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Given:

Crosses pole = 10 seconds and platform = 20 seconds

Platform = 100 m

Formula used:

Speed = Distance / time

1 kmph = 5/18 m/s

Calculation:

Form two equations by assuming length of train L and speed S

⇒ Pole: 10 = D / S

⇒ 10 S = D …..(1)

⇒ Platform: 20 = (D + 100) / S

⇒ 20S = D + 100

⇒ 20S = 10S + 100 ….(From (1))

⇒ 10 S = 100

⇒ S = 10 m/s

Converting m/s to kmph

⇒ 10 m/s = 10 × 18 / 5

⇒ S = 36 kmph

Relative speed = Speed 1 + Speed 2

⇒ Relative speed = 36 + 72

⇒ Relative speed = 108 kmph

∴ Relative speed is 108 kmph

Abstract data types or ADTs are a mathematical specification of a set of data and the set of operations that can be performed on the data. They are abstract in the sense that the focus is on the definitions of the constructor that returns an abstract handle that represents the data, and the various operations with their arguments. The actual implementation is not defined, and does not affect the use of the ADT.

For example, rational numbers (numbers that can be written in the form a/b where a and b are integers) cannot be represented natively in a computer. A Rational ADT could be defined as shown below.

Construction: Create an instance of a rational number ADT using two integers, a and b, where a represents the numerator and b represents the denominator

Operations: addition, subtraction, multiplication, division, exponentiation, comparison, simplify, conversion to a real (floating point) number.

To be a complete specification, each operation should be defined in terms of the data. For example, when multiplying two rational numbers a/b and c/d, the result is defined as ac/bd. Typically, inputs, outputs, preconditions, postconditions, and assumptions to the ADT are specified as well. 

When realized in a computer program, the ADT is represented by an interface, which shields a corresponding implementation. Users of an ADT are concerned with the interface, but not the implementation, as the implementation can change in the future. ADTs typically seen in textbooks and implemented in programming languages (or their libraries) include: 

• String ADT 

• List ADT 

• Stack (last-in, first-out) ADT 

• Queue (first-in, first-out) ADT 

• Binary Search Tree ADT 

• Priority Queue ADT 

• Complex Number ADT (imaginary numbers) 

There is a distinction, although sometimes subtle, between the abstract data type and the data structure used in its implementation. For example, a List ADT can be represented using an array-based implementation or a linked-list implementation. A List is an abstract data type with well-defined operations (add element, remove element, etc.) while a linked-list is a pointer-based data structure that can be used to create a representation of a List. The linked-list implementation is so commonly used to represent a List ADT that the terms are interchanged and understood in common use. 

Similarly, a Binary Search Tree ADT can be represented in several ways: binary tree, AVL tree, red-black tree, array, etc. Regardless of the implementation, the Binary Search Tree always has the same operations (insert, remove, find, etc.) 

Cocktail sort, also known as bidirectional bubble sort, cocktail shaker sort, shaker sort, ripple sort, or shuttle sort, is a stable sorting algorithm that varies from bubble sort in that instead of repeatedly passing through the list from top to bottom, it passes alternately from top to bottom and then from bottom to top. It can achieve slightly better performance than a standard bubble sort.

Complexity in Big O notation is O(n²) for a worst case, but becomes closer to O(n) if the list is mostly ordered at the beginning. 

void cocktail_sort (int A[], int n)

{

int left = 0, right = n;

bool finished; 

do

{

finished = true;

--right;

for (int i = left; i < right; i++)

if (A[i] > A[i+1]) 

{

std::swap(A[i], A[i+1]);

finished = false; 

}

if (finished) return; finished = true;

for (int i = right; i > left; i--)

if (A[i] < A[i-1])

{

std::swap(A[i], A[i-1]);

finished = false; 

}

++left; 

} while (!finished);

}

(i ) Leafcount (T)

{

static int n=0; 

if (T!= NULL)

{leaf count (T→ left); 

if (T→left == NULL && T→right = NULL) 

n++

leafcount (T→right)

}

return (n);}

(ii) Leafcount (T)

{

static int n=0;

if (T! = NULL)

{leaf count (T→ left);

if (T→left!= NULL && T→right! = NULL) 

n++

leafcount (T→right)

}

return (n);} 

The two methods to resolve collision during hashing are: Open addressing and Chaining,

Open addressing: The simplest way to resolve a collision is to start with the hash address and do a sequential search through the table for an empty location. The idea is to place the record in the next available position in the array. This method is called linear probing. An empty record is indicated by a special value called null. The major drawback of the linear probe method is clustering. 

Chaining: In this technique, instead of hashing function value as location we use it as an index into an array of pointers. Each pointer access a chain that holds the element having same location. 

Here List={1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16}

Binary Search for key 2  

(1) Here bottom =1 Top =16 and middle =(1+16)/2 = 8

Since 2 < list(8) 

(2) bottom = 1 Top = middle-1=7 and middle=(1+7)/2 =4

2 < list(4) 

(3) bottom = 1 Top = middle-1=3 and middle=(1+3)/2 = 2

2 = List(2)

So total number of comparisons require = 3

Binary Search for key = 10

(1) Here bottom=1 Top=16 and middle = 8

10 > List(8)

(2) bottom = middle+1=9 Top=16 middle=(9+16)/2=12 

10 < List(12)

(3) bottom =9 Top=middle-1=11 middle=(9+11)/2=10

10 = List(10)

So total no of comparisons = 3

Binary Search for key = 15 

(1) Here bottom=1 Top=16 and middle = 8

15 > List(8) 

(2) bottom = middle+1=9 Top=16 middle=(9+16)/2=12

15 > List(12) 

(3) bottom =middle+1=13 Top=16 middle=(13+16)/2=14

15 > List(14)

(4) bottom = middle+1 =15 Top=16 middle=(15+16)/2=15 

15=List(15)

So total no of comparisons = 4

The proof is by induction on the size n of T .

Let L(T) = No of leaves & D₂(T)= No of nodes of T of degree 2

To Prove D₂(T)=L(T)-1 

Basic Case : n=1 , then T consists of a single node which is a leaf . 

L(T)=1 and D₂(T)=0

So D₂(T)=L(T)-1 

Induction Step : Let n > 1 and assume for all non empty trees T1 of size k<n  that D₂(T`)=L(T`)-1 

Since n >1 , at least one of x=left(T) or y=right(T) is non empty

Assume x is non empty ; the other case is symetric By the induction hypothesis , 

D₂(x) = L(y)-1

IF y is empty , then again by the induction hypothesis ,

D₂(y) =L(y)-1 and

D₂(T) =D₂(x)+D₂(y)+1

=L(x)-1+L(y)-1+1

= L(x)+L(y)-1 

D₂(T) = L(T)-1 Hence Proved

Given:

Bus’s speed from Town R to Town S = 29 kmph

Bus’s speed from Town S to Town R = 26 kmph

Formula Used:

Average Speed = 2 (A × B) / (A + B)

Calculations:

As two equal distance are covered at two different speed

Average Speed = 2 (A × B) / (A + B)

⇒ Average Speed = 2 × 29 × 26 / (29 + 26)

⇒ Average Speed = 1508 / 55

⇒ Average Speed = 27.41 ≈ 27.

The approximate average speed of the bus is 27 kmph.

Given:

Journey time Train A = 12 hours

Journey time Train B = 16 hours

Formula used:

Speed = Distance/time

1 km/h = 5/18 m/s

Calculations:

Let the distance = 12 × 16 = 192 Km

⇒ Speed of Train A = 192/12 = 16 km/hr

⇒ Speed of Train B = 192/16 = 12 km/hr

⇒ Ratio of speed = 16 : 12 

⇒ Ratio of speed = 4 : 3

∴ The Ratio of speed is 4 : 3 

Note:

(If the time taken by taken two trains is given to complete the same distance then we can find directly the ratio of Speed of the train)

Ratio of speed of trains = 1/Ratio of time of train 

Ratio of Time of trains = 12 : 16 = 3 : 4

Ratio of speed of train = 16: 12 = 4 : 3

Let the speed of the stream is x km/hour. 

The speed of the boat in still water is 4 km/hour. 

The upstream speed of the boat is (4 – x) km/hour. 

The downstream speed of the boat is (4 + x) km/hour. 

Distance = 30 km 

We know that,

Time = \(\frac{Distance}{speed}\).

Now,

Time taken by the boat in going 30 km upstream = \(\frac{30}{4-x}\) hours.

And,

Time taken by the boat in going 30 km downstream = \(\frac{30}{4+x}\) hours.

Now,

According to given conditions

3(\(\frac{30}{4+x}\)) = \(\frac{30}{4-x}\)

⇒ \(\frac{3}{4+x}\) = \(\frac{1}{4-x}\)

⇒ 4 + x = 12 − 3x 

⇒ 4x = 12 − 4 = 8 

⇒ x = 2.

Hence, 

The speed of the stream is 2 km/hour.

(ii) Let Jayant answered x number of questions correctly and y number of questions he gives wrong answer. 

∵ Jayant answered 120 questions. 

∴ x + y = 120 … (1) 

∵ Jayant got 348 marks. 

∴ 5 – 2y = 348 … (2) 

(∵ Each correct answer awarded as 5 marks but wrong answer deducted 2 marks) 

Now,

Putting y = 120 – x from equation (1) in to equation (2), we get 

5x – 2 (120 – x) = 348 

⇒ 5x – 240 + 2x = 348 

⇒ 7x = 348 + 240 = 588 

⇒ x = \(\frac{588}{7}\) 

= 84.

Hence,

Jayant answered 84 number of questions correctly.

Given:

Speed to cover 30 km = 15 kmph

Speed to cover 36 km = 36 kmph

Speed to cover 20 km = 5 kmph

Formula used:

Speed = Distance / time

1 kmph = 5/18 m/s

Calculation:

Calculating time taken by each segments

⇒ Segment 1: T1 = 30 / 15

⇒ T1 = 2 hours

⇒ Segment 2: T2 = 36 / 36

⇒ T2 = 1 hour

⇒ Segment 3: T3 = 20/5

⇒ T3 = 4 hours

Total time taken = 2 + 1 + 4

∴ Total time taken is 7 hours

Correct option - (C) right, root, left

Given:

Distance is travelled by a person equally with the speed of 4 km/hr, 6 km/hr and 8 km/hr respectively

Total time taken = 32.5 minutes

Formula used:

Time = Distance/Speed

1 hour = 60 minutes

Calculation:

Let the equal distance be D km

Then total distance = 3D

Now, the total time is taken = 32.5 minutes

⇒ (D/4) + (D/6) + (D/8) = (32.5/60) hrs

⇒ (6D + 4D + 3D)/24 = 32.5/60

⇒ 13D = 32.5 × 24/60

⇒ D = (32.5 × 24)/(60 × 13)

⇒ D = 1 km

Total distance = 3 × 1

⇒ 3 km.

∴ The total distance travelled by him is 3 km 

Given:

Speed1 = 12km/h

Speed2 = 18km/h

Speed3 = 24km/h

Total time taken = 13 hours

Formula used:

Speed × times = Distance

⇒ Distance/speed = times taken

Calculation:

∵ Mohan travels 3 equal distance with different speed.

 Let, the equal distance travel = a km

∴ (a/12) + (a/18) + (a/24) = 13

Taking LCM of 12, 18 and 24, we get 72.

∴ (6a/72) + (4a/72) + (3a/72) = 13

⇒ (6a + 4a + 3a)/72 = 13

⇒ 13a/72 = 13

⇒ a = (13 × 72)/13

⇒ a = 72 km

∴ Total distance travelled by him = a + a + a

= 3 × 72 = 216 km

GIVEN:

Raj covers the same distances with a speed of 2 kmph, 4 kmph, and 8 kmph. He takes 105 min for a whole journey.

FORMULA USED:

Speed = Distance/Time

CALCULATION:

Suppose distances are of ‘x’ km.

So,

x/2 + x/4 + x/8 = 105/60

⇒ 7x/8 = 105/60

⇒ x = 2

Given:

The man covers 36 km distance at the speed = 12 km/hr. 

He covers 40 km distance at the speed = 10 km/hr. 

He covers 54 km distance at the speed = 9 km/hr. 

Formula:

Average speed = (Total distance/Total time)

Calculation:

Time taken by man to cover 36 km distance at the speed of 6 km/hr. = 36/12 = 3 hrs.

Time taken by man to cover 40 km distance at the speed of 5 km/hr. = 40/10 = 4 hrs.

Time taken by man to cover 54 km distance at the speed of 9 km/hr. = 54/9 = 6 hrs.

Total distance = 36 + 40 + 54 = 130 km

Total time = 3 + 4 + 6 = 13 hrs.

∴ Average speed = 130/13 = 10 km/hr. 

Given:

Namrata Time taken

Walking + vehicle = 4 hrs 15 mins

Both ways walking = 5 hrs 30 mins

Calculation:

Namrata time taken for walking for one ways

⇒ (5 hrs 30 mins)/2

⇒ 2 hrs 45 mins

Namrata time taken from vehicle for one ways

⇒ (4 hrs 15 mins) - (2 hrs 45 mins)

⇒ 1 hrs 30 mins

Namrata time taken from vehicle for both ways

⇒ (1 hrs 30 mins) × 2

⇒ 3 hrs.

∴ Namrata time taken from vehicle for both ways is 3 hrs.

Given:

Speed of river flow = 4 km/hr

Speed of boat in downstream = 3 × Speed of boat in upstream

Concept used:

Speed of boat in downstream = Speed of boat in still water + speed of river flow

Speed of boat in upstream = Speed of boat in still water - speed of river flow

Calculation:

Let the speed of boat in still water be 'x' km/hr

Let the speed of boat in downstream and upstream be 'd' km/hr and 'u' km/hr respectively

Then, as per question

d = 3 × u

⇒ x + 4 = 3×(x - 4)

⇒ x + 4 = 3x - 12

⇒ x = 8

∴ Speed of boat in still water is 8 km/hr.

Given:

Rajneesh speed 6 km/h, 4 km/h, 8 km/h.

Total time = 32.5 minutes

Formula used:

Speed = Distance/Time

Calculation:

Let the distance be 'x' km.

Total time taken

⇒ (x/6) + (x/4) + (x/8) hr = 32.5 minutes

⇒ (13x/24) hr = 32.5 minutes

⇒ (13x/24) × 60 minutes = 32.5 minutes

⇒ x = 1 km

Total distance covered 

⇒ 1 + 1 + 1

⇒ 3 km

∴ The total distance is 3 km.

Given:

The sum of digits of the number = 8.

Calculation:

Let the two-digit number be 10x + y 

Where the unit digit and the tenth place digit of the number is y and x respectively.

After reversing the digits the number becomes 10y + x.

According to the 1^st condition,

x + y = 8    ----(1)

According to the 2^nd condition,

10x + y - 18 = 10y + x

⇒ 9x - 9y = 18

⇒ 9(x - y) = 18

⇒ x - y = 2    ----(2)

by adding equation (1) and equation (2), we get

2x = 10

⇒ x = 5

On putting the value of x in equation (1), we get

5 + y = 8

⇒ y = 8 - 5

⇒ y = 3

So, the two-digit number is 10 × 5 + 3

∴ The two-digut number is 53.

141x + 93y = 189

93x + 141y = 45

To Find:

value of x and y

Solution:

we have,

141x+93y=189  -------------------(i)

93x+141y=45  --------------------(ii)

multiply eq(i) by 93 and eq(ii) by 141

⇒ 93(141x+93y=189)

⇒ 141(93x+141y=45)

⇒ 13113x+8649y=17577

⇒ 13113x+19881y=6345

subtracting the equations

we get,

⇒ -11232y = 11232

⇒ y = -1

putting the value of y = -1  in equation (i)

⇒ 141x+93y=189

⇒ 141x+93*(-1)=189

⇒ 141x-93=189

⇒ 141x=189+93

⇒ 141x=282

⇒ x=2

So, x = 2 and y = -1.

Let Rs. x be monthly fixed charge of hostel & Rs.y be the cost of food per day

For student A, x  + 20 y = 1300---(1)

For student B, x + 25y = 1500----(2)

Substract (1) from (2), we get

(x + 25y) - (x + 20y) = 1500 - 1300

⇒ 5y = 200

⇒ y = 40

∴ x + 800 = 1300 (From (1))

⇒ x = 1300 - 800 = 500

Hence, hostal fixed charge is Rs. 500 and the cost of food Rs. 40 per day

Charu asked me where I had been?

Correct option is A) at/in

Sufism the liberal and mystic movement of Islam, reached India in the 11^th century.

Given:

Length of two trains is 85 m and 155 m

Speed of two trains is 62 km/h and 82 km/h.

Formula Used:

Speed = Distance/Time

Speed adds in opposite direction.

To convert km/h into m/s multiply by 5/18

Calculation:

Let the two trains be t1 and t2

with length l1 = 85 m and l2 = 155 m

and speed s1 = 62 km/h and s2 = 82 km/h.

Total length = l1 + l2

⇒ 85 + 155 ⇒ 240 m

Total speed = s1 + s2

⇒ 62 + 82 ⇒ 144 km/h

now, convert speed in m/s

⇒ 144 × 5/18

⇒ 40 m/s

Speed = Distance/Time

⇒ 40 = 240/Time

⇒ Time = 240/40

⇒ Time = 6s

∴ The time taken by trains to cross each other in 6 seconds.

Given:

Two trains, A and B, start simultaneously from two stations X and Y that

are 600 km apart and cross each other at a distance of 225 km from X.

If it took train B 8 hours to complete the journey

Formula used:

Velocity = Distance/time

Calculation:

Speed of B = 600/8

⇒ 75 km/hr

When the point X both trains cross each other

So, distance cover by B

⇒ 600 – 225

⇒ 375 km

And A is 225 km

Time taken by B

⇒ 375/75

⇒ 5 hour

In 5 hour B covered by 375 and A covered by 225

So, speed of A

⇒ 225/5

⇒ 45 km/hr

Then time to complete journey by A

⇒ 600/45

⇒ 13(1/3)

∴ 13(1/3) hours did train A complete it

Given:

5 years ago, the ratio of the ages of Kaif and Raj was 8: 5.

3 years hence, the ratio of their ages will be 15: 10

Calculation

Let the ages of Kaif and Raj 5 years ago be 8x and 5x years

Present age of Kaif = 8x + 5

Present age of Raj = 5x + 5

According to the question,

⇒ ((8x + 5) + 3)/((5x + 5) + 3) = 15/10

⇒ 80x + 50 + 30 = 75x + 75 + 45

⇒ 5x = 120 – 80

⇒ x = 8.

Raj's present age = (5x + 5) = 45 years

∴ Raj’s present age is 45 years

Quantity A:

Salary decrease by 25%.

(r/100 – r) × 100

⇒ {25/ (100 – 25)} × 100 = 33.33%

Quantity B: 33%

∴ Quantity A > Quantity B

Quantity A:

Salary of Karan is 125% higher than that of Kusum

{(Amount of increased salary/original salary)} × 100

Let the salary of Kusum is Rs. 100

∴ Salary of Karan is Rs. 225

⇒ Percentage of salary of Kusum is less than Karan = [{(225 – 100)}/225] × 100

⇒ 55.55 %

Quantity B: 60%

Quantity A:

Let the ages of A, B and C be 6x, 4x and 7x respectively

According to the question

6x + 4x + 7x = 51

x = 3

B’s age = 4 × 3 = 12 years

Quantity B: 12 years

Quantity A:

Amount invested by Mrs Sonia is Rs. 60,000.

Ratio of profit between the partners is 3 : 5.

Ratio of profit = Amount invested by one partner × time period of investment ⦂ Amount invested by other partner × time period of investment

Let the required amount which should be invested by Mrs. Anita is Rs x.

⇒ Ratio of share of profit = Sonia : Anita

⇒ Amount invested by Sonia × time period of investment ⦂ Amount invested by Anita × time period of investment

⇒ 3 : 5 = 60,000 × 1 : x × 1

⇒ x = (60,000 × 5)/3

⇒ x = Rs 1, 00,000

Quantity B: Rs. 95000

∴ Quantity A > Quantity B

Given

Six years ago, the ratio of the ages of Ramesh and Sundar was 6 : 5

Four years hence, the ratio of their ages will be 11 : 10

Calculation

Let present age of Ramesh is x years and

Present age of Sundar is y years 

Now,

Six year ago age of Ramesh and Sundar 

⇒ Ramesh age six year ago = (x - 6) years

⇒ Sundar age six year ago = (y - 6) years

From question 

⇒ (x - 6) ∶ (y - 6) = 6 ∶ 5

⇒ (x - 6)/(y - 6) = (6/5)

⇒ 5x - 30 = 6y - 36

⇒ 6y - 5x = 6    ....(1)

Now, four years hence

⇒ Ramesh age four years hence = (x + 4)

⇒ Sundar age four years hence = (y + 4)

From question 

⇒ (x + 4) ∶ (y + 4) =11 ∶ 10

⇒ (x + 4)/(y + 4) = 11/10

⇒ 10x + 40 = 11y + 44

⇒ 10x - 11y = 4    ....(2)

Multiply equation (1) on both side by 2 

⇒ (6y - 5x = 6) × 2

⇒ 12y - 10x = 12    ....(3)

Now add equation (2) and equation (3)

⇒ 10x - 11y + 12y - 10x = 12 + 4

⇒ y = 16

⇒ Sundar present age = 16 years

∴ Sundar present age = 16 years

Given:

Eight years ago ages of A and B is 9 : 10.

Four years now the ratio of Ages of A and B is 12 ∶ 13

The age of C is 4 years less than the present age of A.

Concept used:

By ratio method.

Calculation:

Let eight years ago age of A be 9x years

Eight years ago age of B be 10x years

Present age,

A's age = 9x + 8

B's age = 10x + 8

After four year,

A's age = 9x + 8 + 4 = 9x + 12

B's age = 10x + 8 + 4 = 10x + 12

According to the question,

(9x + 12)/(10x + 12) = 12/13
⇒ 13(9x + 12) = 12(10x + 12)

⇒ 117x + 156 = 120x + 144

⇒ 3x = 12

⇒ x = 4

So, Present age of A = 9x + 8 = 9 × 4 + 8 = 44

The present age of C = 4 years less than the present age of A

⇒ Present age of C = 44 – 4 = 40

∴  The Present age of C is 40 years.

Given:

Dividend = (172826 - 2156)

Divisor = 16

Concept Used:

Dividend = (divisor × quotient) + remainder 

Calculation:

(172826 - 2156) = (12^3×(26) - 2¹⁵⁶)/2⁴

⇒ (123×(26) - 2156)/24 = (2^{(6 × 26)} × 3^{(3 × 26)} - 2¹⁵⁶) /2⁴

⇒ (2(6 × 26) × 3(3 × 26) - 2156) /24 = 2¹⁵⁶(3⁷⁸ - 1)/2⁴

⇒ 2156(378 - 1)/24 = 2^{(156 -4)}(3⁷⁸ - 1) = 2(152)(378 - 1)

The divisor 16 completely divides the number (172826 - 2156), so remainder is 0

∴ (172826 - 2156) is divided by 16, the remainder is 0

Given:

Sum of the ages of a father and his daughter = 100 yr

5yrs ago ratio of father and daughter = 2 : 1

Calculations:

Let the father’s present age be x.

Let the daughter’s present age be (100 - x).

∴ According to the question

⇒ (x - 5)/(100 - x - 5) = 2/1

⇒ x - 5 = 200 - 2x - 10

⇒ 3x = 195

⇒ x = 195/3

⇒ x = 65

∴ After 10 years,

Age of father/Age of daughter = (65 + 10)/(35 + 10) = 75/45 = 5/3

∴ Required ratio is 5 : 3.

The ratio of the ages of father and daughter after 10 years is 5 : 3. 

Given:

Ronit is twice as efficient as Ajay.

He can finish work in 42 less days than Ajay.

Formula used:

Total work done = Efficiency × Total number of days

Concept used:

The ratio of efficiency of Ronit and Ajay = 2 : 1

The ratio of time of Ronit and Ajay = 1 : 2

Let the time taken by Ronit and Ajay be 1x and 2x respectively.

2x - x = 42

⇒ x = 42

Time taken by Ajay = 2x

⇒ 2 × 42

⇒ 84 days

∴ Ajay can finish the work in 84 days.

Given:

The ratio of 2 numbers = 4 ∶ 5

LCM = 700

Calculation:

Let assume that first number = 4x 

and, second number = 5x

Factors of number

4x = 4 × x,

5x = 5 × x

LCM = x × 4 × 5

HCF = x

According to question

⇒ x × 4 × 5 = 700

⇒ 20x = 700

⇒ x = 35

∴ HCF of numbers is 35 

The correct option is 1 i.e. 35

38 years.

Given:

4 years ago the ratio of ages,

⇒ Prit: Priti = 2 : 1

6 years from present age, the ratio will be

⇒ Prit: Priti = 3 : 2

Calculation:

For 4 years ago,

Let the ages of Prit and Priti = 2x : x

∴ In the present the ages will be = 2x + 4 : x + 4

For 6 years from now the ages will be = 2x + 4 + 6 : x + 4 +6

⇒2x + 10: x + 10

∴ \(\frac{{2{\rm{x}} + 10}}{{{\rm{x}} + 10}} = {\rm{\;}}\frac{{3}}{{2}}\)

⇒ 4x + 20 = 3x + 30

⇒ x = 10

The present age = 2x + 4 : x + 4

∴ Sum of present ages = (2×10 + 4) + (1×10 +4)

⇒ 24 + 14

⇒ 38yrs

12 years.

Given:

Ratio of ages of Son : Mother = 3 : 7

Ratio of ages of Son : Father = 2 : 5

Age _(M) + Age _(F) = 58 years

Calculation:

Let the ages be 6x, 14x and 15x.

According to  the question,

Age _(M) + Age _(F) = 58yrs.

⇒14x + 15x = 58yrs.

⇒ 29x = 58

⇒ x = 2

⇒ Age of a Son = 6x

⇒ 6 × 2 = 12 years

∴ Age of a son is 12 years.

Given:

Ratio of the age of mother to her son is 7 : 3

Sum of the ages of mother and the son = 60 yrs

Calculation:

Let the age of the mother be 7x

⇒ Age of her son = 3x

According to the question:

7x + 3x = 60 

⇒ x = 6

7x - 3x = 4x = 24

∴ Difference in their ages = 24 years

Given:

The ratio of the numbers = 5 ∶ 4

HCF = 6

Concept used:

Two number which is in the form of mx and my, where m is HCF.

So, their LCM can be written as mxy

Calculation:

Let the number be 5x and 4x, where x is HCF.

HCF = 6

LCM = HCF × 5 × 4

⇒ LCM = 6 × 20

⇒ LCM = 120

∴ LCM is 120.

Given:

Ratio of the numbers = 2 ∶ 3

LCM of numbers = 180

HCF = ?

Formula used:

Product of LCM and HCF of two numbers is equal to the product of the numbers

LCM × HCF = a × b (a, b are the two numbers)

Let, numbers = 2x and 3x

∴ LCM = 2 × 3 × x = 6x

∴ 6x = 180

∴ x = 180 / 6 = 30

⇒ Numbers are 30 × 2 = 60 and 30 × 3 = 90

Now, LCM × HCF = ab (a, b are the numbers)

∴ 180 × X = 60 × 90

∴ X = 5400 / 180 = 30

Therefore, HCF = 30

Correct option is E) sorry

Correct option is D) You are welcome

Correct option is D) That's all right.

Correct option is A) it's easy to do, ok

Correct option is B) Yes, please