[Solutions] Real Numbers Exercise 1.1 NCERT CBSE Class 10 Maths Question & Answers

Accepted Answer:

1. For 140:
   $140 = 2 \times 2 \times 5 \times 7$
   $= 2^2 \times 5 \times 7$
2. For 156:
   $156 = 2 \times 2 \times 3 \times 13 $
   $= 2^2 \times 3 \times 13$
3. For 3825:
   $3825 = 3 \times 5 \times 5 \times 7 \times 11 $
   $= 3 \times 5^2 \times 7 \times 11$
4. For 5005:
   $5005 = 5 \times 7 \times 11 \times 13 $
   $= 5 \times 7 \times 11 \times 13$
5. For 7429:
   $7429 = 7 \times 7 \times 11 \times 11 $
   $= 7^2 \times 11^2$

Accepted Answer:

We are given that the army contingent has 616 members and the army band has 32 members. Both groups are to march in the same number of columns.

To find the maximum number of columns in which they can march, we need to find the greatest common divisor (GCD) of 616 and 32.

Using Euclid's Division Algorithm:
$616 = 19 \times 32 + 8$
$32 = 4 \times 8 + 0$

Hence, the GCD of 616 and 32 is 8.

Therefore, the maximum number of columns in which they can march is 8.

Accepted Answer:

(i) For 12, 15, and 21:
Prime factorization:
$12 = 2^2 \times 3$
$15 = 3 \times 5$
$21 = 3 \times 7$

HCF is the product of the common prime factors with the lowest powers:
$HCF(12, 15, 21) = 3$

LCM is the product of all the prime factors, each raised to the highest power:
$LCM(12, 15, 21) $
$= 2^2 \times 3 \times 5 \times 7 = 420$

(ii) For 17, 23, and 29:
Since 17, 23, and 29 are all prime numbers, their prime factorizations are:
$17 = 17$
$23 = 23$
$29 = 29$

HCF is the product of the common prime factors:
$HCF(17, 23, 29) = 1$ (since there are no common prime factors)

LCM is the product of all prime factors:
$LCM(17, 23, 29) $
$= 17 \times 23 \times 29 = 11131$

(iii) For 8, 9, and 25:
Prime factorization:
$8 = 2^3$
$9 = 3^2$
$25 = 5^2$

HCF is the product of the common prime factors with the lowest powers:
$HCF(8, 9, 25) = 1$ (since there are no common prime factors)

LCM is the product of all prime factors, each raised to the highest power:
$LCM(8, 9, 25) $
$= 2^3 \times 3^2 \times 5^2 = 1800$

Accepted Answer:

We are given that $HCF(306, 657) = 9$.

We know the relationship between HCF and LCM:
$LCM(a, b) = \frac{a \times b}{HCF(a, b)}$

Substituting the given values:
$LCM(306, 657) = \frac{306 \times 657}{9}$

First, calculate the product of 306 and 657:
$306 \times 657 = 201762$

Now, divide by 9:
$LCM(306, 657) = \frac{201762}{9} = 22418$

Therefore, the LCM of 306 and 657 is 22418.

Accepted Answer:

We are asked to check whether $6^n$ can end with the digit 0 for any natural number $n$.

For a number to end with the digit 0, it must be divisible by 10. We know that:
$10=2\times 5$

Thus, for $6^n$ to end in 0, it must be divisible by both 2 and 5.

Now, observe the prime factorization of 6:
$6=2\times 3$

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

For $6^n$ to be divisible by 5, it must have 5 as one of its prime factors. However, the prime factorization of $6^n$ contains only the primes 2 and 3, but not 5. Therefore, $6^n$ can never be divisible by 5, and thus it cannot end in 0.

Conclusion:
$6^n$ can never end with the digit 0 for any natural number $n$.

Accepted Answer:

To prove that the given expressions are composite numbers, we show that they have factors other than 1 and themselves.

### First Expression: $7 \times 11 \times 13 + 13$

Factor out the common term $13$:

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

Simplify inside the parentheses:

$ = 13 \times (77 + 1) $

$ = 13 \times 78 $

Since $13$ and $78$ are both greater than 1, the expression is composite.

### Second Expression: $7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1 + 5$

Rewrite as:

$ 7! + 5 $

Factor out $5$:

$ 7! + 5 = 5 \times \left(\frac{7!}{5} + 1 \right) $

Since $7! = 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1$ is divisible by $5$, the term inside the parentheses is an integer greater than 1.

Thus, the expression is composite.

Final Answer:

Both given expressions are composite numbers.

Accepted Answer:

Sonia takes $18$ minutes for one round, and Ravi takes $12$ minutes for one round.

Step 1: Find the prime factorization of $18$ and $12$:
$18=2\times 3^2$
$12=2^2\times 3$

Step 2: LCM is the product of all prime factors, each raised to the highest power:
$LCM(18,12)=2^2\times 3^2=36$

Thus, Sonia and Ravi will meet again at the starting point after $36$ minutes.