These Class 8 Maths Extra Questions Chapter 2 Linear Equations in One Variable will help students prepare well for the exams.
Class 8 Maths Chapter 2 Extra Questions Linear Equations in One Variable
Class 8 Maths Linear Equations in One Variable Extra Questions
Question 1.
Solve 17 + 6p = p.
Solution:
17 + 6p = p
6p – p = – 17
5p = – 17
∴ p = \(\frac{-17}{5}\)
Question 2.
Sum of two numbers is 75. If one exceeds the other by 25. Find the numbers?
Solution:
Let the smallest number = x
If one exceeds the other by 25
∴ The other number = x + 25
Sum of the two numbers = x + (x + 25)
75 = x + (x + 25)
75 = 2x + 25
∴ 2x = 75 – 25 = 50
∴ x = \(\frac{50}{2}\) = 25
∴ Smallest number = 25
Question 3.
Find the value of the variable in the equation 2x + 3 = 3x + 2.
Solution:
2x + 3 = 3x + 2
2x – 3x = 2 – 3
– x = – 1
∴ x = 1
Question 4.
Look at the equation given below : 4x + 3 – 2x + 7
What value of x makes the equation TRUE ?
Solution:
4x + 3 = 2x + 7
4x – 2x = 7 – 3
2x = 4
x = \(\frac{4}{2}\)
x = 2
Question 5.
Nancy saved Rs. 800 MORE in March than in February.
The money she saved in March was 5 times that of what she saved in February. How much did she save in February ?
Solution:
Amount saved in February = ₹ x
5x = 800
x = \(\frac{800}{5}\)
x = 160
∴ Amount saved in February = ₹ 160
Question 6.
24 times of a number is equal to 4 times the sum of 4 and double of the same. Find that number.
Solution:
Let the number required = x
24 × x = 4(4 + 2x)
24x = 16 + 8x
24x – 8x = 16
16x = 16
x = \(\frac{16}{16}\)
x = 1
∴ Required number = 1
Linear Equations in One Variable Extra Questions Class 8
Question 1.
The present age of Vinay’s mother is three times the present age of Vinay. After 5 years, sum of their ages will be 70 years. Find their present ages.
Solution:
Let present age of Vinay be ‘x’ years
Then the present age of his mother = 3x years
After 5 years Vinay’s age = (x + 5) years
After 5 years Vinay’s mother age = (3x + 5) years
After 5 years sum of their ages will be 70 years
∴ (x + 5) + (3x + 5) = 70
4x + 10 = 70
4x = 60
x = \(\frac{60}{4}\)
x = 15
∴ Present Vinay’s age x = 15y
Extra Questions of Linear Equations in One Variable Class 8
Question 1.
Divide Rs. 510 among three workers in such a way that each worker will get in the ratio of 2 : 3 : 5. Find the share of each person.
Solution:
Total amount having three persons = ₹ 510
Share of three persons = 2 : 3 : 5
∴ Total share of three persons = 2 + 3 + 5 = 10
∴ Share of 1st person = 2
∴ \(\frac{2}{10}\) × 510 = 2 × 51 = ₹ 102
Share of 2nd person = 3
∴ \(\frac{3}{10}\) × 510 = 3 × 51 = ₹ 153
Share of 3rd person = 5
∴ \(\frac{5}{10}\) × 510 = 5 × 51 = ₹ 255
Question 2.
Solve 3x + \(\frac{5}{2}\) = \(\frac{x}{2}\) + 15.
Solution:
3x + \(\frac{5}{2}\) = \(\frac{x}{2}\) + 15
Linear Equations in One Variable Class 8 Extra Questions
Question 1.
Find the solution of 2x – 3 = 7.
Solution:
Step 1 : Add 3 to both sides.
2x – 3 + 3 = 7 + 3 (The balance is not disturbed)
or 2x = 10
Step 2 : Next divide both sides by 2.
\(\frac{2 x}{2}\) = \(\frac{10}{2}\)
or x = 5 (required solution)
Question 2.
Solve 2y + 9 = 4
Solution:
Transposing 9 to RHS
2y = 4 – 9
or 2y= -5
Dividing both sides by 2,
y = \(\frac{-5}{2}\) (Solution)
To check the answer :
LHS = 2(\(\frac{-5}{2}\)) + 9 = – 5 + 9 = 4 = RHS (as required)
Question 3.
Solve \(\frac{x}{3}\) + \(\frac{5}{2}\) = –\(\frac{3}{2}\)
Solution:
Transposing \(\frac{5}{2}\) to the RHS, we get
\(\frac{x}{3}\) = \(\frac{-3}{2}\) – \(\frac{5}{2}\) = –\(\frac{8}{2}\)
or \(\frac{x}{3}\) = -4
Multiply both sides by 3,
x = – 4 × 3
or x = – 12 (Solution)
Check:
LHS = –\(\frac{12}{3}\) + \(\frac{5}{2}\) = -4 + \(\frac{5}{2}\)
= \(\frac{- 8 + 5}{2}\) = \(\frac{-3}{2}\) = RHS (as required)
Question 4.
Solve\(\frac{15}{4}\) – 7x = 9
Solution:
We have \(\frac{15}{4}\) – 7x = 9
or – 7x = 9 – \(\frac{15}{4}\) (transposing \(\frac{15}{4}\) to RHS)
or – 7x = \(\frac{21}{4}\)
or x = \(\frac{21}{4 \times(-7)}\) (dividing both sides by – 7)
or x = – \(\frac{3 \times 7}{4 \times 7}\)
or x = – \(\frac{3}{4}\) (Solution)
Check : LHS = \(\frac{15}{4}\) – 7(\(\frac{-3}{4}\))
= \(\frac{15}{4}\) + \(\frac{21}{4}\) = \(\frac{36}{4}\)
= 9 = RHS (as required)
Question 5.
What should be added to twice the rational number \(\frac{-7}{3}\) to get \(\frac{3}{7}\) ?
Solution:
Twice the rational number \(\frac{-7}{3}\) is 2 × (\(\frac{-7}{3}\)) = \(\frac{-14}{3}\). Suppose x added to this number gives \(\frac{3}{7}\); i.e.,
x + (\(\frac{-14}{3}\)) = \(\frac{3}{7}\)
or x – \(\frac{14}{3}\) = \(\frac{3}{7}\) or x = \(\frac{3}{7}\) + \(\frac{14}{3}\) (transposing – \(\frac{14}{3}\) to RHS)
= \(\frac{(3 \times 3)+(14 \times 7)}{21}\) = \(\frac{9 + 98}{21}\) = \(\frac{107}{21}\)
Thus \(\frac{107}{21}\) should be added to 2 × (\(\frac{-7}{3}\)) to give \(\frac{3}{7}\).
Question 6.
The perimeter of a rectangle is 13 cm and its width is 2\(\frac{3}{4}\) cm. Find its length.
Solution:
Assume the length of the rectangle to be x cm.
The perimeter of the rectangle
= 2 × (length + width)
= 2 × (x + 2\(\frac{3}{4}\)) = 2(x + \(\frac{11}{4}\))
The perimeter is given to be 13 cm. Therefore,
2(x + \(\frac{11}{4}\)) = 13
or x + \(\frac{11}{4}\) = \(\frac{13}{2}\) (dividing both sides by 2)
or x = \(\frac{13}{2}\) – \(\frac{11}{4}\) = \(\frac{26}{4}\) – \(\frac{11}{4}\) = \(\frac{15}{4}\) = 3\(\frac{3}{4}\)
The length of the rectangle is 3\(\frac{3}{4}\) cm.
Question 7.
The present age of Sahil’s mother is three times the present age of Sahil. After 5 years their ages will add to 66 years. Find their present ages.
Solution:
Let Sahil’s present age be x years.
Sahil | Mother | Sum | |
Present age | x | 3x | |
Age 5 years later | x + 5 | 3x + 5 | 4x +10 |
It is given that this sum is 66 years.
Therefore, 4x + 10 = 66
This equation determines Sahil’s present age which is x years. To solve the equation,
we transpose 10 to RHS,
4x = 66 – 10 or 4x = 56
or x = \(\frac{56}{4}\) = 14 (Solution)
Thus, Sahil’s present age is 14 years and his mother’s age is 42 years.
Question 8.
Bansi has 3 times as many two – rupee coins as he has five – rupee coins. If he has in all a sum of ₹ 77, how many coins of each denomination does he have ?
Solution:
Let,the number of five – rupee coins that Bansi has be x. Then the number of two – rupee coins he has is 3 times x or 3x.
The amount Bansi has :
i) from 5 rupee coins, ₹ 5 × x = ₹ 5x
ii) from 2 rupee coins, ₹ 2 × 3x = ₹ 6x
Hence the total money he has = ₹ 11x
But this is given to be ₹ 77; therefore,
11x = 77 or x = \(\frac{77}{11}\) = 7
Thus, number of five – rupee coins = x = 7
and number of two – rupee coins = 3x = 21 (Solution)
Question 9.
The sum of three consecutive multiples of 11 is 363. Find these multiples.
Solution:
If x is a multiple of 11, the next multiple ix x + 11.
The next to this is x + 11 + 11 or x + 22.
So we can take three consecutive multiples of 11 as x, x + 11 and x + 22.
It is given that the sum of these consecutive multiples of 11 is 363. This will give the following equation :
x – (x + 11) + (x + 22) = 363
or x + x + 11 + x + 22 = 363
or 3x + 33 = 363
or 3x = 363 – 33
or 3x = 330
or x = \(\frac{330}{3}\)
= 110
(or)
Alternatively, we may think of the multiple of 11 immediately before x. This is (x – 11). Therefore, we may take three consecutive multiples of 11 as x – 11, x, x + 11.
In this case we arrive at the equation
(x – 11) + x + (x + 11) = 363
or 3x = 363
or x = \(\frac{363}{3}\) = 121. Therefore,
x = 121, x – 11 = 110, x + 11 = 132
Hence, the tree consecutive multiples are 110, 121, 132.
Hence, the three consecutive multiples are 110, 121, 132.
Question 10.
The difference between two whole numbers is 66. The ratio of the two numbers is 2 : 5. What are the two numbers ?
Solution:
Since the ratio of the two numbers is . 2 : 5, we may take one number to be 2x
and the other to be 5x. (Note that 2x : 5x is same as 2 : 5.)
The difference between the two numbers is (5x – 2x). It is given that the difference is 66. Therefore,
5x – 2x = 66
or 3x = 66
or x = 22
Since the numbers are 2x and 5x, they are 2 × 22 or 44 ; and 5 × 22 or 110, respectively.
The difference between the two numbers is 110 – 44 = 66 as desired.
Question 11.
Deveshi has a total of ₹ 590 as currency notes in the denominations of ₹ 50, ₹ 20 and ₹ 10. The ratio of the number of ₹ 50 notes and ₹ 20 notes is 3 : 5. If she has a total of 25 notes, how many notes of each denomination she has ?
Solution:
Let the number of ₹ 50 notes and ₹ 20 notes be 3x and 5x, respectively.
But she has 25 notes in total.
Therefore, the number of ₹ 10 notes
= 25 – (3x + 5x) = 25 – 8x
The amount she has
from ₹ 50 notes : 3x × 50 = ₹ 150x
from ₹ 20 notes: 5x × 20 = ₹ 100x
from ₹ 10 notes : (25 – 8x) × 10
= ₹ (250 – 80x)
Hence the total money she has
= 150x+ 100x + (250 – 80x)
= ₹ (170x + 250)
But she has ₹ 590.
Therefore,
170x + 250 = 590
or 170x = 590 – 250 = 340
or x = \(\frac{1}{2}\) = 2 .
The number of ₹ 50 notes she has = 3x = 3 × 2 = 6
The number of ₹ 20 notes She has = 5x = 5 × 2 =10
The number of ₹ 10 notes she has
= 25 – 8x
= 25 – (8 × 2) = 25 – 16 = 9
Question 12.
Solve 2x – 3 = x + 2
Solution:
We have 2x = x + 2 + 3
or 2x = x + 5
or 2x – x = x + 5 – x (subtracting x from both sides)
or x = 5 (solution)
Question 13.
Solve 5x + \(\frac{7}{2}\) = \(\frac{3}{2}\)x – 4
Solution:
Multiply both sides of the equation by 2. We get
2 × (5x + \(\frac{7}{2}\)) = 2 × (\(\frac{3}{2}\)x – 14)
(2 × 5x) + (2 × \(\frac{7}{2}\)) = (2 × \(\frac{3}{2}\)x) – (2 × 14)
or 10x + 7 = 3x – 28
or 10x – 3x + 7 = – 28 (transposing + 3x to LHS)
or 7x + 7 = – 28
or 7x = – 28 – 7
or 7x = – 35
or x = \(\frac{-35}{7}\) or x = -5 (solution)
Question 14.
The digits of a two-digit number differ by 3. If the digits are interchanged, and the resulting number is added to the original number, we get 143. What can be the original number ?
Solution:
Take, for example, a two – digit number, say 56. It can be written as 56 = (10 × 5) + 6.
If the digits in 56 are interchanged, we get 65, which can be written as (10 × 6) + 5.
Let us take the two digit number such that the digit in the units place is b. The digit in the tens place differs from b by 3.
Let us take it as b + 3. So the two – digit number is 10 (b + 3) + b = 10b + 30 + b
= 11b + 30.
With interchange of digits, the resulting two-digit number will be
10b + (b + 3) = 11b + 3
If we add these two two-digit numbers, their sum is (11b + 30) + (11b + 3)
= 11b + 11b + 30 + 3 = 22b + 33
It is given that the sum is 143.
Therefore, 22b + 33 = 143
or 22b = 143 – 33
or 22b = 110
or b = \(\frac{110}{222}\)
or b = 5
The units digit is 5 and therefore the tens digit is 5 + 3
which is 8. The number is 85.
Chec : On interchange of digits the number, we get is 58.
The sum of 85 and 58 is 143 as given.
Question 15.
Aijun is twice as old as Shriya. Five years ago his age was three times Shriya’s age. Find their present ages.
Solution:
Let us take Shriya’s present age to be x years.
Then Arjun’s present age would be 2x years.
Shriya’s age five years ago was (x – 5) years.
Arjun’s age five years ago was (2x – 5) years.
It is given that Arjun’s age five years ago was three times Shriya’s age.
Thus, 2x – 5 = 3(x – 5)
or 2x – 5 = 3x – 15
or 15 – 5 = 3x – 2x
or 10 = x
So, Shriya’s present age = x = 10 years.
Therefore, Arjun’s present age
= 2x = 2 × 10 = 20 years.
Question 16.
Solve \(\frac{6 x+1}{3}\) + 1 = \(\frac{x-3}{6}\)
Solution:
Multiplying both sides of the equation
by 6, \(\frac{6(6 x+1)}{3}\) + 6 × 1 = \(\frac{6(x-3)}{6}\)
or 2 (6x + 1) + 6 = x – 3
or 12x + 2 + 6 = x – 3 (opening the brackets)
or 12x + 8 = x – 3
or 12x – x + 8 = -3
or 11x + 8 = – 3 – 8
or 11x = -11
or x = – 1 (required solution)
Checking:
LHS = \(\frac{6(-1)+1}{3}\) + 1 = \(\frac{-6+1}{3}\) + 1
= \(\frac{-5}{3}\) + \(\frac{3}{3}\) = \(\frac{-5+3}{3}\) = \(\frac{-2}{3}\)
RHS = \(\frac{(-1)-3}{6}\) = \(\frac{-4}{6}\) = \(\frac{-2}{3}\)
LHS = RHS. (as required)
Question 17.
Solve 5x – 2(2x – 7) = 2(3x – 1) + \(\frac{7}{2}\)
Solution:
Let us open the brackets,
LHS = 5x – 4x + 14 = x + 14
RHS = 6x – 2 + \(\frac{7}{2}\) = 6x – \(\frac{4}{2}\) + \(\frac{7}{2}\) = 6x + \(\frac{3}{2}\)
The equation is x + 14 = 6x + \(\frac{3}{2}\)
or 14 = 6x – x + \(\frac{3}{2}\)
or 14 = 5x + \(\frac{3}{2}\)
or 14 – \(\frac{3}{2}\) = 5x (transposing \(\frac{3}{2}\))
or \(\frac{28 – 3}{2}\) = 5x
or \(\frac{25}{2}\) = 5x
or x = \(\frac{25}{2}\) × \(\frac{1}{5}\) = \(\frac{5 \times 5}{2 \times 5}\) = \(\frac{5}{2}\)
Therefore, required solution is x = \(\frac{5}{2}\).
Check :
Question 18.
Solve \(\frac{x+1}{2 x+3}\) = \(\frac{3}{8}\)
Solution:
Observe that the equation is not a linear equation, since the expression on its LHS is not linear. But we can put it into the form of a linear equation. We multiply both sides of the equation by (2x + 3),
(\(\frac{x+1}{2 x+3}\)) × (2x + 3) = \(\frac{3}{8}\) × (2x + 3)
Notice that (2x + 3) gets cancelled on the LHS we have then,
x + 1 = \(\frac{3(2 x+3)}{8}\)
We have now a linear equatioh which we know how to solve.
Multiplying both sides by 8
8 (x + 1) = 3 (2x + 3)
or 8x + 8 = 6x + 9
or 8x = 6x + 9 – 8
or 8x = 6x + 1
or 8x – 6x = 1
or 2x = 1 or x = \(\frac{1}{2}\)
The solution is x = \(\frac{1}{2}\).
Check:
Numerator of LHS = \(\frac{1}{2}\) + 1 = \(\frac{1+2}{2}\) = \(\frac{3}{2}\)
Denominator of LHS = 2x + 3
2 × \(\frac{1}{2}\) + 3 = 1 + 3 = 4
LHS = numerator ÷ denominator
= \(\frac{3}{2}\) ÷ 4 \(\frac{3}{2}\) × \(\frac{1}{4}\) = \(\frac{3}{8}\)
LHS = RHS.
Question 19.
Present ages of Anu and Raj are in the ratio 4 : 5. Eight years from now the ratio of their ages will be 5 : 6. Find their present ages.
Solution:
Let the present ages of Anu and Raj be 4x years and 5x years respectively. After eight years, Anu’s age = (4x + 8) years.
After eight years, Raj’s age = (5x + 8) years.
Therefore, the ratio of their ages after eight years = \(\frac{4 x+8}{5 x+8}\)
This is given to be 5 : 6
Therefore, \(\frac{4 x+8}{5 x+8}\) = \(\frac{5}{6}\)
Cross – multiplication gives
6 (4x + 8) = 5 (5x + 8)
or 24x + 48 = 25x + 40
or 24x + 48 – 40 = 25x
or 24x + 8 = 25x
or 8 = 25x – 24x
or 8 = x
Therefore,
Anu’s present age = 4x = 4 × 8 = 32 years
Raj’s present age = 5x = 5 × 8 = 40 years