CBSE Class 8

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.5 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.5

Solve the following linear equations.

Question 1.
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Solution:
\(\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4}\)
Transposing \(-\frac{1}{5}\) to R.H.S. and \(\frac{\mathrm{x}}{3}\) to L.H.S
\(\frac{x}{2}-\frac{x}{3}=\frac{1}{4}+\frac{1}{5}\)
\(\frac{3 x-2 x}{6}=\frac{5+4}{20}\)
\(\frac{x}{6}=\frac{9}{20}\)
By cross-multiplication, we get
20(x) = 9 × 6
20x = 54
Dividing both sides by 20
\(\frac{20 \mathrm{x}}{20}=\frac{54}{20}\)
x = \(\frac{27}{10}\)

Question 2.
\(\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21\)
Solution:
\(\frac{\mathrm{n}}{2}-\frac{3 n}{4}+\frac{5 \mathrm{n}}{6}=21\)
L.C.M of 2, 4 and 6 = 12
Multiplying each term by 12, we get
\(12 \times \frac{\mathrm{n}}{2}-12 \times \frac{3 \mathrm{n}}{4}+12 \times \frac{5 \mathrm{n}}{6}=21 \times 12\)
6n – 9n + 10n = 252
7n = 252
Dividing both sides by 7
\(\frac{7 \mathrm{n}}{7}=\frac{252}{7}\)
n = 36

Question 3.
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Solution:
\(x+7-\frac{8 x}{3}=\frac{17}{6}-\frac{5 x}{2}\)
Transposing 7 to R.H.S. and \(\frac{-5 x}{2}\) to L.H.S
\(x-\frac{8 x}{3}+\frac{5 x}{2}=\frac{17}{6}-7\)
L.C.M of 3, 2 and 6 = 6
Multiplying each term by 6, we get
\(6 \times x-6 \times \frac{8 x}{3}+6 \times \frac{5 x}{2}=6 \times \frac{17}{6}-6 \times 7\)
6x – 16x + 15x = 17 – 42
5x = -25
Dividing both sides by 5
\(\frac{5 x}{5}=-\frac{25}{5}\)
x = -5

Question 4.
\(\frac{x-5}{3}=\frac{x-3}{5}\)
Solution:
\(\frac{x-5}{3}=\frac{x-3}{5}\)
By cross-multiplication we get
5(x – 5) = 3(x – 3)
5x – 25 = 3x – 9
Transposing -25 to R.H.S and 3x to L.H.S.
5x – 3x = 25 – 9
2x = 16
Dividing both sides by 2
\(\frac{2 \mathrm{x}}{2}\) = \(\frac{16}{2}\)
x = 8

Question 5.
\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}\)
Solution:
\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}=\frac{2}{3}-\mathrm{t}\)
Transposing -t to L.H.S., we get
\(\frac{3 \mathrm{t}-2}{4}-\frac{2 \mathrm{t}+3}{3}+\mathrm{t}=\frac{2}{3}\)
L.C.M of 4 and 3 is 12
Multiplying each term by 12
\(12 \frac{(3 \mathrm{t}-2)}{4}-12 \frac{(2 \mathrm{t}+3)}{3}+12 \mathrm{t}=12 \times \frac{2}{3}\)
3(3t – 2) – 4(2t + 3) + 12t = 8
9t – 6 – 8t – 12 + 12t = 8
13t – 18 = 8
Transposing -18 to R.H.S., we get
13t = 8 + 18
13t = 26
Dividing both sides by 13
\(\frac{13 t}{13}=\frac{26}{13}\)
t = 2

Question 6.
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Solution:
\(m-\frac{m-1}{2}=1-\frac{m-2}{3}\)
Transposing \(\left(\frac{m-2}{3}\right)\) to L.H.S. we get
\(\mathrm{m}-\frac{\mathrm{m}-1}{2}+\frac{\mathrm{m}-2}{3}=1\)
L.C.M of 2 and 3 is 6
Multiplying each term by 6, we get
6m – \(\frac{6(m-1)}{2}+\frac{6(m-2)}{3}\) = 1 × 6
6m – 3(m – 1) + 2(m – 2) = 6
6m – 3m + 3 + 2m – 4 = 6
5m – 1 = 6
Transposing -1 to the R.H.S., we get
5m = 6 + 1
5m = 7
Dividing both sides by 5, we get
\(\frac{5 m}{5}=\frac{7}{5}\)
m = \(\frac{7}{5}\)

Simplify and solve the following linear equations.

Question 7.
3(t – 3) = 5 (2t + 1)
Solution:
3(t – 3) = 5 (2t + 1)
Opening the brackets, we get
3t – 9 = 10t + 5
Transposing -9 to R.H.S. and 10t to L.H.S.
3t – 10t = 5 + 9
-7t = 14
Dividing both sides by -7, we get
\(\frac{-7 t}{-7}=\frac{14}{-7}\)
t = -2

Question 8.
15(y – 4) – 2(y – 9) + 5 (y + 6) = 0
Solution:
15(y – 4) – 2(y – 9) + 5(y + 6) = 0
Opening the brackets, we get
15y – 60 – 2y + 18 + 5y + 30 = 0
15y – 2y + 5y – 60 + 18 + 30 = 0
18y – 12 = 0
Transposing -12 to R.H.S.
18y = 12
Dividing both sides by 18, we get.
\(\frac{18 y}{18}=\frac{12}{18}\)
y = \(\frac{2}{3}\)

Question 9.
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Solution:
3(5z – 7) – 2(9z – 11) = 4(8z – 13) – 17
Opening the brackets, we get
15z – 21 – 18z + 22 = 32z – 52 – 17
15z – 18z + 22 – 21 = 32z – 69
-3z + 1 = 32z – 69
Transposing 1 to R.H.S. and 32z to L.H.S.
-3z – 32z = – 69 – 1
-35z = -70
Dividing both sides by -35, we get
\(\frac{-35 z}{-35}=\frac{-70}{-35}\)
z = 2

Question 10.
0.25(4f – 3) = 0.05(10f – 9)
Solution:
0.25(4f – 3) = 0.05(10f – 9)
Opening the brackets, we get
0.25(4f) – 0.25 × 3 = 0.05 × 10f – 0.05 × 9
f – 0.75 = 0.5f – 0.45
f – 0.5f = 0.75 – 0.45
0.5f = 0.3
Dividing both sides by 0.5, we get
\(\frac{0.5 \mathrm{f}}{0.5}=\frac{0.3}{0.5}\)
f = \(\frac{3}{5}\) = 0.6
f = 0.6

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.4 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.4

Question 1.
Amina thinks of numbers and subtracts \(\frac{5}{2}\) from it. She multiplies the results by 8. The result now obtained is 3 times the same number she thought of. What is the number?
Solution:
Let the number be x
By subtracting \(\frac{5}{2}\), we get x – \(\frac{5}{2}\)
According to the given question
\(8\left(x-\frac{5}{2}\right)=3 x\)
8x – \(\frac{8 \times 5}{2}\) = 3x
8x – 20 = 3x
By transposing 3x to L.H.S. and -20 to R.H.S., we get
8x – 3x = 20
5x = 20
Dividing both sides by 5, we get
\(\frac{5 \mathrm{x}}{5}=\frac{20}{5}\)
x = 4
∴ The required number is 4.

Question 2.
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?
Solution:
Let the number be x
The other positive number is 5x
on adding 21 to both numbers, we get (x + 21) and (5x + 21)
According to the question, we get
2(x + 21) = 5x + 21
2x + 42 = 5x + 21
Transposing 42 to R.H.S and 5x to L.H.S., we get
2x – 5x = 21 – 42
-3x = -21
Dividing both sides by -3, we get
\(\frac{-3 x}{-3}=\frac{-21}{-3}\)
x = 7
the other number 5x = 5 × 7 = 35
Thus, the required numbers are 7 and 35.

Question 3.
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.
Solution:
Let the units digit be ‘x’.
the tens digits = 9 – x(sum of the digits is 9)
The original two digit number = 10(9 – x) + x
= 90 – 10x + x
= 90 – 9x
On interchanging the digits,
the new number = 10x + 9 – x = 9x + 9
According to the given question, we get
New number = Original number + 27
9x + 9 = 90 – 9x + 27
9x + 9 = 117 – 9x
Transposing 9 to R.H.S. and -9x to L.H.S., we get
9x + 9x = 117 – 9
18x = 108
Dividing both sides by 18, we get
\(\frac{18 \mathrm{x}}{18}=\frac{108}{18}\)
x = 6
∴ The original number = 90 – 9x = 90 – 54 = 36.

Question 4.
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?
Solution:
Let the digit in the unit place be ‘x’
Then, the digit at tens place = 3x
The number = 10(3x) + x
= 30x + x
= 31x
On interchanging the digits,
The new number = 10x + 3x = 13x
According to the question
31x + 13x = 88
44x = 88
Dividing both sides by 44
\(\frac{44 x}{44}=\frac{88}{44}\)
x = 2
∴ The number = 31x = 31 × 2 = 62.

Question 5.
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?
Solution:
Let Shobos present age be ‘x’ years
Mother’s present age = 6x years
After 5 years
Shobos age = (x + 5) years
Shobos mothers age = (6x + 5) years
According to the question, we get
\(\frac{1}{3}\) (mothers present age) = Shobos age after 5 years
\(\frac{1}{3}\) × 6x = x + 5
2x = x + 5
Transposing x to LHS
2x – x = 5
x = 5
∴ Shobo’s present age = 5 years
Mothers present age = (6 × 5) = 30 years

Question 6.
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 of ₹ 100 per metre it will cost the village panchayat ₹ 75000 to fence the plot. What are the dimensions of the plot?
Solution:
Let the length of the rectangular plot be 11x metres
and the breadth of the rectangular plot be 4x metres
Perimeter of the plot = 2(l + b)
= 2(11x + 4x)
= 2 × 15x
= 30x
Cost of fencing = ₹ 75000
100 × 30x = 75000
3000x = 75000
Dividing both sides by 3000, we get
\(\frac{3000 \mathrm{x}}{3000}=\frac{75000}{3000}\)
x = 25
Length of the plot = 11 × 25 = 275 metres
Breadth of the plot = 4 × 25 = 100 metres

Question 7.
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 metres 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 of ₹ 36,600. How much trouser material did he buy?
Solution:
Let the length of cloth for shirts be ‘3x’ metres
and the length of cloth for trousers be ‘2x’ metres
Cost of shirts cloth = 3x × 50 = ₹ 150x
Cost of trouser cloth = 2x × 90 = ₹ 180x
S.P of shirts cloth at 12% profit
= ₹ \(\frac{112}{100}\) × 150x
= ₹ 168x
S.P. of trousers cloth at 10% profit
= \(\frac{110}{100}\) × 180x
= ₹ 198x
Total S.P = ₹ 36,600
168x + 198x = ₹ 36,600
366x = 36600
Dividing both sides by 366, we get
\(\frac{366 \mathrm{x}}{366}=\frac{36600}{366}\)
x = 100
Material bought for trousers (2 × 100) = 200 metres

Question 8.
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.
Solution:
Let the total number of deer be ‘x’
Number of deer, grazing in the field = \(\frac{\mathrm{x}}{2}\)
Number of deer playing near by = \(\frac{3}{4}\left(x-\frac{x}{2}\right)=\frac{3 x}{4}-\frac{3 x}{8}=\frac{6 x-3 x}{8}=\frac{3 x}{8}\)
Number of deer drinking water = 9
\(\frac{x}{2}+\frac{3 x}{8}\) + 9 = x
Transposing 9 to RHS and x to LHS we get
\(\frac{x}{2}+\frac{3 x}{8}-x=-9\)
\(\frac{4 x+3 x-8 x}{8}=-9\)
\(\frac{-x}{8}=-9\)
Multiplying both sides by -8
\(\frac{-\mathrm{x}}{8}\) × (-8) = -9 × (-8)
x = 72
∴ Number of deer in herd = 72

Question 9.
A grandfather is ten times older than his granddaughter. He is also 54 years older than her. Find their present ages.
Solution:
Let the present age of granddaughter be ‘x’ years
Present age of grandfather = 10x years
According to the given question, we get
10x – x = 54
9x = 54
Dividing both sides by 9, we get
\(\frac{9 \mathrm{x}}{9}=\frac{54}{9}\)
x = 6
Present age of granddaughter = 6 years.
Present age of grandfather = 10 × 6 = 60 years.

Question 10.
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.
Solution:
Let the present age of son be ‘x’ years
Present age of Aman = 3x
Ten years ago Son’s age = (x – 10) years
Aman’s age = (3x – 10) years
According to the given question, we get
3x – 10 = 5(x – 10)
3x – 10 = 5x – 50
Transposing -10 to R.H.S. and 5x to L.H.S
3x – 5x = 10 – 50
-2x = -40
Dividing both sides by -2
\(\frac{-2 x}{2}=\frac{-40}{-2}\)
x = 20
∴ Sons present age = 20 years
Aman’s present age = (3 × 20) = 60 years.

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.3 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.3

Solve the following equations and check your results.

Question 1.
3x = 2x + 18
Solution:
3x = 2x + 18
Transposing 2x from RHS to L.H.S, we get
3x – 2x = 18
x = 18

Check:
Put x = 18 in L.H.S. and R.H.S. of the equation.
L.H.S. = 3x = 3 × 18 = 54
R.H.S. = 2x + 18 = 2(18) + 18 = 36 +18 = 54
∴ L.H.S. = R.H.S

Question 2.
5t – 3 = 3t – 5
Solution:
5t – 3 = 3t – 5
Transposing (-3) to R.H.S. and 3t to L.H.S., we get
5t – 3t = -5 + 3
2t = -2
Dividing both sides by 2, we get
\(\frac{2 t}{2}=\frac{-2}{2}\)
t = -1

Check:
Put t = -1 in L.H.S. and RHS of the equation
L.H.S. = 5t – 3 = 5(-1) – 3 = -5 – 3 = -8
R.H.S. = 3t – 5 = 3(-1) – 5 = -3 – 5 = -8
Hence, L.H.S. = R.H.S.

Question 3.
5x + 9 = 5 + 3x
Solution:
5x + 9 = 5 + 3x
Transposing 9 to R.H.S. and 3x to L.H.S.
5x – 3x = 5 – 9
2x = -4
Dividing both sides by 2, we get
\(\frac{2 x}{2}=\frac{-4}{2}\)
x = -2

Check:
Put x = -2 in L.H.S. and R.H.S. of the equation
L.H.S. = 5x + 9
= 5 (-2) + 9
= -10 + 9
= -1
R.H.S. = 5 + 3x
= 5 + 3(-2)
= 5 – 6 = -1
Hence, L.H.S. = R.H.S.

Question 4.
4z + 3 = 6 + 2z
Solution:
4z + 3 = 6 + 2z
Transposing 3 to R.H.S. and 2z to LHS
4z – 2z = 6 – 3
2z = 3
Dividing both sides by 2, we get
\(\frac{2 z}{2}=\frac{3}{2}\)
z = \(\frac{3}{2}\)

Check:
Put z = \(\frac{3}{2}\) in L.H.S. and R.H.S of the equation
L.H.S. = 4z + 3
= 4\(\left(\frac{3}{2}\right)\) + 3
= 2(3) + 3
= 6 + 3
= 9
R.H.S = 6 + 2z
= 6 + 2\(\left(\frac{3}{2}\right)\)
= 6 + 3
= 9
Hence, L.H.S. = R.H.S.

Question 5.
2x – 1 = 14 – x
Solution:
2x – 1 = 14 – x
Transposing -1 to RHS and -x to L.H.S.
2x + x = 14 + 1
3x = 15
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{15}{3}\)
x = 5

Check:
Put x = 5 in LHS and RHS of the equation
LHS = 2x – 1
= 2(5) – 1
= 10 – 1
= 9
R.H.S = 14 – x
= 14 – 5
= 9
Hence, L.H.S. = R.H.S.

Question 6.
8x + 4 = 3(x – 1) + 7
Solution:
8x + 4 = 3(x – 1) + 7
8x + 4 = 3x – 3 + 7
8x + 4 = 3x + 4
Transposing 4 to R.H.S. and 3x to LHS, we get
8x – 3x = 4 – 4
5x = 0
Dividing both sides by 5
\(\frac{5 x}{5}=\frac{0}{5}\)
x = 0

Check:
Put x = 0 in L.H.S. and R.H.S. of the equation
L.H.S. = 8x + 4
= 8(0) + 4
= 4
R.H.S = 3(x – 1) + 7
= 3(0 – 1) + 7
= -3 + 7
= 4
Hence, L.H.S. = R.H.S.

Question 7.
x = \(\frac{4}{5}\) (x + 10)
Solution:
x = \(\frac{4}{5}\) (x + 10)
Multiplying both sides by 5, we get
5x = 5 × \(\frac{4}{5}\) (x + 10)
5x = 4(x + 10)
5x = 4x + 40
Transposing 4x to L.H.S.
5x – 4x = 40
x = 40

Check:
Put x = 40 in L.H.S. and R.H.S. of the equation
L.H.S. = x = 40
R.H.S. = \(\frac{4}{5}\) (x + 10)
= \(\frac{4}{5}\) (40 + 10)
= \(\frac{4}{5}\) x 50
= 4 × 10
= 40
Hence, L.H.S. = R.H.S

Question 8.
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
Solution:
\(\frac{2 x}{3}+1=\frac{7 x}{15}+3\)
Transposing 1 to R.H.S. and \(\frac{7 \mathrm{x}}{15}\) to L.H.S.
\(\frac{2 x}{3}-\frac{7 x}{15}=3-1\)
\(\frac{10 x-7 x}{15}\) = 2
\(\frac{3 x}{15}\) = 2
Multiplying both sides by 15, we get
\(\frac{3 x}{15}\) × 15 = 2 × 15
3x = 30
Dividing both sides by 3
\(\frac{3 x}{3}=\frac{30}{3}\)
x = 10

Check:
Put x = 10 in L.H.S. and R.H.S. of the equation

Hence, L.H.S. = R.H.S.

Question 9.
\(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Solution:
\(2 y+\frac{5}{3}=\frac{26}{3}-y\)
Transposing \(\frac{5}{3}\) to R.H.S. and -y to L.H.S.
2y + y = \(\frac{26}{3}-\frac{5}{3}\)
3y = \(\frac{26-5}{3}\)
3y = \(\frac{21}{3}\) = 7
Divide both sides by 3, we get
\(\frac{3 y}{3}=\frac{7}{3}\)
y = \(\frac{7}{3}\)

Check:
Put y = \(\frac{7}{3}\) in .L.H.S. and R.H.S of the equation

Hence, L.H.S. = R.H.S

Question 10.
3m = 5m – \(\frac{8}{5}\)
Solution:
3m = 5m – \(\frac{8}{5}\)
Transposing 5m to L.H.S. we get
3m – 5m = \(\frac{-8}{5}\)
-2m = \(\frac{-8}{5}\)
Dividing both sides by -2, we get
\(\frac{-2 m}{-2}=\frac{-8}{5} \div(-2)\)
m = \(\frac{8}{10}=\frac{4}{5}\)

Check:
Put m = \(\frac{4}{5}\) in L.H.S. and R.H.S. of the equation

Hence, L.H.S = R.H.S

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.2 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.2

Question 1.
If you subtract \(\frac{1}{2}\) from a number and multiply the result by \(\frac{1}{2}\), you get \(\frac{1}{8}\). What is the number.
Solution:
Let the number be ‘x’
According to the given condition, we get
\(\left(x-\frac{1}{2}\right) \frac{1}{2}=\frac{1}{8}\)
Multiplying both sides by 2
\(\mathrm{x}-\frac{1}{2}=\frac{1}{8} \times 2=\frac{1}{4}\)
Transposing \(\left(-\frac{1}{2}\right)\) to R.H.S. we get
x = \(\frac{1}{4}+\frac{1}{2}\)
x = \(\frac{1+2}{4}=\frac{3}{4}\)
∴ The required number is \(\frac{3}{4}\).

Question 2.
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 the breadth of the pool?
Solution:
Let the breadth of the pool be x m
∴ Length of the pool = (2 + 2x) m = (2x + 2) m
Perimeter of the pool = 154 m
2(2x + 2 + x) = 154 m
[∵ Perimeter of the rectangle = 2(l + b)]
2(3x + 2) = 154
6x + 4 = 154
Transposing 4 to the R.H.S.
6x = 154 – 4
6x = 150
Dividing both sides by 6,
\(\frac{6 x}{6}=\frac{150}{6}\)
∴ x = 25
∴ Breadth of the pool = 25m
Length of the pool = 2(25) + 2 = 50 + 2 = 52 m

Question 3.
The base of an isosceles triangle is \(\frac{4}{3}\) cm and the perimeter of the triangle is 4\(\frac{2}{15}\) cm. What is the length of the remaining equal sides?
Solution:
Let the length of the equal sides of a triangle be ‘x’
Base of an isosceles triangle = \(\frac{4}{3}\) cm
Perimeter of the triangle = 4\(\frac{2}{15}\) cm

∴ Length of the equal sides is 1\(\frac{2}{5}\) cm.

Question 4.
Sum of two numbers is 95. If one exceeds the other by 15, find the numbers.
Solution:
Let the smaller number be x
then the greater number = x + 15
Sum of two numbers = 95
x + (x + 15) = 95
2x + 15 = 95
Transposing 15 to R.H.S, we get
2x = 95 – 15 = 80
Dividing both sides by 2
x = \(\frac{80}{2}\) = 40
∴ The smaller number = 40
The greater number = (40 + 15) = 55

Question 5.
Two numbers are in the ratio 5 : 3. If they differ by 18, what are the numbers?
Solution:
Let the two number be 5x and 3x
Difference of two numbers = 18
5x – 3x = 18
2x = 18
Dividing both sides by 2, we get
\(\frac{2 \mathrm{x}}{2}=\frac{18}{2}\)
∴ x = 9
The two numbers are (5 × 9) and (3 × 9) i.e. 45 and 27.
Hence, the required numbers are 45 and 27.

Question 6.
Three consecutive integers add up to 51. What are these integers?
Solution:
Let the three consecutive integers be x, x + 1 and x + 2
Sum of three integers = 51
x + x + 1 + x + 2 = 51
3x + 3 = 51
Transposing 3 to RHS, we get
3x = 51 – 3 = 48
Dividing both sides by 3, we get
\(\frac{3 \mathrm{x}}{3}=\frac{48}{3}\)
x = 16
Now x = 16,
x + 1 = 16 + 1 = 17
x + 2 = 16 + 2 = 18
∴ The required three consecutive integers are 16, 17 and 18.

Question 7.
The sum of three consecutive multiple of 8 is 888. Find the multiples.
Solution:
Let the three consecutive multiples of 8 be x, x + 8, x + 16
Sum of three consecutive multiples = 888
x + x + 8 + x + 16 = 888
3x + 24 = 888
Transposing 24 to R.H.S., we get
3x = 888 – 24 = 864
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{864}{3}\)
x = 288
x + 8 = 288 + 8 = 296
x + 16 = 288 + 16 = 304
∴ The required multiples of 8 are 288, 296 and 304.

Question 8.
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.
Solution:
Let the three consecutive integers be x, x + 1 and x + 2
According to the given condition
2(x) + 3(x + 1) + 4(x + 2) = 74
2x + 3x + 3 + 4x + 8 = 74
9x + 11 = 74
Transposing 11 to R.H.S., we get
9x = 74 – 11
9x = 63
Dividing both sides by 9, we get
\(\frac{9 x}{9}=\frac{63}{9}\)
x = 7
x + 1 = 7 + 1 = 8
x + 2 = 7 + 2 = 9
∴ The consecutive integers are 7, 8 and 9.

Question 9.
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?
Solution:
Let the present age of Rahul be ‘5x’
and the present age of Harron be 7x;
4 years later
Rahul’s age = (5x + 4) years
Harron’s age = (7x + 4) years
Sum of their ages = 56 years
5x + 4 + 7x + 4 = 56
12x + 8 = 56
Transposing 8 to R.H.S. we get
12x = 56 – 8 = 48
Dividing both sides by 12
\(\frac{12 \mathrm{x}}{12}=\frac{48}{12}\)
x = 4
Present age of Rahul = 5 × 4 = 20 years
Present age of Haroon = 7 × 4 = 28 years

Question 10.
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.
Solution:
Let the number of boys in the class be 7x
and the number of girls in the class be 5x
As number of boys is 8 more than the number of girls.
7x = 5x + 8
Transposing 5x to L.H.S.
7x – 5x = 8
2x = 8
Dividing both sides by 2
\(\frac{2 \mathrm{x}}{2}=\frac{8}{2}\)
x = 4
∴ Number of boys = 7 × 4 = 28
Number of girls = 5 × 4 = 20
Total class strength = 28 + 20 = 48 students.

Question 11.
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?
Solution:
Let Baichung’s present age be x years
Baichung father’s age = (x + 29) years
Baichung grandfather’s age = (x + 29 +26) years = (x + 55) years
Sum of the ages of all the three =135 years
x + x + 29 + x + 55 = 135
3x + 84 = 135
Transposing 84 to RHS, we have
3x = 135 – 84 = 51
Dividing both sides by 3, we get
\(\frac{3 x}{3}=\frac{51}{3}\)
x = 17
Baichung’s age =17 years
Baichung fathers age = 17 + 29 = 46 years
Baichung grandfather’s age = 17 + 55 = 72 years.

Question 12.
Fifteen years from now, Ravi’s age will be four times his present age. What is Ravi’s present age?
Solution:
Let Ravi’s present age be ‘x’ years
4 times his present age = 4x years
15 years from now, Ravi’s age = (x + 15) years
According to the given condition
x + 15 = 4x
Transposing 15 and 4x, we get
-4x + x = -15
-3x = -15
Dividing both sides by (-3) we get
\(\frac{-3 x}{-3}=\frac{-15}{-3}\)
x = 5
∴ Ravi’s present age = 5 years.

Question 13.
A rational number is such that when you multiply it by \(\frac{5}{2}\) and add \(\frac{2}{3}\) to the product, you get \(\frac{7}{12}\). What is the number?
Solution:
Let the required rational number be ‘x’
According to the given condition

x = \(\frac{-1}{2}\)
∴ The rational number is \(\frac{-1}{2}\).

Question 14.
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?
Solution:
Let the number of ₹ 100 notes be 2x
the number of ₹ 50 notes be 3x
and the number of ₹ 10 notes be 5x
Value of ₹ 100 notes = 2x × 100 = ₹ 200x
Value of ₹ 50 notes = 3x × 50 = ₹ 150x
Value of ₹ 10 notes = 5x × 10 = ₹ 50x
According to the given condition,
₹ 200x + ₹ 150x + ₹ 50x = ₹ 4,00000
400x = 4,00,000
Dividing both sides by 400, we get
\(\frac{400 \mathrm{x}}{400}=\frac{4,00,000}{400}\)
x = 1000
∴ Number of ₹ 100 notes = 2 × 1000 = 2000
Number of ₹ 50 notes = 3 × 1000 = 3000
Numbers of ₹ 10 notes = 5 × 1000 = 5000

Question 15.
I have a total of ₹ 300 in coins of denomination ₹ 1, ₹ 2 and ₹ 5. The number of ₹ 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?
Solution:
Let the number of ₹ 5 coins be x
then number of ₹ 2 coins = 3x
and the number of coins of ₹ 1 = 160 – (x + 3x) = 160 – 4x
Now Value of ₹ 5 coins = ₹ 5 × x = 5x
Value of ₹ 2 coins = ₹ 2 × 3x = 6x
Value of ₹ 1 coins = ₹ 1 × (160 – 4x) = (160 – 4x)
According to the given condition,
160 – 4x + 5x + 6x = 300
160 + 7x = 300
Transposing 160 to the R.H.S.
7x = 300 – 160
7x = 140
Dividing both sides by 7
\(\frac{7 \mathrm{x}}{7}=\frac{140}{7}\)
x = 20
∴ Number of ₹ 5 coins = 20
Number of ₹ 2 coins = (3 × 20) = 60
Number of ₹ 1 coin = 160 – (4 × 20) = 80

Question 16.
The organisers of an essay competition decide that a winner in 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 ₹ 3000. Find the number of winners, if the total number of participants is 63.
Solution:
Let the number of winners be ‘x’
The total number of participants = 63
The number of non-winners = 63 – x
Prize money given to winners = ₹ 100 × x
Prize money given to non-winner participants = ₹ 25(63 – x)
= ₹ 25 × 63 – ₹ 25x
= ₹ 1575 – ₹ 25x
According to the condition given in the question,
100x + 1575 – 25x = 3000
75x + 1575 = 3000
Transposing 1575 to R.H.S. we get
75x = 3000 – 1575
75x = 1425
Dividing both sides by 75
\(\frac{75 \mathrm{x}}{75}=\frac{1425}{75}\)
x = 19
∴ The number of winners = 19.

These NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Ex 2.1 Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 2 Linear Equations in One Variable Exercise 2.1

Solve the following equations.

Question 1.
x – 2 = 7
Solution:
x – 2 = 7
Transposing (-2) to R.H.S., we get
x = 7 + 2
∴ x = 9

Question 2.
y + 3 = 10
Solution:
y + 3 = 10
Transposing 3 to R.H.S., we get
y = 10 – 3
∴ y = 7

Question 3.
6 = z + 2
Solution:
6 = z + 2
Transposing 2 to L.H.S., we get
6 – 2 = z
4 = z
∴ z = 4

Question 4.
\(\frac{3}{7}+x=\frac{17}{7}\)
Solution:
\(\frac{3}{7}+x=\frac{17}{7}\)
Transposing \(\frac{3}{7}\) to R.H.S., we get
x = \(\frac{17}{7}-\frac{3}{7}=\frac{17-3}{7}=\frac{14}{7}=2\)
∴ x = 2

Question 5.
6x = 12
Solution:
6x = 12
Divided by 6 on both sides, we get
\(\frac{6 x}{6}=\frac{12}{6}\)
∴ x = 2

Question 6.
\(\frac{t}{5}=10\)
Solution:
\(\frac{\mathrm{t}}{5}\) = 10
Multiplying both sides by 5, we get
\(\frac{\mathrm{t}}{5}\) × 5 = 10 × 5
∴ t = 50

Question 7.
\(\frac{2 x}{3}=18\)
Solution:
\(\frac{2 x}{3}=18\)
Multiplying both sides, by 3, we get
\(\frac{2 x}{3}\) × 3 = 18 × 3
2x = 18 × 3
x = \(\frac{18 \times 3}{2}\) = 9 × 3 = 27
∴ x = 27

Question 8.
1.6 = \(\frac{\mathrm{y}}{1.5}\)
Solution:
1.6 = \(\frac{\mathrm{y}}{1.5}\)
Multiplying both sides by 1.5, we get
1.6 × 1.5 = \(\frac{\mathrm{y}}{1.5}\) × 1.5
2.4 = y
∴ y = 2.4

Question 9.
7x – 9 = 16
Solution:
7x – 9 = 16
Transposing (-9) to R.H.S., we get
7x = 16 + 9
Dividing both sides by 7, we get
∴ x = \(\frac{25}{7}\)

Question 10.
14y – 8 = 13
Solution:
14y – 8 = 13
Transposing (-8) to R.H.S., we get
14y = 13 + 8
14y = 21
Dividing both sides by 14 we get,
\(\frac{14 y}{14}\) = \(\frac{21}{14}\)
∴ y = \(\frac{3}{2}\)

Question 11.
17 + 6p = 9
Solution:
17 + 6P = 9
Transposing 17 to RHS, we get
6P = 9 – 17
6P = -8
Dividing both sides by 6, we have
\(\frac{6 P}{6}=\frac{-8}{6}\)
∴ P = \(\frac{-4}{3}\)

Question 12.
\(\frac{x}{3}+1=\frac{7}{15}\)
Solution:
\(\frac{x}{3}+1=\frac{7}{15}\)
Transposing 1 to R.H.S, we get
\(\frac{x}{3}=\frac{7}{15}-1\)
\(\frac{x}{3}=\frac{7-15}{15}\)
\(\frac{x}{3}=\frac{-8}{15}\)
Multiplying both sides by 3, we have
\(\frac{x}{3} \times 3=\frac{-8}{15} \times 3\)
∴ x = \(\frac{-8}{5}\)

These NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers InText Questions and Answers are prepared by our highly skilled subject experts.

NCERT Solutions for Class 8 Maths Chapter 1 Rational Numbers InText Questions

Try These (Page No. 4)

Question 1.
Fill in the blanks in the following table.

Solution:
Using the closure property over addition, subtraction, multiplication, and division for rational numbers, integers, whole numbers, and natural numbers, we have:

Try These (Page No. 6)

Question 2.
Complete the following table:

Solution:

Try These (Page No. 9)

Question 3.
Complete the following table:

Solution:

Try These (Page No. 13)

Question 4.
Find using distributivity.

Solution:

Try These (Page No. 17)

Question 5.
Write the rational number for each point labelled with a letter.

Solution:
(i) Here, the rational number for the point A is \(\frac{1}{5}\).
The rational number for the point B is \(\frac{4}{5}\).
The rational number for the point C is \(\frac{5}{5}\).
The rational number for the point D is \(\frac{8}{5}\).
The rational number for the point E is \(\frac{9}{5}\).

(ii) The rational number for:
The point F is \(\frac{-2}{6}\) or \(\frac{-1}{3}\).
The point G is \(\frac{-5}{6}\).
The point H is \(\frac{-7}{6}\).
The point I is \(\frac{-8}{6}\) or \(\frac{-4}{3}\).
The point J is \(\frac{-11}{6}\).