Samacheer Kalvi Class 9 Maths Solutions Chapter 3 Algebra Exercise 3.14

 

Question 1. The sum of a two digit number and the number formed by interchanging the digits is 110. If 10 is subtracted from the first number, the new number is 4 more than 5 times the sums of the digits of the first number. Find the first number.
Answer: Let the ten's digit of the original number be \( x \) and the unit digit be \( y \).
So, the original number is \( 10x + y \).
When the digits are interchanged, the new number is \( 10y + x \).
According to the first condition, the sum of the original number and the interchanged number is 110:
\( (10x + y) + (10y + x) = 110 \)
\( 11x + 11y = 110 \)
Divide the entire equation by 11:
\( x + y = 10 \rightarrow (1) \)

According to the second condition, if 10 is subtracted from the first number, the new number is 4 more than 5 times the sum of its digits:
\( (10x + y) - 10 = 5(x + y) + 4 \)
\( 10x + y - 10 = 5x + 5y + 4 \)
Rearrange the terms to form another linear equation:
\( 10x - 5x + y - 5y = 4 + 10 \)
\( 5x - 4y = 14 \rightarrow (2) \)

Now, we have a system of two linear equations:
1. \( x + y = 10 \)
2. \( 5x - 4y = 14 \)

Multiply equation (1) by 4 to make the coefficient of \( y \) opposite:
\( 4(x + y) = 4(10) \)
\( 4x + 4y = 40 \rightarrow (3) \)

Add equation (2) and equation (3):
\( (5x - 4y) + (4x + 4y) = 14 + 40 \)
\( 9x = 54 \)
Solve for \( x \):
\( x = \frac{54}{9} \)
\( x = 6 \)

Substitute the value of \( x = 6 \) into equation (1):
\( 6 + y = 10 \)
\( y = 10 - 6 \)
\( y = 4 \)

The ten's digit is 6 and the unit digit is 4.
The original number is \( 10x + y = 10(6) + 4 = 60 + 4 = 64 \). This problem showcases how word problems can be translated into solvable algebraic equations.
In simple words: We set up two math equations from the problem's clues about the number and its digits. Then, we solved these equations to find the digits and build the original number, which turned out to be 64.

🎯 Exam Tip: Always define your variables clearly (e.g., ten's digit as x, unit's digit as y). Carefully translate each sentence of the word problem into a mathematical equation.

 

Question 2. The sum of the numerator and denominator of a fraction is 12. If the denominator is increased by 3, the fraction becomes \( \frac{1}{2} \). Find the fraction.
Answer: Let the numerator of the fraction be \( x \) and the denominator be \( y \).
So, the fraction is \( \frac{x}{y} \).

According to the first condition, the sum of the numerator and denominator is 12:
\( x + y = 12 \rightarrow (1) \)

According to the second condition, if the denominator is increased by 3, the fraction becomes \( \frac{1}{2} \):
\( \frac{x}{y + 3} = \frac{1}{2} \)
Cross-multiply to simplify this equation:
\( 2x = 1(y + 3) \)
\( 2x = y + 3 \)
Rearrange the terms to form another linear equation:
\( 2x - y = 3 \rightarrow (2) \)

Now, we have a system of two linear equations:
1. \( x + y = 12 \)
2. \( 2x - y = 3 \)

Add equation (1) and equation (2). The \( y \) terms will cancel out:
\( (x + y) + (2x - y) = 12 + 3 \)
\( 3x = 15 \)
Solve for \( x \):
\( x = \frac{15}{3} \)
\( x = 5 \)

Substitute the value of \( x = 5 \) into equation (1):
\( 5 + y = 12 \)
\( y = 12 - 5 \)
\( y = 7 \)

So, the numerator is 5 and the denominator is 7.
The required fraction is \( \frac{5}{7} \). This method of solving word problems with two variables is fundamental in algebra.
In simple words: We used two pieces of information from the problem to create two simple math equations. By solving these equations together, we found the top and bottom numbers of the fraction, giving us the answer \( \frac{5}{7} \).

🎯 Exam Tip: Remember that fractions are ratios. Convert word problems involving fractions into linear equations using variables for the numerator and denominator.

 

Question 3. ABCD is a cyclic quadrilateral such that \( \angle A = (4y + 20)^\circ \), \( \angle B = (3y -5)^\circ \), \( \angle C = (4x)^\circ \) and \( \angle D = (7x + 5)^\circ \). Find the four angles.
Answer:

We know that in a cyclic quadrilateral, the sum of opposite angles is \( 180^\circ \).
So, we have two pairs of opposite angles:
1. \( \angle A + \angle C = 180^\circ \)
2. \( \angle B + \angle D = 180^\circ \)

Using the first pair of opposite angles:
\( (4y + 20)^\circ + (4x)^\circ = 180^\circ \)
\( 4x + 4y + 20 = 180 \)
\( 4x + 4y = 180 - 20 \)
\( 4x + 4y = 160 \)
Divide the entire equation by 4:
\( x + y = 40 \rightarrow (1) \)

Using the second pair of opposite angles:
\( (3y - 5)^\circ + (7x + 5)^\circ = 180^\circ \)
\( 7x + 3y - 5 + 5 = 180 \)
\( 7x + 3y = 180 \rightarrow (2) \)

Now we have a system of two linear equations:
1. \( x + y = 40 \)
2. \( 7x + 3y = 180 \)

Multiply equation (1) by 3 to eliminate \( y \):
\( 3(x + y) = 3(40) \)
\( 3x + 3y = 120 \rightarrow (3) \)

Subtract equation (3) from equation (2):
\( (7x + 3y) - (3x + 3y) = 180 - 120 \)
\( 4x = 60 \)
Solve for \( x \):
\( x = \frac{60}{4} \)
\( x = 15 \)

Substitute the value of \( x = 15 \) into equation (1):
\( 15 + y = 40 \)
\( y = 40 - 15 \)
\( y = 25 \)

Now, substitute the values of \( x \) and \( y \) back into the angle expressions to find the measure of each angle:

For \( \angle A \):
\( \angle A = (4y + 20)^\circ = (4(25) + 20)^\circ = (100 + 20)^\circ = 120^\circ \)

For \( \angle B \):
\( \angle B = (3y - 5)^\circ = (3(25) - 5)^\circ = (75 - 5)^\circ = 70^\circ \)

For \( \angle C \):
\( \angle C = (4x)^\circ = (4(15))^\circ = 60^\circ \)

For \( \angle D \):
\( \angle D = (7x + 5)^\circ = (7(15) + 5)^\circ = (105 + 5)^\circ = 110^\circ \)

So, the four angles are \( \angle A = 120^\circ \), \( \angle B = 70^\circ \), \( \angle C = 60^\circ \), and \( \angle D = 110^\circ \). It's important to remember that these properties of cyclic quadrilaterals make solving such problems straightforward.
In simple words: Since it's a cyclic quadrilateral, we know that opposite angles add up to 180 degrees. Using this rule, we made two equations with \( x \) and \( y \), solved them to find \( x=15 \) and \( y=25 \), and then put these numbers back into the angle formulas to get all four angles: 120°, 70°, 60°, and 110°.

🎯 Exam Tip: Always remember the key property of cyclic quadrilaterals: opposite angles sum to \( 180^\circ \). This is crucial for setting up the correct equations.

 

Question 4. On selling a T.V. at 5% gain and a fridge at 10% gain, a shopkeeper gains Rs 2000. But if he sells the T.V. at 10% gain and the fridge at 5% loss, he gains Rs.1500 on the transaction. Find the actual price of the T.V. and the fridge.
Answer: Let the cost price of the TV be Rs \( x \) and the cost price of the fridge be Rs \( y \).

From the first condition:
The shopkeeper sells the TV at 5% gain and the fridge at 10% gain, making a total gain of Rs 2000.
Gain on TV \( = \frac{5}{100} \times x \)
Gain on fridge \( = \frac{10}{100} \times y \)
Total gain: \( \frac{5x}{100} + \frac{10y}{100} = 2000 \)
To simplify, multiply the entire equation by 100:
\( 5x + 10y = 200000 \)
Divide the entire equation by 5:
\( x + 2y = 40000 \rightarrow (1) \)

From the second condition:
The shopkeeper sells the TV at 10% gain and the fridge at 5% loss, making a total gain of Rs 1500.
Gain on TV \( = \frac{10}{100} \times x \)
Loss on fridge \( = \frac{5}{100} \times y \)
Total gain (gain - loss): \( \frac{10x}{100} - \frac{5y}{100} = 1500 \)
To simplify, multiply the entire equation by 100:
\( 10x - 5y = 150000 \)
Divide the entire equation by 5:
\( 2x - y = 30000 \rightarrow (2) \)

Now we have a system of two linear equations:
1. \( x + 2y = 40000 \)
2. \( 2x - y = 30000 \)

Multiply equation (2) by 2 to eliminate \( y \):
\( 2(2x - y) = 2(30000) \)
\( 4x - 2y = 60000 \rightarrow (3) \)

Add equation (1) and equation (3):
\( (x + 2y) + (4x - 2y) = 40000 + 60000 \)
\( 5x = 100000 \)
Solve for \( x \):
\( x = \frac{100000}{5} \)
\( x = 20000 \)

Substitute the value of \( x = 20000 \) into equation (1):
\( 20000 + 2y = 40000 \)
\( 2y = 40000 - 20000 \)
\( 2y = 20000 \)
Solve for \( y \):
\( y = \frac{20000}{2} \)
\( y = 10000 \)

Therefore, the actual price of the TV is Rs 20,000, and the actual price of the fridge is Rs 10,000. These kinds of profit and loss problems are common in real-world business calculations.
In simple words: We used the information about the shopkeeper's gains and losses to write two math equations. By solving these equations, we found that the TV cost Rs 20,000 and the fridge cost Rs 10,000.

🎯 Exam Tip: Be careful with profit (gain) and loss percentages. A loss should be represented with a negative sign in the equation for total profit.

 

Question 5. Two numbers are in the ratio 5 : 6. If 8 is subtracted from each of the numbers, the ratio becomes 4 : 5. Find the numbers.
Answer: Let the two numbers be \( x \) and \( y \).

According to the first condition, the numbers are in the ratio 5 : 6:
\( \frac{x}{y} = \frac{5}{6} \)
Cross-multiply to get a linear equation:
\( 6x = 5y \)
\( 6x - 5y = 0 \rightarrow (1) \)

According to the second condition, if 8 is subtracted from each number, the ratio becomes 4 : 5:
\( \frac{x - 8}{y - 8} = \frac{4}{5} \)
Cross-multiply to get another linear equation:
\( 5(x - 8) = 4(y - 8) \)
\( 5x - 40 = 4y - 32 \)
Rearrange the terms:
\( 5x - 4y = -32 + 40 \)
\( 5x - 4y = 8 \rightarrow (2) \)

Now we have a system of two linear equations:
1. \( 6x - 5y = 0 \)
2. \( 5x - 4y = 8 \)

To eliminate one variable, let's multiply equation (1) by 4 and equation (2) by 5:
Multiply equation (1) by 4:
\( 4(6x - 5y) = 4(0) \)
\( 24x - 20y = 0 \rightarrow (3) \)

Multiply equation (2) by 5:
\( 5(5x - 4y) = 5(8) \)
\( 25x - 20y = 40 \rightarrow (4) \)

Subtract equation (3) from equation (4):
\( (25x - 20y) - (24x - 20y) = 40 - 0 \)
\( x = 40 \)

Substitute the value of \( x = 40 \) into equation (1):
\( 6(40) - 5y = 0 \)
\( 240 - 5y = 0 \)
\( 240 = 5y \)
Solve for \( y \):
\( y = \frac{240}{5} \)
\( y = 48 \)

Therefore, the two numbers are 40 and 48. Verifying these numbers with the original ratios helps confirm the solution.
In simple words: We used the two given ratio conditions to set up two equations. By solving these equations for \( x \) and \( y \), we found the two numbers to be 40 and 48, which perfectly match the original ratios.

🎯 Exam Tip: When dealing with ratios like \( a:b \), always write it as a fraction \( \frac{a}{b} \) to easily form an equation through cross-multiplication.

 

Question 6. 4 Indians and 4 Chinese can do a piece of work in 3 days. While 2 Indian and 5 Chinese can finish it in 4 days. How long would it take for 1 Indian to do it? How long would it take for 1 Chinese to do it?
Answer: Let \( x \) be the number of days it takes for 1 Indian to complete the work alone.
Let \( y \) be the number of days it takes for 1 Chinese to complete the work alone.

The amount of work done by 1 Indian in one day is \( \frac{1}{x} \).
The amount of work done by 1 Chinese in one day is \( \frac{1}{y} \).

From the first condition:
4 Indians and 4 Chinese can do the work in 3 days. This means the total work done by them in one day is \( \frac{1}{3} \).
Work done by 4 Indians in one day \( = \frac{4}{x} \).
Work done by 4 Chinese in one day \( = \frac{4}{y} \).
So, \( \frac{4}{x} + \frac{4}{y} = \frac{1}{3} \rightarrow (1) \)

From the second condition:
2 Indians and 5 Chinese can finish the work in 4 days. This means the total work done by them in one day is \( \frac{1}{4} \).
Work done by 2 Indians in one day \( = \frac{2}{x} \).
Work done by 5 Chinese in one day \( = \frac{5}{y} \).
So, \( \frac{2}{x} + \frac{5}{y} = \frac{1}{4} \rightarrow (2) \)

To solve this system of equations, let's substitute \( a = \frac{1}{x} \) and \( b = \frac{1}{y} \).
Equation (1) becomes: \( 4a + 4b = \frac{1}{3} \)
Multiply by 3 to clear the fraction:
\( 12a + 12b = 1 \rightarrow (3) \)

Equation (2) becomes: \( 2a + 5b = \frac{1}{4} \)
Multiply by 4 to clear the fraction:
\( 8a + 20b = 1 \rightarrow (4) \)

Now we have a simpler system of linear equations:
1. \( 12a + 12b = 1 \)
2. \( 8a + 20b = 1 \)

To eliminate \( a \), multiply equation (3) by 2 and equation (4) by 3:
Multiply equation (3) by 2:
\( 2(12a + 12b) = 2(1) \)
\( 24a + 24b = 2 \rightarrow (5) \)

Multiply equation (4) by 3:
\( 3(8a + 20b) = 3(1) \)
\( 24a + 60b = 3 \rightarrow (6) \)

Subtract equation (5) from equation (6):
\( (24a + 60b) - (24a + 24b) = 3 - 2 \)
\( 36b = 1 \)
Solve for \( b \):
\( b = \frac{1}{36} \)

Substitute the value of \( b = \frac{1}{36} \) into equation (3):
\( 12a + 12(\frac{1}{36}) = 1 \)
\( 12a + \frac{1}{3} = 1 \)
\( 12a = 1 - \frac{1}{3} \)
\( 12a = \frac{2}{3} \)
Solve for \( a \):
\( a = \frac{2}{3 \times 12} = \frac{2}{36} = \frac{1}{18} \)

Now, we convert \( a \) and \( b \) back to \( x \) and \( y \):
Since \( a = \frac{1}{x} \):
\( \frac{1}{x} = \frac{1}{18} \implies x = 18 \)

Since \( b = \frac{1}{y} \):
\( \frac{1}{y} = \frac{1}{36} \implies y = 36 \)

So, it would take 1 Indian 18 days to do the work alone, and it would take 1 Chinese 36 days to do the work alone. These problems are good examples of applying inverse proportionality to work rates.
In simple words: We figured out how much work one Indian and one Chinese person do in a single day. Then, using the information about how long groups take to finish the work, we set up two equations. Solving these equations showed that one Indian takes 18 days and one Chinese takes 36 days to finish the entire job alone.

🎯 Exam Tip: Always remember that "work done per day" is the reciprocal of "total days to complete work". This conversion is key to solving work-rate problems.