Let the smaller number be x. Then the larger number is 5x. Both are positive, so x > 0.

After adding 21 to each, the new numbers are x + 21 (smaller) and 5x + 21 (larger). The condition "one becomes twice the other" gives two cases.

Case A: larger = 2 x (smaller)
5x + 21 = 2 (x + 21)

5x + 21 = 2x + 42

3x = 21

⇒ x = 7.

So, the two numbers are 7 and 5(7) = 35.
Case B: smaller = 2 x (larger)

x + 21 = 2 (5x + 21)
x + 21 = 10x + 42

-9x = 21

⇒ \(x=-\frac{7}{3} .\)

This gives a negative x, contradicting the requirement that the numbers be positive. Reject this case.

Hence the two numbers are 7 and 35.

Observe that the amount left on the n^th day will be ₹(100- 5n).

Therefore, on the 12tn day the amount left will be ₹(100 - 12 x 5) = ₹40.

Let the length ofthe shorter piece = x feet.
Then length of longer piece = 4x feet.
Total length = 300 ft:
x + 4x = 300

5x = 300

x = 60

Hence, Shorter piece = 60 feet;
Longer piece = 4 x 60 = 240 feet.

Let Salil's present age = x years.
Then mother's present age = 3x years.
After 5 years: Salil's age = (x + 5);
Mother's age = (3x + 5)
According to the question: (x + 5) + (3x + 5) = 70
4x + 10 = 70

4x = 60x = 15

Hence, Salil's present age = 15 years
Mother's present age = 3 x 15 = 45 years.

p(t) = 2 + t + 2t² - t³

put t = 0

p(0) = 2 + 0 + 2 (0)² - (0)³

p(0) = 2

Similarly,

p(t) = 2 + t + 2t² - t³

put t = 1

p(1) = 2 + (1) + (1)² - (1)³

= 2 + 1 + 2 - 1

P(1) = 4

And,

p(t) = 2 + t + 2t² - t³

put t = 2

P(2) = 2 + (2) + 2(2)² - (2)³

= 2 + 2 + 2 x 4 - 8

p(2) = 4

(A) 4x² - 3x + 7 = 4x² - 3x¹ + 70

Here, powers of the polynomial are 2, 1 and 0. Since, all powers are whole numbers, it is a polynomial.

Now, since there is only one variable x, it is a polynomial in one variable.

(B) \(y+\frac{2}{y}\) \(=y^1+\frac{2}{y^1}\)

= y¹ + 2y^-1 \(\text { [As } \frac{1}{a^m}=a^{-m} \text { ] }\)

Here, the powers of y are 1 and -1.

Since, -1 is not a whole number, it is not a polynomial.

Cost for m matches = ₹ 50 x m = ₹ 50m.
Joining fee = ₹ 200 (one-time, fixed).
Total amount paid = 50m + 200.

The constant term of a polynomial is the term that does not contain any variable. In 9x³ + 5x²- 8x - 10, the term -10has no variable. Hence, the Constant term is -10.

The polynomial 4z³ + 5z² - 11 can be written as 4z³ + 5z² - 11 z⁰. Since the term containing z (ie, z¹) is missing, its coefficient is 0.

In a polynomial, the exponent of every variable must be a whole number (0, 1, 2, 3, ). In \(\sqrt{x}+2\) , the term \(\sqrt{x}\) can be written as \(x^{\frac{1}{2}},\) where the power \(\frac{1}{2}\) is not a whole number. Hence, \(\sqrt{x}+2\) is not a polynomial.

Let the required polynomial be p(x) = ax + b.
Use the parallelism condition. Two lines are parallel if f they have the same slope.
The slope of y = 5x - 7 is 5, so a = 5.
Use the point on the line. The graph passes through (0, 4), so p(0) = 4,giving b = 4.
Therefore, p(x) = 5x + 4.

Given that water decreases at a constant rate, the situation represents linear decay.

Model is given by W(t) = -40t + 800

(A) After 5 minutes:

W(5) = -40(5) + 800 = -200 + 800 = 600 litres

(B) When the tank is empty, W(t) = 0

-40t + 8 00 = 0

-40t + 800 = 0

-40t = -800

t = 20 minutes

(A) p(0) = 5(0) - 3 = 0 - 3 = -3
(B) p(-1) = 5(-1) - 3 = -5 - 3 = -8
(C) p(2) = 5(2) - 3 = 10 - 3 = 7

The degree of a polynomial = highest power of the variable.

(A) In 2x² - 5x + 3, highest power of x is 2, so degree is 2.
(B) In y³ + 2y - 1, highest power of y is 3, so degree is 3.

To find the linear relationship y = ax + b, we note that when x = 5, y = 150. Also, when x = 12, y = 325. We substitute these in y = ax + b to arrive at the following equations. 150 = 5a + b and 325 = 12a + b
We solve these as follows: Let b = 150 - 5a (from the first equation). We substitute this in the second equation to obtain

325 = 12a + (150 - 5a)

Thus, 325 = 7a + 150 or 7a = 175. So, a = 25.
This leads to b = 150- 5a = 150 - 125 = 25.

We substitute the values of a and b in the equation y = ax + b to obtain y = 25x + 25.
Thus, y = 25x + 25 represents the linear relationship between y, the fare amount in Rupees, and x, the distance travelled in km.

(A) Setting up the linear equation:

Since the integers are in the ratio 2 : 5, let the integers be 2x and 5x (where x is a positive integer).

Difference = 5x - 2x = 3x

According to the question:

3x = 63

(B) Linear polynomial used:

p(x) = 3x - 63

This is a polynomial of degree 1 in the variable x.

(C) Solving:

3x = 63

x = 21

Hence, the first integer = 2x = 2(21) = 42 and second integer = 5x = 5(21) = 105.

(D) Verification:

Difference: 105 - 42 = 63

Ratio: 42 : 105 = \(\frac{42}{21}: \frac{105}{21}=2: 5\)

Both conditions are satisfied.