Explore topic-wise InterviewSolutions in Current Affairs.

Let the common ratio between a and b be x.

∴ a = 2x, and b = 3x

XY is a straight line, rays OM and OP stand on it.

∴ ∠XOM + ∠MOP + ∠POY = 180º

b + a + ∠POY = 180º

3x + 2x + 90º = 180º

5x = 90º

x = 18º

a = 2x = 2 × 18 = 36º

b = 3x= 3 ×18 = 54º

∴ b + c = 180º (Linear Pair)

54º + c = 180º

c = 180º − 54º = 126º

∴ c = 126º

Correct option is  D) none

Correct option is (D) none

Let consecutive numbers are x & x+1.

\(\therefore\) x + (x+1) = 10

\(\Rightarrow\) 2x = 10 - 1 = 9

\(\Rightarrow\) x = \(\frac92.\)

We have, A = {x :x∈ W,x< 2} = {0, 1};

B = {x : x ∈ N, 1 <x< 5} = {2, 3,4}; and C= {3, 5}

(i) B∩ C = {3}

A x (B ∩ C) = {0, 1} x {3} = {(0, 3), (1, 3)}

(ii) (B ∪ C) ={2,3,4, 5}

A x (B ∪ C) = {0, 1} x {2, 3,4, 5}

= {(0,2), (0,3), (0,4), (0,5), (1,2), (1,3), (1,4), (1, 5)}

Let the number of correct answers be x

Number of questions answered wrong are (180 – x)

Total score when answered right = 4x

Marks deducted when answered wrong = 1(180 – x) = 180 – x

So,

4x – (180 – x) = 450

4x – 180 + x = 450

5x = 450 + 180

5x = 630

x = 630/5

= 126

∴ 126 questions he answered correctly.

Let the number be ‘x’ say. 

8 times of a number = 8 × x = 8x 

If 10 is reduced from 8x then 8x – 0 

6 times of a number = 6 × x = 6x

If 4 is added to 6x then 6x + 4 

According to the sum, 

8x – 10 = 6x + 4 

⇒ 8x – 6x = 4 + 10 

⇒ 2x = 14

⇒ x = 7 

∴ The required number = 7

If a number is divided into two parts in he ratio of 5 : 3, 

let the parts be 5x, 3x say. 

According to the sum, 

5x = 3x + 10 

⇒ 5x – 3x = 10

⇒ 2x = 10 

∴ x = \(\frac{10}2\)

∴ x = 5 

∴ The required number be 

x + 3x = 8x 

= 8 × 5 = 40 

And the parts of number are 

5 = 5 × 5 = 25 

3 = 3 × 5 = 15

Let a digit of two digit number be x. 

The sum of two digits = 9 

∴ Another digit = 9 – x 

The number = 10 (9 – x) + x 

= 90 – 10x + x 

= 90 – 9x 

If 27 is subtracted from the number its digits are reversed. 

∴ (90 – 9x) – 27 = 10x + (9 – x) 

63 – 9x = 9x + 9

9x + 9x = 63 – 9 

18x = 54 

∴ x = \(\frac{54}{18}\) = 3 

∴ Units digit = 3 

Tens digit = 9 – 3 = 6 

∴ The number = 63

Let us consider,

Number of 25-paisa coins = x and

Number 50-paisa coins = y

Total number of coins = 50

and total amount = ₹ 19.50 or 1950 paisa

x + y = 50 …(1)

25x + 50y = 1950

x + 2y = 78 …(2)

Subtracting (1) from (2),

y = 28

And, x = 50 – y = 50 – 28 = 22

Number of 25-paisa coins = 22

and 50-paisa coins = 28

Let the number of sea shells found by sandy = x

So, the number of sea shells found by Anita = (x + 5).

The number of sea shells found by Shella = 2 (x + 5).

According to the question,

⇒ x + 2(x + 5) = 16

⇒ x + 2x +10 = 16

⇒ 3x + 10 = 16

Subtracting 10 from both sides, we get

⇒ 3x + 10 – 10 = 16 – 10

⇒ 3x = 6

Dividing both sides by 3, we get

⇒ 3x/6 = 6/3

⇒ x = 4

Thus, the number of sea shells found by Sandy = x = 4,

The number of sea shells found by Anita = x + 5 = 4 + 5 = 9,

The number of sea shells found by Shelia = 2 (x + 5) = 2 (4 + 5) = 2(9) = 18.

Let the number of marbles with Pandy = x

So, the number of marbles with Andy = 2x

Thus, the number of marbles with Sandy = (x/2) + (2x/2) = (3x/2)

According to the question:

⇒ 2x + 75 = (3x/2)

By transposing we get

⇒2x – (3x/2) = -75

⇒ (4x – 3x)/2 = -75

⇒ (x/2) = -75

⇒ x = -150

Since number of marbles cannot be negative

Therefore x = 150

So, Pandy has 150 marbles,

Andy has 2x = 2 (150) = 300 marbles,

Sandy has 3x/2 = 225 marbles.

Let the number of 25-paise coins in the purse be x

So, the value of money in the purse = 0.25x.

But 0.25x = 17.50.

Dividing both sides by 0.25, we get

⇒ 0.25x/0.25 = 17.5/0.25

⇒ x = 70

Thus, the number of 25 paise coins in the purse = 70.

Let the number of 50 paise coins = x

So, the money value contribution of 50 paise coins in bag = 0.5x.

The number of 25 paise coins in bag = 4x

The money value contribution of 25 paise coins in bag = 0.25 (4x) = x.

According to the question,

⇒ 0.5x + x = 30

⇒ 1.5x = 30

Dividing both sides by 1.5, we get

⇒ 1.5x/1.5= 30/1.5

⇒ x = 20

Thus, the number of 50 paise coins = x = 20,

The number of 25 paise coins = 4x = 4 (20) = 80.

Given (2x+7)/5-(3x+11)/2 = (2x+8)-15

Now by taking L.C.M for 2 and 5 is 10

15(2(2x+7)-5(3x+11))/10 = (2x+8)/3-5

By transposing the above equation we can write as

33x-20x=123-70

Again by transposing

x=-53/53

x=-1

(i) 3(x + 1) = 12 + 4(x – 1)

3(x + 1)= 12 + 4(x – 1) 

3x + 3 = 12 + 4x – 4 

3x + 3 = 4x + 8 

3x – 4x = 8 – 3 

-x = 5 

x = -5

(ii) 2x – 5 = 3(x – 5)

2x – 5 = 3(x – 5) 

2x – 5 = 3x – 15 

2x – 3x = -15 + 5 

-x = -10 

x= 10

(i) 2x – 1 = 14 – x

2x + x = 14 + 1 

3x = 15 

x = \(\frac{15}{3}\)

x = 5

(ii) 6x + 1 = 3(x -1) + 7

6x + 1 = 3(x —1) + 7 

6x + 1 =3x - 3 + 7 

6x + 1 = 3x + 4 

6x – 3x = 4 – 1

3x = 3 

x = \(\frac{3}{3}\)

x = 1

Given 9x+5=4(x-2) + 8,

By transposing the above equation we can write as

9x+5=4x-8+8

9x-4x=5

Again by transposing

5x=5

x=5/5=1

From the question,Sumitra has ₹34 = 3400 paise

Let 50 paise coins be x

And 25 paise coins be 2x

Then,

= 50x + 25 (2x) = 3400

= 50x + 50x = 3400

= 100x = 3400

Multiplying both side by (1/100).

= 100x × (1/100) = (3400) × (1/100)

= x = 34

∴The number of 50 paise coins is 34

The number of 25 paise coins is 2x = 2 × 34 = 68

(i) 6(1 – 4x) + 7(2 + 5x) = 53

6( 1 – 4x) + 7(2 + 5x) = 53 

6 – 24x+ 14 + 35x = 53 

35x – 24x + 6 + 14 = 53 

11x+ 20 = 53 

11x = 53 – 20 

x = \(\frac{33}{11}\)

x = 3

(ii) 3(x + 6) + 2(x + 3) = 64

3(x + 6) + 2(x + 3) = 64 

3x + 18 + 2x + 6 = 64 

3x + 2x + 18 + 6 = 64 

5x + 24 = 64 

5x = 64 – 24 

5x = 40

x = \(\frac{40}{5}\)

∴ x = 8

Given 2z-1=14-z,

By transposing the above equation we can write as

2z+z=14+1

3z=15

Again by transposing

z=15/3=5

(d) (1/36)

Because,

Transposing – (3/4) to RHS and it becomes (3/4) and 2x to LHS it becomes -2x

= 5x – 2x = – (2/3) + (3/4)

= 3x = (-8+9)/12

= 3x = 1/12

Multiplying both side by (1/3).

= 3x × (1/3) = (1/12) × (1/3)

= x = 1/36

Given (5x-4)/6=4x+1-(3x+10)/2

Now by taking L.C.M for 1 and 2 is 1

5x-4-6(4x+1) + 3(3x+10)/6=0

By cross multiplication we get

5x-4-6(4x+1) + 3(3x+10) = 0

-10x=-20

Again by transposing,

x=20/10=2

Given 5x+7=2x-8,

By transposing the above equation we can write as

5x-2x=-7-8

3x=-15

Again by transposing

x=-15/3=-5

Given x – (x/4) – (1/2) = 3 + (x/4)

Transposing (x/4) to LHS and (1/2) to RHS

x – (x/4) – (x/4) = 3 + (1/2)

(4x – x – x)/4 = (6 + 1)/2

2x/4 = 7/2

x/2 = 7/2

x = 7

Verification:

Substituting x = 7 in the given equation we get

7 – (7/4) – (1/2) = 3 + (7/4)

(28 – 7 – 2)/4 = (12 + 7)/4

19/4 = 19/4

Thus LHS = RHS

Hence, verified.

Given m – (m – 1)/2 = 1 – (m – 2)/3

(2m – m + 1)/2 = (3 – m + 2)/3

(m + 1)/2 = (5 – m)/3

(m + 1)/2 = (5/3) – (m/3)

(m/2) + (1/2) = (5/3) – (m/3)

Transposing (m/3) to LHS and (1/2) to RHS

(m/2) + (m/3) = (5/3) – (1/2)

(3m + 2m)/6 = (10 – 3)/6

5m/6 = (7/6)

5m = 7

Dividing both sides by 5, we get

5m/5 = 7/5

m = 7/5

Verification:

Substituting m = 7/5 on both sides, we get

(7/5) – (7 – 5)/10 = 1 – (7 – 10)/15

(7/5) – (2/10) = (15 + 3)/15

(14 – 2)/10 = (15 + 3)/15

12/10 = 18/15

(6/5) = (6/5)

Thus LHS = RHS

Hence, verified.

(d)(4/3)

Because,

Transposing (8/3) to RHS and it becomes – (8/3) and (1/4)z to LHS it becomes –(1/4)z

= 2z – (1/4)z = 5 – (8/3)

= [(8-1)/4]z = (15-8)/3

= (7/4)z = 7/3

Multiplying both side by (4/7).

= (7/4)z × (4/7) = ((7/3) × (4/7)

= x = 4/3

Given 5x-(1/3) (x+1) = 6(x + (1/30))

Taking L.C.M on both sides, we get

(15x-(x+1))/3=6(30x+1)/30

By cross multiplication,

10(14x-1) = 6(30x+1)

140x-180x=6+10

-40x=16

-x=2/5

Given two numbers are in the ratio 8:3

So let the numbers be 8x and 3x

According to the question we can write as 8x+3x=143

Which implies 11x=143

Again by transposing x=143/11

Therefore x=13

So the number are 8x=8(13) =104 and 3x=3(3)=39

False

We have, P = {1, 2} and n(P) = 2

n(P x P x P) = n(P) x n(P) x n(P) = 2 x 2 x 2

= 8 But given P x P x P has 4 elements.

Given 4-(2(z-4))/2=1/2(2z+5)

Now by taking L.C.M of 1 and 3 is 3

12-2(z-4)/3 = (2z+5)/2

By cross multiplication we get

2(12-2z+8) = 3(2z+5)

40-4z=6z+15

-10z=-25

z=25/10=5/2

Given 6x + 5 = 2x + 17

Transposing 2x to LHS and 5 to RHS, we get

6x – 2x = 17 – 5

4x = 12

Dividing both sides by 4, we get

4x/4 = 12/4

x = 3

Verification:

Substituting x = 3 in the given equation, we get

6 × 3 + 5 = 2 × 3 + 17

18 + 5 = 6 + 17

23 = 23

Therefore LHS = RHS

Hence, verified.

Given (6x – 2)/9 + (3x + 5)/18 = (1/3)

(6x (2) – 2 (2) + 3x + 5)/18 = (1/3)

(12x – 4 + 3x + 5)/18 = (1/3)

(15x + 1)/ 18 = (1/3)

Multiplying both sides by 18 we get

(15x + 1)/18 × 18 = (1/3) × 18

15x + 1 = 6

Transposing 1 to RHS, we get

= 15x = 6 – 1

= 15x = 5

Dividing both sides by 15, we get

= 15x/15 = 5/15

=x = 1/3

Verification:

Substituting x = 1/3 both sides, we get

(6 (1/3) – 2)/9 + (3 (1/3) + 5)/18 = (1/3)

(2 – 2)/9 + (1 + 5)/ 18 = 1/3

(6/18) = (1/3)

(1/3) = (1/3)

Thus LHS = RHS

Hence, verified.

Given (5x – 1)/3 – (2x – 2)/3 = 1

(5x – 1 – 2x – 2)/3 = 1

(3x + 1)/3 = 1

Multiplying both sides by 3 we get

(3x + 1)/3 × 3 = 1 × 3

(3x + 1) = 3

Subtracting 1 from both sides we get

3x + 1 – 1 = 3 – 1

3x = 2

Dividing both sides by 3, we get

3x + 1 – 1 = 3 -1

3x = 2

Dividing both sides by 3, we get

3x/3 = 2/3

x = 2/3

Verification:

Substituting x = 2/3 in LHS, we get

(5 (2/3) – 1)/3 – (2 (2/3) – 2)/3 = 1

(10/3 -1)/3 – (4/3 – 2)/3 = 1

(7/3)/3 – (-2/3)/3 = 1

(7/9) + (2/9) = 1

(9/9) = 1

1 = 1

Thus LHS = RHS

Hence, verified.

Given 0.6x + 4/5 = 0.28x + 1.16

Transposing 0.28x to LHS and 45 to RHS, we get

0.6x – 0.28x = 1.16 – 45

0.32x = 1.16 – 0.8

0.32x = 0.36

Dividing both sides by 0.32, we get

0.32 x 0.32 = 0.360.32

x = 9/8

Verification:

Substituting x = 9/8 on both sides, we get

0.6(9/8) + 45 = 0.28(9/8) + 1.16

5.4/8 + 4/5 = 2.52/8 + 1.16

0.675 + 0.8 = 0.315 + 1.16

1.475 = 1.475

Thus LHS = RHS

Hence, verified.

We have,

(2x-1)/3 – (6x-2)/5 = 1/3

By taking LCM for 3 and 5 which is 15

((2x-1)×5)/15 – ((6x-2)×3)/15 = 1/3

(10x – 5)/15 – (18x – 6)/15 = 1/3

(10x – 5 – 18x + 6)/15 = 1/3

(-8x + 1)/15 = 1/3

By using cross-multiplication we get,

(-8x + 1)3 = 15

-24x + 3 = 15

-24x = 15 – 3

-24x = 12

x = -12/24

= -1/2

∴ x = -1/2

False

We have, R = {(x, y): y = x – 5, x, y ∈ Z}

When x = 5, then y = 5-5=0 Hence, (5, 2) does not belong to R.

Correct option is  C) 120

Correct option is (C) 120

\(\frac{1}{4}\) x = 30

\(\Rightarrow\) x = 30 \(\times\) 4 = 120.

\(\frac{x+1}{2} + \frac{x+2}{3} = \frac{2x-5}{7} + 9\)

21(x + 1) + 14 (x + 2) = 6(2x – 5) + 9 x 42

Multiplying by 42 as LCM of 2, 3 and 7 is 42

⇒ 21x + 21 + 14x + 28 = 12x – 30 + 378

⇒ 21x + 14x + 21 + 28 = 12x – 30 + 378

⇒ 35x + 49 = 12x + 348

⇒ 35x – 12x = 348 – 49

On transposing 12x and 49

⇒ 23x = 299

⇒ 23x/23 = 299/23

On dividing by 23 on both sides

⇒ x = 13

Correct option is  A) -9

\(\frac{3x-2}{5} = 4 - (\frac{x+2}{3})\)

⇒ 3(3x – 2) = 15 × 4 – 5 (x + 2)

Multiplying by 15 as LCM of 5 and 3 is 15

⇒ 9x – 6 = 60 – 5x – 10

⇒ 9x – 6 = – 5x + 60 – 10

⇒ 9x – 6 = – 5x + 50

⇒ 9x + 5x = 50 + 6

On transposing – 6 and – 5x

⇒ 14x = 56

⇒ 14x/14 = 56/14

On dividing by 14 on both sides

⇒ x = 4

Correct option is  C) 3

Correct option is (C) 3

2.45x + 1.5 = 3.7x – 2.25

\(\Rightarrow\) 3.7x - 2.45x = 1.5+2.25

\(\Rightarrow\) 1.25x = 3.75

\(\Rightarrow\) x = \(\frac{3.75}{1.25}\) = 3.

B) x, x + 1, x + 2

0.6x + 0.25x = 0.45x + 1.2

⇒ 6x/10 + 25x/100 = 45x/100 + 12/10

On changing decimal fractions into simple fraction

⇒ 60x + 25x = 45x + 120

On multiplying by 100 since LCM of 10 and 100 is 100

⇒ 85x = 45x + 120

⇒ 85x – 45x = 120

On transposing 45

40x = 120

⇒ 40x/40 = 120/40

On dividing by 40 on both sides

⇒ x = 3

\(\frac{x+2}{2} + \frac{x+4}{3} = \frac{x +6}{4} + \frac{x +8}{5}\)

⇒ 30(x + 2) + 20(x + 4) = 15(x + 6) + 12(x + 8)

On multiplying by LCM of 2, 3, 4 and 5 i.e. 60 on both sides

⇒ 30x + 60 + 20x + 80 = 15x + 90 + 12x + 96

⇒ 30x + 20x + 60 + 80 = 15x + 12x + 90 + 96

⇒ 50x + 140 = 27x + 186

⇒ 50x – 27x = 186 – 140

On transposing 140 and 27x

⇒ 23x = 46

⇒ 23x/23 = 46/23

On dividing by 23 on both sides

⇒ x = 2

(C) (3, 0)

Since, the graph of linear equation 2x + 3y = 6 meets the X-axis.

So, we put y = 0 in 2x + 3y = 6 

=> 2x + 3(0) = 6

=> 2x + 0 = 6

=> x = 6/2 

=> x = 3

Hence, the coordinate on X-axis is (3, 0).

(A) 1st quadrant

We know that, if a line passes through the Ist quadrant, then all solution lying on the line in first quadrant must be positive because the coordinate of all points in the Ist quadrant are positive.

(B) x + y = 0

Let us consider a linear equation ax + by + c = 0 … (i)

Since, (-2,2), (0, 0) and (2, -2) are the solutions of linear equation therefore it satisfies the Eq. (i), we get

At point(-2,2), -2a + 2b + c = 0 …(ii)

At point (0, 0), 0+0 + c = 0 => c = 0 …(iii)

and at point (2, – 2), 2a - 2b + c = 0 …(iv)

From Eqs. (ii) and (iii),

c = 0 and – 2a + 2b + 0 = 0, – 2a = -2b,a = 2b/2

=> a = b

On putting a = b and c = 0 in Eq. (i),

bx + by + 0 = 0

=> bx + by = 0 

=> – b(x + y) = 0 

=> x + y = 0, b ≠ 0

Hence, x + y= 0 is the required form of the linear equation.

The ratio of ages of Rahul and Lakshmi = 5:7 

Let their ages be 5x, 7x say. 

After 4 years Rahuls agt. = 5x + 4

After 4 years Lakshmis age = 7x + 4

According to the sum, 

After 4 years the sum of their ages = 56 

⇒ (5x + 4) + (7x + 4) = 56 

⇒ 12x + 8 = 56 

⇒ 12x = 56 – 8 = 48 

⇒ x = \(\frac{48}{12}\) = 4 

∴ x = 4 

∴ Rahuls present age 

= 5x = 5 × 4 = 20 years 

∴ Lakshmi’s present age 

= 7x = 7 × 4 = 28 years

The main cause of drought is insufficient rainfall.

Rate of leveling and turning a square lawn = 2.50 per m² 

Total cost of leveling and turning = Rs. 13322.50 

Total area of square lawn = \(\frac{13322.50}{2.50}\) 

= 5329 m² 

Side of square lawn = \(\sqrt{5329}\) 

= 73 m 

Total length of lawn = 4 × 73 

= 292 m 

Cost of fencing the lawn at Rs 5 per metre = 292 × 5 

= Rs. 1460

River Kosi is called as the Sorrow of Bihar.