WBBSE Class 10 Maths Chapter 1 Exercise 1.1 Solutions

Question 1

Write the quadratic polynomials from the following polynomials by understanding them.

(i) $x^2 - 7x + 2$

Solution: The highest power (degree) of the variable $x$ is $2$.

Conclusion: It is a quadratic polynomial.

(ii) $7x^3 - x(x + 2)$

Solution: Simplifying the expression:
$$= 7x^3 - x^2 - 2x$$
Since the highest power of the variable $x$ is $3$, it is a cubic expression.

Conclusion: It is NOT a quadratic polynomial.

(iii) $2x(x + 5) + 1$

Solution: Expanding the expression:
$$= 2x^2 + 10x + 1$$
The highest power of the variable $x$ is $2$.

Conclusion: It is a quadratic polynomial.

(iv) $2x - 1$

Solution: The highest power of the variable $x$ is $1$.

Conclusion: It is NOT a quadratic polynomial.

Question 2

Which of the following equations can be written in the form of $ax^2 + bx + c = 0$, where $a, b, c$ are real numbers and $a \neq 0$?

(i) $x - 1 + \frac{1}{x} = 6 \quad (x \neq 0)$

Solution: Take the LCM on LHS:
$$\frac{x^2 - x + 1}{x} = 6$$ $$\implies x^2 - x + 1 = 6x$$ $$\implies x^2 - 6x - x + 1 = 0$$ $$\implies x^2 - 7x + 1 = 0$$

Conclusion: This can be written in the form $ax^2 + bx + c = 0$. Hence, it is a quadratic equation.

(ii) $x + \frac{3}{x} = x^2 \quad (x \neq 0)$

Solution: Take the LCM on LHS:
$$\frac{x^2 + 3}{x} = x^2$$ $$\implies x^2 + 3 = x^3$$ $$\implies x^3 - x^2 - 3 = 0$$

Conclusion: This is a cubic equation ($3^{\text{rd}}$ degree), so it CANNOT be written in the form $ax^2 + bx + c = 0$.

(iii) $x^2 - 6\sqrt{x} + 2 = 0$

Solution: The power of the variable $x$ under the radical sign is a fraction ($\sqrt{x} = x^{1/2}$), which is not a non-negative integer.

Conclusion: It CANNOT be written in the general form. This is NOT a quadratic equation.

(iv) $(x - 2)^2 = x^2 - 4x + 4$

Solution: Expanding the LHS using $(a-b)^2 = a^2 - 2ab + b^2$:
$$x^2 - 4x + 4 = x^2 - 4x + 4$$ $$\implies 0 = 0$$

Conclusion: This is an identity, NOT a quadratic equation.

Question 3

Let us determine the power of the variable for which the equation $x^6 - x^3 - 2 = 0$ will become a quadratic equation.

Solution: Given equation is $x^6 - x^3 - 2 = 0$.
We can rewrite this expression as:
$$(x^3)^2 - 1\cdot(x^3) - 2 = 0$$
If we replace $x^3 = y$, the equation becomes $y^2 - y - 2 = 0$, which is a quadratic equation in $y$.

Answer: The equation becomes a quadratic equation with respect to $x^3$ (i.e., power 3).

Question 4

(i) Let us determine the value of '$a$' for which the equation $(a-2)x^2 + 3x + 5 = 0$ will not be a quadratic equation.

Solution: For the equation to lose its quadratic nature, the coefficient of $x^2$ must be equal to $0$.
$$\implies a - 2 = 0$$ $$\implies a = 2$$

Answer: If $a = 2$, the given equation will not be a quadratic equation.

(ii) If $\frac{x}{4 - x} = \frac{1}{3x}$ ($x \neq 0, x \neq 4$) be expressed in the form of $ax^2 + bx + c = 0$, let us determine the coefficient of $x$.

Solution: By cross-multiplying:
$$x \cdot 3x = 1 \cdot (4 - x)$$ $$\implies 3x^2 = 4 - x$$ $$\implies 3x^2 + x - 4 = 0$$
This can be written as $3x^2 + 1(x) - 4 = 0$.

Answer: The coefficient of $x$ is $1$.

(iii) Let us express $3x^2 + 7x + 23 = (x + 4)(x + 3) + 2$ in the form of the quadratic equation $ax^2 + bx + c = 0$.

Solution: Expand the right-hand side:
$$3x^2 + 7x + 23 = (x^2 + 3x + 4x + 12) + 2$$ $$3x^2 + 7x + 23 = x^2 + 7x + 14$$
Now move all terms to the LHS:
$$3x^2 - x^2 + 7x - 7x + 23 - 14 = 0$$ $$2x^2 + 0\cdot x + 9 = 0$$

Answer: The required quadratic form is $2x^2 + 9 = 0$ (or $2x^2 + 0x + 9 = 0$).

(iv) Let us express the equation $(x + 2)^3 = x(x^2 - 1)$ in the form of $ax^2 + bx + c = 0$ and write the coefficients of $x^2, x,$ and $x^0$.

Solution: Expand LHS using $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and expand RHS:
$$x^3 + 3(x^2)(2) + 3(x)(2^2) + 2^3 = x^3 - x$$ $$x^3 + 6x^2 + 12x + 8 = x^3 - x$$
Eliminating $x^3$ from both sides and shifting $-x$ to the LHS:
$$6x^2 + 12x + x + 8 = 0$$ $$6x^2 + 13x + 8 = 0$$

Answer:
• Coefficient of $x^2$ is $6$
• Coefficient of $x$ is $13$
• Coefficient of $x^0$ (constant term) is $8$

Question 5: Equation Construction

(i) Divide 42 into two parts such that one part is equal to the square of the other part.

Solution: Let one part be $x$. Therefore, the remaining part becomes $42 - x$.
According to the given condition:
$$x^2 = 42 - x$$ $$\implies x^2 + x - 42 = 0$$

Required Quadratic Equation: $x^2 + x - 42 = 0$

(ii) The product of two consecutive positive odd numbers is 143.

Solution: Let the two consecutive positive odd numbers be $(2x - 1)$ and $(2x + 1)$.
According to the condition:
$$(2x - 1)(2x + 1) = 143$$ $$\implies (2x)^2 - (1)^2 = 143$$ $$\implies 4x^2 - 1 = 143$$ $$\implies 4x^2 - 144 = 0$$
Dividing the whole equation by $4$:
$$\implies x^2 - 36 = 0$$

Required Quadratic Equation: $x^2 - 36 = 0$

(iii) The sum of the squares of two consecutive numbers is 313.

Solution: Let the first number be $x$, so the next consecutive number is $x + 1$.
According to the condition:
$$x^2 + (x + 1)^2 = 313$$ $$\implies x^2 + (x^2 + 2x + 1) = 313$$ $$\implies 2x^2 + 2x + 1 - 313 = 0$$ $$\implies 2x^2 + 2x - 312 = 0$$
Dividing the whole equation by $2$:
$$\implies x^2 + x - 156 = 0$$

Required Quadratic Equation: $x^2 + x - 156 = 0$

Question 6: Word Problems

(i) The length of the diagonal of a rectangular area is 15 m and the length exceeds its breadth by 3 m.

Solution: Let the breadth of the rectangle be $x$ meters.
Therefore, its length = $(x + 3)$ meters.
Using Pythagoras theorem ($\text{Length}^2 + \text{Breadth}^2 = \text{Diagonal}^2$):
$$x^2 + (x + 3)^2 = 15^2$$ $$\implies x^2 + x^2 + 6x + 9 = 225$$ $$\implies 2x^2 + 6x + 9 - 225 = 0$$ $$\implies 2x^2 + 6x - 216 = 0$$
Dividing the equation by $2$:
$$\implies x^2 + 3x - 108 = 0$$

Required Quadratic Equation: $x^2 + 3x - 108 = 0$

(ii) One person bought some kg sugar for Rs. 80. If he could get 4 kg more sugar with that money, then the price of 1 kg sugar would be less by Rs. 1.

Solution: Let the quantity of sugar bought be $x$ kg.
Initial price of $1$ kg sugar = $\frac{80}{x}$ Rs.
If he gets $4$ kg more, new quantity = $(x + 4)$ kg.
New price of $1$ kg sugar = $\frac{80}{x + 4}$ Rs.
According to the problem context:
$$\frac{80}{x} - \frac{80}{x + 4} = 1$$ $$\implies 80 \left( \frac{1}{x} - \frac{1}{x + 4} \right) = 1$$ $$\implies 80 \left( \frac{x + 4 - x}{x(x + 4)} \right) = 1$$ $$\implies \frac{320}{x^2 + 4x} = 1$$ $$\implies x^2 + 4x = 320 \implies x^2 + 4x - 320 = 0$$

Required Quadratic Equation: $x^2 + 4x - 320 = 0$

(iii) The distance between the two stations is 300 km. A train went to the second station from the first station with uniform velocity. If the velocity of the train is 5 km/hour more, then the time taken would be less by 2 hours.

Solution: Let the uniform velocity of the train be $x$ km/hr.
Time taken to cover $300$ km = $\frac{300}{x}$ hours.
If speed is increased by $5$ km/hr, new velocity = $(x + 5)$ km/hr.
New time taken = $\frac{300}{x + 5}$ hours.
According to the problem condition:
$$\frac{300}{x} - \frac{300}{x + 5} = 2$$ $$\implies 300 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 2$$ $$\implies \frac{1500}{x^2 + 5x} = 2$$ $$\implies 2(x^2 + 5x) = 1500$$ $$\implies x^2 + 5x = 750 \implies x^2 + 5x - 750 = 0$$

Required Quadratic Equation: $x^2 + 5x - 750 = 0$

(iv) A clock seller sold a clock by purchasing it at Rs. 336. The amount of his profit percentage is as much as the amount with which he bought the clock.

Solution: Let the cost price (CP) of the clock be Rs. $x$.
Therefore, profit percentage = $x\%$.
Profit value = $x\% \text{ of } x = x \cdot \frac{x}{100} = \frac{x^2}{100}$ Rs.
Since $\text{Selling Price (SP)} = \text{CP} + \text{Profit}$:
$$x + \frac{x^2}{100} = 336$$ $$\implies \frac{100x + x^2}{100} = 336$$ $$\implies x^2 + 100x = 33600 \implies x^2 + 100x - 33600 = 0$$

Required Quadratic Equation: $x^2 + 100x - 33600 = 0$

(v) If the velocity of the stream is 2 km/hr, then the time taken by Ratanmajhi to cover 21 km downstream and upstream is 10 hours.

Solution: Let the speed of the boat in still water be $x$ km/hr.
Downstream speed = $(x + 2)$ km/hr. Upstream speed = $(x - 2)$ km/hr.
Total time taken = Downstream time + Upstream time = $10$ hours.
$$\frac{21}{x + 2} + \frac{21}{x - 2} = 10$$ $$\implies 21 \left( \frac{x - 2 + x + 2}{(x + 2)(x - 2)} \right) = 10$$ $$\implies 21 \left( \frac{2x}{x^2 - 4} \right) = 10$$ $$\implies \frac{42x}{x^2 - 4} = 10 \implies 42x = 10x^2 - 40$$ $$\implies 10x^2 - 42x - 40 = 0$$
Dividing the whole equation by $2$:
$$\implies 5x^2 - 21x - 20 = 0$$

Required Quadratic Equation: $5x^2 - 21x - 20 = 0$

(vi) The time taken to clean out the garden of Majid is 3 hours more than Mahim. Both of them together can complete the work in 2 hours.

Solution: Let Mahim alone take $x$ hours to clean the garden.
Then, Majid takes $(x + 3)$ hours.
In $1$ hour, Mahim does $\frac{1}{x}$ part and Majid does $\frac{1}{x+3}$ part of the total work.
Together, their $1$-hour work is $\frac{1}{2}$ part of the total work.
$$\frac{1}{x} + \frac{1}{x + 3} = \frac{1}{2}$$ $$\implies \frac{x + 3 + x}{x(x + 3)} = \frac{1}{2}$$ $$\implies \frac{2x + 3}{x^2 + 3x} = \frac{1}{2}$$ $$\implies 2(2x + 3) = x^2 + 3x$$ $$\implies 4x + 6 = x^2 + 3x \implies x^2 + 3x - 4x - 6 = 0$$ $$\implies x^2 - x - 6 = 0$$

Required Quadratic Equation: $x^2 - x - 6 = 0$

Question 7

The unit digit of a two-digit number exceeds its ten's digit by 6 and the product of two digits is less by 12 from the number.

Solution: Let the digit at the tens place be $x$.
Then, the digit at the units place = $x + 6$.
The value of the two-digit number = $10(x) + 1(x + 6) = 11x + 6$.
According to the given condition:
$$\text{Product of digits} = \text{Number} - 12$$ $$\implies x(x + 6) = (11x + 6) - 12$$ $$\implies x^2 + 6x = 11x - 6$$ $$\implies x^2 + 6x - 11x + 6 = 0$$ $$\implies x^2 - 5x + 6 = 0$$

Required Quadratic Equation: $x^2 - 5x + 6 = 0$

Question 8

There is a road of equal width around the outside of a rectangular playground having a length of 45 m and breadth of 40 m and the area of the road is 450 sq.m.

Solution: Let the uniform width of the path around the outer boundary be $x$ meters.
Inner playground dimensions: $\text{Length} = 45\text{ m}$, $\text{Breadth} = 40\text{ m}$. Area $= 45 \times 40 = 1800\text{ m}^2$.
Outer dimensions including the path: $\text{Length} = 45 + 2x$, $\text{Breadth} = 40 + 2x$.
Total Area with Path $= (45 + 2x)(40 + 2x)$.
$$\text{Area of the path} = \text{Total Outer Area} - \text{Inner Playground Area}$$ $$\implies (45 + 2x)(40 + 2x) - 1800 = 450$$ $$\implies 1800 + 90x + 80x + 4x^2 - 1800 = 450$$ $$\implies 4x^2 + 170x - 450 = 0$$
Dividing the entire equation by $2$:
$$\implies 2x^2 + 85x - 225 = 0$$

Required Quadratic Equation: $2x^2 + 85x - 225 = 0$

Question 9

Let me write by calculating, for what value of k, 1 will be a root of the quadratic equation x² + kx + 3 = 0.

Solution: Since $1$ is given as a root of the equation $x^2 + kx + 3 = 0$, substituting $x = 1$ must balance the equation to $0$.
$$(1)^2 + k(1) + 3 = 0$$ $$\implies 1 + k + 3 = 0$$ $$\implies k + 4 = 0$$ $$\implies k = -4$$

Answer: The value of $k$ is $-4$.

Question 10

I solve and write the two roots of the quadratic equation: $\frac{a}{ax-1} + \frac{b}{bx-1} = a + b$

Solution: Rearrange terms to pair $a$ with $-b$ and $b$ with $-a$ groups:
$$\left(\frac{a}{ax-1} - b\right) + \left(\frac{b}{bx-1} - a\right) = 0$$ $$\implies \left(\frac{a - b(ax - 1)}{ax - 1}\right) + \left(\frac{b - a(bx - 1)}{bx - 1}\right) = 0$$ $$\implies \left(\frac{a + b - abx}{ax - 1}\right) + \left(\frac{a + b - abx}{bx - 1}\right) = 0$$
Factoring out the common numerator $(a + b - abx)$:
$$\implies (a + b - abx) \left[ \frac{1}{ax - 1} + \frac{1}{bx - 1} \right] = 0$$ $$\implies (a + b - abx) \left[ \frac{bx - 1 + ax - 1}{(ax - 1)(bx - 1)} \right] = 0$$ $$\implies (a + b - abx) \left[ \frac{(a+b)x - 2}{(ax - 1)(bx - 1)} \right] = 0$$
Equating each factor to zero:
Case 1: $a + b - abx = 0 \implies abx = a + b \implies x = \frac{a+b}{ab}$
Case 2: $(a+b)x - 2 = 0 \implies (a+b)x = 2 \implies x = \frac{2}{a+b}$

Roots: $x = \frac{2}{a+b}$ and $x = \frac{a+b}{ab}$

Question 11

I solve the quadratic equation: $\frac{x+3}{x-3} + \frac{x-3}{x+3} = 2\frac{1}{2}$

Solution: Find the common denominator and cross-add:
$$\frac{(x + 3)^2 + (x - 3)^2}{(x - 3)(x + 3)} = \frac{5}{2}$$
Using algebra rule $(a+b)^2 + (a-b)^2 = 2(a^2 + b^2)$ in numerator:
$$\frac{2(x^2 + 3^2)}{x^2 - 3^2} = \frac{5}{2}$$ $$\implies \frac{2(x^2 + 9)}{x^2 - 9} = \frac{5}{2}$$ $$\implies 4(x^2 + 9) = 5(x^2 - 9)$$ $$\implies 4x^2 + 36 = 5x^2 - 45$$ $$\implies 5x^2 - 4x^2 = 36 + 45$$ $$\implies x^2 = 81 \implies x = \pm \sqrt{81}$$ $$\implies x = \pm 9$$

Roots: $x = 9$ and $x = -9$