| 5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. Solution: Let x be any positive integer and y = 3.
By Euclid’s division algorithm, then,
x = 3q + r, where q ≥ 0 and r = 0, 1, 2, as r ≥ 0 and r < 3.
Therefore, putting the value of r, we get,
x = 3q
or
x = 3q + 1
or
x = 3q + 2
Now, by taking the cube of all the three above expressions, we get,
Case (i): When r = 0, then,
x2 = (3q)3 = 27q3= 9(3q3) = 9m; where m = 3q3Case (ii): When r = 1, then,
x3 = (3q + 1)3 = (3q)3 + 13 + 3 × 3q × 1(3q + 1) = 27q3+ 1 + 27q2 + 9q
Taking 9 as common factor, we get,
x3 = 9(3q3+3q2+q) + 1
Putting = m, we get,
Putting (3q3 + 3q2+q) = m, we get ,
x3 = 9m + 8
Case (iii): When r = 2, then,
x3 = (3q + 2)3= (3q)3 + 23 + 3 × 3q × 2(3q + 2) = 27q3 + 54q2 + 36q + 8
Taking 9 as common factor, we get,
x3 = 9(3q3 + 6q2 + 4q) + 8
Putting (3q3+ 6q2+ 4q) = m, we get ,
x3 = 9m + 8
Therefore, from all the three cases explained above, it is proved that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8. Exercise – 1.2 | 1. Express each number as a product of its prime factors: (i) 140 Solution: 140
By Taking the LCM of 140, we will get the product of its prime factor.
Therefore, 140 = 2 × 2 × 5 × 7 × 1 = 22 × 5 × 7 (ii) 156 Solution: 156
By Taking the LCM of 156, we will get the product of its prime factor.
Hence, 156 = 2 × 2 × 13 × 3 × 1 = 22× 13 × 3 (iii) 3825 Solution: 3825
By Taking the LCM of 3825, we will get the product of its prime factor.
Hence, 3825 = 3 × 3 × 5 × 5 × 17 × 1 = 32 × 52 ×17 (iv) 5005 Solution: 5005
By Taking the LCM of 5005, we will get the product of its prime factor.
Hence, 5005 = 5 × 7 × 11 × 13 × 1 = 5 × 7 × 11 × 13 (v) 7429 Solution: 7429
By Taking the LCM of 7429, we will get the product of its prime factor.
Hence, 7429 = 17 × 19 × 23 × 1 = 17 × 19 × 23 |
| 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 Solution: 26 and 91
Expressing 26 and 91 as product of its prime factors, we get,
26 = 2 × 13 × 1
91 = 7 × 13 × 1
Therefore, LCM (26, 91) = 2 × 7 × 13 × 1 = 182
And HCF (26, 91) = 13
Verification
Now, product of 26 and 91 = 26 × 91 = 2366
And Product of LCM and HCF = 182 × 13 = 2366
Hence, LCM × HCF = product of the 26 and 91. (ii) 510 and 92 Solution: 510 and 92
Expressing 510 and 92 as product of its prime factors, we get,
510 = 2 × 3 × 17 × 5 × 1
92 = 2 × 2 × 23 × 1
Therefore, LCM (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
And HCF (510, 92) = 2
Verification
Now, product of 510 and 92 = 510 × 92 = 46920
And Product of LCM and HCF = 23460 × 2 = 46920
Hence, LCM × HCF = product of the 510 and 92. (iii) 336 and 54 Solution: 336 and 54
Expressing 336 and 54 as product of its prime factors, we get,
336 = 2 × 2 × 2 × 2 × 7 × 3 × 1
54 = 2 × 3 × 3 × 3 × 1
Therefore, LCM (336, 54) = 3024
And HCF(336, 54) = 2 × 3 = 6
Verification
Now, product of 336 and 54 = 336 × 54 = 18,144
And Product of LCM and HCF = 3024 × 6 = 18,144
Hence, LCM × HCF = product of the 336 and 54. |
| 3. Find the LCM and HCF of the following integers by applying the prime factorisation method. (i) 12, 15 and 21 Solution: 12, 15 and 21
Writing the product of prime factors for all the three numbers, we get,
12 = 2 × 2 × 3
15 = 5 × 3
21 = 7 × 3
Therefore,
HCF(12,15,21) = 3
LCM(12,15,21) = 2 × 2 × 3 × 5 × 7 = 420 (ii) 17, 23 and 29 Solution: 17, 23 and 29
Writing the product of prime factors for all the three numbers, we get,
17 = 17 × 1
23 = 23 × 1
29 = 29 × 1
Therefore,
HCF(17,23,29) = 1
LCM(17,23,29) = 17 × 23 × 29 = 11339 (iii) 8, 9 and 25 Solution: 8, 9 and 25
Writing the product of prime factors for all the three numbers, we get,
8 = 2 × 2 × 2 × 1
9 = 3 × 3 × 1
25 = 5 × 5 × 1
Therefore,
HCF (8,9,25) = 1
LCM (8,9,25) = 2 × 2 × 2 × 3 × 3 × 5 × 5 = 1800 |
| 4. Given that HCF (306, 657) = 9, find LCM (306, 657). Solution: As we know that,
HCF × LCM = Product of the two given numbers
Therefore,
9 × LCM = 306 × 657
LCM = (306×657)/9 = 22338
Hence, LCM(306,657) = 22338 |
| 5. Check whether 6n can end with the digit 0 for any natural number n. Solution: If the number 6n ends with the digit zero (0), then it should be divisible by 5, as we know any number with unit place as 0 or 5 is divisible by 5.
Prime factorization of 6n = (2 × 3)n
Therefore, the prime factorization of 6n doesn’t contain prime number 5.
Hence, it is clear that for any natural number n, 6n is not divisible by 5 and thus it proves that 6n cannot end with the digit 0 for any natural number n. |
| 6. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers. Solution: By the definition of composite number, we know, if a number is composite, then it means it has factors other than 1 and itself. Therefore, for the given expression;
7 × 11 × 13 + 13
Taking 13 as common factor, we get,
= 13(7 × 11 × 1 + 1)
= 13(77 + 1)
= 13 × 78
= 13 × 3 × 2 × 13
Hence, 7 × 11 × 13 + 13 is a composite number.
Now let’s take the other number,
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5
Taking 5 as a common factor, we get,
= 5(7 × 6 × 4 × 3 × 2 × 1 + 1)
= 5(1008 + 1)
= 5 × 1009
Hence, 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 is a composite number. |
| 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: Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when they will be meeting again at the starting point is LCM of 18 and 12.
Therefore, LCM(18,12) = 2 × 3 × 3 × 2 × 1= 36
Hence, Sonia and Ravi will meet again at the starting point after 36 minutes. Exercise – 1.3 | Ncert Solution Class 10th (Chapter – 1) Exercise – 1.3 Question No. 1 1. Prove that √5 is irrational. Solution: Let us assume, that √5 is rational number.
i.e. √5 = x/y (where, x and y are co-primes)
y√5 = x
Squaring both the sides, we get,
(y√5)2 = x2
⇒ 5y2 = x2……………………………….. (1)
Thus, x2 is divisible by 5, so x is also divisible by 5.
Let us say, x = 5k, for some value of k and substituting the value of x in equation (1), we get,
5y2 = (5k)2
⇒ y2 = 5k2
is divisible by 5 it means y is divisible by 5.
Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.
Hence, √5 is an irrational number. |
| Ncert Solution Class 10th (Chapter – 1) Exercise – 1.3 Question No. 2 2. Prove that 3 + 2√5 + is irrational. Solution: Let us assume 3 + 2√5 is rational.
Then we can find co-prime x and y (y ≠ 0) such that 3 + 2√5 = x/y
Rearranging, we get,
Since, x and y are integers, thus,
is a rational number.
Therefore, √5 is also a rational number. But this contradicts the fact that √5 is irrational.
So, we conclude that 3 + 2√5 is irrational. |
| Ncert Solution Class 10th (Chapter – 1) Exercise – 1.3 Question No. 3 3. Prove that the following are irrationals: (i) 1/√2 Solution: 1/√2
Let us assume 1/√2 is rational.
Then we can find co-prime x and y (y ≠ 0) such that 1/√2 = x/y
Rearranging, we get,
√2 = y/x
Since, x and y are integers, thus, √2 is a rational number, which contradicts the fact that √2 is irrational.
Hence, we can conclude that 1/√2 is irrational. (ii) 7√5 Solution: 7√5
Let us assume 7√5 is a rational number.
Then we can find co-prime a and b (b ≠ 0) such that 7√5 = x/y
Rearranging, we get,
√5 = x/7y
Since, x and y are integers, thus, √5 is a rational number, which contradicts the fact that √5 is irrational.
Hence, we can conclude that 7√5 is irrational. (iii) 6 + √2 Solution: 6 +√2
Let us assume 6 +√2 is a rational number.
Then we can find co-primes x and y (y ≠ 0) such that 6 +√2 = x/y⋅
Rearranging, we get,
√2 = (x/y) – 6
Since, x and y are integers, thus (x/y) – 6 is a rational number and therefore, √2 is rational. This contradicts the fact that √2 is an irrational number.
Hence, we can conclude that 6 +√2 is irrational. Exercise – 1.4 | Ncert Solution Class 10th (Chapter – 1) Exercise – 1.4 Question No. 1 1. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion: Note: If the denominator has only factors of 2 and 5 or in the form of 2m ×5n then it has terminating decimal expansion.
If the denominator has factors other than 2 and 5 then it has a non-terminating decimal expansion. (i) 13/3125 Solution: 13/3125
Factorizing the denominator, we get,
3125 = 5 × 5 × 5 × 5 × 5 = 55
Since, the denominator has only 5 as its factor, 13/3125 has a terminating decimal expansion. (ii) 17/8 Solution: 17/8
Factorizing the denominator, we get,
8 = 2×2×2 = 23
Since, the denominator has only 2 as its factor, 17/8 has a terminating decimal expansion. (iii) 64/455 Solution: 64/455
Factorizing the denominator, we get,
455 = 5×7×13
Since, the denominator is not in the form of 2m ×5n thus 64/455 has a non-terminating decimal expansion. (iv) 15/1600 Solution: 15/ 1600
Factorizing the denominator, we get,
1600 = 26×52
Since, the denominator is in the form of 2m ×5n thus 15/1600 has a terminating decimal expansion. (v) 29/343 Solution: 29/343
Factorizing the denominator, we get,
343 = 7×7×7 = 73 Since, the denominator is not in the form of 2m ×5n thus 29/343 has a non-terminating decimal expansion. (vi) 23/2352 Solution: 23/2352
Clearly, the denominator is in the form of 2m ×5n
Hence, 23/2352 has a terminating decimal expansion. (vii) 129/225775 Solution: 129/225775
As you can see, the denominator is not in the form of 2m ×5n
Hence, 129/225775 has a non-terminating decimal expansion. (viii) 6/15 Solution: 6/15
6/15 = 2/5
Since, the denominator has only 5 as its factor, thus, 6/15 has a terminating decimal expansion. (ix) 35/50 Solution: 35/50
35/50 = 7/10
Factorising the denominator, we get,
10 = 2 × 5
Since, the denominator is in the form of 2m ×5n thus, 35/50 has a terminating decimal expansion. (x) 77/210 Solution: 77/210
77/210 = (7× 11)/ (30 × 7) = 11/30
Factorising the denominator, we get,
30 = 2 × 3 × 5
As you can see, the denominator is not in the form of 2m ×5n .Hence, 77/210 has a non-terminating decimal expansion. |
| Ncert Solution Class 10th (Chapter – 1) Exercise – 1.4 Question No. 2 2. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating decimal expansions. (i) 13/3125 Solution: 
13/3125 = 0.00416 (ii) 17/8 Solution: 
17/8 = 2.125 (iii) 64/455 Solution: Non terminating decimal expansion (iv)15/ 1600 Solution: 
15/1600 = 0.009375 (v) 29/ 343 Solution: Non terminating decimal expansion (vi)23/ (2352) Solution: 23/ (2352) = 23/(8×25) = 23/200 
23/ (2352) = 0.115 (vii) 129/ (225775) Solution: Non terminating decimal expansion (viii) 6/15 Solution: 6/15 = 2/5 
(ix) 35/50 Solution: 35/50 = 7/10 
35/50 = 0.7 (x) 77/210 Solution: Non-terminating decimal expansion. |
| Ncert Solution Class 10th (Chapter – 1) Exercise – 1.4 Question No. 3 3. The following real numbers have decimal expansions as given below. In each case, decide whether they are rational or not. If they are rational, and of the form, p q what can you say about the prime factors of q? (i) 43.123456789 Solution: 43.123456789
Since it has a terminating decimal expansion, it is a rational number in the form of p/q and q has factors of 2 and 5 only. (ii) 0.120120012000120000. . . Solution: 0.120120012000120000. . .
Since, it has non-terminating and non- repeating decimal expansion, it is an irrational number. 
Solution: Since it has non-terminating but repeating decimal expansion, it is a rational number in the form of p/q and q has factors other than 2 and 5. |
|
|
|
