SBI provides prestigious jobs in various posts like Clerk, PO, SO etc. Day by day the quality and the pattern of question paper has been changed. To get the proper idea of question paper a candidate has to be focused on previous years question papers. Here we provide SBI clerk mains previous years question paper 2020 of Quantitative Aptitude section.
Direction (1 – 5) : In an examination, six subjects – S1, S2, S3, S4, S5, and S6 have equal maximum marks. The number of marks scored by one particular candidate in subject S1 is 20% less than his marks in subject S6. The ratio of marks scored by the same candidate in subject S2 to that in S3 is 4:5 and that in S4 to S5 is 3:4. The number of marks scored by this candidate in subject S5 is 25% more than that of S6. He scored 65 marks in the subject S3. He scored 436/9% in the examination and the average of marks scored by him in all the subjects is 218/3.
Solution (1-5) :
Let marks scored in subject S6 = P
Marks scored in subject S1 = P × (80/100) = 4P/5
Marks scored in subject S5 = P × (125/100)= 5P/4
Ratio of marks in subject S2 and S3 = 4 : 5
Ratio of marks in subject S4 and S5 = 3 : 4
Here, 4 unit = 5P/4
=> 3 unit = (5P/4) × (3/4) = 15P/16
Marks scored in subject S4 = 15P/16
Marks scored in subject S3 = 65 ( 5 unit)
5 unit = 65
=> 4 unit = 52
Marks scored in subject S2 = 52
Total marks scored in all subjects together = (218/3) × 6 = 436
Now,
S1 + S2 + S3 + S4 + S5 + S6 = 436
=> 4P/5 + 52 + 65 + 15P/16 + 5P/4 + P = 436
=> (64P+75P+100P+80P)/80 = 436 – 117
=> 319P = 319 × 80
=> P = 80
Marks scored in subject S6 = 80
Marks scored in subject S5 = 5P/4 = 100
Marks scored in subject S4 = 15P/16 = 75
Marks scored in subject S3 = 65
Marks scored in subject S2 = 52
Marks scored in subject S1 = 4P/5 = 64
| Subject |
Marks |
| S1 |
64 |
| S2 |
52 |
| S3 |
65 |
| S4 |
75 |
| S5 |
100 |
| S6 |
80 |
The candidate scored marks = 436/9 %
436/9 % = 436
=> 100% = 436 × (9/436) × 100 = 900
∴ Total marks = 900
Que 1. The number of marks obtained by the candidate in the subject S3 was how much percentage less than that of subject S5?
A) 40
B) 35
C) 50
D) 45
E) None of these
Answer : B
Explanation :
∴ Required less percentage = {(100-65)/100} × 100 = 35%
Here the correct option is B.
Que 2. Total marks scored by S2 and S3 is multiple of which of the following pairs.
A) 13 and 17
B) 9 and 17
C) 9 and 13
D) 11 and 13
E) None of these
Answer : C
Explanation :
∴ Total scored in S2 and S3 together =(52+65) = 117
Option C is correct.
Que 3. What percentage of marks the candidate had scored in the subject S5 over the maximum marks of that subject?
A) 40%
B) 50%
C) 66.66%
D) 33.33%
E) None of these
Answer : C
Explanation :
Maximum marks in S5 = 900/6 = 150
∴ Required percentage = (100/150) × 100 = 66.66%
Here the correct option is C.
Que 4. The number of marks obtained by the candidate in the subject S2 was how much less than in the subject S4?
A) 18
B) 21
C) 23
D) 28
E) 33
Answer : C
Explanation :
∴ Required less marks = (75 – 52) = 23
Here the correct option is C.
Que 5. What were the average marks obtained by the candidate in the subject S5 and S6 together?
A) 70
B) 90
C) 75
D) 85
E) 95
Answer : B
Explanation :
∴ Required average marks = (100+80)/2 = 90
Here the correct option is B.
Que 6. A man spends 15%, 20% & 25% of his monthly income on mutual funds, travelling & medicines respectively and remaining of the Income on households. If the difference between expenditures on households & mutual funds is Rs. 16,800 then calculate his monthly income.
A) Rs. 55,300
B) Rs. 58,400
C) Rs. 67,200
D) Rs. 75,200
E) Rs. 62,500
Answer : C
Explanation :
Let Total income = 100 %
Spend on household = 100 – (15+20+25) = 40 %
According to question,
40% – 15% = 16800
=> 25% = 16800
=> 100% = 16800 × (100/25)
=> 100% = 67200
His monthly income = Rs. 67,200
Here the correct option is C.
Que 7. A man has a total of Rs. 5450 in Rs 10, Rs. 20, and Rs. 50 denomination notes. If ratio of number of Rs. 10 notes to Rs. 50 notes is 19:6 and number of notes of Rs. 50 denomination is 80% less than Rs. 20 notes, then find the total value of Rs. 20 notes.
A) Rs. 4500
B) Rs. 3000
C) Rs. 1500
D) Rs. 950
E) Rs. 900
Answer : B
Explanation :
Let the number of notes of Rs.10 and Rs.50 be 19x and 6x respectively.
Number of Rs.20 notes = 6x * (100/20) = 30x
Total value of Rs.10 notes = 19x * 10 =Rs.190x
Total value of Rs.20 notes = 30x * 20 =Rs.600x
Total value of Rs.50 notes = 6x * 50 = Rs.300x
According to question,
190x + 600x + 300x = 5450
=> 1090x = 5450
=> x = 5450/1090
=> x = 5
Now, 600x = 600 * 5 =Rs.3000
Total value of Rs. 20 notes is Rs. 3000
Here the correct option is B.
Que 8. Pipe A takes 10 hours more than Pipe B to fill a tank. Pipe C(which is equally efficient as Pipe A and Pipe B together), fills the same tank in 12 hours. In how many hours will Pipe A, Pipe B and Pipe C fill the tank if Pipe A is accompanied by B and C on alternate hours.
A) 20 hours
B) 30 hours
C) 10 hours
D) 5 hours
E) None of these
Answer : C
Explanation :
Let pipe B can fill the tank in X hours
Pipe A can fill the tank in (X + 10) hours
Pipe (A+B) together fill the tank in hours = 1/X + 1/(X+10)
Pipe C fill the tank in 1 hour = 1/12
As the efficiency of pipe C is equal to Pipe A and Pipe B together, so
1/X + 1/(X+10) = 1/12
=> (X + X + 10)/X(X+10) = 1/12
=> X² + 10X = 24X + 120
=> X² – 14X – 120 = 0
=> X² – 20X + 6X – 120 = 0
=> (X – 20) (X + 6) = 0
=> X = 20, -6
Time can’t be negative so X = 20 is only acceptable.
Time taken by A = (20+10) = 30 hours
Time taken by B = 20 hours
Time taken by C = 12 hours
LCM of 30, 20, 12 = 60 (total capacity)
Pipe A’s efficiency = 60/30 = 2
Pipe B’s efficiency = 60/20 = 3
Pipe C’s efficiency = 60/12 = 5
(A+B)’s together efficiency = (2+3) = 5
(A+C)’s together efficiency = (2+5) = 7
Tank fill in 2 hour = (5 + 7) = 12
Tank fill in 1 hour = 12/2 = 6
Required time taken to fill the tank = 60/6 = 10
Time taken by Pipe A, B and C to fill the tank alternatively in 10 hours.
Here the correct option is C.
Que 9. A and B started a business with some amount ratio of 2:1. After some months they double their share and after 2 months, B leaves the business and C joined the business with an amount Rs.10000. After 4 months of joining, C doubles his money, and B comes back to the business with his initial principal Rs. 2000. At the end of the year, it is seen that the share of A and C are the same. After how many months A and B double their amount?
A) 3 Months
B) 4 Months
C) 2 Months
D) 6 months
E) None of these
Answer : B
Explanation :
Initial investment of B = 2000
Ratio of investment of A and B = 2 : 1
Investment of B = (2×2000) = 4000
Let after P month they double their investment.
After P month investment of A = 8000
Total share of A = (4000×P) + 8000 × (12-P)
After P months C joined with an investment of Rs.10000 and after 4 months C double his investment.
So, Total share of C = (10000 × 4) + 20000 × [12 – (P+2) – 4] = 40000 + 20000 × (6 – P)
According to question,
4000P + 8000×(12 – P) = 40000 + 20000× (6-P)
=> 4000P + 96,000 – 8000P = 40000 + 120000 – 20000P
=> 16000P = 64000
=> P = 4
After 4 months A and B double their amount.
Here the correct option is B.
After 4 months A and B double their amount.
Here the correct option is B.
After 4 months A and B double their amount.
Here the correct option is B.
Que 10. A man sold his article at 35% gain. If he sold articles for Rs. 50 more, he would have gained 45%, If he wants a profit of 56% then what should be the marked price of the article if there is no discount offered?
A) Rs. 780
B) Rs. 886
C) Rs. 892
D) Rs. 986
E) Rs, 850
Answer : A
Explanation :
CP SP
100 135
145
(145-135) unit = 50
=> 10 unit = 50
=> 100 Unit = 500
Cost price (CP) =Rs. 500
If there is no discount on marked price it means Marked price is equal to Selling price.
To get 56% profit ,
Selling price = 500 × (156/100) = 780
The mark price is Rs.780.
Here the correct option is A.
Direction (11-15): Read the following Caselet DI carefully and answer the following questions. 3 cubes P, Q, and R are cut from the cubes A, B, and C respectively.
1) The length of the side of cube P is given by the equation x²-12x+36= 0 and the length of the side of cube A is 2 cm more than cube P.
2) The length of the side of cube Q is given by the equation y²+21y-46=0 and the length of the side of cube B is 3 cm more than cube Q.
3) The length of the side of cube R is given by the equation z²-10z +25= 0 and the length of the side of cube C is 4 cm more than cube R. Here x, y, and z are in centimeter.
Que 11. How many cubes of type A can be made by a cuboid of dimensions 64 cm x 65 cm x 56 cm?
A) 350
B) 375
C) 400
D) 455
E) 505
Answer : D
Explanation :
x² – 12x + 36 = 0
=> (x – 6)² = 0
=> (x – 6) = 0
=> x = 6
Length of Side of P = 6 cm
Length of Side of A = (6+2) = 8 cm
Required number = (64 × 65 × 56)/8³ = 455
The number of cubes of type A can be made from the cuboid = 455
Here the correct option is D
Que 12. The average length of side of cube P, B and C is how much percent more or less than the average length of side of cube A, Q and R ?
A) 16.5% less
B) 20% more
C) 25% less
D) 33.33% more
E) 42.45% less
Answer : D
Explanation :
y² + 21y – 46 = 0
=> y² + 23y – 2y – 46 = 0
=> (y + 23)(y – 2) = 0
=> y = 2, -23
Length can not be negative so y = -23 is not acceptable.
y = 2
Length of Q = 2
Length of B = (2+3) = 5 cm
z² – 10z + 25 = 0
=> (z – 5)² = 0
=> z – 5 = 0
=> z = 5
Length of R = 5 cm
Length of C = (5+4) = 9 cm
Average length of P, B and C = (6+5+9)/3 = 20/3
Average length of A, Q and R = (8+2+5)/3 = 5
Required percentage = {(20/3 – 5)/5}×100 = 33.33%
∴ The average length of side of cube P, B and C is 33.33% more than the average length of side of cube A, Q and R .
Here the correct option is D.
Que 13. What is the ratio of volume of cube A and P together to the volume of cube B and Q together?
A) 10 : 31
B) 100 : 11
C) 104 : 19
D) 19 : 104
E) 11 : 100
Answer : C
Explanation :
Volume of cube A = 6³ = 216 cm³
Volume of cube P = 8³ = 512 cm³
Volume of cube A and P together = (216 + 512) = 728 cm³
Volume of cube Q = 2³ = 8 cm³
Volume of cube B = 5³ = 125 cm³
Volume of cube B and Q together = (125 + 8) = 133
Required ratio = 728 : 133 = 104 : 19
∴ The ratio of volume of cube A and P together to the volume of cube B and Q together is 104 : 19
Here the correct option is C.
Que 14. What is the ratio of total surface area of cube P to the total surface area of cube C ?
A) 4 : 9
B) 9 : 4
C) 2 : 3
D) 3 : 4
E) 5 : 9
Answer : A
Explanation :
Total surface area of cube P = 6 × 6² cm²
Total surface area of cube C = 6 × 9² cm²
Required ratio = (6×6²) : (6×9²) = 4 : 9
∴ The ratio of total surface area of cube P to the total surface area of cube C is 4 : 9
Here the correct option is A.
Que 15. From the given cubes,which two cubes have equal side length?
A) A and P
B) A and R
C) P and C
D) R and B
E) R and P
Answer : D
Explanation :
from the given cubes,
Length of cube A = 8 cm
Length of cube B = 5 cm
Length of cube C = 9 cm
Length of cube P = 6 cm
Length of cube Q = 2 cm
Length of cube R = 5 cm
here length of cube B and cube R is same.
Here the correct option is D.
Directions (16 – 20) : Read the table carefully and answer the question given below. The Table represents the expenditure, saving and monthly income of three-person Arun, Bipin and Chandan.
| Items |
Arun |
Bipin |
Chandan |
|
Expenditure
on food and
entertainment
|
20% of monthly income |
12% of monthly income |
|
| Savings |
Rs.a |
|
Rs.9000 |
| Monthly income |
Rs.50,000 |
Rs.72,000 |
|
Solution (16 – 20) :
Monthly income of Arun = 50,000
Expenditure on food and entertainment by Arun = 50000 × (20/100) = 10000
Savings (a) = (50000 – 10000) = 40,000
Monthly income of Bipin = 72,000
Expenditure on food and entertainment by Bipin = 72000 × (12/100) = 8640
Savings of Bipin = (72,000 – 8640) = 63,360
Savings of Chandan = Rs.9000
Que 16. If the expenditure of Chandan on food and entertainment is 5% more than the expenditure on food by Arun, then find the amount of expenditure on food and entertainment by Chandan.
A) Rs.10,500
B) Rs.11,000
C) Rs.11,500
D) Rs.12,500
E) Rs.15,500
Answer : A
Explanation :
Expenditure on food and entertainment by Chandan = 10000 × (105/100) = 10,500
Here the correct option is A.
Que 17. If saving of Arun is (2/5) of the income left after spend on food and entertainment, then find the value of a.
A) Rs.14,000
B) Rs.16,000
C) Rs.17,500
D) Rs.22,000
E) Rs.34,000
Answer : B
Explanation :
Saving of Arun (a) = 40000 × (2/5) = Rs.16,000
∴ The value of a = 16,000
Here the correct option is B.
Que 18. If the saving of Chandan is 30% of his monthly income then find the monthly income of Chandan.
A) Rs.12,000
B) Rs.17,000
C) Rs.22,500
D) Rs.30,000
E) Rs.36,000
Answer : D
Explanation :
Let monthly income of Chandan = A
According to question,
A × (30/100) = 9000
=> A = 9000 × (100/30)
=> A = 30000
The monthly income of Chandan = Rs. 30,000
Here the correct option is D.
Que 19. After spending the monthly income on food and entertainment, remaining income of Bipin is his saving, then find saving of Bipin is what percentage of saving of Chandan.
A) 108%
B) 502%
C) 704%
D) 600%
E) 10%
Answer: C
Explanation :
Saving of Bipin = 72000 – 72000 × (12/100) = Rs.63,360
Saving of Chandan = 9000
Required percentage = (63360/9000) × 100 =704%
Here the correct option is C.
Que 20. If the savings of Chandan is 30% of his monthly income then find the sum of the monthly income of Arun, Bipin and Chandan.
A) Rs.1,52,000
B) Rs.2,10,000
C) Rs.2,55,000
D) Rs.3,20,000
E) None of these
Answer : A
Explanation :
Monthly income of Chandan = 30,000
Monthly income of Arun, Bipin and Chandan together = (50000 + 72000 + 30000) = 152000
The sum of the monthly income of Arun Bipin and Chandan is = Rs. 152000
Here the correct option is A.
Que 21. A can do a piece of work in x days, B can take 10 more days than A to complete the same work, and C can do the same work alone in 10 days. If A, B and C together can do 50% of the total work In 2(13/31) days, then find the value of x.
A) 12 days
B) 10 days
C) 15 days
D) 18 days
E) 9 days
Answer : C
Explanation :
A, B and C together can do 100%(total) work in 2×(75/31) = 150/31 days
(A, B and C)’s 1-day work = 31/150
Let A can complete the work in x days.
B can complete the work in (x + 10) days
C complete the work in 10 days
A’s 1 day work = 1/x
B’s 1 day work = 1/(x+10)
C’s 1-day work = 1/10
(A+B+C)’s 1 day work = 1/x + 1/(x+10) + 1/10
Now,
1/x + 1/(x+10) + 1/10 = 31/150
=> (x + 10 + x)/x(x+10) = (31-15)/150
=> (2x + 10)/(x² + 10x) = 8/75
=> 8x² + 80x = 150x + 750
=> 8x² – 70x – 750 = 0
=> 4x² – 35x – 375 = 0
=> 4x² – 60x + 25x – 375 = 0
=> (x – 15) (4x + 25) = 0
=> x = 15, – 25/4
Time can’t be negative so (-25/4) value is not acceptable.
x = 15
∴ The required value of x is 15 days.
Here the correct option is C.
Que 22. A mark list of 30 students was prepared by the teacher and average of their marks was 68. But, mark of one student was written as 35 Instead of 53. Find the correct average marks.
A) 67.4
B) 65.6
C) 64.2
D) 68.6
E) 70.2
Answer : D
Explanation :
Average marks of 30 students = 68
Total sum of 30 students marks = 68×30 = 2040
Correct sum = 2040 – 35 + 53 = 2058
Correct average = 2058/30 = 68.6
∴ The correct average marks is 68.6 .
Here the correct option is D.
Que 23. A pair of unbiased dice is rolled simultaneously, find the probability of getting a difference of three.
A) 1/6
B) 1/4
C) 1/3
D) 1/2
E) 1/9
Answer : A
Explanation :
Total possible outcomes = 6 × 6 = 36
Numbers events of getting a difference of three = (1, 4), (2, 5), (3, 6), (4, 1) , (5, 2), (6, 3) = 6
Probability of getting a difference of three = 6/36 = 1/6
∴ The required probability of getting a difference of three is 1/6.
Here the correct option is A.
Que 24. In a family of 4 persons, A and B are husband and wife. C and D are daughter and son of A and B respectively. D’s age 8 years ago was 2 years. The ratio of the present ages of mother and daughter is 7:1. The sum of the present ages of A, B, and C is 80 years. At the time of the birth of C, the age of A was equal to the present age of B. Find the present age of A.
A) 30 years
B) 40 years
C) 60 years
D) 45 years
E) 50 years
Answer : B
Explanation :
Here A = Father, B = Mother , D = Son and C = Daughter
Present age of D = (2 + 8) = 10 years
B : C = 7 : 1
A + B + C = 80 ———-(1)
At the time of birth of C, A’s age = B’s present age ——–(2)
Let the present age of mother (B) = 7x
present age of daughter ( C) = x
Now putting the values of B and C in equation (2), we get
A – x = 7x
=> A = 8x
Now putting the values of A,B and C in equation (1), we get
A + B + C = 80
=> 8x + 7x + x = 80
=> 16x = 80
=> x = 5
∴ Present age of A = 8 × 5 = 40 years.
Here the correct option is B.
Que 25. The average weight of a class of 50 students is 80 kg. If two of the students with weight 40% & 50% of ‘x’ leave the class & two new students with weight ‘x’ & ‘x-10’ Join the class, then the average weight of the class Is Increased by 2 kg. Find the value of ‘x’?
A) 80
B) 110
C) 90
D) 120
E) 100
Answer : E
Explanation :
Total sum of weight = (80×50) = 4000 kgs
New average = (80 + 2) = 82
40% of x = (40/100) * x = .4x
50% of x = (50/100) * x = .5x
According to question,
82 = (4000 – .4x – .5x + x + x – 10)/50
=> 4100 = 3990 – 1.1x
=> 1.1x = 110
=> x = 100
∴ The required value of x = 100
Here the correct option is E.
Direction(26 – 30) : The bar graph shown below indicate the number of books owned and the number of books issued by the 5 different libraries A, B, C, D and E. Based on the given bar graph answer the question given below.
Total books owned = Book available + Book issued
Solution ( 26 – 30) :
| Library |
Books owned |
Books issued |
Books available |
| A |
440 |
360 |
440-360
= 80
|
| B |
520 |
440 |
520-440
= 80
|
| C |
480 |
420 |
480-420
= 60
|
| D |
420 |
320 |
420-320
= 100
|
| E |
540 |
380 |
540-380
= 160
|
Que 26. Find the average number of books available in all 5 library.
A) 80
B) 110
C) 96
D) 120
E) 100
Answer : C
Explanation :
Required average = (80+80+60+100+160)/5 = 96
Here the correct option is C.
Que 27. Find the ratio between the books available in library A, C and D together to books available in library B and E together.
A) 1 : 2
B) 1 : 3
C) 2 : 3
D) 1 : 1
E) 2 : 5
Answer : D
Explanation :
Books available in library A, C and D together =(80+60+100) = 240
Books available in library B and E together =(80+160) = 240
Required ratio = 240 : 240 = 1 : 1
Here the correct option is D.
Que 28. Books available in library A and D together is how much percent more or less than the books available in library B and E together?
A) 20% more
B) 25% less
C) 35% more
D) 12.5% more
E) 10% less
Answer : B
Explanation :
Books available in library A and D together = (80+100) = 180
Books available in library B and E together = (80+160) = 240
Required less percentage = (240-180)/240 ×100 = 25%
Here the correct option is B.
Que 29. Books available in library D is how much percent of books owned by library B and C together?
A) 20%
B) 25%
C) 35%
D) 12.5%
E) 10%
Answer : E
Explanation :
Books owned in library B and C together = (520+480) = 1000
Required percentage = (100/1000) × 100 = 10%
Here the correct option is E.
Que 30. What is the ratio of total books owned by all the libraries together to total Books issue by all the libraries together?
A) 5 : 2
B) 1 : 3
C) 5 : 4
D) 1 : 1
E) 2 : 5
Answer : C
Explanation :
Total books owned = (440+520+480+420+540) =2400
Total books issued = (360+440+420+320+380) =1920
Required ratio = 2400 : 1920 = 5 : 4
Here the correct option is C.
Directions(31 – 35) : Given below are two quantities named I and II. Based on the given Information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.
A) Quantity I ≤ Quantity II
B) Quantity I > Quantity II
C) Quantity I ≥ Quantity II
D) Quantity I < Quantity II
E) Quantity I = Quantity II
Que 31.
Quantity I: The average of four number is 14. If one number is removed from its set then the average is reduced by 1, Find the removed number.
Quantity II: 15
Answer : B
Explanation :
From Quantity I –
Total of four number = 14×4 = 56
Average of (4-1 = 3) number = (14-1) = 13
Total of three number = 13×3 = 39
Removed number = (56 – 39) = 17
Quantity I > Quantity II
Here the correct option is B.
Que 32.
Quantity I : x² – 39x + 378 = 0
Quantity II : y² – 24y + 108 = 0
Answer : C
Explanation :
Quantity I :
x² – 39x + 378 = 0
=> x² – 21x – 18x + 378 = 0
=> (x – 21) (x – 18) = 0
=> x = 21, 18
Quantity II :
y² – 24y + 108 = 0
=> y² – 18y – 6y + 108 = 0
=> (y – 18) (y – 6) = 0
y = 6, 18
Quantity I ≥ Quantity II
Here the correct option is C.
Que 33.
Quantity I : Two dice are thrown simultaneously. What is the probability of getting even number on both the dice?
Quantity II : 1/3
Answer : D
Explanation :
Total outcomes = 6×6 = 36
Numbers of favourable outcomes = 9 [ (2,2) ; (2, 4) ; (2, 6) ; (4, 2) ; (4, 4) ; (4,6) ; (6,2) ; (6,4), (6,6)]
The probability of getting even number on both the dice = 9/36 = 1/4
Quantity I < Quantity II
Here the correct option is D.
Que 34.
Quantity I : 2x² – 22x + 48 = 0
Quantity II : y² – 20y + 96 = 0
Answer : A
Explanation :
Quantity I :
2x² – 22x + 48 = 0
=> 2x² – 16x – 6x + 48 = 0
=> (2x – 6) (x – 8) = 0
=> x = 8, 3
Quantity II :
y² – 20y + 96 = 0
=> y² – 12y – 8y + 96 = 0
=> (y – 12) (y – 8) = 0
=> y = 8, 12
Quantity I ≤ Quantity II
Here the correct option is A.
Que 35.
Quantity I : A downstream speed is 10 km/hr and upstream speed is 5 km/hr. Distance covered by a boat in still water is 15 km. Find the time taken by boat to cover the distance.
Quantity II : 2 hours
Answer : E
Explanation :
Speed of boat = (10+5)/2 = 7.5 km/hr
Distance = Speed × Time
Required time taken by boat to cover the distance = 15/7.5 = 2 hours
Quantity I = Quantity II
Here the correct option is E.
Directions (36 – 40) : Read the given graphs carefully and answer the questions that follows.
During the summer season, people from Delhi are travelling to different holiday locations by booking tickets from Holiday.com. The total revenue generated by the website is Rs.1,20,000.
| City |
Ticket Cancelled |
| Chandigarh |
3 |
| Jaipur |
12 |
| Manali |
10 |
| Shimla |
13 |
| Rishikesh |
15 |
Note : Cancelled tickets have no reimbursement
Solution (36 – 40) :
Total revenue generated by website = 1,20,000
Revenue generated in Chandigarh = 120000 × (21/100) = 25,200
Revenue generated in Jaipur = 120000 × (22/100) = 26,400
Revenue generated in Manali = 120000 × (15/100) = 18,000
Revenue generated in Shimla = 120000 × (24/100) = 28,800
Revenue generated in Rishikesh = 120000 × (18/100) = 21,600
Que 36. If the cost of each ticket on travelling to Manali is Rs.1200, then what will be the number of people who travelled?
A) 5
B) 1
C) 4
D) 9
E) 20
Answer : A
Explanation :
Let the number of people travelled = p
Total ticket cost on travelling = 1200p
Ticket cost for the people who cancelled ticket = 1200 × 10 = 12,000
Now,
1200p + 12,000 = 18000
=> 1200p = 6000
=> p = 5
∴ The number of people who travelled = 5
Here the correct option is A.
Que 37. The total number of people that travelled to Jaipur was 15 and to Shimla was 13 then calculate the average price per ticket.
A) 1005.50
B) 1010.90
C) 1230.40
D) 1041.50
E) 2000.10
Answer : D
Explanation :
Let average price per ticket = x
Total cost for Jaipur and Shimla together= (26400+28800)= 55200
Total cost on travelling tickets = (15+13)x = 26x
Total cost for cancelled ticket = (12+13)x = 25x
According to question,
26x + 25x = 55200
=> 51x = 55200
=> x = 1041.50
∴ The average price per ticket is Rs.1041.50
Here the correct option is D.
Que 38. What the average number of tickets cancelled per city?
A) 5.5
B) 10.6
C) 4.8
D) 9.7
E) 20
Answer : B
Explanation :
Total number of cancelled tickets = (3 + 12 + 10 + 13 + 15) = 53
Average cancelled ticket per city = 53/5 = 10.6
∴ The average number of tickets cancelled per city is 10.6 .
Here the correct option is B.
Que 39. If the cost of each ticket on travelling to Rishikesh is Rs.1440 and no ticket was cancelled, then what will be the number of people who travelled?
A) 15
B) 11
C) 14
D) 19
E) 20
Answer : A
Explanation :
Number of people who travelled = 21600/1440 = 15
Here the correct option is A.
Que 40. What was the average number of tickets booked per city if tickets were sold at Rs.1500 each?
A) 15
B) 16
C) 14
D) 19
E) 20
Answer : B
Explanation :
Total ticket booked = 120000/1500 = 80
Average ticket booked per city = 80/5 = 16
Here the correct option is B.
Directions (41 – 50) :
Each question below is followed by two statements I and II. You have to determine whether the data given in the statements are sufficient for answering the question. You should use the data and your knowledge of Mathematics to choose the best possible answer.
A) If the data in statement I alone is sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question.
B) If the data in statement II alone is sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.
C) If the data either in statement I alone or in statement II alone is sufficient to answer the question.
D) If the data even in both the statements I and II together are not sufficient to answer the question.
E) If the data in both statements I and II together are needed to answer the question.
Que 41. Two Cars were traveling towards each other. What is the distance between them just 15 minutes before they meet?
I. Speed of Car A and Car B is 15 km/hr and 21 km/hr respectively.
II. The Distance between them originally is 72 km.
Answer : A
Explanation :
Statement I :
Hence two cars were traveling towards each other, so
Relative speed = (15+21) = 36 km/hr
In 15 mins they travelled = 36 × (15/60) = 9 km
9 km is the distance between them just 15 minutes before they meet.
Statement I alone is sufficient.
Statement II :
Only distance given. So we can’t calculate the answer.
∴ If the data in statement I alone is sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question.
Here the correct option is A.
Que 42. Find the total amount invested by Mr. X.
I. Total amount received by Mr. Y after 2 years is Rs. 4000 at compound interest.
II. X and Y invested their amount at simple interest and compound interest respectively and the difference between the interests received by both after 3 years is Rs. 1500.
Answer : D
Explanation :
In statement I and statement II , rate of interest is not given.
So we can not find the answer.
∴ If the data even in both the statements I and II together are not sufficient to answer the question.
Here the correct option is D.
Que 43. What is the sum of the ages of A and B?
I. Ratio of Ages of A and B is 2:1 and that of B and C is 3:2.
II. Difference between ages of A and C is 20 years.
Answer : E
Explanation :
Statement I :
A : B = 2 : 1 ——- ×3 = 6 : 3
B : C = 3 : 2 ——- ×1 = 3 : 2
A : B : C = 6 : 3 : 2
Statement II :
(6-2) unit = 20 years
=> 1 unit = 20/4 = 5 years
Age of A = (5×6) = 30 years
Age of B = (5×3) = 15 years
Sum of ages of A and B = (30+15) = 45 years
∴ the data in both statements I and II together are needed to answer the question.
Here the correct option is E.
Que 44. Find the initial quantity of milk in jar A.
I. Ratio of the Milk and water in Jar A and B is 3: 2 and 4:3 respectively.
II. 28 litres of the mixture of B is poured into A and then the ratio of the Milk and water in Jar A becomes 17:12.
Answer : E
Explanation :
Statement I and II:
Milk : Water = 3 : 2 (Jar A)
Milk : water = 4 : 3 ( Jar B)
Quantity of milk in jar B = 28 × (4/7) = 16 litres
Quantity of water in jar B = (28-16) = 12 litres
Let initial quantity of milk in jar A = x
According to question,
(3x + 16)/(2x + 12) = 17/12
=> 36x + 192 = 34x + 204
=> 2x = 12
=> x = 6
∴ the data in both statements I and II together are needed to answer the question.
Here the correct option is E.
Que 45. Find the difference of number of Females in 2020 and the number of Males in 2019?
I. The ratio of number of Males to Females in 2019 is 3:2 and the difference between number of males and females in 2019 is 500.
II. The ratio of number of Males and Females in 2020 is 6:5 and there is 20% growth in the number of Males in 2020 as compared to 2019.
Answer : E
Explanation :
Statement I :
Let number of males and females in 2019 be 3x and 2x
3x – 2x = 500
=> x = 500
Number of males in 2019 = 500×3 = 1500
Statement II :
Let number of males and females in 2020 6x and 5x.
Number of male in 2020 = 1500 × (120/100) = 1800
6x = 1800
=> x = 300
Number of females = (300×5) = 1500
∴ the data in both statements I and II together are needed to answer the question.
Here the correct option is E.
Que 46. In how many days A will complete the work alone?
I. B and C alone can complete the work in 12 and 15 days respectively.
II. Together A, B and C can complete the work in 4 days.
Answer : E
Explanation :
From statement I and II –
LCM of 12, 15 and 4 = 60 (total work)
B’s 1 day work = 60/12 = 5
C’s 1 day work = 60/15 = 4
(A+B+C)’s 1 day work = 60/4 = 15
A’s 1 day work = 15 – (5+4) = 6
A alone complete the work in (60/6)= 10 days.
∴ Both statements I and II together are needed to answer the question.
Here the correct option is E.
Que 47. Find the Marked Price of Article A.
I. The Selling Price of Article A is 25% of the Selling Price of Article B.
II. The cost price of the Article A is Rs. 300. When a discount of 25% is given on the marked price of the Article, the loss incurred is 10%.
Answer : B
Explanation :
Statement I –
Let selling price of B = 100
Selling price of A = 25
Statement II –
Cost price of article A = 300
Loss = 10%
Selling price = 300 × (90/100) = 270
Let market price = P
Selling price = P × (75/100) = 3P/4
Now, 3P/4 = 270
=> P = 270 × (4/3)
=> P = 360
∴ Market price of article A = Rs.360
Statement II alone is sufficient to answer the question, while the data in statement I alone are not sufficient to answer the question.
Here the correct option is B.
Que 48. Find the average of five consecutive even numbers.
I. The sum of the first two numbers is 8 more than the fifth number.
II. The difference of Fourth number and the first number is 12.
Answer : A
Explanation :
Statement I –
Let five consecutive even numbers be X, (X+2), (X+4), (X+6), (X+8)
According to question,
X + (X+2) – (X+8) = 8
=> X = 8 + 6
=> X = 14
Five consecutive even numbers are = 14 , 16 , 18, 20, 22
Average of this numbers = (14+16+18+20+22)/5 = 90/5 = 18
Statement II –
4th number – 1st number = 12
∴ Statement I alone is sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question.
Here the correct option is A.
Que 49. The Average age of A, B, C, D, and E is 36 years. Find the age of C.
I. Age of A is 36 years and E is 4 years elder than B and 12 years smaller than D.
II. Age of B is 12 years less than C and E is 8 years elder than C.
Answer : E
Explanation :
Total age of Five people = (36×5) = 180
Statement I –
A = 36 years
E = B + 4
E = D – 12
Statement II –
B = C – 12
E = C + 8
From statement I and II ,
E = D – 12
=> C + 8 = D – 12
=> D = C + 20
Now, (A + B + C + D + E) = 180
=> 36 + (C – 12) + C + (C + 20) + (C + 8) = 180
=> 52 + 4C = 180
=> 4C = 180 – 52
=> C = 128/4
=> C = 32
Age of C = 32 years.
Both statements I and II together are needed to answer the question.
Here the correct option is E.
Que 50. How long did it take A to reach his office?
I. speed of A is 80 km/hr.
II. If he increases his speed by two times, he would takes 5 minutes earlier.
Answer : E
Explanation :
Let time taken to reach his office = T hours
Statement I :
Distance covered = 80T
Statement II :
Distance covered = 2 × 80 × (T – 5/60)
Now,
160 × (T – 5/60) = 80T
=> 2T – 1/6 = T
=> T = 1/6 × 60 minutes
=> T = 10 minutes
∴ Both statements I and II together are needed to answer the question.
Here the correct option is E.
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