Mathematics · Real Numbers · NCERT Exercises

Exercise 1.1

Complete, independently verified solutions for NCERT Exercise 1.1.

ChapterResourceExercise / Example

Mathematics · Chapter 1 · NCERT Exercises

Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

Question

Question 1

prime factorisation · fundamental theorem of arithmetic

Express each number as a product of its prime factors.

$140$
$156$
$3825$
$5005$
$7429$

Solution

$140=2\times70=2\times2\times35=2^2\times5\times7$
$156=2\times78=2\times2\times39=2^2\times3\times13$
$3825=5\times765=5\times5\times153=5^2\times3\times51=3^2\times5^2\times17$
$5005=5\times1001=5\times7\times143=5\times7\times11\times13$
$7429=17\times437=17\times19\times23$

Question

Question 2

HCF · LCM · prime factorisation

Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF equals the product of the two numbers.

26 and 91
510 and 92
336 and 54

Solution

$26=2\times13,\qquad91=7\times13$
$\operatorname{HCF}=13,\qquad\operatorname{LCM}=2\times7\times13=182$
$13\times182=2366=26\times91$
$510=2\times3\times5\times17,\qquad92=2^2\times23$
$\operatorname{HCF}=2,\qquad\operatorname{LCM}=2^2\times3\times5\times17\times23=23460$
$2\times23460=46920=510\times92$
$336=2^4\times3\times7,\qquad54=2\times3^3$
$\operatorname{HCF}=2\times3=6,\qquad\operatorname{LCM}=2^4\times3^3\times7=3024$
$6\times3024=18144=336\times54$

Question

Question 3

HCF · LCM · prime factorisation

Find the LCM and HCF of the following integers by applying the prime factorisation method.

12, 15 and 21
17, 23 and 29
8, 9 and 25

Solution

$12=2^2\times3,\quad15=3\times5,\quad21=3\times7$
$\operatorname{HCF}=3,\qquad\operatorname{LCM}=2^2\times3\times5\times7=420$
$17=17,\quad23=23,\quad29=29$
The three prime numbers have no common prime factor.
$\operatorname{HCF}=1,\qquad\operatorname{LCM}=17\times23\times29=11339$
$8=2^3,\quad9=3^2,\quad25=5^2$
The three numbers have no common prime factor.
$\operatorname{HCF}=1,\qquad\operatorname{LCM}=2^3\times3^2\times5^2=1800$

Question

Question 4

HCF · LCM

Given that HCF (306, 657) = 9, find LCM (306, 657).

Solution

Use HCF × LCM = product of the two positive integers.

9\times\operatorname{LCM}(306,657)=306\times657

\operatorname{LCM}(306,657)=\frac{306\times657}{9}=34\times657=22338

Final answer

22338

Question

Question 5

prime factorisation · terminal digit

Check whether 6ⁿ can end with the digit 0 for any natural number n.

Solution

6^n=(2\times3)^n=2^n\times3^n

This factorisation has no factor 5, so 6ⁿ is not divisible by 10.

Final answer

6ⁿ cannot end with zero for any natural number n.

Question

Question 6

composite numbers · factorisation

Explain why each expression is a composite number.

$7\times11\times13+13$
$7\times6\times5\times4\times3\times2\times1+5$

Solution

7\times11\times13+13=13(7\times11+1)=13\times78=1014

7\times6\times5\times4\times3\times2\times1+5=5(1008+1)=5\times1009=5045

Each expression has a non-trivial factorisation, so each is composite.

Question

Question 7

LCM · word problem

Sonia takes 18 minutes to complete one round of a circular track, while Ravi takes 12 minutes. If they start together at the same point and move in the same direction, after how many minutes will they meet again at the starting point?

Solution

18=2\times3^2,\qquad12=2^2\times3

\operatorname{LCM}(18,12)=2^2\times3^2=36

Final answer

They meet again at the starting point after 36 minutes.