# Linear Equations in One Variable

## Exercise 2.4 (NCERT)

### Solution of Questions from 1 to 5

Question 1 - Amina thinks of a number and subtracts 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,

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)

â‡’ 5a + 21 = 2a + 42

By transposing '2a' to LHS, we get

â‡’ 5a + 21 2a = 42

Now, after transposing 21 to RHS, we get

â‡’ 5a 2a = 42 21

â‡’ 3a = 21

After dividing both sides by 3, we get

Therefore, another number 5a = 5 x 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)

â‡’ 10a + (9 a) 27 = 10 (9 a) + a

â‡’ 10a + 9 a 27 = 90 10a + a

â‡’ 10a a + 9 27 = 90 9a

â‡’ 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

â‡’18 a = 108

After dividing both sides by 18, we get

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 x 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

â‡’ 44 a = 88

Now, after dividing both sides by 44, we get

By substituting the value of 'a' in original number we get

Since, original number = 31a = 31 x 2 = 62

Thus, the number = 62 Answer

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,

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

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