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(a) Equal denominators too have been added. (b) Numerators and denominators have been added.

Number of small cubes removed = 1 + 1 + 1 + 1 + 1 +1 + 1 = 7

So, required fraction = 7/20

(ii) Required fraction = 7/27

(iii) Required part is

7/27 ÷ 7/27 =  7/27 x 20/7 = 20/27

The correct answer is 2 1/4 m

Correct answer is

1/6 part of work,

5/6 part of work, complete work

Length of cloth bought for shirt = 3.40 

Length of cloth bought for skirt = 1.10 

Total length of cloth bought by Akshara = 3.40 + 1.10 = 4.50 cm

We know that, 

Volume of cuboid = Length × Breadth × Height 

Firstly, we have to find out volume of cardboard box 

Volume of cardboard box = 120 × 72 × 54 

= 466560 cm³ 

Now, 

Volume of each bar of soap = 6 × 4.5 × 4 

= 108 cm³ 

Therefore,

Total number of bars of soap that can be accommodated in that box 

= \(\frac{Volume\,of\,the\,box}{Volume\,of\,each\,soap}\)

= \(\frac{466560}{108}\)

= 4320 bars

Correct answer is 5/12 part

Correct answer is 4 shirts

Length of rectangle = 20 m2 

Breadth of rectangle = 15 m² 

Area of rectangle = 20 ×15 = 300 m² 

Radius of quadrant = 3.5 m² 

Area of quadrant = \(\frac{1}4\timesπr^2\)

Area of quadrant = \(\frac{1}4\times\frac{22}7\times3.5\times3.5\) 

= \(\frac{19.25}2\) m²

Area of 4 quadrant = \(4\times\frac{19.25}2\) m² = 2 ×19.25 = 38.50 m² 

Area of remaining part = (area of rectangle-area of 4 quadrant)

 Area of remaining part = 300 – 38.50 = 261.5 m²

Length of the rectangular cardboard = 14 cm and

Breadth of the rectangular cardboard = 7 cm

Area of cardboard = Area of rectangle = length × breadth = 14 × 7 = 98 cm²

Let the two circles with equal radii and maximum area have a radius r each.

2r = 7 or r = 7/2 cm

Again,

Area of two circular cut outs = 2 × πr² 

= 2 x 22/7 x (7/2)² 

= 77 cm²

Now,

The area of remaining cardboard = Area of cardboard – Area of two circular cut outs

= 98 – 77

= 21

Therefore, area of remaining cardboard is 21 cm².

(c) 10

F₁P + F₂P = 2a 

Here the equation is 16x² + 25y² = 400

(÷ by 400) ⇒ (16x²/400) + (25y²/400) = 1 

⇒ (x²/25) + (y²/16) = 1

a² = 25 => a = 5 

∴ 2a = 10

(i) 1/2 of 12 = 1/2 x 12 = 12/3 = 4

(ii) 2/5 of 15 = 2/5 x 15 = 30/5 = 6

(i) 20.1 × 4

20.1 × 4 = 80.4

(ii) 0.05 × 7

0.05 × 7 = 0.35

(iii) 211.02 × 4

211.02 × 4 = 844.08

(iv) 2 × 0.86

2 × 0.86 = 1.72

Correct option is: (B) 2

(i) Reciprocal of \(\frac 58 = \frac 85\)

(ii) Reciprocal of \(\frac 87= \frac 78\)

(iii) Reciprocal of \(\frac {13}8 =\frac 8{13}\)

(iv) Reciprocal of \(\frac 34 = \frac 43\)

Distance covered by Raja in 1 second = 1 \(\frac 12\)meters

Distance covered by Raja in 15 minutes 15 x 60 x   1 \(\frac 12\) 

= 900 x 3/2 = \(\frac {3}{2} = \frac {900\times3}{21}\) =1350 m

Let the side of an equilateral triangle be x cm. 

Perimeter of an equilateral triangle = 3x.

= 3 × 53/10

= 3/1 x 53/10

= \(\frac {3\times53}{1\times10}\)

∴ Perimeter of triangle = \(\frac {159}{10}\) or \(5\frac{9}{10}\) cm

Given, the length of a garden = m. 

breadth of a garden = 27/m. 

Area of rectangular garden = length × breadth

= \(\frac {27}{2}\times \frac {15}{2}\)

= \(\frac {27\times15}{2\times2}\)

∴ Area of the garden = 405 or 101\(\frac 14\)

Linear Equation

An equation which can be put in the form ax + by + c = 0, where a, b and c are real numbers and both a and b are nonzero is called a linear equation in two variables.

Solution of an Equation

Each solution (x, y) of a linear equation in two variables.

ax + by + c = 0, corresponds to a point on the line representing the equation, and vice-versa.

Correct option is (D) Natural numbers

The equation 2x + 5y = 7 has a unique solution, if x, y are natural numbers.

D) Natural numbers

(b) 3.42

12.28 x 1.5 - \(\frac{36}{2.4}\) = 18.42 – 15 = 3.42

(b) 1, 5, 2

\(= 2+\frac{1}{x+\frac{1}{y+\frac{1}{z}}}=\frac{37}{13} = 2\frac{11}{13}= 2+\frac{11}{13}\)

 \(⇒\frac{1}{x+\frac{1}{y+\frac{1}{z}}}= \frac{11}{13}⇒x+\frac{1}{y+\frac1z}=\frac{13}{11}\)

⇒ \(x+\frac{1}{y+\frac1z}=1+\frac2{11}\)

⇒ x = 1, y = \(\frac1z=\frac{11}{2}=5\frac12=5+\frac12\)

⇒ x = 1, y = 5, z = 2

(i) 2.56 place value of 2 in the decimal number 2.56 = 2 ones place)

(ii) 21.37 place value of 2 in the decimal number 21.37 = 2 × 10 = 20. (2 is in tens place)

(i) 7 > 0.7 (∵ 7 = 7.0)

(ii) 1.49 > 1.37

1. C) 20 Units

2. A) y = 3x

3. D) 60

4. D) 15

5. B) 33.3

(c) \(\frac{(0.011)^2}{1.1^2\times0.012}\)

\(\frac{(0.11)^2}{(1.1)^2\times0.1}\) = \(\frac{0.0121}{1.21\times0.1} = \frac{0.0121}{0.121} = 0.1;\)

\(\frac{(1.1)^2}{11^2\times(0.01)^2}\) = \(\frac{1.21}{121\times0.0001} = \frac{0.01}{0.0001} = 100;\)

\(\frac{(0.011)^2}{1.1^2\times0.012}\) = \(\frac{0.000121}{1.21\times0.0001} = 1;\)

\(\frac{(0.11)^2}{11^2\times0.01}\) = \(\frac{0.0121}{121\times0.01} = \frac{0.0121}{1.21} = 0.01;\)

Hence, option (c) is the correct answer.

A fraction with denominator greater than the numerator is called a Proper fraction.

(d) 100

Given exp. = (7.5)² + 2 × 7.5 × 2.5 + (2.5)² 

= (7.5 + 2.5)² = 10² = 100.

(b) 299

N = \(0.735\overline{45}\) = \(\frac{73545 -735}{99000}=\frac{72810}{99000}=\frac{809}{1100}\)

∴ Required difference = 1100 – 809 = 299.

Yes, The value of fraction is increase

A) Proper fraction

1. Improper fraction

2. 1

3. Like fractions

Clearly (2, 0) and (-1, 3) satisfy the equation y = -x +2

The equation whose graph is given by y = -x +2

Given, 

The point (2,-2) lies on the graph of the linear equation 

5x + ky = 4 

⇒ 5 × 2 – 2k = 4 

⇒ 2k = 6 

⇒ k = 3

From graph, co-ordinates (1, -1) and (-1, 1) are solutions of one of the equations. 

We will put the value of all the co-ordinates in each equation and check which equation satisfy them.

(i) y = x 

Put x = 1 and y = -1,

Thus, 1 ≠ -1 

L.H.S ≠ R.H.S

Putting x = -1 and y = 1, 

-1 ≠ 1 

L.H.S ≠ R.H.S 

Therefore, y = x does not represent the graph in the given figure.

(ii) x + y = 0 

Putting x = 1 and y = -1,

⇒ 1 + (-1) = 0 

⇒ 0 = 0 

L.H.S = R.H.S 

Putting x = -1 and y = 1, 

(-1) + 1 = 0 

0 = 0 

L.H.S = R.H.S 

Thus, the given solutions satisfy this equation. 

(iii) y = 2x 

Putting x = 1 and y = -1 

-1 = 2 (Not True) 

Putting x = -1 and y = 1 

1 = -2 (Not True) 

Thus, the given solutions does not satisfy this equation. 

(iv) 2 + 3y = 7x 

Putting x = 1 and y = -1 

2 – 3 = 7 

-1 = 7 (Not true)

Putting x = -1 and y = 1 

2 + 3 = -7 

5 = -7 (Not True) 

Thus, the given solutions does not satisfy this equation.

Clearly (-1, 1) and (1, -1) satisfy the equation x + y = 0

The equation whose graph is given by x + y = 0

Given: (-1, 3) and (2, 0) are the solution of one of the following given equations.

Check which equation satisfy both the points.

(i) y = x + 2 

Putting, x = -1 and y = 3 

3 ≠ – 1 + 2 

L.H.S ≠ R.H.S 

Putting, x = 2 and y = 0 

0 ≠ 4 

L.H.S ≠ R.H.S 

Thus, this solution does not satisfy the given equation.

(ii) y = x – 2 

Putting, x = -1 and y = 3 

3 ≠ – 1 – 2 

L.H.S ≠ R.H.S 

Putting, x = 2 and y = 0 

0 = 0 

L.H.S = R.H.S 

Thus, the given solutions does not satisfy this equation completely. 

(iii) y = – x + 2 

Putting, x = – 1 and y = 3 

3 = – ( – 1 ) + 2 

L.H.S = R.H.S 

Putting x = 2 and y = 0 

0 = -2 + 2 

0 = 0 

L.H.S = R.H.S 

Therefore, (0, 2) and (-1,3) satisfy this equation. 

Hence, this is the graph for equation y = -x + 2.

The line passes through (-1, 3) and (2, 0) 

(i) y = x + 2 

⇒ 3 = -1 + 2 

⇒ 3 = 1 

Which is not true 

(ii) y = x – 2 

⇒ 3 = -1 – 2 

⇒ 3 = -3 which is not true 

(iii) y = -x + 2 

⇒ 3 = 1 + 2 

⇒ 3 = 3 

Also, 

for (2, 0) 

⇒ 0 = -2 + 2 

⇒ 0 = 0 

Thus, 

y = - x + 2 is the equation 

(iv) x + 2y = 6 

⇒ -1 + 6 = 6 

⇒ 5 = 6 

Which is not true

From the graph, 

The line passes through (1, - 1) and (-1, 1)

(i) y = x 

⇒ 1 = - 1 

Which is not true 

(ii) x + y = 0 

⇒ 1 – 1 = 0 

⇒ 0 = 0 

Also, 

- 1 + 1 = 0 

⇒ 0 = 0 

Thus. 

x + y = 0 is a equation 

(iii) y = 2x 

⇒ -1 = 2 

Which is not true 

(iv) 2 + 3y = 7x 

⇒ 2 – 3 = 7 

⇒ -1 = 7 

Which is not true

The given equations are:

\(\frac{5}x\) + 6y = 13 ……..(i)

\(\frac{3}x\) + 4y = 7 ……..(ii)

Putting \(\frac{1}x\) = u, we get:

5u + 6y = 13 …….(iii) 

3u + 4y = 7 ……(iv) 

On multiplying (iii) by 4 and (iv) by 6, we get: 

20u + 24y = 52 ……..(v) 

18u + 24y = 42 ……..(vi) 

On subtracting (vi) from (v), we get: 

2u = 10 ⇒ u = 5

⇒ \(\frac{1}x\) = 5 ⇒ x = \(\frac{1}5\)

On substituting x = \(\frac{1}5\) in (i), we get:

\(\frac{\frac{5}1}{3}\) + 6y = 13 

25 + 6y = 13 

6y = (13 – 25) = -12 

y = -2 

Hence, the required solution is x = \(\frac{1}5\) and y = -2.

a) 

b) 

c) 

We know that, s_n − s_n = a_n,

which is the n^th term of А.Р.

Here, a = 8, d = 3

∴ a_n = T_n = a + (n − 1)d

= 8 + (n − 1)3

= 3n + 5

Given series 81 + 82 + 83 + …. + 89 + 90

∴ T_n = I = a + (n − 1)d

∴ 90 = 81 + (n − 1). 1

(here, a = 81, I = 90 and d = 1)

⇒ n = 9

Now,

S_n = \(\frac{n}{2}\)[2a + (n − 1)d]

= \(\frac{10}{2}\)[2 × 81 + (10 − 1) × 1]

= 5[162 + 9]

= 5 × 171 = 855

Given 100 can be written as 10² According to the property of perfect square, for any natural number n, we have sum of first n odd natural numbers = n² Expressing 100 as a sum of 10 odd numbers, we get 100 = (10)²

Here n=10

By applying the law 81 = (1+3+5+7+9+11+13+15+17+19)

Given 6 is the smallest number in Pythagorean triplet But we have for every number m>1, we have (2m, m²-1, m²+1) as a Pythagorean triplet Using the above condition we can write as 2m = 6

m=6/2=3

Then m² = 9

m² – 1 = 9 – 1 = 8

m² + 1 = 9 + 1 =10

Therefore the Pythagorean triplet is = (2m, m²-1, m²+1) = (6, 8, 10)

Given 20 is the smallest number in Pythagorean triplet But we have for every number m>1, we have (2m, m²-1, m²+1) as a Pythagorean triplet Using the above condition we can write as 2m = 20

m= 20/2=10

Then m² =100

m² – 1 = 100 – 1 = 99

m² + 1 = 100 + 1 =101

Therefore the Pythagorean triplet is = (2m, m²-1, m²+1) = (20, 99, 101)

Given 16 is the smallest number in Pythagorean triplet But we have for every number m>1, we have (2m, m²-1, m²+1) as a Pythagorean triplet Using the above condition we can write as 2m = 16

m= 16/2=8

Then m² =64

m² – 1 = 64 – 1 = 63

m² + 1 = 64 + 1 =65

Therefore the Pythagorean triplet is = (2m, m²-1, m²+1) = (16, 63, 65)

Given 14 is the smallest number in Pythagorean triplet But we have for every number m>1, we have (2m, m²-1, m²+1) as a Pythagorean triplet Using the above condition we can write as 2m = 14

m= 14/2=7

Then m² =49

m² – 1 = 49 – 1 = 48

m² + 1 = 49 + 1 =50

Therefore the Pythagorean triplet is = (2m, m²-1, m²+1) = (14, 48, 50)

496 के अपवर्तक – 1, 2, 4, 8, 16, 31, 62, 124, 248, 496

सभी गणुनखण्डों का योग = 1 + 2 + 4 + 8 + 16 + 31 + 62 + 124 + 248 + 496 = 992

(496 x 2) अर्थात् 496 का दो गुना है

अतः 496 एक सम्पूर्ण संख्या है।