OP Malhotra Class 9 Maths Solutions Chapter 5 Simultaneous Linear Equations in Two Variable Exercise 5 (B)

Question 1. If one number is thrice the other and their sum is 16, find the numbers.
Answer: Let the first number be \(x\) and the second number be \(y\).
According to the problem, one number is three times the other, so we can write this as \(x = 3y\).
The sum of the two numbers is 16, so \(x + y = 16\).
Now we have two equations:
(i) \(x + y = 16\)
(ii) \(x - 3y = 0\) (derived from \(x = 3y\))
Subtract equation (ii) from equation (i):
\( (x + y) - (x - 3y) = 16 - 0 \)
\( x + y - x + 3y = 16 \)
\( 4y = 16 \)
\( \implies \) \( y = \frac{16}{4} \)
\( \implies \) \( y = 4 \)
Substitute the value of \(y\) back into the equation \(x = 3y\):
\( x = 3 \times 4 \)
\( x = 12 \)
So, the two numbers are 12 and 4. You can check that 12 is thrice 4, and their sum is \(12 + 4 = 16\).
In simple words: We have two unknown numbers. One is three times bigger than the other, and when you add them together, you get 16. By setting up simple equations, we find these numbers are 12 and 4.

🎯 Exam Tip: When dealing with "one number is thrice the other," always represent this with \(x = 3y\) or \(y = 3x\). Ensure your equations clearly reflect both conditions given in the problem.

Question 2. The sum of two numbers is 6 and their difference is 4. Find the numbers.
Answer: Let the first number be \(x\) and the second number be \(y\).
According to the problem, the sum of the two numbers is 6:
(i) \(x + y = 6\)
The difference between the two numbers is 4:
(ii) \(x - y = 4\)
To find the numbers, we can add the two equations:
\( (x + y) + (x - y) = 6 + 4 \)
\( 2x = 10 \)
\( \implies \) \( x = \frac{10}{2} \)
\( \implies \) \( x = 5 \)
Now, substitute the value of \(x\) into equation (i):
\( 5 + y = 6 \)
\( y = 6 - 5 \)
\( y = 1 \)
Alternatively, we could subtract equation (ii) from equation (i):
\( (x + y) - (x - y) = 6 - 4 \)
\( x + y - x + y = 2 \)
\( 2y = 2 \)
\( \implies \) \( y = \frac{2}{2} \)
\( \implies \) \( y = 1 \)
Using \(y=1\) in \(x+y=6\), we get \(x+1=6\), so \(x=5\).
The two numbers are 5 and 1. These types of problems are easily solved using elimination.
In simple words: We have two numbers. When you add them, you get 6. When you subtract them, you get 4. The numbers are 5 and 1.

🎯 Exam Tip: For problems involving the sum and difference of two numbers, the most efficient method is usually to add and subtract the equations to find each variable directly.

Question 3. In a two digit number, the sum of digits is 13. If the number is subtracted from the one obtained by interchanging the digits, the result is 45. What is the number ?
Answer: Let the units digit of the two-digit number be \(x\) and the tens digit be \(y\).
The original number can be written as \(10y + x\).
When the digits are interchanged, the new number is \(10x + y\).
According to the first condition, the sum of the digits is 13:
(i) \(x + y = 13\)
According to the second condition, if the original number is subtracted from the number obtained by interchanging the digits, the result is 45:
\( (10x + y) - (10y + x) = 45 \)
\( 10x + y - 10y - x = 45 \)
\( 9x - 9y = 45 \)
Divide the entire equation by 9:
(ii) \(x - y = 5\)
Now we have a system of two linear equations:
(i) \(x + y = 13\)
(ii) \(x - y = 5\)
Add equation (i) and equation (ii):
\( (x + y) + (x - y) = 13 + 5 \)
\( 2x = 18 \)
\( \implies \) \( x = \frac{18}{2} \)
\( \implies \) \( x = 9 \)
Substitute \(x = 9\) into equation (i):
\( 9 + y = 13 \)
\( y = 13 - 9 \)
\( y = 4 \)
So, the units digit is 9 and the tens digit is 4. The original number is \(10y + x\).
Number = \(10 \times 4 + 9 = 40 + 9 = 49\). This is a classic type of number problem solved using algebra.
In simple words: We are looking for a two-digit number. The two digits add up to 13. If you swap the digits and subtract the original number, you get 45. The number is 49.

🎯 Exam Tip: Remember to correctly represent a two-digit number as \(10 \times (\text{tens digit}) + (\text{units digit})\) and the number with reversed digits as \(10 \times (\text{units digit}) + (\text{tens digit})\). This is a common pitfall.

Question 4. A number consists of two digits whose sum is 5. When the digits are reversed, the number becomes greater by 9. Find the number.
Answer: Let the units digit be \(x\) and the tens digit be \(y\).
The original number is \(10y + x\).
The number obtained by reversing the digits is \(10x + y\).
According to the first condition, the sum of the digits is 5:
(i) \(x + y = 5\)
According to the second condition, when the digits are reversed, the new number is 9 greater than the original number:
\( (10x + y) = (10y + x) + 9 \)
\( 10x + y - 10y - x = 9 \)
\( 9x - 9y = 9 \)
Divide the entire equation by 9:
(ii) \(x - y = 1\)
Now we have a system of two linear equations:
(i) \(x + y = 5\)
(ii) \(x - y = 1\)
Add equation (i) and equation (ii):
\( (x + y) + (x - y) = 5 + 1 \)
\( 2x = 6 \)
\( \implies \) \( x = \frac{6}{2} \)
\( \implies \) \( x = 3 \)
Substitute \(x = 3\) into equation (i):
\( 3 + y = 5 \)
\( y = 5 - 3 \)
\( y = 2 \)
The units digit is 3 and the tens digit is 2. The original number is \(10y + x\).
Number = \(10 \times 2 + 3 = 20 + 3 = 23\). This method helps identify the digits correctly.
In simple words: We are looking for a two-digit number where the digits add up to 5. If you switch the digits, the new number is 9 more than the first one. The number is 23.

🎯 Exam Tip: Pay close attention to phrases like "becomes greater by 9" to correctly set up the equation (New Number = Original Number + 9). A common mistake is to subtract in the wrong order.

Question 5. The sum of a two digit number and the number obtained by reversing the order of its digits is 121 and the two digits differ by 3. Find the numbers.
Answer: Let the units digit be \(x\) and the tens digit be \(y\).
The original number is \(10y + x\).
The number obtained by reversing the digits is \(10x + y\).
According to the first condition, the sum of the original number and the reversed number is 121:
\( (10y + x) + (10x + y) = 121 \)
\( 11x + 11y = 121 \)
Divide the entire equation by 11:
(ii) \(x + y = 11\)
According to the second condition, the two digits differ by 3. This means either \(x - y = 3\) or \(y - x = 3\). The solution assumes \(x - y = 3\). Let's use:
(i) \(x - y = 3\)
Now we have a system of two linear equations:
(i) \(x - y = 3\)
(ii) \(x + y = 11\)
Add equation (i) and equation (ii):
\( (x - y) + (x + y) = 3 + 11 \)
\( 2x = 14 \)
\( \implies \) \( x = \frac{14}{2} \)
\( \implies \) \( x = 7 \)
Substitute \(x = 7\) into equation (ii):
\( 7 + y = 11 \)
\( y = 11 - 7 \)
\( y = 4 \)
The units digit is 7 and the tens digit is 4. The original number is \(10y + x\).
Number = \(10 \times 4 + 7 = 40 + 7 = 47\). If we had assumed \(y-x=3\), we would get 74, which also works.
In simple words: We need a two-digit number. When you add it to the number you get by flipping its digits, you get 121. Also, the two digits are 3 apart. The number is 47.

🎯 Exam Tip: When given that digits "differ by" a certain value, consider both possibilities (\(x-y\) or \(y-x\)) if not specified. In many cases, both numbers (e.g., 47 and 74) would be valid answers if the question doesn't constrain further.

Question 6. Seven times a given two digit number is equal to four times the number obtained by reversing the order of digits. The sum of the digits of the number is 3. Find the number.
Answer: Let the units digit be \(x\) and the tens digit be \(y\).
The original two-digit number is \(10y + x\).
The number obtained by reversing the digits is \(10x + y\).
According to the first condition, seven times the original number is equal to four times the reversed number:
\( 7(10y + x) = 4(10x + y) \)
\( 70y + 7x = 40x + 4y \)
\( 70y - 4y = 40x - 7x \)
\( 66y = 33x \)
Divide both sides by 33:
\( 2y = x \)
So, (ii) \(x = 2y\)
According to the second condition, the sum of the digits is 3:
(i) \(x + y = 3\)
Now substitute equation (ii) into equation (i):
\( (2y) + y = 3 \)
\( 3y = 3 \)
\( \implies \) \( y = \frac{3}{3} \)
\( \implies \) \( y = 1 \)
Substitute \(y = 1\) back into equation (ii):
\( x = 2 \times 1 \)
\( x = 2 \)
The units digit is 2 and the tens digit is 1. The original number is \(10y + x\).
Number = \(10 \times 1 + 2 = 10 + 2 = 12\). This problem tests understanding of place value.
In simple words: We are looking for a two-digit number. The digits add up to 3. If you multiply the number by 7, you get the same answer as multiplying the number with its digits swapped by 4. The number is 12.

🎯 Exam Tip: Always set up the number (10y+x) and reversed number (10x+y) equations correctly before forming the main algebraic relationships. Simplify equations by dividing by common factors to make calculations easier.

Question 7. If three times, the larger of the two numbers is divided by the smaller one. we get 4 as remainder. Also, if seven times the smaller number is divided by the larger one, we get 5 as quotient and 1 as remainder. Find the numbers.
Answer: Let the larger number be \(x\) and the smaller number be \(y\).
Using the division algorithm (Dividend = Divisor \(\times\) Quotient + Remainder):
According to the first condition, three times the larger number (\(3x\)) divided by the smaller number (\(y\)) gives a quotient of 4 and a remainder of 3. (Note: The question states "4 as remainder", but the solution uses 4 as quotient and 3 as remainder, which is a common interpretation for such problem patterns. I will follow the solution's logic for consistency.)
\( 3x = 4y + 3 \)
(i) \(3x - 4y = 3\)
According to the second condition, seven times the smaller number (\(7y\)) divided by the larger number (\(x\)) gives a quotient of 5 and a remainder of 1:
\( 7y = 5x + 1 \)
Rearrange this equation:
(ii) \(5x - 7y = -1\)
Now we have a system of two linear equations. We can solve this by elimination.
Multiply equation (i) by 7 and equation (ii) by 4 to make the coefficients of \(y\) equal:
From (i): \( 7 \times (3x - 4y) = 7 \times 3 \implies 21x - 28y = 21 \)
From (ii): \( 4 \times (5x - 7y) = 4 \times (-1) \implies 20x - 28y = -4 \)
Subtract the second new equation from the first new equation:
\( (21x - 28y) - (20x - 28y) = 21 - (-4) \)
\( 21x - 28y - 20x + 28y = 21 + 4 \)
\( x = 25 \)
Now substitute \(x = 25\) into equation (i):
\( 3(25) - 4y = 3 \)
\( 75 - 4y = 3 \)
\( -4y = 3 - 75 \)
\( -4y = -72 \)
\( \implies \) \( y = \frac{-72}{-4} \)
\( \implies \) \( y = 18 \)
So, the larger number is 25 and the smaller number is 18. This problem requires careful setup using the division algorithm.
In simple words: We have two numbers. If you take three times the bigger number and divide it by the smaller one, you get 4 with 3 left over. If you take seven times the smaller number and divide it by the bigger one, you get 5 with 1 left over. The numbers are 25 and 18.

🎯 Exam Tip: When using the division algorithm, ensure you correctly identify the quotient and remainder from the problem statement. A clear setup of the equations is crucial for accuracy.

Question 8. Divide 36 into four parts so that if 2 is added to the first part, 2 is subtracted from the second part, the third part is multiplied by 2, and the fourth part is divided by 2, then the resulting number is the same in each case.
Answer: Let the equal resulting number in each case be \(k\).
Based on the conditions:
1. If 2 is added to the first part, it equals \(k\). So, First Part = \(k - 2\).
2. If 2 is subtracted from the second part, it equals \(k\). So, Second Part = \(k + 2\).
3. If the third part is multiplied by 2, it equals \(k\). So, Third Part = \( \frac{k}{2} \).
4. If the fourth part is divided by 2, it equals \(k\). So, Fourth Part = \(2k\).
The total sum of these four parts is 36:
\( (k - 2) + (k + 2) + \frac{k}{2} + 2k = 36 \)
Combine the terms with \(k\) and the constant terms:
\( k + k + \frac{k}{2} + 2k = 36 \)
\( 4k + \frac{k}{2} = 36 \)
To combine \(4k\) and \( \frac{k}{2} \), find a common denominator:
\( \frac{8k}{2} + \frac{k}{2} = 36 \)
\( \frac{9k}{2} = 36 \)
Multiply both sides by 2:
\( 9k = 36 \times 2 \)
\( 9k = 72 \)
\( \implies \) \( k = \frac{72}{9} \)
\( \implies \) \( k = 8 \)
Now, find each of the four parts using \(k=8\):
First Part = \(k - 2 = 8 - 2 = 6\)
Second Part = \(k + 2 = 8 + 2 = 10\)
Third Part = \( \frac{k}{2} = \frac{8}{2} = 4\)
Fourth Part = \(2k = 2 \times 8 = 16\)
The four parts are 6, 10, 4, and 16. It's a fun way to use inverse operations.
In simple words: We need to split the number 36 into four parts. Each part is changed in a different way (add 2, subtract 2, multiply by 2, divide by 2), and after these changes, all four parts become the same number. Those four parts are 6, 10, 4, and 16.

🎯 Exam Tip: When a problem states that multiple operations result in the "same number," it's often easiest to introduce a new variable for that common result and express each original part in terms of it using inverse operations.

Question 9. If the numerator of a fraction is increased by 2 and the denominator by 1, it becomes \( \frac{5}{8} \) and if the numerator and the denominator of the same fraction are each increased by 1, the fraction becomes equal to \( \frac{1}{2} \). Find the fraction.
Answer: Let the fraction be \( \frac{x}{y} \), where \(x\) is the numerator and \(y\) is the denominator.
According to the first condition:
If the numerator is increased by 2 (\(x+2\)) and the denominator by 1 (\(y+1\)), the fraction becomes \( \frac{5}{8} \).
\( \frac{x+2}{y+1} = \frac{5}{8} \)
Cross-multiply:
\( 8(x+2) = 5(y+1) \)
\( 8x + 16 = 5y + 5 \)
Rearrange into the standard form \(Ax + By = C\):
(i) \(8x - 5y = 5 - 16 \implies 8x - 5y = -11\)
According to the second condition:
If both the numerator and the denominator are each increased by 1 (\(x+1\) and \(y+1\)), the fraction becomes \( \frac{1}{2} \).
\( \frac{x+1}{y+1} = \frac{1}{2} \)
Cross-multiply:
\( 2(x+1) = 1(y+1) \)
\( 2x + 2 = y + 1 \)
Rearrange into the standard form \(Ax + By = C\):
(ii) \(2x - y = 1 - 2 \implies 2x - y = -1\)
Now we have a system of two linear equations:
(i) \(8x - 5y = -11\)
(ii) \(2x - y = -1\)
From equation (ii), we can express \(y\) in terms of \(x\):
\( y = 2x + 1 \)
Substitute this expression for \(y\) into equation (i):
\( 8x - 5(2x + 1) = -11 \)
\( 8x - 10x - 5 = -11 \)
\( -2x - 5 = -11 \)
\( -2x = -11 + 5 \)
\( -2x = -6 \)
\( \implies \) \( x = \frac{-6}{-2} \)
\( \implies \) \( x = 3 \)
Now substitute \(x = 3\) back into the expression for \(y\):
\( y = 2(3) + 1 \)
\( y = 6 + 1 \)
\( y = 7 \)
So, the numerator is 3 and the denominator is 7. The fraction is \( \frac{3}{7} \). This is a common method for solving fraction problems.
In simple words: We're trying to find a fraction. If you add 2 to its top number and 1 to its bottom number, it becomes \( \frac{5}{8} \). If you add 1 to both its top and bottom numbers, it becomes \( \frac{1}{2} \). The fraction is \( \frac{3}{7} \).

🎯 Exam Tip: Always set up the initial fraction as \(x/y\). Carefully translate the "increased by" or "decreased by" statements into correct algebraic expressions for the numerator and denominator before cross-multiplying to form linear equations.

Question 10. If 1 is added to the denominator of a fraction, the fraction becomes \( \frac{1}{2} \). If 1 is added to the numerator of the fraction, the fraction becomes 1. Find the fraction.
Answer: Let the fraction be \( \frac{x}{y} \), where \(x\) is the numerator and \(y\) is the denominator.
According to the first condition:
If 1 is added to the denominator, the fraction becomes \( \frac{1}{2} \).
\( \frac{x}{y+1} = \frac{1}{2} \)
Cross-multiply:
\( 2x = y + 1 \)
Rearrange into standard form:
(i) \(2x - y = 1\)
According to the second condition:
If 1 is added to the numerator, the fraction becomes 1.
\( \frac{x+1}{y} = 1 \)
Cross-multiply:
\( x+1 = y \)
Rearrange into standard form:
(ii) \(x - y = -1\)
Now we have a system of two linear equations:
(i) \(2x - y = 1\)
(ii) \(x - y = -1\)
Subtract equation (ii) from equation (i) to eliminate \(y\):
\( (2x - y) - (x - y) = 1 - (-1) \)
\( 2x - y - x + y = 1 + 1 \)
\( x = 2 \)
Substitute \(x = 2\) back into equation (ii):
\( 2 - y = -1 \)
\( -y = -1 - 2 \)
\( -y = -3 \)
\( \implies \) \( y = 3 \)
So, the numerator is 2 and the denominator is 3. The fraction is \( \frac{2}{3} \). This approach systematically finds the unknown values.
In simple words: We are looking for a fraction. If you add 1 to its bottom number, it becomes \( \frac{1}{2} \). If you add 1 to its top number, it becomes 1. The fraction is \( \frac{2}{3} \).

🎯 Exam Tip: Always check if your calculated fraction satisfies both original conditions in the problem to ensure accuracy. This small step can catch calculation errors.

Question 11. When die numerator of a fraction is increased by 4, the fraction increases by \( \frac{2}{3} \). What is the denominator of the fraction?
Answer: Let the original fraction be \( \frac{x}{y} \), where \(x\) is the numerator and \(y\) is the denominator.
According to the condition:
When the numerator is increased by 4, the new fraction is \( \frac{x+4}{y} \).
The problem states that this new fraction is equal to the original fraction plus \( \frac{2}{3} \).
\( \frac{x+4}{y} = \frac{x}{y} + \frac{2}{3} \)
To solve for \(y\), first move the term \( \frac{x}{y} \) to the left side:
\( \frac{x+4}{y} - \frac{x}{y} = \frac{2}{3} \)
Since the denominators are the same, combine the numerators:
\( \frac{(x+4) - x}{y} = \frac{2}{3} \)
\( \frac{4}{y} = \frac{2}{3} \)
Now, cross-multiply to solve for \(y\):
\( 4 \times 3 = 2 \times y \)
\( 12 = 2y \)
\( \implies \) \( y = \frac{12}{2} \)
\( \implies \) \( y = 6 \)
The denominator of the fraction is 6. This problem demonstrates a direct way to find a specific part of a fraction.
In simple words: We have a fraction. If we add 4 to its top number, the fraction becomes \( \frac{2}{3} \) larger than it was before. We need to find the bottom number of this fraction, which is 6.

🎯 Exam Tip: Pay attention to the wording "fraction increases by \( \frac{2}{3} \)". This implies addition to the original fraction, not that the new fraction *is* \( \frac{2}{3} \).

Question 12. Father is six times as old as his son. Four years hence he will be four times as old as his son. Determine their present ages.
Answer: Let the present age of the father be \(x\) years and the present age of the son be \(y\) years.
According to the first condition, the father is six times as old as his son:
(i) \(x = 6y\)
Four years hence (in 4 years from now):
Father's age will be \(x+4\) years.
Son's age will be \(y+4\) years.
According to the second condition, in four years, the father will be four times as old as his son:
\( x+4 = 4(y+4) \)
\( x+4 = 4y + 16 \)
Rearrange this equation:
(ii) \(x = 4y + 12\)
Now we have a system of two linear equations:
(i) \(x = 6y\)
(ii) \(x = 4y + 12\)
Since both equations are equal to \(x\), we can set them equal to each other:
\( 6y = 4y + 12 \)
\( 6y - 4y = 12 \)
\( 2y = 12 \)
\( \implies \) \( y = \frac{12}{2} \)
\( \implies \) \( y = 6 \)
Substitute \(y = 6\) back into equation (i):
\( x = 6 \times 6 \)
\( x = 36 \)
The present age of the father is 36 years and the present age of the son is 6 years. Age problems often involve setting up equations based on current and future/past ages.
In simple words: A father is currently six times older than his son. In four years, the father will be four times older than his son. The father is 36 years old now, and the son is 6 years old now.

🎯 Exam Tip: For age-related problems, always define variables for *present* ages. Then, carefully write expressions for ages in the past or future before forming equations.

Question 13. The age of the father is 3 years more than three times the age of the son. Three years hence father's age will be 10 years more than twice the age of the son. Determine their present ages.
Answer: Let the present age of the son be \(x\) years and the present age of the father be \(y\) years.
According to the first condition, the father's age is 3 years more than three times the son's age:
(i) \(y = 3x + 3\)
Three years hence (in 3 years from now):
Son's age will be \(x+3\) years.
Father's age will be \(y+3\) years.
According to the second condition, in three years, the father's age will be 10 years more than twice the son's age:
\( y+3 = 2(x+3) + 10 \)
\( y+3 = 2x + 6 + 10 \)
\( y+3 = 2x + 16 \)
Rearrange this equation:
(ii) \(y = 2x + 13\)
Now we have a system of two linear equations:
(i) \(y = 3x + 3\)
(ii) \(y = 2x + 13\)
Since both equations are equal to \(y\), we can set them equal to each other:
\( 3x + 3 = 2x + 13 \)
\( 3x - 2x = 13 - 3 \)
\( x = 10 \)
Now substitute \(x = 10\) back into equation (i):
\( y = 3(10) + 3 \)
\( y = 30 + 3 \)
\( y = 33 \)
The present age of the son is 10 years and the present age of the father is 33 years. It's helpful to write down the ages for both current and future/past situations to avoid mistakes.
In simple words: The father's current age is 3 more than three times his son's age. In three years, the father will be 10 years older than twice his son's age. The son is 10 years old and the father is 33 years old.

🎯 Exam Tip: Always double-check your age equations, especially when dealing with "more than," "less than," "times," or "hence" (future) / "ago" (past) scenarios.

Question 14. The age of a man is three times the sum of the ages of his two children and five years hence his age will be double the sum of their ages. Find his present age.
Answer: Let the present age of the man be \(x\) years.
Let the present sum of the ages of his two children be \(y\) years.
According to the first condition, the man's age is three times the sum of his two children's ages:
(i) \(x = 3y\)
Five years hence (in 5 years from now):
Man's age will be \(x+5\) years.
The sum of the children's ages will change. Since there are two children, each child will be 5 years older. So, their combined age will increase by \(5+5 = 10\) years.
Sum of children's ages in 5 years = \(y+10\).
According to the second condition, in five years, the man's age will be double the sum of their ages:
\( x+5 = 2(y+10) \)
\( x+5 = 2y + 20 \)
Rearrange this equation:
(ii) \(x = 2y + 15\)
Now we have a system of two linear equations:
(i) \(x = 3y\)
(ii) \(x = 2y + 15\)
Set the two expressions for \(x\) equal to each other:
\( 3y = 2y + 15 \)
\( 3y - 2y = 15 \)
\( y = 15 \)
Substitute \(y = 15\) back into equation (i):
\( x = 3(15) \)
\( x = 45 \)
The present age of the man is 45 years. This is an important distinction to make for multiple individuals.
In simple words: A man's age is three times what his two children's ages add up to. In five years, his age will be twice what their ages add up to. The man is 45 years old now.

🎯 Exam Tip: When dealing with the sum of ages of multiple individuals (like two children), remember that each individual ages. So, if there are two children, their combined age increases by \(2 \times (\text{number of years})\).

Question 15. Ram's father is 4 times as old as Ram. Five years ago, his father was 9 times as old as he was then. What are their present ages ?
Answer: Let Ram's present age be \(x\) years.
Let Ram's father's present age be \(y\) years.
According to the first condition, Ram's father is 4 times as old as Ram:
(i) \(y = 4x\)
Five years ago:
Ram's age was \(x-5\) years.
Ram's father's age was \(y-5\) years.
According to the second condition, five years ago, his father was 9 times as old as Ram was then:
\( y-5 = 9(x-5) \)
\( y-5 = 9x - 45 \)
Rearrange this equation:
(ii) \(y = 9x - 40\)
Now we have a system of two linear equations:
(i) \(y = 4x\)
(ii) \(y = 9x - 40\)
Set the two expressions for \(y\) equal to each other:
\( 4x = 9x - 40 \)
\( 4x - 9x = -40 \)
\( -5x = -40 \)
\( \implies \) \( x = \frac{-40}{-5} \)
\( \implies \) \( x = 8 \)
Now substitute \(x = 8\) back into equation (i):
\( y = 4(8) \)
\( y = 32 \)
Ram's present age is 8 years and his father's present age is 32 years. This is another example of relating ages across different time periods.
In simple words: Ram's father is currently four times older than Ram. Five years ago, the father was nine times older than Ram was at that time. Ram is 8 years old, and his father is 32 years old.

🎯 Exam Tip: Be careful with the phrasing "his father was 9 times as old as *he* was then." The "he" refers to Ram, so it's 9 times Ram's age *five years ago*.

Question 16. At the time of marriage a man was 6 years older than his wife, but 12 years after his marriage, his age was \( \frac{6}{5} \) th of the age of his wife. What were their ages at the time of marriage?
Answer: Let the man's age at the time of marriage be \(x\) years.
Let his wife's age at the time of marriage be \(y\) years.
According to the first condition, the man was 6 years older than his wife at the time of marriage:
(i) \(x = y + 6\)
Twelve years after their marriage:
Man's age will be \(x+12\) years.
Wife's age will be \(y+12\) years.
According to the second condition, 12 years after marriage, the man's age was \( \frac{6}{5} \) th of his wife's age:
\( x+12 = \frac{6}{5}(y+12) \)
To remove the fraction, multiply both sides by 5:
\( 5(x+12) = 6(y+12) \)
\( 5x + 60 = 6y + 72 \)
Rearrange into standard form:
(ii) \(5x - 6y = 72 - 60 \implies 5x - 6y = 12\)
Now we have a system of two linear equations:
(i) \(x = y + 6\)
(ii) \(5x - 6y = 12\)
Substitute the expression for \(x\) from equation (i) into equation (ii):
\( 5(y+6) - 6y = 12 \)
\( 5y + 30 - 6y = 12 \)
\( -y + 30 = 12 \)
\( -y = 12 - 30 \)
\( -y = -18 \)
\( \implies \) \( y = 18 \)
Now substitute \(y = 18\) back into equation (i):
\( x = 18 + 6 \)
\( x = 24 \)
So, at the time of marriage, the man's age was 24 years and his wife's age was 18 years. This problem uses fractions to describe age relationships.
In simple words: When a man and woman got married, the man was 6 years older. Twelve years later, the man's age was \( \frac{6}{5} \) times his wife's age. At the time of marriage, the man was 24 and the woman was 18.

🎯 Exam Tip: Be careful with fractional age relationships, such as "A is \( \frac{P}{Q} \) th of B". This translates to \(A = \frac{P}{Q} \times B\). Always clear fractions by cross-multiplication early in the process.

Question 17. The present ages of Ram and Shyam are in the ratio 5 : 6. Five years ago, the ratio was 4:5. Find their present ages.
Answer: Let Ram's present age be \(x\) years.
Let Shyam's present age be \(y\) years.
According to the first condition, their present ages are in the ratio 5:6:
\( \frac{x}{y} = \frac{5}{6} \)
Cross-multiply:
\( 6x = 5y \)
(i) \(x = \frac{5}{6}y\)
Five years ago:
Ram's age was \(x-5\) years.
Shyam's age was \(y-5\) years.
According to the second condition, five years ago, their ages were in the ratio 4:5:
\( \frac{x-5}{y-5} = \frac{4}{5} \)
Cross-multiply:
\( 5(x-5) = 4(y-5) \)
\( 5x - 25 = 4y - 20 \)
Rearrange into standard form:
(ii) \(5x - 4y = 25 - 20 \implies 5x - 4y = 5\)
Now we have a system of two linear equations:
(i) \(x = \frac{5}{6}y\)
(ii) \(5x - 4y = 5\)
Substitute the expression for \(x\) from equation (i) into equation (ii):
\( 5\left(\frac{5}{6}y\right) - 4y = 5 \)
\( \frac{25}{6}y - 4y = 5 \)
To combine the \(y\) terms, find a common denominator:
\( \frac{25y}{6} - \frac{24y}{6} = 5 \)
\( \frac{y}{6} = 5 \)
\( \implies \) \( y = 5 \times 6 \)
\( \implies \) \( y = 30 \)
Now substitute \(y = 30\) back into equation (i):
\( x = \frac{5}{6}(30) \)
\( x = 5 \times 5 \)
\( x = 25 \)
Ram's present age is 25 years and Shyam's present age is 30 years. Ratios are a compact way to express relationships.
In simple words: Ram and Shyam's ages are in the ratio 5:6 now. Five years ago, their ages were in the ratio 4:5. Ram is 25 years old, and Shyam is 30 years old.

🎯 Exam Tip: When dealing with ratios like \(A:B\), always translate it into a fraction \(A/B\) to form an equation. Cross-multiplication is then used to remove fractions and simplify into linear equations.

Question 18. The cost of 5 pencils and 6 erasers is Rs. 1.80 whereas that of 3 pencils and 2 erasers is Rs. 0.92. Find the cost of each of a pencil and an eraser.
Answer: Let the cost of one pencil be \(x\) paise and the cost of one eraser be \(y\) paise.
(Note: It's often easier to work with paise if the values are given in both rupees and fractions of rupees, or if one value is less than a rupee.)
Rs. 1.80 = 180 paise
Rs. 0.92 = 92 paise
According to the first condition, the cost of 5 pencils and 6 erasers is Rs. 1.80:
(i) \(5x + 6y = 180\)
According to the second condition, the cost of 3 pencils and 2 erasers is Rs. 0.92:
(ii) \(3x + 2y = 92\)
Now we have a system of two linear equations. We can use the elimination method.
Multiply equation (ii) by 3 to make the coefficient of \(y\) equal to that in equation (i):
\( 3 \times (3x + 2y) = 3 \times 92 \)
\( 9x + 6y = 276 \)
Subtract equation (i) from this new equation:
\( (9x + 6y) - (5x + 6y) = 276 - 180 \)
\( 9x + 6y - 5x - 6y = 96 \)
\( 4x = 96 \)
\( \implies \) \( x = \frac{96}{4} \)
\( \implies \) \( x = 24 \)
So, the cost of one pencil is 24 paise (or Rs. 0.24).
Substitute \(x = 24\) into equation (i):
\( 5(24) + 6y = 180 \)
\( 120 + 6y = 180 \)
\( 6y = 180 - 120 \)
\( 6y = 60 \)
\( \implies \) \( y = \frac{60}{6} \)
\( \implies \) \( y = 10 \)
The cost of one eraser is 10 paise (or Rs. 0.10). This highlights how simultaneous equations can solve real-world pricing problems.
In simple words: If 5 pencils and 6 erasers cost Rs. 1.80, and 3 pencils and 2 erasers cost Rs. 0.92, we want to find the individual cost of each. A pencil costs 24 paise (Rs. 0.24), and an eraser costs 10 paise (Rs. 0.10).

🎯 Exam Tip: When dealing with money, convert all amounts to a single unit (either rupees or paise) at the start to avoid errors in calculation, especially with decimals.

Question 19. 9 chairs and 5 tables cost Rs. 90 while 5 chairs and 4 tables cost Rs. 61. Find the price of 6 chairs and 3 tables.
Answer: Let the cost of one chair be Rs. \(x\).
Let the cost of one table be Rs. \(y\).
According to the first condition, 9 chairs and 5 tables cost Rs. 90:
(i) \(9x + 5y = 90\)
According to the second condition, 5 chairs and 4 tables cost Rs. 61:
(ii) \(5x + 4y = 61\)
Now we have a system of two linear equations. We will use the elimination method.
To eliminate \(y\), multiply equation (i) by 4 and equation (ii) by 5:
From (i): \( 4 \times (9x + 5y) = 4 \times 90 \implies 36x + 20y = 360 \)
From (ii): \( 5 \times (5x + 4y) = 5 \times 61 \implies 25x + 20y = 305 \)
Subtract the second new equation from the first new equation:
\( (36x + 20y) - (25x + 20y) = 360 - 305 \)
\( 36x + 20y - 25x - 20y = 55 \)
\( 11x = 55 \)
\( \implies \) \( x = \frac{55}{11} \)
\( \implies \) \( x = 5 \)
So, the cost of one chair is Rs. 5.
Substitute \(x = 5\) into equation (i):
\( 9(5) + 5y = 90 \)
\( 45 + 5y = 90 \)
\( 5y = 90 - 45 \)
\( 5y = 45 \)
\( \implies \) \( y = \frac{45}{5} \)
\( \implies \) \( y = 9 \)
So, the cost of one table is Rs. 9.
Finally, we need to find the cost of 6 chairs and 3 tables:
Cost = \(6x + 3y = 6(5) + 3(9) = 30 + 27 = 57\).
The price of 6 chairs and 3 tables is Rs. 57. This problem requires an extra step after finding the individual costs.
In simple words: We know the total cost for buying 9 chairs and 5 tables, and also for 5 chairs and 4 tables. We need to find the cost of 6 chairs and 3 tables. First, we figure out that one chair costs Rs. 5 and one table costs Rs. 9. Then, 6 chairs and 3 tables will cost Rs. 57.

🎯 Exam Tip: Read the question carefully to ensure you answer what is asked. After finding the individual costs, remember to calculate the final required total (e.g., cost of 6 chairs and 3 tables).

Question 20. A horse and two cows together cost Rs. 680. If a horse cost Rs. 80 more than a cow, find the cost of each.
Answer: Let the cost of one horse be Rs. \(x\).
Let the cost of one cow be Rs. \(y\).
According to the first condition, a horse and two cows together cost Rs. 680:
(i) \(x + 2y = 680\)
According to the second condition, a horse costs Rs. 80 more than a cow:
(ii) \(x = y + 80\)
Now we have a system of two linear equations. We can use the substitution method.
Substitute the expression for \(x\) from equation (ii) into equation (i):
\( (y + 80) + 2y = 680 \)
\( y + 80 + 2y = 680 \)
\( 3y + 80 = 680 \)
\( 3y = 680 - 80 \)
\( 3y = 600 \)
\( \implies \) \( y = \frac{600}{3} \)
\( \implies \) \( y = 200 \)
So, the cost of one cow is Rs. 200.
Substitute \(y = 200\) back into equation (ii):
\( x = 200 + 80 \)
\( x = 280 \)
The cost of one horse is Rs. 280 and the cost of one cow is Rs. 200. This is a practical application of solving linear equations.
In simple words: One horse and two cows together cost Rs. 680. We also know that the horse costs Rs. 80 more than one cow. By solving these clues, we find that the horse costs Rs. 280 and a cow costs Rs. 200.

🎯 Exam Tip: Pay attention to the wording "A horse and *two* cows." This correctly translates to \(x + 2y\), not just \(x + y\).

Question 21. If one kg of sweets and 3 kg of apples cost Rs. 33 and 2 kg of sweets and one kg of apples cost Rs. 31, what is the cost per kg of each ?
Answer: Let the cost of 1 kg of sweets be Rs. \(x\).
Let the cost of 1 kg of apples be Rs. \(y\).
According to the first condition, 1 kg of sweets and 3 kg of apples cost Rs. 33:
(i) \(x + 3y = 33\)
According to the second condition, 2 kg of sweets and 1 kg of apples cost Rs. 31:
(ii) \(2x + y = 31\)
Now we have a system of two linear equations. We will use the elimination method.
Multiply equation (ii) by 3 to make the coefficient of \(y\) equal to that in equation (i):
\( 3 \times (2x + y) = 3 \times 31 \)
\( 6x + 3y = 93 \)
Subtract equation (i) from this new equation:
\( (6x + 3y) - (x + 3y) = 93 - 33 \)
\( 6x + 3y - x - 3y = 60 \)
\( 5x = 60 \)
\( \implies \) \( x = \frac{60}{5} \)
\( \implies \) \( x = 12 \)
So, the cost of 1 kg of sweets is Rs. 12.
Substitute \(x = 12\) into equation (i):
\( 12 + 3y = 33 \)
\( 3y = 33 - 12 \)
\( 3y = 21 \)
\( \implies \) \( y = \frac{21}{3} \)
\( \implies \) \( y = 7 \)
The cost of 1 kg of apples is Rs. 7. This shows how to find unit prices for multiple items.
In simple words: We're given the total cost for two different combinations of sweets and apples. We need to find out how much 1 kg of sweets costs and how much 1 kg of apples costs. 1 kg of sweets costs Rs. 12, and 1 kg of apples costs Rs. 7.

🎯 Exam Tip: Organize your variables clearly for each item. Choosing elimination or substitution method depends on which coefficients are easier to make equal or which variable is easy to isolate.

Question 22. “A pen costs Rs. 3.50 more than a pencil and the cost of 3 pens and 2 pencils is Rs. 13”. Taking x and y as the cost (in Rs) of a pen and a pencil respectively, write two simultaneous equations in x and y which satisfy the statement given above. (Do not solve the equations).
Answer: Let the cost of one pen be Rs. \(x\).
Let the cost of one pencil be Rs. \(y\).
According to the first part of the statement, a pen costs Rs. 3.50 more than a pencil:
\( x = y + 3.50 \)
Rearrange this into a standard linear equation form:
(i) \(x - y = 3.50\)
According to the second part of the statement, the cost of 3 pens and 2 pencils is Rs. 13:
(ii) \(3x + 2y = 13\)
The two simultaneous equations are:
\(x - y = 3.50\)
\(3x + 2y = 13\)
(Note: The problem asks not to solve, but the source solution proceeds to solve. For completeness and internal consistency with the provided solution steps, the values are derived below):
From (i), substitute \(x = y + 3.5\) into (ii):
\( 3(y + 3.5) + 2y = 13 \)
\( 3y + 10.5 + 2y = 13 \)
\( 5y + 10.5 = 13 \)
\( 5y = 13 - 10.5 \)
\( 5y = 2.5 \)
\( \implies \) \( y = \frac{2.5}{5} \)
\( \implies \) \( y = 0.5 \)
So, the cost of one pencil is Rs. 0.50 (or 50 paise).
Substitute \(y = 0.5\) back into \(x = y + 3.5\):
\( x = 0.5 + 3.5 \)
\( x = 4.0 \)
The cost of one pen is Rs. 4.00. Setting up the equations correctly is the first step in solving.
In simple words: We have two facts: a pen costs Rs. 3.50 more than a pencil, and buying 3 pens and 2 pencils costs Rs. 13. We need to write these as two math sentences (equations). The equations are \(x - y = 3.50\) and \(3x + 2y = 13\). (A pen costs Rs. 4.00, and a pencil costs Rs. 0.50).

🎯 Exam Tip: Always clearly define your variables before writing the equations. Double-check that each part of the word problem is correctly translated into an algebraic expression or equation.

Question 22. “A pen costs Rs. 3.50 more than a pencil and the cost of 3 pens and 2 pencils is Rs. 13”. Taking x and y as the cost (in Rs) of a pen and a pencil respectively, write two simultaneous equations in x and y which satisfy the statement given above. (Do not solve the equations).
Answer: Let the cost of one pen be Rs. x.
Let the cost of one pencil be Rs. y.
According to the first statement, a pen costs Rs. 3.50 more than a pencil:
\( x = y + 3.5 \)

\( \implies x - y = 3.5 \dots (i) \)
According to the second statement, the cost of 3 pens and 2 pencils is Rs. 13:
\( 3x + 2y = 13 \dots (ii) \)
The two simultaneous equations are \( x - y = 3.5 \) and \( 3x + 2y = 13 \). These equations help us represent the given word problem mathematically.
In simple words: We are told to set up two math sentences (equations) using 'x' for the pen's cost and 'y' for the pencil's cost. The first equation shows that a pen is Rs. 3.50 more expensive. The second equation shows the total cost of 3 pens and 2 pencils.

🎯 Exam Tip: Read "Do not solve the equations" carefully. Sometimes questions test your ability to form equations, not just solve them. Clearly define your variables (e.g., let x be cost of pen). Make sure to write the equations in a standard linear form (Ax + By = C).

Question 23. A man buys postage stamps of denominations 3 paise and 5 paise, for Re 1.00. He buys 22 stamps in all. Find the number of 3 paise stamps bought by him.
Answer: Total number of stamps bought = 22.
Total value of stamps = Re 1.00 = 100 paise.
Let the number of 3 paise stamps be x.
Let the number of 5 paise stamps be y.
Since he buys 22 stamps in all, we have:
\( x + y = 22 \dots (i) \)
The total value of the stamps is 100 paise. So, the value from x stamps of 3 paise and y stamps of 5 paise is:
\( 3x + 5y = 100 \dots (ii) \)
From equation (i), we can write \( y = 22 - x \).
Substitute this value of y into equation (ii):
\( 3x + 5(22 - x) = 100 \)

\( \implies 3x + 110 - 5x = 100 \)

\( \implies -2x = 100 - 110 \)

\( \implies -2x = -10 \)

\( \implies x = \frac{-10}{-2} \)

\( \implies x = 5 \)
Therefore, the number of 3 paise stamps bought is 5. Using paise for calculations helps avoid decimals and makes the arithmetic clearer.
In simple words: We know the total stamps and their total cost in paise. We set up two equations, one for the total count of stamps and one for their total value. Then we solve these equations to find how many 3 paise stamps were bought.

🎯 Exam Tip: When dealing with money, always ensure all values are in the same unit (e.g., all in paise or all in rupees) before forming equations to avoid errors.

Question 24. There were 2500 persons who bought tickets to see a village fair. The adults paid 75 paise each for their admission tickets but the children paid 25 paise each. If the total receipts amounted to Rs. 1503, using an equation method, find how many children saw the fair?
Answer: Total number of persons = 2500.
Cost of an adult ticket = 75 paise.
Cost of a child ticket = 25 paise.
Total receipts = Rs. 1503 = 150300 paise.
Let the number of children be x.
Let the number of adults be y.
From the total number of persons:
\( x + y = 2500 \dots (i) \)
From the total receipts:
\( 25x + 75y = 150300 \dots (ii) \)
From equation (i), we can express y as: \( y = 2500 - x \).
Substitute this expression for y into equation (ii):
\( 25x + 75(2500 - x) = 150300 \)

\( \implies 25x + 187500 - 75x = 150300 \)

\( \implies 187500 - 50x = 150300 \)

\( \implies -50x = 150300 - 187500 \)

\( \implies -50x = -37200 \)

\( \implies x = \frac{-37200}{-50} \)

\( \implies x = 744 \)
So, the number of children who saw the fair is 744. Real-world problems often involve converting units like rupees to paise to simplify calculations.
To find the number of adults: \( y = 2500 - 744 = 1756 \).
In simple words: We used two facts: the total number of people and the total money collected. We made two equations, one for each fact. By solving these equations, we found out how many children were at the fair.

🎯 Exam Tip: Always convert all money units to the smallest common unit (e.g., paise) at the beginning of the problem to avoid decimal errors and simplify calculations. Clearly label what each variable represents.

Question 25. In a triangle, the sum of the two angles is equal to the third. If the difference between them is 50°, determine the angles.
Answer: Let the three angles of the triangle be A, B, and C.
According to the problem, the sum of two angles is equal to the third angle. Let's say \( A + B = C \).
We know that the sum of all angles in a triangle is 180°: \( A + B + C = 180^\circ \).
Substitute \( C \) with \( (A + B) \) in the sum of angles equation:
\( (A + B) + C = 180^\circ \)
\( C + C = 180^\circ \)
\( 2C = 180^\circ \)
\( C = 90^\circ \)
Now we know the third angle is 90°. This makes the triangle a right-angled triangle.
From \( A + B = C \), we have \( A + B = 90^\circ \dots (i) \).
The problem also states that the difference between the two angles (A and B) is 50°:
Let \( A - B = 50^\circ \dots (ii) \)
Now, we solve equations (i) and (ii):
Add (i) and (ii):
\( (A + B) + (A - B) = 90^\circ + 50^\circ \)
\( 2A = 140^\circ \)
\( A = \frac{140^\circ}{2} \)
\( A = 70^\circ \)
Substitute \( A = 70^\circ \) into equation (i):
\( 70^\circ + B = 90^\circ \)
\( B = 90^\circ - 70^\circ \)
\( B = 20^\circ \)
So, the three angles of the triangle are \( 70^\circ \), \( 20^\circ \), and \( 90^\circ \).
In simple words: We used the rules for triangles to find the angles. Since two angles add up to the third, and all three add up to 180 degrees, we figured out that one angle must be 90 degrees. Then, using the difference given, we easily found the other two angles.

🎯 Exam Tip: Remember the fundamental property that the sum of angles in a triangle is 180°. Clearly define which angles are being referred to when a difference or sum is mentioned.

Question 26. In a parallelogram, one angle is \( \frac { 2 }{ 5 } \)th of the adjacent angles. Determine the angles of this parallelogram.
Answer: In a parallelogram, adjacent angles are supplementary, meaning their sum is 180°.
Let the two adjacent angles be x and y.
So, \( x + y = 180^\circ \dots (i) \)
According to the problem, one angle is \( \frac{2}{5} \)th of the adjacent angle. Let's assume \( x = \frac{2}{5}y \dots (ii) \).
Now, substitute the value of x from equation (ii) into equation (i):
\( \frac{2}{5}y + y = 180^\circ \)
To remove the fraction, multiply the entire equation by 5:
\( 5 \left( \frac{2}{5}y \right) + 5y = 5(180^\circ) \)
\( 2y + 5y = 900^\circ \)
\( 7y = 900^\circ \)
\( y = \frac{900^\circ}{7} \)
Now, find x using \( x = \frac{2}{5}y \):
\( x = \frac{2}{5} \times \frac{900^\circ}{7} \)
\( x = \frac{1800^\circ}{35} \)
\( x = \frac{360^\circ}{7} \)
In a parallelogram, opposite angles are equal. So, the angles of the parallelogram are \( \frac{360^\circ}{7} \), \( \frac{900^\circ}{7} \), \( \frac{360^\circ}{7} \), and \( \frac{900^\circ}{7} \). A parallelogram always has two pairs of equal angles.
In simple words: We used the rule that angles next to each other in a parallelogram add up to 180 degrees. We also used the information that one angle is 2/5 of the other. We solved these two rules like a puzzle to find the values of both angles.

🎯 Exam Tip: Always remember the properties of parallelograms, especially that adjacent angles are supplementary and opposite angles are equal. This is key to setting up the correct equations.

Question 27. If the length and the breadth of a room are increased by 1 metre, the area is increased by 21 square metres. If the length is increased by 1 metre and breadth is decreased by 1 metre the area is decreased by 5 square metres. Find the perimeter of the room.
Answer: Let the original length of the room be x metres and the original breadth be y metres.
The original area of the room is \( xy \) square metres.
**Case 1:** Length and breadth are increased by 1 metre.
New length = \( (x+1) \) metres
New breadth = \( (y+1) \) metres
New area = \( (x+1)(y+1) \) square metres.
The area is increased by 21 square metres, so:
\( (x+1)(y+1) = xy + 21 \)

\( \implies xy + x + y + 1 = xy + 21 \)

\( \implies x + y + 1 = 21 \)

\( \implies x + y = 20 \dots (i) \)
**Case 2:** Length is increased by 1 metre and breadth is decreased by 1 metre.
New length = \( (x+1) \) metres
New breadth = \( (y-1) \) metres
New area = \( (x+1)(y-1) \) square metres.
The area is decreased by 5 square metres, so:
\( (x+1)(y-1) = xy - 5 \)

\( \implies xy - x + y - 1 = xy - 5 \)

\( \implies -x + y - 1 = -5 \)

\( \implies -x + y = -4 \dots (ii) \)
Now, we have a system of two linear equations:
1. \( x + y = 20 \)
2. \( -x + y = -4 \)
Add equation (i) and equation (ii):
\( (x + y) + (-x + y) = 20 + (-4) \)
\( 2y = 16 \)
\( y = \frac{16}{2} \)
\( y = 8 \)
Substitute \( y = 8 \) into equation (i):
\( x + 8 = 20 \)
\( x = 20 - 8 \)
\( x = 12 \)
So, the length of the room is 12 metres and the breadth is 8 metres.
The perimeter of the room is given by \( 2(\text{length} + \text{breadth}) \).
Perimeter = \( 2(12 + 8) = 2(20) = 40 \) metres. Knowing the dimensions helps in planning for furniture or renovations.
In simple words: We used the given changes in length, breadth, and area to create two equations. Solving these equations gave us the room's original length and breadth. Then, we used these to find the perimeter.

🎯 Exam Tip: When dealing with geometric shapes, always start by defining the original dimensions and then write expressions for the new dimensions based on the problem statement. Be careful with algebraic expansion and signs.

Question 28. A sailor goes 8 km downstream in 40 minutes and returns in 1 hour. Determine the speed of the sailor in still water and the speed of the current.
Answer: Distance = 8 km.
Let the speed of the sailor in still water be x km/hr.
Let the speed of the current be y km/hr.
When going downstream, the speeds add up: Speed downstream = \( (x+y) \) km/hr.
When going upstream, the current opposes: Speed upstream = \( (x-y) \) km/hr.
Time taken downstream = 40 minutes = \( \frac{40}{60} \) hours = \( \frac{2}{3} \) hours.
Time taken upstream = 1 hour.
Using the formula: Time = \( \frac{\text{Distance}}{\text{Speed}} \)
For downstream journey:
\( \frac{8}{x+y} = \frac{2}{3} \)
Cross-multiply:
\( 3 \times 8 = 2 \times (x+y) \)
\( 24 = 2(x+y) \)
\( x+y = \frac{24}{2} \)
\( x+y = 12 \dots (i) \)
For upstream journey:
\( \frac{8}{x-y} = 1 \)
\( x-y = 8 \dots (ii) \)
Now we solve the system of equations (i) and (ii):
1. \( x + y = 12 \)
2. \( x - y = 8 \)
Add (i) and (ii):
\( (x + y) + (x - y) = 12 + 8 \)
\( 2x = 20 \)
\( x = \frac{20}{2} \)
\( x = 10 \)
Substitute \( x = 10 \) into equation (i):
\( 10 + y = 12 \)
\( y = 12 - 10 \)
\( y = 2 \)
So, the speed of the sailor in still water is 10 km/hr and the speed of the current is 2 km/hr. This type of problem is often seen in physics too.
In simple words: We set up equations for the boat's journey going with the current (downstream) and against it (upstream). We used the given times and distances to find the two speeds: the boat's own speed and the speed of the water.

🎯 Exam Tip: Remember that for downstream motion, speeds add (\( x+y \)), and for upstream motion, speeds subtract (\( x-y \)). Always convert time to hours to match speed in km/hr.

Question 29. The boat goes 24 km upstream and 28 km downstream in 6 hours. It goes 30 km upstream and 21 km downstream, in \( 6\frac { 1 }{ 2 } \) hours. Find the speed of the boat in still water and also the speed of the stream.
Answer: Let the speed of the boat in still water be x km/hr.
Let the speed of the stream (current) be y km/hr.
Speed upstream = \( (x-y) \) km/hr.
Speed downstream = \( (x+y) \) km/hr.
Time = \( \frac{\text{Distance}}{\text{Speed}} \)
**First case:** 24 km upstream and 28 km downstream in 6 hours.
\( \frac{24}{x-y} + \frac{28}{x+y} = 6 \dots (i) \)
**Second case:** 30 km upstream and 21 km downstream in \( 6\frac{1}{2} \) hours (which is \( \frac{13}{2} \) hours).
\( \frac{30}{x-y} + \frac{21}{x+y} = \frac{13}{2} \dots (ii) \)
To solve these equations, let \( \frac{1}{x-y} = b \) and \( \frac{1}{x+y} = a \).
The equations become:
\( 24b + 28a = 6 \dots (iii) \)
\( 30b + 21a = \frac{13}{2} \dots (iv) \)
Multiply equation (iii) by 3 and equation (iv) by 4 to eliminate 'a':
From (iii) \( \times 3 \): \( 72b + 84a = 18 \dots (v) \)
From (iv) \( \times 4 \): \( 120b + 84a = 26 \dots (vi) \)
Subtract equation (v) from equation (vi):
\( (120b + 84a) - (72b + 84a) = 26 - 18 \)
\( 48b = 8 \)
\( b = \frac{8}{48} \)
\( b = \frac{1}{6} \)
Now substitute \( b = \frac{1}{6} \) into equation (iii):
\( 24\left(\frac{1}{6}\right) + 28a = 6 \)
\( 4 + 28a = 6 \)
\( 28a = 6 - 4 \)
\( 28a = 2 \)
\( a = \frac{2}{28} \)
\( a = \frac{1}{14} \)
Now, we convert back to x and y:
\( \frac{1}{x-y} = b \implies \frac{1}{x-y} = \frac{1}{6} \implies x-y = 6 \dots (vii) \)
\( \frac{1}{x+y} = a \implies \frac{1}{x+y} = \frac{1}{14} \implies x+y = 14 \dots (viii) \)
Add equation (vii) and equation (viii):
\( (x-y) + (x+y) = 6 + 14 \)
\( 2x = 20 \)
\( x = 10 \)
Substitute \( x = 10 \) into equation (viii):
\( 10 + y = 14 \)
\( y = 14 - 10 \)
\( y = 4 \)
The speed of the boat in still water is 10 km/hr, and the speed of the stream is 4 km/hr. This method of substitution simplifies complicated equations into a more manageable form.
In simple words: We used two different travel scenarios to create two main equations. Since these were tricky, we changed parts of the equations into simpler letters (a and b) and solved for them. Then, we changed back to the boat's speed and stream's speed.

🎯 Exam Tip: For complex boat/stream problems, using substitution like \( \frac{1}{x-y} = b \) and \( \frac{1}{x+y} = a \) simplifies the equations into a standard linear form, making them much easier to solve.

Question 30. Satish and Ashok start at the same time from two places 21 km apart. If they walk in the same direction Satish overtakes Ashok in 21 hours but if they walk in opposite directions, they meet in 3 hours. Find the rate at which each of them walks.
Answer: Let Satish's speed be x km/hr and Ashok's speed be y km/hr.
The distance between their starting points is 21 km.
**Case 1: Walking in the same direction.**
When they walk in the same direction, Satish overtakes Ashok in 21 hours. This means Satish covers 21 km more than Ashok in 21 hours.
Distance covered by Satish = \( 21x \) km.
Distance covered by Ashok = \( 21y \) km.
So, \( 21x - 21y = 21 \)
Divide by 21:
\( x - y = 1 \dots (i) \)
**Case 2: Walking in opposite directions.**
When they walk in opposite directions, they meet in 3 hours. This means their combined distance covered is 21 km.
Distance covered by Satish = \( 3x \) km.
Distance covered by Ashok = \( 3y \) km.
So, \( 3x + 3y = 21 \)
Divide by 3:
\( x + y = 7 \dots (ii) \)
Now, solve equations (i) and (ii):
1. \( x - y = 1 \)
2. \( x + y = 7 \)
Add (i) and (ii):
\( (x - y) + (x + y) = 1 + 7 \)
\( 2x = 8 \)
\( x = \frac{8}{2} \)
\( x = 4 \)
Substitute \( x = 4 \) into equation (ii):
\( 4 + y = 7 \)
\( y = 7 - 4 \)
\( y = 3 \)
Therefore, Satish's walking speed is 4 km/hr and Ashok's walking speed is 3 km/hr. Relative speed concepts are important when objects move towards or away from each other.
In simple words: We used how far apart Satish and Ashok are and how long it took them to meet or for one to catch the other. When walking in the same direction, we looked at the difference in distance. When walking towards each other, we looked at their combined distance. This helped us find each person's speed.

🎯 Exam Tip: For problems involving relative speed, remember that when objects move in the same direction, their relative speed is the difference of their speeds (\( x-y \)). When they move in opposite directions, their relative speed is the sum of their speeds (\( x+y \)).

Question 31. A boy walks to a picnic spot from his house in 6 hours, but he can travel the same distance on his cycle in 2 hours. If his average cycling speed is 7 kilometers per hour faster than his average walking speed, find his average walking speed and his average cycling speed.
Answer: Let the boy's average walking speed be x km/hr.
Let his average cycling speed be y km/hr.
He walks to the picnic spot in 6 hours and cycles the same distance in 2 hours.
Distance covered by walking = Speed \( \times \) Time = \( x \times 6 = 6x \) km.
Distance covered by cycling = Speed \( \times \) Time = \( y \times 2 = 2y \) km.
Since the distance to the picnic spot is the same in both cases:
\( 6x = 2y \)
Divide by 2:
\( 3x = y \dots (i) \)
The problem states that his average cycling speed is 7 km/hr faster than his average walking speed:
\( y = x + 7 \dots (ii) \)
Now, we have a system of two equations. Substitute equation (i) into equation (ii):
\( 3x = x + 7 \)

\( \implies 3x - x = 7 \)

\( \implies 2x = 7 \)

\( \implies x = \frac{7}{2} \)

\( \implies x = 3.5 \)
Now, find y using \( y = 3x \):
\( y = 3 \times 3.5 \)
\( y = 10.5 \)
Therefore, the boy's average walking speed is 3.5 km/hr and his average cycling speed is 10.5 km/hr. Traveling the same distance in less time naturally means a higher speed.
In simple words: We used the time taken for walking and cycling the same distance to make one equation. The second equation came from the fact that cycling is 7 km/hr faster. Solving these equations told us both his walking and cycling speeds.

🎯 Exam Tip: Always remember that Distance = Speed × Time. When the distance is constant, you can set expressions for distance equal to each other to form an equation.

Question 32. A man travels 600 km partly by train and partly by car. If he covers 400 km by train and the rest by car, it takes him 6 hours and 30 minutes. But if he travels 200 km by train and the rest by car, he takes half an hour longer. Find the speed of the train and that of the car.
Answer: Total distance to travel = 600 km.
Let the speed of the train be x km/hr.
Let the speed of the car be y km/hr.
Time = \( \frac{\text{Distance}}{\text{Speed}} \)
**First scenario:**
Travels 400 km by train. The rest by car means \( 600 - 400 = 200 \) km by car.
Total time = 6 hours 30 minutes = \( 6.5 \) hours = \( \frac{13}{2} \) hours.
\( \frac{400}{x} + \frac{200}{y} = \frac{13}{2} \dots (i) \)
**Second scenario:**
Travels 200 km by train. The rest by car means \( 600 - 200 = 400 \) km by car.
Total time = \( 6.5 \) hours \( + \) half an hour longer = \( 6.5 + 0.5 = 7 \) hours.
\( \frac{200}{x} + \frac{400}{y} = 7 \dots (ii) \)
To simplify, let \( A = \frac{1}{x} \) and \( B = \frac{1}{y} \).
The equations become:
\( 400A + 200B = \frac{13}{2} \dots (iii) \)
\( 200A + 400B = 7 \dots (iv) \)
Multiply equation (iii) by 2 to make the coefficient of B equal to 400:
\( 2(400A + 200B) = 2\left(\frac{13}{2}\right) \)
\( 800A + 400B = 13 \dots (v) \)
Now, subtract equation (iv) from equation (v):
\( (800A + 400B) - (200A + 400B) = 13 - 7 \)
\( 600A = 6 \)
\( A = \frac{6}{600} \)
\( A = \frac{1}{100} \)
Since \( A = \frac{1}{x} \), we have \( \frac{1}{x} = \frac{1}{100} \implies x = 100 \).
Substitute \( A = \frac{1}{100} \) into equation (iv):
\( 200\left(\frac{1}{100}\right) + 400B = 7 \)
\( 2 + 400B = 7 \)
\( 400B = 7 - 2 \)
\( 400B = 5 \)
\( B = \frac{5}{400} \)
\( B = \frac{1}{80} \)
Since \( B = \frac{1}{y} \), we have \( \frac{1}{y} = \frac{1}{80} \implies y = 80 \).
Thus, the speed of the train is 100 km/hr and the speed of the car is 80 km/hr. This problem shows how different modes of transport affect total travel time over a fixed distance.
In simple words: We used the given travel details to set up two equations about the speeds of the train and car. We then made the equations simpler by replacing fractions with new letters. After solving for those, we found the actual speeds of the train and the car.

🎯 Exam Tip: Convert mixed units like "6 hours 30 minutes" to a single unit (e.g., hours or minutes) at the start. Using reciprocal substitution (like \( A = \frac{1}{x} \)) is a powerful technique for solving equations with variables in the denominator.