Linear Equation In One Variable

We have the Linear Equation as follows:

4x + 7 = x + 2

⇒ 4x-x = 2-7(transpose x to LHS and 7 to RHS)

⇒ 3x = -5

⇒

x is equal to which is neither a fraction nor an integer.

Option(a) is incorrect because:

Explanation

We have the Linear Equation as follows:

3x + 2 = 5x + 2

⇒ 3x-5x = 2-2(transpose 5x to LHS and 2 to RHS)

⇒ -2x = 0

⇒

⇒

Here x is 0 and it is an integer.

Option (b) is incorrect because:

Explanation

We have the Linear Equation as follows:

4x-18 = 2

⇒ 4x = 2 + 18 (transpose 18 to RHS)

⇒ 4x = 20

⇒

⇒ x = 5

Here x is 5 and it is an integer.

Option (d) is incorrect because:

Explanation

We have the Linear Equation as follows:

5x-8 = x + 4

⇒ 5x-x = 8 + 4(transpose x to LHS and 8 to RHS)

⇒ 4x = 12

⇒

⇒ x = 3

Here x is 3 which is an integer.

The given equation is

ax + b = 0

_{⇒ax = 0-b}

_{⇒ax = -b}

_⇒

Other options are incorrect.

Given equation is :

8x-3 = 25 + 17x

⇒ 8x-17x = 25 + 3 (transpose 17x to LHS and 3 to RHS)

(divide both sides by -9)

Hence, x is a rational number.

The shifting of a number from one side of an equation to other is called Transposition.

If we have an equation, say x + a = 0, then x = -a. Here we have shifted number ‘a’ from left hand side to right hand side.

The given equation is

⇒

⇒ (transpose to LHS)

⇒ (calculate the LCM of LHS)

⇒

⇒

⇒ (divide both sides by 19)

Hence the numerical value of 2x-7.

⇒ (place the value of x)

⇒

The numerical value of x is

The given expressions 3x-4 and 2x + 1 are equal.

In this case 3x – 4 = 2x + 1

⇒ 3x - 2x = 1 + 4 (transpose 2x to LHS and 4 to RHS)

⇒ x = 5

Hence the value of x is 5.

The given equation is ax = b, i.e. ,thus x becomes a

positive integer, as both a and b are positive integers.

Linear equation in one variable has only one variable with power 1.

For example,2x + 1 = 0,4y + 3 = 9,z + 1 = 2

The algebraic expression with one variable which has the highest power of the variable as 1 is known as the Linear Expression. Hence ,1 + z is the only expression which has the power of the variable z as 1.

A linear equation in one variable has only one solution.

For example, The value of x for the equation px + q = 0

becomes unique, i.e.

Given:
To Find: s
Rearranging the values we get,

Now, take L.C.M of 5 and 3
L.C.M(5, 3) = 15
Now, the R.H.S becomes,

Given:

⇒ (cross multiplication)

⇒ y = ²

Hence the value of y is ².

Let the digit at unit’s place be b.

Then the digit at ten’s place = (3 + b)

The Number = 10(3 + b) + b = 30 + 10b + b = 11b + 30

Given: Shilpa’s age three years ago was x years.

Hence Shilpa’s present age is (x + 3)

Arpita’s present age is Shilpa’s present age = 3(x + 3)

Let the 3 consecutive multiples of 7 be 7, (7x + 7),(7x + 14)where x is a natural number.

As per the question,

7 + (7x + 7) + (7x + 14) = 357

⇒ 21x + 21 = 357

⇒ 21(x + 1) = 357

⇒ (divide both sides by 21)

⇒ x + 1 = 17

⇒ x = 17-1(transpose 1 to RHS)

Hence the smallest multiple of 7 is ,i.e. 112.

Graph sample of linear equations.

A linear equation is an algebraic equation in which every term is either a constant or the product of a constant and a single variable.

Example: 2x + y = 6 is an equation in two variables.

The highest power of both the variables is 1.

3x – 4 = 1 – 2 x

Or,3x + 2 x = 1 + 4=5

Or, 5 x=5

Or, x=1

2y = 5y –

solving the equation consist of determining which values of the variable make the equality true this value of variable is called solution of the equation

Let the number is m

So,

Putting the value of x=-2,

So the number is 3

Let the three numbers is x ,(x + 1)(x + 2)

From question we get

So the numbers are 3,4,5

Let the B’s share=x and A’s share=(x + 8)

From question we get,

The A’s share is =8 + 8.5=16.5

The process of transposing of a term of an equation is called transposition

So the term can be change to the other side of the equation by changing its sign

Like if x + a=2 is an equation and we want to transpose a then

X=2-a

So here the sign of a changes from positive to negative

From the equation,

We get,

+ 30 = 18

=

=

Let the number be X

from the question,

,

Let the present age of swarnim is a years

After 18 years it will be = years

4 times of age =4a

From the question ,

4 times of x is 4x

Let the numerator is x and denominator=x + 10

So the fraction is=

If the numerator is increased by 1 the and denominator is decreased by 1 the fraction become

So the denominator (x + 9)

Let the two multiples of 10 is 10X and 10x + 10

From the question, we get,

So the smaller multiple of 10 is=10*10=100

True.

As given three years ago, the age of boy = y

∴ the present age of the boy = (y + 3) years.

⇒ The age of boy 2 years ago = (y-2) years.

Hence, it is true.

False

As given shikha’s present age is p years.

∴ Reemu’s present age will 4p

⇒ After 5 years Reemu’s age will be (4p + 5) years.

⇒ Hence, it is not true.

false

let us assume 2-digit number be ‘yx’

‘x’ is at unit place and ‘y’ is at tenth place.

∴ according to the question

⇒ y + x = 9

∴y = 9-x

⇒ number will be (10y + x)

⇒ required number = 10(9-x) + x

⇒ required number = 90-10x + x

⇒ required number = 90-9x

⇒ Hence, it is not true.

True

let us assume Anju’s mother present age is ‘x’ years.

given Anju’s present age is ‘y’ years.

According to the question.

⇒ x + y = 65

⇒ x = (65-y)

⇒ Before 5 years Anju’s age was (y-5) years and Anju’s mother’s age was (x-5) years

⇒ Anju’s mother age before 5 years = (65-y)-5

= (60-y) years.

⇒ Hence, it is true.

False

let number of boys be 5x and girls be 4x.

∴ according to the question

⇒ 5x = 4x + 9

⇒ 5x-4x = 9

⇒ x = 9

∴ number of boys = 5x = 5 × 9 = 45

⇒ And number of girls = 4x = 4 × 9 = 36

⇒ Hence, it is not true.

true

let us assume A’s age is ‘x’ years

⇒ then B’s age will (90 – x) years

⇒ before 5 years A’s age = (x-5)years

⇒ and B’s age = (90 – x – 5) = (85 – x) years.

⇒ Hence, it is true.

False

let us assume two different equation and verify,

⇒ 6x + 6 = 12 ----(i)

⇒ 6x = 12-6

⇒ 6x = 6

⇒ x = 1

⇒ 6x-30 = -24 ----(ii)

⇒ 6x = -24 + 30

⇒ 6x = 6

⇒ x = 1

⇒ Thus we can see that two different equation can have same answer.

⇒ Hence, it is not true.

False

given equation is,

⇒ 3x-3 = 9

⇒ Transposing – 3 to RHS

⇒ 3x = 9 + 3

⇒ 3x = 12

⇒ x = 4

⇒ We get 3x = 12 on transposing – 3 to RHS.

⇒ Hence, it is not true.

False

Given equation is,

⇒ 2x = 4-x

⇒ Transposing – x to LHS

⇒ 2x + x = 4

⇒ 3x = 4

⇒ Thus we can see that on transposing – x to LHS we get .

⇒ Hence, it is not true.

False

given equation is,

⇒

⇒ (transposing to RHS)

⇒ Hence, it is not true.

False

given equation is,

⇒ Hence, it is not true.

True

given equation is,

⇒ 6x = 18

Multiplying both sides by 3

⇒ 6x × 3 = 18 × 3

⇒ 18x = 54

False

given that,

cross multiplying both sides.

∴x = 11 × 15

⇒ Hence, it is not true.

False

we know that even numbers are 2,4,6,8,10,12,….so on.

here we see that –

⇒ first even number is 2

⇒ then next is (2 + 2) = 4

⇒ then (4 + 2) = 6

⇒ then (6 + 2) = 8

⇒ so we can say that if ‘x’ is an even number then next even number be (x + 2).

⇒ Hence, it is not true.

True

let number be ‘x’

then consecutive number be (x + 1)

now according to the question.

⇒ x + x + 1 = 93

⇒ 2x + 1 = 93

⇒ 2x = 93-1 (transposing 1 to RHS)

⇒ 2x = 92

⇒ So one number is 46

∴ other consecutive number be (46 + 1) = 47

⇒ Other number =

= 47

⇒ Hence, it is true.

let one number be ‘x’

⇒ Then other number be (40-x)

⇒ Let us suppose (40 – x)>x

⇒ According to the question. When each number is increased by 8. Then,

⇒ (40-x) + 8 = 3 × (x + 8)

⇒ 48-x = 3x + 24

⇒ -x-3x = 24-48

⇒ -4x = -24

⇒ So, one number is 6

⇒ and other number = (40-6)

= 34

⇒ Difference between two numbers =

= 28

⇒ But in the question given that two differs by 40

⇒ 28≠40.

⇒ Hence, it is not true.

(x = 8)

Given equation is,

⇒ cross multiplying equations

⇒ 3x-8 = 2x

⇒ 3x-2x = 8 (transposing 2x to LHS and – 8 to RHS)

⇒ X = 8.

⇒ Hence value of x is 8.

(x = -2)

Given equation is,

⇒ cross multiplying both sides

⇒ 5x = 4x-2

⇒ 5x-4x = -2 (transposing 4x to LHS)

⇒ x = -2

⇒ Hence value of x is -2.

(x = 7)

Given equation is,

⇒ Cross multiplying both sides

⇒ 6x-9 = 4x + 5

⇒ 6x-4x = 5 + 9 (transposing 4x to LHS and 9 to RHS)

⇒ 2x = 14

⇒ Hence value of x is 7.

()

Given equation is,

⇒ Cross multiplying both sides,

⇒ 8(x-1) = 5x

⇒ 8x-8 = 5x

⇒ 8x-5x = 8 (transposing 5x to LHS and 8 to RHS)

⇒ Hence value of x is .

Given equation is,

cross multiplying both sides

⇒ 5-5x + 3 + 3x = 8-16x

⇒ -5x + 3x + 16x = 8-8

⇒ 14x = 0

⇒ Hence value of x is 0.

Given equation is,

Cross multiplying both sides.

⇒ x + 25 = 7x-6

⇒ x-7x = -6-25 (transposing 7x to LHS and 25 to RHS)

⇒ -6x = -31

⇒ Hence value of x is .

Given equation is,

Cross multiplying both sides

⇒ 20y-20 = -2y-3

⇒ 20y + 2y = -3 + 20 (transposing 2y to LHS and 20 to RHS)

⇒ 22y = 17

⇒ Hence value of y is .

Given equation is,

Cross multiplying both sides

⇒ 6x = 5x-5

⇒ 6x-5x = -5 (transposing 5x to LHS)

⇒ x = -5

⇒ Hence value of x is -5.

(x = 2)

Given equation is,

⇒ 0.4(3x-1) = 0.5x + 1

⇒ 1.2x-0.4 = 0.5x + 1

⇒ 1.2x-0.5x = 1 + 0.4 (transposing 0.5x to LHS and 0.4 to RHS)

⇒ Hence value of x is 2.

Given equation is,

⇒ 8x-7-3x = 6x-2x-3

⇒ 5x-7 = 4x-3

⇒ 5x-4x = -3 + 7 (transposing – 7 to RHS and 4x to LHS)

⇒ x = 4

⇒ Hence value of x is 4.

Given equation is,

⇒ 10x-5-7x = 5x + 15-8

⇒ 3x-5 = 5x + 7

⇒ 3x-5x = 7 + 5 (transposing 5x to LHS and 5 to RHS)

⇒ -2x = 12

⇒ Hence value of x is -6.

Given equation is,

⇒ 4t-3-(3t + 1) = 5t-4

⇒ 4t-3-3t-1 = 5t-4

⇒ t-4 = 5t-4

⇒ t-5t = -4 + 4 (transposing 5t to LHS and – 4 to RHS)

⇒ -4t = 0

⇒ Hence value of t is 0.

Given equation is,

⇒ 5(x-1)-2(x + 8) = 0

⇒ 5x-5-2x-16 = 0

⇒ 3x-21 = 0

⇒ 3x = 21 (transposing 21 to RHS)

⇒ x = 21/3

⇒ x = 7

⇒ Hence value of x is 7.

Given equation is,

(∵ LCM of 2,4,6, is 12)

Cross multiplying both sides.

⇒ Hence value of x is 2.

Given equation is,

(LCM of 2,3,12 is 12)

Cross multiplying both sides

⇒ Hence value of x is .

Given equation is,

Cross multiplying both sides.

⇒ 3x + 3 = 4x-8

⇒ 3x-4x = -8-3 (transposing 4x to LHS and – 3 to RHS)

⇒ -x = -11

⇒ x = 11

⇒ Hence value of x is 11.

Given equation is,

Cross multiplying both sides

⇒ 6x-3 = 15x + 5

⇒ 6x-15x = 5 + 3 (transposing 15x to LHS and – 3 to RHS)

⇒ Hence value of x is .

Given: 1 – (x – 2) – [(x - 3) – (x - 1)] = 0

⇒ 1 - x + 2 - [x - 3 - x + 1] = 0

⇒ 3 - x - x + 3 + x - 1 = 0

⇒ 3 - x + 2 = 0

⇒ 5 - x = 0

⇒ -x = -5 (transposing 5 to RHS)

⇒ x = 5

⇒ Hence value of x is 5.

Given equation is,

(In LHS, LCM is 3 and in RHS, LCM is 4)

Cross multiplying both sides

⇒ 32x + 8 = 51-3x

⇒ 32x + 3x = 51-8 (transposing 3x to LHS and 8 to RHS)

⇒ 35x = 43

⇒ Hence value of x is .

Given equation is,

(in LHS, LCM is 4)

Cross multiplying both sides.

⇒ 15t + 5 = 16t-12

⇒ 15t-16t = -12-5 (transposing 16t to LHS and 5 to RHS)

⇒ -t = -17

⇒ t = 17

⇒ Hence value of t is 17.

Given equation is,

(LCM is 4 in both sides)

multiplying by 4 in both sides

⇒ 4 × (-4y + 7)/4 = 4 × (4y + 3)/4

⇒ -4y + 7 = 4y + 3

⇒ -4y-4y = 3-7 (transposing 4y to LHS and 7 to RHS)

⇒ Hence value of y is .

(x = 37)

Given equation is,

⇒ 0.25(4x-5) = 0.75x + 8

⇒ x-1.25 = 0.75x + 8

⇒ x-0.75x = 8 + 1.25(transposing 0.75x to LHS and 1.25 to RHS)

⇒ 0.25x = 9.25

⇒ Hence value of x is 37.

Given equation is,

Cross multiplying both sides,

⇒ 45-15y = 8-72y

⇒ 45-8 = -72y + 15y (transposing 8 to LHS and – 15y to RHS)

⇒ 37 = -57y

⇒ y = -37/57

⇒ Hence value of y is .

Given equation is,

Cross multiplying both sides,

⇒ 12x + 8 = -6x + 9

⇒ 12x + 6x = 9-8 (transposing 8 to RHS and – 6x to LHS)

⇒ Hence value of x is .

Given equation is,

Cross multiplying both sides.

⇒ 15x + 3 = -2x

⇒ 15x + 2x = -3 (transposing – 2x to LHS and 3 to RHS)

⇒ 17x = -3

⇒ Hence value of x is .

Given equation is,

(taking LCM 6 in both sides and adding)

Multiplying both sides by 6

⇒ 12t + 5 = 6t + 7

⇒ 12t-6t = 7-5 (transposing 6t to LHS and 5 to RHS)

⇒ 6t = 2

⇒ t = 2/6

⇒ t = 1/3

⇒ Hence value of t is .

Given equation is,

(taking LCM 2 in LHS and 3 in RHS)

Cross multiplying both sides.

(transposing – 2m to LHS and 3 to RHS)

⇒ Hence value of m is .

Given equation is,

⇒ 4(3p + 2)-5(6p-1) = 2(p-8)-6(7p-4)

⇒ 12p + 8-30p + 5 = 2p-16-42p + 24

⇒ -18p + 13 = -40p + 8

⇒ -18p + 40p = 8-13 (transposing – 40p to LHS and 13 to RHS)

⇒ 22p = -5

⇒ Hence value of p is .

Given equation is,

⇒ 3(5x-7) + 2(9x-11) = 4(8x-7)-111

⇒ 15x-21 + 18x-22 = 32x-28-111

⇒ 33x-43 = 32x-139

⇒ 33x-32x = -139 + 43 (taking 32x to LHS and – 43 to RHS)

⇒ x = -96

⇒ Hence value of x is -96.

Given equation is,

⇒ 0.16(5x-2) = 0.4x + 7

⇒ 0.8x-0.32 = 0.4x + 7

⇒ 0.8x-0.4x = 7 + 0.32 (moving 0.4x to LHS and – 0.32 to RHS)

⇒ x = 18.3

⇒ Hence value of x is 18.3.

Let the total number of flowers Radha had at first be x

After visiting first temple number of flowers remaining with

her = )

=

After visiting second temple number of flowers remaining

her =

=

After visiting third temple number of flowers remaining

her =

=

At last flowers remaining are 3

i.e.,

⇒

(take L.C.M on left side and solve)

⇒ 2x-x = 3 × 8

(multiply by 8 on both the sides)

⇒ x = 24

∴Number of flowers she took to temple at first is 24

Let the amount received by Kiran be rupees x

Then the amount received by Salma will be 1000 more than

Kiran = x + 100

The amount received by Jenifer will be 500 more than

Kiran = x + 500

Given that total amount = 13500

⇒ x + x + 1000 + x + 500 = 13500

(adding all the amounts all three having and equating to the total amount)

⇒ 3x + 1500 = 13500

⇒ 3x = 13500-1500

⇒ 3x = 12000

⇒

⇒ x = 4000

∴The amount received by Jenifer = x + 500

= 4000 + 500

= 4500

∴The amount received by Salma = x + 1000

= 4000 + 1000

= 5000

Let the volume of water in the second tank be

The volume of water in the first tank be

According to the question

⇒ 2x-25 = x + 25

⇒ 2x-x = 25 + 25

⇒ x = 50

∴The volume of water in the second tank is 50l

Volume of water in the first tank = 2x

= 2 × 50

= 100

∴The volume of water in the first tank is 100l

Let Anushka and Aarushi have equal amount of money in

their pocket, which is Rs.

After giving of the money of Anushka to Aarushi

Amount of Aarushi

According to the question

⇒

⇒

(By taking common)

(taking L.C.M. and solving on left)

(multiply by 6 on both the sides)

4x = 9600

x = 2400

(divide by 4 on both the sides)

∴The equal amount of money in Anushka and Aarushi pocket is Rs.2400

Money gifted by Anushka = □(1/3 of 2400)

= □(1/3 × 2400)

Hence money gifted by Anushka = Rs.800

Given, ratio of the remaining flowers to that of flowers in

The beginning = 3:5

Number of remaining flowers = 3x

Number of flowers or total flowers at the beginning = 5x

Number of total flower Karstubh had = 60(given)

5x = 60

x = 12

(divide by 5 on both the sides)

Number of remaining flowers after offering in the temple = 3 × 12

= 36

Number of flowers offered by Karstubh in the temple = 60-32

= 24

Let us write three consecutive even number series as 3

Consecutive even numbers = x,x + 2,x + 4

Sum of consecutive even numbers = x + x + 2 + x + 4

48 = 3x + 6

-3x = 6-48

-3x = -42

x = 14

(divide by 3 on both sides)

∴3 consecutives even numbers = x,x + 2,x + 4

= 14,14 + 2,14 + 4

= 14,16,18

Hence the above series biggest number

Let the 3 consecutive odd numbers be x,x + 2,x + 4

According to the question

x + x + 2 + x + 4 = 69

3x + 6 = 69

3x = 69-6

3x = 63

x = 21

(divide by 3 on both the sides)

The numbers are x = 21

x + 2 = 21 + 2 = 23

x + 4 = 21 + 4 = 25

Among 21,23 and 25 it is found that 23 is a prime number

∴Prime number = 23

Let the 3 consecutive numbers be x,x + 1,x + 2

According to the given question

x + x + 1 + x + 2 = 156

3x + 3 = 156

3x = 156-3

3x = 153

x = 51

(divide by 3 on both the sides)

So the numbers are x,x + 1 + 2

51,51 + 1,51 + 2

51,52,53

∴52 is the multiple of 3

Let the number be

Its fifth part is

Its fourth part is =
According to the question

(taking L.C.M. of left side)

(multiply by 20 on both sides)

Hence, fifth part

Fourth part

Let one part be

The other part will be

According to the question

(by taking L.C.M. on left side)

(multiply by 7 on both sides)

(divide by 9 on both the sides)

Hence the 2 parts are 42 and 12

Let the number be

Then

According to the question

10x + y = (10y + x)-9

10x + y = 10y + x-

10x-x + y-10y = -9

9x-9y = -9

x-y = -1⋯(ii)

Adding both equations (i) and (ii)

x + y + x-y = 11-1

2x = 10

x = 5

Put value of x in equation (i)

x + y = 11

5 + y = 11

y = 11-5

y = 6

Hence the number is xy = 56

Let the third side be

Then (3x-4) be one of the equal sides

According to the question

55 = x + 2(3x-4)

55 = x + 6x-8

55 = 7x-8

55 + 8 = 7x

63 = 7x

9 = x

(divide by 7 on both the sides)

∴The third side = x = 9m

Hence the equal two sides = (3x-4) = (3 × 9-4) = (27-4)

= 23 m each

Let his present age be x

4 years ago his age = x-4 years

After 12 years his age = 12 + x years

According to the question

12 + x = 3(x-4)

12 + x = 3x-12

x-3x = -12-12

-2x = -24

x = 12

(divide by -2 on both the sides)

∴The present age of the person = 12 years

Let the donated property be

The total property

According to the question

(by taking all y’s one side and getting L.C.M. on left side)

y = 6x

(multiply by -6 on both the sides)

y = 6 × 100000

(x = 100000)

y = Rs.6,00,000

∴She had property worth Rs. 6,00,000

Let the number be

According to the question

(by taking L.C.M on left side)

8x-4 = 10

(multiply by 2 on both the sides)

8x = 10 + 4

8x = 14

x = 7/4

(divide by 8 on both the sides)

Let the four consecutive integers be x,x + 1,x + 2,x + 3

According to the question

x + x + 1 + x + 2 + x + 3 = 266

4x + 6 = 266

4x = 266-6 = 260

x = 65

(divide by 4 on both the sides)

Hence the 4 consecutive integers are 65,65 + 1,65 + 2,65 + 3

= 65,66,67,68

Let the weight of box A be x kg

Then,

Weight of Box A = Weight of Box B +

Weight of Box A = Weight of Box B +

⇒ Weight of Box B =

And weight of Box C = weight of Box B +

⇒ weight of Box C =

∴ Weight of box C =

Now,

The total weight of three boxes = =

Now,

Total weight = Weight of boxes (A+B+C)

⇒

⇒

⇒

⇒

⇒

Hence, the weight of Box A =

Perimeter of the given rectangle = 240 cm

Let the length of the rectangle = x cm

Let the breadth of the rectangle = y cm

According to the question

2(x + y) = 240

x + y = 240⋯(i)

(divide by 2 on both the sides)

Let the new length be

Let the new breath be

According to the question

(divide by 2 on both the sides)

From

(take L.C.M on both side and solve)

x = 2y

(multiply by 10 on both the sides and solve)

x-2y = 0⋯(iii)

subtracting (iii)from (i)

x-2y-x-y = -120

-3y = -120

y = 40

Put the value of y in (i)

x + y = 120

x = 120-40

x = 80

The breath = 40 cm

The length = 80 cm

Hence, length of the 2^nd rectangle

Breath of the 2^nd rectangle

Let age of B = x

Let age of A = x + 5

5 years ago, the ratio was 3:2

According to the question

(by cross multiplication)

3x = 2x + 10

3x-2x = 10

x = 10

Age of B = 10 years

Age of A = 10 + 5 = 15 years

5 years later,

Age of B = x + 5 = 10 + 5 = 15 years

Age of A = x + 5 + 5 = 10 + 10 = 20 years

Suppose the denominator of the rational number is x.

Then, according to the question, numerator = Denominator – 2

= x – 2

Now,

On cross multiplying we get,

2(x-2-1) = x -1

⇒ 2(x-3) = x-1

⇒ 2x – 6 = x- 1

⇒ x = 5

Hence, the denominator = 5

And numerator = 5-2 = 3

∴ the rational number =

Let us assume, is the tenth-place digit and is the unit

Place digit of a two – digit number

The two digit number = 10x + y and reversed number = 10y + x

According to the question

y = 2x⋯(i)

Given,

10x + y + 27 = 10y + x

9y-9x = 27

y-x = 3

(divide by 9 on both the sides)

y = 3 + x⋯(ii)

Substitute the value of y from (i) in(ii)

2x = 3 + x

2x-x = 3

x = 3

∴y = 2x = 2 × 3 = 6

Hence the two digit number = 10x + y = 10 × 3 + 6 = 36

Number of days in the month of February in 2009 were 28

Days

His total salary of being present is 28 × 500

= 14000

Salary received by him is 9100

So number of days he was absent = 14000-9100

= 4900

So the number of days he was absent were 49 days

Let the steamer speed be x

Downstream speed = (x + 3)

Distance between the points = 3(x + 3)⋯(i)

Upstream speed = (x-3)

Distance between the points = 5(x-3)⋯(ii)

Equate (i) and (ii)

3(x + 3) = 5(x-3)

3x + 9 = 5x-15

-2x = -6

(divide by -2 on both the sides)

x = 3

Let 500 notes be x

Let 1000 notes be y

According to the question

x + y = 175⋯(i)

According to the question

500x + 1000y = 100000⋯(ii)

500(x + 2y) = 100000

x + 2y = 20

subtrate (i)-(ii)

x + y-x-2y = 175-200

y = 25

Putting the value of y in (i)

x + y = 175

x = 175-25

x = 150

Let the number of passengers bought Rs. 3 tickets = x

Let the number of passengers bought Rs. 10 tickets = y

Given, there are 40 passengers in a bus

x + y = 40⋯(i)

Again, total collection from the passengers is Rs. 295

3x + 10y = 295⋯(ii)

substract (i) from (ii)

Multiply by 10 in (i)

10x + 10y = 400⋯(iii)

substract (iii) and (ii)

10x-3x + 10y-10y = 400-295

7x = 105

So, the number of passengers bought Rs.3 tickets is 15

Let the numerator be x

Denominator = x-4

According to the question

x + 6 = 3(x-4)
x + 6 = 3x-12

x-3x = -12-6

-2x = -18

x = 9

(divide by -2 on both the sides)

The numerator = 9

The denominator = x-4 = 9-4 = 5

Total days he worked = 430

Money he received daily = 120

Fine for per day absent = 10

He got a total money = 2300

In 30 days it will be = 120 × 30 = 3600

But he received = 2300

He work = 2300/120 = 19.1 or 19(approx.) = 19days

∴He did not work for = 30-19 = 11 days

Let Jusum purchased x chocolates

Total cost of chocolates = Rs. 10x

Similarly, she purchased x candies

Total cost of candies = Rs.5x

According to the question

Profit on chocolates = 20% of 10x = 20/100 × 10x = Rs. 2x

Profit on candies = 8% of 5x = 8/100 × 5x = Rs. 0.4x

∴Total profit = 2x + 0.4x = Rs. 2.4x

But it is given that total profit is Rs. 240

According to the question

2.4x = 240

x = 100

(divide by 2.4 on both the sides)

Hence she purchased 100 chocolates

Let the steamer speed will be x

Downstream speed = (x + 1)

Distance between the points = 5(x-1)⋯(i)

Upstream speed = (x-1)

Distance between the points = 6(x-1)⋯(ii)

on solving (i) and (ii)

5(x + 1) = 6(x-1)

5x + 5 = 6x-6

5x-6x = -6-5

-x = -11

x = 1

∴The steamer speed will be 11km/hr

= 18 km/hr

∴It is one car and speed of another car is 18-8 = 10km/hr

Let the carpenter worked for x hr.

Given, labour charge = Rs. 200 per hr

Then, total labour charge = Rs. 200x

As, amount charged by carpenter for making a bed = cost of materials + total labour charge

∴2500 = 1100 + 200x

200x = 2500-1100

200x = 1400

Hence, the carpenter worked for 7hrs

Divide the give fig. into two

Perimeter of first shape = 2a + 2b

= 2(x + 2) + 2(2x + 2)

Perimeter of second shape = 2a + b

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

Total perimeter = 77 cm

2(x + 2) + 2(2x + 2) + 2(x + 1) + (2x + 2) = 77

2x + 4 + 4x + 4 + 2x + 2 + 2x + 2 = 77

10x + 12 = 77

10x = 77-12

10x = 65

x = 65/10 = 6.5cm

Perimeter = 2a + 2b

186 = 2(5x + 6) + 2(2x + 66)

186 = 10x + 12 + 4x + 132

186 = 14x + 144

186-144 = 14x

42 = 14x

x = 3 cm

Let A’s share be x

Then B’s share will be (200-x)

According to the question

3(200-x)-2x = 200

600-3x-2x = 200

600-5x = 200

5x = 400

x = 80

(divide by 5 on both the sides)

Hence, A’s share is Rs. 80

B’s share is 200-80 = Rs. 120

Let the number be x

According to the question

2x + 20 = 4(25)

2x = 100-20

2x = 80

x = 40