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Answer is (C)

Because this relationship is only transitive

If “x is less than or equal to y” then this relation is reflexive and anti-symmetric.

Answer is (A)

Given sets are

A = {2, 3, 4} and B = {3, 4, 5, 6, 7, 8}

Relation R from A to B is defined as

“x divides y” ∀ x ∈ A, y ∈ B

So, when x = 2 ∈ A

Then 2 divides the element 4, 6, 8 of B.

So, (2, 4), (2, 6), (2, 8) ∈ R

when x = 3 ∈ A, then 3 divides elements 3, 6 of B

So, (3, 3), (3, 6) ∈ R

when x = 4 ∈ A then 4 divides the elements 4, 8 of B.

So, (4, 4), (4, 8) ∈ R

R = {(2, 4), (2, 6), (2, 8), (3, 3), (3, 6), (4, 4), (4, 8)}

R^-1 = {(4, 2), (6, 2), (8, 2), (3, 3), (6, 3), (4, 4), (8, 4)}

Given that, f : A → B and g : B → C be the bijective functions.

Let A = {1,3,4}, B ={2,5,1} and C = {3,1,2}

f : A → B is bijective function.

∴ f = {(1, 2), (3, 5), (4, 1)

f^-1 = {(2,1),(5,3),(1,4)}

g : B → C is bijective function.

∴ g = {(2, 3), (5, 1), (1, 4)}

g^-1 ={(3,2),(1,5),(4,1)}

Now,

gof (1) = g(f(1)) = g(2) = 3

gof (3) = g(f(3)) = g(5) = 1

gof (4) = g(f(4)) = g(1) = 4

∴ gof = {(1,3),(3,1),(4,4)} (1)

(gof)^-1 = {(3,1),(1,3),(4,4)} (2)

fog (2) = f(g(2)) = f(3) = 5

fog (5) = f(g(5)) = f(1) = 2

fog (1) = f(g(1)) = f(4) = 1

∴ fog = {(2,5),(5,2),(1,1)} (3)

(fog)^-1 = {(5,2),(2,5),(1,1)} (4)

f^-1og^-1 (3) = f^-1(g^-1(3)) = f^-1(2) = 1

f^-1og^-1 (1) = f^-1(g^-1(1)) = f^-1(5) = 3

f^-1og^-1 (4) = f^-1(g^-1(4)) = f^-1(1) = 4

∴ f^-1og^-1 = {(3,1),(1,3),(4,4)} (5)

g^-1of^-1 (2) = g^-1(f^-1(2)) = g^-1(1) = 5

g^-1of^-1 (5) = g^-1(f^-1(5)) = g^-1(3) = 2

g^-1of^-1 (1) = g^-1(f^-1(1)) = g^-1(4) = 1

∴ g^-1of^-1 = {(2,5),(5,2),(1,1)} (6)

On comparing 1,2,3,4,5 and 6 we observe that 2 and 5 are same.

i.e (g o f)^–1 = f^-1og^-1

The material to be used for making an electromagnet should have high permerability and low retentivity. Moreover, less energy should be utilised for magnetisation of the material.

Thus, material B should be used for making the core of an electromagnet

The material to be used for making an electromagnet should have high permerability and low retentivity. Moreover, less energy should be utilised for magnetisation of the material.

Thus, material B should be used for making the core of an electromagnet.

Not in direct proportion. (As the speed increases, time taken by the car reduces)

Direct proportion. (As the number of pens increases, their total cost also increases)

(A) 2 : 5

The comparison of two numbers or quantities by division is known as the ratio. Symbol ‘:’ is used to denote ratio.

Ratio of 8 books to 20 books = 8/20

Divide both numerator and denominator by 4.

= 2/5

Therefore, ratio of 8 books to 20 books = 2 : 5

Cost of 15 chairs = Rs 7500

Cost of 1 chair = Rs 7500/15 = Rs 500

Number of chairs that can be purchased for Rs 12000 = 12000/500

= 24 chairs

Comparing 1/4th of an inch and 3/8th of an inch.

1/4 = 2/8 < 3/8

Second plant grew more.

Pages read in 2 hours = 20

Then pages read in 1 hour = 20/2 = 10 pages.

Pages will be read in 8 hours = 8 × 10 = 80 pages.

Number of pages will be read in 8 hrs = 80 pages

We know that as number of note books increases, the cost also increases such that the ratio of number of note books and the ratio of their costs will remain the same. 

That means here number of note books and the cost are in direct proportion. 

Let the cost of 7 note books be ‘x’. 

Then, 4 : 80 = 7 : x 

If the ratios are equal, the product of means = The product of the extremes 4 × x = 80 × 7

⇒ x = 80 x 7/4 = ₹ 140

Thus, the cost of 7 note book is equal to ₹ 140.

Given the cost of 4 note books = ₹ 80 

Cost of 1 note book = 80 ÷ 4 = ₹ 20 

Cost of 7 note books = ₹ 20 × 7

= ₹ 140

(i) True

(ii) False

The correct answer is (A) 2:5

The correct answer is 3 : 1

(b) 3 : 4

Triangle has three sides and a rectangle has four sides.

The correct answer is (D) 1:3

The correct answer is (D) 1:8 

The correct answer is (a) 5 : 9 (b) 3 : 10 77.

(D) 1 : 8

From the question it is given that,

Width of a picture = 60 cm

Length of a picture = 1.8 m

We know that, 1 m = 100 cm

So, 1.8 m = 180 cm

Perimeter of rectangle = 2 (length + breadth)

= 2 (180 + 60)

= 2 (240)

= 480

Therefore, The ratio of its width to its perimeter in lowest form = 60/480

Divide both numerator and denominator by 20.

= 3/24

Again, divide both numerator and denominator by 3.

= 1/8

= 1 : 8

Given sides of a triangle are in the ratio 1: 2: 3

Perimeter = 36 cm

Sum of the terms of the ratio = 1 + 2 + 3 = 6

First side = (1/6) × 36

= 6 cm

Second side = (2/6) × 36

= 2 × 6

= 12 cm

Third side = (3/6) × 36

= 6 × 3

= 18 cm

The correct answer is 36 and 64

Sum of the ratios 3 + 4 + 5 = 12 

W. K. T sum of the angles of a triangle = 75° 

∴ \(\frac{3}{12}\) × 180 = 45°, 12 

\(\frac{3}{12}\) × 180 = 60°, 

\(\frac{3}{12}\) × 180 = 75° 

∴ The three angles of a triangle are 45°, 60°, 75°

(i) 2 : 5 (ii) 2 : 1 (iii) 1 : 2 (iv) 2 : 5

An article is sold at 40% gain

100 + 40 = 140

If the cost price is 100 then the selling price 100 + expenditure is 4x – 1000 + 40 = 140 

∴ \(\frac{SP}{CP} = \frac{140}{100} \)

= \(\frac{7}{5}\)

∴ SP : CP = 7 : 5

Correct option is (A) 8

Let the amount taken be Rs P.

\(\therefore\) S.I. = \(\frac{16}{25}\times\) P = \(\frac{16P}{25}\)

R = T

\(\therefore\) S.I. = \(\frac{PRT}{100}\)

\(\Rightarrow\) \(\frac{16P}{25}\) = \(\frac{P\times R\times R}{100}\)  \((\because\)  R = T)

\(\Rightarrow\) \(R^2=\) \(\frac{16P}{25}\times\frac{100}P\) \(=16\times4=64=8^2\)

\(\Rightarrow\) R = 8

\(\Rightarrow\) T = R = 8

Correct option is  A) 8

Correct option is (C) ₹ 2166.66

Let the amount be P.

R = \(12\frac{1}{2}\%=\frac{25}{2}\%\)

T = 1 year

Amount after 1 year \(=P(1+\frac R{100})^1\)

\(\Rightarrow\) \(P(1+\frac R{100})\) = 2437.50   (Given)

\(\Rightarrow\) \(P(1+\frac{25}{200})\) = 2437.50

\(\Rightarrow\) \(P(1+\frac18)\) = 2437.50

\(\Rightarrow\) \(\frac{9P}8\) = 2437.50

\(\Rightarrow\) P = 2437.50 \(\times\frac89=\frac{19500}9\) = 2166.66

\(\therefore\) Amount is Rs 2166.66

Correct option is  C) ₹ 2166.66

Correct option is (A) 7%

P = 500, T = 3 years. Let rate of interest be R

Amount after 3 years = P + S.I.

= P + \(\frac{PRT}{100}\)

\(=500+\frac{500\times R\times3}{100}\)

= 500 + 15R

Given that amount after 3 years is Rs 605.

\(\therefore\) 500 + 15R = 605

\(\Rightarrow\) 15R = 605 - 500 = 105

\(\Rightarrow\) R = \(\frac{105}{15}\) = 7%

Correct option is  A) 7%

Last year 

Cost of 1000 articles = ₹ 5000

∴ Cost of one article = \(\frac {5000}{1000} = ₹ 5\)

This year 

Cost of 1000 articles = ₹ 4000

∴ Cost of one article = \(\frac {4000}{1000} = ₹ 4\)

Decrease in price = ₹ 5 – ₹ 4 = ₹ 1 

Decrease In percentage = 1/5  × 100% = 20%

Let the cost price be ₹ 100. 

Then the marked price = 100 + 20 = ₹ 120 

Discount = 10%, so discount = x 120 = 12%  

SP = Marked price – Discount = ₹ 120 – ₹ 12 = ₹ 108 

Gain = 8/100 × 100 = 8% 

The shopkeeper gains 8% after discount.

Correct option is (B) ₹ 1800

P = Rs 5000

R = 12%

T = 3 years

S.I. = \(\frac{PRT}{100}\) \(=\frac{5000\times12\times3}{100}\)

= Rs 1800

Correct option is  B) ₹ 1800

Correct option is  A) ₹ 832

Correct option is (A) ₹ 832

\(C.I.=P(1+\frac R{100})^n-P\)

\(=5000(1+\frac8{100})^2-5000\)  \((\because\) P = Rs 5000, R = 8% and n = 2 years)

\(=5000(1+\frac{16}{100}+\frac{64}{10000}-1)\)

\(=5000\times\frac{1600+64}{10000}\)

\(=\frac{1664}2=832\)

\(\therefore\) Required C.I. = Rs 832

A) Loss percentage 1%

Correct option is (A) Loss percentage 1%

Let cost price of first cycle is Rs x and cost price of second cycle is Rs y.

\(\therefore\) Selling price of first cycle = cost price of first cycle + profit

= Rs (x + 10% of x)

\(=Rs(x+\frac x{10})=Rs\,\frac{11x}{10}\)

And selling price of second cycle = cost price of second cycle - loss

= Rs (y - 10% of y)

\(=Rs(y-\frac y{10})=Rs\,\frac{9y}{10}\)

Cost price of both cycles = Rs (x + y)

Selling price of both cycles \(=Rs\,(\frac{11x}{10}+\frac{9y}{10})\)

\(\because\) Given that both cycles are sold on equal price.

\(\therefore\) \(\frac{11x}{10}=\frac{9y}{10}\)

\(\Rightarrow\) 11x = 9y

\(\Rightarrow\) \(y=\frac{11x}9\)    ______________(1)

\(\therefore\) Profit in percent \(=\frac{\text{Selling price - cost price}}{\text{cost price}}\times100\%\)

\(=\cfrac{(\frac{11x}{10}+\frac{9y}{10})-(x+y)}{x+y}\times100\%\)

\(=\cfrac{\frac{x}{10}-\frac{y}{10}}{x+y}\times100\%\)

\(=\cfrac{x-\frac{11x}9}{10(x+\frac{11x}9)}\times100\%\)

\(=\cfrac{\frac{9x-11x}9}{\frac{9x+11x}9}\times10\%\)

\(=\frac{-2x}{20x}\times10\%\)

= -1%

Hence, overall loss 1%.

The correct answer is 23 : 47 

z/4 = - 1.5

z = - 1.5 x 4

z = - 6.0

(a) Kalidasa

(c) Copper coin

(a) Gold coin

(A) 65°27'16"

The perimeter of a certain sector of a circle is equal to the length of the arc of a semicircle having the same radius. The angle of the sector in degrees is 65°27'16"

This is due to the Newton's third law of motion. When the man jumps out of the boat, he applies a force on it in the backward direction and in turn, the reaction of the boat on the man pushes him out of the boat.

The radius of curvature of a mirror is 20 cm the focal length is 10 cm.

• Newton’s third law of motion states that ‘Every action force has an equal and opposite reaction force which acts simultaneously.’ 
• When the fuel in a rocket is ignited, it bums as a result of chemical reaction. 
• The exhaust gases escape with a great force in the backward direction. 
• It exerts an equal and opposite reaction force on the rocket, due to which the rocket moves in the forward direction. 
• Thus, the principle of launching of rocket is based on Newton’s third law of motion.

The unit of power of lens is diopter.

(a) Banyan Tree has roots called prop roots which are hanging structures that help in support.

(b) Turnip has tap roots which get swollen and store food.

(c) Mangrove trees are found in marshy area. The roots get modified into pneumatic structures providing extra passage to allow additional oxygen to the plant.

Root in some plants change their shape and structure and become modified to perform functions other than absorption and conduction of water and minerals. This change is called as the modification of root.

(C) mitochondria

D. Lm = 5/6 Lg

Correct answer is (A) Epistasis