Long-Answer Word Problems
Solution: Let the smaller positive whole number be $x$.
Therefore, the other number is $x + 3$.
According to the problem statement:
$$x^2 + (x + 3)^2 = 117$$
$$\implies x^2 + x^2 + 6x + 9 = 117$$
$$\implies 2x^2 + 6x + 9 - 117 = 0$$
$$\implies 2x^2 + 6x - 108 = 0$$
Dividing the entire equation by $2$:
$$\implies x^2 + 3x - 54 = 0$$
Factorizing by splitting the middle term ($9 \times -6 = -54$ and $9 - 6 = 3$):
$$\implies x^2 + 9x - 6x - 54 = 0$$
$$\implies x(x + 9) - 6(x + 9) = 0$$
$$\implies (x - 6)(x + 9) = 0$$
• Either $x - 6 = 0 \implies x = 6$
• Or $x + 9 = 0 \implies x = -9$
Since the question explicitly mentions positive whole numbers, $x = -9$ is not acceptable.
Solution: Let the height of the triangle be $x$ meters.
Therefore, the base of the triangle $= (2x + 18)$ meters.
The mathematical formula for the area of a triangle is $\frac{1}{2} \times \text{base} \times \text{height}$:
$$\frac{1}{2} \times (2x + 18) \times x = 360$$
$$\implies \frac{1}{2} \times 2(x + 9) \times x = 360$$
$$\implies x(x + 9) = 360$$
$$\implies x^2 + 9x - 360 = 0$$
Factorizing by splitting the middle term ($24 \times -15 = -360$ and $24 - 15 = 9$):
$$\implies x^2 + 24x - 15x - 360 = 0$$
$$\implies x(x + 24) - 15(x + 24) = 0$$
$$\implies (x - 15)(x + 24) = 0$$
• Either $x - 15 = 0 \implies x = 15$
• Or $x + 24 = 0 \implies x = -24$
Since height cannot be negative, $x = -24$ is rejected.
Solution: Let the positive whole number be $x$.
According to the problem statement:
$$2x^2 - 5x = 3$$
$$\implies 2x^2 - 5x - 3 = 0$$
Splitting the middle term:
$$\implies 2x^2 - 6x + x - 3 = 0$$
$$\implies 2x(x - 3) + 1(x - 3) = 0$$
$$\implies (x - 3)(2x + 1) = 0$$
• Either $x - 3 = 0 \implies x = 3$
• Or $2x + 1 = 0 \implies 2x = -1 \implies x = -\frac{1}{2}$ (not possible for a whole positive number).
Solution: Let the speed of the Jeep car be $x$ km/hr.
Therefore, the speed of the motor car $= (x + 5)$ km/hr.
Time taken by the Jeep to travel 200 km $= \frac{200}{x}$ hours.
Time taken by the motor car to travel 200 km $= \frac{200}{x + 5}$ hours.
According to the problem:
$$\frac{200}{x} - \frac{200}{x + 5} = 2$$
$$\implies 200 \left( \frac{1}{x} - \frac{1}{x + 5} \right) = 2$$
$$\implies 200 \left( \frac{x + 5 - x}{x(x + 5)} \right) = 2$$
$$\implies \frac{1000}{x^2 + 5x} = 2 \implies 2(x^2 + 5x) = 1000$$
$$\implies x^2 + 5x = 500 \implies x^2 + 5x - 500 = 0$$
Splitting the middle term ($25 \times -20 = -500$):
$$\implies x^2 + 25x - 20x - 500 = 0$$
$$\implies x(x + 25) - 20(x + 25) = 0$$
$$\implies (x - 20)(x + 25) = 0$$
• Either $x - 20 = 0 \implies x = 20$
• Or $x + 25 = 0 \implies x = -25$ (rejected because speed cannot be negative).
Speed of the Jeep car is 20 km/hr. Therefore, the speed of the motor car is $20 + 5 = 25$ km/hr.
Solution: Let the length of the land be $x$ meters.
Since Area $= \text{Length} \times \text{Breadth} = 2000$, the breadth $= \frac{2000}{x}$ meters.
The perimeter of a rectangle is $2(\text{Length} + \text{Breadth}) = 180$ m:
$$2\left(x + \frac{2000}{x}\right) = 180$$
$$\implies x + \frac{2000}{x} = 90$$
$$\implies \frac{x^2 + 2000}{x} = 90 \implies x^2 + 2000 = 90x$$
$$\implies x^2 - 90x + 2000 = 0$$
Splitting the middle term ($-50 - 40 = -90$ and $-50 \times -40 = 2000$):
$$\implies x^2 - 50x - 40x + 2000 = 0$$
$$\implies x(x - 50) - 40(x - 50) = 0$$
$$\implies (x - 50)(x - 40) = 0$$
• Either $x = 50$ (Length $= 50$m, Breadth $= \frac{2000}{50} = 40$m)
• Or $x = 40$ (Length $= 40$m, Breadth $= \frac{2000}{40} = 50$m)
Conventionally, length is taken to be greater than breadth.
Solution: Let the unit place digit be $x$.
Then, the tens place digit $= x - 3$.
The original two-digit number $= 10(x - 3) + x = 10x - 30 + x = 11x - 30$.
According to the condition specified:
$$\text{Number} - (\text{Product of digits}) = 15$$
$$\implies (11x - 30) - x(x - 3) = 15$$
$$\implies 11x - 30 - x^2 + 3x - 15 = 0$$
$$\implies -x^2 + 14x - 45 = 0$$
Multiplying by $-1$:
$$\implies x^2 - 14x + 45 = 0$$
$$\implies x^2 - 9x - 5x + 45 = 0$$
$$\implies x(x - 9) - 5(x - 9) = 0 \implies (x - 9)(x - 5) = 0$$
• Either $x = 9$ or $x = 5$. All provide real numbers since tens digit $(x-3)$ will remain positive ($9-3=6$ or $5-3=2$).
Solution: Let the faster pipe take $x$ minutes to fill the tank completely.
Then the slower pipe takes $(x + 5)$ minutes.
In 1 minute, both pipes together can fill:
$$\frac{1}{x} + \frac{1}{x + 5} \text{ part of the tank.}$$
Total time given together is $11\frac{1}{9} = \frac{100}{9}$ minutes. Therefore, total work completed equals 1:
$$\left( \frac{1}{x} + \frac{1}{x + 5} \right) \times \frac{100}{9} = 1$$
$$\implies \left( \frac{x + 5 + x}{x(x + 5)} \right) = \frac{9}{100}$$
$$\implies \frac{2x + 5}{x^2 + 5x} = \frac{9}{100}$$
Cross multiplying:
$$\implies 9(x^2 + 5x) = 100(2x + 5)$$
$$\implies 9x^2 + 45x = 200x + 500$$
$$\implies 9x^2 + 45x - 200x - 500 = 0$$
$$\implies 9x^2 - 155x - 500 = 0$$
Using splitting middle term ($-180 \times 25 = -4500$ and $-180 + 25 = -155$):
$$\implies 9x^2 - 180x + 25x - 500 = 0$$
$$\implies 9x(x - 20) + 25(x - 20) = 0$$
$$\implies (x - 20)(9x + 25) = 0$$
• Either $x - 20 = 0 \implies x = 20$
• Or $9x + 25 = 0 \implies x = -\frac{25}{9}$ (Time cannot be negative, so rejected).
Solution: Let Pijush alone complete the total work in $y$ days.
Then Porna alone takes $(y + 6)$ days.
Combined 1-day work output:
$$\frac{1}{y} + \frac{1}{y + 6} = \frac{1}{4}$$
$$\implies \frac{y + 6 + y}{y(y + 6)} = \frac{1}{4}$$
$$\implies \frac{2y + 6}{y^2 + 6y} = \frac{1}{4}$$
Cross multiplying:
$$\implies y^2 + 6y = 4(2y + 6)$$
$$\implies y^2 + 6y = 8y + 24$$
$$\implies y^2 - 2y - 24 = 0$$
$$\implies y^2 - 6y + 4y - 24 = 0$$
$$\implies y(y - 6) + 4(y - 6) = 0 \implies (y - 6)(y + 4) = 0$$
• Either $y = 6$ or $y = -4$ (rejected).
Pijush takes 6 days. Porna takes $6 + 6 = 12$ days.
Solution: Let the original price of 1 dozen (12 pens) be Rs. $x$.
Reduced price of 1 dozen pens = Rs. $(x - 6)$.
Number of pens bought initially for Rs. 30 $= \frac{30}{x} \times 12$ pens.
Number of pens bought after price drop for Rs. 30 $= \frac{30}{x - 6} \times 12$ pens.
According to the problem context:
$$\frac{360}{x - 6} - \frac{360}{x} = 3$$
Dividing the entire expression by 3:
$$\frac{120}{x - 6} - \frac{120}{x} = 1$$
$$\implies 120 \left( \frac{x - (x - 6)}{x(x - 6)} \right) = 1$$
$$\implies \frac{120 \times 6}{x^2 - 6x} = 1$$
$$\implies x^2 - 6x = 720 \implies x^2 - 6x - 720 = 0$$
Splitting middle terms ($-30 \times 24 = -720$):
$$\implies x^2 - 30x + 24x - 720 = 0$$
$$\implies x(x - 30) + 24(x - 30) = 0 \implies (x - 30)(x + 24) = 0$$
• Either $x = 30$ or $x = -24$ (price cannot be negative).
1. $(x - 3)^2 = x^2 - 6x + 9$ is a quadratic equation. FALSE
(Reason: It is an algebraic identity, valid for all values of x, hence an identity, not an equation)
2. The equation $x^2 = 25$ has only one root. FALSE
(Reason: It yields two distinct solutions, $x = 5$ and $x = -5$)
- If $a = 0$ and $b \neq 0$ in the standard expression $ax^2 + bx + c = 0$, then it reduces to a Linear equation.
- If two roots of a quadratic equation are equal and match $1$, then the explicit equation form is $x^2 - 2x + 1 = 0$.
- The roots of the incomplete equation $x^2 - 6x = 0$ are 0 and 6.
Solution: Substitute root $x = 1$ into the quadratic formula:
$$(1)^2 + a(1) + 3 = 0$$
$$\implies 1 + a + 3 = 0 \implies a + 4 = 0 \implies a = -4$$
Solution: We know that the product of roots in a quadratic equation $x^2 - \text{S}x + \text{P} = 0$ is equal to the constant term.
Here, Product of roots $= 6$.
Given one root $= 2$. Let the other root be $\alpha$.
$$2 \times \alpha = 6 \implies \alpha = 3$$
Solution: For equation $2x^2 + kx + 4 = 0$, dividing by 2 gives $x^2 + \frac{k}{2}x + 2 = 0$.
Product of roots $= \frac{\text{Constant term}}{\text{Coefficient of } x^2} = \frac{4}{2} = 2$.
Given one root $= 2$. Let the second root be $\beta$.
$$2 \times \beta = 2 \implies \beta = 1$$