Applications of Algebraic Equations in Word Problems

1. Read and Understand: We are looking for two numbers. Their sum is given as 25, and we know the relationship between their values.

2. Define Variables: Let the smaller of the two numbers be represented by the variable $x$. The problem states that the other number is 5 more than the smaller one. So, the larger number can be represented as $x + 5$.

3. Translate to Equations: The sum of the two numbers is 25. This translates to the equation:

$$ (\text{Smaller number}) + (\text{Larger number}) = 25 $$ $$ x + (x + 5) = 25 \quad \text{... (i)} $$

4. Solve the Equations: Now, we solve the linear equation for $x$.

$$ x + x + 5 = 25 $$

Combine like terms:

$$ 2x + 5 = 25 $$

Subtract 5 from both sides to isolate the term with $x$:

$$ 2x = 20 $$

Divide both sides by 2 to solve for $x$:

$$ x = 10 $$

So, the smaller number is $x = 10$. The larger number is $x + 5 = 10 + 5 = 15$.

5. Check and Interpret: The two numbers are 10 and 15. Let's check the conditions in the original problem:

• Is the sum of the two numbers 25? $10 + 15 = 25$. Yes, this is correct.
• Is one number 5 more than the other? 15 is 5 more than 10. Yes, this is correct.

The solution is consistent with the problem statement. The two numbers are 10 and 15.

1. Read and Understand: We need to find the current ages of two people: a father and his son. We are given how their current ages are related and how their ages will be related in 10 years.

2. Define Variables: Let the son's current age be $x$ years. The father is 30 years older than the son, so the father's current age is $x + 30$ years.

Now, consider their ages in 10 years:

Son's age in 10 years will be $x + 10$ years.

Father's age in 10 years will be $(x + 30) + 10 = x + 40$ years.

3. Translate to Equations: The problem states that in 10 years, the father's age will be three times the son's age. This translates to the equation:

$$ (\text{Father's age in 10 years}) = 3 \times (\text{Son's age in 10 years}) $$ $$ x + 40 = 3(x + 10) \quad \text{... (ii)} $$

4. Solve the Equations: Now, we solve the linear equation for $x$.

$$ x + 40 = 3(x + 10) $$

Distribute the 3 on the right side:

Collect terms involving $x$ on one side and constant terms on the other side. Subtract $x$ from both sides and subtract 30 from both sides:

$$ 10 = 2x $$

Divide both sides by 2 to solve for $x$:

$$ \frac{10}{2} = \frac{2x}{2} $$ $$ x = 5 $$

So, the son's current age is $x = 5$ years. The father's current age is $x + 30 = 5 + 30 = 35$ years.

5. Check and Interpret: Let's check the conditions in the original problem with the ages Son: 5 years, Father: 35 years.

• Is the father currently 30 years older than the son? $35 - 5 = 30$. Yes, correct.
• In 10 years, will the father's age be three times the son's age? In 10 years, the son will be $5 + 10 = 15$ years old, and the father will be $35 + 10 = 45$ years old. Is $45 = 3 \times 15$? Yes, $45 = 45$, correct.

The solution is consistent. The current ages are Son: 5 years, Father: 35 years.

1. Read and Understand: We need to find the price per kilogram for apples and the price per kilogram for grapes. We are given two pieces of information about the total cost for different quantities of apples and grapes.

2. Define Variables: Let the cost of 1 kg of apples be $\textsf{₹}x$. Let the cost of 1 kg of grapes be $\textsf{₹}y$.

3. Translate to Equations:

The first condition states that the cost of 2 kg of apples and 1 kg of grapes is $\textsf{₹}160$. This translates to:

$$ 2 \times (\text{Cost per kg of apples}) + 1 \times (\text{Cost per kg of grapes}) = 160 $$

The second condition states that the cost of 4 kg of apples and 2 kg of grapes is $\textsf{₹}300$. This translates to:

$$ 4 \times (\text{Cost per kg of apples}) + 2 \times (\text{Cost per kg of grapes}) = 300 $$

We now have a system of two linear equations in two variables $x$ and $y$:

$$ 2x + y = 160 $$ $$ 4x + 2y = 300 $$

4. Solve the Equations: We can use the elimination method to solve this system. Let's try to eliminate $y$. The coefficient of $y$ in the first equation is 1, and in the second equation is 2. Multiply equation (iii) by 2 to make the coefficient of $y$ equal to 2 in both equations:

$$ 2 \times (2x + y) = 2 \times 160 $$

Now we have the system:

$$ 4x + 2y = 320 \quad \text{... (v)} $$ $$ 4x + 2y = 300 \quad \text{... (iv)} $$

Subtract equation (iv) from equation (v) to eliminate $x$ and $y$:

$$ (4x + 2y) - (4x + 2y) = 320 - 300 $$ $$ 4x + 2y - 4x - 2y = 20 $$ $$ 0 = 20 $$

This result, $0 = 20$, is a contradiction. This means that there are no values of $x$ and $y$ that can satisfy both equations simultaneously. The system of equations is inconsistent.

5. Check and Interpret: The inconsistent result indicates that the scenario described in the problem is mathematically impossible under the assumption of constant pricing. There are no fixed prices for apples and grapes that would make both statements true. This might suggest an error in the reported costs in the original problem statement.

Answer: The system of equations is inconsistent, which means there is no solution for the cost per kg of apples and grapes that satisfies both conditions. The problem as stated has no solution.

1. Read and Understand: Same goal - find the cost per kg for apples and grapes.

2. Define Variables: Same variables: Let the cost of 1 kg of apples be $\textsf{₹}x$ and the cost of 1 kg of grapes be $\textsf{₹}y$.

3. Translate to Equations:

From the first condition: $2x + y = 160$ $\quad \text{... (iii)'}$

From the second condition: $4x + 3y = 420$ $\quad \text{... (iv)'}$

We have the system:

$$ 2x + y = 160 $$ $$ 4x + 3y = 420 $$

4. Solve the Equations: We can use the elimination method again. Multiply equation (iii)' by 2 to eliminate $x$:

$$ 2 \times (2x + y) = 2 \times 160 $$

Now, subtract equation (v)' from equation (iv)' to eliminate $x$:

$$ (4x + 3y) - (4x + 2y) = 420 - 320 $$ $$ 4x + 3y - 4x - 2y = 100 $$ $$ y = 100 $$

Now that we have the value of $y$, substitute $y=100$ into equation (iii)' (or (iv)') to solve for $x$. Using equation (iii)' since it's simpler:

$$ 2x + y = 160 $$ $$ 2x + 100 = 160 $$

Subtract 100 from both sides:

$$ 2x = 160 - 100 $$ $$ 2x = 60 $$

Divide both sides by 2:

$$ x = \frac{60}{2} $$ $$ x = 30 $$

So, the cost of 1 kg of apples is $\textsf{₹}30$, and the cost of 1 kg of grapes is $\textsf{₹}100$.

5. Check and Interpret: Let's check if these values satisfy the original conditions:

• Condition 1: Cost of 2 kg apples + 1 kg grapes = $2(30) + 1(100) = 60 + 100 = 160$. Correct.
• Condition 2: Cost of 4 kg apples + 3 kg grapes = $4(30) + 3(100) = 120 + 300 = 420$. Correct.

The solution is consistent with the problem statement. The cost of apples is $\textsf{₹}30$ per kg and the cost of grapes is $\textsf{₹}100$ per kg.

1. Read and Understand: We have a rectangle and a triangle. The problem asks for the dimensions (length and breadth) of the rectangle. We are given how the breadth relates to the length and how the area of the rectangle relates to the area of the triangle. The triangle's base is related to the rectangle's breadth, and its height is given.

2. Define Variables: Let the length of the rectangular park be $l$ metres. The breadth is 3 m less than the length, so the breadth is $(l - 3)$ metres.

For the triangular park: its base is equal to the breadth of the rectangle, so the base is $(l - 3)$ metres. Its height is given as 12 metres.

3. Translate to Equation:

Calculate the areas in terms of $l$:

Area of the rectangular park $= \text{length} \times \text{breadth} = l \times (l - 3) = l^2 - 3l$ square metres.

Area of the triangular park $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times (l - 3) \times 12$ square metres.

$$ \text{Area of triangle} = 6 \times (l - 3) = 6l - 18 \text{ square metres.} $$

The problem states that the area of the rectangular park is 4 square metres more than the area of the triangular park. Translate this relationship into an equation:

$$ (\text{Area of rectangle}) = (\text{Area of triangle}) + 4 $$ $$ l^2 - 3l = (6l - 18) + 4 $$ $$ l^2 - 3l = 6l - 14 $$

Now, rearrange this equation to bring all terms to one side and get it into the standard quadratic form $al^2 + bl + c = 0$:

$$ l^2 - 3l - 6l + 14 = 0 $$ $$ l^2 - 9l + 14 = 0 \quad \text{... (vii)} $$

4. Solve the Quadratic Equation: We need to solve the equation $l^2 - 9l + 14 = 0$. This is a quadratic equation with $a=1, b=-9, c=14$. We can try to solve this by factorization.

We look for two numbers that multiply to $c=14$ and add up to $b=-9$. The pairs of factors of 14 are (1, 14), (2, 7), (-1, -14), (-2, -7). The pair that sums to -9 is -2 and -7.

So, we can factor the quadratic expression as $(l - 2)(l - 7) = 0$.

$$ (l - 2)(l - 7) = 0 $$

For this product to be zero, at least one of the factors must be zero. This gives two possible solutions for $l$:

$$ l - 2 = 0 \implies l = 2 $$ $$ l - 7 = 0 \implies l = 7 $$

5. Check and Interpret: We have two potential values for the length $l$: 2 m and 7 m. We must check if these values make sense in the context of the problem, particularly for the dimensions of the rectangle.

Case 1: If the length $l = 2$ metres.

The breadth is $b = l - 3 = 2 - 3 = -1$ metre. A negative length or breadth is not physically possible for a real park. So, $l=2$ is an extraneous solution and is not valid for this problem.

Case 2: If the length $l = 7$ metres.

The breadth is $b = l - 3 = 7 - 3 = 4$ metres. A length of 7 m and a breadth of 4 m are positive and valid dimensions.

Let's check if these dimensions satisfy the area condition:

Area of the rectangular park $= l \times b = 7 \times 4 = 28$ square metres. Base of the triangular park $= \text{breadth of rectangle} = 4$ metres. Height of triangle $= 12$ metres. Area of the triangular park $= \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \times 12 = 2 \times 12 = 24$ square metres.

Is the area of the rectangle 4 square metres more than the area of the triangle? $28 = 24 + 4$. Yes, this condition is satisfied.

The valid dimensions for the rectangular park are length 7 m and breadth 4 m.

Answer: The length of the rectangular park is 7 m and its breadth is 4 m.

1. Read and Understand: We need to find the original constant speed of the train. We are given the distance traveled and how the time taken changes if the speed is reduced.

2. Define Variables: Let the uniform speed of the train be $v$ km/h. We know the distance $D = 480$ km. The fundamental relationship is $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.

Original time taken for the journey, $T_1 = \frac{D}{v} = \frac{480}{v}$ hours.

If the speed had been 8 km/h less, the new speed would be $(v - 8)$ km/h.

The new time taken for the same distance would be $T_2 = \frac{D}{v - 8} = \frac{480}{v - 8}$ hours.

Note that speed must be positive, so $v > 0$. Also, the reduced speed must be positive for the journey to be completed, so $v - 8 > 0$, which means $v > 8$. So, we expect a solution $v > 8$.

3. Translate to Equation: The problem states that the new time taken ($T_2$) is 3 hours more than the original time taken ($T_1$). This gives the equation:

$$ T_2 = T_1 + 3 $$ $$ \frac{480}{v - 8} = \frac{480}{v} + 3 \quad \text{... (viii)} $$

4. Solve the Quadratic Equation: We need to solve this equation for $v$. First, clear the denominators by multiplying the entire equation by the least common multiple of the denominators, which is $v(v-8)$. Since we know $v > 8$, $v$ and $v-8$ are non-zero, so this multiplication is valid.

$$ v(v-8) \left( \frac{480}{v - 8} \right) = v(v-8) \left( \frac{480}{v} + 3 \right) $$

Distribute $v(v-8)$ on the right side:

$$ 480v = v(v-8) \frac{480}{v} + v(v-8)3 $$

Simplify:

$$ 480v = 480(v-8) + 3v(v-8) $$

Expand the right side:

$$ 480v = 480v - 480 \times 8 + 3v^2 - 24v $$ $$ 480v = 480v - 3840 + 3v^2 - 24v $$

Move all terms to one side to get the quadratic equation in standard form $av^2 + bv + c = 0$. Subtract $480v$ from both sides:

$$ 0 = 3v^2 - 24v - 3840 $$

Divide the entire equation by 3 to simplify the coefficients:

$$ \frac{0}{3} = \frac{3v^2}{3} - \frac{24v}{3} - \frac{3840}{3} $$ $$ 0 = v^2 - 8v - 1280 \quad \text{... (ix)} $$

We now solve the quadratic equation $v^2 - 8v - 1280 = 0$ for $v$. This is in the form $av^2 + bv + c = 0$ with $a=1, b=-8, c=-1280$. Let's use the quadratic formula $v = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$.

Calculate the discriminant $\Delta = b^2 - 4ac$:

$$ \Delta = (-8)^2 - 4(1)(-1280) $$ $$ \Delta = 64 - (-5120) $$ $$ \Delta = 64 + 5120 $$ $$ \Delta = 5184 $$

The discriminant is positive, so there are two distinct real roots.

Now find the square root of the discriminant: $\sqrt{5184}$. We can use prime factorization or estimation. $70^2 = 4900$ and $80^2 = 6400$. The number ends in 4, so its square root must end in 2 or 8. Let's try 72. $72 \times 72 = 5184$.

$$ \sqrt{5184} = 72 $$

Apply the quadratic formula:

$$ v = \frac{-(-8) \pm \sqrt{5184}}{2(1)} $$ $$ v = \frac{8 \pm 72}{2} $$

This gives two potential solutions for $v$:

$$ v_1 = \frac{8 + 72}{2} = \frac{80}{2} = 40 $$ $$ v_2 = \frac{8 - 72}{2} = \frac{-64}{2} = -32 $$

5. Check and Interpret: The variable $v$ represents the speed of the train. Speed in this context must be a positive value. Therefore, $v = -32$ km/h is not a valid solution for this word problem.

Consider the other solution, $v = 40$ km/h. This is a positive value. We also established earlier that $v > 8$ for the reduced speed to be positive, and $40 > 8$, which is true.

Let's check if $v=40$ satisfies the time condition:

Original speed $= 40$ km/h. Original time $T_1 = \frac{480 \text{ km}}{40 \text{ km/h}} = 12$ hours. Reduced speed $= 40 - 8 = 32$ km/h. New time $T_2 = \frac{480 \text{ km}}{32 \text{ km/h}} = \frac{480}{32}$ hours. $$ \frac{480}{32} = \frac{\cancel{16} \times 30}{\cancel{16} \times 2} = \frac{30}{2} = 15 \text{ hours.} $$

The new time is 15 hours. Is this 3 hours more than the original time (12 hours)? $15 = 12 + 3$. Yes, this condition is met.

The uniform speed of the train is 40 km/h.

Answer: The uniform speed of the train is 40 km/h.

1. Read and Understand: We are mixing two types of pulses with different prices per kg to create a specific total quantity of mixture with a target average price per kg. We need to determine how much of each type of pulse is needed.

2. Define Variables: Let $x$ be the quantity (in kg) of the pulse variety costing $\textsf{₹}50$ per kg.

Let $y$ be the quantity (in kg) of the pulse variety costing $\textsf{₹}75$ per kg.

3. Translate to Equations:

Equation based on total quantity: The total quantity of the mixture is 50 kg. The sum of the quantities of the two pulse varieties must equal this total.

$$ x \text{ kg} + y \text{ kg} = 50 \text{ kg} $$

Equation based on total cost: The total cost of the mixture is the total quantity (50 kg) multiplied by the cost per kg of the mixture ($\textsf{₹}60$). So, total cost = $50 \text{ kg} \times \textsf{₹}\$60/\text{kg} = \textsf{₹}\$3000$. The total cost of the mixture is also the sum of the costs of the individual quantities of each pulse variety:

$$ (\text{Cost of } x \text{ kg of } \textsf{₹}50/\text{kg pulse}) + (\text{Cost of } y \text{ kg of } \textsf{₹}75/\text{kg pulse}) = \text{Total cost} $$ $$ (x \text{ kg} \times \textsf{₹}\$50/\text{kg}) + (y \text{ kg} \times \textsf{₹}\$75/\text{kg}) = \textsf{₹}\$3000 $$ $$ 50x + 75y = 3000 $$

This equation can be simplified by dividing all terms by 25:

$$ \frac{50x}{25} + \frac{75y}{25} = \frac{3000}{25} $$ $$ 2x + 3y = 120 \quad \text{... (xii)} $$

We now have a system of two linear equations in two variables:

$$ x + y = 50 $$ $$ 2x + 3y = 120 $$

4. Solve the Equations: We can use either the substitution or elimination method. Let's use substitution. From equation (xi), we can express $x$ in terms of $y$:

$$ x = 50 - y $$

Substitute this expression for $x$ into equation (xii):

$$ 2x + 3y = 120 $$ $$ 2(50 - y) + 3y = 120 $$

Distribute the 2:

$$ 100 - 2y + 3y = 120 $$

Combine the $y$ terms:

$$ 100 + y = 120 $$

Subtract 100 from both sides to solve for $y$:

$$ y = 120 - 100 $$ $$ y = 20 $$

Now substitute the value of $y=20$ back into the expression for $x$ ($x = 50 - y$):

$$ x = 50 - 20 $$ $$ x = 30 $$

So, the quantity of the first pulse variety is 30 kg, and the quantity of the second pulse variety is 20 kg.

5. Check and Interpret:

Quantity of $\textsf{₹}50/\text{kg}$ pulse = 30 kg. Quantity of $\textsf{₹}75/\text{kg}$ pulse = 20 kg.

Total quantity: $30 \text{ kg} + 20 \text{ kg} = 50$ kg. This matches the given total quantity.

Total cost: Cost of 30 kg at $\textsf{₹}50/\text{kg} = 30 \times 50 = \textsf{₹}1500$. Cost of 20 kg at $\textsf{₹}75/\text{kg} = 20 \times 75 = \textsf{₹}1500$. Total cost $= \textsf{₹}1500 + \textsf{₹}1500 = \textsf{₹}3000$.

Cost of mixture per kg: $\textsf{₹}3000 / 50 \text{ kg} = \textsf{₹}60/\text{kg}$. This matches the given cost of the mixture per kg.

The solution is consistent with the problem statement. The shopkeeper should mix 30 kg of the pulse costing $\textsf{₹}50$ per kg and 20 kg of the pulse costing $\textsf{₹}75$ per kg.

Answer: The quantity of the pulse costing $\textsf{₹}50$ per kg is 30 kg, and the quantity of the pulse costing $\textsf{₹}75$ per kg is 20 kg.

1. Read and Understand: We are given two scenarios of a boat traveling both upstream and downstream, with the total time for each scenario. We need to find the boat's speed in still water and the stream's speed.

2. Define Variables: Let the speed of the boat in still water be $v_b$ km/h.

Let the speed of the stream be $v_s$ km/h.

Speed downstream $= v_b + v_s$ km/h.

Speed upstream $= v_b - v_s$ km/h.

Using Time = Distance / Speed:

• Time to go 30 km upstream $= \frac{30}{v_b - v_s}$ hours.
• Time to go 44 km downstream $= \frac{44}{v_b + v_s}$ hours.
• Time to go 40 km upstream $= \frac{40}{v_b - v_s}$ hours.
• Time to go 55 km downstream $= \frac{55}{v_b + v_s}$ hours.

3. Translate to Equations:

The first condition states that the total time for 30 km upstream and 44 km downstream is 10 hours:

$$ (\text{Time upstream}) + (\text{Time downstream}) = \text{Total time} $$

The second condition states that the total time for 40 km upstream and 55 km downstream is 13 hours:

$$ (\text{Time upstream}) + (\text{Time downstream}) = \text{Total time} $$

This is a system of two equations with variables in the denominators. To transform it into a system of linear equations, we can introduce new variables for the reciprocal terms. Let $u = \frac{1}{v_b - v_s}$ and $w = \frac{1}{v_b + v_s}$. Substituting these into equations (xiii) and (xiv), we get a linear system in $u$ and $w$:

$$ 30u + 44w = 10 \quad \text{... (xiii)'} $$ $$ 40u + 55w = 13 \quad \text{... (xiv)'} $$

4. Solve the Equations: We can use elimination to solve for $u$ and $w$. To eliminate $u$, find the LCM of 30 and 40, which is 120. Multiply equation (xiii)' by 4 and equation (xiv)' by 3:

$$ 4 \times (30u + 44w) = 4 \times 10 \implies 120u + 176w = 40 \quad \text{... (xv)} $$ $$ 3 \times (40u + 55w) = 3 \times 13 \implies 120u + 165w = 39 \quad \text{... (xvi)} $$

Subtract equation (xvi) from equation (xv):

$$ (120u + 176w) - (120u + 165w) = 40 - 39 $$ $$ 120u + 176w - 120u - 165w = 1 $$ $$ 11w = 1 $$ $$ w = \frac{1}{11} $$

Now substitute the value of $w = \frac{1}{11}$ back into equation (xiii)' (or (xiv)') to find $u$. Using equation (xiii)' since the numbers are smaller:

$$ 30u + 44w = 10 $$ $$ 30u + 44 \left(\frac{1}{11}\right) = 10 $$ $$ 30u + 4 = 10 $$

Subtract 4 from both sides:

$$ 30u = 10 - 4 $$ $$ 30u = 6 $$

Divide by 30:

$$ u = \frac{6}{30} = \frac{1}{5} $$

Now that we have the values for $u$ and $w$, substitute back into the definitions $u = \frac{1}{v_b - v_s}$ and $w = \frac{1}{v_b + v_s}$ to find $v_b$ and $v_s$:

$$ u = \frac{1}{v_b - v_s} \implies \frac{1}{v_b - v_s} = \frac{1}{5} \implies v_b - v_s = 5 \quad \text{... (xvii)} $$ $$ w = \frac{1}{v_b + v_s} \implies \frac{1}{v_b + v_s} = \frac{1}{11} \implies v_b + v_s = 11 \quad \text{... (xviii)} $$

We now have a new system of two linear equations in $v_b$ and $v_s$. Add equation (xvii) and equation (xviii):

$$ (v_b - v_s) + (v_b + v_s) = 5 + 11 $$ $$ v_b - v_s + v_b + v_s = 16 $$ $$ 2v_b = 16 $$ $$ v_b = \frac{16}{2} = 8 $$

Substitute the value of $v_b = 8$ into equation (xviii) to find $v_s$:

$$ v_b + v_s = 11 $$ $$ 8 + v_s = 11 $$ $$ v_s = 11 - 8 $$ $$ v_s = 3 $$

So, the speed of the boat in still water is 8 km/h, and the speed of the stream is 3 km/h.

5. Check and Interpret:

Speed of boat in still water = 8 km/h. Speed of stream = 3 km/h.

Speed downstream $= 8 + 3 = 11$ km/h.

Speed upstream $= 8 - 3 = 5$ km/h.

Check condition 1 (30 km upstream and 44 km downstream in 10 hours): Time upstream $= \frac{30}{5} = 6$ hours. Time downstream $= \frac{44}{11} = 4$ hours. Total time $= 6 + 4 = 10$ hours. Correct.

Check condition 2 (40 km upstream and 55 km downstream in 13 hours): Time upstream $= \frac{40}{5} = 8$ hours. Time downstream $= \frac{55}{11} = 5$ hours. Total time $= 8 + 5 = 13$ hours. Correct.

The solution is consistent. The speed of the boat in still water is 8 km/h, and the speed of the stream is 3 km/h.

Answer: The speed of the boat in still water is 8 km/h and the speed of the stream is 3 km/h.

1. Read and Understand: We have two pipes filling a tank. We know their combined filling time and the difference between their individual filling times. We need to find the time each pipe takes to fill the tank alone.

2. Define Variables: Let the time taken by the faster pipe to fill the tank separately be $x$ minutes.

The other pipe takes 5 minutes more than the faster one to fill the tank separately, so the time taken by the slower pipe is $(x + 5)$ minutes.

Based on these times, we can determine their rates:

Rate of the faster pipe = $\frac{1}{x}$ of the tank filled per minute.

Rate of the slower pipe = $\frac{1}{x+5}$ of the tank filled per minute.

The combined time taken by both pipes working together is given as $11\frac{1}{9}$ minutes. Convert this mixed number to an improper fraction: $11\frac{1}{9} = \frac{11 \times 9 + 1}{9} = \frac{99 + 1}{9} = \frac{100}{9}$ minutes.

The combined rate is the reciprocal of the combined time:

$$ \text{Combined rate} = \frac{1}{\text{Combined time}} = \frac{1}{\frac{100}{9}} = \frac{9}{100} \text{ of the tank filled per minute.} $$

We must assume $x > 0$ since time cannot be negative. Also, $x+5 > 0$, which is true if $x > 0$.

3. Translate to Equation: The combined rate of the two pipes working together is the sum of their individual rates.

$$ (\text{Rate of faster pipe}) + (\text{Rate of slower pipe}) = (\text{Combined rate}) $$

This equation involves algebraic fractions. We need to solve for $x$.

Combine the terms on the left side by finding a common denominator, which is $x(x+5)$:

$$ \frac{1 \times (x+5)}{x(x+5)} + \frac{1 \times x}{(x+5)x} = \frac{9}{100} $$ $$ \frac{(x+5) + x}{x(x+5)} = \frac{9}{100} $$ $$ \frac{2x+5}{x^2 + 5x} = \frac{9}{100} $$

Now, cross-multiply to eliminate the denominators:

$$ 100(2x+5) = 9(x^2 + 5x) $$

Distribute the numbers on both sides:

$$ 200x + 500 = 9x^2 + 45x $$

Rearrange all terms to one side to get a standard quadratic equation $ax^2 + bx + c = 0$. Subtract $200x$ and 500 from both sides:

$$ 0 = 9x^2 + 45x - 200x - 500 $$ $$ 0 = 9x^2 - 155x - 500 \quad \text{... (xx)} $$

4. Solve the Quadratic Equation: We need to solve the equation $9x^2 - 155x - 500 = 0$. This is a quadratic equation in the form $ax^2 + bx + c = 0$ with $a=9, b=-155, c=-500$. We will use the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ because factoring might be difficult with these coefficients.

Calculate the discriminant $\Delta = b^2 - 4ac$:

$$ \Delta = (-155)^2 - 4(9)(-500) $$ $$ \Delta = 24025 - (-18000) $$ $$ \Delta = 24025 + 18000 $$ $$ \Delta = 42025 $$

The discriminant is positive, so there are two distinct real roots.

Now find the square root of the discriminant: $\sqrt{42025}$. Let's estimate. $200^2 = 40000$. $210^2 = 44100$. The number ends in 25, so its square root must end in 5. Let's try 205. $205 \times 205 = 42025$.

$$ \sqrt{42025} = 205 $$

Apply the quadratic formula to find the possible values of $x$:

$$ x = \frac{-(-155) \pm \sqrt{42025}}{2(9)} $$ $$ x = \frac{155 \pm 205}{18} $$

This gives two possible solutions for $x$:

$$ x_1 = \frac{155 + 205}{18} = \frac{360}{18} $$ $$ x_1 = 20 $$ $$ x_2 = \frac{155 - 205}{18} = \frac{-50}{18} = -\frac{25}{9} $$

5. Check and Interpret: The variable $x$ represents the time taken by a pipe to fill a tank, which must be a positive value. Therefore, $x = -\frac{25}{9}$ minutes is not a valid solution in this context.

Consider the valid solution $x = 20$ minutes. This is the time taken by the faster pipe.

The time taken by the slower pipe is $x + 5 = 20 + 5 = 25$ minutes.

Let's check if these individual times result in the correct combined time. Rate of faster pipe $= \frac{1}{20}$ tank per minute.

Rate of slower pipe $= \frac{1}{25}$ tank per minute.

Combined rate $= \frac{1}{20} + \frac{1}{25}$. Find a common denominator (LCM of 20 and 25 is 100):

$$ \text{Combined rate} = \frac{1 \times 5}{20 \times 5} + \frac{1 \times 4}{25 \times 4} = \frac{5}{100} + \frac{4}{100} = \frac{9}{100} \text{ tank per minute.} $$

The time taken by both pipes together is the reciprocal of the combined rate:

$$ \text{Time together} = \frac{1}{\text{Combined rate}} = \frac{1}{\frac{9}{100}} = \frac{100}{9} \text{ minutes.} $$

Convert $\frac{100}{9}$ minutes to a mixed number: $100 \div 9 = 11$ with a remainder of 1. So, $\frac{100}{9} = 11\frac{1}{9}$ minutes. This matches the given combined time.

The solution is consistent. The faster pipe fills the tank in 20 minutes, and the slower pipe fills it in 25 minutes.

Answer: The two pipes would fill the tank separately in 20 minutes and 25 minutes.

1. Read and Understand: We are given the new price of an article after a percentage decrease. We need to find the original price before the decrease.

2. Define Variables: Let the original price of the article be $\textsf{₹}x$.

3. Translate to Equation: The price is decreased by 10%. This means the amount of decrease is 10% of the original price. In decimal form, 10% is $0.10$. So, the decrease amount is $0.10x$.

The new price is obtained by subtracting the decrease amount from the original price:

$$ \text{New Price} = \text{Original Price} - \text{Decrease Amount} $$ $$ 450 = x - 0.10x $$

On the right side, we can factor out $x$:

$$ 450 = (1 - 0.10)x $$

4. Solve the Equation: Now, we solve the linear equation $450 = 0.90x$ for $x$. Divide both sides by $0.90$:

$$ x = \frac{450}{0.90} $$

To perform the division, we can rewrite $0.90$ as a fraction or multiply the numerator and denominator by 100 to remove the decimal:

$$ x = \frac{450}{\frac{90}{100}} = 450 \times \frac{100}{90} $$ $$ x = \frac{450 \times 100}{90} $$

Simplify the fraction by canceling common factors (e.g., 90 divides 450 exactly 5 times):

$$ x = \frac{\cancel{450}^{5} \times 100}{\cancel{90}^{1}} $$ $$ x = 5 \times 100 = 500 $$

The original price is $\textsf{₹}500$.

5. Check and Interpret: The original price was $\textsf{₹}500$. A 10% decrease on $\textsf{₹}500$ is $10\%$ of $500 = \frac{10}{100} \times 500 = \frac{1}{10} \times 500 = 50$. The decrease amount is $\textsf{₹}50$. The new price is the original price minus the decrease: $\textsf{₹}500 - \textsf{₹}50 = \textsf{₹}450$. This matches the new price given in the problem. The solution is consistent.

Answer: The original price of the article was $\textsf{₹}500$.

1. Read and Understand: We are looking for a two-digit number. We have two pieces of information: the sum of its digits and how the number changes when 27 is added (the digits swap places).

2. Define Variables: Let the ten's digit of the two-digit number be $t$, and let the unit's digit be $u$. Both $t$ and $u$ must be integers, and $t$ must be a non-zero digit (1-9) for it to be a two-digit number, while $u$ can be any digit from 0-9.

The value of a two-digit number with ten's digit $t$ and unit's digit $u$ is $10 \times t + 1 \times u = 10t + u$.

When the digits are reversed, the new number has $u$ in the ten's place and $t$ in the unit's place. The value of the reversed number is $10 \times u + 1 \times t = 10u + t$.

3. Translate to Equations:

Equation based on the sum of digits: The sum of the digits is 9.

Equation based on adding 27: If 27 is added to the original number, the digits are reversed.

$$ (\text{Original Number}) + 27 = (\text{Reversed Number}) $$

Rearrange equation (xxi) to group terms with $t$ and $u$ on one side and constants on the other:

$$ 10t - t + u - 10u = -27 $$ $$ 9t - 9u = -27 $$

This equation can be simplified by dividing all terms by 9:

$$ \frac{9t}{9} - \frac{9u}{9} = \frac{-27}{9} $$ $$ t - u = -3 \quad \text{... (xxi)'} $$

We now have a system of two linear equations in two variables $t$ and $u$:

$$ t + u = 9 $$ $$ t - u = -3 $$

4. Solve the Equations: We can use the elimination method. Notice that the coefficients of $u$ are $+1$ and $-1$. Adding the two equations will eliminate $u$:

$$ (t + u) + (t - u) = 9 + (-3) $$ $$ t + u + t - u = 9 - 3 $$ $$ 2t = 6 $$

Divide by 2 to solve for $t$:

$$ t = \frac{6}{2} $$ $$ t = 3 $$

Now substitute the value of $t = 3$ into equation (xx) ($t + u = 9$) to find $u$:

$$ 3 + u = 9 $$

Subtract 3 from both sides:

$$ u = 9 - 3 $$ $$ u = 6 $$

The ten's digit is $t = 3$, and the unit's digit is $u = 6$.

The original two-digit number is $10t + u = 10(3) + 6 = 30 + 6 = 36$.

5. Check and Interpret:

The number is 36. The digits are 3 and 6. Their sum is $3 + 6 = 9$. This satisfies the first condition.

If 27 is added to the number: $36 + 27 = 63$.

The reversed number (digits of 36 swapped) is 63. Since $36 + 27 = 63$, the second condition is also satisfied.

The solution is consistent with the problem statement. The number is 36.

Answer: The number is 36.