NCERT Solutions Class 8 Maths Chapter 2

Q.1 Solve the following equations.

$\begin{array}{l}1.\text{\hspace{0.33em}x}-2=7\\ 2.\text{\hspace{0.33em}y}+3=10\\ 3.6=\text{z}+2\end{array}$

$\begin{array}{l}4.\frac{3}{7}+\text{x}=\frac{17}{7}\\ 5.6\text{x}=12\\ 6.\frac{\text{t}}{5}=10\end{array}$

$\begin{array}{l}7.\frac{2\text{x}}{3}=18\\ 8.1.6=\frac{\text{y}}{1.5}\\ 9.7\text{x}-9=16\end{array}$

$\begin{array}{l}10.14\text{y}-8=13\\ 11.17+6\text{p}=9\\ 12.\frac{\text{x}}{3}+1=\frac{7}{15}\end{array}$

Ans.

$\begin{array}{l}1.\\ \text{x}-2=7\\ \text{Transposing 2 to R.H.S,we obtain}\\ \Rightarrow \text{x}=7+2\\ \Rightarrow \text{x}=9\\ \\ 2.\text{}\\ \text{y}+3=10\\ \text{Transposing 3 to R.H.S,we obtain}\\ \Rightarrow \text{y}=10-3\\ \Rightarrow \text{y}=7\\ \\ 3.\\ \text{}6=\text{z}+2\\ \text{Transposing 2 to L.H.S,we obtain}\\ \Rightarrow 6-2=\text{z}\\ \Rightarrow \text{z}=4\\ \\ 4.\text{}\\ \frac{3}{7}+\text{x}=\frac{17}{7}\\ \text{Transposing}\frac{3}{7}\text{to R.H.S,we obtain}\\ \Rightarrow \text{x}=\frac{17}{7}-\frac{3}{7}\\ \Rightarrow \text{x}=\frac{17-3}{7}\\ \Rightarrow \text{x}=\frac{14}{7}\\ \Rightarrow \text{x}=2\\ \\ 5.\text{}\\ 6\text{x}=12\\ \text{Dividing both sides by 6,we get}\\ \Rightarrow \frac{6\text{x}}{6}=\frac{12}{6}\\ \Rightarrow \text{x}=2\\ \\ 6.\\ \frac{\text{t}}{5}=10\\ \text{Multiplying both the sides by}\text{5,we get}\\ \Rightarrow \frac{\text{t}}{5}\times 5=10\times 5\\ \Rightarrow \text{t}=50\\ \\ 7.\\ \text{\hspace{0.17em}}\frac{2\text{x}}{3}=18\\ \text{Multiplying both the sides by}\frac{3}{2}\text{,we get}\\ \Rightarrow \frac{2\text{x}}{3}\times \frac{3}{2}=18\times \frac{3}{2}\\ \Rightarrow \text{x}=27\\ \\ 8.\\ \text{}1.6=\frac{\text{y}}{1.5}\\ \text{Multiplying both the sides by}1.5\text{,}\text{we get}\\ \Rightarrow 1.6\times 1.5=\frac{\text{y}}{1.5}\times 1.5\\ \Rightarrow \text{y}=2.4\\ \\ 9.\\ 7\text{x}-9=16\\ \text{Transposing 9 to R.H.S,we get}\\ \Rightarrow 7\text{x}=16+9\\ \Rightarrow 7\text{x}=25\\ \text{Dividing both the sides by 7}\\ \Rightarrow \frac{7\text{x}}{7}=\frac{25}{7}\\ \Rightarrow \text{x}=\frac{25}{7}\\ \\ 10.\text{}\\ 14\text{y}-8=13\\ \text{Transposing 8 to R.H.S,we get}\\ \Rightarrow 14\text{y}=13+8\\ \Rightarrow 14\text{y}=21\\ \text{Dividing both the sides by 14}\\ \Rightarrow \frac{14\text{y}}{14}=\frac{21}{14}\\ \Rightarrow \text{y}=\frac{3}{2}\\ \\ 11.\\ \text{}17+6\text{p}=9\\ \text{Transposing 17 to R.H.S,we get}\\ \Rightarrow 6\text{p}=9-17\\ \Rightarrow 6\text{p}=-8\\ \text{Dividing both sides by 6}\\ \Rightarrow \frac{6\text{p}}{6}=\frac{-8}{6}\\ \Rightarrow \text{p}=\frac{-4}{3}\\ \\ 12.\\ \text{}\frac{\text{x}}{3}+1=\frac{7}{15}\\ \text{Transposing1 to R.H.S,we get}\\ \Rightarrow \frac{\text{x}}{3}=\frac{7}{15}-1\\ \Rightarrow \frac{\text{x}}{3}=\frac{7-15}{15}\\ \Rightarrow \frac{\text{x}}{3}=\frac{-8}{15}\\ \text{Multiplying both the sides by 3}\\ \Rightarrow \frac{\text{x}}{3}\times 3=\frac{-8\times 3}{15}\\ \Rightarrow \text{x}=\frac{-8}{5}\end{array}$

Q.2

$\begin{array}{l}\mathrm{If}\mathrm{you}\mathrm{subtract}\frac{1}{2}\mathrm{from}\mathrm{a}\mathrm{number}\mathrm{and}\mathrm{multiply}\mathrm{the}\mathrm{result}\\ \mathrm{by}\frac{1}{2},\mathrm{you}\mathrm{get}\frac{1}{8}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{number}?\end{array}$

Ans.

$\begin{array}{l}\text{Let the number be x}.\\ \mathrm{}\text{According to the question the equation becomes},\\ (\mathrm{x}-\frac{1}{2})\times \frac{1}{2}=\frac{1}{8}\\ \text{On multiplying both the sides by 2,we get}\\ \Rightarrow (\mathrm{x}-\frac{1}{2})\times \frac{1}{2}\times 2=\frac{1}{8}\times 2\\ \Rightarrow \mathrm{x}-\frac{1}{2}=\frac{1}{4}\\ \text{On transposing}\frac{\text{1}}{\text{2}}\text{to R.H.S, we get}\\ \Rightarrow \mathrm{x}=\frac{1}{4}+\frac{1}{2}\\ \Rightarrow \mathrm{x}=\frac{1+2}{4}\\ \Rightarrow \mathrm{x}=\frac{3}{4}\end{array}$

Q.3 The perimeter of a rectangular swimming pool is 154 m. Its length is 2 m more than twice its breadth. What are the length and breadth of the pool?

Ans. The perimeter of a rectangular swimming pool is 154 m.
Let the breadth be x m. The length will be (2x + 2) m.
According to the question the equation becomes,

$\begin{array}{l}\text{2}\left(\text{2}x+\text{2}+x\right)=\text{154}\\ \Rightarrow \text{2}\left(\text{3}x+\text{2}\right)=\text{154}\\ \text{Dividing both sides by 2, we obtain}\\ \Rightarrow \frac{\text{2}\left(\text{3}x+\text{2}\right)}{2}=\frac{\text{154}}{2}\\ \Rightarrow x+\text{2}=\text{77}\\ \text{On transposing 2 to R.H.S, we obtain}\\ \Rightarrow \text{3}x=\text{77}-\text{2}\\ \Rightarrow \text{3}x=\text{75}\\ \text{On dividing both sides by 3, we obtain}\\ \frac{\text{3}x}{3}=\frac{\text{75}}{3}\\ x=\text{25}\\ \therefore \text{2}x+\text{2}=\text{2}\times \text{25}+\text{2}=\text{52}\end{array}$

Hence, the length and breadth of the pool are 52 m and 25 m.

Q.4

$\begin{array}{l}\mathrm{The}\mathrm{base}\mathrm{of}\mathrm{an}\mathrm{is}\mathrm{osceles}\mathrm{triangleis}\frac{4}{3}\mathrm{cm}.\mathrm{The}\mathrm{perimeter}\\ \mathrm{of}\mathrm{the}\mathrm{triangle}\mathrm{is}4\frac{2}{15}\mathrm{cm}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{length}\mathrm{of}\mathrm{either}\\ \mathrm{of}\mathrm{the}\mathrm{remaining}\mathrm{equal}\mathrm{sides}?\end{array}$

Ans.

$\begin{array}{l}\text{Let the length of equal sides be x cm.}\\ \text{Perimeter =xcm+xcm+Base=4}\frac{\text{2}}{\text{15}}\text{cm}\\ \Rightarrow 2\mathrm{x}+\frac{4}{3}=\frac{62}{15}\\ \text{On transposing}\frac{\text{4}}{\text{3}}\text{to R.H.S, we obtain}\\ \Rightarrow 2\mathrm{x}=\frac{62}{15}-\frac{4}{3}\\ \Rightarrow 2\mathrm{x}=\frac{62-20}{15}\\ \Rightarrow 2\mathrm{x}=\frac{42}{15}\\ \text{On dividing both sides by 2, we obtain}\\ \Rightarrow \frac{2\mathrm{x}}{2}=\frac{42}{15\times 2}\\ \Rightarrow \mathrm{x}=\frac{7}{5}\\ \text{Therefore, the length of equal sides is}\frac{\text{7}}{\text{5}}\text{cm or 1}\frac{\text{2}}{\text{5}}\text{.}\end{array}$

Q.5 Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.

Ans. Let one number be x.

Therefore, the other number will be x + 15.

According to the question,

x + x + 15 = 95

2x + 15 = 95

On transposing 15 to R.H.S, we obtain

2x = 95 − 15

2x = 80

On dividing both sides by 2, we get

x = 40

Therefore, x+15 = 40+15 = 55

Hence the numbers are 40 and 55.

Q.6 Two numbers are in the ratio 5:3. If they differ by 18, what are the numbers?

Ans. Let the ratio between these numbers be x.
Therefore, the numbers will be 5x and 3x respectively.
Difference between these numbers = 18
According, to the question the equation becomes
5x − 3x = 18
2x = 18

$\begin{array}{l}\text{Dividing both sides by 2,we get}\\ \frac{2\mathrm{x}}{2}=\frac{18}{2}\\ \Rightarrow \mathrm{x}=9\end{array}$

Therefore, the first number is 5x=5×9=45 and,
The second number is 3x=3×9=27.

Q.7 Three consecutive integers add up to 51. What are these integers?

Ans. Let three consecutive integers be x, x + 1, and x + 2.

Sum of these numbers = x+ x + 1 + x + 2 = 51

3x + 3 = 51

On transposing 3 to R.H.S, we obtain

3x = 51 − 3

3x = 48

$\begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3,\mathrm{we}\mathrm{get}\\ \frac{3\mathrm{x}}{3}=\frac{48}{3}\\ \Rightarrow \mathrm{x}=16\\ \therefore \mathrm{x}+1=17\\ \mathrm{x}+2=18\end{array}$

Hence, the integers are 16, 17 and 18.

Q.8 The sum of three consecutive multiples of 8 is 888. Find the multiples.

Ans. Let the three consecutive multiples of 8 be 8x, 8(x + 1), 8(x + 2).

Sum of these numbers = 8x + 8(x + 1) + 8(x + 2) = 888

8(x + x + 1 + x + 2) = 888

8(3x + 3) = 888

\begin{array}{l}\text{On dividing both sides by 8, we get}\\ \Rightarrow \frac{8(3x+3)}{8}=\frac{888}{8}\\ \Rightarrow 3x+3=111\\ \text{On transposing 3 to RHS,we get}\\ \Rightarrow 3x=111-3\\ \Rightarrow 3x=108\\ \text{On dividing both sides by 3,we get}\\ \Rightarrow \frac{3x}{3}=\frac{108}{3}\\ \Rightarrow x=36\end{array}

Therefore, 8x = 8×36 = 288
8(x+1) = 8(36+1) = 8×37 = 296
8(x+2) = 8(36+2) = 8×38 = 304
Hence, the numbers are 288,296 and 304.

Q.9 Three consecutive integers are such that when they are taken in increasing order and multiplied by 2, 3 and 4 respectively, they add up to 74. Find these numbers.

Ans. Let three consecutive integers be x, x + 1, x + 2. According to the question,

2x + 3(x + 1) + 4(x + 2) = 74

2x + 3x + 3 + 4x + 8 = 74

9x + 11 = 74

On transposing 11 to R.H.S, we obtain

9x = 74 − 11

9x = 63

\begin{array}{l}\text{On dividing both sides by 9,we get}\\ \Rightarrow \frac{9x}{9}=\frac{63}{9}\\ \Rightarrow x=7\\ \therefore x+1=8\\ x+2=9\\ \text{Hence,the numbers are 7, 8 and 9}\text{.}\end{array}

Q.10 The ages of Rahul and Haroon are in the ratio 5:7. Four years later the sum of their ages will be 56 years. What are their present ages?

Ans. Let the ratio between Rahul’s age and Haroon’s age be x.

Therefore, the age of Rahul and Haroon will be 5x years and 7x years.

Four years later, the age of Rahul and Haroon will be (5x + 4) years and (7x + 4) years.

According to the given question, the equation becomes,

(5x + 4 + 7x + 4) = 56

12x + 8 = 56

On transposing 8 to R.H.S, we obtain

12x = 56 − 8

12x = 48

$$\begin{gathered} \begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}12,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{12\mathrm{x}}{12}=\frac{48}{12}\\ \Rightarrow \mathrm{x}=4\\ \therefore \mathrm{Rahul}‘\mathrm{s}\mathrm{age}=5\mathrm{x}=5\times 4=20\mathrm{years} \\ \mathrm{Haroon}‘\mathrm{s}\mathrm{age}\mathrm{is}=7\mathrm{x}\mathrm{years}=7\times 4=28\mathrm{years}\end{array} \end{gathered}$$

Q.11 The number of boys and girls in a class are in the ratio 7:5. The number of boys is 8 more than the number of girls. What is the total class strength?

Ans. Let the ratio between the number of boys and numbers of girls be x.

Then, number of boys = 7x

Number of girls = 5x

According to the given question,

Number of boys = Number of girls + 8

7x = 5x + 8

On transposing 5x to L.H.S, we obtain

7x − 5x = 8

2x = 8

$\begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}2,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{2x}{2}=\frac{8}{2}\\ \Rightarrow x=4\\ \therefore \mathrm{Number}\mathrm{of}\mathrm{boys}=7\mathrm{x}=7\times 4=28\\ \mathrm{Number}\mathrm{of}\mathrm{girls}=\mathrm{x}=5\times 4=20\\ \mathrm{Hence},\mathrm{the}\mathrm{total}\mathrm{strength}\mathrm{of}\mathrm{class}=28+20=48\mathrm{students}\end{array}$

Q.12 Baichung’s father is 26 years younger than Baichung’s grandfather and 29 years older than Baichung. The sum of the ages of all the three is 135 years. What is the age of each one of them?

Ans. Let Baichung’s father’s age be x years.

Therefore, Baichung’s age will be (x − 29) years and Baichung’s grandfather’s age will be (x + 26) years.

According to the given question, the equation becomes:

x + x − 29 + x + 26 = 135

3x − 3 = 135

On transposing 3 to R.H.S, we obtain

3x = 135 + 3

3x = 138

$$\begin{gathered} \begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{3\mathrm{x}}{3}=\frac{138}{3}\\ \Rightarrow \mathrm{x}=46\\ \\ \therefore \mathrm{Baichung}‘\mathrm{s}\mathrm{father}‘\mathrm{s}\mathrm{age}=\mathrm{x}\mathrm{years}=46\mathrm{years}\\ \mathrm{Baichung}‘\mathrm{s}\mathrm{age}=(\mathrm{x}-29)\mathrm{years}=(46-29)=17\mathrm{years}\\ \mathrm{Baichung}‘\mathrm{s}\mathrm{grandfather}‘\mathrm{s}\mathrm{age}=(\mathrm{x}+26)\mathrm{years}\\ =(46+26)\mathrm{years} \\ =72\mathrm{years}\end{array} \end{gathered}$$

Q.13 Fifteen years from now Ravi’s age will be four times his present age. What is Ravi’s present age?

Ans. Let Ravi’s present age be x years.

Fifteen years later, Ravi’s age = 4 × His present age

x + 15 = 4x

On transposing x to R.H.S, we obtain

15 = 4x − x

15 = 3x

$\begin{array}{l}\mathrm{On}\mathrm{dividing}\mathrm{both}\mathrm{sides}\mathrm{by}3,\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{15}{3}=\frac{3\mathrm{x}}{3}\\ \Rightarrow \mathrm{x}=5\\ \\ \therefore \mathrm{Ravi}‘\mathrm{s}\mathrm{present}\mathrm{age}=5\mathrm{years}\end{array}$

Q.14

$\begin{array}{l}\mathrm{A}\mathrm{rational}\mathrm{number}\mathrm{is}\mathrm{such}\mathrm{that}\mathrm{when}\mathrm{you}\mathrm{multiply}\mathrm{it}\mathrm{by}\frac{5}{2}\\ \mathrm{and}\mathrm{add}\frac{2}{3}\mathrm{to}\mathrm{the}\mathrm{product},\mathrm{you}\mathrm{get}\frac{-7}{12}.\mathrm{What}\mathrm{is}\mathrm{the}\mathrm{number}?\end{array}$

Ans.

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{number}\mathrm{be}\mathrm{x}.\\ \text{According to the question,the equation becomes}\\ \frac{5}{2}\mathrm{x}+\frac{2}{3}=-\frac{7}{12}\\ \text{On transposing}\frac{2}{3}\text{to R.H.S,}\text{we get}\\ \Rightarrow \frac{5}{2}\mathrm{x}=-\frac{7}{12}-\frac{2}{3}\\ \Rightarrow \frac{5}{2}\mathrm{x}=\frac{-7-8}{12}\\ \Rightarrow \frac{5}{2}\mathrm{x}=\frac{-15}{12}\\ \text{On multiplying both sides by}\frac{2}{5},\mathrm{we}\mathrm{get}\\ \Rightarrow \frac{5\mathrm{x}}{2}\times \frac{2}{5}=\frac{-15}{12}\times \frac{2}{5}\\ \Rightarrow \mathrm{x}=-\frac{1}{2}\\ \text{Hence,}\text{the number is}-\frac{1}{2}.\end{array}$

Q.15 Lakshmi is a cashier in a bank. She has currency notes of denominations ₹ 100, ₹ 50 and ₹ 10, respectively. The ratio of the number of these notes is 2:3:5. The total cash with Lakshmi is ₹ 4,00,000. How many notes of each denomination does she have?

Ans. Let the ratio between the numbers of notes of different denominations be x.

Therefore, numbers of ₹ 100 notes, ₹ 50 notes, and ₹ 10 notes will be 2x, 3x, and 5x respectively.

Amount of ₹ 100 notes = ₹ (100×2x) = ₹ 200x

Amount of ₹ 50 notes = ₹ (50×3x) = ₹ 150x

Amount of ₹ 10 notes = ₹ (10×5x) = ₹ 50x

Total cash = ₹ 400000.

Therefore,

200x + 150x + 50x = 400000

⇒ 400x = 400000

On dividing both sides by 400, we obtain

x = 1000

Therefore,

Number of ₹ 100 notes = 2x = 2 × 1000 = 2000 notes

Number of ₹ 50 notes = 3x = 3 × 1000 = 3000 notes

Number of ₹ 10 notes = 5x = 5 × 1000 = 5000 notes

Q.16 I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of Rs 2 coins is 3 times the number of ₹ 5 coins. The total number of coins is 160. How many coins of each denomination are with me?

Ans. Let the number of ₹ 5 coins be x.

Since the number of ₹ 2 coins is 3 times the number of ₹ 5 coins, so the number of ₹ 2 coins = 3x

Therefore, the number of ₹1 coins = 160 − (Number of coins of ₹ 5 and of ₹ 2)

= 160 − (3x + x) = 160 − 4x

Now, Amount of ₹ 1 coins = ₹ [1 × (160 − 4x)] = ₹ (160 − 4x)

Amount of ₹ 2 coins = ₹ (2 × 3x) = ₹ 6x

Amount of ₹ 5 coins = ₹ (5 × x) = ₹ 5x

Total amount is ₹300.

Therefore, 160 − 4x + 6x + 5x =300

160 + 7x = 300

On transposing 160 to R.H.S., we obtain

7x = 300 – 160 = 140

On dividing both sides by 7, we get

x = 20No. of ₹ 1 coins = 160 – 4x = 160 – 4×20 = 80

No. of ₹ 2 coins = 3x = 3×20 = 60

No. of ₹ 5 coins = x = 20

Therefore, number of ₹ 1 coins is 80, ₹ 2 coins is 60 and ₹ 5 coins is 20.

Q.17 The organisers of an essay competition decide that a winner of the competition gets a prize of ₹ 100 and a participant who does not win gets a prize of ₹ 25. The total prize money distributed is ₹ 3,000. Find the number of winners, if the total number of participants is 63.

Ans. Let the number of winners be x.

Therefore, the number of participants who did not win will be 63 − x.

Amount given to the winners = ₹ (100 × x) = ₹ 100x

Amount given to the participants who did not win = ₹ [25(63 − x)]

= ₹ (1575 − 25x)

According to the given question,

100x + 1575 − 25x = 3000

On transposing 1575 to R.H.S, we obtain

75x = 3000 − 1575

75x = 1425

$\begin{array}{l}\text{On dividing both sides by 75},\text{we obtain}\\ \Rightarrow \frac{75\mathrm{x}}{75}=\frac{1425}{75}\\ \Rightarrow \mathrm{x}=19\\ \\ \mathrm{Hence},\text{the number of winners}=19\end{array}$

Q.18

$\begin{array}{l}\text{Solve the following equations and check your results.}\\ 1.3\text{x}=2\text{x}+18\\ 2.5\text{t}-3=3\text{t}-5\\ 3.5\text{x}+9=5+3\text{x}\\ 4.4\text{z}+3=6+2\text{z}\\ 5.2\text{x}-1=14-\text{x}\\ 6.8\text{x}+4=3\left(\text{x}-1\right)+7\\ 7.\text{\hspace{0.33em}x}=\frac{4}{5}\left(\text{x}+10\right)\\ 8.\frac{2\text{x}}{3}+1=\frac{7\text{x}}{15}+3\\ 9.2\text{y}+\frac{5}{3}=\frac{26}{3}-\text{y}\\ 10.3\text{m}=5\text{m}-\frac{8}{5}\end{array}$

Ans.

$\begin{array}{l}1.3\mathrm{x}=2\mathrm{x}+18\\ \text{On transposing 2x to L.H.S,}\mathrm{we}\mathrm{get}\\ \Rightarrow 3\mathrm{x}-2\mathrm{x}=18\\ \Rightarrow \mathrm{x}=18\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=3\mathrm{x}\\ =3\times 18\\ =54\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=2\mathrm{x}+18\\ =2\times 18+18\\ =36+18=54\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$

$\begin{array}{l}2.5\mathrm{t}-3=3\mathrm{t}-5\\ \text{On transposing 3t to L.H.S and}-\text{3 to R.H.S,}\text{we get}\\ \Rightarrow 5\mathrm{t}-3\mathrm{t}=3-5\\ \Rightarrow 2\mathrm{t}=-2\\ \Rightarrow \mathrm{t}=-1\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=5\mathrm{t}-3\\ =5\times (-1)-3\\ =-5-3\\ =-8\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=3\mathrm{t}-5\\ =3\times (-1)-5\\ =-3-5\\ =-8\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}3.5\mathrm{x}+9=5+3\mathrm{x}\\ \text{On transposing 3x to L.H.S and}9\text{to R.H.S,}\text{we get}\\ \Rightarrow 5\mathrm{x}-3\mathrm{x}=5-9\\ \Rightarrow 2\mathrm{x}=-4\\ \text{On dividing both sides by 2,we get}\\ \Rightarrow \frac{2\mathrm{x}}{2}=-\frac{4}{2}\\ \Rightarrow \mathrm{x}=-2\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=5\mathrm{x}+9\\ =5\times (-2)+9\\ =(-10)+9\\ =-1\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=5+3\mathrm{x}\\ =5+3\times (-2)\\ =5+(-6)\\ =-1\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}4.4\mathrm{z}+3=6+2\mathrm{z}\\ \text{On transposing 2z to L.H.S and}3\text{to R.H.S,}\text{we get}\\ \Rightarrow 4\mathrm{z}-2\mathrm{z}=6-3\\ \Rightarrow 2\mathrm{z}=3\\ \text{On dividing both sides by 2,}\text{we get}\\ \Rightarrow \frac{2\mathrm{z}}{2}=\frac{3}{2}\\ \Rightarrow \mathrm{z}=\frac{3}{2}\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=4\mathrm{z}+3\\ =4\times \left(\frac{3}{2}\right)+3\\ =6+3\\ =9\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=6+2\mathrm{z}\\ =6+2\times \frac{3}{2}\\ =6+3\\ =9\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}5.\mathrm{}2\mathrm{x}-1=14-\mathrm{x}\\ \text{On transposing x to L.H.S and}1\text{to R.H}.\mathrm{S},\mathrm{we}\mathrm{get}\\ \Rightarrow 2\mathrm{x}+\mathrm{x}=14+1\\ \Rightarrow 3\mathrm{x}=15\\ \text{On dividing both sides by 3,we get}\\ \Rightarrow \frac{3\mathrm{x}}{3}=\frac{15}{3}\\ \Rightarrow \mathrm{x}=5\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=2\mathrm{x}-1\\ =2\times 5-1\\ =10-1\\ =9\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=14-\mathrm{x}\\ =14-5\\ =9\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}6.8\mathrm{x}+4=3(\mathrm{x}-1)+7\\ \Rightarrow 8\mathrm{x}+4=3\mathrm{x}-3+7\\ \Rightarrow 8\mathrm{x}+4=3\mathrm{x}+4\\ \text{On transposing 3x to L.H.S and}4\text{to R.H.S,we get}\\ \Rightarrow 8\mathrm{x}-3\mathrm{x}=4-4\\ \Rightarrow 5\mathrm{x}=0\\ \Rightarrow \mathrm{x}=0\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=8\mathrm{x}+4\\ =8\times 0+4\\ =4\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=3(\mathrm{x}-1)+7\\ =3(0-1)+7\\ =-3+7\\ =4\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}7.\mathrm{}\mathrm{x}=\frac{4}{5}(\mathrm{x}+10)\\ \mathrm{Multiplying}\mathrm{both}\mathrm{sides}\mathrm{by}5,\mathrm{we}\mathrm{get}\\ \Rightarrow 5\mathrm{x}=4(\mathrm{x}+10)\\ \Rightarrow 5\mathrm{x}=4\mathrm{x}+40\\ \mathrm{Transposing}4\mathrm{x}\mathrm{to}\mathrm{L}.\mathrm{H}.\mathrm{S},\mathrm{we}\mathrm{get}\\ \Rightarrow 5\mathrm{x}-4\mathrm{x}=40\\ \Rightarrow \mathrm{x}=40\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{x}\\ =40\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{4}{5}(\mathrm{x}+10)\\ =\frac{4}{5}(40+10)\\ =\frac{4}{5}\left(50\right)\\ =\frac{4}{5}\times \left(50\right)\\ =4\times 10]\\ =40\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}8.\frac{2\mathrm{x}}{3}+1=\frac{7\mathrm{x}}{15}+3\\ \text{On transposing}\frac{7\mathrm{x}}{15}\text{to L.H.S and}1\text{to R.H.S,we get}\\ \Rightarrow \frac{2\mathrm{x}}{3}-\frac{7\mathrm{x}}{15}=3-1\\ \text{Taking LCM of 3 and 15,we get}\\ \Rightarrow \frac{10\mathrm{x}-7\mathrm{x}}{15}=2\\ \Rightarrow \frac{3\mathrm{x}}{15}=2\\ \Rightarrow 3\mathrm{x}=30\\ \Rightarrow \mathrm{x}=\frac{30}{3}=10\\ \mathrm{L}.\mathrm{H}.\mathrm{S}=\frac{2\mathrm{x}}{3}+1\\ =\frac{2\times 10}{3}+1\\ =\frac{20}{3}+1\\ =\frac{20+3}{3}=\frac{23}{3}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{7\mathrm{x}}{15}+3\\ =\frac{7\mathrm{x}+45}{15}\\ =\frac{7\times 10+45}{15}\\ =\frac{115}{15}\\ =\frac{23}{3}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}9.2\mathrm{y}+\frac{5}{3}=\frac{26}{3}-\mathrm{y}\\ \text{On transposing}\mathrm{y}\text{to L.H.S and}\frac{5}{3}\text{to R.H.S,we get}\\ \Rightarrow 2\mathrm{y}+\mathrm{y}=\frac{26}{3}-\frac{5}{3}\\ \Rightarrow 3\mathrm{y}=\frac{21}{3}\\ \Rightarrow \mathrm{y}=\frac{21}{9}\\ \Rightarrow \mathrm{y}=\frac{7}{3}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=2\mathrm{y}+\frac{5}{3}\\ =2\times \frac{7}{3}+\frac{5}{3}\\ =\frac{14}{3}+\frac{5}{3}\\ =\frac{14+5}{3}=\frac{19}{3}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=\frac{26}{3}-\mathrm{y}\\ =\frac{26}{3}-\frac{7}{3}\\ =\frac{26-7}{3}\\ =\frac{19}{3}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}10.3\mathrm{m}=5\mathrm{m}-\frac{8}{5}\\ \text{On transposing}5\mathrm{m}\text{to L.H.S ,we get}\\ \Rightarrow 3\mathrm{m}-5\mathrm{m}=-\frac{8}{5}\\ \Rightarrow -2\mathrm{m}=-\frac{8}{5}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\Rightarrow \mathrm{m}=-\frac{8}{5\times (-2)}\\ \Rightarrow \mathrm{m}=\frac{4}{5}\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\mathrm{L}.\mathrm{H}.\mathrm{S}=3\mathrm{m}\\ =3\times \frac{4}{5}\\ =\frac{12}{5}\\ \mathrm{R}.\mathrm{H}.\mathrm{S}=5\mathrm{m}-\frac{8}{5}\\ =5\times \frac{4}{5}-\frac{8}{5}\\ =4-\frac{8}{5}\\ =\frac{20-8}{5}\\ =\frac{12}{5}\\ \therefore \mathrm{L}.\mathrm{H}.\mathrm{S}=\mathrm{R}.\mathrm{H}.\mathrm{S}\end{array}$ \begin{array}{l}\\ \end{array}

Q.19

$\begin{array}{l}\mathrm{Amina}\mathrm{thinks}\mathrm{of}\mathrm{a}\mathrm{number}\mathrm{and}\mathrm{subtracts}\frac{5}{2}\mathrm{from}\mathrm{it}.\mathrm{She}\\ \mathrm{multiplies}\mathrm{the}\mathrm{result}\mathrm{by}8.\mathrm{The}\mathrm{result}\mathrm{now}\mathrm{obtained}\mathrm{is}\\ 3\mathrm{times}\mathrm{the}\mathrm{same}\mathrm{number}\mathrm{she}\mathrm{thought}\mathrm{of}.\mathrm{What}\mathrm{is}\mathrm{the}\\ \mathrm{number}?\end{array}$

Ans. Let the number be x.
According to the given question,

$8(\mathrm{x}-\frac{5}{2})=3\mathrm{x}$

8x − 20 = 3x
Transposing 3x to L.H.S and −20 to R.H.S, we obtain
8x − 3x = 20
5x = 20
On dividing both sides by 5, we get
x = 4
Hence, the number is 4.

Q.20 A positive number is 5 times another number. If 21 is added to both the numbers, then one of the new numbers becomes twice the other new number. What are the numbers? What is the sum of the measures of the angles of a convex quadrilateral? Will this property hold if the quadrilateral is not convex? (Make a non-convex quadrilateral and try!)

Ans. Let the numbers be x and 5x.

$\begin{array}{l}\text{According to the question},\text{the equation becomes}\\ \text{21}+\text{5}\mathrm{x}=\text{2}(\mathrm{x}+\text{21})\\ \Rightarrow \text{21}+\text{5}\mathrm{x}=\text{2}\mathrm{x}+\text{42}\\ \\ \text{Transposing 2}\mathrm{x}\text{to L}.\text{H}.\text{S and 21 to R}.\text{H}.\text{S},\text{we obtain}\\ \Rightarrow \text{5}\mathrm{x}-\text{2}\mathrm{x}=\text{42}-\text{21}\\ \text{3}\mathrm{x}=\text{21}\\ \\ \text{Dividing both sides by 3},\text{we obtain}\\ \mathrm{x}=\text{7}\\ \text{5}\mathrm{x}=\text{5}\times \text{7}=\text{35}\\ \\ \text{Hence},\text{the numbers are 7 and 35 respectively}.\end{array}$

Q.21 Sum of the digits of a two-digit number is 9. When we interchange the digits, it is found that the resulting new number is greater than the original number by 27. What is the two-digit number?

Ans. Let the digits at tens place and ones place be x and 9 − x respectively.
Therefore, original number = 10x + (9 − x) = 9x + 9
On interchanging the digits, the digits at ones place and tens place will be x and 9 − x respectively.

Therefore, new number after interchanging the digits = 10(9 − x) + x
= 90 − 10x + x
= 90 − 9x

According to the given question,
New number = Original number + 27
90 − 9x = 9x + 9 + 27
90 − 9x = 9x + 36

Transposing 9x to R.H.S and 36 to L.H.S, we obtain
90 − 36 = 18x
54 = 18x

Dividing both sides by 18, we obtain
3 = x and 9 − x = 6

Hence, the digits at tens place and ones place of the number are 3 and 6 respectively.

Therefore, the two-digit number is 9x + 9
= 9 × 3 + 9
= 36

Q.22 One of the two digits of a two digit number is three times the other digit. If you interchange the digits of this two-digit number and add the resulting number to the original number, you get 88. What is the original number?

Ans.

$\begin{array}{l}\text{Let the digits at tens place and ones place be ‘}\mathrm{x}‘\text{and ‘3}\mathrm{x}‘.\\ \therefore \mathrm{the}\text{original number}=\text{1}0\mathrm{x}+\text{3}\mathrm{x}=\text{13}\mathrm{x}\\ \\ \text{On interchanging the digits},\text{the digits at ones place will be ‘x’and}\\ \text{tens place will be ‘3}\mathrm{x}‘.\\ \\ \therefore \text{New number}=\text{1}0\times \text{3}\mathrm{x}+\mathrm{x}=\text{3}0\mathrm{x}+\mathrm{x}=\text{31}\mathrm{x}\\ \\ \text{According to the given question},\mathrm{the}\mathrm{equation}\mathrm{becomes}\\ \text{8813}\mathrm{x}+\text{31}\mathrm{x}=\text{88}\\ \Rightarrow \text{44}\mathrm{x}=\text{88}\end{array}$

$\begin{array}{l}\text{Dividing both sides by 44},\text{we obtain}\\ \Rightarrow \frac{\text{44}\mathrm{x}}{44}=\frac{\text{88}}{44}\\ \Rightarrow \mathrm{x}=\text{2}\\ \\ \therefore ,\text{original number}=\text{13}\mathrm{x}=\text{13}\times \text{2}=\text{26}\\ \\ \text{Hence},\text{the two}-\text{digit number may be 26 or 62}.\end{array}$

Q.23 Shobo’s mother’s present age is six times Shobo’s present age. Shobo’s age five years from now will be one third of his mother’s present age. What are their present ages?

Ans. Let Shobo’s age be x years.
His mother’s age will be 6x years.
According to the given question, the equation becomes
:

$\mathrm{x}+5=\frac{6\mathrm{x}}{3}$

Multiplying both sides by 3

$\begin{array}{l}(\mathrm{x}+5)\times 3=\frac{6\mathrm{x}}{3}\times 3\\ \Rightarrow 3\mathrm{x}+15=6\mathrm{x}\\ \text{Transposing 3x to R.H.S,we get}\\ \Rightarrow \text{15}=\text{6x}-\text{3x}\\ \Rightarrow \text{15}=3\mathrm{x}\\ \text{Dividing both sides by 3, we get}\\ \Rightarrow \frac{\text{15}}{3}=\frac{3\mathrm{x}}{3}\\ \Rightarrow \mathrm{x}=5\\ \\ \therefore 6\mathrm{x}=6\times 5=30\end{array}$

Therefore, the present ages of Shobo and his mother will be 5 years and 30 years.

Q.24 There is a narrow rectangular plot, reserved for a school, in Mahuli village. The length and breadth of the plot are in the ratio 11:4. At the rate ₹100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?

Ans. Let the ratio between the length and breadth of the rectangular plot be x.

Hence, the length and breadth of the rectangular plot will be 11x m and 4x m respectively.

Perimeter of the plot = 2(Length + Breadth)

$\begin{array}{l}=\left[2(11\mathrm{x}+4\mathrm{x})\right]\mathrm{m}\\ =30\mathrm{x}\end{array}$

It is given that the cost of fencing the plot at the rate of ₹ 100 per metre is ₹ 75, 000.

100 × Perimeter = 75000

100 × 30x = 75000

3000x = 75000

Dividing both sides by 3000, we obtain

\frac{\text{3}000x}{3000}=\frac{\text{75}000}{3000}

x = 25

Therefore, Length = 11x m = (11 × 25) m = 275 m

and Breadth = 4x m = (4 × 25) m = 100 m

Hence, the dimensions of the plot are 275 m and 100 m.

Q.25 Hasan buys two kinds of cloth materials for school uniforms, shirt material that costs him ₹ 50 per metre and trouser material that costs him ₹ 90 per metre. For every 3 meters of the shirt material he buys 2 metres of the trouser material. He sells the materials at 12% and 10% profit respectively. His total sale is Rs 36,600. How much trouser material did he buy?

Ans. Let the trouser and shirt material bought by Hasan be 2x and 3x .
The selling price of trouser material is

$\begin{array}{l}=₹(90+\frac{90\times 12}{100})\\ =₹100.80\end{array}$

The selling price of Shirt material is

$\begin{array}{l}=₹\left(50+\frac{50\times 10}{100}\right)\\ =₹55\end{array}$

Total sale =36,600

$\begin{array}{l}\text{1}00.\text{8}0\times \left(\text{2}\mathrm{x}\right)+\text{55}\times \left(\text{3}\mathrm{x}\right)=\text{36,60}0\\ \text{2}0\text{1}.\text{6}0\mathrm{x}+\text{165}\mathrm{x}=\text{36,60}0\\ \text{366}.\text{6}0\mathrm{x}=\text{3660}0\\ \text{Dividing both sides by 366}.\text{6}0,\text{we obtain}\\ \Rightarrow \frac{\text{366}.\text{6}0\mathrm{x}}{366.60}=\frac{\text{3660}0}{366.60}\\ \Rightarrow \mathrm{x}=99.836\\ \Rightarrow \text{x}=\text{1}00\\ \\ \therefore \text{Trouser material}=\text{2}\mathrm{x}\text{m}\\ =(\text{2}\times \text{1}00)\text{m}\\ =\text{2}00\text{m}\end{array}$

Q.26 Half of a herd of deer are grazing in the field and three fourths of the remaining are playing nearby. The rest 9 are drinking water from the pond. Find the number of deer in the herd.

Ans. Let the number of deer be x.

Number of deer grazing in the field = x / 2

Remaining deer= x – (x / 2)

Therefore, number of deer playing nearby is

$\begin{array}{l}=\frac{3}{4}\times (\mathrm{x}-\frac{\mathrm{x}}{2})\\ =\frac{3}{4}\times \frac{\mathrm{x}}{2}\\ =\frac{3\mathrm{x}}{8}\end{array}$

The rest 9 are drinking water from the pond, so the equation becomes:

$\begin{array}{l}\mathrm{x}-(\frac{\mathrm{x}}{2}+\frac{3\mathrm{x}}{8})=9\\ \Rightarrow \mathrm{x}-\left(\frac{4\mathrm{x}+3\mathrm{x}}{8}\right)=9\\ \Rightarrow \mathrm{x}-\frac{7\mathrm{x}}{8}=9\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\Rightarrow \frac{8\mathrm{x}-7\mathrm{x}}{8}=9\\ \Rightarrow \frac{\mathrm{x}}{8}=9\end{array}$ $\begin{array}{l}\end{array}$

$\begin{array}{l}\text{Mutiplying both sides by 8,we get}\\ \Rightarrow \frac{\mathrm{x}}{8}\times 8=9\times 8\\ \Rightarrow \mathrm{x}=72\end{array}$

Therefore, the total number of deer in the herd is 72.

Q.27 A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.

Ans. Let the granddaughter’s age be x years.

Therefore, grandfather’s age will be 10x years.

According to the question, the equation becomes:

10x = x + 54

Transposing x to L.H.S, we obtain

10x − x = 54

9x = 54

x = 6

Granddaughter’s age = x years = 6 years

Grandfather’s age = 10x years

= (10 × 6) years

= 60 years

Therefore, the grandfather’s age is 60 years and granddaughter’s age is 6 years.

Q.28 Aman’s age is three times his son’s age. Ten years ago he was five times his son’s age. Find their present ages.

Ans. Let Aman’s son’s age be x years.
Therefore, Aman’s age will be 3x years.
Ten years ago, their age was (x − 10) years and (3x − 10) years.
According to the question, the equation is:

$\begin{array}{l}3\mathrm{x}-10=5(\mathrm{x}-10)\\ \Rightarrow 3\mathrm{x}-10=5\mathrm{x}-50\\ \mathrm{On}\mathrm{transposing}\text{5x to L.H.S and 10 to R.H.S,we get}\\ \Rightarrow 3\mathrm{x}-5\mathrm{x}=10-50\\ \Rightarrow -2\mathrm{x}=-40\\ \Rightarrow 2\mathrm{x}=40\end{array}$

$\begin{array}{l}\text{On dividing both the sides by 2,we get}\\ \Rightarrow \frac{2\mathrm{x}}{2}=\frac{40}{2}\\ \Rightarrow \mathrm{x}=20\end{array}$

Therefore,
Aman’s son’s age is 20 years and
Aman’s age = 3 × 20 years = 60 years.

Q.29

$\begin{array}{l}\mathrm{Solve}\mathrm{the}\mathrm{following}\mathrm{linear}\mathrm{equations}.\\ 1.\frac{\mathrm{x}}{2}-\frac{1}{5}=\frac{\mathrm{x}}{3}+\frac{1}{4}\\ 2.\frac{\mathrm{n}}{2}-\frac{3\mathrm{n}}{4}+\frac{5\mathrm{n}}{6}=21\\ 3.\mathrm{x}+7-\frac{8\mathrm{x}}{3}=\frac{17}{6}-\frac{5\mathrm{x}}{2}\end{array}$

$\begin{array}{l}4.\frac{\mathrm{x}-5}{3}=\frac{\mathrm{x}-3}{5}\\ 5.\frac{3\mathrm{t}-2}{4}-\frac{2\mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}\\ 6.\mathrm{m}-\frac{\mathrm{m}-1}{2}=1-\frac{\mathrm{m}-2}{3}\\ \mathrm{Simplify}\mathrm{and}\mathrm{solve}\mathrm{the}\mathrm{following}\mathrm{linear}\mathrm{equations}.\\ 7.3(\mathrm{t}-3)=5(2\mathrm{t}+1)\\ 8.15(\mathrm{y}-4)-2(\mathrm{y}-9)+5(\mathrm{y}+6)=0\\ 9.3(5\mathrm{z}-7)-2(9\mathrm{z}-11)=4(8\mathrm{z}-13)-17\\ 10.0.25(4\mathrm{f}-3)=0.05(10\mathrm{f}-9)\end{array}$

Ans.

$\begin{array}{l}1.\frac{\mathrm{x}}{2}-\frac{1}{5}=\frac{\mathrm{x}}{3}+\frac{1}{4}\\ \text{Taking L.C.M on both the sides ,we get}\\ \Rightarrow \frac{5\mathrm{x}-2}{10}=\frac{4\mathrm{x}+3}{12}\\ \text{Multiplying both the sides by10,we get}\\ \Rightarrow \frac{5\mathrm{x}-2}{10}\times 10=\frac{4\mathrm{x}+3}{12}\times 10\\ \Rightarrow (5\mathrm{x}-2)=\frac{4\mathrm{x}+3}{12}\times 10\end{array}$

$\begin{array}{l}\text{Multiplying both the sides by12,we get}\\ \Rightarrow (5\mathrm{x}-2)\times 12=\frac{4\mathrm{x}+3}{12}\times 10\times 12\\ \Rightarrow 60\mathrm{x}-24=40\mathrm{x}+30\\ \text{Transposing 40x to L.H.S and 24 to R.H.S,we get}\\ \Rightarrow 60\mathrm{x}-40\mathrm{x}=30+24\\ \Rightarrow 20\mathrm{x}=54\\ \text{Dividing both sides by 20 ,we get}\\ \Rightarrow \frac{20\mathrm{x}}{20}=\frac{54}{20}\\ \Rightarrow \mathrm{x}=2.7\end{array}$ \begin{array}{l}\mathrm{}\end{array}

$\begin{array}{l}2.\frac{\mathrm{n}}{2}-\frac{3\mathrm{n}}{4}+\frac{5\mathrm{n}}{6}=21\\ \text{Taking L.C.M of 2,4 and 6,we get}\\ \Rightarrow \frac{6\mathrm{n}-9\mathrm{n}+10\mathrm{n}}{12}=21\\ \Rightarrow \frac{7\mathrm{n}}{12}=21\end{array}$ \begin{array}{l}\end{array}

$\begin{array}{l}\text{Multiplying bith sides by 12,}\text{we get}\\ \Rightarrow \frac{7\mathrm{n}}{12}\times 12=21\times 12\\ \Rightarrow 7\mathrm{n}=21\times 12\\ \Rightarrow 7\mathrm{n}=252\\ \text{Dividing both sides by 7,}\text{we get}\\ \Rightarrow \frac{7\mathrm{n}}{7}=\frac{252}{7}\\ \Rightarrow \mathrm{n}=\frac{252}{7}\\ \Rightarrow \mathrm{n}=36\end{array}$

$\begin{array}{l}3.\mathrm{x}+7-\frac{8\mathrm{x}}{3}=\frac{17}{6}-\frac{5\mathrm{x}}{2}\\ \text{Transposing}\frac{5\mathrm{x}}{2}\text{\hspace{0.17em}to L.H.S and 7 to R.H.S,}\text{we get}\\ \Rightarrow \mathrm{x}-\frac{8\mathrm{x}}{3}+\frac{5\mathrm{x}}{2}=\frac{17}{6}-7\\ \text{Taking LCM both the sides, we get}\end{array}$

$\begin{array}{l}\Rightarrow \frac{6\mathrm{x}-16\mathrm{x}+15\mathrm{x}}{6}=\frac{17-42}{6}\\ \Rightarrow \frac{5\mathrm{x}}{6}=\frac{-25}{6}\\ \text{Multiplying both the sides by 6}\\ \Rightarrow \frac{5\mathrm{x}}{6}\times 6=\frac{-25}{6}\times 6\\ \Rightarrow 5\mathrm{x}=-25\\ \text{Dividing both the sides by 5}\\ \Rightarrow \frac{5\mathrm{x}}{5}=\frac{-25}{5}\\ \Rightarrow \mathrm{x}=-5\end{array}$ \begin{array}{l}\mathrm{}\end{array}