# 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