# Equations and Inequations Questions and Answers updated daily – Aptitude

Equations and Inequations Questions: Solved 316 Equations and Inequations Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

## Equations and Inequations Questions

21. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives corresponding answer.

Quantity I: If a shopkeeper to make 20% profit at 40% discount, what is the profit he will make if he gives a 40% discount?

Quantity II: Four-fifth of two-fifth of 30%of the number is 60. What is 4% of that number?

Quantity I: If a shopkeeper to make 20% profit at 40% discount, what is the profit he will make if he gives a 40% discount?

Quantity II: Four-fifth of two-fifth of 30%of the number is 60. What is 4% of that number?

SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Quantity I:

MP = x

SP = x * 60/100 = 0.6x

CP = 0.6x * 100/120 = 0.5x

Profit = (0.6 - 0.5)x/0.5x * 100 = 20%

Quantity II:

4/5 * 2/5 * 30/100 * x = 60

x = 625

4/100 * 625 = 25

Hence, Quantity II > Quantity I

MP = x

SP = x * 60/100 = 0.6x

CP = 0.6x * 100/120 = 0.5x

Profit = (0.6 - 0.5)x/0.5x * 100 = 20%

Quantity II:

4/5 * 2/5 * 30/100 * x = 60

x = 625

4/100 * 625 = 25

Hence, Quantity II > Quantity I

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22. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives corresponding answer.

Quantity I: In how many different ways can the letters of the word HOLIDAY can be arranged so that vowels occupy odd positions?

Quantity II: In how many different ways can the letters of the word LEADING can be arranged so that vowels always come together?

Quantity I: In how many different ways can the letters of the word HOLIDAY can be arranged so that vowels occupy odd positions?

Quantity II: In how many different ways can the letters of the word LEADING can be arranged so that vowels always come together?

SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Quantity I:

Required no. of ways = 4P

4*3*2*4! = 576 ways

Quantity II:

LEADING = 7 letter

3vowels (EAI) supposed to form one letter

Now remaining letter arranged in 5!

Vowels arranged in 3!

Therefore 5!*3! = 720 ways

Hence, Quantity I < Quantity II

Required no. of ways = 4P

_{3}* 4P_{4}4*3*2*4! = 576 ways

Quantity II:

LEADING = 7 letter

3vowels (EAI) supposed to form one letter

Now remaining letter arranged in 5!

Vowels arranged in 3!

Therefore 5!*3! = 720 ways

Hence, Quantity I < Quantity II

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23. In the following question, there are two equations. Solve the equations and answer accordingly:

I. 3x

II. 28y

I. 3x

^{2}- 13x +14 = 0II. 28y

^{2}+ 11y + 1 = 0SHOW ANSWER

Correct Ans:x > y

Explanation:

I. 3x

3x

3x(x - 2) - 7(x - 2) = 0

(3x - 7)(x - 2) = 0

x = (7/3, 2)

II. 28y

28y

y = (-1/4, -1/7)

Hence, x > y

^{2}- 13x + 14 = 03x

^{2}- 6x - 7x + 14 = 03x(x - 2) - 7(x - 2) = 0

(3x - 7)(x - 2) = 0

x = (7/3, 2)

II. 28y

^{2}+ 11y + 1 = 028y

^{2}+ 7y + 4y + 1 = 0y = (-1/4, -1/7)

Hence, x > y

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24. In the following question, two equations I and II are given. You have to solve both the equations and give Answer as,

I. x

II. y

I. x

^{2}- x - 272 = 0II. y

^{2}+ y - 156 = 0SHOW ANSWER

Correct Ans:x = y or relationship cannot be determined.

Explanation:

I. x

x

(x - 17) (x + 16) = 0

x = 17, -16

II. y

y

(y + 13) (y - 12) = 0

y = -13, 12

Hence the relationship cannot be determined.

^{2}- x - 272 = 0x

^{2}- 17x + 16x - 272 = 0(x - 17) (x + 16) = 0

x = 17, -16

II. y

^{2}+ y - 156 = 0y

^{2}+ 13y - 12y - 156 = 0(y + 13) (y - 12) = 0

y = -13, 12

Hence the relationship cannot be determined.

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25. In the following question, two equations I and II are given. You have to solve both the equations and give Answer as,

I. 5x = 7y + 21

II. 11x + 4y + 109 = 0

I. 5x = 7y + 21

II. 11x + 4y + 109 = 0

SHOW ANSWER

Correct Ans:x > y

Explanation:

5x - 7y = 21 -------- (1)

11x +4y = -109 -------- (2)

Multiply equ (1) by 11

55x - 77y = 231 --------- (3)

Multiply equ (2) by 5

55x + 20y = -545 --------- (4)

By solving the equation (3) & (4), we get,

-97y = 776

y = -776/97 = -8

Sub y = -8 in equ (1)

5x - 7(-8) = 21

5x + 56 = 21

5x = 21 - 56

5x = -35

x = -35/5 = -7

x = -7, y = -8

Hence, x > y

11x +4y = -109 -------- (2)

Multiply equ (1) by 11

55x - 77y = 231 --------- (3)

Multiply equ (2) by 5

55x + 20y = -545 --------- (4)

By solving the equation (3) & (4), we get,

-97y = 776

y = -776/97 = -8

Sub y = -8 in equ (1)

5x - 7(-8) = 21

5x + 56 = 21

5x = 21 - 56

5x = -35

x = -35/5 = -7

x = -7, y = -8

Hence, x > y

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26. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives corresponding answer.

Vessel A contains 60 liters mixture of milk and water and Vessel B contains x liters mixture of milk and water in the ratio of 3:2.

Quantity I: If the quantity of water in Vessel A is equal to the quantity of the milk in vessel B and the difference between quantity of the milk and water in vessel B is 8 liters, then the quantity of water in vessel A is what percent of the quantity of milk in vessel A?

Quantity II: If quantity of the water in vessel B is 16 liters, then the total quantity of the mixture in vessel B is what percent of the total quantity of the mixture in vessel A?

Vessel A contains 60 liters mixture of milk and water and Vessel B contains x liters mixture of milk and water in the ratio of 3:2.

Quantity I: If the quantity of water in Vessel A is equal to the quantity of the milk in vessel B and the difference between quantity of the milk and water in vessel B is 8 liters, then the quantity of water in vessel A is what percent of the quantity of milk in vessel A?

Quantity II: If quantity of the water in vessel B is 16 liters, then the total quantity of the mixture in vessel B is what percent of the total quantity of the mixture in vessel A?

SHOW ANSWER

Correct Ans:Quantity I = Quantity II or No relation

Explanation:

Quantity I:

3x - 2x = 8 liters

Quantity of the milk in vessel B = 3 * 8 = 24 liters

Quantity of the water in vessel A = 24 liters

Quantity of the milk in vessel A = 60 - 24 = 36 liters

Required percentage = 24/36 * 100 = 66.67%

Quantity II:

Water quantity in vessel B = 16 liters

Total quantity of vessel B = 5/2 * 16 = 40 liters

Required percentage = 40/60 * 100 = 66.67%

Quantity I = Quantity II

3x - 2x = 8 liters

Quantity of the milk in vessel B = 3 * 8 = 24 liters

Quantity of the water in vessel A = 24 liters

Quantity of the milk in vessel A = 60 - 24 = 36 liters

Required percentage = 24/36 * 100 = 66.67%

Quantity II:

Water quantity in vessel B = 16 liters

Total quantity of vessel B = 5/2 * 16 = 40 liters

Required percentage = 40/60 * 100 = 66.67%

Quantity I = Quantity II

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27. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives corresponding answer.

Quantity I: ‘X’: The cost of 5 bags and 12 notes is Rs. 3100 and the cost of 7 bags and 18 notes is Rs. 4400. What is the cost of 3 notes?

Quantity II: ‘Y’: Y

Quantity I: ‘X’: The cost of 5 bags and 12 notes is Rs. 3100 and the cost of 7 bags and 18 notes is Rs. 4400. What is the cost of 3 notes?

Quantity II: ‘Y’: Y

^{2}= 22500SHOW ANSWER

Correct Ans:Quantity I ≥ Quantity II

Explanation:

Quantity I:

According to the question:

5b + 12n = 3100 —– (i)

7b + 18n = 4400 —– (ii)

After solving equation (i) and (ii),

We get, b= 500 and n= 50

Cost of 3 notes = 50×3= 150

Quantity II:

Y

Y = ±150

Quantity I ≥ Quantity II

According to the question:

5b + 12n = 3100 —– (i)

7b + 18n = 4400 —– (ii)

After solving equation (i) and (ii),

We get, b= 500 and n= 50

Cost of 3 notes = 50×3= 150

Quantity II:

Y

^{2}= 22500Y = ±150

Quantity I ≥ Quantity II

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28. In the following question, there are two equations. Solve the equations and answer accordingly.

I. x

II. y

I. x

^{2}+ 23x + 120 = 0II. y

^{2}+ 13y + 42 = 0SHOW ANSWER

Correct Ans:x < y

Explanation:

I. x

x

x(x + 8) + 15(x + 8) = 0

(x + 15)(x + 8) = 0

x = -15, -8

II. y

y

y(y + 7) + 6(y + 7) = 0

(y + 6)(y + 7) = 0

y = -6, -7

Hence x < y

^{2}+ 23x + 120 = 0x

^{2}+ 8x + 15x + 120 = 0x(x + 8) + 15(x + 8) = 0

(x + 15)(x + 8) = 0

x = -15, -8

II. y

^{2}+ 13y + 42 = 0y

^{2}+ 7y + 6y + 42 = 0y(y + 7) + 6(y + 7) = 0

(y + 6)(y + 7) = 0

y = -6, -7

Hence x < y

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29. In the following question, there are two equations. Solve the equations and answer accordingly.

I. 3x

II. 2y

I. 3x

^{2}+ 26x + 56 = 0II. 2y

^{2}+ 15y + 28 = 0SHOW ANSWER

Correct Ans:x ≤ y

Explanation:

I. 3x

3x

3x(x + 4) + 14(x + 4) = 0

(3x + 14) (x + 4) = 0

x = -14/3, -4 = -4.667, -4

II. 2y

2y

2y(y + 4) + 7 (y + 4) = 0

(2y + 7) (y + 4) = 0

y = -7/2, -4 = -3.5, -4

Hence, x ≤ y

^{2}+ 26x + 56 = 03x

^{2}+ 12x + 14x + 56 = 03x(x + 4) + 14(x + 4) = 0

(3x + 14) (x + 4) = 0

x = -14/3, -4 = -4.667, -4

II. 2y

^{2}+ 15y + 28 = 02y

^{2}+ 8y + 7y + 28 = 02y(y + 4) + 7 (y + 4) = 0

(2y + 7) (y + 4) = 0

y = -7/2, -4 = -3.5, -4

Hence, x ≤ y

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30. In the following question, there are two equations. Solve the equations and answer accordingly.

I. 20x² - 119x + 176 = 0

II. 45x² + 200x + 155 = 0

I. 20x² - 119x + 176 = 0

II. 45x² + 200x + 155 = 0

SHOW ANSWER

Correct Ans:x > y

Explanation:

I. 20x² - 119x + 176 = 0

20x² - 64x - 55x + 176 = 0

4x(5x - 16) - 11(5x - 16) = 0

(4x - 11)(5x - 16) = 0

x = 11/4, 16/5

II. 45x² + 200x + 155 = 0

45x² + 45x + 155x + 155 = 0

45x(x + 1) + 155(x + 1) = 0

(45x + 155)(x + 1) = 0

x = -155/45, -1

Hence, x > y

20x² - 64x - 55x + 176 = 0

4x(5x - 16) - 11(5x - 16) = 0

(4x - 11)(5x - 16) = 0

x = 11/4, 16/5

II. 45x² + 200x + 155 = 0

45x² + 45x + 155x + 155 = 0

45x(x + 1) + 155(x + 1) = 0

(45x + 155)(x + 1) = 0

x = -155/45, -1

Hence, x > y

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31. Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer

Quantity 1: Compound interest earned on a certain sum at 10% for 2 years is Rs. 1260. Find the sum.

Quantity 2: Simple interest earned on a certain sum at 12% for 3 years is Rs. 2160. Find the sum.

Quantity 1: Compound interest earned on a certain sum at 10% for 2 years is Rs. 1260. Find the sum.

Quantity 2: Simple interest earned on a certain sum at 12% for 3 years is Rs. 2160. Find the sum.

SHOW ANSWER

Correct Ans:Quantity 1 = Quantity 2

Explanation:

**Quantity 1:**

---> Let, the sum of money = Rs. x

---> Rate of interest = 10%

---> No. of years = 2

---> According to problem,

---> x (1+(10/100))

^{2}- x = 1260

---> 1.21x â€“ x = 1260

---> 0.21x = 1260

---> x = (1260/0.21)

---> x = 6000

---> The sum of money = Rs. 6000

---> Quantity 1 = 6000

**Quantity 2:**

---> Let, the sum of money = Rs. y

---> Rate of interest = 12%

---> Time = 3 years

---> According to problem,

---> y * 12 * (3/100) = 2160

---> 0.36 * y = 2160

---> y = 6000

---> The sum of money = Rs. 6000

---> Quantity 2 = 6000

**Hence the answer is Quantity 1 = Quantity 2**

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32. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.

Manoj's monthly salary is 25% more than Mohit's salary. Mayank's monthly salary is Rs. 1750 more than Mohit's monthly salary. Sum of Manoj's, Mayank's and Mohit's yearly salary is Rs. 3,33,000.

Quantity I: Sum of monthly salary of Manoj and Mohit together

Quantity II: Rs. 20,000

Manoj's monthly salary is 25% more than Mohit's salary. Mayank's monthly salary is Rs. 1750 more than Mohit's monthly salary. Sum of Manoj's, Mayank's and Mohit's yearly salary is Rs. 3,33,000.

Quantity I: Sum of monthly salary of Manoj and Mohit together

Quantity II: Rs. 20,000

SHOW ANSWER

Correct Ans:Quantity II > Quantity I

Explanation:

Let the Mohit's monthly salary be Rs. X.

Manoj's monthly salary is 25% more than Mohit's salary.

Manoj's monthly salary = 1.25X

Mayank's monthly salary = X + 1750

Mohit's monthly salary + Manoj's monthly salary + Mayank's monthly salary = Total monthly salary

X + 1.25X + X + 1750 = (333000/12)

3.25X = 27750 - 1750

X = 26000/3.25

X = Rs. 8000

Mohit's monthly salary = Rs. 8000

Manoj's monthly salary = Rs. 10,000

Mayank's monthly salary = Rs. 9750

Quantity I:

Sum of monthly salary of Manoj and Mohit = 10000 + 8000 = Rs. 18,000

Hence, Quantity II > Quantity I.

Manoj's monthly salary is 25% more than Mohit's salary.

Manoj's monthly salary = 1.25X

Mayank's monthly salary = X + 1750

Mohit's monthly salary + Manoj's monthly salary + Mayank's monthly salary = Total monthly salary

X + 1.25X + X + 1750 = (333000/12)

3.25X = 27750 - 1750

X = 26000/3.25

X = Rs. 8000

Mohit's monthly salary = Rs. 8000

Manoj's monthly salary = Rs. 10,000

Mayank's monthly salary = Rs. 9750

Quantity I:

Sum of monthly salary of Manoj and Mohit = 10000 + 8000 = Rs. 18,000

Hence, Quantity II > Quantity I.

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33. Direction: 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.

Quantity I: If a shopkeeper sold 5 oranges at the cost price of 6 oranges, then his profit percentage.

Quantity II: A shopkeeper sold an article for Rs. 276. If the cost price of the article is Rs. 240, then his profit percentage.

Quantity I: If a shopkeeper sold 5 oranges at the cost price of 6 oranges, then his profit percentage.

Quantity II: A shopkeeper sold an article for Rs. 276. If the cost price of the article is Rs. 240, then his profit percentage.

SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Quantity I:**

---> SP of 5 oranges = CP of 6 oranges

---> (SP/CP) = (6/5)

---> So, profit percentage gained = (1/5) * 100 = 20 %

**Quantity II:**

---> Selling price of the article = Rs. 276

---> Cost price of the article = Rs. 240

---> Profit percentage = ((276 - 240) /240) * 100 = 15 %

---> So, Quantity I > Quantity II

**Hence the answer is : Quantity I > Quantity II**

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34. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.

Quantity I: Distance travelled by the bus to reach point B from point A if a car travels the same distance in 5 hrs and the speed of the bus is 120 km/hr which is 120% of the speed of the car.

Quantity II: Distance travelled by a boat to reach point D from point C if the speed of the boat in still water is 15 km/hr and speed of current is 3 km/hr. It goes from point C to D downstream and return back from point D to C upstream in 25 hrs.

Quantity I: Distance travelled by the bus to reach point B from point A if a car travels the same distance in 5 hrs and the speed of the bus is 120 km/hr which is 120% of the speed of the car.

Quantity II: Distance travelled by a boat to reach point D from point C if the speed of the boat in still water is 15 km/hr and speed of current is 3 km/hr. It goes from point C to D downstream and return back from point D to C upstream in 25 hrs.

SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

Quantity I:

WKT, Distance = Speed x Time

Time taken by car = 5 hrs

Speed of bus = 120 km/hr

Given that speed of the bus is 120% of the speed of the car.

120 = (120/100)*Speed of car

Speed of car = 100 km/hr

Distance = 100 x 5 = 500 km.

Quantity II:

Let the distance travelled be 'd' km.

Speed of the boat in still water = 15 km/hr

Speed of current = 3 km/hr

As per question,

[d/(15 + 3)] + [d/(15 - 3)] = 25

(d/18) + (d/12) = 25

(2 + 3)d/36 = 25

5d/36 = 25

d = (25 x 36)/5

d = 180 km

Hence, Quantity I > Quantity II.

WKT, Distance = Speed x Time

Time taken by car = 5 hrs

Speed of bus = 120 km/hr

Given that speed of the bus is 120% of the speed of the car.

120 = (120/100)*Speed of car

Speed of car = 100 km/hr

Distance = 100 x 5 = 500 km.

Quantity II:

Let the distance travelled be 'd' km.

Speed of the boat in still water = 15 km/hr

Speed of current = 3 km/hr

As per question,

[d/(15 + 3)] + [d/(15 - 3)] = 25

(d/18) + (d/12) = 25

(2 + 3)d/36 = 25

5d/36 = 25

d = (25 x 36)/5

d = 180 km

Hence, Quantity I > Quantity II.

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35. Direction: 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.

Amit can complete a piece of work in 60 days whereas Akash and Sumit working together can complete it in 15 days. When Amit and Sumit alternately work for a day each the work gets completed in 40 days.

Quantity 1:No. of days in which Akash will complete twice the work.

Quantity 2:No. of days in which Sumit will complete twice the work.

Amit can complete a piece of work in 60 days whereas Akash and Sumit working together can complete it in 15 days. When Amit and Sumit alternately work for a day each the work gets completed in 40 days.

Quantity 1:No. of days in which Akash will complete twice the work.

Quantity 2:No. of days in which Sumit will complete twice the work.

SHOW ANSWER

Correct Ans:Quantity I = Quantity II or No relation

Explanation:

**Quantity I:**

---> Let total units of work be 60 units.

---> Then units done by Amit in one day = 1 unit

---> For 40 days, Amit and Sumit work alternately.

---> Amit does 20 * 1 = 20 units and Sumit does 40 units in 20 days

---> Sumit does 2 units/day. Akash and Sumit do all the units in 15 days.

---> Which means Sumit does 15 * 2 = 30 units and Akash does remaining 30 units.

---> Hence, Akash does 2 units/day.

---> Efficiency of Akash & Sumit is same.

**Hence, Quantity 1 = Quantity 2**

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36. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.

Total surface area of a cylinder is 200% more than that of its sum of area of base and top of cylinder. Volume of cylinder is 2156 cm

Quantity I: Volume of cone, whose base radius and height is same as that of radius and height of cylinder respectively

Quantity II: Volume of hemisphere, whose radius is same as that of radius of cylinder.

Total surface area of a cylinder is 200% more than that of its sum of area of base and top of cylinder. Volume of cylinder is 2156 cm

^{3}.Quantity I: Volume of cone, whose base radius and height is same as that of radius and height of cylinder respectively

Quantity II: Volume of hemisphere, whose radius is same as that of radius of cylinder.

SHOW ANSWER

Correct Ans:Quantity I = Quantity II or Relation cannot be established

Explanation:

WKT, TSA of cylinder = 2π r(r + h)

Sum of area of base and top of cylinder = 2π r

TSA of cylinder is 200% more than sum of area of base and top of cylinder.

TSA of cylinder : Sum of area of base and top of cylinder = 300 : 100

TSA of cylinder/Sum of area of base and top of cylinder = 3/1

2π r(r + h)/2π r

r + h = 3r

h = 2r

Volume of cylinder = 2156 cm

π r

π (2r

r

r = 7 cm

h = 2(7) = 14 cm

Quantity I:

Volume of cone = (1/3)π r

= (1/3)*(22/7)*(7)

= 2156/3

Quantity II:

Volume of hemisphere = (2/3)π r

= (2/3)*(22/7)*(7)

= 2156/3

Hence, Quantity I = Quantity II.

Sum of area of base and top of cylinder = 2π r

^{2}TSA of cylinder is 200% more than sum of area of base and top of cylinder.

TSA of cylinder : Sum of area of base and top of cylinder = 300 : 100

TSA of cylinder/Sum of area of base and top of cylinder = 3/1

2π r(r + h)/2π r

^{2}= 3/1r + h = 3r

h = 2r

Volume of cylinder = 2156 cm

^{3}π r

^{2}h = 2156π (2r

^{3}) = 2156r

^{3}= (2156 x 7)/(22 x 2)r = 7 cm

h = 2(7) = 14 cm

Quantity I:

Volume of cone = (1/3)π r

^{2}h= (1/3)*(22/7)*(7)

^{2}*(14)= 2156/3

^{3}Quantity II:

Volume of hemisphere = (2/3)π r

^{3}= (2/3)*(22/7)*(7)

^{3}= 2156/3

^{3}Hence, Quantity I = Quantity II.

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37. Read the following information carefully & establish a relation between quantity I & quantity II.

There are five munch, four 5- stars and three KitKat chocolate in a bag. A boy take out four chocolates randomly from the bag.

There are five munch, four 5- stars and three KitKat chocolate in a bag. A boy take out four chocolates randomly from the bag.

**Quantity I:**Probability of selecting two munch and two five stars**Quantity II:**Probability of two munch one five store and one KitKat.SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Quantity I:

Total number of chocolates = 12

Required probability = [5C2 x 4C2]/12C4

= [(5 x 4)/2 x (4 x 3)/2]/[(12 x 11)/(2 x 3 x 4)]

= 60/495

= 4/33

Quantity II:

Required probability = [5C2 x 4C1 x 3C1]/12C4

= [(5 x 4)/2 x 4 x 3]/[(12 x 11)/(2 x 3 x 4)]

= 120/495

= 8/33

Hence, Quantity I < Quantity II.

Total number of chocolates = 12

Required probability = [5C2 x 4C2]/12C4

= [(5 x 4)/2 x (4 x 3)/2]/[(12 x 11)/(2 x 3 x 4)]

= 60/495

= 4/33

Quantity II:

Required probability = [5C2 x 4C1 x 3C1]/12C4

= [(5 x 4)/2 x 4 x 3]/[(12 x 11)/(2 x 3 x 4)]

= 120/495

= 8/33

Hence, Quantity I < Quantity II.

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38. Read the following information carefully & establish a relation between quantity I & quantity II.

A, B and C enter into a partnership. 'A' invests Rs. 4000 for the whole year, 'B' puts in Rs. 6000 at the first and increasing to Rs. 8000 at the end of 4

months, whilst C puts in at first Rs. 8000 but withdraw Rs. 2000 at the end of 9 months. Total annual profit is Rs. 56,500.

**Quantity I:**Profit share of 'A'.A, B and C enter into a partnership. 'A' invests Rs. 4000 for the whole year, 'B' puts in Rs. 6000 at the first and increasing to Rs. 8000 at the end of 4

months, whilst C puts in at first Rs. 8000 but withdraw Rs. 2000 at the end of 9 months. Total annual profit is Rs. 56,500.

**Quantity II:**Amount which when lend on C.I. at 20% interest being compounded annually for 3 years, gives total interest equal to Rs.9100.SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Quantity I:

WKT, Ratio of Profit = {A's capital * time} : {B's capital * time} : {C's capital * time}

Ratio of Profit = (4000 x 12) : [(6000 x 4) + (8000 x 8)] ; [(8000 x 9) + (6000 x 3)]

= 48000 : 88000 : 90000

= 24 : 44 : 45

Share of A = (24/113)*56500 = Rs. 12,000

Quantity II:

WKT, Amount = CI + P

Amount = P[1 + (r /100)

CI + P = P[1 + (r /100)

9100 + P = P[1 + (20/100)]

9100 + P =P[1728/1000]

9100 + P = 1.728P

9100 = 0.728P

P = Rs. 12,500.

Hence, Quantity I < Quantity II.

WKT, Ratio of Profit = {A's capital * time} : {B's capital * time} : {C's capital * time}

Ratio of Profit = (4000 x 12) : [(6000 x 4) + (8000 x 8)] ; [(8000 x 9) + (6000 x 3)]

= 48000 : 88000 : 90000

= 24 : 44 : 45

Share of A = (24/113)*56500 = Rs. 12,000

Quantity II:

WKT, Amount = CI + P

Amount = P[1 + (r /100)

^{n}]CI + P = P[1 + (r /100)

^{n}]9100 + P = P[1 + (20/100)]

^{3}9100 + P =P[1728/1000]

9100 + P = 1.728P

9100 = 0.728P

P = Rs. 12,500.

Hence, Quantity I < Quantity II.

Workspace

39. In the following question, there are two equations. Solve the equations and answer accordingly.

I. 3x - √2x = 4

II. (9y - 4)/3 = √2y

I. 3x - √2x = 4

II. (9y - 4)/3 = √2y

SHOW ANSWER

Correct Ans:x ≥ y

Explanation:

I. 3x - √2x = 4

(3x - 4)

9x

9x

9x

9x(x - 2) - 8(x - 2) = 0

(9x - 8) (x - 2) = 0

x = 8/9; 2

II. (9y - 4)/3 = √2y

(9y - 4)

81y

81y

81y

9y(9y - 2) - 8(x - 2) = 0

(9y - 2) (9y - 8) = 0

y = 2/9; 8/9

Hence, x ≥ y.

(3x - 4)

^{2}= (√2x)^{2}9x

^{2}+ 16 - 24x = 2x9x

^{2}- 26x + 16 = 09x

^{2}- 18x - 8x + 16 = 09x(x - 2) - 8(x - 2) = 0

(9x - 8) (x - 2) = 0

x = 8/9; 2

II. (9y - 4)/3 = √2y

(9y - 4)

^{2}= (3√2y)^{2}81y

^{2}+ 16 - 72y = 18y81y

^{2}- 90y + 16 = 081y

^{2}- 72y - 18y + 16 = 09y(9y - 2) - 8(x - 2) = 0

(9y - 2) (9y - 8) = 0

y = 2/9; 8/9

Hence, x ≥ y.

Workspace

40. In the following question, there are two equations. Solve the equations and answer accordingly.

I. x

II. y

I. x

^{2}- 5√3x + 18 = 0II. y

^{2}- 3√3y - 30 = 0SHOW ANSWER

Correct Ans:x = y or no relation can be established between x and y

Explanation:

I. x

x

(x - 2√3x) (x - 3√3x) = 0

x = 2√3x ; 3√3x

II. y

y

(y + 2√3y) (y - 5√3y) = 0

y = -2√3y ; 5√3y

(x1, y1) = (2√3x ; -2√3y) = x > y

(x2, y1) = (3√3x ; -2√3y) = x > y

(x1, y2) = (2√3x ; 5√3y) = y > x

(x2, y2) = (3√3x ; 5√3y) = y > x

Hence, no relation can be established.

^{2}- 5√3x + 18 = 0x

^{2}- 2√3x - 3√3x + 18 = 0(x - 2√3x) (x - 3√3x) = 0

x = 2√3x ; 3√3x

II. y

^{2}- 3√3y - 30 = 0y

^{2}- 5√3y + 2√3y - 30 = 0(y + 2√3y) (y - 5√3y) = 0

y = -2√3y ; 5√3y

(x1, y1) = (2√3x ; -2√3y) = x > y

(x2, y1) = (3√3x ; -2√3y) = x > y

(x1, y2) = (2√3x ; 5√3y) = y > x

(x2, y2) = (3√3x ; 5√3y) = y > x

Hence, no relation can be established.

Workspace

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