Write the terminating decimal 15 . 265 in the form of p q , q ≠ 0 and p , q are co-primes. Write the denominators in 2 n 5 m form.

Given, 15 . 265 Let us express the given terminating decimal to p q form by multiplying the numerator and denominator by 1000 as there are three digits after the decimal point. So, 15 . 265 = 15 . 265 × 1000 1 × 1000 = 15265 1000 = 3053 200 Here, 3053 and 200 are co-prime numbers. 2 200 2 100 2 50 5 25 5 We have, 3053 200 = 3053 2 × 2 × 2 × 5 × 5 = 3053 2 3 × 5 2 Therefore, the denominator 200 in 2 n 5 m form is 2 3 × 5 2 and the p q form of the given terminating decimal is 3053 200 .

Write the terminating decimal 0 . 1255 in the form of p q , q ≠ 0 and p , q are co-primes. Write the denominators in 2 n 5 m form.

Given, 0 . 1255 Let us express the given terminating decimal to p q form by multiplying the numerator and denominator by 10000 as there are four digits after the decimal point. So, 0 . 1255 = 0 . 1255 × 10000 1 × 10000 = 1255 10000 = 251 2000 Here, 251 and 2000 are co-prime numbers. 2 2000 2 1000 2 500 2 250 5 125 5 25 5 We have, 251 2000 = 251 2 × 2 × 2 × 2 × 5 × 5 × 5 = 251 2 4 × 5 3 Therefore, the denominator 2000 in 2 n 5 m form is 2 4 × 5 3 and the p q form of the given terminating decimal is 251 2000 .

Write the terminating decimal 0 . 4 in the form of p q , q ≠ 0 and p , q are co-primes. Write the denominators in 2 n 5 m form.

Given, 0 . 4 Let us express the given terminating decimal to p q form by multiplying the numerator and denominator by 10 as there is only one digit after the decimal point. So, 0 . 4 = 0 . 4 × 10 1 × 10 = 4 10 = 2 5 Here, 2 and 5 are co-prime numbers. We have, 2 5 = 2 1 × 5 = 2 2 0 × 5 1 Therefore, the denominator 5 in 2 n 5 m form is 2 0 × 5 1 and the p q form of the given terminating decimal is 2 5 .

Write the terminating decimal 23 . 34 in the form of p q , q ≠ 0 and p , q are co-primes. Write the denominators in 2 n 5 m form.

Given, 23 . 34 Let us express the given terminating decimal to p q form by multiplying the numerator and denominator by 100 as there are two digits after the decimal point. So, 23 . 34 = 23 . 34 × 100 1 × 100 = 2334 100 = 1167 50 Here, 1167 and 50 are co-prime numbers. We have, 1167 50 = 1167 2 × 5 × 5 = 1167 2 1 × 5 2 Therefore, the denominator 50 in 2 n 5 m form is 2 1 × 5 2 and the p q form of the given terminating decimal is 1167 50 .

Write the terminating decimal 1215 . 8 in the form of p q , q ≠ 0 and p , q are co-primes. Write the denominators in 2 n 5 m form.

Given, 1215 . 8 Let us express the given terminating decimal to p q form by multiplying the numerator and denominator by 10 as there is only one digit after the decimal point. We have, 1215 . 8 = 1215 . 8 × 10 1 × 10 = 12158 10 = 6079 5 Here, 6079 and 5 are co-prime numbers. Again, we have, 6079 5 = 6079 1 × 5 = 6079 2 0 × 5 1 Therefore, the denominator 5 in 2 n 5 m form is 2 0 × 5 1 and the p q of the given terminating decimal is 6079 5 .

E M B I B E

Mathematics Class 10>Chapter 1 - Real Numbers>Do This

Telangana Board Solutions for Chapter: Real Numbers, Exercise 9: Do This

Author:Telangana Board

Telangana Board Mathematics Solutions for Exercise - Telangana Board Solutions for Chapter: Real Numbers, Exercise 9: Do This

Attempt the practice questions on Chapter 1: Real Numbers, Exercise 9: Do This with hints and solutions to strengthen your understanding. Mathematics Class 10 solutions are prepared by Experienced Embibe Experts.