Explore topic-wise InterviewSolutions in Current Affairs.

The given equation is 2x – 5y = 10 

On substituting (0, 2), the L.H.S becomes 

2(0)-5(2) = 0-10 = -10 

R.H.S = 10 

L.H.S ≠ R.H.S 

∴ (0, 2) is not a solution.

Substituting (5, 0) in the 

L.H.S of 2x – 5y = 10, 

we get 

2(5)-5(0) = 10-0 = 10 = R.H.S 

∴ (5, 0) is a solution.

3x + 4y = k 

If x = -1, y = 2 is a solution of the equation, then 

⇒ 3 × -1 + 4 × 2 = k 

⇒ k = 5

Correct answer = (a) parallel

If a pair of linear equations in two variables is inconsistent, then no solution exists as they have no common point. And, since there is no common solution, their graph lines do not intersect. 

Hence, they are parallel.

Correct answer = (d) 1/x - 1/y = 0

Given: 

x = -y and y > 0 

Now, we have: 

(i) x² y 

On substituting x = -y, we get: 

(-y)² y = y³ > 0 (∵ y > 0) 

This is true. 

(ii) x + y 

On substituting x = -y, we get: 

(-y) + y = 0 

This is also true. 

(iii) xy 

On substituting x = -y, we get:

(-y)y = -y² (∵ y > 0) 

This is again true. 

(iv) 1/x - 1/y = 0

⇒ y−x/xy = 0 

On substituting x = -y, we get: 

y−(−y) (−y)y = 0 ⇒ 2y/−y² = 0 ⇒ 2y = 0 ⇒ y = 0.

The given equations are:

\(\frac{7-4x}3\) = y

⇒ 4x + 3y = 7 ……..(i) 

and 2x + 3y + 1 = 0 

⇒ 2x + 3y = -1 ……….(ii) 

On subtracting (ii) from (i), we get: 

2x = 8 

⇒ x = 4 

On substituting x = 4 in (i), we get: 

16x + 3y = 7 

⇒ 3y = (7 – 16) = - 9 

⇒ y = - 3 

Hence, the solution is x = 4 and y = - 3.

(i) (3, 0) 

⇒ 2 × 3 – 0 = 6 

⇒ 6 = 6 

Thus, 

(3, 0) is a solution

(ii) (0, 6) 

⇒ 2 × 0 – 6 = 6 

⇒ - 6 = 6

This is not true, thus (0, 6) is not a solution

(iii) (2, -2) 

⇒ 2 × 2 + 2 = 6 

⇒ 6 = 6 

Thus, 

(2, -2) is a solution.

(iv) (√3, 0) 

⇒ 2√3 – 0 = 6 

⇒ 2√3 = 6 

This is not true, thus (2√3, 0) is not a solution.

(v) (1/2,- 5) 

⇒ (2/2) – (-5) = 6 

⇒ 6 = 6 

Thus, 

(1/2, -5) is a solution.

Given equation is 2x – 5y = 10 

Put x = 1/2 and y = 2 in the given equation.

Then 2 (1/2) -5 (2) = 10

1-10 = 10

-9 = 10 false

∴ (1/2, 2) is not a solution.

The given equations are: 

2x – 5y = \(\frac{8}3\).....(i)

3x – 2y = \(\frac{5}6\)

On multiplying (i) by 2 and (ii) by 5, we get: 

4x - 10y = \(\frac{16}3\).....(iii)

15x – 10y = \(\frac{25}6\).....(iv)

On adding (iii) and (iv), we get:

19x = \(\frac{57}6\)

⇒ x = \(\frac{57}{6\times19}\) = \(\frac{3}6=\frac{1}2\)

On substituting x = \(\frac{1}2\) in (i), we get:

2 x \(\frac{1}2\) + 5y = \(\frac{8}3\)

⇒ 5y = \((\frac{8}3-1)\) = \(\frac{5}3\)

⇒ y = \(\frac{5}{3\times5}\) = \(\frac{1}3\)

Hence, the solution is x = \(\frac{1}2\) and y = \(\frac{1}3\).

(i) Check for (3, 0) 

Put x = 3 and y = 0 in equation 2x – y = 6

2(3) – (0) = 6 

6 = 6 

True statement. 

(3,0) is a solution of 2x – y = 6. 

(ii) Check for (0, 6)

Put x = 0 and y = 6 in 2x – y = 6 

2 x 0 – 6 = 6 -6 = 6

False statement. 

⇒ (0, 6) is not a solution of 2x – y = 6. 

(iii) Check for (2, -2) 

Put x = 0 and y = 6 in 2x – y = 6 

2 x 2 – (-2) = 6 

4 + 2 = 6 

6 = 6 

True statement. 

⇒ (2,-2) is a solution of 2x – y = 6. 

(iv) Check for (√3, 0)

Put x = √3 and y = 0 in 2x – y = 6 

2 x √3 – 0 = 6 

2√3 = 6 

False statement. 

⇒(√3, 0) is not a solution of 2x – y = 6. 

(v) Check for (1/2, -5) 

Put x = 1/2 and y = -5 in 2x – y = 6 

2 x (1/2) – (-5) = 6 

1 + 5 = 6 

6 = 6 

True statement. 

⇒ (1/2, -5) is a solution of 2x – y = 6.

(i) Given: 5x – 2y = 10 

Substituting x = 0 ⇒ 5 × 0 – 2y = 10 ⇒ – 2y = 10 ⇒ – y = 10/2 ⇒ y = – 5 

Thus x =0 and y = -5 is the solution of 5x - 2y = 10 

Substituting y = 0 ⇒ 5x – 2 x 0 = 10 ⇒ 5x = 10 ⇒ x = 2 

Thus x =2 and y = 0 is a solution of 5x – 2y = 10 

(ii) Given, – 4x + 3y = 12 

Substituting x = 0 ⇒ -4 × 0 + 3y = 12 ⇒ 3y = 12 ⇒ y = 4 

Thus x = 0 and y = 4 is a solution of the -4x + 3y = 12 

Substituting y = 0 ⇒ -4 x + 3 x 0 = 12 ⇒ – 4x = 12 ⇒ x = -3 

Thus x = -3 and y = 0 is a solution of -4x + 3y = 12 

(iii) Given, 2x + 3y = 24 

Substituting x = 0 ⇒ 2 x 0 + 3y = 24 ⇒ 3y =24 ⇒ y = 8 

Thus x = 0 and y = 8 is a solution of 2x+ 3y = 24 

Substituting y = 0 ⇒ 2x + 3 x 0 = 24 ⇒ 2x = 24 ⇒ x =12 

Thus x = 12 and y = 0 is a solution of 2x + 3y = 24

Correct option is (C) (2, 4)

\(\because\) \(5\times2+3\times4-22\) = 10 + 12 - 22

= 22 - 22 = 0

\(\therefore\) (2, 4) satisfies the equation 5x + 3y – 22 = 0

\(\therefore\) (2, 4) is a solution of 5x + 3y – 22 = 0.

Correct option is  C) (2, 4)

The given system of equation is: 

4x - 3y = 8 ……(i)

6x - y = \(\frac{29}3\)........(ii)

On multiplying (ii) by 3, we get: 

18x – 3y = 29 ….(iii) 

On subtracting (iii) from (i) we get: 

-14x = -21

x = \(\frac{21}{14}=\frac{3}2\)

Now, substituting the value of x = \(\frac{3}2\) in (i), we get:

4 x \(\frac{3}2\)- 3y = 8

⇒ 6 – 3y = 8 

⇒ 3y = 6 – 8 = -2

y = \(\frac{-2}3\)

Hence, the solution x = \(\frac{3}2\) and y = \(\frac{-2}3\).

Let the larger number be x and the smaller number be y. Then, we have: 

x –y = 26 ………….(i) 

x = 3y …………(ii) 

On substituting x = 3y in (i), we get: 

3y –y = 26 

⇒ 2y = 26 

⇒ y = 13 

On substituting y = 13 in (i), we get: 

x – 13 = 26 

⇒ x = 26 + 13 = 39 

Hence, the required numbers are 39 and 13.

The given equations are:

 \(\frac{x}3+\frac{y}4= 11 \)

⇒ 4x + 3y = 132 …….(i)

and \(\frac{5x}6-\frac{y}3+7= 0 \)

⇒ 5x – 2y = -42 ……..(ii) 

On multiplying (i) by 2 and (ii) by 3, we get: 

8x + 6y = 264 …...(iii) 

15x – 6y = -126 ….(iv) 

On adding (iii) and (iv), we get: 

23x = 138 

⇒ x = 6 

On substituting x = 6 in (i), we get: 

24 + 3y = 132 

⇒ 3y = (132 – 24) = 108 

⇒ y = 36 

Hence, the solution is x = 6 and y = 36.

The given system of equation is: 

x + y = 3…….(i) 

4x – 3y = 26 ……(ii) 

On multiplying (i) by 3, we get:

3x + 3y = 9 ….(iii) 

On adding (ii) and (iii), we get: 

7x = 35 

⇒ x = 5 

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

5 + y = 3 

⇒ y = (3 – 5) = -2 

Hence, the solution is x = 5 and y = -2 

The given system of equation is: 

x + y = 3…….(i) 

4x – 3y = 26 ……(ii) 

On multiplying (i) by 3, we get:

3x + 3y = 9 ….(iii) 

On adding (ii) and (iii), we get: 

7x = 35 

⇒ x = 5 

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

5 + y = 3 

⇒ y = (3 – 5) = -2 

Hence, the solution is x = 5 and y = -2

Option : (B)

x – 2 = 0 

x = 2 is a point on the number line

x + y = 3 ……..(1)

4x – 3y = 26 ……(2)

Isolate x from equation (1), we get

x = 3 – y

Substituting the value of x in equation (2),

4(3 – y) – 3y = 26

12 – 4y – 3y = 26

-7y = 14

y = -2

Solve for x:

x = 3 – y = 3 – (-2) = 5

Answer: x = 5 and y = -2

The given system of equation can be written as: 

9x - 2y = 108 ……(i) 

3x + 7y = 105 ……(ii) 

On multiplying (i) by 7 and (ii) by 2, we get: 

63x + 6x = 108 × 7 + 105 × 2 

⇒69x = 966 

⇒x = 966/69 = 14 

Now, substituting x = 14 in (i), we get: 

9 × 14 – 2y = 108 

⇒2y = 126 – 108 

⇒y = 18/2 = 9 

Hence, x = 14 and y = 9.

The given system of equation can be written as: 

9x - 2y = 108 ……(i) 

3x + 7y = 105 ……(ii) 

On multiplying (i) by 7 and (ii) by 2, we get: 

63x + 6x = 108 × 7 + 105 × 2 

⇒ 69x = 966

⇒x = \(\frac{996}{69}= 14\)

Now, substituting x = 14 in (i), we get: 

9 × 14 – 2y = 108 

⇒ 2y = 126 – 108

⇒ y = \(\frac{18}{2}= 9\)

Hence, x = 14 and y = 9.

Option : (D)

Infinitely many equations satisfy x = 2 and y = 3 as infinitely many lines pass through a single point.

The given system of equation is: 

2x – y + 3 = 0…….(i) 

3x – 7y + 10 = 0 ……(ii) 

From (i), write y in terms of x to get 

y = 2x + 3 

Substituting y = 2x + 3 in (ii), we get 

3x – 7(2x + 3) + 10 = 0 

⇒ 3x – 14x – 21 + 10 = 0 

⇒ -7x = 21 – 10 = 11

x = - \(\frac{11}7\)

Now substituting x = – \(\frac{11}7\) 7 in (i), we have

- \(\frac{22}7\) - y + 3 = 0

y = 3 - \(\frac{22}7\) = - \(\frac{1}7\)

Hence, x = – \(\frac{11}7\) and y = - \(\frac{1}7\).

The given system of equation is: 

2x + 3y = 0 ……(i) 

3x + 4y = 5 ……(ii) 

On multiplying (i) by 4 and (ii) by 3, we get: 

8x + 12y = 0 ……(iii) 

9x + 12y = 15 …….(iv) 

On subtracting (iii) from (iv) we get:

x = 15 

On substituting the value of x = 15 in (i), we get: 

30 + 3y = 0 

⇒ 3y = - 30 

⇒ y = -10 

Hence, the solution is x = 15 and y = -10.

Correct option is  B) Line

The given system of equation is: 

3x - 5y - 19 = 0 ……(i) 

- 7x + 3y + 1 = 0 ……(ii) 

On multiplying (i) by 3 and (ii) by 5, we get: 

9x - 15y = 57 ……(iii) 

- 35x + 15y = -5 …….(iv) 

On subtracting (iii) from (iv) we get: 

- 26x = (57 – 5) = 52 

⇒ x = - 2 

On substituting the value of x = - 2 in (i), we get:

–6 – 5y – 19 = 0 

⇒ 5y = (–6 – 19) = -25 

⇒ y = -5 

Hence, the solution is x = -2 and y = - 5.

The given system of equation is: 

2x - 3y = 13 ……(i) 

7x - 2y = 20 ……(ii) 

On multiplying (i) by 2 and (ii) by 3, we get: 

4x - 6y = 26 ……(iii) 

21x - 6y = 60 …….(iv) 

On subtracting (iii) from (iv) we get: 

17x = (60 – 26) = 34 

⇒ x = 2 

On substituting the value of x = 2 in (i), we get: 

4 – 3y = 13 

⇒ 3y = (4 – 13) = -9 

⇒ y = -3 

Hence, the solution is x = 2 and y = -3.

A) (0, 4), (6, 0)

Correct option is  D) 86°

Correct option is (C) Infinite

A linear equation will have infinitely many solutions.

\(\therefore\) Number of solutions of 7x – 3y – 15 = 0 is infinite.

Correct option is  C) Infinite

Correct option is (D) 0

6x = 7y

\(\Rightarrow\) 6x - 7y = 0

By comparing above equation with linear equation ax+by+c = 0, we get c = 0.

Correct option is  D) 0

Correct option is (D) 2

\(\frac{-x}{8}=\frac{y}{4}\)

\(\Rightarrow\) \(-x=\frac{8y}{4}\)

\(\Rightarrow\) -x = 2y

\(\Rightarrow\) x+2y = 0

Comparing above equation with linear form, we get b = 2.

Correct option is  D) 2

Correct option is (D) 26

\(\because\) \((\sqrt{5},-3)\) is a solution of \(\sqrt5x-7y=k\)

\(\therefore\) \(\sqrt5\times\sqrt5-7\times-3=k\)

\(\therefore\) k = 5+21 = 26

Correct option is  D) 26

Correct option is (A) y = –3

A line parallel to X–axis is y = c, where c is constant. Among all given lines y = –3 is a line parallel to X-axis.

Correct option is  A) y = – 3

Correct option is (D) –10

5x – 6y = 10

\(\Rightarrow\) 5x - 6y - 10 = 0

By comparing above equation with ax+by+c = 0, we get c = -10.

Correct option is  D) – 10

Correct option is (A) 1

\(\because\) x = 3, y = 2 is a solution of the equation 5x – 7y = k.

\(\therefore\) \(5\times3-7\times2=k\)

\(\Rightarrow\) k = 15 - 14 = 1

Correct option is  A) 1

Correct option is (A) 12

\(\because\) (2, – 1) is a solution of 4x + by + 4 = 0

\(\therefore\) \(4\times2+b\times-1+4=0\)

\(\Rightarrow\) b = 8+4 = 12

Correct option is  A) 12

Correct option is (B) y = -2

Line parallel to X – axis is y = c.

Since, required line is passing through (7, –2).

\(\therefore\) c = -2

Hence, equation of required line is y = -2.

Correct option is  B) y = -2

Correct option is (A) (-1, –1)

y = –x + 2

\(\therefore\) (-1, –1) does not lie on the line y = –x + 2.

A) (- 1, – 1)

Correct option is (A) 3x – 4y – 5 = 0

3x – 4y = 5

\(\Rightarrow\) 3x – 4y - 5 = 0

A) 3x – 4y – 5 = 0

Correct option is (C) –10x + 3y – 5 = 0

–2x = \(\frac{-3}{5}\)y + 1

\(\Rightarrow\) -2x = \(\frac{-3y+5}{5}\)

\(\Rightarrow\) -10x = -3y+5

\(\Rightarrow\) -10x + 3y - 5 = 0 is a linear form of given line.

C) – 10x + 3y – 5 = 0

The given system of equations is 

0.3x + 0.5y = 0.5 …….(i) 

0.5x + 0.7y = 0.74 …….(ii) 

Multiplying (i) by 5 and (ii) by 3 and subtracting (ii) from (i), we get 

2.5y - 2.1y = 2.5 - 2.2 

⇒ 0.4y = 0.28

⇒ y = \(\frac{0.28}{0.4}\) = 0.7

Now, substituting y = 0.7 in (i), we have 

0.3x + 0.5 × 0.7 = 0.5 

⇒ 0.3x = 0.50 – 0.35 = 0.15

⇒ x = \(\frac{0.15}{0.3}\) = 0.5 

Hence, x = 0.5 and y = 0.7.

Let the shape of plot be quadrilateral ABCD. Itwaari agrees to give land as follows: 

Diagonal AC of a quadrilateral is joined. BA side of land ABCD is produced to E. 

CE and DE || AC are drawn. 

Now, ∆ACD and ∆ACE are on same base aC and in between AC || DE. 

∴ ar.(∆ACD) = ar.(∆ACE) 

Subtracting ar.(∆AOC) on both sides 

ar (∆ACD) – ar(∆AOC) = ar(∆ACE) – ar(∆AOC) 

ar.(∆DOC) = ar.(∆AOC). 

Now, Itwaari agrees to give ADOC to construct a Health Centre. 

Area of ∆AOE and ∆CEB is remains with Itrwaari.

Consider △ DGA and △ AED

We know that both the triangles have the same base AD and lie between the parallel lines AD and EG.

So we get

Area of △ DGA = Area of △ AED ……. (1)

Consider △ DBC and △ BFD

We know that both the triangles have the same base DB and lie between the parallel lines BD and CF.

So we get

Area of △ DBF = Area of △ BCD ……. (2)

By adding both the equations

Area of △ DGA + Area of △ DBF = Area of △ AED + Area of △ BCD

By adding △ ABD both sides

Area of △ DGA + Area of △ DBF + Area of △ ABD = Area of △ AED + Area of △ BCD + Area of △ ABD

So we get

Area of △ DGF = Area of pentagon ABCDE

Therefore, it is proved that ar (pentagon ABCDE) = ar (△ DGF).

Consider △ MCQ and △ MPB

From the figure we know that

∠ QCM and ∠ PBM are alternate angles

So we get

∠ QCM = ∠ PBM

We know that M is the midpoint of BC

CM = BM

∠ CMQ and ∠ PBM are vertically opposite angles

∠ CMQ = ∠ PBM

By ASA congruence criterion

△ MCQ ≅ △ MPB

So we get

Area of △ MCQ = Area of △ MPB

We know that

Area of ABCD = Area of APQD + Area of DMPB – Area of △ MCQ

So we get

Area of ABCD = Area of APQD

Therefore, it is proved that ar (ABCD) = ar (APQD).

Data: ABCDE is a pentagon. A line through B parallel to AC meets DC produced at F. 

To Prove: (i) ar.(∆ACB) = ar.(∆ACF) 

(ii) ar.(AEDF) = ar.(ABCDE) 

Proof: (i) ∆ACB and ∆HCF are on base AC and between AC || BF. 

∴ ar.(∆ACB) = ar.(∆ACF) 

(ii) Now, W(∆ACB) = W(∆ACF) proved, 

adding ACDE on both sides,

 ar.(∆ACB) + ar.(ACDE) = ar.(∆ACF) + ar.(ACDE) 

∴ ar.(ABCDE) = ar.(AEDF)

Data: Diagonals AC and BD of a trapezium ABCD with AB || DC, intersect each other at ‘O’. 

To Prove: ar.(∆AOD) = ar.(∆BOC) 

Proof: ∆ABD and ∆ABC are on base AB and between AB || DC. 

∴ ar. (∆ABD) = ar. (∆ABC) 

Subtracting ar.(∆ABO) on both sides, 

ar.(∆ABD) – ar.(∆ABO) = ar.(∆ABC) – ar.(∆ABO) 

∴ ar.(∆AOD) = ar.(∆BOC).

(i) Radius of cone(r) = 6 cm 

Height of cone(h) = 7 cm 

We know, 

Volume of a right circular cone = \(\frac{1}{3}\)πr²h 

= \(\frac{1}{3}\) x 3.14 x 6² x 7 

= 264 

Volume of a right circular cone is 264 cm³ 

(ii) Radius of cone(r) = 3.5 cm 

Height of cone(h) = 12 cm 

Volume of a right circular cone = \(\frac{1}{3}\)πr²h 

= \(\frac{1}{3}\) x 3.14 x 3.5² x 12 

= 154

Volume of a right circular cone is 154 cm³ 

(iii) Height of cone(h) = 21 cm 

Slant height of cone(l) = 28 cm 

We know, 

l² = r² + h² 

28² = r² + 21² 

or r = 7√7 

Now, 

Volume of a right circular cone = \(\frac{1}{3}\)πr²h 

= \(\frac{1}{3}\) x 3.14 x (7√7)² x 21 

= 7546 

Volume of a right circular cone is 7546 cm³

Curved surface area = 792 cm² 

The radius and slant height of a cone are in the ratio 4:7 

Let 4x be the radius and 7x be the height of cone. 

Now, 

Curved surface area (C.S.A.) = πrl 

So, 22/7 x (4x) x (7x) = 792 

or x² = 9 

or x = 3 

Therefore, Radius = 4x = 4(3) cm = 12 cm