Exercise 2.2 Practice
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Overview
This page provides comprehensive Ch 2: Polynomials - Exercise 2.2 Practice. Practice linear and quadratic polynomials evaluation, along with linear equations in one variable word problems. Free step-by-step NCERT solutions and explanations.
Evaluating Polynomials & Simple Equations
Q1: Evaluate Linear Polynomial
Find the value of the linear polynomial $5x - 3$ if:
(i) $x = 0$
(ii) $x = -1$
(iii) $x = 2$
(i) $x = 0$
(ii) $x = -1$
(iii) $x = 2$
Let the polynomial be $p(x) = 5x - 3$.
(i) When $x = 0$:
Substitute $x = 0$ in $p(x)$: $$p(0) = 5(0) - 3 = 0 - 3 = -3$$
Substitute $x = 0$ in $p(x)$: $$p(0) = 5(0) - 3 = 0 - 3 = -3$$
(ii) When $x = -1$:
Substitute $x = -1$ in $p(x)$: $$p(-1) = 5(-1) - 3 = -5 - 3 = -8$$
Substitute $x = -1$ in $p(x)$: $$p(-1) = 5(-1) - 3 = -5 - 3 = -8$$
(iii) When $x = 2$:
Substitute $x = 2$ in $p(x)$: $$p(2) = 5(2) - 3 = 10 - 3 = 7$$
Substitute $x = 2$ in $p(x)$: $$p(2) = 5(2) - 3 = 10 - 3 = 7$$
(i) $-3$, (ii) $-8$, (iii) $7$
Q2: Evaluate Quadratic Polynomial
Find the value of the quadratic polynomial $7s^2 - 4s + 6$ if:
(i) $s = 0$
(ii) $s = -3$
(iii) $s = 4$
(i) $s = 0$
(ii) $s = -3$
(iii) $s = 4$
Let the polynomial be $p(s) = 7s^2 - 4s + 6$.
(i) When $s = 0$:
Substitute $s = 0$ in $p(s)$: $$p(0) = 7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6$$
Substitute $s = 0$ in $p(s)$: $$p(0) = 7(0)^2 - 4(0) + 6 = 0 - 0 + 6 = 6$$
(ii) When $s = -3$:
Substitute $s = -3$ in $p(s)$: $$p(-3) = 7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81$$
Substitute $s = -3$ in $p(s)$: $$p(-3) = 7(-3)^2 - 4(-3) + 6 = 7(9) + 12 + 6 = 63 + 12 + 6 = 81$$
(iii) When $s = 4$:
Substitute $s = 4$ in $p(s)$: $$p(4) = 7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102$$
Substitute $s = 4$ in $p(s)$: $$p(4) = 7(4)^2 - 4(4) + 6 = 7(16) - 16 + 6 = 112 - 16 + 6 = 102$$
(i) $6$, (ii) $81$, (iii) $102$
Q3: Age Problem
The present age of Salil’s mother is three times Salil’s present age. After 5 years, their ages will add up to 70 years. Find their present ages.
Let Salil's present age be $x$ years.
Then, Salil's mother's present age is $3x$ years.
Then, Salil's mother's present age is $3x$ years.
After 5 years:
Salil's age = $x + 5$ years.
Mother's age = $3x + 5$ years.
Salil's age = $x + 5$ years.
Mother's age = $3x + 5$ years.
According to the problem, the sum of their ages after 5 years is 70:
$$(x + 5) + (3x + 5) = 70$$
$$4x + 10 = 70$$
$$4x = 60$$
$$x = 15$$
So, Salil's present age is $15$ years.
Salil's mother's present age is $3(15) = 45$ years.
Salil's mother's present age is $3(15) = 45$ years.
Salil's age = 15 years, Mother's age = 45 years
Q4: Integer Difference and Ratio
The difference between two positive integers is 63. The ratio of the two integers is 2:5. Find the two integers.
Since the ratio of the two integers is $2:5$, let the smaller integer be $2k$ and the larger integer be $5k$, where $k$ is a positive constant.
According to the problem, the difference between them is 63:
$$5k - 2k = 63$$
$$3k = 63$$
$$k = 21$$
Now find the two integers:
Smaller integer = $2k = 2(21) = 42$
Larger integer = $5k = 5(21) = 105$
Smaller integer = $2k = 2(21) = 42$
Larger integer = $5k = 5(21) = 105$
The two integers are 42 and 105
Q5: Value of Coins
Ruby has 3 times as many two-rupee coins as she has five rupee-coins. If she has a total ₹88, how many coins does she have of each type?
Let the number of five-rupee coins Ruby has be $x$.
Then, the number of two-rupee coins she has is $3x$.
Then, the number of two-rupee coins she has is $3x$.
Calculate the value of the coins:
Value of five-rupee coins = $5 \times x = 5x$ rupees.
Value of two-rupee coins = $2 \times (3x) = 6x$ rupees.
Total value = $5x + 6x = 11x$ rupees.
Value of five-rupee coins = $5 \times x = 5x$ rupees.
Value of two-rupee coins = $2 \times (3x) = 6x$ rupees.
Total value = $5x + 6x = 11x$ rupees.
Given that the total amount is ₹88:
$$11x = 88$$
$$x = 8$$
Therefore:
Number of five-rupee coins = $x = 8$
Number of two-rupee coins = $3x = 3(8) = 24$
Number of five-rupee coins = $x = 8$
Number of two-rupee coins = $3x = 3(8) = 24$
8 five-rupee coins and 24 two-rupee coins
Q6: Cutting Fence
A farmer cuts a 300 feet fence into two pieces of different sizes. The longer piece is four times as long as the shorter piece. How long are the two pieces?
Let the length of the shorter piece be $x$ feet.
Then, the length of the longer piece is $4x$ feet.
Then, the length of the longer piece is $4x$ feet.
The total length of the fence is 300 feet:
$$x + 4x = 300$$
$$5x = 300$$
$$x = 60$$
Therefore:
Shorter piece = $60$ feet
Longer piece = $4 \times 60 = 240$ feet
Shorter piece = $60$ feet
Longer piece = $4 \times 60 = 240$ feet
Shorter piece = 60 ft, Longer piece = 240 ft
Q7: Rectangle Dimensions
If the length of a rectangle is three more than twice its width and its perimeter is 24 cm, what are the dimensions of the rectangle?
Let the width of the rectangle be $w$ cm.
Then, the length of the rectangle is $l = 2w + 3$ cm.
Then, the length of the rectangle is $l = 2w + 3$ cm.
The perimeter $P$ of a rectangle is given by the formula:
$$P = 2(l + w)$$
Given that the perimeter is 24 cm:
$$2((2w + 3) + w) = 24$$
$$2(3w + 3) = 24$$
$$3w + 3 = 12$$
$$3w = 9$$
$$w = 3$$
Now find the dimensions:
Width = $3$ cm
Length = $2(3) + 3 = 9$ cm
Width = $3$ cm
Length = $2(3) + 3 = 9$ cm
Width = 3 cm, Length = 9 cm