Exercise 2.4 ,Question: 2

Exercise 2.4

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 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

we get a=7

Therefore, another number 5a = 5 x 7 = 35

Thus, required numbers are 7 and 35 Answer