# SBI Clerk Mains Quantitative Aptitude Question Paper 2021

SBI provides jobs in various posts like Clerk, PO, SO etc. Day by day the quality and the pattern of question paper has been changed and modified. 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 2021 of Quantitative Aptitude section.

**Directions( Q 1-5): Read the following information carefully and answer the questions that follow: **

**Five members of a family viz. Mr. Rahul, his wife Sushmita, their two children Varnali and Vishal and their daughter-in-law monika spent different amount from their respective salaries in a month. The total expenditure of the family per month is Rs. 3,30,000 which is 75% of the total income of the family per month. Sushmita’s saving is 1/3rd of the savings of her daughter-in-law who earns Rs. 60,000 a month and spent 60% of her salary. The salary earned by Varnali is the highest in the family i.e. Rs. 1,20,000 per month and her expenditure is twice the expenditure of Monika. Sushmita saves 10% of her income and her husband saves 20% of his income. Mr. Rahul earns Rs. 90,000 per month. The total saving of the family is Rs. 1,10,000. **

**Solution (1-5) : **

Total Expenditure = 3,30,000

Total Savings = 1,10,000

75% = 3,30,000

100% = 3,30,000 Ã— (100/75) = 4,40,000

Total Income = 4,40,000

Monika’s Income = 60,000

Monika’s Expenditure = 60,000 Ã— (60/100) = 36,000

Monika’s Saving = (60,000 – 36,000) = 24,000

Sushmita’s Saving =(1/3) Ã— 24,000 = 8,000

Varnali’s Income = 1,20,000

Varnali’s Expenditure = 2 Ã— 36,000 = 72,000

Varnali’s Saving = (1,20,000 – 72,000) = 48,000

10% = 8,000

100% = 80,000

Sushmita’s Income = 80,000

Sushmita’s Expenditure = (80,000 – 8,000) = 72,000

Rahul’s Income = 90,000

Rahul’s Saving = (20/100) Ã— 90,000 = 18,000

Rahul’s Expenditure = (90,000 – 18,000) = 72,000

Vishal’s Income = 4,40,00 = 90,000

Vishal’s Saving = 1,10,00 – (24,000 + 8,000 + 48,000 + 18,000) = 12,000

Vishal’s Expenditure = (90,000 – 12,000) = 78,000

Members | Income | Expenditure | Saving |
---|---|---|---|

Mr.Rahul | 90,000 | 72,000 | 18,000 |

Sushmita | 80,000 | 72,000 | 8,000 |

Varnali | 1,20,000 | 72,000 | 48,000 |

Vishal | 90,000 | 78,000 | 12,000 |

Monika | 60,000 | 36,000 | 24,000 |

**Que 1. Income of which of the following two members of the family is same? **

**A) Mr.Rahul and Sushmita **

**B) Mr.Rahul and Vishal **

**C) Varnali and Monika **

**D) Sushmita and Vishal **

**E) None of these **

**Answer : B **

**Explanation : **

Mr.Rahul and Vishal have same income of Rs.90,000.

**Here the correct option is B.**

**Que 2. Salary earned by Monika is what percentage of salary earned by Sushmita? **

**A) 75% **

**B) 35% **

**C) 45% **

**D) 65% **

**E) None of these **

**Answer : A **

**Explanation : **

Required percentage = (60,000/80,000) Ã— 100 = 75%

**Here the correct option is A.**

**Que 3. What is the difference between the amount spent by Sushmita and Varnali in a month? **

**A) Rs.4000 **

**B) Rs.150 **

**C) Rs.2000 **

**D) Rs.300 **

**E) None of these **

**Answer : E **

**Explanation : **

Required difference = (72,000 – 72,000) = 0

**Option E is the correct answer. **

**Que 4. What is the average salary of the family per month?**

**A) Rs.45,000 **

**B) Rs.55,000 **

**C) Rs.68,000 **

**D) Rs.88,000 **

**E) None of these **

**Answer : D **

**Explanation : **

Required average salary = (4,40,000/5) = Rs 88,000

**Here the correct option is D.**

**Que 5. What is the ratio of the salary earned to salary spent by Monika in a month? **

**A) 5 : 3 **

**B) 3 : 5 **

**C) 3 : 8 **

**D) 7 : 8 **

**E) None of these **

**Answer : A **

**Explanation : **

Required ratio = 60000 : 36000 = 5 : 3

**Here the correct option is A.**

**Que 6. The distance between the two trains is 150 km and moving in the opposite direction on the same track. The speed of one train is 65 km/hr and the speed of another train is 55 km/hr. A bird starts flying at a speed of 60 km/hr at the location of the faster train. When it reaches the slower train, it turns around and flies in the opposite direction at the same speed. When it reaches the faster train again it turns around and so on. When the train collide, how far has the bird flown (in km)?**

**A) 50 **

**B) 65 **

**C) 75 **

**D) 64 **

**E) None of these **

**Answer : C **

**Explanation : **

Distance between two trains = 150 km

Since both the trains are running in opposite directions, so

Relative speed = (65 + 55) = 120 km/hr

They collide each other in = 150/120 hours = 1.25 hours

The bird covered distance in 1.25 hours = 60 Ã— 1.25 = 75 km

The Bird flew 75 km.

Here the correct option is C.

**Que 7. Chemical mixtures X consist of chemicals A and B in ratio of 1:3, Chemical Y consist of chemicals A and B in ratio of 3:2. Chemical X Is present in vessel whose capacity is 2/5 times that of vessel in which chemical Y is present. Both the chemicals X and Y are mixed in a large container. What is % of concentration is present in final mixture as A? **

**A) 30% **

**B) 40% **

**C) 50% **

**D) 60% **

**E) None of these **

**Answer : C **

**Explanation : **

Chemical X = (2/5) Ã— Chemical Y

(Chemical X)/(Chemical Y) = 2/5

Let Chemical X = 2P and Chemical Y = 5P

Total quantity in final mixture =(2P + 5P) = 7P

Parts of A in Chemical X = 2P Ã— (1/4) = P/2

Parts of A in Chemical Y = 5P Ã— (3/5) = 3P

Parts of A in final mixture = (P/2 + 3P) = 7P/2

Required percentage = [(7P/2)/7P] Ã— 100 % = 50%

The 50% concentration is present in the final mixture as A.

Here the correct option is C.

**Que 8. Income of Karan is Rs.x and income of Arjun is Rs.y. Expenditure on food by both is 10% of their income. The expenditure on entertainment by Karan is twice the expenditure on food. The saving of Karan is Rs.35,000. Expenditure on entertainment by Arjun is Rs.12,000. The saving of Arjun is 75% of his income. Find their income? **

**A) Rs.72,000 & Rs.50,000 **

**B) Rs.80,000 & Rs.50,000 **

**C) Rs.80,000 & Rs.65,000 **

**D) Rs.70,000 & Rs.45,000 **

**E) None of these **

**Answer : B **

**Explanation : **

__For Karan : __

Income = Rs. x

Expenditure on food = x Ã— (10/100) = x/10

Expenditure on entertainment = 2x/10 = x/5

Saving = x – (x/10 + x/5)

According to the question,

x – (x/10 + x/5) = 35000

=> 10x – x – 2x = 350000

=> 7x = 350000

x = 50,000

âˆ´ Income of Karan = 50,000

__For Arjun : __

Income = y

Saving = y Ã— (75/100) = 3y/4

Expenditure on food = y Ã— (10/100) = y/10

Expenditure on entertainment = 12,000

Saving = y – (y/10 + 12000)

Now,

3y/4 = y – (y/10 + 12000)

=> y – y/10 – 3y/4 = 12000

=> 20y – 2y – 15y = 12000 Ã— 20

=> 3y = 240000

y = 80,000

Income of Arjun =80,000

Income of Karan =50,000

Here the correct option is B.

**Que 9. A, B and C invest together in business wherein they receive profit of Rs 63,000. A reinvests his share of profit in a scheme giving 10% rate of interest compounded annually after two years and B reinvests his share of profit which gives him 20% rate of interest compounded annually at end of two years. If interest received by A and B after 2 years is Rs 3969 and Rs 8316 then find compound interest earned by C if he reinvests his share in same scheme as A did for 2 years.**

**A) Rs.5292 **

**B) Rs.5590 **

**C) Rs.4650 **

**D) Rs.6570 **

**E) None of these**

**Answer : A **

**Explanation : **

Let A’s investment in a scheme = Rs. 100

Rate = 10%

Now,

Amount = 100 Ã— ( 1 + .10)Â²

=> Amount = 121

C.I = (121-100) = 21

21 unit = 3969

100 unit = (3969/21) Ã— 100 = 18,900

Investment of A in a scheme = Rs.18,900

Let B’s investment in a scheme = 100

Rate = 20%

Amount = 100 Ã— (1 + 0.2)Â²

=> Amount = 144

C.I = (144 – 100) = 44

44 unit = 8316

100 unit = (8316/44) Ã— 100 = Rs.18,900

Investment of B = Rs.18,900

Profit share of C = 63000 – (2 Ã— 18900) = Rs.25200

Amount of C = 25200 Ã— 1.10 Ã— 1.10 = 30492

Interest = (30492 – 25200) = 5292

**âˆ´**** Interest earned by C = Rs.5292 **

**Here the correct option is A.**

**Que 10. Kiran sells two types of milk. In the first type of milk, there is 40% milk is present and the rest is water, whereas, in the second types only 40% water is present. He takes 30 litres and 20 litres of mixture from both types respectively. What is the percentage of milk present in the new mixture?**

**A) 36% **

**B) 25% **

**C) 48% **

**D) 72% **

**E) None of these **

**Answer : C **

**Explanation : **

__For first type__ :

Milk = 40% and Water = 60%

Milk present in 30 litres of mixture = 30 Ã— (40/100) = 12 litres

__For the 2nd type__ :

Water = 40% and Milk = 60%

Milk present in 20 litres of mixture = 20 Ã— ( 60/100) = 12 litres

After adding first type and second type mixture ,the total mixture = (30 + 20) = 50 litres

Pure milk in the final mixture = (12 + 12) = 24 L

Required percentage = (24/50) Ã— 100 = 48%

The percentage of milk present in the new mixture is =48%

Here the correct option is C.

**Direction (11-13) : What approximate value should come in the place of question mark (?) in the following question?**

**Que 11. 12.97Â³ x 3.99 Ã· 52.08+ 107.99 Ã· Â³âˆš1727.98 Ã—6.93=? **

**A) 232 **

**B) 250 **

**C) 278 **

**D) 195 **

**E) 205**

**Answer : A**

**Explanation : **

12.97Â³ x 3.99 Ã· 52.08 + 107.99 Ã· Â³âˆš1727.98 Ã— 6.93 = ?

=> 13Â³ Ã— 4 Ã— (1/52) + 108 Ã— (1/12) Ã— 7 = ?

=> 2197 Ã— (1/13) + 63 = ?

=> 169 + 63 = ?

âˆ´ ? = 232

The required approx value =232

Here the correct option is A.

**Que 12. (24.99% of 250 Ã— 3.95) + (21.05% of 300 – 49.99) = (?) – 19.99% of 40 **

**A) 230 **

**B) 252 **

**C) 271 **

**D) 305 **

**E) 351**

**Answer : C **

**Explanation : **

(24.99% of 250 Ã— 3.95) + (21.05% of 300 – 49.99) = (?) – 19.99% of 40

=> (25% of 250 Ã— 4) + (21% of 300 – 50) = ? – 20% of 40

=> {(25/100) Ã— 250 Ã— 4} + {(21/100) Ã— 300 – 50)} = ? – (20/100) Ã— 40

=> 250 + 63 – 50 = ? – 8

=> ? = 263 + 8

âˆ´ ? = 271

The required approx value =271

Here the correct option is C.

**Que 13. (345.97 + 129.88 – 45.03) + (34.87 Ã· 6.96 Ã— 2.99) = ? – âˆš1521 **

**A) 604 **

**B) 592 **

**C) 515 **

**D) 485 **

**E) None of these**

**Answer : D **

**Explanation : **

(345.97 + 129.88 – 45.03) + (34.87 Ã· 6.96 Ã— 2.99) = ? – âˆš1521

=> (346 + 130 – 45) + (35 Ã· 7 Ã— 3) = ? – âˆš1521

=> 431 + 15 = ? – 39

=> ? = 446 + 39

âˆ´ ? = 485

The required value =485

Here the correct option is D.

**Que 14. In the following number series, a wrong number is given. Find out the wrong number. **

**12, 12, 16, 34, 80, 182, 362 **

**A) 16 **

**B) 34 **

**C) 80 **

**D) 182 **

**E) 362 **

**Answer : C **

**Explanation : **

12 + (1Â³ – 1Â²) = 12

12 + (2Â³ – 2Â²) = 16

16 + (3Â³ – 3Â²) = 34

34 + (4Â³ – 4Â²) = **82**

82 + (5Â³ – 5Â²) = 182

182 +(6Â³ – 6Â²) = 362

The wrong number is 80.

Here the correct option is C.

**Que 15. In the following number series, a wrong number is given. Find out the wrong number. **

**74, 125, 76, 122, 79, 117 **

**A) 79 **

**B) 122 **

**C) 125 **

**D) 117 **

**E) 76**

**Answer : A **

**Explanation : **

74 + 51 = 125

125 – 49 = 76

76 + 46 = 122

122 – 42 = **80**

80 + 37 = 117

The wrong number is 79

Here the correct option is A.

**Que 16.** **The following question are accompanied by three statements (I), (II) and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the following question.**

**There are four members P, Q, R, and S partners in the business. What is the profit share of Q? **

**I: P and Q started the business with an investment of Rs x and Rs 3x respectively and after 6 months, R and S joined them with an investment of Rs. (x + 5000) and Rs. 4x respectively.**

**II: At the end of the year profit share of R is Rs. 16000.**

**III: At the end of one year, the profit ratio of R and S is 1:3. **

**A) Only II and III **

**B) Any two of them **

**C) Only I and II **

**D) All of them **

**E) None **

**Answer : D **

**Explanation : **

Profit = Investment Ã— Time

Statement I :

Profit ratio of P, Q, R and S = x Ã— 12 : 3x Ã— 12 : (x + 5000) Ã— 6 : 4x Ã— 6 = 12x : 36x : (x + 5000) Ã— 6 : 24x

From statement I, we do not know the value of x, so Statement I alone is not sufficient to answer the question.

Statement II and III:

The profit share of R = Rs.16000

Profit ratio of R and S = 1 : 3

Now, (6x + 30000)/24x = 1/3

=> 18x + 90,000 = 24x

=> 6x = 90,000

=> x = 15,000

Profit ratio = (12 Ã— 15000) : (36 Ã— 15000) : (6 Ã— 15000 + 30000) : (24 Ã— 15000)

= 12 : 36 : 8 : 24

= 3 : 9 : 2 : 6

Here 2 unit = 16000

=> 9 unit = (16000/2) Ã— 9 = Rs.72,000

All statements are necessary to answer the question.

Here the correct option is D.

**Que 17.** **The given question consists of 2 statements, I and II respectively. Please read the questions carefully and decide which of the statement (s) is/are sufficient/necessary to answer the question.**

**Find the ratio of spirit and water in the new mixture if the content of the two containers is mixed. **

**I: Two equal containers contain spirit and water in the ratio of 2:3 and 4: 5 respectively. Both the mixtures are mixed and formed a new mixture.**

**II: Containers contain 10 L and 12 L spirit respectively.**

**A) The data in statement I alone is sufficient to answer the question, while the data in statement Il alone is not sufficient to answer the question. **

**B) The data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question. **

**C) Either statement I or statement II alone is sufficient to answer the question. **

**D) The data in both the statements I and II are not sufficient to answer the questions. **

**E) The data in both the statements I and II together are necessary to answer. **

**Answer : A **

**Explanation : **

Spirit in first container = 2/5

Water in first container = 3/5

Spirit in 2nd container = 4/9

Water in 2nd container = 5/9

Total Spirit in new mixture = (2/5 + 4/9) =38/45

Total water in new mixture =(3/5 + 5/9) =52/45

Ratio of Spirit and water in new mixture

= 38/45 : 52/45 = 19 : 26

**âˆ´** **The data in statement I alone is sufficient to answer the question, while the data in statement Il alone is not sufficient to answer the question. **

**Here the correct option is A.**

**Que 18.** **The following question are accompanied by three statements (1), (II) and (III). You have to determine which statements(s) is/are sufficient/necessary to answer the following question.**

**Sandeep and Prateek are travelling toward each other at certain speeds from point A and point B respectively. At what time will both meet each other at point C if both start travelling at 3:00 pm? **

**1. Sandeep takes 4 hours to reach point B from point C while Prateek takes 9 hours to reach point A from point C.**

**II. Distance between point A and point B is 1200 km, and the ratio of the speed of Sandeep and Prateek is 2:3.**

**III. The difference between the speed of Sandeep and Prateek is 40 km/hr. **

**A) Only I is sufficient to answer **

**B) Only II and III are sufficient to answer **

**C) Either I or II and III together are sufficient to answer **

**D) All are necessary to answer **

**E) None of them is sufficient to answer **

**Answer : C **

**Explanation :**

From Statement I :

Let, Speed of Sandeep = M km/hr and Speed of Prateek = N km/hr

Time taken by them to meet each other = T hrs

Sandeep takes 4 hours to reach point B from point C while Prateek takes 9 hours to reach point A from point C.

Distance between A to C = 9N km

Distance between B to C = 4M km

We know, Time = Distance/Speed

So,

9N/M = T ——–(1)

4M/N = T ———(2)

Multiplying (1) and (2) , we get

TÂ² = (9N/M) Ã— (4M/N)

=> TÂ² = 36

=> T = 6 hours

(3 pm + 6 hours) = 9 pm

Sandeep and Prateek both will meet each other at 9 pm.

Statement alone is sufficient to answer the question.

From Statement II and III :

Distance = 1200 km

Let Speed of Sandeep and Prateek be 2P and 3P respectively.

Both are moving opposite direction to each other .

So relative speed = (2P + 3P) = 5P

Time taken to meet each other = 1200/5P

Statement II alone is not sufficient.

From Statement III,

Difference between the speed of Sandeep and Prateek is 40 km/hr

(3P-2P) = 40

=> P = 40 km/hr

Speed of Sandeep = 2 Ã— 40 = 80 km/hr

Speed of Prateek = 3 Ã— 40 = 120 km/hr

Relative speed = (80 + 120) = 200 km/hr

Time taken to meet each other = 1200/200 =6 hour

(3 pm + 6) = 9 pm

Statement II and III together are sufficient to answer the question.

**âˆ´ Either I or II and III together are sufficient to answer. **

**Here the correct option is C.**

**Que 19. 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.**

**What is the amount received by Shimona? **

**1. Samarth and Shimona started a business with an investment in the ratio of 5: 4. After 4 months Sujoy joined their business with an amount 20% more than that of Samarth after 2 more months Shimona took out 25% of her investment. The total profit obtained after the end of year is Rs.25000.**

**II. Samarth invested Rs.7000 more than that of Shimona. Sujoy invested 9000 more than Samarth and the difference between the profit earned by them is Rs.2000. **

**A) The data in statement I alone is sufficient to answer the question, while the data in statement Il alone is not sufficient to answer the question. **

**B) The data in statement II alone is sufficient to answer the question, while the data in statement I alone is not sufficient to answer the question. **

**C) Either statement I or statement II alone is sufficient to answer the question. **

**D) The data in both the statements I and II are not sufficient to answer the questions. **

**E) The data in both the statement I and II together are necessary to answer. **

**Answer : A **

**Explanation : **

Statement I :

Let investment of Samarth and Shimona be 5P and 4P respectively.

Investment of Sujoy = 5P Ã— (120/100) = 6P

Ratio of profit share of Samarth, Shimona and Sujoy = (5P Ã— 12) : [(4P Ã— 6) + (4P Ã— (75/100) Ã— 6] : (6P Ã— 8) = 60P : 42P : 48P = 10 : 7 : 8

Total profit = Rs.25,000

Amount received by Shimona = 25000 Ã— (7/25) = 7000

Statement I alone is sufficient to answer the question.

From Statement II :

Let investment of Shimona = X

Investment of Samarth = (X + 7000)

Investment of Sujoy = (X + 7000) + 9000 =X+16000

Difference between the profit earned by them is Rs.2000.

Here Statement II alone is not sufficient to answer the question.

**âˆ´ The data in statement I alone is sufficient to answer the question, while the data in statement Il alone is not sufficient to answer the question. **

**Here the correct option is A.**

**Que 20. Priyanka and Kailash Invested Rs.3000 together. Priyanka invested her amount at compound Interest for ‘x/10’ years at x% rate interest and Kailash invested half of his amount at compound interest and remaining half at simple interest at the same rate of interest for the same time period. [It is given that ‘x/10’ is a natural number] What is the value of ‘x’?**

**Statement I:** **Ratio of investment of Priyanka and Kailash is 1:2 respectively and total sum received by them together after a particular time at x% rate of interest is Rs. 4280. **

**Statement II: Simple interest received by kailash is Rs. 40 less than compound interest received by his.**

**Statement III: Sum of simple interest and compound interest received by Kailash after a certain time at x% rate of interest is Rs. 2500 Which of the above statement is redundant to find the answer of the given question? **

**A) Only Statement I alone **

**B) Only Statement II alone **

**C) Only Statement III alone **

**D) Only either statement I and II together or statement II and III together. **

**E) None of these **

**Answer : E **

**Explanation : **

Let Amount invested by Kailash = M

Investment of Priyanka = (3000 – M)

CI for Priyanka =(3000 – M) Ã— [1 + (x/100)^(x/10) – 1]

CI for Kailash =M/2 Ã— x Ã—(X/10)/100= xÂ²M/2000

statement I :

let time = t years

Investment of Kailash (M) = (2/3) Ã— 3000 =2000

Investment of Priyanka = (3000 – 2000) = 1000

CI for Priyanka = 1000 Ã— [1 + (x/100)^t – 1]

CI for Kailash = 1000 Ã— x Ã— [1 + (x/100)^t – 1]

SI for Kailash = (1000 Ã— x Ã— t)/100 = 10xt

According to question,

(4280 – 3000) = 1000Ã—[1+(x/100)^t – 1] + 1000 Ã— x Ã— [1+(x/100)^t – 1] + 10xt

In this equation, there are two variables.so we can not find the value of x.

Statement II :

According to statement II,

40 = (M/2)[{1 + (x/100)^(x/10) – 1} – (xÂ²/1000)]

from this equation we can not find the value of x.

Statement III:

Let time = T years

SI + CI = 2500

2500 = [(M/2) Ã— x Ã— T]/100 + [ (M/2) Ã— {1 + (x/100)^T – 1]

From this equation we can not find the value of x.

**âˆ´** **In statement I,II and III we found four unknown variables ( M, t , x and T) .So we can not find the value of x. **

**Here the correct option is E.**** **

**Direction (21-23) : Read the data given in paragraph carefully and answer the question. The paragraph given below shows the information about a milkman who has five vessels A, B, C, D and E out of which volume of A, B and C is 30 liters each and volume of vessels of D and E is 50 liters each. Cost of pure milk is Rs. 40 per liter and the cost of water is 0. In each vessel, a milkman has a mixture of milk and water. When the milkman sold the mixture from vessel A at Rs. 50 per liter he earned a total profit of 50%. Quantity of water in vessel B is 240% of the quantity of water in vessel A. Total profit earned by milkman when he sold the mixture from vessel D at the cost of pure milk is (300/7)% and total loss occurred to milkman when he sold the mixture from vessel E at Rs. 25.2 per liter is 25%. Total quantity of pure milk in vessel C is one-third of the total quantity of pure milk in vessel B and vessel E together. **

**Solution (21 – 23) : **

__For Vessel A__ :

Let Pure milk in vessel A = x litres

Water = (30 – x) litres

Cost price = 40x

Selling price = 30 Ã— 50 = 1500

Profit = (1500 – 40x )

Profit % = [(1500 – 40x)/40x] * 100

According to question,

[(1500 – 40x)/40x] * 100 = 50

=> (1500 – 40x)/40x = 50/100

=> 3000 – 80x = 40x

=> 120x = 3000

=> x = 25

Quantity of pure milk in vessel A = 25 litres

Quantity of water in vessel A = (30-25) = 5 litres

__For Vessel B__ :

Water = 240% of 5 = (240/100) Ã— 5 = 12 litres

Milk = (30 – 12) = 18 litres

__For Vessel D__ :

Let Quantity of Pure milk = y

Water = (50 – y)

Cost price = 40y

Selling price = 50Ã—40 = 2000

Profit % = [(2000 – 40y)/40y] Ã— 100

According to question,

[(2000 – 40y)/40y] Ã— 100 = 300/7

=> (50 – y)/y = 300/700

=> 350 – 7y = 3y

=> 10y = 350

=> y = 35

Quantity of milk = 35 litres

Quantity of water = (50-35) = 15 litres

__For Vessel E__ :

Let Quantity of pure milk = z

Water = (50 – z)

Cost price = 40z

Selling price = (50 Ã— 25.2) = 1260

Loss % = [(40z – 1260)/40z] Ã— 100

According to question,

[(40z – 1260)/40z] Ã— 100 = 25

=> (2z – 63)/2z = 25/100

=> 8z – 252 = 2z

=> 6z = 252

=> Z = 42

Quantity of milk = 42 litres

Quantity of water = (50-42) = 8 litres

__For Vessel C__ :

Quantity of pure milk = (1/3) Ã— (18 + 42) =20 L

Quantity of water = (30 – 20) = 10 litres

Vessel |
Total Quantity |
Milk | Water |
Ratio Milk : Water |
---|---|---|---|---|

A | 30 | 25 | 5 | 5 : 1 |

B | 30 | 18 | 12 | 3 : 2 |

C | 30 | 20 | 10 | 2 : 1 |

D | 50 | 35 | 15 | 7 : 3 |

E | 50 | 42 | 8 | 21 : 4 |

**Que 21. Milkman sold 3 litres mixture from vessel C and added the same amount of water, next time he sold 5 litres from the same vessel and added the same amount of water to it. Finally, he removes 2 litres of mixture and adds 2 litres of water. What is the ratio of milk to water in vessel C after these operations? **

**A) 2 : 3 **

**B) 7 : 8 **

**C) 8 : 7 **

**D) 5 : 3 **

**E) None of these **

**Answer : B **

**Explanation : **

Quantity of pure milk after replacement= 20 Ã— (1 – 3/30) Ã— (1 – 5/30) Ã— (1 – 2/30)= 20 Ã— (27/3) Ã— (25/30) Ã— (28/30) = 14 litres

Quantity of water after replacement = (30-14) =16 litres

**Required ratio = 14 : 16 = 7 : 8**

**Here the correct option is B.**** **

**Que 22. If milkman mixes the mixture of vessel B with mixture of vessel E and sells this mixture at cost price of pure milk, then what is the profit percent of milkman? **

**A) 24â…–% **

**B) 33â…“% **

**C) 16â…”% **

**D) 45Â¼% **

**E) None of these**

**Answer : B **

**Explanation : **

Quantity of milk in Vessel B and E together =(18 + 42) = 60 litres

Quantity of water in Vessel B and E together = (12 + 8) = 20 litres

Here profit = Quantity of water

Profit % = (20/60) Ã— 100 = 33(â…“) %

Required profit percentage =33(â…“) %

Here the correct option is B.

**Que 23. If the milkman uses the same cane to measure milk, once from vessel A and once from vessel D, then what is the amount of pure milk purchased by the customer out of 3 litres mixture? **

**A) 1.2 litres **

**B) 2.3 litres **

**C) 1.5 litres **

**D) 2.2 litres **

**E) None of these **

**Answer : B **

**Explanation : **

Ratio of Milk and water in vessel A= 5 : 1

Ratio of milk and water in vessel D = 7 : 3

Milk in vessel A = 5/6

Milk in vessel D = 7/10

Here the milkman using the same cane to measure so that its ratio be same i.e, 1 : 1 .

Let part of milk in the mixture purchased by customer = p

Applying the allegation method,

(5/6) (7/10)

p

1 1

=>(7/10) – p =1 =>p – (5/6) = 1

Now,

(7/10) – p = p – (5/6)

=> 2p = (7/10) + (5/6)

=> 2p = 46/30

=> p = 23/30

**âˆ´** **Amount of pure milk in the mixture purchased by customer = 3 Ã— (23/30) = 2.3 litres. **

**Here the correct option is B.**

**Direction (24-28) : Read the following data carefully and answer the questions given below. **

**The following pie chart shows the member of geeksforgeeks family in degree in 2020. **

**Table shows the ratio of male and female in every department. **

**Total member of geeksforgeeks family in 2020=720 **

Department of geeksforgeeks family | Male : Female |
---|---|

H.R | 2 : 3 |

Content | 4 : 1 |

Tech. | 3 : 7 |

Marketing | 3 : 2 |

Product | 1 : 1 |

Freelance | 3 : 7 |

**Solution (24 – 28) : **

Total member of geeksforgeeks family in 2020 = 720

From pie chart,

Total HR member = 720 Ã— (5/360) = 10

Total Content member = 720 Ã— (200/360) = 400

Total Tech. member = 720 Ã— (25/360) = 50

Total Marketing member = 720 Ã— (60/360) =120

Total Product member = 720 Ã— (20/360) = 40

Total Freelance member = 720 Ã— (50/360) =100

Now from table chart,

Male in HR = 10 Ã— (2/5) = 4

Female in HR = (10 – 4) = 6

Male in Content = 400 Ã— (4/5) = 320

Female in Content = (400 – 320) = 80

Male in Tech. = 50 Ã— (3/10) = 15

Female in Tech. = (50 – 15) = 35

Male in Marketing = 120 Ã— (3/5) = 72

Female in Marketing = (120 – 72) = 48

Male in Product = 40 Ã— (1/2) = 20

Female in Product = (40 – 20) = 20

Male in Freelance = 100 Ã— (3/10) = 30

Female in Freelance = (100 – 30) = 70

Department | Total member | Male | Female |
---|---|---|---|

HR | 10 | 4 | 6 |

Content | 400 | 320 | 80 |

Tech. | 50 | 15 | 35 |

Marketing | 120 | 72 | 48 |

Product | 40 | 20 | 20 |

Freelance | 100 | 30 | 70 |

**Que 24. What is the percentage of the number of male in content department to the number of female in tech. department? **

**A) 950.50% **

**B) 914.29% **

**C) 934.59% **

**D) 890.90% **

**E) None of these **

**Answer : B **

**Explanation : **

âˆ´ Required percentage = (320/35) Ã— 100 = 914.29 %

**Here the correct option is B.**

**Que 25. If every male earns Rs.5000 per day in tech. department, then find the total income of males in tech. department for one month. (in lakh) **

**A) 21.5 **

**B) 22.5 **

**C) 30 **

**D) 33.5 **

**E) 36.8 **

**Answer : B **

**Explanation : **

Total earning per day in tech department by male = (15 Ã— 5000) = Rs.75,000

**âˆ´** Total earning in a month = (30Ã—75000) = Rs 22.5 lakh

**Here the correct option is B.**

**Que 26. Find the number of total males in all departments in geeksforgeeks family. **

**A) 461 **

**B) 400 **

**C) 380 **

**D) 350 **

**E) 332 **

**Answer: A **

**Explanation : **

**âˆ´** Required total males in all departments = (4 + 320 + 15 + 72 + 20 + 30) = 461

**Here the correct option is A.**

**Que 27. The number of freelance members will be increased by 80% and the number of marketing members will be decreased by 50% in 2021 but the ratio of male and female remains same as 2020. Find the ratio of the number of males in freelance department and the number of females in marketing department in 2021. **

**A) 4 : 9 **

**B) 9 : 4 **

**C) 2 : 3 **

**D) 3 : 2 **

**E) 3 : 5**

**Answer : B **

**Explanation : **

New no. of freelance member = 100 Ã— (180/100) =180

New no.of Marketing member = 120 Ã— (50/100) = 60

Ratio of male and female remains same as before. So

Now male in freelance = 180 Ã— (3/10) = 54

Female in Marketing = 60 Ã— (2/5) = 24

**âˆ´** **Required ratio = 54 : 24 = 9 : 4 **

**Here the correct option is B.**

**Que 28. How many members should be added in H.R. department to become the same number of product department? **

**A) 15 **

**B) 20 **

**C) 25 **

**D) 30 **

**E) 10**

**Answer : D **

**Explanation : **

Total member in HR department = 10

Total member in Product department = 40

(40 – 10) = 30

**âˆ´ 30 members should be added in HR department to become the same number of Product o. **

**Here the correct option is D.**

**Direction (29-33) : In the given question,two equations numbered I and II are given. Solve both the equations and mark the appropriate answer. **

**A) x > y **

**B) x â‰¥ y **

**C) x < y **

**D) x â‰¤ y **

**E) x = y or relationship can not be established. **

**Que 29. **

**I. xÂ² + 39x + 368 = 0 **

**II. yÂ² + 42y + 425 = 0 **

**Answer : E**

**Explanation : **

xÂ² + 39x + 368 = 0

=> xÂ² + 23x + 16x + 368 = 0

=> (x + 23) (x + 16) = 0

=> x = -23 , -16

yÂ² + 42y + 425 = 0

=> yÂ² + 25y + 17y + 425 = 0

=> (y + 25) ( y + 17) = 0

y = -25 , -17

**âˆ´** **Relationship can not be established. **

**Here the correct option is E.**

**Que 30. **

**I. 3xÂ² – 18x + 15 = 0 **

**II. 4yÂ² – 84y + 320 = 0 **

**Answer : D**

**Explanation : **

3xÂ² – 18x + 15 = 0

=> 3xÂ² – 15x – 3x + 15 = 0

=> (3x – 3) (x – 5) = 0

=> x = 5 , 1

4yÂ² – 84y + 320 = 0** **

=> 4yÂ² – 20y – 64y + 320 = 0

=> (4y – 64) (y – 5) = 0

=> y = 5, 16

**x â‰¤ y **

**Here the correct option is D.**

**Que 31. **

**I. 3xÂ² – (2+12âˆš13)x + 8âˆš13 = 0 **

**II. 18yÂ² – (10+9âˆš13)y + 5âˆš13 = 0 **

**(Take âˆš13 = 3.60) **

**Answer : E **

**Explanation : **

3xÂ² – (2+12âˆš13)x + 8âˆš13 = 0

=> 3xÂ² – 2x – 12âˆš13x + 8âˆš13 = 0

=> (x – 4âˆš13) (3x – 2) = 0

=> x = 4âˆš13 , 2/3

18yÂ² – (10+9âˆš13)y + 5âˆš13 = 0

=> 18yÂ² – 10y – 9âˆš13y + 5âˆš13 = 0

=> (9y – 5) (2y – âˆš13) = 0

=> y = (5/9), (âˆš13/2)

**âˆ´** **Relationship can not be established. **

**Here the correct option is E.**

**Que 32. **

**I. 5xÂ² – 91x + 102 = 0 **

**II. 7yÂ² – 130y + 187 = 0 **

**Answer : E **

**Explanation : **

I. 5xÂ² – 91x + 102 = 0

=> 5xÂ² – 85x – 6x + 102 = 0

=> (5x – 6)(x – 17) = 0

=> x = 17, (6/5)

II. 7yÂ² – 130y + 187 = 0

=> 7yÂ² – 119y – 11y + 187 = 0

=> (7y – 11) (y – 17) = 0

=> y = 17 , (11/7)

**âˆ´**** x = y or relationship can not be established. **

**Here the correct option is E.**

**Que 33. **

**I. 50xÂ² – 45x + 9 = 0 **

**II. 25yÂ² – 35y + 12 = 0 **

**Answer : D **

**Explanation :**

50xÂ² – 45x + 9 = 0

=> 50xÂ² – 30x – 15x + 9 = 0

=> (10x – 3)(5x – 3) = 0

=> x = 3/10, 3/5

x = 0.3, 0.6

25yÂ² – 35y + 12 = 0

=> 25yÂ² – 20y – 15y + 12 = 0

=> (5y – 3)(5y – 4) = 0

=> y = 3/5, 4/5

y = 0.6, 0.8

**x â‰¤ y. **

**Here the correct option is D.**

**Que 34. Amit wants to cross a river and reach a point opposite to where he is now on other side of river. He needs to travel north but flow of river is in west direction. Amit’s boat speed is 5 km/hr and stream speed is 4 km/hr. River is 15 km wide. Find the time to reach the point where he intends to reach If he walks at the speed of 2 km/hr after reaching shore on the other side(in hour)**

**A) 8 ****B) 9 ****C) 12 ****D) 10 ****E) 6** **Answer : B ****Explanation : **

Distance = Speed Ã— Time

Time taken to cross the river = 15/5 = 3 hours

Distance travelled in 3 hours towards west = (3Ã—4) = 12 km

By foot time taken to travel 12 km distance = 12/2 = 6 hours

**âˆ´** **Required total time = (6+3) = 9 hours **

**Here the correct option is B.**

**Que 35. Saima Invested Rs. 50000 In gold for the whole year. After 5 months of Saima, her brother joined her and invested Rs 60000. Next year Salma invested Rs. 10000 more and her brother withdrew Rs. 5000 and at the end of two years profit earned by Saima is Rs. 42000. Find the total profit earned if they distributed half of the total profit equally and rest in the capital ratio.**

**A) Rs.60,000 ****B) Rs.75,000 ****C) Rs.80,000 ****D) Rs.89,000 ****E) None of these**

**Answer : C ****Explanation : **

Profit ratio = Investment Ã— Time

Capital ratio of them

= [(50000 Ã— 12) + (50000 + 10000) Ã— 12] : [(60000 Ã— 7) + (60000 – 5000) Ã— 12]

= (600000 + 720000) : (420000 + 660000)

= 11 : 9

Let total profit = x

Half of the total profit = x/2

Equally they both get = x/4

Saima got from her investment ratio = (x/2) Ã— (11/20) = 11x/40

According to question,

x/4 + 11x/40 = 42,000

=> 21x/40 = 42000

=> x = 80,000

**âˆ´** **Total profit earned by them = Rs.80,000 **

** Here the correct option is C.**

**Que 37. A broadband provider marks up the cost price of an internet plan by 50% and offers a discount of 20%. He asks the customer to pay GST of 18% on the selling price. The customer refuses to pay the tax due to which the provider himself pays the GST. Find his profit or loss percentage.**

**A) 2â…—% Loss**

**B) 1â…—% Profit**

**C) 2â…—% Profit**

**D) No profit no loss**

**E) 1â…—% Lose**

**Answer : E ****Explanation : **

Let cost price of the internet plan = x

Marked price = x Ã— (150/100) = 3x/2

Selling price = (3x/2) Ã— (80/100) = 6x/5

After paid 18% paid by the provider on selling price,

Selling price = (6x/5) Ã— (82/100) = 123x/125

Loss = x – (123x/125) = 2x/125

Loss % = {(2x/125)/x} Ã— 100 = 1â…—%

Required loss percentage = 1â…—%

Here the correct option is E.

**Que 37. A piece of work is completed by 3 men and 4 boys in 7 days. The same piece of work is completed by 5 men and 3 boys in 6 days. Find the time taken by 1 man and 2 boys to complete the 4 times of the same work?**

**A) 55 days**

**B) 66 days**

**C) 59 days**

**D) 69 days**

**E) 60 days**

**Answer : B ****Explanation : **

(3M + 4B) Ã— 7 = (5M + 3B) Ã— 6

=> 21M + 28B = 30M + 18B

=> 9M = 10B

Now,

3 Men = 10/3 boys

3 Men + 4 boys = (10/3 + 4) boys = 22/3 boys

1 man + 2 boys = (10/9 + 2) boys = 28/9 days

22/3 boys complete the work in 7 days

1 boy complete the work = 7 Ã— (22/3) days

28/9 boys complete the work = 7 Ã— (22/3) Ã— (9/28) = 66/4 days

28/9 boys complete the 4 times of the total work in = (66/4) Ã— 4 = 66 days

**âˆ´** **Required time = 66 days **

**Here the correct option is B.**

**Que 38. Clinton is having two varieties of tea. First variety of tea costing Rs. 25 per kg while second variety of tea is Rs. 32 per kg. Find how much quantity of 1st variety should he added to 30 kg of 2nd variety such that the cost of the mixture of both varieties be worth Rs. 27 per kg?**

**A) 70 kg**

**B) 75 kg**

**C) 80 kg**

**D) 72 kg**

**E) 77 kg**

**Answer : B ****Explanation : **

Applying alligation method,

First variety 2nd variety

25 32

27

(32 – 27)= 5 (27 – 25) =2

Ratio of first variety and 2nd variety tea = 5 : 2

2 unit = 30 kg

5 unit = (30/2) Ã— 5 = 75 kg

**âˆ´** **Required Quantity = 75 kg **

**Here the correct option is B.**

**Que 39. The sum of the ages of 5 members of a family 3 years ago was 108 years. Today, when the daughter has been married off and replaced by a daughter-in-law, the sum of their ages is 96. Assuming that there has been no other change in the family structure and all the people are alive, what is the difference in the age of the daughter and the daughter-in-law?**

**A) 27 years**

**B) 16 years**

**C) 22 years**

**D) 25 years**

**E) None of the above**

**Answer : A ****Explanation : **

Let the present ages of five members of the family = A, B, C, D, E (D= daughter’s age)

According to question,

(A – 3) + (B – 3) + (C – 3) + ( D – 3) + (E – 3) = 108

=> A + B + C + D + E = 123 ———(1)

Let the age of daughter in law = P

Now, according to question,

A + B + C + P + E = 96 ————(2)

Subtracting equation (1) and (2), we get

(A + B + C +D + E) – (A + B +C +P + E) = 123 – 96

=> D – P = 27

**âˆ´** **The required age difference = 27 years **

**Here the correct option is A.**

**Que 40. In a factory, toys are manufactured, out of which some are defected. Toys are packed in a box. Three toys are picked from box by inspector. If two of the toys out of three are perfect, box passes otherwise it fails. Find the probability of box getting passed.**

**A) 0.25**

**B) 0.5**

**C) 0.75**

**D) 0.35**

**E) 0.85**

**Answer : B ****Explanation : **

Let X and Y denotes the Perfect toy and defective toys respectively.

Sample space = {XXX, XXY, XYX, YXX, YYY, YYP, YXY, XYY}

Sample space = 8

Atleast 2 perfect toys = {XXX, XXY, XYX, YXX}

Sample space(atleast 2 perfect toys) = 4

Box passes = atleast two perfect toys

= 4/8 = 0.5

**âˆ´** **Required Probability of bos passing = 0.5 **

**Here the correct option is B.**

**Directions (41-45) : Read the following table carefully and answer the given questions: The following table shows No of hours per day,****No of Days in Total, No of Days on leave, Income of three persons A, B, and C.**

No of hours Per day | No of Day in total | No of Days on leave | Income | |
---|---|---|---|---|

A | Z | 48 | b | |

B | Y | X | 7 | |

C | 8 | a | 14 |

**1. Work has been done in March (31) and April (30), total days = 61 days**

**2. Wages rate for per hour on a working day is Rs. 500**

**3. Wage rate for Non-working Day is Rs.1500**

**4. Assuming either the work is completed in a given period of time or no work has been done i.e. on Idle day.**

**Que 41. For B, If the sum of values of x and y is equal to 60, then Find the Income of B (In Rs).**

**A) 209000 ****B) 172500 ****C) 185000 ****D) 200000 ****E) 252500**

**Answer : B ****Explanation : **

Total number of days = 61 days

Leave days of B = 7 days.

Working days of B =(61 – 7) = 54 days

According to question,

x + y = 60

=> 54 + y = 60

=> y = 6

Now Income = (1500 Ã— 7) + (500 Ã— 6 Ã— 54) =172500

Income of B =Rs. 172500

Here the correct option is B.

**Que 42. Find the number of working hours per day for A, if the income of A is Rs.259500.**

**A) 10 ****B) 6 ****C) 7 ****D) 13 ****E) 19**

**Answer : A ****Explanation : **

Total number of days = 61 days

Work days of A = 48 days

Leave days of A = (61 – 48) = 13 days

Let number of hours per day = H

Now According to the question,

(13 Ã— 1500) + (H Ã— 48 Ã— 500) = 259500

=> 19500 + 24000 H = 259500

=> 24000 H = 259500 – 19500

=> H = (240000/24000)

=> H = 10

The required number of working hours per day for A =10 hours

Here the correct option is A.

The required number of working hours per day for A =10 hours

Here the correct option is A.

**Que 43. Find the income earned by C.**

**A) 189000 ****B) 209000 ****C) 201000 ****D) 179000 ****E) 215000**

**Answer : B ****Explanation : **

Total number of days = 61 days

Leave days for C = 14 days

Work days for C = (61 – 14) = 47 days

Income = (14 Ã— 1500) + (8 Ã— 47 Ã— 500) = 209,000

The income earned by C =209,000

Here the correct option is B.

**Que 44. If the number of working hours per days for A is 6, find the income of A(in Rs.)**

**A) 160000 ****B) 163500 ****C) 171000 ****D) 177500 ****E) 185000**

**Answer : B ****Explanation : **

Working hours per day for A = 6 hours

Work days for A = 48 days

Leave days for A = (61 – 48) = 13 days

Income = (13 Ã— 1500) + (6 Ã— 48 Ã— 500) = 163,500

The required income of A = Rs.163500

Here the correct option is B.

**Que 45. If a new person is D, and his working days are equal to leave days of C and working hours are 4 more than working hours of C, then find the income of D(in Rs.)**

**A) 154500 ****B) 125600 ****C) 130000 ****D) 175900 ****E) 150000**

**Answer : A ****Explanation : **

Working days of D = leave days of C = 14

Working hours of D = (8 + 4) = 12 hours

Leave days of D = (61 – 14) = 47 days

Income = (47 Ã— 1500) + (12 Ã— 14 Ã— 500) = 154,500

The required income of D =Rs.154,500

Here the correct option is A.

**Directions (46-50) : Study the chart carefully and the questions that follow.**

**The pie charts show the distribution of the total number of cookies sold by companies A, B, C, and D and the distribution of chocolate flavor cookies sold by companies A, B, C, and D respectively. Also, Total Cookies = Chocolate Cookies + Butterscotch flavor cookies. Number of Butterscotch flavor cookies sold by A = 110 **

** Number of Butterscotch flavor cookies sold by B = 30 **

**Solution (46-50) : **

From pie chart I,

(N + 5) + N + (N + 20) + 2N = 100

=> 5N + 25 = 100

=> N = 15

From pie chart II,

M + 20 + (M + 10) + 40 = 100

=> 2M + 70 = 100

=> M = 15

Let, Total number of cookies sold = x

Cookies sold by A = (15+5) = 20% of total cookies

Total number of cookies sold by A = (20/100) Ã— x = x/5

Total number of Butterscotch flavor cookies sold by A = 110

Therefore, Total number of Chocolate flavor cookies sold by A = (x/5 – 110)

Let, Total number of Chocolate flavor cookies sold = P

According to question,

(x/5 – 110) = (15/100) Ã— P ——–(1)

Total number of cookies sold by B = (15/100) Ã— x = 3x/20

Total number of Butterscotch flavor cookies sold by B = 30

Total number of Chocolate flavor cookies sold by B = (3x/20 – 30)

According to question,

(3x/20 – 30) = (20/100) Ã— P ———-(2)

Multiplying equation (1) by 3 and equation (2) by 4 and then subtracting them, we get

(3x/5 – 330) – (3x/5 – 120) = 45P/100 – 20P/25

=> – 210 = – 35P/100

=> P = 210 Ã— (100/35)

=> P = 600

Total number of Chocolate flavor cookies sold = 600

Chocolate flavor cookies sold by A = 600 Ã— (15/100) = 90

Chocolate flavor cookies sold by B = 600 Ã— (20/100) = 120

Chocolate flavor cookies sold by C = 600 Ã— (25/100) = 150

Chocolate flavor cookies sold by D. = 600 Ã— (40/100) = 240

Total cookies = Butterscotch Cookies + Chocolate cookies

Total cookies sold by A = (110 + 90) = 200

From the total cookies chart,

200 = (20/100) Ã— total number of cookies

=> Total number of cookies = 1000

Cookies sold by A = 200

Cookies sold by B = (120 + 30) = 150

Cookies sold by C = (35/100) Ã— 1000 = 350

Cookies sold by D = (30/100) Ã— 1000 = 300

Chocolate cookies sold by C = (25/100) Ã— 600 = 150

Chocolate cookies sold by D = (40/100) Ã— 600 = 240

Butterscotch sold by C = (350 – 150) = 200

Butterscotch Sold by D = (300 – 240) = 60

Company |
Chocolate Cookies |
Butterscotch Cookies |
Total Cookie |
---|---|---|---|

A | 90 | 110 | 200 |

B | 120 | 30 | 150 |

C | 150 | 200 | 350 |

D | 240 | 60 | 300 |

**Que 46. What is the ratio of chocolate flavour cookies sold by company C to the total cookies sold by D?**

**A) 1 : 2 ****B) 2 : 1 ****C) 2 : 3 ****D) 3 : 4 ****E) None of these**

**Answer : A ****Explanation : **

Required ratio = 150 : 300 = 1 : 2

**âˆ´ The ratio of chocolate flavour cookies sold by company C to the total cookies sold by D is 1 : 2 **

**Here the correct option is A.**

**Que 47. Total cookies sold by A and B together is how much more percentage than the butterscotch cookies sold by C?**

**A) 25% ****B) 40% ****C) 50% ****D) 75% ****E) none of these**

**Answer : D ****Explanation : **

Total cookies sold by A and B together = (200 + 150) = 350

Required more percentage = {(350 – 200)/200} Ã— 100 = 75%

**âˆ´** **Total cookies sold by A and B together is 75% more percentage than the butterscotch cookies sold by C **

**Here the correct option is D.**

**Que 48. If A sold (x+15) chocolates flavour cookies and B sold (y – 25) chocolate flavor cookies, then what the ratio of x : y ?**

**A) 15 : 29 ****B) 15 : 31 ****C) 1 : 2 ****D) 29 : 15 ****E) 31 : 15 **

**Answer : A ****Explanation : **

x + 15 = 90

=> x = 75

y – 25 = 120

=> y = 145

x : y = 75 : 125 =15 : 29

The required ratio of x : y = 15 : 29

Here the correct option is A.

**Que 49. The total number of Butterscotch cookies sold by all companies together is approximate how much percentage more or less than the total number of chocolate cookies sold by all the companies together?**

**A) 50%**

**B) 33.33%**

**C) 60%**

**D) 30%**

**E) None of these **

**Answer : B ****Explanation : **

Total number of Butterscotch cookies sold by all companies together = (110 + 30 + 200 + 60) =400

Total number of Chocolate flavor cookies sold by all companies together = 600

Required less percentage ={(600-400)/600} Ã— 100 = 33.33%

**âˆ´ ****The total number of Butterscotch cookies sold by all companies together is approximate 33.33% less percentage than the total number of chocolate cookies sold by all the companies together **

**Here the correct option is B.**

**Que 50. Butterscotch cookies sold by D is approximate what percent of chocolate cookies sold by B and D together?**

**A) 6.67%**

**B) 6.25%**

**C) 8.33%**

**D) 16.67%**

**E) None of these**

**Answer : D ****Explanation : **

Chocolate flavor cookies sold by B and D together = (120 + 240) = 360

Required percentage = (60/360) Ã— 100 = 16.67%

**âˆ´ ****Butterscotch cookies sold by D is approximate 16.67% of chocolate cookies sold by B and D together. **

**Here the correct option is D.**

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