200 employees in an office is 58 years, where 3/5 employees are males and the remaining employees are females, and also the ratio of the average age of male employees to female employees is 5 : 7. Find the average age of male employees.
A) 50 years
B) 70 years
C) 35 years
D) 45 years
E) None of these
Answer : A
Explanation :
Total age of 200 employees = 200 × 58 = 11600
Number of male employees = 200 × (3/5) = 120
Number of female employees = (200 – 120) = 80
Let, average age of male employees = 5x
Average age of females employees = 7x
Total age of male employees = 120 × 5x
Total age of female employees = 80 × 7x
Now, (120 × 5x) + (80 × 7x) = 11600
=> 600x + 560x = 11600
=> x = 11600/1160
=> x = 10
∴ The average age of male employees = 5x = 50
Que 17. A person invested a sum of Rs.8000 at the rate of 12.5% annually at simple interest for 3 years after getting interest from an amount, he again invested the total amount including simple interest at the rate of 20% compounded annually for the same years. Find the total amount.
Answer : B
Explanation :
Simple Interest= (P × R × T)/100
=> S.I = (8000 × 12.5 × 3)/100
=> S I = 3000
Amount = principal + simple interest = (8000+3000) = 11000
This amount will be used as a principal for Compound interest.
20% = 1/5
Principal Amount
5³ 6³ (because time is 3 years given)
125 216
Here, 125 unit = 11000
=> 216 unit = (11000/125) × 216 = 19008
∴ The required amount = 19008.
Que 18. 60 men can complete a work in some days, to complete the same work by 40 men took extra 12 days. Then find the days taken by 12 men to complete the 3/5th of the total work.
A) 72 days
B) 84 days
C) 60 days
D) 50 days
E) 92 days
Answer : A
Explanation :
M1 × D1 = M2 × D2
=> 60 × D1 = 40 (D1 + 12)
=> 60 D1 – 40 D1 = 480
=> D1 = 480/20
D1 = 24 (total days)
Total work = men × time taken = 60 × 24 = 1440 units
3/5 part of 1440 = 864 units
Now, 12 × D = 864
=> D = 864/12 = 72
∴ The required number of days = 72
Que 19. The below question is followed by two statements labelled I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below :
I) A running track outside of a square park is given.
II) Area of the path outside the square park is 112 cm² and the width of the path is 2 cm.
Find the area of the park.
A) Statement I alone is sufficient to answer the question
B) Statement II alone is sufficient to answer the question.
C) Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
D) Either statement I or statement II alone is sufficient to answer the question.
E) Neither statement I nor statement II is sufficient to answer the question.
Answer: B
Explanation :
Statement I :
A running track outside of the park is given.
Statement I alone is not sufficient.
Statement II :
Let, the side of the square park be x.
Side of the path = (x + 2 + 2) = ( x + 4)
According to question, (x+4)² – x² = 112
16 + 8x = 112
x = 12.
Area of the square = 12 × 12 = 144 cm2
Statement II alone is sufficient.
Que 20. The below question is followed by two statements labelled I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below :
I) There is a total of 6 numbers of which 3 are consecutive even and 3 consecutive odd numbers.
II) Difference between the middle term of each odd and even number is 7.
Find the average of six numbers.
A) Statement I alone is sufficient to answer the question.
B) Statement II alone is sufficient to answer the question.
C) Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
D) Either statement I or statement II alone is sufficient to answer the question.
E) Neither statement I nor statement II is sufficient to answer the question.
Answer : E
Explanation :
Statement I :
Let, even numbers = 2x, (2x + 2), (2x + 4) Odd numbers = 2y, (2y + 1), (2y + 3)
Required average = (6x + 6 + 6y + 4)/6
here we don’t know about the value of x and y si one can not find the exact value of the average.
Statement I alone is not sufficient.
Statement II :
(2y+ 1) – (2x+2) = 7
(y – x) = 8
Statement II alone is not sufficient.
∴ Neither statement I nor statement II is sufficient to answer the question.
Que 21. There are 3 clubs A, B and C and the ratio of the total population visit in the club is 10 : 12 : 9 respectively. If the number of males visit in club A is 35%, in club B is 45% and in club C is 40% of the total population, then find the ratio of total males and females visit in all the clubs together.
A) 29 : 33
B) 25 : 33
C) 22 : 37
D) 22 : 35
E) 25 : 37
Answer : E
Explanation :
Let, Total population visit in club A = 10x
Total population visit in club B = 12x
Total population visit in club C = 9x
Males visit in club A = 10x × (35/100) = 3.5x
Female visit in club A= (10x – 3.5x) = 6.5x
Males visit in club B = 12x × (45/100) = 5.4x
Female visit in club B = (12x – 5.4x) = 6.6x
Males visit in club C = 9x × (40/100) = 3.6x
Females visit in club C = (9x – 3.6x) = 5.4x
Total males in all clubs together =(3.5x + 5.4x + 3.6x) = 12.5x
Total females in all clubs together =(6.5x + 6.6x + 5.4x) = 18.5x
Required ratio = 12.5 : 18.5 = 25 : 37
∴ The ratio of total males and females visit in all the clubs together = 25 : 37.
Que 22. A container contained 160 litres of milk. Few litres of milk is replaced by water then the quantity of milk will be 3 times the quantity of water. Again 20 litres of mixture is replaced by 15 litres of water then find the quantity of water in the final mixture?
A) 92 litres
B) 50 litres
C) 45 litres
D) 57 litres
E) None of these
Answer :B
Explanation :
Milk : Water = 3 : 1
Milk = 160 × (3/4) = 120 litres
Water = (160 – 120) = 40 litres
After 20 litres of mixture taken out ,the remaining mixture = (160 – 20) = 140 litres
Now, milk = 140 × 3/4 = 105 litres
Water = (140 – 105) = 35 litres
After adding 15 litres of water then the quantity of water = (35 + 15) = 50 litres.
∴ The quantity of water in the final mixture = 50 litres.
Que 23. Aditya takes thrice the time taken by Harshit and Vaibhav together to do a job. Harshit takes four times as time taken by Aditya and Vaibhav together to do the work. If all three together can complete the job in 24 days, then the number of days, Aditya alone will take to finish the job is –
A) 90 days
B) 96 days
C) 100 days
D) 80 days
E) 110 days
Answer: B
Explanation :
Aditya takes thrice the time taken by Harshit and Vaibhav together to do a job.
The ratio of the efficiency of Aditya to the combined efficiency of Harshit and Vaibhav = 1 : 3
In this case total efficiency = 4 units.
Now, Harshit takes four times as time taken by Aditya and Vaibhav together to do the work
The required ratio of their efficiency = 1 : 4
Total efficiency = 5 units.
We know that the combined efficiency of all of them must be equal in each case so,
Aditya’ efficiency : (Harshit + Vaibhav)’s efficiency = 5 : 15
Harshit’s efficiency : (Aditya + Vaibhav)’s efficiency = 4 : 16
Efficiency of Aditya = 5 units/ day.
Efficiency of Harshit = 4 units/day.
Efficiency of Vaibhav = 20 – 5 – 4 = 11 units/day
Total work = (20) × 24 = 480 units.
Time taken by Adits to complete the whole work = 480/5 = 96 days.
∴ Aditya alone can do the job in 96 days.
Que 24. Three friends invested Rs.48000, Rs.52000 and Rs.36000 respectively in a business. The partnership condition is that each will get 8% per annum on the capital and the remaining profit will be divided in the ratio of their capital. If at the end of the 1 year the total profit is Rs.32640, then find the share of the first friend in the profit.
A) Rs.11340
B) Rs.11000
C) Rs.12500
D) Rs.11520
E) None of these
Answer : D
Explanation :
Each will get 8% on their Capitals.
A gets = 48000 × (8/100) = 3840
B gets = 52000 × (8/100) = 4160
C gets = 36000 × (8/100) = 2880
(A + B + C) together get = (3840 + 4160 + 2880) = 10880
Remaining profit = (32640-10880) = 21760
Ratio of investment of A,B and C = 48000 : 52000 : 36000 = 12 : 13 : 9
First friend gets = 21760 × (12/34) = 7680
∴ First friend got total profit = (7680 + 3840) =11520
Que 25. Lalit bought second-hand Motorcycles and spent Rs.500 on their repairs. Then he sold it to Rajiv at a profit of 10% and then he sold it to Muneesh at a loss of 20%, Finally, Muneesh sold it for Rs.38720 at a profit of 10%. How much amount did Lalit pay initially for the motorcycles?
A) Rs.39000
B) Rs.38900
C) Rs.40000
D) Rs.39500
E) None of these
Answer : D
Explanation :
Let, the price of motor cycle = X
Repair cost = 500
Total cost price for lalit = (X + 500)
Now,
(X + 500) × 110/100 × 80/100 × 110/100 = 38720
=> X + 500 = 38720 × (1000/11×8×11)
=> X + 500 = 40000
X = 39500
∴ Amount paid by Lalit initially for the motor cycles is Rs.39500.
Que 26. A shopkeeper marked a price of an article at x% above the cost price and he sold it at a discount of 0.5x%, then he earns 8% profit, Find the value of x, if x>20.
A) 90
B) 80
C) 70
D) 60
E) 50
Answer : B
Explanation :
Successive discount = x – 0.5x – (x * 0.5x)/100 = 0.5x – x²/200
According to question,
0.5x – x²/200 = 8
=> x²/200 – 0.5x + 8 = 0
=> x² – 100x + 1600 = 0 (multiplying both sides by 200)
=> x² – 20x – 80x + 1600 = 0
=> (x – 20)(x -80) = 0
=> x = 20, 80
∵ x is greater than 20, then the value of x =80.
Directions (27-29): Below question is followed by two statements labeled I and II. Decide if these statements are sufficient to conclusively answer the question. Choose the appropriate answer from the options given below:
Que 27.
(I) Raman takes double the time of Sunita to complete (1/3) part of the work. (II) Sunita completes the work in 5 days.
Find the total time taken by Raman and Sunita working together to complete the work.
A) Statement I alone is sufficient to answer the question.
B) Statement II alone is sufficient to answer the question.
C) Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
D) Either statement I or statement II alone is sufficient to answer the question.
E) Neither statement I nor statement II is sufficient to answer the question.
Answer: C
Explanation :
Statement I :
Let, Sunita takes D days to complete the work.
Raman takes 2D days to complete 1/3 work.
Raman takes 6D days to complete the work.
Statement I alone is not sufficient.
Statement II :
Sunita takes 5 days to complete the work.
Raman takes 30 days to complete the work.
Hence now we can calculate easily the total days taken by together to complete the whole work.
Statement II alone is not sufficient.
∴ Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
Que 28.
I) A car travel at 20 km/hr for the first two hours, then it travels at 40 km/hr for the next 9/5 hours and completed the 8/11 part of its journey.
II) Car completed the remaining journey in the next 6/5 hours.
Find the average speed of the entire journey.
A) Statement I alone is sufficient to answer the question.
B) Statement II alone is sufficient to answer the question.
C) Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
D) Either statement I or statement II alone is sufficient to answer the question.
E) Neither statement I nor statement II is sufficient to answer the question.
Answer : C
Explanation :
Statement I :
Distance cover in first two hours=(2 × 20) = 40 km
Distance cover in next 9/5 hours=(40 × 9/5) = 72 km
8/11 part of journey = (40 + 72) = 112 km.
Total distance = 112 × 11/8 = 154 km
∴ The statement I alone is not sufficient.
Statement II :
Remaining time = 6/5 hours
Remaining journey=(154 – 112) =42 km
Speed = 42/(6/5) = 35 km/hr
Statement II alone is not sufficient.
∴ Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
Que 29.
I) A man is standing at a bridge. Distance from starting point is 250 m and from end of the bridge is 330 m.
II) If a train takes 5 minutes to cross the whole bridge at 2 m/s,
Find the length of the train.
A) Statement I alone is sufficient to answer the question.
B) Statement II alone is sufficient to answer the question.
C) Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
D) Either statement I or statement II alone is sufficient to answer the question.
E) Neither statement I nor statement II is sufficient to answer the question.
Answer: C
Explanation :
Statement I :
Total length of the bridge=(250 + 330) = 580 m
In this case, we don’t know about the speed of the train
∴ The statement (I) alone is not sufficient to answer the question.
Statement II :
Total distance covered by train=(5 × 60 × 2) = 600m
Here we don’t know about the length of the bridge.
∴ Statement II alone is not sufficient.
By combining I and II we get,
Length of train = (600 – 580) = 20 m.
∴ Statement I and II together are sufficient, but neither of the two alone is sufficient to answer the question.
Que 30. A circle inside a hexagon, the perimeter of the hexagon is 84 cm. Find the difference between the area of the circle and the hexagon. (Take, √3=1.73)
A) 46.62 cm²
B) 50.50 cm²
C) 42 cm²
D) 45.05 cm²
E) 57.46 cm²
Answer : A
Explanation :
Let the side of the hexagon = a
Perimeter of hexagon = 6a
6a = 84
a = 14 cm.
Area of hexagon = 6×(√3/4)×a²
=6×(1.73/4)×14² =508.62 cm²
From the above figure,
∠BOC = 360°/6 = 60°
OA is a bisector of ∠BOC, So
∠BOA = 60°/2 = 30°
AB = (1/2) BC = (1/2)×14 = 7 cm
Now from ∆BOA,
tan30° = AB/OA
=> 1/√3 = 7/OA
∴ OA = 7√3
Area of circle =πr² =(22/7)×(7√3)² =462 cm²
∴ Required difference =(508.62-462) =46.62 cm²
Directions (31-35): In a shooting competition, 150 shooters participated. Every participant picks at least one of the three guns namely A,B,C. The number of participants who pick all three guns is 26. Participants who pick gun B are 71. The number of participants who pick exactly two guns is 64. 48 participants pick guns A and C. The number of participants who pick only A gun is 27. The number of participants who pick gun A but not C is 50. The bullets fired from each of the guns either hit the target or missed the target. The bullets missed from gun A only is 70% more than that of the bullets that hit the target from gun A only. The bullet that hit the target from gun C only are 100% more than that of the bullets that missed the target.
Solution (31-35):
Here it is given that every participant picks at least one of the three guns namely A, B, and C. It means that no one is there who doesn’t pick a gun.
Total participants = 150
Number of participants who pick all three guns=26
Number of participants who pick gun B= 71
Number of participants who pick guns A & C=48
Number of participants who pick exactly two guns=64
Number of participants who pick gun A only = 27
Number of participants who pick gun A but not gun C = 50
Here from the figure,
V = 26
S+ V + Q + T = 71
U + V = 48 then U = 48 – 26 = 22
S + T + U = 64
U = 22
P = 27
P + S = 50
=> S = 50 – P = 50 – 27 = 23
S = 23
S + T + U = 64
=> 23 + T + 22 = 64
T= 19
S + V + Q + T = 71
=> 23 + 26 + Q + 19 = 71
Q = 3
Now, R = 150 – (P+Q+S+T+U+V)
R = 150 – (27 + 3 + 23 + 19 + 22 + 26)
R = 30
Que 31. What is the number of participants who pick both guns B and C?
Answer : A
Explanation :
Participants who pick both guns B and C = (V + T) = (26 + 19) = 45
Que 32. How many participants pick gun C?
A) 107
B) 112
C) 97
D) 92
E) 90
Answer : C
Explanation :
Participants who pick gun C = (R + T + U + V) = (30 + 19 + 22 + 26) = 97
Que 33. How many participants pick only gun B?
A) 12
B) 3
C) 7
D) 10
E) 21
Answer: B
Explanation :
Participants who pick only gun B = 3
Que 34. What is the number of participants who pick guns B and C but not A?
A) 52
B) 45
C) 60
D) 50
E) 42
Answer : A
Explanation :
Number of participants who pick guns B and C but not A = (19+3+30) = 52
Que 35. What is the number of participants who pick at most two guns?
A) 100
B) 105
C) 112
D) 124
E) 136
Answer : D
Explanation :
Number of participants who pick at most two guns = (150-26) = 124
Que 36. Match the following column (1) and column (2) :
| Quadratic equation | Relation between Roots |
|---|---|
| I) x² – 4x – 12 | a) Twice of the bigger root will be 6 less than 20 |
| II) y² – 5y – 14 | b) The sum of both the roots will be 4. |
| III) z² – 11z + 30 | c) Multiplication of both the roots will be 30. |
A) (i) – b , (ii) – a , (iii) – c
B) (i) – a , (ii) – b , (iii) – c
C) (i) – c , (ii) – a , (iii) – b
D) (i) – a , (ii) – c , (iii) – b
E) (i) – b , (ii) – c , (iii) – a
Answer: A
Explanation :
I) x² – 4x – 12 = 0
=> x² – 6x + 2x – 12 = 0
=> (x-6)(x+2) = 0
=> x = -2 , 6
Sum of the roots = 6 + (-2) = 4(b)
II) y² – 5y – 14 = 0
=> y² – 7y + 2y – 14 = 0
=> (y-7)(y+2) = 0
=> y = -2 , 7
Here , (7 × 2) = 14 that means 6 lesser than 20(a)
III) z² – 11z + 30 = 0
=> z² – 6z – 5z + 30 = 0
=> (z-6)(z-5) = 0
z = 5 , 6
Multiplication = 5×6 = 30(c)
Directions (37-41): Study the given data carefully and answer the following questions. There are three bikes A, B and C. They can travel some distance at different speeds.
Bike A: It travels some distance which is 33⅓% more than the distance travelled by car B.
Bike B: It travels at speed of 50 km/hr for 7½ hours.
Bike C: It goes a distance of 400 km at 20% less speed than car B.
Solution (37-41):
We know, Distance = Speed × Time
Distance travelled by Bike B = 50×7½ = 375 km
Distance travelled by Bike A = (375×133⅓)/100 =500 km
20% of 50 km/hr = 50×(20/100) = 10 km
Speed of Bike C = (50 – 10) = 40 km/hr
Time taken to travel 400 by Bike C = 400/40 =10 h
Que 37. Find the difference between the distance travelled by Bike B and Bike C.
A) 21 km
B) 25 km
C) 30 km
D) 32 km
E) 45 km
Answer : B
Explanation :
Required difference = (400 – 375) = 25 km.
Que 38. If Bike A takes 25 hours to travel the distance, then the speed of Bike A is how much less than the speed of Bike B?
A) 60%
B) 45%
C) 50%
D) 35%
E) 65%
Answer : A
Explanation :
Speed of bike A = 500/25 = 20 km/hr
Required percentage = (30/50) × 100 = 60%
Que 39. If Bike C starts to travel first and after 2 hours bike B starts travelling in the same direction, then find after how many hours bike B meets bike C?
A) 5 hours
B) 8 hours
C) 11 hours
D) 15 hours
E) 16 hours
Answer: B
Explanation :
Relative speed = (50 – 40) = 10 km/hr [ two bikes are travelling in same direction]
Distance travelled by C in 2 hours = 2 × 40 = 80 km.
Time taken when bike B to meet bike C = 80/10 =8 hours.
Que 40. Find the time taken by Bike C to travel the distance of 400 km.
A) 2 hours
B) 4 hours
C) 8 hours
D) 10 hours
E) 13 hours
Answer : D
Explanation :
Speed of bike C = 40 km/hr
Time taken by Bike C to cover 400 km = 400/40 = 10 hours.
Que 41. If Bike A takes 8 hours to travel the distance of 500 km, then find the speed of bike A.
A) 62.5 km/hr
B) 68 km/hr
C) 48 km/hr
D) 50.75 km/hr
E) 85 km/hr
Answer : A
Explanation :
Speed of bike A = 500/8 = 62.5 km/hr
Directions (42-46): Study the given data carefully and answer the following question.
There are four persons A, B, C, and D. Each person has different types of a denomination.
| Person | Total amount(in Rs) | Denomination | ||
|---|---|---|---|---|
| A | 1440 | 500 | 200 | 20 |
| B | 1525 | 200 | 100 | 5 |
| C | 3710 | 500 | 10 | 20 |
| D | 1250 | 20 | 5 | 100 |
Note: The number of notes of each denomination is the same for each person.
Solution (42-46):
Amount = Denomination of notes × number of notes
For A :
Let, the number of notes of each denomination be P.
Now, (500 × P + 200 × P + 20 × P) = 1440
=> 720P = 1440
=> P = 2
For B :
Let, the number of notes of each denomination be Q.
(200 × Q + 100 × Q + 5 × Q) = 1525
=> 305 Q = 1525
=> Q = 5
For C :
Let, the number of notes of each denomination be R.
(500 × R + 10 × R + 20 × R) = 3710
=> 530 R = 3710
=> R = 7
For D :
Let, the number of notes of each denomination be S.
(20 × S + 5 × S + 100 × S) = 1250
=> 125 S = 1250
=> S = 10
Que 42. Find the number of notes of each denomination for person C.
A) 5
B) 7
C) 10
D) 12
E) 2
Answer : B
Explanation :
Required number of notes = 7
Que 43. Find the average of the amount of denomination of 200 for person A and the amount of denomination of 100 for person B.
A) 200
B) 250
C) 450
D) 500
E) 560
Answer : C
Explanation :
The amount of denomination of 200 for person A = 200 × 2 = Rs.400
The amount of denomination of 100 for person B = 100 × 5 = 500
Required Average = (400 + 500)/2 = 450
Que 44. By what percentage the total amount of person B is more than that of person D?
A) 18%
B) 21%
C) 22%
D) 25%
E) 26%
Answer : C
Explanation :
Required percentage = (1525 – 1250)/1250 ×100 = 22%
Que 45. Find the difference between the number of notes of each denomination for person B and person C?
A) 10
B) 2
C) 5
D) 6
E) 12
Answer : B
Explanation :
Required difference = (7 – 5) = 2
Que 46. Find the sum of the amount of denomination of 500 for person A and person C.
A) 4500
B) 6000
C) 5000
D) 3500
E) 4200
Answer : A
Explanation :
Amount of denomination of 500 for person A = 500 × 2 = 1000
Amount of denomination of 500 for person C = 500 × 7 = 3500
The required sum = (1000 + 3500) = Rs 4500.
Directions (47-50): Read the following bar graph and table and answer the following questions.
The following bar graph shows the number of people on which a drug is tested in five centres.
The following table shows the number of persons who developed nausea after taking drugs-
| Centre | No. of persons developed Nausea |
|---|---|
| A | 40 |
| B | 80 |
| C | 110 |
| D | 75 |
| E | 30 |
Note :
I) 50% of the total volunteers on which the drug is tested in each centre don’t show any symptoms.
II) Out of the persons who showed symptoms (either Nausea or Placebo) some developed Nausea while the remaining developed Placebo.
Solution (47-50) :
Total number of drug tested people = (480 + 620 + 520 + 360 + 260) = 2240
Total number of drug tested people who have not any symptoms = 2240 × 50% = 1120
For A
Total drug tested people= 480
Total drug tested people without any symptoms =480 × 50% = 240
Total drug tested people with symptoms = 240
Number of Nausea developed person = 40
Number of Placebo developed people =(240 – 40) =200
For B
Number of drug tested people = 620
Number of persons with symptoms = 620 × (50/100) = 310
Number of Nausea developed persons = 80
Number of Placebo developed persons = (310 – 80) =230
For C
Total drug tested person = 520
Number of persons with symptoms =520 × 50% =260
Number of Nausea developed persons = 110
Number of Placebo developed persons = (260 – 110) = 150
For D
Total drug tested person = 360
Total drug tested person with symptoms =360 × 50% = 180
Total number of Nausea developed persons = 75
Total number of Placebo developed persons = (180 – 75) = 105
For E
Total number of drug tested person= 260
Total number of drug tested person with symptoms = 260 × 50% = 130
Total number of Nausea developed persons = 30
Total number of Placebo developed persons = (130-30) = 100
| Centre | Nausea developed persons | Placebo developed persons |
|---|---|---|
| A | 40 | 200 |
| B | 80 | 230 |
| C | 110 | 150 |
| D | 75 | 105 |
| E | 30 | 100 |
Que 47. What is the ratio between the number of persons who developed Nausea in centre C and the number of persons who developed Nausea in centre B and E together?
A) 1 : 1
B) 2 : 1
C) 1 : 2
D) 3 : 1
E) 1 : 3
Answer : A
Explanation :
Required ratio = 110 : (80 + 30) = 1 : 1
Que 48. What is the average number of persons who developed Placebo?
A) 123
B) 135
C) 157
D) 167
E) 175
Answer : C
Explanation :
Required average = (200 + 230 + 150 + 105 + 100)/5 =785/5 =157
Que 49. The number of persons who developed Nausea in centres A, D and E is what percentage less than the number of persons who developed Placebo in centres C and E together?
A) 23%
B) 42%
C) 47%
D) 33%
E) 51%
Answer : B
Explanation :
Nausea developed persons in A,D and E together = (40 + 75 + 30) = 145
Placebo developed persons in C and E together = (150 + 100) = 250
Required less % = (250 – 145)/250 × 100 = 42%.
Que 50. What is the difference between the number of persons who developed Placebo in centres C and D together and the number of persons who developed Nausea in the same centre?
A) 70
B) 60
C) 45
D) 75
E) 110
Answer : A
Explanation :
Placebo developed persons in C and D together =(150+105) = 255
Nausea developed persons in C and D together = (110+75) =185
Required difference =(255 – 185) = 70.
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