NCERT Solutions for Class 8 Maths Chapter 4 - Linear Equation In One Variable

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Answer:

Case I: 3x + 2 = 5x + 2
⇒ 2 – 2 = 5x – 3x       [transposing 3x to LHS and 2 to RHS]
$\Rightarrow \frac{2x}{2}=\frac{0}{2}$                 [dividing both ides by 2]
⇒ x = 0
Thus, it's an integer.

Case II:  4x – 18 = 2
⇒ 4x = 2 + 18           [transposing –18 to RHS]
$\Rightarrow \frac{4x}{4}=\frac{20}{4}$              [dividing both ides by 2]
$\Rightarrow x = \frac{20}{4}=5$
Thus, it's an integer

Case III: 4x + 7 = x + 2
⇒ 4x – x = 2 – 7          [transposing x to LHS and 7 to RHS]
$\Rightarrow \frac{3x}{3}=\frac{-5}{3}$                [dividing both ides by 2]
$\Rightarrow x = \frac{-5}{3}=-1.6666.....$
Thus, it's neither a fraction nor an integer

Case IV: 5x – 8 = x + 4
⇒ 5x – x = 8 + 4           [transposing x to LHS and –8 to RHS] 
$\Rightarrow \frac{4x}{4}=\frac{12}{4}$                   [dividing both ides by 4]
$\Rightarrow x=\frac{12}{4}=3$
Thus, it is an integer.
Hence, the correct answer is option C.

 

Answer:

Given: ax + b = 0
$\Rightarrow ax=–b$           [transposing b to RHS]
$\Rightarrow \frac{ax}{a}=\frac{–b}{a}$          [dividing both sides by a]
$\Rightarrow x=\frac{-b}{a}$
Hence, the correct answer is option C.

Answer:

Given: 8x – 3 = 25 + 17x
⇒ 8x – 17x = 25 + 3       [transposing 17x on LHS and –3 on RHS] 
$\Rightarrow \frac{–9x}{–9}=\frac{28}{–9}$                  [dividing both sides by –9]
$\Rightarrow x=\frac{-28}{9}$
Thus, it is a rational number
Hence, the correct answer is option C.

Answer:

The shifting of a number from one side of an equation to other side is called transposition.
Say for example: x + 7 = 0 is the equation ⇒ x = –7 Here, the number ‘7’ shifts from left hand side to right hand side.
Hence, the correct answer is option A. 

Answer:

Given: $\frac{5x}{3}-4=\frac{2x}{5}$
$\Rightarrow \frac{5x}{3}–\frac{2x}{5}=4$            [transposing $\frac{2x}{5}$ on LHS and –4 on RHS]
$\Rightarrow \frac{25x –6x}{15}=4$            [simplification of equation]
$\Rightarrow \frac{19x}{19}=\frac{60}{19}$               [dividing both sides by 19]
$\Rightarrow x=\frac{60}{19} ...\left(1\right)$
To find value of 2x – 7, putting the value from (1)
$\Rightarrow 2\times \frac{60}{19}–7$
$\Rightarrow \frac{120-7\times 19}{19}$              [simplification of equation]
$$\begin{gathered} \Rightarrow \frac{120-133}{19} \\ \Rightarrow \frac{-13}{19} \end{gathered}$$
Thus the value of 2x –7, when $x=\frac{60}{19}$is $\frac{–13}{19}$
Hence, the correct answer is option B.


 

Answer:

Given:  3x – 4 = 2x + 1 
⇒  3x – 2x = 4 + 1           [transposing 2x on LHS and –4 on RHS]
⇒ x = 5 
Thus, for x = 5 the expressions 2x – 4 and 2x + 1 becomes equal.
Hence, the correct answer is option C.

Answer:

Given:  ax = b
$\Rightarrow \frac{ax}{a}=\frac{b}{a}$        [dividing both sides by a]
$\Rightarrow x=\frac{b}{a}$
As, a and b both are positive thus the solution of ax = b will always be positive.
Hence, the correct answer is option A.

Answer:

Only one variable of power 1 exists in a linear equation in one variable. For example: 5x – 1 = 2, 2p – 9 = 3 and z + 20 = – 8 are the linear equations in one variable.
Hence, the correct answer is option C.

Answer:

The linear expression is defined as an algebraic expression in one variable with the maximum power of the variable equal to 1.
Because the power of the variable z is 1, the only linear expression is 1 + z.
Hence, the correct answer is option D.

Answer:

There is only one solution to a linear equation with one variable.
For Example: the linear equation ax + b = 0 has unique solution.
$\Rightarrow x=\frac{-b}{a}$
Hence, the correct answer is option A.
 

Answer:

Given:  $\frac{1}{3}+$ $\mathrm{S}=\frac{2}{5}$
$\Rightarrow \mathrm{S}=\frac{2}{5}-\frac{1}{3}$           [transposing $\frac{1}{3}$ to RHS]
$\Rightarrow \mathrm{S}=\frac{6-5}{15}$               [taking LCM in RHS]
$\Rightarrow \mathrm{S}=\frac{1}{15}$
Hence, the correct answer is option B.

Answer:

Given: $\frac{-4}{3}y=-\frac{3}{4}$
$\Rightarrow y=\left(\frac{-3}{4}\right)\times \left(\frac{-3}{4}\right)$         [By cross multiplication]
$\Rightarrow y={\left(\frac{3}{4}\right)}^2$
Thus, the value of y is ${\left(\frac{3}{4}\right)}^2$
Hence, the correct answer is option C.

Answer:

The digit at unit’s place is b.
Then, digit at ten’s place = (3 + b)
Therefore, the number = 10 (3 + b) + b 
                                     = 30 + 10b + b
                                     = 11b + 30
Hence, the correct answer is option A.

Answer:

Given, Shilpa’s age three years ago = x
Then, Shilpa’s present age = (x + 3)
Arpita’s present age = 3 x Shilpa’s present age = 3 (x + 3)
Hence, the correct answer is option D.

Answer:

Let the 3 consecutive multiple of 7 be 7x, (7x + 7) and (7x + 14), where x is a natural number
Then, 7x + 7x + 7 + 7x + 14 = 357
⇒ 21x + 21 = 357
⇒ 21(x + 1) = 357
⇒ $\frac{21(x+1)}{21}=357$          [dividing both sides by 21]
⇒ x + 1 = 17
⇒ x = 17 – 1                  [transposing 1 to RHS]
∴ x = 16
Thus, the smallest multiplying of 7 is 7 × 16 = 112
Hence, the correct answer is option A.


 

Answer:

highest

Answer:

3x – 4 = 1 – 2 x 
⇒ 3x + 2x = 1 + 4         [transposing like terms together]
⇒ 5x = 5                       [dividing both sides by 5]
⇒ $\frac{5x}{5}=\frac{5}{5}$
⇒ x = 1
∴ solution of equation is x = 1.

Answer:

Given, 2y = 5y – $\frac{18}{5}$ 
$\Rightarrow \frac{18}{5}=5y–2y$          [transposing like terms together]
$\Rightarrow \frac{18}{5}=3y$
$\Rightarrow \frac{18}{5\times 3}=\frac{3y}{3}$          [dividing both side by 3]
$\Rightarrow \frac{6}{5}=y$
∴ solution of equation ⇒ $y=\frac{6}{5}.$
 

Answer:

solution

Answer:

9x – _______ = –21         ...(1)
where the solution of equation is x = –2.
Putting x = –2 in (1),
9(–2) −_______ = –21
⇒ –18 – ________ = –21
⇒ –18 + 21 = ________           (Transposing)
⇒ 3 = ________
∴ ans = 3.

Answer:

3, 4, 5
as (3 + 4) + 5 = 7 + 5

Answer:

A gets ₹8 more than B.
Let us suppose that B has ₹b.
⇒ A has = ₹(b + 8)
Now, the sum of the amounts of money they have = ₹25
⇒ money with A + money with B = ₹25
⇒ ₹(b + 8) + ₹b = ₹25
⇒₹b + ₹8 + ₹b = ₹25
⇒ ₹2b + ₹8  = 25
⇒ ₹2b = 25 – 8
      b = ₹8.5

Answer:

sign

Answer:

x – 8 = 2
⇒ x = 2 + 8     (Transposing like terms together)

Answer:

$\frac{x}{5}+30=18$
$\Rightarrow \frac{x}{5}=18–30$
$\Rightarrow \frac{x}{5}=–12$
cross multiplying,
⇒ x × 1 = –12 × 5
⇒ x = –60.
∴ solution of equation ⇒ x = –60.

Answer:

Let the number be x
$\Rightarrow \frac{x}{8}=–3$
cross multiplying, we get
⇒ x × 1 = –3 × 8
⇒ x = –24
Hence, the number is –24.

Answer:

Given, 4p – 9 = 11

$$\begin{gathered} \Rightarrow 4p=11+9 \left[\mathrm{transposing} -9 \mathrm{from} \mathrm{LHS} \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow \frac{4p}{4}=\frac{20}{4} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 4\right] \\ \Rightarrow p=5 \end{gathered}$$

Hence, the value of p is 5.

Answer:

Given, $\frac{2x}{5}-2=5-\frac{3x}{5}$
$$\begin{gathered} \Rightarrow \frac{2x-10}{5}=\frac{25-3x}{5} \left[\mathrm{Taking} \mathrm{LCM} \mathrm{on} \mathrm{RHS} \mathrm{and} \mathrm{LHS}\right] \\ \Rightarrow \left(\frac{2x-10}{5}\right)\times 5=\left(\frac{25-3x}{5}\right)\times 5 \left[\mathrm{Multiplying} \mathrm{RHS} \mathrm{and} \mathrm{LHS} \mathrm{by} 5\right] \\ \Rightarrow 2x-10=25-3x \\ \Rightarrow 2x+3x=25+10 \left[\mathrm{transposing} -3x \mathrm{to} \mathrm{LHS} \mathrm{and} -10 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow 5x=35 \\ \Rightarrow \frac{5x}{5}=\frac{35}{5} \\ \Rightarrow x=7 \end{gathered}$$

Answer:

Let the present age be x.

Age after 18 years = x + 18
Given,$$\begin{gathered} 4x=x+18 \\ \Rightarrow 4x-x=18 \left[\mathrm{transposing} x \mathrm{on} \mathrm{LHS}\right] \\ \Rightarrow \frac{3x}{3}=\frac{18}{3} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 3\right] \\ \Rightarrow x=6 \end{gathered}$$

Hence, present age is 6 years.

Answer:

Given, add 15 to 4x which equals 39
         ⇒ 4x + 15 = 39
Hence, the equation is 4x + 15 = 39.
 

Answer:

$$\begin{gathered} \mathrm{Given}, \mathrm{D}=\mathrm{N}+10 ..............\left(1\right) \\ \mathrm{New} \mathrm{numerator} \mathrm{be} {\mathrm{N}}^1=\mathrm{N}+1\Rightarrow \mathrm{N}={\mathrm{N}}^1-1 \\ \mathrm{New} \mathrm{denominator} \mathrm{be} {\mathrm{D}}^1=\mathrm{D}-1\Rightarrow \mathrm{D}={\mathrm{D}}^1+1 \\ \mathrm{Now}, \mathrm{putting} \mathrm{these} \mathrm{values} \mathrm{in} \left(1\right), \mathrm{we} \mathrm{get} \\ {\mathrm{D}}^1+1={\mathrm{N}}^1-1+10 \\ \Rightarrow {\mathrm{D}}^1+1={\mathrm{N}}^1+9 \\ \Rightarrow {\mathrm{D}}^1={\mathrm{N}}^1+9-1 \left[\mathrm{transposing} 1 \mathrm{to} \mathrm{RHS}\right] \\ \therefore {\mathrm{D}}^1={\mathrm{N}}^1+8 \end{gathered}$$

Answer:

Let the consecutive multiple be x and x + 10.

Given, $$\begin{gathered} x+x+10=210 \\ \Rightarrow 2x+10=210 \\ \Rightarrow 2x=210-10 \left[\mathrm{transposing} 10 \mathrm{on} \mathrm{RHS}\right] \\ \Rightarrow \frac{2x}{2}=\frac{200}{2} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 2\right] \\ \Rightarrow x=100 \end{gathered}$$

The smaller multiple is 100.

Answer:

Statement is False
Given, 3 year ago, age of boy = y
Then, present age of boy = (y + 3)

Hence, 2 year ago age of boy = y + 3 – 2 = y + 1

Answer:

Statement is False
Given, Shikha’s present age = p year
Then, Reemu’s present age = 4x (Shikha’s present age) = 4p year
Therefore after 5 year, Reemu’s age is = (4p + 5) year

Answer:

The statement is False
Given, unit’s digit = x
and sum of digits = 9
Ten’s digit = 9 – x
Hence, the number = 10 (9 – x) + x
                             = 90 – 10x + x
                             = 90 – 9x

Answer:

The statement is True
Given, Anju’s present age = y year
Then, Anju’s mother age = (65 – y) year
5-year before, Anju’s mother age = 65 – y – 5
                                                 = (60 – y) year

Answer:

The statement is False
Let the number of boys be 5x and the number of girls be 4x.
According to the question, 5x = 4x + 9

$\Rightarrow$ 5x – 4x = 9 [Transposing 4x to LHS]
$\Rightarrow$x = 9

Hence, number of boys = 5 × 9 = 45

Answer:

The statement is True
Let the age of A be x year
Then, age of B = (90 – x) year
Five years ago, the age of A = (x – 5) year
The age of B = 90 – x – 5 = (85 – x) year

Hence, the ages of A and B five years back would be (x – 5) year and (85 – x) year.

Answer:

The statement is False
Two different equations may have the same answer.
e.g. 3x + 6 = 9 and 4x – 3 = 1 are the two linear equations whose solution is 1.

Answer:

The statement is False
Given, 3x – 3 = 9

$\Rightarrow$ 3x = 9 + 3 [transposing –3 to RHS]
$\Rightarrow$ 3x = 12

Answer:

The statement is False
Given, 2x = 4 – x

⇒ 2x + x = 4           [transposing – x to LHS]
⇒ 3x = 4

Answer:

The statement is False
$$\begin{gathered} \mathrm{Given}, \frac{15}{8}-7x=9 \\ \Rightarrow -7x=9-\frac{15}{8} \end{gathered}$$

Answer:

The statement is False
Given, $$\begin{gathered} \frac{x}{3}+1=\frac{7}{15} \\ \Rightarrow \frac{x}{3}=\frac{7}{15}-1 \left[\mathrm{transposing} 1 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow \frac{x}{3}=\frac{7-15}{15} \\ \Rightarrow \frac{x}{3}=\frac{-8}{15} \end{gathered}$$

Answer:

The statement is True
Given, 6x = 18
           $$\begin{gathered} \Rightarrow 3\times 6x=18\times 3 \left[\mathrm{Multiplying} \mathrm{both} \mathrm{sides} \mathrm{by} 3\right] \\ \Rightarrow 18x=54 \end{gathered}$$

Answer:

The statement is False

Given, $$\begin{gathered} \frac{x}{11}=15 \\ \Rightarrow x=11\times 15 \left[\mathrm{Cross} \mathrm{multiplication}\right] \end{gathered}$$

Answer:

The statement is False
Given, x is an even number.
Then, the next even number is (x + 2).

Answer:

The statement is True
Given, one of the consecutive number is = x
Then, the next consecutive number = x + 1
So, $$\begin{gathered} x+x+1=93 \\ \Rightarrow 2x+1=93 \\ \Rightarrow 2x=93-1 \left[\mathrm{transposing} 1 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow \frac{2x}{2}=\frac{92}{2} \left[\mathrm{Dividing} \mathrm{both} \mathrm{side} \mathrm{by} 2\right] \\ \therefore x=46 \end{gathered}$$

Hence, the other consecutive number = 46 + 1 = 47
                                                         ⇒ 93 – 46 = 93 – x

Answer:

The statement is False.
Given, one number = x
other number = 40 – x
Let (40 – x) > x

So, $$\begin{gathered} 40-x+8=3\left(x+8\right) \\ \Rightarrow 48-x=3x+24 \\ \Rightarrow -x-3x=24-48 \left[\mathrm{transposing} 3x \mathrm{to} \mathrm{LHS} \mathrm{and} 48 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow -4x=-24 \\ \Rightarrow \frac{-4x}{-4}=\frac{-24}{-4} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} -4\right] \\ \therefore x=6 \end{gathered}$$

Hence, one number is 6.
other number is = 40 – 6 = 34
The difference between the numbers = 34 – 6 = 28 ≠ 40.
 

Answer:

Given, $$\begin{gathered} \frac{3x-8}{2x}=1 \\ \Rightarrow 3x-8=2x \left[\mathrm{cross} \mathrm{multiplication}\right] \\ \Rightarrow 3x-2x=8 \left[\mathrm{transposing} 2x \mathrm{to} \mathrm{LHS} \mathrm{and} 8 \mathrm{to} \mathrm{RHS}\right] \\ \therefore x=8 \end{gathered}$$

Answer:

Given, $$\begin{gathered} \frac{5x}{2x-1}=2 \\ \Rightarrow 5x=2\left(2x-1\right) \left[\mathrm{cross} \mathrm{multiplication}\right] \\ \Rightarrow 5x=4x-2 \\ \Rightarrow 5x-4x=-2 \left[\mathrm{transposing} 4x \mathrm{to} \mathrm{LHS}\right] \\ \therefore x=-2 \end{gathered}$$

Answer:

Given, $$\begin{gathered} \frac{2x-3}{4x+5}=\frac{1}{3} \\ \Rightarrow 3\left(2x-3\right)=4x+5 \left[\mathrm{cross} \mathrm{multiplication}\right] \\ \Rightarrow 6x-9=4x+5 \\ \Rightarrow 6x-4x=9+5 \left[\mathrm{transposing} 4x \mathrm{to} \mathrm{LHS} \mathrm{and} -9 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow 2x=14 \\ \Rightarrow \frac{2x}{2}=\frac{14}{2} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 2\right] \\ \therefore x=7 \end{gathered}$$

Answer:

Given, $$\begin{gathered} \frac{8}{x}=\frac{5}{x-1} \\ \Rightarrow 8\left(x-1\right)=5x \left[\mathrm{cross} \mathrm{multiplication}\right] \\ \Rightarrow 8x-8=5x \\ \Rightarrow 8x-5x=8 \left[\mathrm{transposing} 5x \mathrm{to} \mathrm{LHS} \mathrm{and} -8 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow 3x=8 \\ \Rightarrow \frac{3x}{3}=\frac{8}{3} \left[\mathrm{dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 3\right] \\ \Rightarrow x=\frac{8}{3} \end{gathered}$$

Answer:

Given, $$\begin{gathered} \frac{5\left(1-x\right)+3\left(1+x\right)}{1-2x}=8 \\ \Rightarrow 5\left(1-x\right)+3\left(1+x\right)=8\left(1-2x\right) \\ \Rightarrow 5-5x+3+3x=8-16x \\ \Rightarrow 8-2x=8-16x \\ \Rightarrow 16x-2x=8-8 \left[\mathrm{transposing} -16 \mathrm{to} \mathrm{LHS} \mathrm{and} 8 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow 14x=0 \\ \Rightarrow \frac{14x}{14}=\frac{0}{14} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 14\right] \\ \therefore x=0 \end{gathered}$$

Answer:

Given, $$\begin{gathered} \frac{0.2x+5}{3.5x-3}=\frac{2}{5} \\ \Rightarrow 5\left(0.2x+5\right)=2\left(3.5x-3\right) \left[\mathrm{cross} \mathrm{multiplication}\right] \\ \Rightarrow x+25=7x-6 \\ \Rightarrow x-7x=-6-25 \left[\mathrm{transposing} 7x \mathrm{to} \mathrm{LHS} \mathrm{and} 25 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow -6x=-31 \\ \Rightarrow \frac{-6x}{-6}=\frac{-31}{-6} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} -6\right] \\ \therefore x=\frac{31}{6} \end{gathered}$$

Answer:

Given, $$\begin{gathered} \frac{y-\left(4-3y\right)}{2y-\left(3+ 4y\right)}=\frac{1}{5} \\ \Rightarrow 5\left(y-4+3y\right)=2y-3-4y \left[\mathrm{cross} \mathrm{multiplication}\right] \\ \Rightarrow 5\left(4y-4\right)=-3-2y \\ \Rightarrow 20y-20=-3-2y \\ \Rightarrow 20y+2y=20-3 \left[\mathrm{transposing} -20 \mathrm{to} \mathrm{RHS} \mathrm{and} -2\mathrm{y} \mathrm{to} \mathrm{LHS}\right] \\ \Rightarrow 22y=17 \\ \Rightarrow \frac{22y}{22}=\frac{17}{22} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 22\right] \\ \Rightarrow y=\frac{17}{22} \end{gathered}$$

Answer:

Given, $$\begin{gathered} \frac{x}{5}=\frac{x-1}{6} \\ \Rightarrow 6x=5\left(x-1\right) \left[\mathrm{cross} \mathrm{multiplication}\right] \\ \Rightarrow 6x=5x-5 \\ \Rightarrow 6x-5x=-5 \left[t\mathrm{ransposing} 5x \mathrm{to} \mathrm{LHS}\right] \\ \Rightarrow x=-5 \end{gathered}$$

Answer:

Given, 0.4(3x –1) = 0.5x + 1
           $$\begin{gathered} \Rightarrow 1.2x-0.4=0.5x+1 \\ \Rightarrow 1.2x-0.5x=1+0.4 \left[\mathrm{transposing} 0.5 \mathrm{to} \mathrm{LHS} \mathrm{and} -0.4 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow 0.7x=1.4 \\ \Rightarrow \frac{0.7x}{0.7}=\frac{1.4}{0.7} \left[\mathrm{dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 0.7\right] \\ \therefore x=2 \end{gathered}$$

Answer:

$$\begin{gathered} \mathrm{Given}, 8x-7-3x=6x-2x-3 \\ \Rightarrow 8x-3x-6x+2x=-3+7 \\ \left[\mathrm{transposing} 6x,-2x t\mathrm{o} \mathrm{LHS} \mathrm{and} -7 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow 10x-9x=4 \\ \therefore x=4 \end{gathered}$$

Answer:

Given, 10x – 5 – 7x = 5x + 15 – 8
$$\begin{gathered} \Rightarrow 10x-7x-5x=5+15-8 \\ \left[\mathrm{transposing} 5x \mathrm{to} \mathrm{LHS} \mathrm{and} -5 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow -2x=12 \\ \Rightarrow \frac{-2x}{-2}=\frac{12}{-2} \left[\mathrm{dividing} \mathrm{both} \mathrm{sides} \mathrm{by} -2\right] \\ \therefore x=-6 \end{gathered}$$

Answer:

Given, 4t – 3 – (3t +1) = 5t – 4
$$\begin{gathered} \Rightarrow 4t-3-3t-1=5t-4 \\ \Rightarrow t-4=5t-4 \\ \Rightarrow t-5t=4-4 \left[\mathrm{transposing} 5t \mathrm{to} \mathrm{LHS} \mathrm{and} -4 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow -4t=0 \\ \Rightarrow \frac{-4t}{-4}=\frac{0}{-4} \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} -4\right] \\ \therefore t=0 \end{gathered}$$

Answer:

Given, 5(x – 1) – 2(x + 8) = 0
$$\begin{gathered} \Rightarrow 5x-5-2x-16=0 \\ \Rightarrow 3x-21=0 \\ \Rightarrow 3x=21 \left[\mathrm{transposing} -21 \mathrm{to} \mathrm{RHS}\right] \\ \Rightarrow \frac{3x}{3}=\frac{21}{3} \left[\mathrm{dividing} \mathrm{both} \mathrm{sides} \mathrm{by} 3\right] \\ \therefore x=7 \end{gathered}$$

Answer:

$$\begin{gathered} \frac{x}{2}-\frac{1}{4}\left(x-\frac{1}{3}\right)=\frac{1}{6}\left(x+1\right)+\frac{1}{12} \\ \Rightarrow \frac{x}{2}-\frac{x}{4}+\frac{1}{12}=\frac{x}{6}+\frac{1}{6}+\frac{1}{12} \\ \Rightarrow \frac{x}{4}=\frac{x+1}{6} \\ \Rightarrow \frac{x}{4}=\frac{x+1}{6} \\ \Rightarrow 6x=4x+4 \\ \Rightarrow 2x=4 \\ \Rightarrow x=2 \end{gathered}$$

Answer:

$$\begin{gathered} \frac{1}{2}\left(x+1\right)+\frac{1}{3}\left(x-1\right)=\frac{5}{12}\left(x-2\right) \\ \Rightarrow \frac{x}{2}+\frac{1}{2}+\frac{x}{3}-\frac{1}{3}=\frac{5x}{12}-\frac{10}{12} \\ \Rightarrow \frac{5x}{6}+\frac{1}{6}=\frac{5x}{12}-\frac{5}{6} \\ \Rightarrow \frac{5x}{6}-\frac{5x}{12}=-\frac{5}{6}-\frac{1}{6} \\ \Rightarrow 5x\left(\frac{6}{72}\right)=-1 \\ \Rightarrow 5x=-12 \\ \Rightarrow x=\frac{-12}{5} \end{gathered}$$

Answer:

$$\begin{gathered} \frac{x+1}{4}=\frac{x-2}{3} \\ \Rightarrow 3x+3=4x-8 \\ \Rightarrow x=11 \end{gathered}$$

Answer:

$$\begin{gathered} \frac{2x-1}{5}=\frac{3x+1}{3} \\ \Rightarrow 6x-3=15x+5 \\ \Rightarrow 9x=-8 \\ \Rightarrow x=\frac{-8}{9} \end{gathered}$$

Answer:

$$\begin{gathered} 1-\left(x-2\right)-\left[\left(x-3\right)-\left(x-1\right)\right]=0 \\ \Rightarrow 1-x+2-\left[x-3-x+1\right]=0 \\ \Rightarrow 3-x+2=0 \\ \Rightarrow x=5 \end{gathered}$$
 

Answer:

$$\begin{gathered} 3x-\frac{x-2}{3}=4-\frac{x-1}{4} \\ \Rightarrow \frac{9x-x+2}{3}=\frac{16-x+1}{4} \\ \Rightarrow \frac{8x+2}{3}=\frac{17-x}{4} \\ \Rightarrow 32x+8=51-3x \\ \Rightarrow 35x=43 \\ \Rightarrow x=\frac{43}{35} \end{gathered}$$

Answer:

$$\begin{gathered} \frac{3t+5}{4}-1=\frac{4t-3}{5} \\ \Rightarrow \frac{3t+5-4}{4}=\frac{4t-3}{5} \\ \Rightarrow \frac{3t+1}{4}=\frac{4t-3}{5} \\ \Rightarrow 15t+5=16t-12 \\ \Rightarrow t=17 \end{gathered}$$

Answer:

$$\begin{gathered} \frac{2y-3}{4}-\frac{3y-5}{2}=y+\frac{3}{4} \\ \Rightarrow \frac{4y-6-12y+20}{8}=\frac{4y+3}{4} \\ \Rightarrow \frac{-8y+14}{8}=\frac{4y+3}{4} \\ \Rightarrow -32y+56=32y+24 \\ \Rightarrow 64y=32 \\ \Rightarrow y=\frac{1}{2} \end{gathered}$$

Answer:

$$\begin{gathered} 0.25\left(4x-5\right)=0.75x+8 \\ \Rightarrow x-1.25=0.75x+8 \\ \Rightarrow 0.25x=9.25 \\ \Rightarrow x=37 \end{gathered}$$

Answer:

$$\begin{gathered} \frac{9-3y}{1-9y}=\frac{8}{5} \\ \Rightarrow 45-15y=8-72y \\ \Rightarrow 57y=-37 \\ \Rightarrow y=\frac{-37}{57} \end{gathered}$$

Answer:

$$\begin{gathered} \frac{3x+2}{2x-3}=-\frac{3}{4} \\ \Rightarrow 12x+8=-6x+9 \\ \Rightarrow 18x=1 \\ \Rightarrow x=\frac{1}{18} \end{gathered}$$

Answer:

$$\begin{gathered} \frac{5x+1}{2x}=-\frac{1}{3} \\ \Rightarrow 15x+3=-2x \\ \Rightarrow 17x=-3 \\ \Rightarrow x=\frac{-3}{17} \end{gathered}$$

Answer:

$$\begin{gathered} \frac{3t-2}{3}+\frac{2t+3}{2}=t +\frac{7}{6} \\ \Rightarrow \frac{6t-4+6t+9}{6}=\frac{6t+7}{6} \\ \Rightarrow 12t+5=6t+7 \\ \Rightarrow 6t=2 \\ \Rightarrow t=\frac{1}{3} \end{gathered}$$

Answer:

$$\begin{gathered} m-\frac{m-1}{2}=1-\frac{m-2}{3} \\ \Rightarrow \frac{2m-m+1}{2}=\frac{3-m+2}{3} \\ \Rightarrow \frac{m+1}{2}=\frac{5-m}{3} \\ \Rightarrow 3m+3=10-2m \\ \Rightarrow 5m=7 \\ \Rightarrow m=\frac{7}{5} \end{gathered}$$
 

Answer:

$$\begin{gathered} 4(3p+2)–5(6p–1)=2(p–8)–6(7p–4) \\ \Rightarrow 12p+8-30p+5=2p-16-42p+24 \\ \Rightarrow -18p+13=-40p+8 \\ \Rightarrow 22p=-5 \\ \Rightarrow p=\frac{-5}{22} \end{gathered}$$

Answer:

$$\begin{gathered} 3\left(5x–7\right)+2\left(9x–11\right)=4\left(8x–7\right)–111 \\ \Rightarrow 15x-21+18x-22=32x-28-111 \\ \Rightarrow 33x-43=32x-139 \\ \Rightarrow x=-96 \end{gathered}$$

Answer:

$$\begin{gathered} 0.16(5x–2)=0.4x+7 \\ \Rightarrow 0.8x-0.32=0.4x+7 \\ \Rightarrow 0.4x=7.32 \\ \Rightarrow x=\frac{7.32}{0.4} \\ \Rightarrow x=18.3 \end{gathered}$$

Answer:

Let the number of flowers be x.
The number of flowers offered in first temple = $\frac{x}{2}$
The number of flowers offered in second temple = $\frac{x}{4}$
The number of flowers offered in third temple = $\frac{x}{8}$
According to the question,
$$\begin{gathered} x-\left(\frac{x}{2}+\frac{x}{4}+\frac{x}{8}\right)=3 \\ \Rightarrow \frac{8x-\left(4x+2x+x\right)}{8}=3 \\ \Rightarrow 8x-7x=24 \\ \Rightarrow x=24 \end{gathered}$$

Hence, total number of flowers in the beginning were 24.

Answer:

Let the money received by Kiran be x.
Then, the money received by Salma = x + 1000
And, the money received by Jenifer = x + 500
According to the question,
x + x + 500 + x + 1000 = 13500
$$\begin{gathered} \Rightarrow 3x+1500=13500 \\ \Rightarrow 3x=12000 \\ \Rightarrow x=4000 \end{gathered}$$

So, money received by Jenifer is â‚¹4500.

Answer:

Let the volume of the water in second tank be x.
Then, the volume of the water in first tank will be 2x.
According to the question,
$$\begin{gathered} x+25=2x-25 \\ \Rightarrow x=50 \end{gathered}$$

Hence, volume of water in one tank is 50 L and in the other tank is 100 L.

Answer:

Let Anushka and Arushi have equal amount of money in their pocket i.e. x.
After giving $\frac{1}{3}$ of Anushka's money to Arushi, the amount with Arushi = $x+\frac{x}{3}$.
According to the question,
$$\begin{gathered} \left(x+\frac{x}{3}\right)-\frac{1}{2}\left(x+\frac{x}{3}\right)=1600 \\ \Rightarrow \left(x+\frac{x}{3}\right)\left(1-\frac{1}{2}\right)=1600 \\ \Rightarrow \left(x+\frac{x}{3}\right)\times \frac{1}{2}=1600 \\ \Rightarrow \frac{3x+x}{3}=3200 \\ \Rightarrow 4x=9600 \\ \Rightarrow x=2400 \end{gathered}$$

Hence, money gifted by Anushka is â‚¹800.

Answer:

Let the number of flowers offered by Kaustubh in the temple be x.
Then, the flowers remaining = 60 − x
According to the question,
$$\begin{gathered} \frac{\mathrm{Number} \mathrm{of} \mathrm{remaining} \mathrm{flowers}}{\mathrm{Number} \mathrm{of} \mathrm{flowers} \mathrm{in} \mathrm{the} \mathrm{beginning}}=\frac{3}{5} \\ \Rightarrow \frac{60-x}{60}=\frac{3}{5} \\ \Rightarrow 300-5x\mathit{=}180 \\ \Rightarrow 5x=120 \\ \Rightarrow x=24 \end{gathered}$$

Hence, the number of flowers offered by him in the temple is 24.

Answer:

Let the three consecutive even natural numbers be x, x + 2, x + 4.
According to the question,
$$\begin{gathered} x+x+2+x+4=48 \\ \Rightarrow 3x+6=48 \\ \Rightarrow 3x=42 \\ \Rightarrow x=14 \end{gathered}$$

Therefore, the three consecutive even natural numbers are 14, 16, 18.
Hence, the greatest of the three consecutive numbers is 18.

Answer:

Let the three consecutive odd natural numbers be x, x + 2, x + 4.
According to the question,
$$\begin{gathered} x+x+2+x+4=69 \\ \Rightarrow 3x+6=69 \\ \Rightarrow 3x=63 \\ \Rightarrow x=21 \end{gathered}$$

The three consecutive odd natural numbers are 21, 23, 25.
Hence, the prime number is 23.

Answer:

Let the three consecutive natural numbers be x, x + 1, x + 2.
According to the question,
$$\begin{gathered} x+x+1+x+2=156 \\ \Rightarrow 3x+3=156 \\ \Rightarrow 3x=153 \\ \Rightarrow x=51 \end{gathered}$$

The three consecutive natural numbers are 51, 52, 53.
Hence, the multiple of 13 is 52.

Answer:

Let the number be x.
According to the question,
$$\begin{gathered} \frac{x}{5}+30=\frac{x}{4}-30 \\ \Rightarrow \frac{x}{5}-\frac{x}{4}=-30-30 \\ \Rightarrow \frac{-x}{20}=-60 \\ \Rightarrow x=1200 \end{gathered}$$

Hence, the required number is 1200.

Answer:

Let the other part be x. Then, the first part will be $\frac{2x}{7}$.
According to the question,
$$\begin{gathered} x+\frac{2x}{7}=54 \\ \Rightarrow \frac{7x+2x}{7}=54 \\ \Rightarrow \frac{9x}{7}=54 \\ \Rightarrow x=42 \end{gathered}$$

Hence, the two parts are 12 and 42.

Answer:

Let the units digit be x. Then, the tens digit will be 11 − x.
So, the number will be $10\left(11-x\right)+x$ = 110 − 9x.
By interchanging the digits, we get $10x+\left(11-x\right)$ = 9x + 11.
According to the question,
$$\begin{gathered} 9x+11-\left(110-9x\right)=9 \\ \Rightarrow 9x+11-110+9x=9 \\ \Rightarrow 18x=9-11+110 \\ \Rightarrow 18x=108 \\ \Rightarrow x=6 \end{gathered}$$

So, units digit is 6 and tens digit is 5.
Hence, the number so formed is 56.

Answer:

Let the third side be x.
Then, the two other sides = 3x − 4.
Perimeter of the triangle = 55
$$\begin{gathered} \Rightarrow x+3x-4+3x-4=55 \\ \Rightarrow 7x-8=55 \\ \Rightarrow 7x=63 \\ \Rightarrow x=9 \end{gathered}$$

Hence, the dimensions of the triangle are 9 m, 23 m and 23 m.

Answer:

Let Kanwar's present age be x years.
After 12 years, Kanwar's age = (x + 12) years
And, four years ago, Kanwar's age = (x − 4) years
According to the question,
$$\begin{gathered} x+12=3\left(x-4\right) \\ \Rightarrow x+12=3x-12 \\ \Rightarrow 3x-x=24 \\ \Rightarrow 2x=24 \\ \Rightarrow x=12 \end{gathered}$$

Hence, the present age is 12 years.

Answer:

Let Anima's property be x.
Property left for daughter = $\frac{x}{2}$

Remaining property = $x-\frac{x}{2}=\frac{x}{2}$

Property remaining for the son = $\frac{1}{3}\times \mathrm{Remaining} \mathrm{property}=\frac{1}{3}\times \frac{x}{2}=\frac{x}{6}$

Remaining property = $x-\left(\frac{x}{2}+\frac{x}{6}\right)=\frac{x}{3}$

Now,
Remaining property = donation money 
$$\begin{gathered} \Rightarrow \frac{x}{3}=100000 \\ \Rightarrow x=300000 \end{gathered}$$

Hence, Anima had â‚¹300000.

Answer:

Let the number be x.
According to the question,
$$\begin{gathered} 4\left(x-\frac{1}{2}\right)=5 \\ \Rightarrow 4x-2=5 \\ \Rightarrow 4x=7 \\ \Rightarrow x=\frac{7}{4} \end{gathered}$$

Hence, the required number is $\frac{7}{4}$.

Answer:

Let the four consecutive integers be x, x + 1, x + 2, x + 3.
According to the question,
$$\begin{gathered} x+x+1+x+2+x+3=266 \\ \Rightarrow 4x+6=266 \\ \Rightarrow 4x=260 \\ \Rightarrow x=65 \end{gathered}$$

Hence, the four consecutive integers are 65, 66, 67 and 68.

Answer:

Let the weight of box A be x kg.

Then, weight of box B = $x-\frac{5}{2}$ kg

And, weight of box C = $x-\frac{5}{2}+\frac{41}{4}=x+\frac{31}{4}$ kg

According to the question,
$$\begin{gathered} x+x-\frac{5}{2}+x+\frac{31}{4}=\frac{195}{4} \\ \Rightarrow 3x+\frac{31-10}{4}=\frac{195}{4} \\ \Rightarrow 3x=\frac{195}{4}-\frac{21}{4} \\ \Rightarrow 3x=\frac{174}{4} \\ \Rightarrow x=\frac{29}{2} \\ \Rightarrow x=14\frac{1}{2} \end{gathered}$$

Hence, box A weighs $14\frac{1}{2}$ kg.

Answer:

Perimeter of rectangle = 240
⇒ 2(Length + Breadth) = 240
⇒ Length + Breadth = 120
Let the length of the rectangle be x.
⇒ Breadth = 120 − x
New length of rectangle = $x+10\% \mathrm{of} x=\frac{110}{100}x$
New breadth of rectangle = $\left(120-x\right)-20\% \mathrm{of} \left(120-x\right)=\left(120-x\right)\left(1-\frac{20}{100}\right)=\frac{80}{100}\left(120-x\right)$

According to the question,
$$\begin{gathered} 2\left(\mathrm{New} \mathrm{Length}+\mathrm{New} \mathrm{Breadth}\right)=240 \\ \Rightarrow 2\left[\frac{110}{100}x+\frac{80}{100}\left(120-x\right)\right]=240 \\ \Rightarrow \frac{110x+9600-80x}{100}=120 \\ \Rightarrow 30x+9600=12000 \\ \Rightarrow 30x=2400 \\ \Rightarrow x=80 \end{gathered}$$

Hence, length is 80 cm and breadth is 40 cm.

Answer:

Let the present age of B be $x$ years.
Then, the present age of A will be $\left(x+5\right)$ years.
Five years ago,
age of A = $\left(x+5-5\right)=x$ years
age of B = $\left(x-5\right)$ years
According to the question,
$$\begin{gathered} \frac{x}{x-5}=\frac{3}{2} \\ \Rightarrow 2x=3\left(x-5\right) \\ \Rightarrow 2x=3x-15 \\ \Rightarrow x=15 \end{gathered}$$

Hence, the present age of B is 15 years and that of A is 20 years.