Class 8 Maths

# Linear Equations

## Exercise 2.4 Part 1

Question 1: Amina thinks of a number and subtracts 5/2 from it. She multiplies the result by 8. The result now obtained is 3 times the same number she thought of. What is the number?

Solution: Let the number thought by Anamika = a

According to question

(text(Number)-5/2)xx8=3xx\text(Number)

Or, (a-5/2)xx8=3xxa

Or, (a-5/2)xx8=3a

After removing the bracket we get:

8a-5/2xx8=3a

Or, 8a-(40)/(2)=3a

Or, 8a-20=3a

After transposing 3a to LHS we get:

8a-20-3a=0

Or, 8a-3a-20=0

Or, 5a-20=0

After transposing -20 to RHS we get:

5a=20

After dividing both sides by 5 we get:

(5a)/(5)=(20)/(5)

Or, a=4

Thus, the required number = 4 Answer

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 given positive number = a
Therefore, another number which is 5 times of it = 5a

Now, after adding 21 to both of the number,
First number = a + 21
Second number = 5a + 21

According to question, one new number becomes twice of the other new number
Therefore,
Second number = 2 x first number
i.e. 5a + 21 = 2 (a + 21)
⇒ 5a + 21 = 2a + 42

By transposing ‘2a’ to LHS, we get
⇒ 5a + 21 – 2a = 42
Now, after transposing 21 to RHS, we get
⇒ 5a – 2a = 42 – 21
⇒ 3a = 21

After dividing both sides by 3, we get

(3a)/(3)=(21)/(3)

Or, a=7

Therefore, another number 5a = 5 xx 7 = 35
Thus, required numbers are 7 and 35 Answer

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 number at ones place of two digit number = a
According to question, the sum of digits of given two digit number = 9
i.e. Digit at tens place + digit at ones place = 9
Or, Digit at tens place + a = 9

By transposing ‘a’ to RHS, we get
Digit at tens place = 9 – a
Thus, the number = 10(9 – a) + a
After interchange of digit, the number = 10a + (9 – a)

Since, number obtained after interchange of digit is greater than the original number by 27
Therefore, New number – 27 = Original number

Here, we have original number = 10(9 – a) + a
And, new number = 10a + (9 – a)
⇒ 10a + (9 – a) – 27 = 10 (9 – a) + a
⇒ 10a + 9 – a – 27 = 90 – 10a + a
⇒ 10a – a + 9 – 27 = 90 – 9a
⇒ 9a – 18 = 90 – 9a

By transposing 18 to RHS, we get
9a = 90 – 9a + 18
By transposing – 9a to LHS, we get
9a + 9a = 90 + 18
⇒18 a = 108

After dividing both sides by 18, we get

(18a)/(18)=(108)/(18)

Or, a=(108)/(18)=6

Since, digit at tens place = 9 –a
Thus, by substituting the value of a, we get
The number at tens place = 9 – a = 9 – 6 = 3
Thus, number at tens place = 3
And number at ones place = a = 6

Thus, the number = 36 Answer

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 one of the digit , which is at ones place, of a two digit number = a
Therefore, other digit of the two digit number = 3a
Therefore, number = (10 xx 3a) + a = 30a + a = 31a
After interchange, the digit = 10a + 3a = 13a

Now, since sum of the original and resulting number = 88
Therefore, 31a + 13a = 88
⇒ 44 a = 88
Now, after dividing both sides by 44, we get

(44a)/(44)=(88)/(44)

Or, a=2

By substituting the value of ‘a’ in original number we get
Since, original number = 31a = 31 xx 2 = 62
Thus, the number = 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 the present age of Shobo = a
Therefore, Shobo’s mother’s present age = 6a
After five years, the age of Shobo = a + 5
As per question,

Present age of Shobo’s mother ÷ 3 = Age of Shobo after 5 year

Or, (6a)/(3)=a+5

Or, 2a=a+5

By transposing a to LHS we get:

2a-a=5

Or, a=5

Thus, present age of Shobo = 5 year
Since, present age of Shobo’s mother = 6a
Thus, present age of Shobo’s mother = 6 x 5 = 30 year
Therefore, Present age of Shobo = 5 year and present age of Shobo’s mother = 30 year Answer