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Question 1:

Express each number as a product of its prime factors: (i) 140 (ii) 156 (iii) 3825 (iv) 5005 (v) 7429

Solution 1:

$(i) 140=2\times 2\times 5\times 7$ $= 2^{2}\times 5\times 7$

$(ii)156=2\times 2\times 3\times 13=2^{2}\times 3\times 13$

$(iii) 3825=3\times 3\times 5\times 5\times 17=3^{2}\times 5^{2}\times 17$

(iv) $5005=5\times 7\times 11\times 13$

$(iv)7429=17\times 19\times 23$

Comments

Tasneem

Question 2:

Find the LCM and HCF of the following pairs of integers and verify that LCM*HCF = product of the two numbers. (i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54

Solution 2:

$(i) 26=2\times 13$

$91= 7\times 13$

HCF $=13. LCM=2\times 7\times 13=182$

Now, HCF X LCM = $183\times 13=2366.......(i)$

Product of two numbers $=26\times 91=2366...(ii)$

From (i) and (ii) it implies that $HCF\times LCM=$ Product of the 2 numbers

Hence verified.

$(ii) 510=2\times 3\times 5\times 17$

$92=2\times 2\times 2\times 23$

HCF = $2, LCM=2^{2}\times 3\times 5\times 17=2\times 23460$

Now, $HCF\times LCM=2\times 23460=46920...(i)$

Product of the numbers, $510\times 92=46920$ ...... $(ii)$

From (i) and (ii) it implies that $HCF\times LCM=$ Product of the 2 numbers

Hence verified.

$(ii) 336=2^{4}\times 3\times 7$

$54=2\times 3\times 3$

HCF=6, LCM = $2^{4}\times 3^{3}\times 7$ $=3024$

NOw, HCF X LCM $=3024\times 6=18144.....(i)$

Product of the numbers,

$336\times 54=18144.....(ii)$

From (i) and (ii) it implies that $HCF\times LCM=$ Product of the 2 numbers

Hence verified.

Comments

Tasneem

Question 3:

Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25

Solution 3:

$(i) 12=2\times 2\times 3$

$15=3\times 5$

$21=3\times 7$

HCF $=3$ , LCM= $2\times 2\times 3\times 5\times 7$ $=420$

$(ii) 17=17\times 1$

$23= 23\times 1$

$29=29\times 1$

HCF $=1$ , LCM $=17\times 23\times 29$ $=11339$

$(iii) 8=2\times 2\times 2\times 2\times 1$

$9=3\times3\times 1$

$25=5\times 5\times 1$

HCF $=1$ , LCM $=2^{3}\times 3^{2}\times 5^{2}$ $=1800$

Comments

Tasneem

Question 4:

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

Solution 4:

Product of HCF and LCM of 2 numbers=product of these numbers

HCF=9

LCM=?

first number=306

second number=657

so,

9 X LCM=306 X 657

LCM= $\frac{306\times 657}{9}$

LCM =22338

Comments

Anish Tiwari

Question 5:

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

Solution 5:

f any number ends with the digit 0, it should be divisible by 10 or in other words its prime factorisation must include primes 2 and 5 both
Prime factorisation of 6^_n = (2 x 3)^_n

By Fundamental Theorem of Arithmetic Prime factorisation, every number is unique. So 5 is not a prime factor of 6^_n.
Hence, for any value of n, 6n will not be divisible by 5.
Therefore, $6^{n}$ cannot end with the digit 0 for any natural number n.

Comments

Tasneem

Question 6:

Explain why 7*11*13+13 and 7*6*5*4*3*2*1+5 are composite numbers.

Solution 6:

Observe that,
7 x 11 x 13 + 13 = 13 x (7 x 11 + 1) = 13 x (77 + 1)
= 13 x 78
= 13 x 13 x 6

The given expression has 6 and 13 as its factors. Therefore, it is a composite number.
7 x 6 x 5 x 4 x 3 x 2 x 1 + 5 = 5 x (7 x 6 x 4 x 3 x 2 x 1 + 1)
= 5 x (1008 + 1)
= 5 x 1009

1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, it is a composite number.

Comments

Tasneem

Question 7:

There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the same direction. After how many minutes will they meet again at the starting point?

Solution 7:

Ravi and Sonia donot take the same amount of time.

Ravi takes lesser time than Sonia for completing 1 round of the circular path.

As they are going in the same direction, they meet again at the same time when Ravi will have completed 1 round of that circular path with respect to Sonia.
i.e When Sonia completes one round then Ravi completes 1.5 rounds.

The minimum time when Sonia and Ravi meet again will be the LCM Of 18 minutes and 12 minutes
Now
18 = 2 x 3 x 3 = 2 x 3^₂
And, 12 = 2 x 2 x 3 = 2^₂ x 3

LCM of 12 and 18 = product of factors raised to highest exponent =
2² x 3² = 36

Therefore, Ravi and Sonia will meet together at the starting point after 36 minutes.

Comments

Tasneem

All Exercises - Chapter 1 - Real Numbers

Exercise 1.1

Exercise 1.2

Exercise 1.3

Exercise 1.4

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Learn Exercise 1.2 with Free Lessons & Tips

All Exercises - Chapter 1 - Real Numbers

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