Samacheer Kalvi Class 8 Maths Solutions Chapter 3 Algebra Exercise 3.7

Question 1. Fill in the blanks:
(i) The solution of the equation \( ax + b = 0 \) is \( -\frac{b}{a} \).
(ii) If a and b are positive integers then the solution of the equation \( ax = b \) has to be always Positive.
(iii) One-sixth of a number when subtracted from the number itself gives 25. The number is 30.
(iv) If the angles of a triangle are in the ratio 2:3:4 then the difference between the greatest and the smallest angle is 40°.
(v) In an equation \( a + b = 23 \). The value of a is 14 then the value of b is 9.

Answer:
(i) To solve the equation \( ax + b = 0 \) for x, we first move b to the right side by subtracting it from both sides. This gives \( ax = -b \). Next, to find x, we divide both sides by a, so \( x = -\frac{b}{a} \). This is a basic step in solving linear equations.
(ii) If a and b are positive numbers, and the equation is \( ax = b \), then \( x = \frac{b}{a} \). When you divide a positive number by another positive number, the result is always positive. Therefore, x will always be a positive value.
(iii) Let the unknown number be \( x \). The problem states that one-sixth of this number (\( \frac{x}{6} \)) when subtracted from the number itself (\( x \)) equals 25. So, the equation is \( x - \frac{x}{6} = 25 \). To solve this, find a common denominator, which is 6. This means \( \frac{6x}{6} - \frac{x}{6} = 25 \), which simplifies to \( \frac{5x}{6} = 25 \). To find \( x \), multiply both sides by 6 and then divide by 5: \( 5x = 25 \times 6 \), so \( 5x = 150 \), and \( x = \frac{150}{5} = 30 \). So the number is 30.
(iv) Let the angles of the triangle be \( 2x \), \( 3x \), and \( 4x \). We know that the sum of angles in any triangle is 180°. So, \( 2x + 3x + 4x = 180^\circ \). Combining these, we get \( 9x = 180^\circ \). Dividing by 9, we find \( x = \frac{180^\circ}{9} = 20^\circ \). Now we can find each angle: \( 2x = 2 \times 20^\circ = 40^\circ \), \( 3x = 3 \times 20^\circ = 60^\circ \), and \( 4x = 4 \times 20^\circ = 80^\circ \). The greatest angle is \( 80^\circ \) and the smallest is \( 40^\circ \). The difference between them is \( 80^\circ - 40^\circ = 40^\circ \). This method helps find specific values from ratios.
(v) We are given the equation \( a + b = 23 \) and that \( a = 14 \). To find b, we substitute the value of a into the equation: \( 14 + b = 23 \). Then, subtract 14 from both sides: \( b = 23 - 14 \). This gives \( b = 9 \). Thus, the value of b is 9.
In simple words: Each part involves solving a basic algebra problem. For (i), move the constant and then divide to find \( x \). For (ii), remember that dividing two positive numbers gives a positive result. For (iii), set up an equation where a number minus its sixth part equals 25, then solve. For (iv), use the fact that triangle angles add up to 180 degrees to find the value of \( x \), then calculate the angles and their difference. For (v), simply replace 'a' with 14 in the equation and solve for 'b'.

🎯 Exam Tip: Always show your working steps clearly, especially for algebraic solutions, as partial credit can be awarded for correct steps even if the final answer is slightly off.

Question 2. Say True or False
(i) "Sum of a number and two times that number is 48” can be written as \( y + 2y = 48 \).
(ii) \( 5(3x + 2) = 3(5x - 7) \) is a linear equation in one variable.
(iii) \( x = 25 \) is the solution of one third of a number is less than 10 the original number.

Answer:
(i) True. If 'y' is the number, then "sum of a number and two times that number" means \( y + 2y \). "is 48" means equals 48. So, \( y + 2y = 48 \) is the correct way to write the statement. This is a direct translation of the English phrase into an algebraic equation.
(ii) True. A linear equation in one variable means an equation where the highest power of the variable (here 'x') is 1, and there is only one variable present. When you expand \( 5(3x + 2) = 3(5x - 7) \), you get \( 15x + 10 = 15x - 21 \). Although the \( 15x \) terms cancel out, the original equation, before simplification, is indeed a linear equation in one variable, x. It simplifies to \( 10 = -21 \), which is a contradiction, meaning there is no solution, but it is still a linear equation.
(iii) False. The statement "one third of a number is less than 10 the original number" seems to be trying to say "one third of a number is 10 less than the original number". If we assume this interpretation, the equation would be \( \frac{x}{3} = x - 10 \). Solving this:
\( \frac{x}{3} = x - 10 \)
Multiply by 3: \( x = 3x - 30 \)
Subtract \( 3x \) from both sides: \( x - 3x = -30 \)
\( -2x = -30 \)
Divide by -2: \( x = 15 \).
Since the solution is \( x = 15 \) and not \( x = 25 \), the statement is false. The initial wording might be slightly ambiguous, but the derived solution shows 25 is not correct for the likely intended meaning.
In simple words: For (i), the equation correctly shows a number plus twice itself. For (ii), the equation is linear because 'x' is not raised to any power higher than one. For (iii), the given solution of \( x = 25 \) is wrong when you correctly set up and solve the problem as "one third of a number is 10 less than the original number", which gives \( x = 15 \).

🎯 Exam Tip: Always read the question carefully, especially for "True or False" statements, and try to rephrase ambiguous sentences into a clear algebraic equation before solving.

Question 3. One number is seven times another. If their difference is 18, find the numbers.
Answer: Let the two numbers be \( x \) and \( y \).
According to the problem, one number is seven times another, so we can write this as \( x = 7y \).
Also, their difference is 18, which means \( x - y = 18 \).
Now, we can substitute the first equation (\( x = 7y \)) into the second equation:
\( 7y - y = 18 \)
This simplifies to:
\( 6y = 18 \)
To find \( y \), divide both sides by 6:
\( y = \frac{18}{6} \)
\( y = 3 \)
Now that we have \( y = 3 \), we can find \( x \) using \( x = 7y \):
\( x = 7 \times 3 \)
\( x = 21 \)
So, the two numbers are 3 and 21.
In simple words: We have two numbers. One is 7 times the other. When you subtract the smaller number from the larger one, you get 18. By using a little algebra, we found the numbers are 3 and 21.

🎯 Exam Tip: When dealing with "one number is N times another" and "their difference is M", always set up two equations and use substitution to solve for both unknown numbers.

Question 4. The sum of three consecutive odd numbers is 75. Which is the largest among them?
Answer: Let the first odd number be \( x \).
Since odd numbers differ by 2, the next consecutive odd number will be \( x + 2 \).
The third consecutive odd number will be \( (x + 2) + 2 = x + 4 \).
The problem states that the sum of these three numbers is 75. So, we set up the equation:
\( x + (x + 2) + (x + 4) = 75 \)
Combine the like terms:
\( 3x + 6 = 75 \)
Now, subtract 6 from both sides:
\( 3x = 75 - 6 \)
\( 3x = 69 \)
Divide by 3 to find \( x \):
\( x = \frac{69}{3} \)
\( x = 23 \)
So, the first odd number is 23. This method is generally helpful for finding unknown consecutive numbers. The three consecutive odd numbers are:
First number: \( x = 23 \)
Second number: \( x + 2 = 23 + 2 = 25 \)
Third number: \( x + 4 = 23 + 4 = 27 \)
The largest among these numbers is 27.
In simple words: We know three odd numbers in a row add up to 75. We called the first one \( x \). Since odd numbers are two apart, the others are \( x+2 \) and \( x+4 \). Adding them up and solving the equation showed the numbers are 23, 25, and 27. The biggest one is 27.

🎯 Exam Tip: For consecutive odd or even numbers, always represent them as \( x \), \( x+2 \), \( x+4 \), etc. For consecutive integers, use \( x \), \( x+1 \), \( x+2 \).

Question 5. The length of a rectangle is \( \frac{1}{3} \)rd of its breadth. If its perimeter is 64 m, then find the length and breadth of the rectangle.
Answer: Let the length of the rectangle be \( l \) and the breadth be \( b \).
According to the problem, the length is \( \frac{1}{3} \)rd of its breadth, so we have:
\( l = \frac{1}{3}b \)
This can also be written as \( b = 3l \) (Equation 1).
The perimeter of a rectangle is given by the formula \( \text{Perimeter} = 2 \times (l + b) \).
We are given that the perimeter is 64 m, so:
\( 2 \times (l + b) = 64 \)
Divide by 2:
\( l + b = 32 \)
Now, substitute the value of \( b \) from Equation 1 (\( b = 3l \)) into this equation:
\( l + 3l = 32 \)
Combine the like terms:
\( 4l = 32 \)
Divide by 4 to find \( l \):
\( l = \frac{32}{4} \)
\( l = 8 \text{ m} \)
Now, use the value of \( l \) to find \( b \) using Equation 1:
\( b = 3l = 3 \times 8 \)
\( b = 24 \text{ m} \)
So, the length of the rectangle is 8 m and the breadth is 24 m. This approach ensures consistency between the given conditions and the geometric formula.
In simple words: The length of a rectangle is one-third of its width. The total distance around it (perimeter) is 64 meters. We found that the length is 8 meters and the width is 24 meters.

🎯 Exam Tip: Always write down the formulas for perimeter or area first, then substitute the given values and relationships between length and breadth to solve for the unknowns.

Question 6. A total of 90 currency notes, consisting only of Rs 5 and Rs 10 denominations, amount to Rs 500. Find the number of notes in each denomination.
Answer: Let \( x \) be the number of Rs 5 notes and \( y \) be the number of Rs 10 notes.
We are given that there are a total of 90 notes:
\( x + y = 90 \) (Equation 1)
The total value of the notes is Rs 500. The value from Rs 5 notes is \( 5x \) and from Rs 10 notes is \( 10y \). So:
\( 5x + 10y = 500 \) (Equation 2)
Now we have a system of two linear equations. We can solve by substitution or elimination.
Let's use elimination. Multiply Equation 1 by 5 to make the x coefficients match:
\( 5 \times (x + y) = 5 \times 90 \)
\( 5x + 5y = 450 \) (Equation 3)
Now, subtract Equation 3 from Equation 2:
\( (5x + 10y) - (5x + 5y) = 500 - 450 \)
\( 5y = 50 \)
Divide by 5 to find \( y \):
\( y = \frac{50}{5} \)
\( y = 10 \)
Now substitute \( y = 10 \) back into Equation 1 to find \( x \):
\( x + 10 = 90 \)
\( x = 90 - 10 \)
\( x = 80 \)
So, there are 80 notes of Rs 5 and 10 notes of Rs 10. This method of setting up and solving simultaneous equations is key for such problems.
In simple words: We had 90 notes in total, either 5-rupee or 10-rupee notes. The total money was 500 rupees. We figured out there are 80 five-rupee notes and 10 ten-rupee notes.

🎯 Exam Tip: For problems involving different denominations or items with total quantities and total values, always set up two equations, one for quantity and one for value, then solve them simultaneously.

Question 7. At present, Thenmozhi's age is 5 years more than that of Murali's age. Five years ago, the ratio of Thenmozhi's age to Murali's age was 3 : 2. Find their present ages.
Answer: Let Thenmozhi's present age be \( t \) and Murali's present age be \( m \).
From the first statement, "Thenmozhi's age is 5 years more than that of Murali's age":
\( t = m + 5 \) (Equation 1)
Now consider "Five years ago":
Thenmozhi's age five years ago was \( t - 5 \).
Murali's age five years ago was \( m - 5 \).
The ratio of their ages five years ago was 3 : 2. So:
\( \frac{t-5}{m-5} = \frac{3}{2} \)
To solve this, cross-multiply:
\( 2(t - 5) = 3(m - 5) \)
\( 2t - 10 = 3m - 15 \) (Equation 2)
Now, substitute Equation 1 (\( t = m + 5 \)) into Equation 2:
\( 2(m + 5) - 10 = 3m - 15 \)
\( 2m + 10 - 10 = 3m - 15 \)
\( 2m = 3m - 15 \)
Move \( 2m \) to the right side and 15 to the left side:
\( 15 = 3m - 2m \)
\( m = 15 \)
So, Murali's present age is 15 years. This systematic substitution helps find the ages accurately. Now, find Thenmozhi's age using Equation 1:
\( t = m + 5 \)
\( t = 15 + 5 \)
\( t = 20 \)
Therefore, Thenmozhi's present age is 20 years and Murali's present age is 15 years.
In simple words: Thenmozhi is 5 years older than Murali. Five years ago, if you compared their ages, Thenmozhi was 3 parts for every 2 parts of Murali's age. By setting up equations for their current and past ages, we found Thenmozhi is 20 and Murali is 15 years old now.

🎯 Exam Tip: When solving age problems, always define variables for present ages and then express past or future ages relative to these variables. Pay close attention to keywords like "more than", "less than", and "ratio".

Question 8. A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Find the original number.
Answer: Let the tens digit of the original number be \( t \) and the units digit be \( u \).
The original number can be written as \( 10t + u \).
The problem states that the sum of the digits is 9:
\( t + u = 9 \) (Equation 1)
If the digits are interchanged, the new number will be \( 10u + t \).
The problem also states that if 27 is subtracted from the original number, the digits are interchanged:
\( (10t + u) - 27 = 10u + t \)
Now, let's simplify this equation by moving terms around:
\( 10t - t + u - 10u = 27 \)
\( 9t - 9u = 27 \)
We can divide the entire equation by 9:
\( \frac{9t}{9} - \frac{9u}{9} = \frac{27}{9} \)
\( t - u = 3 \) (Equation 2)
Now we have a system of two linear equations:
1) \( t + u = 9 \)
2) \( t - u = 3 \)
We can solve this by adding the two equations:
\( (t + u) + (t - u) = 9 + 3 \)
\( 2t = 12 \)
Divide by 2:
\( t = \frac{12}{2} \)
\( t = 6 \)
Now substitute \( t = 6 \) back into Equation 1 to find \( u \):
\( 6 + u = 9 \)
\( u = 9 - 6 \)
\( u = 3 \)
So, the tens digit is 6 and the units digit is 3. The original number is \( 10t + u = 10(6) + 3 = 60 + 3 = 63 \). This step-by-step process is crucial for solving digit-based problems.
In simple words: We had a two-digit number. Its digits added up to 9. When 27 was taken away from the number, the digits swapped places. We found that the original number was 63.

🎯 Exam Tip: For two-digit number problems, always represent the original number as \( 10t + u \) and the number with interchanged digits as \( 10u + t \), where \( t \) is the tens digit and \( u \) is the units digit.

Question 9. The denominator of a fraction exceeds its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get \( \frac{3}{2} \). Find the original fraction.
Answer: Let the numerator of the original fraction be \( n \) and the denominator be \( d \).
The first statement says "The denominator of a fraction exceeds its numerator by 8":
\( d = n + 8 \) (Equation 1)
The second statement describes what happens when the fraction is changed: "If the numerator is increased by 17 and the denominator is decreased by 1, we get \( \frac{3}{2} \)":
The new numerator is \( n + 17 \).
The new denominator is \( d - 1 \).
So, the new fraction is \( \frac{n + 17}{d - 1} \), and this is equal to \( \frac{3}{2} \):
\( \frac{n + 17}{d - 1} = \frac{3}{2} \)
To solve this, cross-multiply:
\( 2(n + 17) = 3(d - 1) \)
\( 2n + 34 = 3d - 3 \) (Equation 2)
Now, we substitute Equation 1 (\( d = n + 8 \)) into Equation 2:
\( 2n + 34 = 3(n + 8) - 3 \)
\( 2n + 34 = 3n + 24 - 3 \)
\( 2n + 34 = 3n + 21 \)
Move \( 2n \) to the right side and 21 to the left side:
\( 34 - 21 = 3n - 2n \)
\( 13 = n \)
So, the numerator \( n = 13 \). This substitution helps find the numerator efficiently. Now, use Equation 1 to find the denominator \( d \):
\( d = n + 8 \)
\( d = 13 + 8 \)
\( d = 21 \)
The original fraction is \( \frac{n}{d} = \frac{13}{21} \).
In simple words: We had a fraction where the bottom number was 8 bigger than the top number. When the top number went up by 17 and the bottom number went down by 1, the new fraction became 3/2. We worked it out and found the original fraction was 13/21.

🎯 Exam Tip: Always define your numerator and denominator as variables, then carefully translate the word problem into two algebraic equations. Substitution is often the most straightforward method for solving such systems.

Question 10. If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.
Answer: Let the distance to be covered by the train be \( d \) km.
Let the usual time taken to cover the distance be \( t \) hours.
We use the formula: \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

**Case 1:** Speed = 60 km/hr
The train is late by 15 minutes. We need to convert minutes to hours: \( 15 \text{ minutes} = \frac{15}{60} \text{ hours} = \frac{1}{4} \text{ hour} \).
So, the time taken in this case is \( t + \frac{1}{4} \) hours.
Using the formula: \( \frac{d}{60} = t + \frac{1}{4} \)
To eliminate the fraction, multiply the entire equation by 60:
\( d = 60t + 60 \times \frac{1}{4} \)
\( d = 60t + 15 \) (Equation 1)

**Case 2:** Speed = 85 km/hr
The train is late by 4 minutes. Convert to hours: \( 4 \text{ minutes} = \frac{4}{60} \text{ hours} = \frac{1}{15} \text{ hour} \).
So, the time taken in this case is \( t + \frac{1}{15} \) hours.
Using the formula: \( \frac{d}{85} = t + \frac{1}{15} \)
Multiply the entire equation by 85:
\( d = 85t + 85 \times \frac{1}{15} \)
\( d = 85t + \frac{17}{3} \) (Equation 2)

Now we have two equations for \( d \). We can equate them to solve for \( t \):
\( 60t + 15 = 85t + \frac{17}{3} \)
Gather the \( t \) terms on one side and the constants on the other:
\( 15 - \frac{17}{3} = 85t - 60t \)
To subtract the fractions, find a common denominator for 15 and \( \frac{17}{3} \): \( \frac{45}{3} - \frac{17}{3} = 25t \)
\( \frac{28}{3} = 25t \)
To find \( t \), divide both sides by 25:
\( t = \frac{28}{3 \times 25} \)
\( t = \frac{28}{75} \) hours.

Finally, substitute the value of \( t \) back into Equation 1 to find \( d \):
\( d = 60t + 15 \)
\( d = 60 \times \frac{28}{75} + 15 \)
\( d = \frac{1680}{75} + 15 \)
\( d = 22.4 + 15 \)
\( d = 37.4 \text{ km} \)
The distance to be covered by the train is 37.4 km. Ensuring all units are consistent (hours for time, km/hr for speed, km for distance) is crucial for accurate calculations.
In simple words: A train takes longer if it goes slow (60 km/hr, 15 minutes late) and a little less longer if it goes faster (85 km/hr, 4 minutes late). We set up two equations based on time, speed, and distance, and solved them. The total distance the train needs to cover is 37.4 kilometers.

🎯 Exam Tip: Always convert all time units to hours (or consistent units) before plugging them into speed-distance-time formulas to avoid calculation errors. Define the usual time as a variable to handle "late by" scenarios correctly.

Question 11. Sum of a number and its half is 30 then the number is
(a) 15
(b) 20
(c) 25
(d) 40
Answer: (b) 20
In simple words: If you add a number and half of that number together, you get 30. We found out that the number must be 20, because 20 plus half of 20 (which is 10) equals 30.

🎯 Exam Tip: When dealing with fractions of a number, it's often easiest to write it as an equation and then multiply by the common denominator to remove fractions before solving.

Question 12. The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is ___________.
(a) 62°
(b) 72°
(c) 78°
(d) 68°
Answer: (a) 62°
In simple words: The exterior angle of a triangle is always equal to the sum of its two opposite interior angles. If the exterior angle is 120° and one opposite interior angle is 58°, the other opposite interior angle must be \( 120^\circ - 58^\circ = 62^\circ \). This is a basic rule about angles in a triangle.

🎯 Exam Tip: Remember the theorem: "An exterior angle of a triangle is equal to the sum of its interior opposite angles." This is a fundamental property to solve such problems quickly.

Question 13. What sum of money will earn 500 as simple interest in 1 year at 5% per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer: (c) 10000
In simple words: We want to find the starting amount of money (principal) that will give 500 rupees as simple interest in 1 year, at an interest rate of 5% each year. Using the simple interest formula, we found the principal amount is 10,000 rupees.

🎯 Exam Tip: The formula for simple interest is \( \text{S.I.} = \frac{P \times R \times T}{100} \). Always clearly identify which values you have (S.I., R, T) and which one you need to find (P) before arranging the formula.

Question 14. The product of LCM and HCF of two numbers is 24. If one of the number is 6, then the other number is ___________.
(a) 6
(b) 2
(c) 4
(d) 8
Answer: (c) 4
In simple words: There's a rule that says if you multiply the LCM and HCF of two numbers, you get the same answer as multiplying the two numbers themselves. We knew the product of LCM and HCF was 24, and one number was 6, so the other number must be 4, because \( 6 \times 4 = 24 \).

🎯 Exam Tip: Remember the fundamental property: "Product of two numbers = (Their HCF) × (Their LCM)". This property is extremely useful for solving problems involving HCF, LCM, and two numbers.

Question 15. The largest number of the three consecutive numbers is \( x + 1 \), then the smallest number is ___________.
(A) \( x \)
(B) \( x + 1 \)
(C) \( x + 2 \)
(D) \( x - 1 \)
Answer: (D) \( x - 1 \)
In simple words: If you have three numbers that follow each other (like 5, 6, 7), and the biggest one is \( x+1 \), then the middle number would be \( x \), and the smallest number would be \( x-1 \). This is how consecutive numbers work.

🎯 Exam Tip: For consecutive integers, numbers increase by 1. So, if the largest is \( x+1 \), the one before it is \( (x+1)-1 = x \), and the one before that is \( x-1 \).

Question 1. Fill in the blanks:
(i) The solution of the equation \( ax + b = 0 \) is \( \underline{ \left( -\frac{b}{a} \right) } \).
(ii) If a and b are positive integers then the solution of the equation \( ax = b \) has to be always Positive.
(iii) One-sixth of a number when subtracted from the number itself gives 25. The number is 30.
(iv) If the angles of a triangle are in the ratio 2:3:4 then the difference between the greatest and the smallest angle is 40°.
(v) In an equation \( a + b = 23 \). The value of \( a \) is 14 then the value of \( b \) is 9.
Answer:
(i) The solution is found by isolating x. Start with \( ax + b = 0 \). Subtract b from both sides: \( ax = -b \). Then, divide by a: \( x = -\frac{b}{a} \).
(ii) If \( a \) and \( b \) are both positive integers, then the solution \( x = \frac{b}{a} \) will always be positive because a positive number divided by a positive number yields a positive result. This maintains the directional property of division.
(iii) Let the number be \( x \). The equation is \( x - \frac{x}{6} = 25 \). To solve this, find a common denominator: \( \frac{6x - x}{6} = 25 \). This simplifies to \( \frac{5x}{6} = 25 \). Multiply both sides by 6: \( 5x = 150 \). Divide by 5: \( x = 30 \).
(iv) Let the angles of the triangle be \( 2x, 3x, \) and \( 4x \). Since the sum of angles in a triangle is \( 180^\circ \), we have \( 2x + 3x + 4x = 180^\circ \). This means \( 9x = 180^\circ \). Dividing by 9, we get \( x = 20^\circ \). The angles are then \( 2 \times 20^\circ = 40^\circ \), \( 3 \times 20^\circ = 60^\circ \), and \( 4 \times 20^\circ = 80^\circ \). The difference between the greatest and smallest angle is \( 80^\circ - 40^\circ = 40^\circ \). A triangle always has three angles that sum to 180 degrees.
(v) Given the equation \( a + b = 23 \) and \( a = 14 \). Substitute the value of \( a \) into the equation: \( 14 + b = 23 \). To find \( b \), subtract 14 from both sides: \( b = 23 - 14 \). So, \( b = 9 \).
In simple words: We solved different types of equations. For simple linear equations, we use inverse operations. For ratios in triangles, we use the sum of angles property. For word problems, we turn the words into math equations to find the unknown values.

🎯 Exam Tip: For fill-in-the-blanks or short answer questions, always write down the full, correct answer. For algebraic problems, show your steps clearly to get full marks, even if the final answer is simple.

Question 2. Say True or False
(i) "Sum of a number and two times that number is 48” can be written as \( y + 2y = 48 \).
(ii) \( 5(3x + 2) = 3(5x - 7) \) is a linear equation in one variable.
(iii) \( x = 25 \) is the solution of one third of a number is less than 10 the original number.
Answer:
(i) True. If 'y' is a number, then "two times that number" is \( 2y \). Their sum is \( y + 2y \). So, \( y + 2y = 48 \) correctly represents the statement. The process of translating word problems into algebraic expressions is crucial.
(ii) True. The equation \( 5(3x + 2) = 3(5x - 7) \) simplifies to \( 15x + 10 = 15x - 21 \). When solved, it becomes \( 10 = -21 \), which is a contradiction, but the equation itself still involves only one variable, 'x', raised to the power of 1, making it a linear equation. The key is the variable's highest power.
(iii) False. Let the number be \( x \). "One third of a number is 10 less than the original number" means \( \frac{x}{3} = x - 10 \). To solve this: multiply by 3 to clear the fraction: \( x = 3(x - 10) \implies x = 3x - 30 \). Rearrange the terms: \( 30 = 3x - x \implies 30 = 2x \). Divide by 2: \( x = 15 \). Since \( x = 15 \), the statement that \( x = 25 \) is the solution is false.
In simple words: We checked if mathematical statements match their algebraic forms. An equation is linear if its variable's highest power is one. We also solved an equation to confirm if a given value was truly its solution.

🎯 Exam Tip: When dealing with True/False questions involving equations, either translate the word problem into an equation or simplify the given equation to verify its properties or solution.

Question 3. One number is seven times another. If their difference is 18, find the numbers.
Answer:
Let the two numbers be \( x \) and \( y \).
From the first statement, "One number is seven times another," we can write \( x = 7y \).
From the second statement, "If their difference is 18," we have \( x - y = 18 \).
Now, substitute the first equation into the second one:
\( 7y - y = 18 \)
\( 6y = 18 \)
\( y = \frac{18}{6} \)
\( y = 3 \)
To find \( x \), substitute the value of \( y \) back into \( x = 7y \):
\( x = 7 \times 3 \)
\( x = 21 \)
So, the two numbers are 3 and 21. When you have two unknown values, setting up a system of equations helps to find both simultaneously.
In simple words: We used two clues to find the numbers. One number is 7 times bigger than the other, and if you subtract them, you get 18. We figured out the numbers are 3 and 21.

🎯 Exam Tip: When solving word problems with two unknowns, always define your variables clearly and set up two equations based on the given information. Then use substitution or elimination to solve them.

Question 4. The sum of three consecutive odd numbers is 75. Which is the largest among them?
Answer:
Let the three consecutive odd numbers be represented by \( x \), \( x+2 \), and \( x+4 \). This is because consecutive odd numbers always differ by 2.
The problem states that their sum is 75:
\( x + (x+2) + (x+4) = 75 \)
Combine the like terms:
\( 3x + 6 = 75 \)
Subtract 6 from both sides:
\( 3x = 75 - 6 \)
\( 3x = 69 \)
Divide by 3:
\( x = \frac{69}{3} \)
\( x = 23 \)
Now, find the three numbers:
First number: \( x = 23 \)
Second number: \( x+2 = 23+2 = 25 \)
Third number: \( x+4 = 23+4 = 27 \)
The largest among these numbers is 27.
In simple words: We found three odd numbers that add up to 75. The biggest one is 27.

🎯 Exam Tip: When representing consecutive odd or even numbers algebraically, always remember they differ by 2 (e.g., \( x, x+2, x+4 \)), whereas consecutive integers differ by 1 (e.g., \( x, x+1, x+2 \)).

Question 5. The length of a rectangle is \( \frac{1}{3} \) rd of its breadth. If its perimeter is 64 m, then find the length and breadth of the rectangle.
Answer:
Let the length of the rectangle be \( l \) and the breadth be \( b \).
According to the problem, the length is \( \frac{1}{3} \) of its breadth:
\( l = \frac{1}{3} b \)
This can also be written as \( b = 3l \) .......... (1)
The perimeter of a rectangle is given by the formula \( P = 2(l + b) \).
We are given that the perimeter \( P \) is 64 m:
\( 2(l + b) = 64 \)
Now, substitute the expression for \( b \) from equation (1) into the perimeter equation:
\( 2(l + 3l) = 64 \)
\( 2(4l) = 64 \)
\( 8l = 64 \)
Divide both sides by 8 to find \( l \):
\( l = \frac{64}{8} \)
\( l = 8 \) m
Now, use the value of \( l \) to find \( b \):
\( b = 3l = 3 \times 8 \)
\( b = 24 \) m
Thus, the length of the rectangle is 8 m and the breadth is 24 m. This problem demonstrates how to apply algebraic substitution in geometric problems.

🎯 Exam Tip: Always draw a simple diagram for geometry problems. This helps visualize the given information and apply the correct formulas for perimeter or area.

Question 6. A total of 90 currency notes, consisting only of Rs. 5 and Rs. 10 denominations, amount to Rs. 500. Find the number of notes in each denomination.
Answer:
Let \( x \) be the number of Rs. 5 notes and \( y \) be the number of Rs. 10 notes.
Based on the total number of notes, we form the first equation:
\( x + y = 90 \) .......... (1)
Based on the total value of the notes, we form the second equation:
\( 5x + 10y = 500 \) .......... (2)
To solve this system of equations, we can use the elimination method. Multiply equation (1) by 5:
\( 5(x + y) = 5 \times 90 \)
\( 5x + 5y = 450 \) .......... (3)
Now, subtract equation (3) from equation (2):
\( (5x + 10y) - (5x + 5y) = 500 - 450 \)
\( 5x + 10y - 5x - 5y = 50 \)
\( 5y = 50 \)
Divide by 5 to find \( y \):
\( y = \frac{50}{5} \)
\( y = 10 \)
Substitute the value of \( y \) back into equation (1) to find \( x \):
\( x + 10 = 90 \)
\( x = 90 - 10 \)
\( x = 80 \)
Therefore, there are 80 Rs. 5 notes and 10 Rs. 10 notes. Setting up clear equations for each piece of information is key.
In simple words: We have 90 money notes, either Rs. 5 or Rs. 10, totaling Rs. 500. We found there are 80 notes of Rs. 5 and 10 notes of Rs. 10.

🎯 Exam Tip: For problems involving different denominations or items, set up two equations: one for the total count and one for the total value. Solve using substitution or elimination method.

Question 7. At present, Thenmozhi's age is 5 years more than that of Murali's age. Five years ago, the ratio Thenmozhi's age to Murali's age was 3 : 2. Find their present ages.
Answer:
Let Thenmozhi's present age be \( t \) and Murali's present age be \( m \).
From the first statement, "Thenmozhi's age is 5 years more than that of Murali's age":
\( t = m + 5 \) .......... (1)
Now, consider their ages five years ago:
Thenmozhi's age five years ago = \( t - 5 \)
Murali's age five years ago = \( m - 5 \)
The ratio of their ages five years ago was 3 : 2:
\( \frac{t-5}{m-5} = \frac{3}{2} \)
Cross-multiply to remove the fractions:
\( 2(t-5) = 3(m-5) \)
\( 2t - 10 = 3m - 15 \)
Now, substitute \( t = m + 5 \) from equation (1) into this equation:
\( 2(m + 5) - 10 = 3m - 15 \)
\( 2m + 10 - 10 = 3m - 15 \)
\( 2m = 3m - 15 \)
Rearrange the terms to solve for \( m \):
\( 15 = 3m - 2m \)
\( m = 15 \)
Now, substitute \( m = 15 \) back into equation (1) to find \( t \):
\( t = 15 + 5 \)
\( t = 20 \)
Therefore, Thenmozhi's present age is 20 years and Murali's present age is 15 years. Age problems often involve setting up equations for different points in time.
In simple words: We used clues about ages. Thenmozhi is 5 years older than Murali. Five years ago, Thenmozhi was 1.5 times Murali's age. We found Thenmozhi is 20 and Murali is 15 now.

🎯 Exam Tip: For age-related problems, always define variables for present ages first. Then, express ages in the past or future relative to these present ages, forming equations to solve.

Question 8. A number consists of two digits whose sum is 9. If 27 is subtracted from the original number, its digits are interchanged. Find the original number.
Answer:
Let the tens digit of the number be \( t \) and the units digit be \( u \).
The original two-digit number can be expressed as \( 10t + u \). For example, if a number is 21, it is \( 2 \times 10 + 1 \).
If the digits are interchanged, the new number would be \( 10u + t \).
From the first condition, "the sum of the two digits is 9":
\( t + u = 9 \) .......... (1)
From the second condition, "If 27 is subtracted from the original number, its digits are interchanged":
\( (10t + u) - 27 = 10u + t \)
Now, let's simplify and rearrange this equation to group like terms:
\( 10t - t + u - 10u = 27 \)
\( 9t - 9u = 27 \)
Divide the entire equation by 9:
\( t - u = 3 \) .......... (2)
Now we have a system of two linear equations:
1) \( t + u = 9 \)
2) \( t - u = 3 \)
Add equation (1) and equation (2):
\( (t + u) + (t - u) = 9 + 3 \)
\( 2t = 12 \)
Divide by 2 to find \( t \):
\( t = \frac{12}{2} \)
\( t = 6 \)
Substitute the value of \( t = 6 \) back into equation (1) to find \( u \):
\( 6 + u = 9 \)
\( u = 9 - 6 \)
\( u = 3 \)
The tens digit is 6 and the units digit is 3. Therefore, the original number is \( 10t + u = 10(6) + 3 = 60 + 3 = 63 \). This method is widely used for solving problems involving two-digit numbers.
In simple words: We found a two-digit number where the digits add up to 9. If you take 27 away from the number, the digits swap places. The number is 63.

🎯 Exam Tip: For problems involving two-digit numbers, always represent the number as \( 10t + u \) (where \( t \) is the tens digit and \( u \) is the units digit) and the reversed number as \( 10u + t \). This setup is fundamental for algebraic solutions.

Question 9. The denominator of a fraction exceeds Its numerator by 8. If the numerator is increased by 17 and the denominator is decreased by 1, we get \( \frac{3}{2} \). Find the original fraction.
Answer:
Let the numerator of the fraction be \( n \) and the denominator be \( d \).
From the first condition, "The denominator of a fraction exceeds its numerator by 8":
\( d = n + 8 \) .......... (1)
From the second condition, "If the numerator is increased by 17 and the denominator is decreased by 1, we get \( \frac{3}{2} \)":
New numerator \( = n + 17 \)
New denominator \( = d - 1 \)
So, the new fraction is \( \frac{n+17}{d-1} = \frac{3}{2} \)
To solve this, cross-multiply:
\( 2(n+17) = 3(d-1) \)
\( 2n + 34 = 3d - 3 \)
Rearrange this equation to group \( n \) and \( d \) terms:
\( 2n - 3d = -3 - 34 \)
\( 2n - 3d = -37 \) .......... (2)
Now, substitute the expression for \( d \) from equation (1) into equation (2):
\( 2n - 3(n+8) = -37 \)
\( 2n - 3n - 24 = -37 \)
\( -n - 24 = -37 \)
Add 24 to both sides:
\( -n = -37 + 24 \)
\( -n = -13 \)
Multiply by -1 to find \( n \):
\( n = 13 \)
Now, substitute \( n = 13 \) back into equation (1) to find \( d \):
\( d = 13 + 8 \)
\( d = 21 \)
Therefore, the original fraction is \( \frac{n}{d} = \frac{13}{21} \). Fractional problems are often simplified by eliminating the denominators through cross-multiplication.
In simple words: We found a fraction where the bottom number is 8 more than the top. When we changed the top and bottom numbers, the fraction became 3/2. The original fraction was 13/21.

🎯 Exam Tip: When setting up equations for word problems involving fractions, always define your numerator and denominator variables clearly. Cross-multiplication is an effective first step to simplify equations with fractions on both sides.

Question 10. If a train runs at 60 km/hr it reaches its destination late by 15 minutes. But, if it runs at 85 kmph it is late by only 4 minutes. Find the distance to be covered by the train.
Answer:
Let the distance to be covered by the train be \( D \) km, and let the usual time it should take be \( T \) hours.
The fundamental formula for speed, distance, and time is \( \text{Time} = \frac{\text{Distance}}{\text{Speed}} \).

Case 1: Speed is 60 km/hr and the train is 15 minutes late.
First, convert 15 minutes to hours: \( 15 \text{ minutes} = \frac{15}{60} \text{ hours} = \frac{1}{4} \text{ hour} \).
So, the time taken in this case is \( T + \frac{1}{4} \) hours.
Using the formula: \( \frac{D}{60} = T + \frac{1}{4} \)
Multiply both sides by 60 to eliminate the denominator and isolate \( D \):
\( D = 60 \left( T + \frac{1}{4} \right) \)
\( D = 60T + 15 \) .......... (1)

Case 2: Speed is 85 km/hr and the train is 4 minutes late.
Convert 4 minutes to hours: \( 4 \text{ minutes} = \frac{4}{60} \text{ hours} = \frac{1}{15} \text{ hour} \).
So, the time taken in this case is \( T + \frac{1}{15} \) hours.
Using the formula: \( \frac{D}{85} = T + \frac{1}{15} \)
Multiply both sides by 85 to eliminate the denominator and isolate \( D \):
\( D = 85 \left( T + \frac{1}{15} \right) \)
\( D = 85T + \frac{85}{15} \)
Simplify the fraction \( \frac{85}{15} \) by dividing by 5: \( D = 85T + \frac{17}{3} \) .......... (2)

Now, we have two expressions for the distance \( D \). We can equate them to solve for \( T \):
\( 60T + 15 = 85T + \frac{17}{3} \)
Rearrange the equation to group \( T \) terms on one side and constant terms on the other:
\( 15 - \frac{17}{3} = 85T - 60T \)
Calculate the left side: \( \frac{15 \times 3 - 17}{3} = 25T \)
\( \frac{45 - 17}{3} = 25T \)
\( \frac{28}{3} = 25T \)
Solve for \( T \):
\( T = \frac{28}{3 \times 25} \)
\( T = \frac{28}{75} \) hours.

Finally, substitute the value of \( T \) back into equation (1) to find the distance \( D \):
\( D = 60T + 15 \)
\( D = 60 \times \frac{28}{75} + 15 \)
Simplify \( \frac{60}{75} \) by dividing both by 15: \( \frac{4}{5} \).
\( D = 4 \times \frac{28}{5} + 15 \)
\( D = \frac{112}{5} + 15 \)
\( D = 22.4 + 15 \)
\( D = 37.4 \) km
The distance to be covered by the train is 37.4 km. Speed, distance, and time problems often require careful unit conversion and setting up simultaneous equations.
In simple words: A train takes a certain time to cover a distance. If it goes at 60 km/hr, it's 15 mins late. If it goes at 85 km/hr, it's 4 mins late. By comparing these situations, we found the total distance is 37.4 km.

🎯 Exam Tip: Always ensure all units (time, speed, distance) are consistent before applying formulas. Convert minutes to hours early in the problem to avoid errors. Also, remember that 'late by X minutes' means (usual time + X minutes).

Question 11. Sum of a number and its half is 30 then the number is
(a) 15
(b) 20
(c) 25
(d) 40
Answer: (b) 20
Let the number be \( x \).
According to the problem, the sum of the number and its half is 30:
\( x + \frac{x}{2} = 30 \)
To eliminate the fraction, multiply every term in the equation by 2:
\( 2 \times x + 2 \times \frac{x}{2} = 2 \times 30 \)
\( 2x + x = 60 \)
Combine the \( x \) terms:
\( 3x = 60 \)
Divide by 3 to solve for \( x \):
\( x = \frac{60}{3} \)
\( x = 20 \)
The number is 20. Checking the answer: \( 20 + \frac{20}{2} = 20 + 10 = 30 \), which is correct.
In simple words: We found a number where if you add it to half of itself, you get 30. That number is 20.

🎯 Exam Tip: When an equation contains fractions, multiply all terms by the least common multiple (LCM) of the denominators to clear the fractions, making the equation easier to solve.

Question 12. The exterior angle of a triangle is 120° and one of its interior opposite angle 58°, then the other opposite interior angle is
(a) 62°
(b) 72°
(c) 78°
(d) 68°
Answer: (a) 62°
According to the Exterior Angle Theorem, the measure of an exterior angle of a triangle is equal to the sum of the measures of its two remote (opposite) interior angles.
Let the unknown interior opposite angle be \( x \).
Given: Exterior angle \( = 120^\circ \)
One interior opposite angle \( = 58^\circ \)
So, we can set up the equation:
\( 120^\circ = x + 58^\circ \)
To find \( x \), subtract \( 58^\circ \) from both sides:
\( x = 120^\circ - 58^\circ \)
\( x = 62^\circ \)
Thus, the other opposite interior angle is \( 62^\circ \). This property is a fundamental concept in geometry.

🎯 Exam Tip: Clearly identify the exterior angle and its two remote interior angles. The theorem simplifies complex angle-finding problems significantly.

Question 13. What sum of money will earn 500 as simple interest in 1 year at 5% per annum?
(a) 50000
(b) 30000
(c) 10000
(d) 5000
Answer: (c) 10000
The formula for Simple Interest (SI) is:
\( SI = \frac{P \times R \times T}{100} \)
Where:
\( P \) = Principal amount (the sum of money)
\( R \) = Rate of interest per annum
\( T \) = Time in years
Given values are:
\( SI = \text{Rs. } 500 \)
\( R = 5\% \)
\( T = 1 \text{ year} \)
We need to find \( P \). Substitute the given values into the formula:
\( 500 = \frac{P \times 5 \times 1}{100} \)
Now, solve for \( P \):
Multiply both sides by 100:
\( 500 \times 100 = P \times 5 \)
\( 50000 = 5P \)
Divide by 5:
\( P = \frac{50000}{5} \)
\( P = 10000 \)
So, the sum of money (principal) that will earn Rs. 500 as simple interest is Rs. 10,000. Understanding how to rearrange formula is essential.
In simple words: We want to know how much money (principal) you need to put away to earn Rs. 500 interest in one year, if the interest rate is 5% each year. The answer is Rs. 10,000.

🎯 Exam Tip: Always write down the formula first and list all known values. Then, substitute and solve for the unknown. Be careful with unit conversions, especially for time periods.

Question 14. The product of LCM and HCF of two numbers is 24. If one of the number is 6, then the other number is
(a) 6
(b) 2
(c) 4
(d) 8
Answer: (c) 4
A fundamental property of numbers states that for any two positive integers, the product of the numbers is equal to the product of their Least Common Multiple (LCM) and Highest Common Factor (HCF).
So, \( \text{First Number} \times \text{Second Number} = \text{LCM} \times \text{HCF} \).
Given information:
Product of LCM and HCF \( = 24 \)
One of the numbers \( = 6 \)
Let the other number be \( x \).
Using the property:
\( 6 \times x = 24 \)
To find \( x \), divide both sides by 6:
\( x = \frac{24}{6} \)
\( x = 4 \)
Therefore, the other number is 4. This property is very useful in number theory problems.
In simple words: If you multiply two numbers, you get the same result as multiplying their LCM and HCF. Given that the LCM times HCF is 24 and one number is 6, the other number must be 4.

🎯 Exam Tip: Memorize the key property: \( \text{Product of two numbers} = \text{LCM of numbers} \times \text{HCF of numbers} \). This property is crucial for solving problems involving LCM and HCF efficiently.

Question 15. The largest number of the three consecutive numbers is \( x+1 \), then the smallest number is
(A) \( x \)
(B) \( x+1 \)
(C) \( x+2 \)
(D) \( x-1 \)
Answer: (D) \( x-1 \)
When we talk about consecutive numbers, they follow each other in order, with a difference of 1 between them.
If the largest of the three consecutive numbers is \( x+1 \), we can find the other two by subtracting 1 repeatedly:
The largest number = \( x+1 \)
The middle number = \( (x+1) - 1 = x \)
The smallest number = \( x - 1 \)
Thus, the three consecutive numbers are \( x-1, x, \) and \( x+1 \). The pattern for consecutive numbers is always consistent.
In simple words: If the biggest of three numbers in a row is \( x+1 \), then the smallest number will be \( x-1 \).

🎯 Exam Tip: Understand how consecutive numbers are represented algebraically. If \( x \) is a number, consecutive integers are \( x-1, x, x+1 \). If you are given the largest, subtract to find the smaller ones.