# 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

41. In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly.

I. √x - [(18)

II. √y - [(19)

I. √x - [(18)

^{(15/2)}/x^{2}] = 0II. √y - [(19)

^{(9/2)}/y] = 0SHOW ANSWER

Correct Ans:x < y

Explanation:

Given equation I. √x - [(18)

---> [√x * x

----> [√x * x

----> x

---> x

---> x

--->

Given equation II. √y - [(19)

---> [√y * y - (19)

---> √y * y - (19)

---> y

---> y

---> y

--->

Here,

^{(15/2)}/x^{2}] = 0---> [√x * x

^{2}- (18)^{(15/2)}]/ x^{2}= 0----> [√x * x

^{2}- (18)^{(15/2)}] = 0----> x

^{[(1/2) + 2]}= (18)^{(15/2)}---> x

^{(5/2)}= (18)^{(15/2)}---> x

^{(5/2)}= (18^{3})^{(5/2)}--->

**x = (18**^{3}) = 5832Given equation II. √y - [(19)

^{(9/2)}/y] = 0---> [√y * y - (19)

^{(9/2)}] /y = 0---> √y * y - (19)

^{(9/2)}= 0---> y

^{[(1/2) + 1]}= (19)^{(9/2)}---> y

^{(3/2)}= (19)^{(9/2)}---> y

^{(3/2)}= (19^{3})^{(3/2)}--->

**y = (19**^{3}) = 6859Here,

**x < y**
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42. Read the following information carefully & establish a relation between quantity I & quantity II.

Train A cross a platform of length 520 meters in 22.8 sec and a man in 7.2 sec.

Train A cross a platform of length 520 meters in 22.8 sec and a man in 7.2 sec.

**Quantity I:**If train A cross train B running in same direction at 96 km/hr in 63 seconds then find the length of train B.**Quantity II:**What is length of train C having speed of 90 km/hr and cross train A in 7.2 sec running in opposite direction.SHOW ANSWER

Correct Ans:Quantity I = Quantity II

Explanation:

Let the length of the train be 'l' meter.

WKT, Speed = Distance/Time

Speed of train A = (l + 520)/22.8

Also train cross a man, Speed of train = l/7.2

As per the question,

l/7.2 = (l + 520)/22.8

22.8l = 7.2l + 3744

15.6l = 3744

l = 240 m

Speed of train A = (240 + 520)/22.8 = 100/3 m/s

Quantity I:

Let the length of train B be 'b' meter.

[(100/3) - (96 x (5/18))] = (240 + b)/63

20/3 = (240 + b)/63

b = 420 - 240 = 180 m.

Quantity II:

Let the length of the train C be 'c' meter.

[(100/3) - (90 x (5/18))] = (240 + c)/7.2

1260 = 720 + 3c

c = 180 m.

Hence, Quantity I = Quantity II.

WKT, Speed = Distance/Time

Speed of train A = (l + 520)/22.8

Also train cross a man, Speed of train = l/7.2

As per the question,

l/7.2 = (l + 520)/22.8

22.8l = 7.2l + 3744

15.6l = 3744

l = 240 m

Speed of train A = (240 + 520)/22.8 = 100/3 m/s

Quantity I:

Let the length of train B be 'b' meter.

[(100/3) - (96 x (5/18))] = (240 + b)/63

20/3 = (240 + b)/63

b = 420 - 240 = 180 m.

Quantity II:

Let the length of the train C be 'c' meter.

[(100/3) - (90 x (5/18))] = (240 + c)/7.2

1260 = 720 + 3c

c = 180 m.

Hence, Quantity I = Quantity II.

Workspace

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

**Quantity I:**The percentage increase in area of a circle, when the radius of the circle is increased by 100%.**Quantity II:**The percentage increase in area of a rectangle when length is increased by 150% and breadth of the rectangle is increased by 160%.SHOW ANSWER

Correct Ans:Quantity I = Quantity II

Explanation:

Quantity I:

Let the radius of circle be 'r' and the area of circle of π r

When the radius of the circle is increased by 100%, radius becomes 2r.

Then area of circle = 4π r

Percentage increase in area of circle = [(4π r

= {π r

= 300%.

Quantity II:

Let length and breadth of rectangle be x and y respectively.

Area of rectangle = xy

When length is increased by 150%,

Length of rectangle = (150/100)x = 2.5x

When breadth of the rectangle is increased by 160% ,

Breadth of rectangle = (160/100)y = 1.6y

Area of rectangle = (2.5x)(1.6y) = 4xy

Percentage increase in area of rectangle = [(4xy - xy)/xy]*100 = 300%

Hence, Quantity I = Quantity II.

Let the radius of circle be 'r' and the area of circle of π r

^{2}.When the radius of the circle is increased by 100%, radius becomes 2r.

Then area of circle = 4π r

^{2}Percentage increase in area of circle = [(4π r

^{2}- π r^{2})/π r^{2}]*100= {π r

^{2}[4 - 1]/π r^{2}}*100= 300%.

Quantity II:

Let length and breadth of rectangle be x and y respectively.

Area of rectangle = xy

When length is increased by 150%,

Length of rectangle = (150/100)x = 2.5x

When breadth of the rectangle is increased by 160% ,

Breadth of rectangle = (160/100)y = 1.6y

Area of rectangle = (2.5x)(1.6y) = 4xy

Percentage increase in area of rectangle = [(4xy - xy)/xy]*100 = 300%

Hence, Quantity I = Quantity II.

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

I. 4 = 3/x + 5/2x

II. 14y = 6/y + 5

I. 4 = 3/x + 5/2x

^{2}= 0II. 14y = 6/y + 5

SHOW ANSWER

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

Explanation:

I. 4 = 3/x + 5/2x

4 = (6x + 5)/2x

8x

8x

2x(4x - 5) + (4x - 5) = 0

(2x + 4) (4x - 5) = 0

x = -1/2; 5/4

II. 14y = 6/y + 5

14y = (6 + 5y)/y

14y

14y

2y(7y - 6) + 1(7y - 6) = 0

(2y + 1) (7y - 6) = 0

y = -1/2; 6/7

(x1, y1) = (-1/2; -1/2) = x = y

(x2, y1) = (5/4; -1/2) = x > y

(x1, y2) = (-1/2; 6/7) = y > x

(x2, y2) = (5/4; 6/7) = x > y

Hence, no relation can be established.

^{2}= 04 = (6x + 5)/2x

^{2}8x

^{2}- 6x - 5 = 08x

^{2}- 10x + 4x - 5 = 02x(4x - 5) + (4x - 5) = 0

(2x + 4) (4x - 5) = 0

x = -1/2; 5/4

II. 14y = 6/y + 5

14y = (6 + 5y)/y

14y

^{2}- 5y - 6 = 014y

^{2}- 12y + 7y - 6 = 02y(7y - 6) + 1(7y - 6) = 0

(2y + 1) (7y - 6) = 0

y = -1/2; 6/7

(x1, y1) = (-1/2; -1/2) = x = y

(x2, y1) = (5/4; -1/2) = x > y

(x1, y2) = (-1/2; 6/7) = y > x

(x2, y2) = (5/4; 6/7) = x > y

Hence, no relation can be established.

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

I. 12m

II. 7n

I. 12m

^{2}- 126m + 294 = 0II. 7n

^{2}+ 123n - 504 = 0SHOW ANSWER

Correct Ans:m > n

Explanation:

I. 12m

12m

12m(m - 7) - 42(m - 7) = 0

(12m - 42) (m - 7) = 0

m = 7, 7/2

II. 7n

7n

7n(n + 21) - 24(n + 21) = 0

(n + 21) (7n - 24) = 0

n = -21; 24/7.

Hence, m > n.

^{2}- 126m + 294 = 012m

^{2}- 84m - 42m + 294 = 012m(m - 7) - 42(m - 7) = 0

(12m - 42) (m - 7) = 0

m = 7, 7/2

II. 7n

^{2}+ 123n - 504 = 07n

^{2}+ 147n - 24n - 504 = 07n(n + 21) - 24(n + 21) = 0

(n + 21) (7n - 24) = 0

n = -21; 24/7.

Hence, m > n.

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

I. √784x + 1234 = 1486

II. √1089y + 2081 = 2345

I. √784x + 1234 = 1486

II. √1089y + 2081 = 2345

SHOW ANSWER

Correct Ans:x > y

Explanation:

I. √784x + 1234 = 1486

28x + 1234 = 1486

28x = 252

x = 9

II. √1089y + 2081 = 2345

33y + 2081 = 2345

33y = 264

y = 8

Hence, x > y.

28x + 1234 = 1486

28x = 252

x = 9

II. √1089y + 2081 = 2345

33y + 2081 = 2345

33y = 264

y = 8

Hence, x > y.

Workspace

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

Train A of length 140 m can cross a platform of length 260 m in 25 second.

Train A of length 140 m can cross a platform of length 260 m in 25 second.

**Quantity I:**The ratio of speed of train A and Train B is 4 : 7. Find the half length of Train B if it can cross a pole in 12 seconds.**Quantity II:**172 mSHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Given:

Length of train A = 140 m

Length of platform = 260 m

Time = 25 seconds

WKT, Speed of train = Distance/Time

Speed of train A = (140 + 260)/25 = 16 m/s

Quantity I:

Given, ratio of speeds = 4 : 7

Speed of train B = (7/4)*16 = 28 m/s

Half length of train B = (1/2)*28*12 = 168 m

Quantity II: 172

Hence, Quantity I < Quantity II.

Length of train A = 140 m

Length of platform = 260 m

Time = 25 seconds

WKT, Speed of train = Distance/Time

Speed of train A = (140 + 260)/25 = 16 m/s

Quantity I:

Given, ratio of speeds = 4 : 7

Speed of train B = (7/4)*16 = 28 m/s

Half length of train B = (1/2)*28*12 = 168 m

Quantity II: 172

Hence, Quantity I < Quantity II.

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

**Quantity I:**A started a business with Rs. 25000 and is joined afterwards by B with Rs. 33333.33. After how many months did B join if the profits at the end of the year are divided equally.**Quantity II:**A began a business with Rs. 12000. He was joined afterwards by B with Rs. 4000 for how much period does B join. If the profits at the end of the year are divided in the ratio of 4: 1.SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

Quantity I:

Lets assume B joined after x months.

Its given that profit are equal,

capital investment of A = capital investment of B

25000*12 = 33333.33*(12 - x)

9 = 12 - x

x = 3

So, B joined after 3 months.

Quantity II:

Lets assume B joined for x months.

Given, ratio of profit = 4 : 1

WKT,

Ratio of capital investment of A/Ratio of capital investment of B = Ratio of profit of A/Ratio of profit of B

(12000*12)/(4000*x) = 4/1

x = 9

So, B joined for 6 months.

Hence, Quantity I < Quantity II.

Lets assume B joined after x months.

Its given that profit are equal,

capital investment of A = capital investment of B

25000*12 = 33333.33*(12 - x)

9 = 12 - x

x = 3

So, B joined after 3 months.

Quantity II:

Lets assume B joined for x months.

Given, ratio of profit = 4 : 1

WKT,

Ratio of capital investment of A/Ratio of capital investment of B = Ratio of profit of A/Ratio of profit of B

(12000*12)/(4000*x) = 4/1

x = 9

So, B joined for 6 months.

Hence, Quantity I < Quantity II.

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49. In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly.

I. 48/xÂ² âˆ’ 14/x + 1 = 0

II. 45/yÂ² + 1/y = 2

I. 48/xÂ² âˆ’ 14/x + 1 = 0

II. 45/yÂ² + 1/y = 2

SHOW ANSWER

Correct Ans:x > y

Explanation:

I. 48/xÂ² âˆ’ 14/x + 1 = 0

xÂ² - 14x + 48 = 0

xÂ² - 8x - 6x + 48 = 0

x(x - 8) - 6(x - 8) = 0

(x - 6) (x - 8) = 0

x = 6, 8

II. 45/yÂ² + 1/y = 2

2yÂ² - y - 45 = 0

2yÂ² - 10y + 9y - 45 = 0

2y(y - 5) + 9(y - 5) = 0

(2y + 9) (y - 5) = 0

y = -9/2, 5

Hence, x > y.

xÂ² - 14x + 48 = 0

xÂ² - 8x - 6x + 48 = 0

x(x - 8) - 6(x - 8) = 0

(x - 6) (x - 8) = 0

x = 6, 8

II. 45/yÂ² + 1/y = 2

2yÂ² - y - 45 = 0

2yÂ² - 10y + 9y - 45 = 0

2y(y - 5) + 9(y - 5) = 0

(2y + 9) (y - 5) = 0

y = -9/2, 5

Hence, x > y.

Workspace

50. In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly.

I. y

II. [(x/23) + (1/4)] = 117/92x

I. y

^{3/2}+ 3(y)^{1/2}- 54(y)^{-1/2}= 0II. [(x/23) + (1/4)] = 117/92x

SHOW ANSWER

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

Explanation:

I. y

y

(y

y

y

y(y + 9) - 6(y + 9) = 0

(y - 6) (y + 9) = 0

y = 6, -9

II. [(x/23) + (1/4)] = 117/92x

(x/23) + (1/4) - (117/92x) = 0

(4x

4x

4x

4x(x + 9) - 13(x + 9) = 0

(4x - 13) (x + 9) = 0

x = 13/4; -9

(x1, y1) = (13/4, 6) = y > x

(x2, y1) = (-9, 6) = y > x

(x1, y2) = (13/4, -9) = x > y

(x2, y2) = (-9, -9) = y = x

Hence, no relation can be established.

^{3/2}+ 3(y)^{1/2}- 54(y)^{-1/2}= 0y

^{3/2}+ 3(y)^{1/2}- 54/(y)^{1/2}= 0(y

^{2}+ 3y - 54)/y^{1/2}= 0y

^{2}+ 3y - 54 = 0y

^{2}+ 9y - 6y - 54 = 0y(y + 9) - 6(y + 9) = 0

(y - 6) (y + 9) = 0

y = 6, -9

II. [(x/23) + (1/4)] = 117/92x

(x/23) + (1/4) - (117/92x) = 0

(4x

^{2}+ 23x - 117)/92x = 04x

^{2}+ 23x - 117 = 04x

^{2}+ 36x - 13x - 117 = 04x(x + 9) - 13(x + 9) = 0

(4x - 13) (x + 9) = 0

x = 13/4; -9

(x1, y1) = (13/4, 6) = y > x

(x2, y1) = (-9, 6) = y > x

(x1, y2) = (13/4, -9) = x > y

(x2, y2) = (-9, -9) = y = x

Hence, no relation can be established.

Workspace

51. In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly.

I. (10/x

II. (14/y

I. (10/x

^{2}) - (13/x) + 4 = 0II. (14/y

^{2}) + 2 = (11/y)SHOW ANSWER

Correct Ans:x ≤ y

Explanation:

Given, I. (10/x

---> (10 - 13x + 4x

---> (10 - 13x + 4x

---> 4x

---> 4x

---> 4x(x - 2) - 5(x - 2) = 0

---> (x - 2) (4x - 5) = 0

--->

II. (14/y

---> (14/y

---> (14 + 2y

---> (14 + 2y

---> 2y

---> 2y

---> 2y(y - 2) - 7(y - 2) = 0

---> (y - 2) (2y - 7) = 0

--->

Now, Comparing the values of x and y, we get

For x = 2, y = 2 ---> x = y

For x = 2, y = 7/2 ---> x < y

For x = 5/4, y = 2 ---> x < y

For x = 5/4, y = 7/2 ---> x < y

Hence,

^{2}) - (13/x) + 4 = 0---> (10 - 13x + 4x

^{2}) / x^{2}= 0---> (10 - 13x + 4x

^{2}) = 0---> 4x

^{2}- 13x + 10 = 0---> 4x

^{2}- 8x - 5x + 10 = 0---> 4x(x - 2) - 5(x - 2) = 0

---> (x - 2) (4x - 5) = 0

--->

**x = 2, 5/4**II. (14/y

^{2}) + 2 = (11/y)---> (14/y

^{2}) + 2 - (11/y) = 0---> (14 + 2y

^{2}- 11y) / y^{2}) = 0---> (14 + 2y

^{2}- 11y) = 0---> 2y

^{2}- 11y + 14 = 0---> 2y

^{2}- 4y - 7y + 14 = 0---> 2y(y - 2) - 7(y - 2) = 0

---> (y - 2) (2y - 7) = 0

--->

**y = 2, 7/2**Now, Comparing the values of x and y, we get

For x = 2, y = 2 ---> x = y

For x = 2, y = 7/2 ---> x < y

For x = 5/4, y = 2 ---> x < y

For x = 5/4, y = 7/2 ---> x < y

Hence,

**x ≤ y**
Workspace

52. In the following questions, two equations numbered I and II are given. You have to solve both the equations and give answer

I. 4x + 3y = (1600)

II. 6x - 5y = (484)

X & Y both are natural numbers.

I. 4x + 3y = (1600)

^{1/2}II. 6x - 5y = (484)

^{1/2}X & Y both are natural numbers.

SHOW ANSWER

Correct Ans:if x > y

Explanation:

4x + 3y = 40 ------- (i)

6x â€“ 5y = 22 --------- (ii)

Multiply equ (i) by 6

24x + 18y = 240 ------- (iii)

Multiply equ (ii) by 4

24x â€“ 20 = 88 ------- (iv)

Sub equ (iii) and (iv)

38y = 152

y = 4

sub y = 4 in equ (i)

4x = 40 - 12

4x = 28

x = 7

So, x > y

6x â€“ 5y = 22 --------- (ii)

Multiply equ (i) by 6

24x + 18y = 240 ------- (iii)

Multiply equ (ii) by 4

24x â€“ 20 = 88 ------- (iv)

Sub equ (iii) and (iv)

38y = 152

y = 4

sub y = 4 in equ (i)

4x = 40 - 12

4x = 28

x = 7

So, x > y

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53. The solution of the pair of equation (x/2) + y = 0.8 and 7/[x + (y/2)] = 10 is

SHOW ANSWER

Correct Ans:x = 2/5; y = 3/5

Explanation:

Given:

(x/2) + y = 0.8

x + 2y = 1.6

10x + 20y = 16 ....(i)

7/[x + (y/2)] = 10

x + (y/2) = 7/10

20x + 10y = 14 ....(ii)

By solving (i) and (ii),

x = 2/5; y = 3/5.

(x/2) + y = 0.8

x + 2y = 1.6

10x + 20y = 16 ....(i)

7/[x + (y/2)] = 10

x + (y/2) = 7/10

20x + 10y = 14 ....(ii)

By solving (i) and (ii),

x = 2/5; y = 3/5.

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54. What is the value of x + y in the solution of the following equations?

(x/4) + (y/3) = (5/12) and (x/2) + y = 1

(x/4) + (y/3) = (5/12) and (x/2) + y = 1

SHOW ANSWER

Correct Ans:(3/2)

Explanation:

Given, (x/4) + (y/3) = (5/12) and (x/2) + y = 1

Multiplying the first equation with 12 and second equation with 4, we get

3x + 4y = 5 -------> eqn (i)

2x + 4y = 4 -------> eqn (ii)

Now, eqn (i) - eqn (ii), we get

x = 1

Then from eqn (i), 4y = 5 - 3

---> 4y = 2

---> 2y = 1

---> y = 1/2

Hence, the value of

=

Multiplying the first equation with 12 and second equation with 4, we get

3x + 4y = 5 -------> eqn (i)

2x + 4y = 4 -------> eqn (ii)

Now, eqn (i) - eqn (ii), we get

x = 1

Then from eqn (i), 4y = 5 - 3

---> 4y = 2

---> 2y = 1

---> y = 1/2

Hence, the value of

**x + y**= 1 + (1/2)=

**3/2**
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55. Read the following information carefully & establish a relation between quantity I & quantity II:

**Quantity I:**X^{2}= 6084**Quantity II:**Y = √6084SHOW ANSWER

Correct Ans:quantity I ≤ quantity II

Explanation:

Let find the value of Quantity I , Quantity II

Quantity I : X

X = ±78

Quantity II : Y = √6084

Y = 78

Quantity I : X

^{2}= 6084X = ±78

Quantity II : Y = √6084

Y = 78

**Hence, Quantity I ≤ Quantity II**
Workspace

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

A horse and a cow were sold for Rs. 540, making a profit of 25% on the horse and 20% on the cow. By selling for Rs. 538, the profit would be 20% on the horse and 25% on the cow.

A horse and a cow were sold for Rs. 540, making a profit of 25% on the horse and 20% on the cow. By selling for Rs. 538, the profit would be 20% on the horse and 25% on the cow.

**Quantity I:**CP of one cow**Quantity II:**CP of one horseSHOW ANSWER

Correct Ans:quantity I < quantity II

Explanation:

Let CP of a cow and a horse be x and y respectively

For Case I,

SP of both = 125% of y + 120 % of x

Simplifying it,

25y + 24x = 540*20

24x = 540*20 - 25y

x = (540*20 - 25y)/24

x = (540*20)/24 - 25y/24

x = 450 - 25y/24 ----------- (i)

For case II,

SP of both = 120% of y + 125% of x

Simplifying it,

24y + 25x = 538*20

25x = 538*20 - 24y

x = (538*20 - 24y)/25

x = 2152/5 - 24y/25 ----------- (ii)

Solving Equ (i) and (ii),

450 - 25y/24 = 2152/5 - 24y/25

450 - 2152/5 = 25y/24 - 24y/25

(2250 - 2152)/5 = (625y - 576y)/600

98/5 = 49/120

y = 240

Sub y = 240 in equ (i)

x = 450 - (25*240)24

x = 450 - 250 = 200

x = 200, y = 240

Hence, Quantity I < Quantity II

For Case I,

SP of both = 125% of y + 120 % of x

Simplifying it,

25y + 24x = 540*20

24x = 540*20 - 25y

x = (540*20 - 25y)/24

x = (540*20)/24 - 25y/24

x = 450 - 25y/24 ----------- (i)

For case II,

SP of both = 120% of y + 125% of x

Simplifying it,

24y + 25x = 538*20

25x = 538*20 - 24y

x = (538*20 - 24y)/25

x = 2152/5 - 24y/25 ----------- (ii)

Solving Equ (i) and (ii),

450 - 25y/24 = 2152/5 - 24y/25

450 - 2152/5 = 25y/24 - 24y/25

(2250 - 2152)/5 = (625y - 576y)/600

98/5 = 49/120

y = 240

Sub y = 240 in equ (i)

x = 450 - (25*240)24

x = 450 - 250 = 200

x = 200, y = 240

Hence, Quantity I < Quantity II

Workspace

57. Find the appropriate relation for quantity 1 and quantity 2 in the following question:

Two train going in the opposite direction cross each other in 12 sec.

Two train going in the opposite direction cross each other in 12 sec.

**Quantity 1:**Length of train 1 if it crosses the pole in 9 sec**Quantity 2:**Length of train 2 if it crosses the pole in 24 secSHOW ANSWER

Correct Ans:quantity 1 > quantity 2

Explanation:

Speed = distance/time

Let the speed and length of 1st train be â€˜S

Let the speed and length of 2nd train be â€˜S

Given, train going in the opposite direction cross each other in 12 sec.

Relative speed between the trains going in opposite direction = S

Total distance travelled = L

âˆ´ S

Quantity I : Length of train 1 if it crosses the pole in 9 sec

â‡’ S

Quantity II : Length of train 2 if it crosses the pole in 24 sec

â‡’ S

Substituting value of S

L

L

L

L

L

L

L

So, quantity 1 > quantity 2

Let the speed and length of 1st train be â€˜S

_{1}â€™ and â€˜L_{1}â€™ respectively.Let the speed and length of 2nd train be â€˜S

_{2}â€™ and â€˜L_{2}â€™ respectively.Given, train going in the opposite direction cross each other in 12 sec.

Relative speed between the trains going in opposite direction = S

_{1}+ S_{2}Total distance travelled = L

_{1}+ L_{2}âˆ´ S

_{1}+ S_{2}= (L_{1}+ L_{2})/12 -------- (1)Quantity I : Length of train 1 if it crosses the pole in 9 sec

â‡’ S

_{1}= L_{1}/9Quantity II : Length of train 2 if it crosses the pole in 24 sec

â‡’ S

_{2}= L_{2}/24Substituting value of S

_{1}and S_{2}in eq1L

_{1}/9 + L_{2}/24 = L_{1}/12 + L_{2}/12L

_{1}/9 - L_{1}/12 = L_{2}/12 - L_{2}/24L

_{1}/36 = L_{2}/24L

_{1}/3 = L_{2}/2L

_{1}/L_{2}= 3/2L

_{1}: L_{2}= 3 : 2L

_{1}> L_{2}So, quantity 1 > quantity 2

Workspace

58. Find the appropriate relation for quantity 1 and quantity 2 in the following question:

**Quantity I:**Vimal's Farm has only Ostriches and Giant marsupial (4 legs mammal), total count of legs was 14 less than 4 times the total count of heads. How many Ostriches are there in total?**Quantity II:**Ten politicians are made to stand in a row for any purpose in Parliament. Two politicians are selected at random from the given group. Find the probability that the politician thus selected were positioned adjacent to each other.SHOW ANSWER

Correct Ans:Quantity I > Quantity II

Explanation:

**Quantity I:**

We know that, Ostriches has 1 head and 2 legs.

Giant marsupial has 1 head and 4 legs

Let the number of Ostriches and that of Giant marsupial be x and y respectively.

According to question,

Total count of legs = 4 times the total count of heads - 14

--> 2x + 4y = [4 * (x + y)] - 14

---> 2x + 4y = 4x + 4y - 14

--> 2x -14 = 0

--> x = 7

Thus, the

**number of Ostriches = x = 7**

**Quantity II:**

Two Politicians can be selected at random from a group of 10 politician in

_{10}C

_{2}ways

----> (10*9)/(2*1) = 45 ways

Two politicians positioned

**adjacent to each other**can be selected from a group of 10 politician standing in a row in 9 ways

Hence,

**required probability**= 9/45 = 1/5 =

**0.2**

Here,

**Quantity I > Quantity II**

Workspace

59. Find the appropriate relation for quantity 1 and quantity 2 in the following question:

**Quantity I:**A gave one-fifth of the amount he had to B. B in turn gave half of what he received from A to C. If the difference between the remaining amount with A and the amount received by C is Rs. 700, how much money did B receive from A?**Quantity II:**Rs 250SHOW ANSWER

Correct Ans:Quantity I < Quantity II

Explanation:

**Quantity I:**

Let, initially A had Rs. x

Then, amount received by B (from A) = Rs. (x/5)

Now the remaining amount with A = x - (x/5) = 4x/5

Now, amount received by C (from B) = Rs. (x/5) * (1/2) = Rs. (x/10)

Given, remaining amount with A - amount received by C = Rs. 700

---> (4x/5) - (x/10) = 700

---> (8x - x)/10 = 700

---> 7x/10 = 700

--->

**x = Rs. 1000**--> which is the amount that initially A had.

Then,

**amount received by B (from A)**= Rs. (x/5) = (1000/5) =

**Rs. 200**

Given,

**Quantity II: Rs 250**

Here,

**Quantity I < Quantity II**

Workspace

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

Quantity I. p

Quantity II. q

Quantity I. p

^{2}= 81Quantity II. q

^{2}+ 19q + 90 = 0SHOW ANSWER

Correct Ans:quantity I ≥ quantity II

Explanation:

Let find the value of Quantity I , Quantity II

Quantity I. p

^{2}= 81

p = + 9, - 9;

Quantity II. q

^{2}+ 19q + 90 = 0

q

^{2}+ 10q + 9q + 90 = 0

(q+9)(q+10) = 0

q = -10, - 9;

**Hence, Quantity I ≥ Quantity II**

Workspace

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