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We have seen that the shaded region and origin are on the same side of the line 3x + 4y = 12.

For (0,0) we have 0 + 0 - 12 < 0. So the shaded region satisfies the inequality 3x + 4y \(\le\)12.

We have seen that the shaded region and origin are on the same side of the line 4x + 3y =12

For (0,0) we have 0 + 0 -12 < 0. So the shaded region satisfies the inequality 4x + 3y \(\le\)12.

Also , the region lies in the first quadrent Therefore x \(\ge\) 0 and y \(\ge\)0

Thus the linear inequation comprising the given solution set are +4y \(\le\)12, 4x + 3y \(\le\) 12, x \(\ge\) 0, y \(\ge\)0

True

Since, given graph is a line parallel to y-axis at a distance 3 units to the right of the origin. Hence, it represents a linear equation x = 3.

True.

Justification:

We have the equation, x + y = 0.

x + y = 0

x = – y

from the graph, we get the points ( – 3, 3) and ( – 1, 1),

Considering the point ( – 3, 3)

x = – 3 and y = 3

Hence, substituting ( – 3, 3) in equation,

We get,

– 3 = 3 which satisfies the equation x = – y

Considering the point ( – 1, 1)

x = – 1 and y = 1

Hence, substituting ( – 1, 1) in equation,

We get,

– 1 = 1 which satisfies the equation x = – y

Therefore, the given solution: ( – 3, 3) and ( – 1, 1) satisfies the given equation x = – y.

Hence, the given graph represents the linear equation x + y = 0.

Given: 

x + 2y = -1 ……. (1)

2x – 3y = 12 …….. (2) 

From equation (1) 

x + 2y = -1

⇒ x = – 1 – 2y 

Substituting x = – 1 – 2y in equation (2), we get

2 (- 1 – 2y) – 3y = 12

– 2 – 4y – 3y = 12 

– 2 – 7y = 12 

7y = – 2 – 12 

∴ y = \(\frac{-14}{7}\)= -2 

Substituting y = – 2 in equation (1), we get 

x + 2 (- 2) = – 1 

x = – 1 + 4 

x = 3 

 ∴ The solution is (3, – 2)

x - y + z = 4

x + y + z = 2

2x + y - 3z = 0

Adding eq 1 and eq2

⇒ 2(x + z) = 6

⇒ x = 3 – z

(3) – (2) – (1)

⇒ y – 5z = - 4

⇒ y = - 4 + 5z

Substituting in eq2

⇒ 3 – z - 4 + 5z + z = 2

⇒ 5z = 3

⇒ z = 3/5

Thus,

x = 3 – z

⇒ x = 12/5 Y = 5z – 4

⇒ y = -1

The given pair of equations are: 

x + 2y = \(\frac{3}{2}\) …………(i) 

2x + y = \(\frac{3}{2}\)………….(ii) 

Let us eliminate y from the given equations. The coefficients of y in the given equations are 2 and 1 respectively. The L.C.M of 2 and 1 is 2. So, we make the coefficient of y equal to 2 in the two equations.

Multiplying equation (i) x 1 and (ii) x 2 

⇒ x + 2y = \(\frac{3}{2}\) ……………(iii) 

4x + 2y = 3 ………………. (iv) 

Subtracting equation (iii) from (iv) 

(4x – x) + (2y - 2y) = 3x = 3 – (\(\frac{3}{2}\)) 

⇒ 3x = \(\frac{3}{2}\) 

⇒ x = \(\frac{1}{2}\)

Putting x = \(\frac{1}{2}\) in equation (iv) 

4(\(\frac{1}{2}\)) + 2y = 3 

⇒ 2 + 2y = 3 

∴ y= \(\frac{1}{2}\)

The solution of the system of equation is x = \(\frac{1}{2}\) and y = \(\frac{1}{2}\).

The given pair of equations are:

x + y/2 = 4 …………(i) 

2y + x/3 = 5…………(ii) 

From (i) we get, 

x + y/2 = 4 

⇒ 2x + y = 8 [After taking LCM] 

⇒ y = 8 – 2x …..(iv) 

From (ii) we get, 

x + 6y = 15 …………(iii) [After taking LCM] 

Substituting y in (iii), we get 

x + 6(8 – 2x) = 15 

⇒ x + 48 – 12x = 15 

⇒ -11x = 15 – 48 

⇒ -11x = -33

⇒ x = 3 

Putting x = 3 in (iv), we get 

y = 8 – (2 x 3) 

∴ y = 8 – 6 = 2 

Hence, the solution of the given system of equation are x = 3 and y = 2 respectively.

Given eq, is 3x + 5y - 2 = 0 ....(i) 

Taking L.H.S. = 3x + 5y - 2 = 3 × 2 + 5 × (-1) - 2 = 6 - 5 - 2 = -1 ≠ 0 

Here L.H.S. ≠ R.H.S. therefore x = 2, y = -1 is not a solution of given equations. 

Taking \(\frac{1}{x}\) = u 

Then the two equation becomes, 

4u + 3y = 8…………(i) 

6u – 4y = -5……….(ii) 

From (i), we get 

4u = 8 – 3y 

⇒ u = \(\frac{(8 − 3y)}{4}\) …….. (iii) 

Substituting u in (ii) 

[6\(\frac{(8 − 3y)}{4}\)] – 4y = -5 

⇒  [\(\frac{3(8−3y)}{2}\)] − 4y = −5 

⇒ 24 − 9y −8y = −5 x 2 [After taking LCM] 

⇒ 24 – 17y = -10 

⇒ -17y =- 34 

⇒ y = 2 

Putting y = 2 in (iii) we get, 

u = \(\frac{(8 − 3(2))}{4}\)

⇒ u = \(\frac{(8 − 6)}{4}\) 

⇒ u = \(\frac{2}{4}\) = \(\frac{1}{2}\) 

⇒ x = \(\frac{1}{u}\) = 2 

∴ x = 2 

So, the solution of the pair of equations given is x=2 and y =2.

The given pair of equations are: 

x/3 + y/4 = 11………(i)

5x/6 − y/3 = −7…….(ii) 

From (i), we get 

x/3 + y/4 = 11 

⇒ 4x + 3y = (11 x 12) [After taking LCM] 

⇒ 4x =132 – 3y 

⇒ x = \(\frac{(132 – 3y)}{4}\)……. (iv) 

From (ii), we get 

5x/6 − y/3 = −7 

⇒ 5x – 2y = -42 ………(iii) [After taking LCM] 

Now, substituting x in equation (iii) we get, 

5[\(\frac{(132 − 3y)}{4}\)] – 2y = -42 

⇒ 660 – 15y – 8y = -42 x 4 [After taking LCM] 

⇒ 660 + 168 = 23y 

⇒ 23y = 828 

⇒ y = 36 

Now, putting the value of y in the equation (iv) 

x = \(\frac{(132 – 3(36))}{4}\)

⇒ x = \(\frac{(132 − 108)}{4}\) = \(\frac{24}{4}\) 

∴ x = 6 

Thus, the value of x and y obtained are 6 and 36 respectively.

The given pair of equations are:

x/7 + y/3 = 5……..(i)

x/2 – y/9 = 6…….(ii) 

From (i), we get 

x/7 + y/3 = 5 

⇒ 3x + 7y = (5 x 21) [After taking LCM] 

⇒ 3x = 105 – 7y 

⇒ x = \(\frac{(105 – 7y)}{3}\)……. (iv) 

From (ii), we get 

x/2 – y/9 = 6 

⇒ 9x – 2y = 108 …………(iii) [After taking LCM] 

Now, substituting x in equation (iii) we get, 

9[\(\frac{(105 − 7y)}{3}\)] – 2y = 108 

⇒ 945 – 63y – 6y = 324 [After taking LCM] 

⇒ 945 – 324 = 69y 

⇒ 69y = 621 

⇒ y = 9 

Now, putting the value of y in the equation (iv) 

x = \(\frac{(105 − 7(9))}{3}\)

⇒ x = \(\frac{(105 − 63)}{3}\) = \(\frac{42}{3}\) 

∴ x = 14 

Thus, the value of x and y obtained are 14 and 9 respectively.

The given two linear equations are 

2x – 7y = 3 …………. (1) 

4x + y = 21 …………..(2) 

From the equation (2) we get 

y = 21 – 4x now we substitute this Y 

value in equation (1) 

We get 

2x – 7(21 – 4x) = 3 

⇒ 2x- 147 + 28 x – 3 

⇒ 30 x = 147 + 3 = 150 

then x = 150/30 ∴ x = 5

 Now put x = 5 in equation (2) we get 

4(5) + y = 21 

20 + y = 21 

y = 21 – 20 = 1 

So x = 5 and y = 1 are the solutions of the system.

The given pair of equations are: 

7(y+3) – 2(x+2) = 14……..(i)

4(y-2) + 3(x-3) = 2…….. (ii) 

From (i), we get 

7y + 21 – 2x – 4 = 14 

7y = 14 + 4 – 21 + 2x

⇒ y = \(\frac{(2x − 3)}{7}\)

From (ii), we get 

4y – 8 + 3x – 9 = 2 

4y + 3x – 17 – 2 = 0 

⇒ 4y + 3x – 19 = 0 ……(iii) 

Now, substituting y in equation (iii) 

4[\(\frac{(2x − 3)}{7}\)] + 3x – 19 = 0 

8x – 12 + 21x – (19 x 17) = 0 [after taking LCM] 

29x = 145 

⇒ x = 5 

Now, putting the value of x and in the equation (ii) 

4(y - 2) + 3(5 - 3) = 2

⇒ 4y - 8 + 6 = 2

⇒ 4y = 4 

∴ y = 1 

Thus, the value of x and y obtained are 5 and 1 respectively.

 \(\frac{10}{x-y}\)+\(\frac{2}{x-y}\) = 4

\(\frac{15}{x+y}\) - \(\frac{9}{x-y}\) = - 2

Multiplying eq1 by 9 and eq2 by 2 and adding

⇒ 120/(x + y) = 32

⇒ x + y = 15/4 ---- (1)

Multiplying eq1 by 3 and eq2 by 2 and subtracting

⇒ 24/(x – y) = 16

⇒ x – y = 3/2 ---- (2)

Adding (1) and (2)

⇒ 2x = 21/4

⇒ x = 21/8

Thus,

21/8 – y = 3/2

⇒ y = 9/8

 \(\frac{x}{2}\)+ y = 0.8

\(\cfrac{7}{x+\frac{y}{2}}\) = 10

⇒ x + y/2 = 0.7 

Multiplying eq1 by 2 and subtracting eq2 from it.

⇒ x + 2y – x – y/2 = 1.6 – 0.7

⇒ 3y/2 = 0.9

⇒ y = 0.6

Thus,

x/2 + 0.6 = 0.8

⇒ x = 0.4

The given pair of equations are: 

\(\frac{x}{2}\) + y = 0.8 

⇒ x + 2y = 1.6…… (a)

7/(x + \(\frac{ y}{2}\)) = 10 

⇒7 = 10(x +\(\frac{ y}{2}\)) 

⇒7 = 10x + 5y 

Let’s, multiply LHS and RHS of equation (a) by 10 for easy calculation 

So, we finally get 

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

And, 10x + 5y = 7 ……(ii) 

Now, subtracting two equations we get, 

⇒ (i) – (ii) 

15y = 9 

⇒ y = \(\frac{3}{5}\) 

Next, putting the value of y in the equation (i) we get, 

x = [16 − 20(3/5)]/10

⇒  \(\frac{(16 – 12)}{10}\) = \(\frac{4}{10}\) 

∴ x = \(\frac{2}{5}\) 

Thus, the value of x and y obtained are \(\frac{2}{5}\) and \(\frac{3}{5}\) respectively.

The given pair of equations are: 

3x – 7y + 10 = 0 ……….(i)

y – 2x – 3 = 0 …………..(ii) 

From (ii) 

y – 2x – 3 = 0 

y = 2x + 3 ………………(iii) 

Now, substituting y in equation (i) we get, 

⇒ 3x – 7(2x + 3) + 10 = 0

⇒ 3x – 14x – 21 + 10 = 0

⇒ -11x = 11 

⇒ x = -1 

Next, putting the value of x in the equation (iii) we get, 

⇒ y = 2(-1) + 3 

∴ y= 1 

Thus, the value of x and y is found to be -1 and 1 respectively.

 \(\frac{5}{x+y}\) - \(\frac{2}{x-y}\) = - 1

\(\frac{15}{x+y}\) + \(\frac{7}{x-y}\) = 10

Multiplying eq1 by 3 and subtracting from eq2

⇒ 13/(x – y) = 13

⇒ x – y = 1 ------ (3)

Multiplying eq1 by 7 and eq2 by 2 and adding

⇒ 65/(x + y) = 13

⇒ x + y = 5 ------- (4)

Thus,

2x = 6

⇒ x = 3

y = x – 1

⇒ y = 2

The given pair of equations are: 

(x + y)/xy = 2 

⇒ 1/y + 1/x = 2……. (i) 

(x – y)/xy = 6 

⇒ 1/y – 1/x = 6………(ii) 

Let 1/x = u and 1/y = v, so the equation (i) and (ii) becomes 

v + u = 2……. (iii) 

v – u = 6……..(iv) 

Adding (iii) and (iv), we get 

2v = 8 

⇒ v = 4

⇒ y = 1/v = \(\frac{1}{4}\) 

Substituting v = 4 in (iii) to find x, 

4 + u = 2 

⇒ u = -2 

⇒ x = 1/u = -\(\frac{1}{2}\) 

Hence, the solution is x = -\(\frac{1}{2}\) and y = \(\frac{1}{4}\).

7(y+3) - 2(x+2) = 14

7y + 21 - 2x - 4 = 14

7y - 2x = 14 - 21 + 4

⇒ 7y – 2x = -3 ---- (1)

4(y - 2) +3(x - 3)  = 2

4y - 8 + 3x - 9 = 2

4y + 3x = 2 + 8 + 9

⇒ 4y + 3x = 19 ------ (2)

Multiplying eq1 by 3 and eq2 by 2 and adding them

3(7y – 2x) + 2(4y + 3x ) = - 3 (3) + 19(2)

⇒ 21y – 6x + 8y + 6x = - 9 + 38

⇒ 29 y = 29

⇒ y = 1

Put the value of y in (1) to get,

7 – 2x = - 3- 2x = - 3 - 7 - 2x = - 10

⇒ x = 5

The given pair of equations are: 

0.4x + 0.3y = 1.7 

0.7x – 0.2y = 0.8 

Let’s, multiply LHS and RHS by 10 to make the coefficients as an integer 

4x + 3y = 17 …………(i) 

7x – 2y = 8 …………..(ii) 

From (ii) 

7x – 2y = 8 

x = \(\frac{(8 + 2y)}{7}\)…………(iii) 

Now, substituting x in equation (i) we get,

⇒ 4[\(\frac{(8 + 2y)}{7}\)] + 3y = 17 

⇒ 32 + 8y + 21y = (17 x 7) 

⇒ 29y = 87 

⇒ y = 3 

Next, putting the value of y in the equation (iii) we get, 

⇒ x = \(\frac{(8 + 2(3))}{7}\) 

⇒ x = \(\frac{14}{7}\) 

∴ x = 2 

Thus, the value of x and y is found to be 2 and 3 respectively.

Let unit’s digit = y

and the ten’s digit = x

So, the original number = 10x + y

The sum of the number = 10x + y

The sum of the digit = x + y

According to the question,

x + y = 15 …(i)

After interchanging the digits, the number = x + 10y

and 10x + y + 9 = x + 10y

⇒ 10x + y + 9 = x + 10y

⇒ 10x – x + y – 10y = – 9

⇒ 9x – 9y = – 9

⇒ x – y = – 1 …(ii)

On adding Eq. (i) and (ii) , we get

x + y + x – y = 15 – 1

⇒ 2x = 14

⇒ x = 7

On substituting the value of x = 5 in Eq. (i), we get

x + y = 15

⇒ 7 + y = 15

⇒ y = 8

So, the Original number = 10x + y

= 10×7 + 8

= 70 + 8

= 78

Hence, the two digit number is 78.

(i) Mode = 7 Since 7 occurs 4 times 

(ii) Mode = 11 Since it occurs 4 times

Mode is 122 cm because it occur maximum number of times. i.e. frequency is 18.

In ∆ABC

AD is the bisector of ∠BAC

⇒ ∠1 = ∠2 …(i)

Also in ∆ADC

∠3 = ∠2 + ∠C (ext. angle is equal to sum of opposite interior angles)

⇒ ∠3 > ∠2 (ext. angle is greater than one of the interior angles)

⇒ But ∠1 = ∠2

⇒ ∠3 > ∠1

⇒ AB > BD

Hence proved.

Answer is (D) 3.4 cm

Answer is (A) AB

Answer is (C) less

∠ABC = 180° – 135° = 45°,

∠ACB = 65°

i.e. ∠ACB > ∠ABC,

⇒ AB > AC.

Answer is (A) greater

Answer is (D) less

Answer is (A) greater

Since, maximum frequency 4 corresponds to value 7.2, then mode = 7.2

Given ratio of incomes of two persons = 9 : 7 

So let the incomes of each = Rs. 9x and Rs. 7x and ratio of expenditures = 4 : 3 

So let the expenditures of each = 4y and 3y then earnings of each = (income – expenditure) of each 

⇒ 9x – 4y = Rs. 2000 and 7x – 3y = 200 

∴ 9x – 4y = 7x – 3y = 2000 

⇒ 9x – 7x = 4y – 3y 

⇒ y = 2x 

now putting y = 2x in 9x – 4y = 2000 

we get 9x – 4(2x) = 2000 ⇒ x = 2000 

∴ Income of each = 9x = 9(2000) = 18000 and 7x = 7(2000) = 14,000

Let us denote the incomes of the two person by Rs 9x and Rs 7x and their expenditures by Rs 4y and Rs 3y respectively. Then the equations formed in the situation is given by :

9x – 4y = 2000 (1)

and 7x – 3y = 2000 (2)

Step 1: Multiply Equation (1) by 3 and Equation (2) by 4 to make the coefficients of y equal. Then we get the equations:

27x – 12y = 6000     .........(3)

28x – 12y = 8000   .........(4)

Step 2: Subtract Equation (3) from Equation (4) to eliminate y, because the coefficients of y are the same. So, we get

(28x – 27x) – (12y – 12y) = 8000 – 6000

i.e., x = 2000

Step 3: Substituting this value of x in (1), we get 

9(2000) – 4y = 2000

i.e., y = 4000

So, the solution of the equations is x = 2000, y = 4000. Therefore, the monthly incomes of the persons are Rs 18,000 and Rs 14,000, respectively.

Verification : 18000 : 14000 = 9 : 7. Also, the ratio of their expenditures =

18000 – 2000 : 14000 – 2000 = 16000 : 12000 = 4 : 3

Let the ten’s and the unit’s digits in the first number be x and y, respectively.

So, the first number may be written as 10 x + y in the expanded form (for example, 56 = 10(5) + 6).

When the digits are reversed, x becomes the unit’s digit and y becomes the ten’s digit. This number, in the expanded notation is 10y + x (for example, when 56 is reversed, we get 65 = 10(6) + 5).

According to the given condition.

(10x + y) + (10y + x) = 66

i.e., 11(x + y) = 66

i.e., x + y = 6  ..... (1)

We are also given that the digits differ by 2, therefore,

either x – y = 2   ........(2)

or y – x = 2   ..........(3)

If x – y = 2, then solving (1) and (2) by elimination, we get x = 4 and y = 2.

In this case, we get the number 42.

If y – x = 2, then solving (1) and (3) by elimination, we get x = 2 and y = 4.

In this case, we get the number 24.

Thus, there are two such numbers 42 and 24.

3375 x 512 = 3√(3375 x 512)

= 3√(15 x 15 x 15) x (8 x 8 x 8)

= 15 x 8 = 120

Work done in an electric field is independent of the path and depends only on the initial and final positions.

Here initial and final points are coincident.

Work =q × (V_final – V_initial)

= q (VA – VA)

So, net work done is zero.

(c) decreases because the charge moves along the electric field.

The amount of charge on the capacitor plates is 8 μ C

(c) In A : V remains same and hence Q changes.

(d) In B : Q remains same and hence V changes.

sphere is more.

There is attractive force between an electron and a proton, therefore when they come nearer, the work is done by the system itself and so the potential energy of system decreases.

(1) Its mass,

(2) Distribution of mass about the axis of rotation,

(3) Its shape and size, and 

(4) The position and orientation of the axis of rotation.