CBSE Class 10 Mathematics Linear Equations Assignment Set 11

Q1. The sum of the numerator and the denominator of a fraction is 3 less than twice the denominator. If each of the numerator and denominator is decreased by 1, the fraction becomes \(\frac{1}{2}\). Find the fraction.
Answer: Let the numerator of the fraction be \(x\) and the denominator be \(y\).
The fraction is \(\frac{x}{y}\).
According to the first condition, the sum of the numerator and the denominator is 3 less than twice the denominator:
\(x + y = 2y - 3\)
\(x - y = -3\)     (Equation 1)
According to the second condition, if each of the numerator and denominator is decreased by 1, the fraction becomes \(\frac{1}{2}\):
\(\frac{x - 1}{y - 1} = \frac{1}{2}\)
\(2(x - 1) = y - 1\)
\(2x - 2 = y - 1\)
\(2x - y = 1\)     (Equation 2)
Subtracting Equation 1 from Equation 2:
\((2x - y) - (x - y) = 1 - (-3)\)
\(x = 4\)
Substitute \(x = 4\) into Equation 1:
\(4 - y = -3\)
\(y = 7\)
Thus, the numerator is 4 and the denominator is 7.
The required fraction is \(\frac{4}{7}\).
In simple words: Set up two algebraic equations based on the problem statements and solve them simultaneously to find the numerator and denominator.

Exam Tip: Always double check your final fraction by applying the given conditions directly to it.

Q2. Find a two digit number which is three times the sum of the digits. If 45 is added to the number, the digits change their places. Find the number. [Ans: 27]
Answer: Let the tens digit be \(x\) and the units digit be \(y\).
The two-digit number is \(10x + y\).
According to the first condition, the number is three times the sum of its digits:
\(10x + y = 3(x + y)\)
\(10x + y = 3x + 3y\)
\(7x - 2y = 0\)     (Equation 1)
According to the second condition, if 45 is added to the number, the digits interchange their places:
\(10x + y + 45 = 10y + x\)
\(9x - 9y = -45\)
\(x - y = -5\)     (Equation 2)
Multiply Equation 2 by 2:
\(2x - 2y = -10\)     (Equation 3)
Subtract Equation 3 from Equation 1:
\((7x - 2y) - (2x - 2y) = 0 - (-10)\)
\(5x = 10\)
\(x = 2\)
Substitute \(x = 2\) into Equation 2:
\(2 - y = -5\)
\(y = 7\)
Therefore, the two-digit number is \(10(2) + 7 = 27\).
In simple words: Express a two-digit number as \(10x + y\), form equations using the sum and interchanged digits, and solve for \(x\) and \(y\).

Exam Tip: Remember that any two-digit number with tens digit \(x\) and units digit \(y\) is written algebraically as \(10x + y\).

Q3. 12 men and 16 boys can do a piece of work in 5 days whereas 13 men and 24 boys can do it in 4 days. How long will it take to finish the work 1 man or 1 boy? [Ans: 1 Man in 100 days, 1 boy in 200 days]
Answer: Let 1 man alone finish the work in \(x\) days, so 1 man's 1 day work is \(\frac{1}{x}\).
Let 1 boy alone finish the work in \(y\) days, so 1 boy's 1 day work is \(\frac{1}{y}\).
According to the first condition, 12 men and 16 boys complete the work in 5 days, so their 1 day work is \(\frac{1}{5}\):
\(\frac{12}{x} + \frac{16}{y} = \frac{1}{5}\)     (Equation 1)
According to the second condition, 13 men and 24 boys complete the work in 4 days, so their 1 day work is \(\frac{1}{4}\):
\(\frac{13}{x} + \frac{24}{y} = \frac{1}{4}\)     (Equation 2)
Let \(u = \frac{1}{x}\) and \(v = \frac{1}{y}\). The equations become:
\(12u + 16v = \frac{1}{5}\)
\(13u + 24v = \frac{1}{4}\)
Multiply Equation 1 by 3 and Equation 2 by 2:
\(36u + 48v = \frac{3}{5}\)
\(26u + 48v = \frac{1}{2}\)
Subtract the second modified equation from the first:
\(10u = \frac{3}{5} - \frac{1}{2} = \frac{6 - 5}{10} = \frac{1}{10}\)
\(u = \frac{1}{100}\)
Thus, \(x = 100\).
Substitute \(u = \frac{1}{100}\) into \(12\left(\frac{1}{100}\right) + 16v = \frac{1}{5}\):
\(\frac{12}{100} + 16v = \frac{1}{5}\)
\(16v = \frac{1}{5} - \frac{3}{25} = \frac{5 - 3}{25} = \frac{2}{25}\)
\(v = \frac{2}{25 \times 16} = \frac{1}{200}\)
Thus, \(y = 200\).
1 man can finish the work in 100 days and 1 boy can finish the work in 200 days.
In simple words: Convert the problem into the amount of work done in one day by one man and one boy, use substitution to solve, and invert to find total days.

Exam Tip: Work and time problems are best solved by taking the reciprocal of the rates (1 day work) and using substitution for simpler linear equations.

Q4. 5 years hence a father will be 3 times as old as the son and 5 years ago he was seven times as old as the son. How old are they now? [Ans: 40 yrs, 10 yrs]
Answer: Let the present age of the father be \(x\) years and the present age of the son be \(y\) years.
5 years hence (after 5 years):
Father's age = \(x + 5\)
Son's age = \(y + 5\)
According to the first condition:
\(x + 5 = 3(y + 5)\)
\(x + 5 = 3y + 15\)
\(x - 3y = 10\)     (Equation 1)
5 years ago:
Father's age = \(x - 5\)
Son's age = \(y - 5\)
According to the second condition:
\(x - 5 = 7(y - 5)\)
\(x - 5 = 7y - 35\)
\(x - 7y = -30\)     (Equation 2)
Subtract Equation 2 from Equation 1:
\((x - 3y) - (x - 7y) = 10 - (-30)\)
\(4y = 40\)
\(y = 10\)
Substitute \(y = 10\) into Equation 1:
\(x - 3(10) = 10\)
\(x - 30 = 10\)
\(x = 40\)
The present age of the father is 40 years and the present age of the son is 10 years.
In simple words: Define present ages as variables, write expressions for their ages in the future and past, and solve the resulting equations.

Exam Tip: In age-related problems, always define variables for present ages first before adding or subtracting years.

Q5. A boat goes upstream 42 km and downstream 72 km in 13 hours. It also goes upstream 36 km and downstream 48 km in 10 hours. Find the speed of the boat (streamer) and stream. Ans (9 km, 3 km)
Answer: Let the speed of the boat in still water be \(u\) km/h and the speed of the stream be \(v\) km/h.
Upstream speed = \((u - v)\) km/h.
Downstream speed = \((u + v)\) km/h.
According to the first condition:
\(\frac{42}{u - v} + \frac{72}{u + v} = 13\)     (Equation 1)
According to the second condition:
\(\frac{36}{u - v} + \frac{48}{u + v} = 10\)     (Equation 2)
Let \(X = \frac{1}{u - v}\) and \(Y = \frac{1}{u + v}\). The equations become:
\(42X + 72Y = 13\)
\(36X + 48Y = 10\)
Multiply Equation 1 by 2 and Equation 2 by 3:
\(84X + 144Y = 26\)
\(108X + 144Y = 30\)
Subtract the first modified equation from the second:
\(24X = 4\)
\(X = \frac{4}{24} = \frac{1}{6}\)
So, \(u - v = 6\)     (Equation 3)
Substitute \(X = \frac{1}{6}\) into \(42X + 72Y = 13\):
\(42\left(\frac{1}{6}\right) + 72Y = 13\)
\(7 + 72Y = 13\)
\(72Y = 6\)
\(Y = \frac{6}{72} = \frac{1}{12}\)
So, \(u + v = 12\)     (Equation 4)
Adding Equation 3 and Equation 4:
\(2u = 18\)
\(u = 9\)
Subtracting Equation 3 from Equation 4:
\(2v = 6\)
\(v = 3\)
Thus, the speed of the boat in still water is 9 km/h and the speed of the stream is 3 km/h.
In simple words: Use upstream and downstream speeds as combined variables, solve for them using substitution, and then find the individual boat and stream speeds.

Exam Tip: Remember upstream speed is \((u - v)\) and downstream speed is \((u + v)\), where boat speed \(u\) is always greater than stream speed \(v\).

Q6. In a triangle ABC, angle C = 3 angle B = 2 (angle A + angle B). Find the three angles. [Ans: angle A = 20 degrees, angle B = 40 degrees, angle C = 120 degrees]
Answer: Let the angles of triangle ABC be \(A\), \(B\), and \(C\).
We know that the sum of angles in a triangle is 180 degrees:
\(A + B + C = 180^\circ\)     (Equation 1)
From the given relation:
\(C = 3B\)
Also:
\(3B = 2(A + B)\)
\(3B = 2A + 2B\)
\(2A = B\)
\(A = \frac{B}{2}\)
Substitute \(A = \frac{B}{2}\) and \(C = 3B\) into Equation 1:
\(\frac{B}{2} + B + 3B = 180^\circ\)
\(\frac{B}{2} + 4B = 180^\circ\)
\(\frac{9B}{2} = 180^\circ\)
\(9B = 360^\circ\)
\(B = 40^\circ\)
Now find angle \(A\) and angle \(C\):
\(A = \frac{40^\circ}{2} = 20^\circ\)
\(C = 3(40^\circ) = 120^\circ\)
Thus, angle \(A = 20^\circ\), angle \(B = 40^\circ\), and angle \(C = 120^\circ\).
In simple words: Express all angles in terms of a single variable (like angle B) using the given equations, substitute them into the angle sum property of triangles, and solve.

Exam Tip: Always verify that your calculated triangle angles add up to exactly 180 degrees.