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obscure : difficult to understand 

mystic : spiritual 

prophetic : predictive 

paraphernalia : belongings 

mutter : incoherent/incomprehensible speech 

gratified : thankful

Answer is: Witch.

(B) Remains the same

By property, if we multiply or divide both sides of a linear equation with a non-zero number, then the solution of the linear equation remains the same i.e., the solution of the linear equation is remains unchanged.

Let the number is  x

According to the question:

Three-fourth of the number is = \(\frac{3x}4\)

Fourth-fifth of the number is = \(\frac{4x}5\)

\(\frac{4x}5\) - \(\frac{3x}4\) = 4

LCM of 5 and 4 is 20

\(\frac{16x-15x}{20}=4\)

x = 80

Therefore number is 80

Correct option is C) Rs. 1400

Given:

The larger of two supplementary angles exceeds the smaller by 18 degrees.

To find:

The measure of both angles.

Solution:

Let the larger angle be ‘a’ and the smaller angle be ‘b’.

Given, larger of two supplementary angles exceeds the smaller by 18 degrees.

⇒ a = b + 18

⇒ a - b = 18 ...... (1)

The sum of supplementary angles is 180°

⇒ a + b = 180.....(2)

Adding the equations 1 and 2 we get,

a - b + a + b = 18 + 180

⇒ 2a = 198

⇒ a = 99°

Put the value of b in eq. 1 to get,

99 - b = 18

⇒ - b = 18 - 99

⇒ - b = - 81

⇒ b = 81

Thus,

b = 81°

Hence the measure of smaller angle is 81° and larger angle is 99°.

Let the stake money be ‘a’ and ‘b’ respectively.

Given, to one he said; if your bird wins, than you give me your stake - money, but if you do not win, I shall give you two third of that. Going to the other, he promised in the same way to give three fourths.

From both of them his gain would be only 12 gold coins

If the 1^st one wins

⇒ a - 3b/4 = 12 ----- (1)

If the 2^nd one wins

⇒ b – 2a/3 = 12 ----- (2)

Equating 1 and 2

⇒ 5a/3 = 7b/4

⇒ a = 21b/20

Thus,

21b/20 – 3b/4 = 12

⇒ 6b/20 = 12

⇒ b = 40

Thus,

a = 21b/20 = Rs. 42

Let the length of a rectangle = x m

and the breadth of a rectangle = y m

Then, Area of rectangle = xy m²

Condition I :

Area is reduced by 8m², when length = (x – 5) m and breadth = (y + 3) m

Then, area of rectangle = (x – 5)×(y + 3) m²

According to the question,

xy – (x – 5)×(y + 3) = 8

⇒ xy – (xy + 3x – 5y – 15) = 8

⇒ xy – xy – 3x + 5y + 15 = 8

⇒ – 3x + 5y = 8 – 15

⇒ 3x – 5y = 7 …(i)

Condition II:

Area is increased by 74m², when length = (x + 3) m and breadth = (y + 2) m

Then, area of rectangle = (x + 3)×(y + 2) m²

According to the question,

(x + 3)×(y + 2) – xy = 74

⇒ (xy + 3y + 2x + 6) – xy = 74

⇒ xy + 2x + 3y + 6 – xy = 74

⇒ 2x + 3y = 74 – 6

⇒ 2x + 3y = 68 …(ii)

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

6x – 10y = 14 …(iii)

6x + 9y = 204 …(iv)

On subtracting Eq. (i) from Eq. (ii), we get

6x + 9y – 6x + 10y = 204 – 14

⇒ 19y = 190

⇒ y = 10

On putting the value of y = 10 in Eq. (i), we get

3x – 5 (10) = 7

⇒ 3x – 50 = 7

⇒ 3x = 57

⇒ x = 19

Hence, the length of the rectangle is 19 m and the breadth of a rectangle is 10 m

Correct option is B) 32, 30

Let’s assume, 

The speed of the train be x km/hr 

The speed of the car = y km/hr 

From the question, it’s understood that there are two parts 

# Part 1: When the man travels 400 km by train and the rest by car. 

# Part 2: When Ramesh travels 200 km by train and the rest by car. 

Part 1, 

Time taken by the man to travel 400km by train = 400/x hrs [∵ time = distance/ speed] 

Time taken by the man to travel (600 – 400) = 200 km by car = 200/y hrs 

Time taken by a man to cover 600 km = 400/x hrs + 200/y hrs 

Total time taken for this journey = 6 hours + 30 mins = 6 + 1/2 = 13/2 

So, by equations its 

400/x + 200/y = 13/2 

400/x + 200/y = 13/2 

400/x + 200/y = 13/2 

200 (2/x + 1/y) = 13/2 

2/x + 1/y = 13/400 .…(i) 

Part 2, 

Time taken by the man to travel 200 km by train = 200/x hrs. [∵ time = distance/ speed] 

Time taken by the man to travel (600 – 200) = 400 km by car = 200/y hrs 

For the part, the total time of the journey is given as 6 hours 30 mins + 30 mins that is 7 hrs, 

200/x + 400/y = 7 

200 (1/x + 2/y) = 7 

1/x + 2/y = 7/200 …..(ii) 

Taking 1/x = u, and 1/y = v, 

So, the equations (i) and (ii) becomes, 

2u + v = 13/400 ….. (iii) 

u + 2v = 7/200 ……. (iv) 

Solving (iii) and (iv), by 

(iv) x 2 – (iii)

⇒ 3v = 14/200 – 13/400 

3v = 1/400 x (28 – 13) 

3v = 15/400 

v = 1/80 

⇒ y = 1/v = 80 

Now, using v in (iii) we find u, 

2u + (1/80) = 13/400 

2u = 13/400 – 1/80 

2u = 8/400 

u = 1/100 

⇒ x = 1/u = 100 

Hence, the speed of the train is 100 km/hr and the speed of the car is 80 km/hr.

Correct option is B) 36 days

Let the speed of her rowing in still water be ‘a’ and speed of current be ‘b.’

We know, speed = \(\frac{distance}{time}\)

Relative speed of boat going upstream = a – b

Relative speed of boat going downstream = a + b

Given,

Ritu can row downstream 20 km in 2 hours, and upstream 4 km in 2 hours

⇒ a + b = 20/2 = 10 ---- (1)

and a – b = 4/2 = 2 ------ (2)

Adding eq1 and eq2

⇒ 2a = 12

⇒ a = 6 km/hr

Also,

b = a – 2 = 4 km/hr

Therefore,

Her speed of rowing in still water is 6 km/h and the speed of the current is 4 km/h

Let the speed of car from A be ‘a’ and of car from B be ‘b’

Speed = distance/time

Relative speed of cars when moving in same direction = a + b

Relative speed of cars when moving in opposite direction = a – b

Given,

Places A and B are 100 km apart on a highway.

One car starts from A and another from B at the same time.

If the cars travel in the same direction at different speeds, they meet in 5 hours.

If they travel towards each other, they meet in 1 hour

⇒ a – b = 100/5 = 20 (1)

Also,

a + b = 100/1 = 100 (2)

Adding (1) and (2)

a - b + a + b = 20 + 100

⇒ 2a = 120

⇒ a = 60 km/hr

Putting value of a in (1) we get,

Thus,

b = 60 – 20 = 40 km/hr

Speed of two cars are 60 km/h and 40 km/h

Let’s assume, 

The speed of the train be x km/hr 

The speed of the car = y km/hr 

From the question, it’s understood that there are two parts 

# Part 1: When Ramesh travels 160 Km by train and the rest by car. 

# Part 2: When Ramesh travels 240 Km by train and the rest by car. 

Part 1, 

Time taken by Ramesh to travel 160 km by train = 160/x hrs [∵ time = distance/ speed] 

Time taken by Ramesh to travel the remaining (760 – 160) km i.e., 600 km by car =600/y hrs 

So, the total time taken by Ramesh to cover 760Km = 160/x hrs + 600/y hrs 

It’s given that, 

Total time taken for this journey = 8 hours 

So, by equations its 

160/x + 600/y = 8 

20/x + 75/y = 1 [on dividing previous equation by 8] …………………… (i) 

Part 2, 

Time taken by Ramesh to travel 240 km by train = 240/x hrs 

Time taken by Ramesh to travel (760 – 240) = 520 km by car = 520/y hrs 

For this journey, it’s given that Ramesh will take a total of is 8 hours and 12 minutes to finish. 

240/x + 520/y = 8 hrs 12 mins = 8 + (12/60) = 41/5 hr 

240/x + 520/y = 41/5 

6/x + 13/y = 41/200 ………. (ii) 

Solving (i) and (ii), we get the required solution 

Let’s take 1/x = u and 1/y = v, 

So, (i) and (ii) becomes, 

20u + 75v = 1 ……….. (iii) 

6u + 13v = 41/200 ……. (iv) 

On multiplying (iii) by 3 and (iv) by 10, 

60u + 225v = 3 

60u + 130v = 41/20 

Subtracting the above two equations, we get 

(225 – 130)v = 3 – 41/20 

95v = 19/ 20 

⇒ v = 19/ (20 x 95) = 1/100 

⇒ y = 1/v = 100 

Using v = 1/100 in (iii) to find v, 

20u + 75(1/100) = 1 

20u = 1 – 75/100 

⇒ 20u = 25/100 = 1/4 

⇒ u = 1/80 

⇒ x = 1/u = 80 

So, the speed of the train is 80 km/hr and the speed of car is 100 km/hr.

Let’s consider the car starting from point A as X and its speed as x km/hr. 

And, the car starting from point B as Y and its speed as y km/hr. 

It’s seen that there are two cases in the question: 

# Case 1: Car X and Y are moving in the same direction 

# Case 2: Car X and Y are moving in the opposite direction 

Let’s assume that the meeting point in case 1 as P and in case 2 as Q. 

Now, solving for case 1: 

The distance travelled by car X = AP 

And, the distance travelled by car Y = BP 

As the time taken for both the cars to meet is 8 hours, 

The distance travelled by car X in 7 hours = 8x km [∵ distance = speed x time] 

⇒ AP = 8x 

Similarly, 

The distance travelled by car Y in 8 hours = 8y km 

⇒ BP = 8Y 

As the cars are moving in the same direction (i.e. away from each other), we can write 

AP – BP = AB 

So, 8x – 8y = 80 

⇒ x – y = 10 ……… (i) [After taking 8 common out] 

Now, solving for case 2: 

In this case as it’s clearly seen that, 

The distance travelled by car X = AQ 

And, The distance travelled by car Y = BQ 

As the time taken for both the cars to meet is 1 hour and 20 min, 

⇒ 1 + (20/60) = 4/3 hr

The distance travelled by car x in 4/3 hour = 4x/3 km 

⇒ AQ = 4x/3 

Similarly, 

The distance travelled by car y in 4/3 hour = 4y/3 km 

⇒ BQ = 4y/3 

Now, since the cars are moving in the opposite direction (i.e. towards each other), we can write 

AQ + BQ = AB 

⇒ 4x/3 + 4y/3 = 80 

⇒ 4x + 4y = 240 

⇒ x + y = 60 …………… (ii) [After taking LCM] 

Hence, by solving (i) and (ii) we get the required solution 

From (i), we have x = 10 + y……. (iii) 

Substituting this value of x in (ii). 

⇒ (10 + y) + y = 60 

⇒ 2y = 50 

⇒ y = 25 

Now, using y = 30 in (iii), we get 

⇒ x = 35 

Therefore, 

– Speed of car X = 35 km/hr. 

– Speed of car Y = 25 km/hr.

Let the required fraction be x/y . 

Then, we have: 

x + y = 8 ……(i) 

And, x+3/y+3 = 3/4 

⇒4(x + 3) = 3(y + 3) 

⇒4x + 12 = 3y + 9

⇒ 4x – 3y = -3 ……(ii) 

On multiplying (i) by 3, we get: 

3x + 3y = 24 

On adding (ii) and (iii), we get: 

7x = 21 

⇒ x = 3 

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

3 + y = 8 

⇒ y = (8 – 3) = 5 

∴ x = 3 and y = 5 

Hence, the required fraction is 3 5 .

Let one of the number be ‘x’

The other number as 184 – x

So, One-third of one part may exceed one-seventh of another part by 8.

x/3 – (184-x)/7 = 8

LCM for 3 and 7 is 21

(7x – 552 + 3x)/21 = 8

By cross-multiplying we get,

(7x – 552 + 3x) = 8(21)

10x – 552 = 168

10x = 168 + 552

10x = 720

x = 720/10

= 72

∴ One of the number is 72 and other number is 184 – x => 184 – 72 = 112.

Let the two parts be x and (184 – x)Then,

= (1/3)x – (1/7) (184 – x) = 8

= (1/3)x – (184/7) + (x/7) = 8

Transposing – (184/7) to LHS and it becomes (184/7)

= (1/3) x + (x/7) = 8 + (184/7)

= [(7 + 3)/21]x = (56 +184)/7

= (10/21) x = 240/7

Multiplying both side by (21/10)

= (10/21)x × (21/10) = (240/7) × (21/10)

= x = (240 × 21)/ (7 × 10)

= x = (24 × 3)/ (1 × 1)

= x = 72

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

⇒ (8x + 3) (x – 5) = 4x (2x – x)

⇒ 8x² – 40x + 3x – 15 = 8x² – 16x

⇒ – 40x + 3x + 16x = 15

⇒ – 21x = 15

⇒ x = - 15/21

⇒ x = - 5/7

4x – [2 + {x – (3 – x)}] = 3x + 6
⇒ 4x – [2 + (x – 3 + x)] = 3x + 6
⇒ 4x – [2 + 2x – 3] = 3x + 6
⇒ 4x – (2x – 1) = 3x + 6
⇒ 4x – 2x + 1 = 3x + 6
⇒ 2x + 1 = 3x + 6
⇒ 2x – 3x = 6 – 1
⇒ – x = 5
⇒ x = – 5

3(x + 5) = 4x + 9

⇒ 3x + 15 = 4x + 9

⇒ 3x – 4x = 9 – 15

On transposing 4x and 15

⇒ – x = – 6

-x/-1 = - 9/1

On dividing by – 1 on both sides

⇒ x = 6

Let the actual speed of the man be x km/hr and y be the actual time taken by him in hours. 

So, we know that 

Distance covered = speed x distance

⇒ Distance = x × y = xy …………(i) 

First condition from the question says that, 

If the speed of the man increase by 1/2 km/hr, the journey time will reduce by 1 hour. 

Showing this using variables, we have 

⇒ When speed is (x + 1/2) km/hr, time of journey = y – 1 hours 

Now, 

Distance covered = (x + 1/2) x (y – 1) km 

Since the distance is the same i.e xy we can equate it, [from (i)] xy = (x + 1/2) x (y – 1) 

And we finally get, 

-2x + y – 1 = 0 ………………(ii) 

From the second condition of the question, we have 

If the speed reduces by 1 km/hr then the time of journey increases by 3 hours.

⇒ When speed is (x-1) km/hr, time of journey is (y+3) hours 

Since, the distance covered = xy [from (i)] 

xy = (x-1)(y+3)

⇒ xy = xy – 1y + 3x – 3 

⇒ xy = xy + 3x – 1y – 3 

⇒ 3x – y – 3 = 0 ……………… (iii) 

From (ii) and (iii), the value of x can be calculated by 

(ii) + (iii)

⇒ x – 4 = 0 

x = 4 

Now, y can be obtained by using x = 4 in (ii) 

-2(4) + y – 1 = 0 

⇒ y = 1 + 8 = 9 

Hence, putting the value of x and y in equation (i), we find the distance 

Distance covered = xy 

= 4 × 9 

= 36 km 

Thus, the distance is 36 km and the speed of walking is 4 km/hr.

Let the required fraction be x/y . 

Then, we have: 

y = x + 11 

⇒ y – x = 11 ……(i) 

Again, x+8/y+8 = 3/4 

⇒4(x + 8) = 3(y + 8) 

⇒4x + 32 = 3y + 24 

⇒ 4x – 3y = -8 ……(ii) 

On multiplying (i) by 4, we get: 

4y – 4x = 44 

On adding (ii) and (iii), we get: 

y = (-8 + 44) = 36 

On substituting y = 36 in (i), we get: 

36 – x = 11 

⇒ x = (36 – 11) = 25 

∴ x = 25 and y = 36 

Hence, the required fraction is 25/36 .

Correct option is (A) \((1,\frac{-15}{2})\)

Put x = 1 in equation \(\frac{x}{2}-\frac{y}{3}=3,\) we get

\(\frac{1}{2}-\frac{y}{3}=3\)

\(\Rightarrow\) \(\frac{y}{3}=\frac{1}{2}-3=\frac{1-6}{2}=\frac{-5}2\)

\(\therefore\) \(y=\frac{-5}2\times3=\frac{-15}2\)

Hence, \((1,\frac{-15}{2})\) is a solution of equation \(\frac{x}{2}-\frac{y}{3}=3.\)

Correct option is  A) (1, \(\frac{-15}{2}\))

Let the number of notes of Rs 10 are  x

Number of notes of Rs 20 are 50 - x

Amount due to Rs 10 notes = 10x

Amount due to Rs 20 notes = 20 \(\times\) (50 - x) = 1000 - 20x

According to the question total amount = Rs 800

10x +1000 - 20x = 800

-10x = 800 - 1000

x = 20

Therefore the number of notes of Rs 10 are 20

Number of notes of Rs 20 are 50 - 20 =30

Let the larger number be x and the smaller number be y. 

Then, we have: 

x + y = 16 ……(i) 

And, 1/x + 1/y = 1/3 ……(ii) 

⇒3(x + y) = xy 

⇒3 × 16 = xy [Since from (i), we have: x + y = 16] 

∴ xy = 48 …….(iii) 

We know: 

(x – y)² = (x + y)² – 4xy 

(x – y)² = (16)² – 4 × 48 = 256 – 192 = 64 

∴ (x – y) = ±√64 = ±8 

Since x is larger and y is smaller, we have: 

x – y = 8 ………(iv) 

On adding (i) and (iv), we get: 

2x = 24 

⇒x = 12 

On substituting x = 12 in (i), we get: 

12 + y = 16 

⇒ y = (16 – 12) = 4 

Hence, the required numbers are 12 and 4

Let the required number be x.

Then, according to question,

\(\frac{5+x}{13+x} = \frac{3}{5}\)

⇒ 5(5 + x) = 3(13 + x)

⇒ 25 + 5x = 39 + 3x

⇒ 5x – 3x = 39 – 25

⇒ 2x = 14

⇒ x = 14/2

⇒ x = 7

Hence, the required number be 7.

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

⇒ 6(4x + 8) = 5(5x + 8)

By cross multiplication

⇒ 24x + 48 = 25x + 40

⇒ 24x – 25x = 40 – 48

On transposing 48 and 25x

⇒ – x = – 8

⇒ -x/-1 = -8/-1

On dividing by – 1 on both sides

⇒ x = 8

A boat goes 32 km upstream and 36 km downstream in 7 hours. In t hours, it can go 40 km upstream and 48 km downstream. If x represents the speed of the boat in still water in km/hr and y represents the speed of the stream in km/hr, then x + y = 12, x - y = 6.

Let the larger number be x and the smaller number be y. 

Then, we have: 

x + y = 16 ……(i) 

And, 1/x + 1/y = 1/3 ……(ii) 

⇒3(x + y) = xy 

⇒3 × 16 = xy [Since from (i), we have: x + y = 16] 

∴ xy = 48 …….(iii) 

We know: 

(x – y)² = (x + y)² – 4xy 

(x – y)² = (16)² – 4 × 48 = 256 – 192 = 64 

∴ (x – y) = ±\(\sqrt{64}\) = ± 8 

Since x is larger and y is smaller, we have: 

x – y = 8 ………(iv) 

On adding (i) and (iv), we get: 

2x = 24 

⇒x = 12 

On substituting x = 12 in (i), we get: 

12 + y = 16 ⇒ y = (16 – 12) = 4 

Hence, the required numbers are 12 and 4.

Let the required fraction be x/y . 

As per the question 

x + y = 4 + 2x 

⇒ y – x = 4 ……(i) 

After changing the numerator and denominator 

New numerator = x + 3 

New denominator = y + 3 

Therefore x+3/ y+3 = 2/3 

⇒3(x + 3) = 2(y + 3) 

⇒3x + 9 = 2y + 6 

⇒ 2y – 3x = 3 ……(ii) 

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

3y – 2y = 12 – 3 

⇒y = 9 

Now, putting y = 9 in (i), we get: 

9 – x = 4

⇒ x = 9 – 4 = 5 

Hence, the required fraction is 5/9 .

Let the required fraction be x/y . 

As per the question 

x + y = 4 + 2x 

⇒ y – x = 4 ……(i) 

After changing the numerator and denominator 

New numerator = x + 3 

New denominator = y + 3 

Therefore 

x+3/y+3 = 2/3 

⇒3(x + 3) = 2(y + 3) 

⇒3x + 9 = 2y + 6 

⇒ 2y – 3x = 3 ……(ii) 

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

3y – 2y = 12 – 3 

⇒y = 9 

Now, putting y = 9 in (i), we get: 

9 – x = 4 ⇒ x = 9 – 4 = 5 

Hence, the required fraction is 5/9 .

Let the required fraction be x/y. 

Then, we have: 

x+1/y+1 = 4/5 

⇒ 5(x + 1) = 4(y + 1) 

⇒ 5x + 5 = 4y + 4 

⇒ 5x – 4y = -1 ………(i) 

Again, we have: 

x−5/y−5 = 1/2 

⇒ 2(x – 5) = 1(y – 5) 

⇒ 2x – 10 = y – 5 

⇒ 2x –y = 5 ………(ii) 

On multiplying (ii) by 4, we get: 

8x – 4y = 20 ……..(iii) 

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

3x = (20 – (-1) )=20 + 1 = 21 

⇒ 3x=21 

⇒ x=7 

On substituting x = 7 in (i), we get 

5 × 7 – 4y = -1 

⇒ 35 – 4y = -1 

⇒ 4y = 36 

⇒ y = 9 

∴ x = 7 and y = 9 

Hence, the required fraction is 7/9.

Let the speed of steamer be x km/hr.

Speed of stream = 1 km/hr.

Downstream speed = (x + 1) km/hr.

Upstream speed = (x – 1) km/hr.

By using the formula

Distance = speed × time

= (x + 1) × 9 and

= (x – 1) × 10

9x + 9 = 10x – 10

9x – 10x = -10 -9

-x = -19

x = 19 km/hr.

∴ The speed of the steamer in still water is 19 km/hr.

Distance between the ports is 9(x + 1) = 9(19+1) = 9(20) = 180 km.

Le the numerator be x and the denominator be (2x – 2)

By using the formula

Fraction = numerator/denominator

= x / (2x – 2)

So, the numerator and denominator are increased by 3, then fraction is 2/3

(x + 3)/(2x – 2 + 3) = 2/3

(x + 3)/(2x + 1) = 2/3

By cross-multiplying we get,

3(x + 3) = 2(2x + 1)

3x + 9 = 4x + 2

3x – 4x = 2 – 9

-x = -7

x = 7

∴ The numerator is x = 7, denominator is (2x – 2) = (2(7) – 2) = 14-2 = 12

And the fraction is numerator/denominator = 7/12

Let the speed of steamer = x km/hr

Speed of stream = 1 km/hr

Downstream speed = (x + 1) km/hr

Upstream speed = (x – 1) km/hr

Distance = speed × time

⇒ 9 (x + 1)

= 10 (x – 1)9 x + 9

= 10 x – 10x

= 10 + 9 = 19 km/hr

Therefore speed of the steamer is 19 km/hr

Distance travelled = 9(x + 1) = 9 × 20 = 180 km.

Let the speed of one train be x km/hr.

Speed of other train be (x + 5) km/hr.

Total distance between two stations = 340 km

By using the formula

Distance = speed × time

So, Distance covered by one train in 2 hrs. Will be x×2 = 2x km

Distance covered by other train in 2 hrs. Will be 2(x + 5) = (2x + 10) km

Distance between the trains is 30 km

2x + 2x + 10 + 30 = 340

4x + 40 = 340

4x = 340 – 40

4x = 300

x = 300/4

= 75

∴ The speed of one train is x = 75 km/hr.

Speed of other train is (x + 5) = 75 + 5 = 80 km/hr.

Let the present age of the man be x years and that of his son be y years. 

After 5 years man’s age = x + 5 

After 5 years ago son’s age = y + 5 

As per the question 

x + 5 = 3(y + 5) 

⇒ x – 3y = 10 ……………(i) 

5 years ago man’s age = x – 5 

5 years ago son’s age = y – 5 

As per the question 

x – 5 = 7(y – 5) 

⇒ x – 7y = -30 …….(ii) 

Subtracting (ii) from (i), we have 

4y = 40 ⇒ y = 10 

Putting y = 10 in (i), we get 

x – 3 × 10 = 10 

⇒ x = 10 + 30 = 40

Hence, man’s present age = 40 years and son’s present age = 10 years

Let the present age of the man be x years and that of his son be y years. 

After 5 years man’s age = x + 5 

After 5 years ago son’s age = y + 5 

As per the question 

x + 5 = 3(y + 5) ⇒ x – 3y = 10 ……………(i) 

5 years ago man’s age = x – 5 

5 years ago son’s age = y – 5 

As per the question 

x – 5 = 7(y – 5) ⇒ x – 7y = -30 …….(ii)

Subtracting (ii) from (i), we have 

4y = 40 

⇒ y = 10 

Putting y = 10 in (i), we get 

x – 3 × 10 = 10 

⇒ x = 10 + 30 = 40

Hence, man’s present age = 40 years and son’s present age = 10 years.

Home is the happiest place in the world for all of us. No other place gives us comfort and protection as the home gives us. No other place makes us happy as the home does.

Oliver Twist and his companions suffered the tortures of slow starvation. They became wild with hunger. A tall boy of them announced that he would eat the boy who slept next to him unless he had enough food to eat. A council was held and it was decided that Oliver should ask the master for more. That evening after they had eaten the served gruel, Oliver went to the master and requested him for more. Thus, the hunger of the tall boy compelled him to say this. His hunger and misery too compelled him to say this.

Moment of inertia of a body depends on position and orientation of the axis of rotation. It also depends on shape, size of the body and also on the distribution of mass of the body about the given axis.

Rotational analogue of mass of a body is moment of inertia of the body.

As x-axis lies on xy plane and xz.

So its distance from xy and xz plane is 0.

∴ By basic definition of 3-D coordinate we can say that y-coordinate and z–coordinate are 0.

As, perpendicular is drawn from point P(3, 4, 5) to x-axis , so distance of point of intersection of this line from yz plane remains the same.

∴ x-coordinate of the new point(L) = 3

Or we can say that coordinates of L are (3, 0, 0)

Hence, option(A) is the only correct choice.

The co-ordinates of point L are (6, 7, 0).

(d) \(\frac{52!}{(16!)^3\times(3!)}\)

First player can get 16 cards in ⁵²C₁₆ ways 

Second player can get 16 cards in ³⁶C₁₆ ways 

Third player can get 16 cards in ²⁰C₁₆ ways 

Fourth player can get 4 cards in ⁴C₄ ways 

But the first three sets can be interchanged in 3! ways 

∴ Required number of ways = ⁵²C₁₆ × ³⁶C₁₆ × ²⁰C₁₆ × ⁴C₄ × \(\frac1{3!}\)

= \(\frac{52!}{(16!)^3\times(3!)}\)

We have, 

Number of ways of choosing 4 cards from a pack of 52 cards is the number of combinations of 52 different things taken 4 at time 

∴ Required number of ways = ⁵²C₄

= (52 x 51 x 50 x 49)/(4 x 3 x 2 x 1) = 270725

(i) There are four suits, namely diamond, club, spade, heart and each suit has 13 cards. We have to choose 4 cards of the same suit so 4 diamond cards out of 13 diamond cards can be selected in ¹³C₄ ways. Similarly,there are ¹³C₄ ways of choosing 4 spades and ¹³C₄ ways of choosing 4 hearts. 

∴ Required number of ways 

= ¹³c₄ + ¹³c₄ + ¹³c₄ + ¹³c₄

= 4x¹³C₄ = 4 x (13 x 12 x 11 x 10)/(4 x 3 x 2 x 1) = 2860

(ii) There are 13 cards in each suit we have to select 4 cards belonging to 4 different suits. There are ¹³C₁ ways of choosing one card from 13 diamond cards, ¹³C₁ ways of choosing 1 card from 13 cards of spades, ¹³C₁ ways of choosing 1 card from 13 cards of club and ¹³C₁ ways of choosing 1 card from 13 cards of hearts. By multiplication principle, the required number of ways 

= ¹³C₁ x ¹³C₁ x ³C₁ x ³C₁ 

= 13 x 13 x 13 x 13 = 134  =28561

(iii) Face cards means Kings, Queen, Jack there are 4 suits, each suit has 3 face cards. Therefore there are 12 face cards and 4 are to be selected out of 12 cards, in 12C4 ways Required number of ways = 12C4

= (12 x 11 x 10 x 9)/(4 x 3 x 2 x 1) = 495

(iv) There are 26 red cards and 26 black cards. 

∴ We have select 2 red and 2 black cards. This can be done in ²⁶C₂ x ²⁶C₂ ways. 

= (26 x 25)/(2 x 1) x (26 x 25)/(2 x 1)

= 325 x 325 = 105625

∴ Required number of ways = ²⁶C₂ x ²⁶C₂

= (26 x 25)/(2 x 1) x (26 x 25)/(2 x 1)

= 325 x 325 = 105625

(v) There are 26 red and 26 black cards

∴ 4 red cards can be selected in ²⁶C₄ ways 4 black cards can be selected in ²⁶C₄ ways 

∴ Required number of ways = ²⁶C₄ + ²⁶C₄

= 2x²⁶C₄ = (26 x 25 x 24 x 26)/(4 x 3 x 2 x 1) = 29900

Total letters = 9, M = 2, T = 2, E = 2 Total number of permutations \(\frac{9!}{(2!)^3}\)

(a) Number of permutations which being in with T and end with T is given by \(\frac{7!}{(2!)^2}\)

(b) Vowels are together can be taken as 1 unit.

4 vowels = 1 and remaining 5, totally 6 can be permuted (I 0 E)(112) in \(\frac{6!}{(2!)^2}\) and the vowels they themselves can be done in \(\frac{4!}{(2!)}\) ways

∴ Total number of ways \(\frac{6!}{(2!)^2}\) . \(\frac{4!}{(2!)}\)

(c) No two vowels are together: Should be in between the consonants, i.e. -C – M – M – T – T. 

– There are 6 vacant places in between consonants to arrange 4 vowels. This can be arranged in \(\frac{^6P_4}{2!}\) and 5 consonants can be arranged in \(\frac{5!}{(2!)^2}\)

∴ Total number of ways =  \(\frac{^6P_4}{2!}\) x \(\frac{5!}{(2!)^2}\)

(d) End with MITE: The remaining 5 letters can be arranged in 5! ways.