NCERT Class 10 Maths Exercise 1.1 Solution – Real Numbers

Solution:

1. $$\begin{array}{rl}2& 140\\ 2& 70\\ 5& 35\\ 7& 7\\ & 1\end{array}$$

   Divide 140 by the smallest prime, 2. You get 70. Divide by 2 again to get 35. Now 2 does not divide 35, so use the next prime, 5. You get 7, and 7 is itself a prime.

2. $$140=2\times 2\times 5\times 7$$

   So 140 = 2 × 2 × 5 × 7.

3. $$140=2^2\times 5\times 7$$

   In powers, this is 2² × 5 × 7. This is our answer.

Solution:

1. $$\begin{array}{rl}2& 156\\ 2& 78\\ 3& 39\\ 13& 13\\ & 1\end{array}$$

   Divide 156 by 2 to get 78. Divide by 2 again to get 39. The sum of the digits of 39 is 12, which is in the table of 3, so divide by 3 to get 13. And 13 is a prime.

2. $$156=2^2\times 3\times 13$$

   So 156 = 2² × 3 × 13.

Solution:

1. $$3+8+2+5=18\Rightarrow 3\text{ divides it}$$

   The sum of the digits of 3825 is 3 + 8 + 2 + 5 = 18. Since 18 is in the table of 3, 3825 must be divisible by 3.

2. $$\begin{array}{rl}3& 3825\\ 3& 1275\\ 5& 425\\ 5& 85\\ 17& 17\\ & 1\end{array}$$

   3825 ÷ 3 = 1275. Divide by 3 again to get 425. The last digit is now 5, so divide by 5 to get 85. Divide by 5 again to get 17. And 17 is a prime.

3. $$3825=3^2\times 5^2\times 17$$

   So the answer is 3² × 5² × 17.

Solution:

1. $$5005=5\times 1001$$

   The last digit of 5005 is 5, so first divide by 5. You get 1001.

2. $$1001=7\times 11\times 13$$

   1001 is a famous number. Remember that 1001 = 7 × 11 × 13. It is useful very often in the exam.

3. $$5005=5\times 7\times 11\times 13$$

   So 5005 = 5 × 7 × 11 × 13. All four are different primes.

Solution:

1. $$\text{Not divisible by 2, 3 or 5}(7+4+2+9=22)$$

   7429 is odd, so 2 does not divide it. The sum of its digits is 22, so 3 does not divide it. The last digit is 9, so 5 does not divide it. Now let us try the next primes.

2. $$7:7\times 1061=7427\times 11:7-4+2-9=-4\times 13:13\times 571=7423\times$$

   Try 7. It does not divide. For 11, use the trick and find the alternate digit difference: 7 – 4 + 2 – 9 = -4. This is not a multiple of 11, so 11 does not divide it. 13 also does not divide it.

3. $$\begin{array}{rl}17& 7429\\ 19& 437\\ 23& 23\\ & 1\end{array}$$

   Now try 17. 17 × 437 = 7429. It works! Then divide 437 by 19 to get 23, which is a prime.

4. $$7429=17\times 19\times 23$$

   So 7429 = 17 × 19 × 23.

Solution:

1. $$\text{Floor }26m\times 91m\Rightarrow \text{largest square tile}=\text{HCF}$$

   Here is a real-life example. To put the largest square tile on a floor of 26 metres by 91 metres, the side of the tile must be the HCF of these two numbers. Let us find it.

2. $$26=2\times 13,91=7\times 13$$

   First, find the prime factorisation of both numbers. 26 = 2 × 13 and 91 = 7 × 13.

3. $$\text{HCF}=13$$

   The only common prime is 13, so HCF = 13.

4. $$\text{LCM}=2\times 7\times 13=182$$

   The LCM has all the primes: 2, 7 and 13. Multiplying them gives 182.

5. $$\text{LCM}\times \text{HCF}=182\times 13=2366$$

   Now let us check. 182 × 13 = 2366.

6. $$26\times 91=2366\mathrm{✓}$$

   And 26 × 91 is also 2366. Both are equal, so the answer is verified!

Solution:

1. $$510=2\times 3\times 5\times 17,92=2^2\times 23$$

   510 = 2 × 3 × 5 × 17 and 92 = 2² × 23.

2. $$\text{HCF}=2$$

   The only common prime is 2, and its smaller power is 2¹. So HCF = 2.

3. $$\text{LCM}=2^2\times 3\times 5\times 17\times 23=23460$$

   The LCM has all the primes, each with its bigger power: 2², 3, 5, 17, 23. Multiplying them gives 23460.

4. $$\text{LCM}\times \text{HCF}=23460\times 2=46920$$

   Check: 23460 × 2 = 46920.

5. $$510\times 92=46920\mathrm{✓}$$

   And 510 × 92 is also 46920. Verified!

Solution:

1. $$336=2^4\times 3\times 7,54=2\times 3^3$$

   336 = 2⁴ × 3 × 7 and 54 = 2 × 3³.

2. $$\text{HCF}=2\times 3=6$$

   The common primes are 2 and 3. Take the smaller power of each: 2¹ and 3¹. So HCF = 6.

3. $$\text{LCM}=2^4\times 3^3\times 7=3024$$

   The LCM has the bigger power of 2, which is 2⁴, the bigger power of 3, which is 3³, and 7. Multiplying them gives 3024.

4. $$6\times 3024=18144=336\times 54\mathrm{✓}$$

   Check: 6 × 3024 = 18144, and 336 × 54 is also 18144. Verified!

Solution:

1. $$12=2^2\times 3,15=3\times 5,21=3\times 7$$

   Factorise all three. 12 = 2² × 3, 15 = 3 × 5, and 21 = 3 × 7.

2. $$\text{HCF}=3$$

   The only prime common to all three is 3. So HCF = 3.

3. $$\text{LCM}=2^2\times 3\times 5\times 7=420$$

   The LCM has all the primes with their bigger powers: 2², 3, 5, 7. Multiplying them gives 420.

Solution:

1. $$17,23,29\text{ all three are prime}$$

   Look carefully: 17, 23 and 29 are all prime numbers.

2. $$\text{HCF}=1$$

   There is no common prime in the three, so HCF = 1.

3. $$\text{LCM}=17\times 23\times 29=11339$$

   The LCM is the product of the three, which is 11339.

Solution:

1. $$8=2^3,9=3^2,25=5^2$$

   8 = 2³, 9 = 3², and 25 = 5².

2. $$\text{HCF}=1$$

   There is no common prime, so HCF = 1.

3. $$\text{LCM}=8\times 9\times 25=1800$$

   All three are co-prime, so LCM = product = 8 × 9 × 25 = 1800.

Solution:

1. $$\text{LCM}\times \text{HCF}=306\times 657$$

   Here our golden formula helps: LCM × HCF = product of the two numbers.

2. $$\text{LCM}={\displaystyle \frac{306\times 657}{9}}$$

   So LCM = (306 × 657) ÷ 9.

3. $$=306\times 73$$

   657 ÷ 9 = 73. So we divide first, which makes the calculation shorter.

4. $$\text{LCM}=22338$$

   306 × 73 = 22338. This is our answer.

Solution:

1. $$\text{Think first! Which digit does it end in?}$$

   First, stop and think: when does a number end in 0? Take a moment.

2. $$\text{Ends in 0}\Rightarrow \text{number }10\text{ ’s multiple}\Rightarrow 2\text{ and }5\text{ both are factors}$$

   A number ends in 0 only when it is a multiple of 10. So its prime factorisation must have both 2 and 5.

3. $$6^n=(2\times 3{)}^n=2^n\times 3^n$$

   Now break up 6ⁿ. Since 6 = 2 × 3, we get 6ⁿ = 2ⁿ × 3ⁿ.

4. $$\text{The only prime factors are 2 and 3. There is no prime 5!}$$

   See, the only primes here are 2 and 3. There is no 5 at all.

5. $$\therefore 6^n\text{ can never end in }0\text{ .}$$

   By the Fundamental Theorem of Arithmetic, the factorisation is unique. So a 5 cannot come from anywhere. Therefore 6ⁿ can never end in 0.

Solution:

1. $$7\times 11\times 13+13=13(7\times 11+1)$$

   Here, 13 is common to both parts. So take 13 out.

2. $$=13\times 78=1014$$

   7 × 11 = 77, plus 1 is 78. So the number = 13 × 78.

3. $$1014=13\times 78\Rightarrow \text{13 and 78 are also factors, besides 1 and itself}$$

   This is a product of two numbers, and both are greater than 1. So it has factors other than 1 and itself. Therefore it is a composite number.

Solution:

1. $$7\times 6\times 5\times 4\times 3\times 2\times 1+5=5(7\times 6\times 4\times 3\times 2\times 1+1)$$

   Here the first part has a 5, and we also add 5. So take 5 out as a common factor.

2. $$=5\times 1009=5045$$

   7 × 6 × 4 × 3 × 2 = 1008. Adding 1 gives 1009. So the number = 5 × 1009.

3. $$\text{5 and 1009 are both}>1\Rightarrow \text{composite}$$

   5 and 1009 are both greater than 1. So this number is composite.

Solution:

1. $$\text{Keyword: ’meet again’ }\text{=}\text{ LCM}$$

   Look for the key idea here. When a question asks after how much time they will meet again, it is always an LCM question.

2. $$18=2\times 3^2,12=2^2\times 3$$

   18 = 2 × 3² and 12 = 2² × 3.

3. $$\text{LCM}(18,12)=2^2\times 3^2=36$$

   The LCM has the bigger power of 2, which is 2², and the bigger power of 3, which is 3². Multiplying them gives 36.

4. $$\text{After }\mathbf{36}\text{ minute}\text{ , they meet at the starting point}$$

   So Sonia and Ravi will meet again at the starting point after 36 minutes.