CBSE Class 10 Maths HOTs Pair Of Linear Equations In Two Variables Set 04

Very short Questions

Question. Write the relationship between the coefficients, if the following pair of equations are inconsistent.
\( ax + by + c = 0; a'x + b'y + c' = 0 \).
Answer: The required relationship is:
\[ \frac{a}{a'} = \frac{b}{b'} \neq \frac{c}{c'} \]

Question. When will the system \( kx - y = 2 \) and \( 6x - 2y = 3 \) has a unique solution only? 
Answer: A pair of linear pair has unique solution only when,
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Then, \( \frac{k}{6} \neq \frac{-1}{-2} \)
So, \( k \neq 3 \).

Question. Find the solution of \( x + y = 3 \) and \( 7x + 6y = 2 \).
Answer: \( x + y = 3 \) gives, \( y = 3 - x \) ...(i)
So, \( 7x + 6y = 2 \) gives \( 7x + 6(3 - x) = 2 \)

\( \implies \) \( 7x + 18 - 6x = 2 \)
i.e. \( x = -16 \)
From (i), \( y = 3 + 16 = 19 \)
Thus, \( x = -16 \) and \( y = 19 \) is the required solution.

Short Answer (SA-I) Type Questions

Question. Find the value(s) of \( k \) for which the pair of equations \( \begin{cases} kx + 2y = 3 \\ 3x + 6y = 10 \end{cases} \) has a unique solution. 
Answer: Given: pair of equation is \( kx + 2y = 3 \)
\( 3x + 6y = 10 \)
For a unique solution, \( \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \)
Here, \( a_1 = k, b_1 = 2, a_2 = 3, b_2 = 6 \)
\( \therefore \frac{k}{3} \neq \frac{2}{6} \)

\( \implies \) \( k \neq 1 \)
Hence, the pair of equation has a unique solution for all real values of \( k \) except 1.

Question. The larger of two supplementary angles exceeds the smaller by \( 18^\circ \). Find the angles. 
Answer: Let, the smaller angle be \( x \) and the larger angle be \( y \).
According to the given conditions:
\( y = x + 18^\circ \)
or \( -x + y = 18^\circ \) ...(i)
and \( x + y = 180^\circ \) ...(ii)
(Sum of the supplementary angles is \( 180^\circ \))
Now, on adding equation (i) and (ii), we get:
\( -x + y = 18^\circ \)
\( x + y = 180^\circ \)
\( 2y = 198 \)

\( \implies \) \( y = 99^\circ \)
Put the value of \( y \) in equation (i), we get
\( -x + 99 = 18^\circ \)

\( \implies \) \( x = 99^\circ - 18^\circ \)
\( x = 81^\circ \)
Hence, the two supplementary angles are \( 81^\circ \) and \( 99^\circ \).

Question. In a \( \Delta ABC, \angle A = x^\circ, \angle B = 3x^\circ \) and \( \angle C = y^\circ \). If \( 3y^\circ - 5x^\circ = 30^\circ \) prove that the triangle is right angled. 
Answer: We know that,
\( \angle A + \angle B + \angle C = 180^\circ \)
[Sum of interior angles of triangle ABC is \( 180^\circ \)]

\( \implies \) \( x + 3x + y = 180^\circ \)

\( \implies \) \( 4x + y = 180^\circ \) ...(i)
and \( 3y - 5x = 30 \) [Given] ...(ii)
Multiply equation (i) by 3
\( 12x + 3y = 540^\circ \) ...(iii)
Subtracting (ii) from (iii), we get
\( 17x = 510 \)
\( x = 30^\circ \)
Putting value of \( x \) in equation (i), we get
\( 4 \times 30^\circ + y = 180^\circ \)
\( y = 60^\circ \)
\( \therefore \angle A = 30^\circ, \angle B = 3 \times 30^\circ = 90^\circ \) and \( \angle C = 60^\circ \).
Hence, \( \Delta ABC \) is right angled triangle at B.

Question. Find \( c \) if the system of equations \( cx + 3y + (3 - c) = 0; 12x + cy - c = 0 \) has infinitely many solutions? 
Answer: Given equation is:
\( cx + 3y + (3 - c) = 0 \)
\( 12x + cy - c = 0 \)
Condition for equations to have infinitely many solutions is:
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
Here, \( a_1 = c, b_1 = 3, c_1 = 3 - c \)
\( a_2 = 12, b_2 = c, c_2 = -c \)
\( \therefore \frac{c}{12} = \frac{3}{c} = \frac{3 - c}{-c} \)

\( \implies \) \( c^2 = 36 \implies c = 6 \) or \( c = -6 \) ...(i)
Also, \( -3c = 3c - c^2 \)

\( \implies \) \( c = 6 \) or \( c = 0 \) ...(ii)
From (i) and (ii), we get, \( c = 6 \).
Hence, the value of \( c = 6 \).

Question. For what value of \( k \), does the system of linear equations \( 2x + 3y = 7 \), \( (k - 1) x + (k + 2) y = 3k \) have an infinite number of solutions? 
Answer: The given system of linear equation is
\( 2x + 3y = 7 \)
\( (k - 1)x + (k + 2)y = 3k \)
For infinitely many solutions
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
\( a_1 = 2, b_1 = 3, c_1 = 7 \)
and \( a_2 = (k - 1), b_2 = (k + 2), c_2 = 3k \)

\( \implies \) \( \frac{2}{k - 1} = \frac{3}{k + 2} = \frac{7}{3k} \)

\( \implies \) \( 2(k + 2) = 3(k - 1) \); \( 3 (3k) = 7(k + 2) \)

\( \implies \) \( 2k + 4 = 3k - 3 \); \( 9k = 7k + 14 \)

\( \implies \) \( k = 7; k = 7 \)
Hence, the value of \( k \) is 7.

Question. If \( 2x + y = 23 \) and \( 4x - y = 19 \), find the values of \( 5y - 2x \) and \( \frac{y}{x} - 2 \).
Answer: The given equations are
\( 2x + y = 23 \) ...(i)
\( 4x - y = 19 \) ...(ii)
On adding both equations, we get

\( \implies \) \( 6x = 42 \)

\( \implies \) \( x = 7 \)
Putting the value of \( x \) in eq. (i), we get

\( \implies \) \( 2(7) + y = 23 \)

\( \implies \) \( y = 23 - 14 \)

\( \implies \) \( y = 9 \)
We have \( 5y - 2x = 5(9) - 2(7) = 45 - 14 = 31 \)
and \( \frac{y}{x} - 2 = \frac{9}{7} - 2 = -\frac{5}{7} \)
Hence, the values of \( (5y - 2x) \) and \( \frac{y}{x} - 2 \) are 31 and \( -\frac{5}{7} \) respectively.

Question. Write an equation for a line passing through the point representing solution of the pair of linear equations \( x + y = 2 \) and \( 2x - y = 1 \). How many such lines can we find?
Answer: The given equations are
\( x + y = 2 \) ...(i)
\( 2x - y = 1 \) ...(ii)
Adding eq. (i) and (ii), we have
\( 3x = 3 \implies x = 1 \)
Substituting \( x = 1 \) in eq. (i), we have
\( y = 1 \)
So, the solution is \( x = 1 \) and \( y = 1 \) and the point that represents the solution is \( (1, 1) \).
We also know that an infinite number of lines can pass through a given point, say \( (1, 1) \).
Hence, infinite lines can pass through the intersection point of the linear equations \( x + y = 2 \) and \( 2x - y = 1 \) i.e., \( (1, 1) \).

Question. A fraction becomes \( \frac{1}{3} \) when 1 is subtracted from the numerator and it becomes \( \frac{1}{4} \) when 8 is added to its denominator. Find the fraction. 
Answer: Let the fraction be \( \frac{a}{b} \).
Then, according to the question,
\( \frac{a - 1}{b} = \frac{1}{3} \) and \( \frac{a}{b + 8} = \frac{1}{4} \)

\( \implies \) \( 3a - b = 3 \) and \( 4a - b = 8 \)
On solving these equations, we get:
\( a = 5, b = 12 \).
So, the fraction is \( \frac{5}{12} \).

Short Answer (SA-II) Type Questions

Question. For which value(s) of \( \lambda \) do the pair of linear equations \( \lambda x + y = \lambda^2 \) and \( x + \lambda y = 1 \) have
(a) no solution?
(b) infinitely many solutions?
(c) a unique solution? [NCERT]
Answer: The given pair of linear equations is
\( \lambda x + y - \lambda^2 = 0 \)
and \( x + \lambda y - 1 = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = \lambda, b_1 = 1, c_1 = -\lambda^2 \);
\( a_2 = 1, b_2 = \lambda, c_2 = -1 \)
\[ \frac{a_1}{a_2} = \frac{\lambda}{1}; \frac{b_1}{b_2} = \frac{1}{\lambda}; \frac{c_1}{c_2} = \frac{-\lambda^2}{-1} = \frac{\lambda^2}{1} \]
(A) For no solution,
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2} \]
\( \frac{\lambda}{1} = \frac{1}{\lambda} \neq \frac{\lambda^2}{1} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} \) and \( \frac{1}{\lambda} \neq \frac{\lambda^2}{1} \)
\( \lambda^2 - 1 = 0 \) and \( \lambda^2 \neq \lambda \)
\( (\lambda - 1)(\lambda + 1) = 0 \) and \( \lambda^2 - \lambda \neq 0 \)
\( (\lambda - 1)(\lambda + 1) = 0 \) and \( \lambda(\lambda - 1) \neq 0 \)
\( \lambda = 1, -1 \) and \( \lambda \neq 0, 1 \)
Here, we take only \( \lambda = -1 \).
Hence for \( \lambda = -1 \), the pair of linear equations has no solution.
(B) For infinitely many solutions,
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]
\( \frac{\lambda}{1} = \frac{1}{\lambda} = \frac{\lambda^2}{1} \)
\( \frac{\lambda}{1} = \frac{1}{\lambda} \) and \( \frac{1}{\lambda} = \frac{\lambda^2}{1} \)
\( \lambda^2 - 1 = 0 \) and \( \lambda^2 = \lambda \)
\( (\lambda - 1)(\lambda + 1) = 0 \) and \( \lambda^2 - \lambda = 0 \)
\( (\lambda - 1)(\lambda + 1) = 0 \) and \( \lambda(\lambda - 1) = 0 \)
\( \lambda = 1, -1 \) and \( \lambda = 0, 1 \)
\( \lambda = 1 \) satisfies both the equations.
Hence, for \( \lambda = 1 \), the pair of linear equations has infinitely many solutions.
(C) For a unique solution,
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
\( \lambda \neq \frac{1}{\lambda} \)
\( \lambda^2 - 1 \neq 0 \)
\( (\lambda - 1)(\lambda + 1) \neq 0 \)
\( \lambda \neq 1, -1 \)
Hence, for all real values of \( \lambda \) except \( \pm 1 \), the given pair of equations has a unique solution.

Question. For which values of \( a \) and \( b \) will the following pair of linear equations have infinitely many solutions?
\( x + 2y = 1 \)
\( (a - b)x + (a + b)y = a + b - 2 \) [CBSE 2013, 11]
Answer: The given pair of linear equations is
\( x + 2y = 1 \)
and \( (a - b)x + (a + b)y = a + b - 2 \)
\( x + 2y - 1 = 0 \)
and \( (a - b)x + (a + b)y - (a + b - 2) = 0 \)
Comparing with \( a_1x + b_1y + c_1 = 0 \) and \( a_2x + b_2y + c_2 = 0 \), we have
\( a_1 = 1, b_1 = 2, c_1 = -1 \)
\( a_2 = (a - b), b_2 = (a + b), c_2 = -(a + b - 2) \)
\[ \frac{a_1}{a_2} = \frac{1}{a - b}; \frac{b_1}{b_2} = \frac{2}{a + b}; \frac{c_1}{c_2} = \frac{-1}{-(a + b - 2)} = \frac{1}{a + b - 2} \]
For infinitely many solutions,
\[ \frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2} \]

\( \implies \) \( \frac{1}{a - b} = \frac{2}{a + b} = \frac{1}{a + b - 2} \)
Taking the first two parts
\( \frac{1}{a - b} = \frac{2}{a + b} \)

\( \implies \) \( a + b = 2(a - b) \)

\( \implies \) \( 2a - a = 2b + b \)

\( \implies \) \( a = 3b \) ...(i)
Taking the last two parts,
\( \frac{2}{a + b} = \frac{1}{a + b - 2} \)

\( \implies \) \( 2(a + b - 2) = (a + b) \)

\( \implies \) \( 2a + 2b - 4 = a + b \)

\( \implies \) \( a + b = 4 \) ...(ii)
Putting the value of \( a \) from eq. (i) in eq. (ii), we get

\( \implies \) \( 3b + b = 4 \)

\( \implies \) \( 4b = 4 \)

\( \implies \) \( b = 1 \)
Putting the value of \( b \) in eq. (i), we get
\( a = 3(1) = 3 \)
The values \( (a, b) = (3, 1) \) satisfies all the parts.
Hence, the required values of \( a \) and \( b \) are 3 and 1 respectively for which the given pair of linear equations has infinitely many solutions.

Question. Write a pair of linear equations which has the unique solution \( x = -1, y = 3 \). How many such pairs can you write? 
Answer: We know that the condition for the pair of system to have a unique solution is
\[ \frac{a_1}{a_2} \neq \frac{b_1}{b_2} \]
Let the equations be
\( a_1x + b_1y + c_1 = 0 \)
\( a_2x + b_2y + c_2 = 0 \)
It is given that \( x = -1 \) and \( y = 3 \) is the unique solution of these two equations, then it must satisfy the above equations.

\( \implies \) \( a_1(-1) + b_1(3) + c_1 = 0 \)

\( \implies \) \( -a_1 + 3b_1 + c_1 = 0 \) ...(i)
and \( a_2(-1) + b_2(3) + c_2 = 0 \)

\( \implies \) \( -a_2 + 3b_2 + c_2 = 0 \) ...(ii)
The restricted values of \( a_1, a_2 \) and \( b_1, b_2 \) are only
\( \frac{a_1}{a_2} = \frac{b_1}{b_2} \) ...(iii)
So, all the real values of \( a_1, a_2, b_1, b_2 \) except those which satisfy eq. (iii) and satisfy eq. (i), and eq. (ii) will have the solution \( x = -1 \) and \( y = 3 \).
Hence, infinitely many pairs of linear equations are possible.

Question. Solve the pair of equations: \( \frac{2}{x} + \frac{3}{y} = 11, \frac{5}{x} - \frac{4}{y} = -7 \). Hence, find the value of \( 5x - 3y \). 
Answer: Given equations are
\( \frac{2}{x} + \frac{3}{y} = 11 \) ...(i)
and \( \frac{5}{x} - \frac{4}{y} = -7 \) ...(ii)
Eq (i) \( \times \) 5 and eq (ii) \( \times \) 2 give
\( \frac{10}{x} + \frac{15}{y} = 55 \) ...(iii)
and \( \frac{10}{x} - \frac{8}{y} = -14 \) ...(iv)
On subtracting eq (iv) from eq (iii), we have
\( \frac{23}{y} = 69 \)

\( \implies \) \( y = \frac{1}{3} \)
On substituting this value \( y = \frac{1}{3} \) in eq (i), we have:
\( \frac{2}{x} + 9 = 11 \)
i.e., \( \frac{2}{x} = 2 \)
or \( x = 1 \)
Thus, \( x = 1, y = \frac{1}{3} \) is the required solution.
Hence, \( 5x - 3y = 5(1) - 3(\frac{1}{3}) = 5 - 1 = 4 \).

Question. Find the solution of the pair of equations \( \frac{x}{10} + \frac{y}{5} - 1 = 0 \) and \( \frac{x}{8} + \frac{y}{6} = 15 \). Hence, find \( \lambda \), if \( y = \lambda x + 5 \).
Answer: The given pair of equations is
\( \frac{x}{10} + \frac{y}{5} - 1 = 0 \)

\( \implies \) \( x + 2y - 10 = 0 \)

\( \implies \) \( x + 2y = 10 \) ...(i)
and \( \frac{x}{8} + \frac{y}{6} = 15 \)

\( \implies \) \( 3x + 4y - 360 = 0 \)

\( \implies \) \( 3x + 4y = 360 \) ...(ii)
Multiplying eq. (i) by 2 and subtracting it from eq. (ii), we get
\( (3x + 4y) - (2x + 4y) = 360 - 20 \)

\( \implies \) \( x = 340 \)
Putting the value of \( x \) in eq. (i), we get
\( 340 + 2y = 10 \)

\( \implies \) \( 2y = -330 \)

\( \implies \) \( y = -165 \)
It is given that \( y = \lambda x + 5 \)
Putting the values of \( x \) and \( y \) in the above equation, we get
\( y = \lambda x + 5 \)

\( \implies \) \( -165 = \lambda(340) + 5 \)

\( \implies \) \( -\lambda(340) = 5 + 165 \)

\( \implies \) \( -340\lambda = 170 \)

\( \implies \) \( \lambda = \frac{-170}{340} = -\frac{1}{2} \)
Hence, the solution of the pair of equations is \( x = 340, y = -165 \) and the required value of \( \lambda \) is \( -\frac{1}{2} \).

Question. The present age of a father is three years more than three times the age of his son. Three years hence the father's age will be 10 years more than twice the age of the son. Determine their present ages. 
Answer: Let '\( x \)' (in years) be the present age of the father and '\( y \)' (in years) be the present age of the son.
Then, according to the given question:
\( x = 3y + 3 \) or \( x - 3y = 3 \) ...(i)
After 3 years, Father's age \( = x + 3 \), son's age \( = y + 3 \)
and \( x + 3 = 2 (y + 3) + 10 \)
or \( x - 2y = 13 \) ...(ii)
On solving the two equations, we get:
\( y = 10, \) and \( x = 33 \).
Thus, the father's present age is 33 years and the son's present age is 10 years.

Question. Taxi charges in a city consist of fixed charges and the remaining charges depend upon the distance travelled. For a journey of 10 km, the charge paid is Rs. 75 and for a journey of 15 km, the charge paid is Rs. 110. Find the the fixed charge and charges per km. Also, find the charges of covering a distance of 35 km.
Answer: Let the fixed charge be Rs. \( x \) and charges for per km be Rs. \( y \).
As per the question,
\( x + 10y = 75 \) and \( x + 15y = 110 \)
On solving the two equations, we get
\( x = 5, y = 7 \)
Thus, the fixed charge is Rs. 5 and the charge per km is Rs. 7.
Hence, charge for 35 km is Rs. \( [5 + 35(7)] \), i.e., Rs. 250.

Question. A man can row a boat downstream 20 km in 2 hours and upstream 4 km in 2 hours. Find his speed of rowing in still water. Also find the speed of the stream. 
Answer: Let the speed of the stream be \( x \) km/h and the speed of rowing in still water be \( y \) km/h.
Then, the speed of rowing in the downstream is '\( x + y \)' km/h.
And the speed of rowing in the upstream is \( y - x \) km/h
As per the question:
\( \frac{20}{y + x} = 2 \) and \( \frac{4}{y - x} = 2 \)

\( \implies \) \( x + y = 10 \) and \( y - x = 2 \)
On solving the two equations, we get:
\( x = 4 \) and \( y = 6 \)
Thus, the speed of rowing in still water is 6 km/h, and the speed of the stream is 4 km/hr.

Question. The angles of a triangle are \( x, y \) and \( 40^\circ \). The difference between the two angles \( x \) and \( y \) is \( 30^\circ \). Find \( x \) and \( y \). 
Answer: It is given that \( x, y \) and \( 40^\circ \) are the angles of a triangle.
We know that the sum of all the angles of a triangle is \( 180^\circ \)

\( \implies \) \( x + y + 40 = 180 \)

\( \implies \) \( x + y = 140 \) ...(i)
Also, it is given that the difference of angles \( x \) and \( y \) is
\( x - y = 30 \) ...(ii)
Adding eq. (i) and (ii), we get
\( (x + y) + (x - y) = 140 + 30 \)

\( \implies \) \( 2x = 170 \)

\( \implies \) \( x = 85 \)
Putting the value of \( x \) in eq. (i), we get

\( \implies \) \( 85 + y = 140 \)

\( \implies \) \( y = 140 - 85 \)
\( y = 55 \)
Hence, the required values of \( x \) and \( y \) are \( 85^\circ \) and \( 55^\circ \) respectively.

Question. A part of monthly hostel charges in a college hostel are fixed and the remaining depends on the number of days one has their meals in the mess. When a student A takes food for 25 days, he has to pay Rs. 4,500, whereas a student B who takes food for 30 days has to pay Rs. 5,200. Find the fixed charges per month and the cost of food per day. 
Answer: Let, the fixed charge per student \( = \) Rs. \( x \)
Cost of food per day per student \( = \) Rs. \( y \)
According to the given condition,
\( x + 25y = 4500 \) ...(i)
\( x + 30y = 5200 \) ...(ii)
On subtracting equation (i) from equation (ii), we get
\( 5y = 700 \)

\( \implies \) \( y = \frac{700}{5} = 140 \)
Put the value of \( y \) in equation (i), we get:
\( x + 25 \times 140 = 4500 \)

\( \implies \) \( x + 3500 = 4500 \)

\( \implies \) \( x = 1000 \)
Hence, the fixed charge per student is Rs. 1,000 and cost of food per day is Rs. 140.

Question. There are some students in two examination halls A and B. To make the number of students equal in each hall, 10 students are sent from A to B. But, if 20 students are sent from B to A, the number of students in A becomes double the number of students in B. Find the number of students in the two halls. 
Answer: Let the number of students in hall A and B be \( x \) and \( y \) respectively.
By the given condition, to make the number of students equal in each hall, 10 students are sent from A to B

\( \implies \) \( x - 10 = y + 10 \)

\( \implies \) \( x - y = 20 \) ...(i)
Also, it is given that if 20 students are sent from B to A, the number of students in A becomes double the number of students in B

\( \implies \) \( (x + 20) = 2(y - 20) \)

\( \implies \) \( x - 2y = -60 \) ...(ii)
Subtracting eq. (ii) from eq. (i), we get
\( (x - y) - (x - 2y) = 20 - (-60) \)

\( \implies \) \( y = 80 \)
Putting the value of \( y \) in eq. (i), we get
\( x - 80 = 20 \)

\( \implies \) \( x = 100 \)
Hence, 100 students are in hall A and 80 students are in hall B.

Question. In a competitive examination, one mark is awarded for each correct answer, while \( 1/2 \) mark is deducted for every wrong answer. Rahul answered 120 questions and got 90 marks. How many questions did he answer correctly? 
Answer: Let \( x \) be the number of correct answers,
Then, marks awarded for correct answer \( = x \times 1 = x \)
Total no. of questions attempted \( = 120 \)
Number of wrong answers \( = (120 - x) \)
Marks deducted for wrong answers \( = (120 - x) \times \frac{1}{2} = \frac{120 - x}{2} \)
Total marks awarded to Rahul \( = 90 \)

\( \implies \) \( x - \left(\frac{120 - x}{2}\right) = 90 \)

\( \implies \) \( x + \frac{x}{2} - 60 = 90 \)

\( \implies \) \( \frac{3x}{2} = 150 \)

\( \implies \) \( x = \frac{150 \times 2}{3} \)

\( \implies \) \( x = 100 \)
Hence, Rahul answered 100 questions correctly.

Question. A father’s age is three times the sum of the ages of his children. After 5 years, his age will be two times the sum of their ages. Find the present age of the father. 
Answer: Let the sum of the ages of two children be \( x \) years and father's age be \( y \) years.
According to the given condition:
\( y = 3x \) ...(i)
After 5 years:
Father’s age \( = (y + 5) \) years
Sum of the ages of children \( = (x + 5 + 5) \) years.
Then, \( y + 5 = 2(x + 10) \)
or \( y - 2x - 15 = 0 \) ...(ii)
On subtracting equation (i) from equation (ii), we get:
\( y - 2x - 15 = 0 \)
\( y - 3x = 0 \)
\( x - 15 = 0 \)
\( x = 15 \)
If we put the value of \( x \) in equation (i), we get
\( y - 3 \times 15 = 0 \)

\( \implies \) \( y = 45 \)
Hence, the present age of the father is 45 years.

Question. Solve the following system of equations:
\( \frac{21}{x} + \frac{47}{y} = 110 \)
\( \frac{47}{x} + \frac{21}{y} = 162, x, y \neq 0 \) 
Answer: Let \( \frac{1}{x} = a \) and \( \frac{1}{y} = b \)

\( \implies \) \( 21a + 47b = 110 \) and \( 47a + 21b = 162 \)
Adding and subtracting the two equations, we get
\( a + b = 4 \) and \( a - b = 2 \)
Solving the above two equations, we get
\( a = 3 \) and \( b = 1 \)
\( \therefore x = \frac{1}{3} \) and \( y = 1 \)

Question. The sum of reciprocals of a child’s age (in years) 3 years ago and 5 years from now is \( 1/3 \). Find his present age.
Answer: Let the present age of the child in years be \( x \).
Then,
\( \frac{1}{x - 3} + \frac{1}{x + 5} = \frac{1}{3} \)

\( \implies \) \( \frac{(x + 5) + (x - 3)}{(x - 3)(x + 5)} = \frac{1}{3} \)

\( \implies \) \( 3(2x + 2) = (x - 3)(x + 5) \)

\( \implies \) \( 6x + 6 = x^2 + 2x - 15 \)

\( \implies \) \( x^2 - 4x - 21 = 0 \)

\( \implies \) \( (x - 7)(x + 3) = 0 \)

\( \implies \) \( x - 7 = 0 \) or \( x + 3 = 0 \)

\( \implies \) \( x = 7 \) or \( x = -3 \)
(\( x = -3 \) is a not possible)
Thus, her present age is 7 years.

Question. A man wished to give Rs. 12 to each person and found that he fell short of Rs. 6 when he wanted to give to all the persons present. He, therefore, distributed Rs. 9 to each person and found that Rs. 9 were left over. How much money did he have and how many persons were there? 
Answer: Let, number of persons \( = x \)
Money share per person \( = y \)
Therefore, total money \( = \text{Rs. } xy \)
According to the question,
\( 12 \times x = xy + 6 \)
\( 12x - 6 = xy \) ...(i)
and \( 9x = xy - 9 \)
\( 9x + 9 = xy \) ...(ii)
Equating (i) and (ii), we get
\( 12x - 6 = 9x + 9 \)
\( 3x = 15 \)
\( x = 5 \)
Put the value of \( x \) in equation (i). Then
\( 12 \times 5 - 6 = xy \)

\( \implies \) \( xy = 54 \)
So, he have Rs. 54 and there were 5 persons.

Question. Find the solution of the pair of equations: \( \frac{3}{x} + \frac{8}{y} = -1 \); \( \frac{1}{x} - \frac{2}{y} = 2, x, y \neq 0 \). 
Answer: Given, pair of equation is—
\( \frac{3}{x} + \frac{8}{y} = -1 \) ...(i)
\( \frac{1}{x} - \frac{2}{y} = 2 \) ...(ii)
If we multiply equation (ii) by 3 and subtract it from equation (i), we get
\( \frac{3}{x} + \frac{8}{y} = -1 \)
\( \frac{3}{x} - \frac{6}{y} = 6 \)
\( \frac{14}{y} = -7 \)

\( \implies \) \( y = -2 \)
Put the value of \( y = -2 \) in equation (i), we get
\( \frac{3}{x} + \left(\frac{8}{-2}\right) = -1 \)

\( \implies \) \( \frac{3}{x} = -1 + 4 \)

\( \implies \) \( x = 1 \)
Hence, the value of \( x \) and \( y \) are 1 and -2 respectively.

Question. Ratio between the girls and boys in a class of 40 students is \( 2:3 \). Five new students joined the class. How many of them must be boys so that the ratio between girls and boys becomes \( 4:5 \)?
Answer: Let number of girls \( = 2x \)
and number of boys \( = 3x \)
Total students, \( 2x + 3x = 40 \)
\( x = 8 \)
So, number of girls \( = 2 \times 8 = 16 \)
and number of boys \( = 3 \times 8 = 24 \)
Let out of 5 students, \( y \) denotes number of boys.
Then, number of girls \( = 5 - y \)
According to question,
\( \frac{16 + 5 - y}{24 + y} = \frac{4}{5} \)
\( 5(21 - y) = 4(24 + y) \)
\( 105 - 5y = 96 + 4y \)
\( 9y = 105 - 96 \)
\( 9y = 9 \)
\( y = 1 \)
So, there must be one boy among five new students.

Question. Sumit is 3 times as old as his son. Five years later, he shall be two and a half times as old as his son. How old is Sumit at present? 
Answer: Let the present age of Sumit's son be \( x \) years and the present age of Sumit be \( y \) years.
According to the given conditions:
\( y = 3x \) or \( -3x + y = 0 \) ...(i)
Five years later
Sumit's son's age \( = (x + 5) \) years
Sumit's age \( = (y + 5) \) years
\( \therefore (y + 5) = 2 \frac{1}{2} (x + 5) \)

\( \implies \) \( 2(y + 5) = 5(x + 5) \)

\( \implies \) \( 2y + 10 = 5x + 25 \)

\( \implies \) \( -5x + 2y = 15 \) ...(ii)
If we multiply equation (i) by 2 and subtract the equation (ii) from (i), we get
\( -6x + 2y = 0 \)
\( -5x + 2y = 15 \)
\( -x = -15 \) or \( x = 15 \)
If we put the value of \( x \) in equation (i), we get
\( -3 \times 15 + y = 0 \)

\( \implies \) \( y = 45 \)
Hence, Sumit's present age is 45 years and Sumit's son's present age is 15 years.

Question. A two digit number is 4 times the sum of the digits. It is also equal to 3 times the product of the digits. Find the number.
Answer: Let the digit at unit’s place be \( x \) and at ten’s place be ‘\( y \)’.
Then, the number is \( 10y + x \)
According to the question,
\( (10y + x) = 4(x + y) \)
\( 10y + x = 4x + 4y \)
\( 6y - 3x = 0 \)
\( -x + 2y = 0 \)

\( \implies \) \( x = 2y \) ...(i)
and \( 10y + x = 3xy \)
\( \therefore 10y + 2y = 3 \times 2y \times y \) [from (i)]
\( 12y = 6y^2 \)

\( \implies \) \( y = 2 \)
\( \therefore x = 4 \)
Hence, the number is 24.

Question. A and B each has a certain number of mangoes. A says to B, "If you give 30 of your mangoes, I will have twice as many as left with you." B replies "If you give me 10, I will have thrice as many as left with you". How many mangoes does each have? 
Answer: Let number of mangoes with A be \( x \).
Number of mangoes with B be \( y \).
According to the question,
\( x + 30 = 2(y - 30) \)
\( x + 30 = 2y - 60 \)
\( x - 2y = -90 \) ...(i)
and \( y + 10 = 3(x - 10) \)
\( y + 10 = 3x - 30 \)
\( 3x - y = 40 \) ...(ii)
Multiplying equation (ii) by 2 and subtracting from equation (i), we get
\( -5x = -170 \)
\( x = 34 \)
Putting \( x = 34 \) in equation (i), we get
\( 34 - 2y = -90 \)
\( -2y = -90 - 34 \)
\( -2y = -124 \)
\( y = 62 \)
So, number of mangoes with A are 34 and number of mangoes with B are 62.
 

Long Answer Type Questions 

Question. Determine, algebraically, the vertices of the triangle formed by the lines \( 3x - y = 3, 2x - 3y = 2 \) and \( x + 2y = 8 \). 
Answer: The given equation of lines are:
\( 3x - y = 3 \) ...(i)
\( 2x - 3y = 2 \) ...(ii)
\( x + 2y = 8 \) ...(iii)
Let lines (i), (ii) and (iii) represent the side of a \( \Delta ABC \) i.e., AB, BC and CA respectively.
On solving lines (i) and (ii), we will get the intersection point B.
Multiplying eq. (i) by 3 and then subtracting eq. (ii), we get
\( (9x - 3y) - (2x - 3y) = 9 - 2 \)

\( \implies \) \( 7x = 7 \implies x = 1 \)
Putting the value of \( x \) in eq. (i), we get \( 3 \times 1 - y = 3 \implies y = 0 \).
Hence, vertex B is \( (1, 0) \).
On solving lines (ii) and (iii), we get point C.
Multiplying eq. (iii) by 2 and then subtracting eq. (ii), we get
\( (2x + 4y) - (2x - 3y) = 16 - 2 \)

\( \implies \) \( 7y = 14 \implies y = 2 \).
Putting value of \( y \) in eq. (iii), we get \( x + 2(2) = 8 \implies x = 4 \).
Hence, vertex C is \( (4, 2) \).
On solving lines (iii) and (i), we get point A.
Multiplying eq. (i) by 2 and then adding eq. (iii), we get
\( (6x - 2y) + (x + 2y) = 6 + 8 \)

\( \implies \) \( 7x = 14 \implies x = 2 \).
Putting value of \( x \) in eq. (i), we get \( 3 \times 2 - y = 3 \implies y = 3 \).
Hence, vertex A is \( (2, 3) \).
Vertices of \( \Delta ABC \) are \( A(2, 3), B(1, 0) \) and \( C(4, 2) \).

Question. It can take 12 hours to fill a swimming pool using two pipes. If the pipe of larger diameter is used for four hours and the pipe of smaller diameter for 9 hours, only half of the pool can be filled. How long would it take for each pipe to fill the pool separately? 
Answer: Let two pipes A and B of diameter \( d_1 \) and \( d_2 \) (\( d_1 > d_2 \)) take \( x \) and \( y \) hours to fill the pool, respectively. Then,
\( \frac{1}{x} + \frac{1}{y} = \frac{1}{12} \) ...(i)
and \( \frac{4}{x} + \frac{9}{y} = \frac{1}{2} \) ...(ii)
Let \( \frac{1}{x} = u \) and \( \frac{1}{y} = v \).
\( u + v = \frac{1}{12} \) ...(iii)
and \( 4u + 9v = \frac{1}{2} \) ...(iv)
If we multiply equation (iii) by 4 and subtract it from (iv):
\( 4u + 4v = \frac{1}{3} \)
\( 4u + 9v = \frac{1}{2} \)
\( -5v = \frac{-1}{6} \implies v = \frac{1}{30} \).
And \( u = \frac{1}{20} \).
\( \therefore x = 20, y = 30 \).
Thus, the larger pipe takes 20 hours and smaller pipe takes 30 hours alone.

Question. Ankita travels 14 km to her home partly by rickshaw and partly by bus. She takes half an hour if she travels 2 km by rickshaw and the remaining distance by bus. On the other hand, if she travels 4 km by rickshaw and the remaining distance by bus, she takes 9 minutes longer. Find the speed of the rickshaw and of the bus. 
Answer: Let speed of rickshaw be \( x \) km/hr and bus be \( y \) km/hr.
Case I: Rickshaw travel \( = 2 \) km, Bus travel \( = 12 \) km.
\( \frac{2}{x} + \frac{12}{y} = \frac{1}{2} \) ...(i)
Case II: Rickshaw travel \( = 4 \) km, Bus travel \( = 10 \) km.
\( \frac{4}{x} + \frac{10}{y} = \frac{1}{2} + \frac{9}{60} = \frac{13}{20} \) ...(ii)
Let \( \frac{1}{x} = u, \frac{1}{y} = v \).
\( 2u + 12v = \frac{1}{2} \) and \( 4u + 10v = \frac{13}{20} \).
Solving these, we get \( v = \frac{1}{40} \) and \( u = \frac{1}{10} \).
Rickshaw speed \( = 10 \) km/hr, Bus speed \( = 40 \) km/hr.

Question. A motorboat can travel 30 km upstream and 28 km downstream in 7 hrs. It can travel 21 km upstream and return in 5 hrs. Find the speed of the boat in still water and the speed of the stream.
Answer: Let speed of boat \( = x \) km/hr, speed of stream \( = y \) km/hr.
Upstream speed \( = x - y \), Downstream speed \( = x + y \).
Case I: \( \frac{30}{x - y} + \frac{28}{x + y} = 7 \) ...(i)
Case II: \( \frac{21}{x - y} + \frac{21}{x + y} = 5 \) ...(ii)
Let \( p = \frac{1}{x - y}, q = \frac{1}{x + y} \).
\( 30p + 28q = 7 \) and \( 21p + 21q = 5 \implies p + q = \frac{5}{21} \).
Solving, \( p = \frac{1}{6}, q = \frac{1}{14} \).
\( x - y = 6 \) and \( x + y = 14 \).
Adding: \( 2x = 20 \implies x = 10 \); then \( y = 4 \).
Speed of boat is 10 km/hr and stream speed is 4 km/hr.

Question. A shopkeeper sells a saree at a profit of 8% and a sweater at a discount of 10%, thereby getting a sum Rs. 1008. If she had sold the saree at a profit of 10% and the sweater at a discount of 8%, she would have got Rs. 1028. Find the cost of the saree and the list price of the sweater. 
Answer: Let cost of saree \( = x \) and list price of sweater \( = y \).
Case I: \( 108\% \text{ of } x + 90\% \text{ of } y = 1008 \implies 1.08x + 0.90y = 1008 \implies 6x + 5y = 5600 \).
Case II: \( 110\% \text{ of } x + 92\% \text{ of } y = 1028 \implies 1.1x + 0.92y = 1028 \implies 55x + 46y = 51400 \).
Solving these: \( x = 600, y = 400 \).
Cost of saree is Rs. 600 and sweater list price is Rs. 400.

Question. Ruhi invested a certain amount of money in two schemes A and B, which offer interest at the rate of 8% per annum and 9% per annum, respectively. She received Rs. 1860 as annual interest. However, had she interchanged the amount of investments in the two schemes, she would have received Rs. 20 more as annual interest. How much money did she invest in each scheme?
Answer: Let investment in A \( = x \) and B \( = y \).
Case I: \( 0.08x + 0.09y = 1860 \implies 8x + 9y = 186000 \) ...(i)
Case II: \( 0.09x + 0.08y = 1880 \implies 9x + 8y = 188000 \) ...(ii)
Solving these: \( x = 12000, y = 10000 \).
Ruhi invested Rs. 12000 in scheme A and Rs. 10000 in scheme B.

Question. Two water taps together can fill a tank in \( 1 \frac{7}{8} \) hours. The tap with a larger diameter takes 2 hours less than the tap with the smaller one to fill the tank separately. Find the time in which each tap can fill the tank. 
Answer: Let smaller tap fill in \( x \) hours, larger tap in \( (x - 2) \) hours.
\( \frac{1}{x} + \frac{1}{x - 2} = \frac{8}{15} \implies \frac{2x - 2}{x^2 - 2x} = \frac{8}{15} \)
\( 8x^2 - 16x = 30x - 30 \implies 4x^2 - 23x + 15 = 0 \).
\( (4x - 3)(x - 5) = 0 \implies x = 5 \) (since \( x = 3/4 \) not possible).
Smaller tap fills in 5 hours, larger tap in 3 hours.

Question. Rahul had some bananas and he divided them into two lots A and B. He sold the first lot at the rate of Rs. 2 for 3 bananas and the second lot at the rate of Rs. 1 per banana and got a total of Rs. 400. If he had sold the first lot at the rate of Rs. 1 per banana and the second lot at the rate of Rs. 4 for 5 bananas, his total collection would have been Rs. 460. Find the total number of bananas he had.
Answer: Let bananas in lot A \( = x \) and B \( = y \).
Case I: \( \frac{2x}{3} + y = 400 \implies 2x + 3y = 1200 \) ...(i)
Case II: \( x + \frac{4y}{5} = 460 \implies 5x + 4y = 2300 \) ...(ii)
Solving these: \( x = 300, y = 200 \). Total bananas \( = 300 + 200 = 500 \).

Question. The angles of a cyclic quadrilateral ABCD are \( \angle A = (6x + 10)^\circ, \angle B = (5x)^\circ, \angle C = (x + y)^\circ \) and \( \angle D = (3y - 10)^\circ \). Find x and y and hence the values of the four angles.
Answer: In a cyclic quadrilateral, opposite angles sum to \( 180^\circ \).
\( \angle A + \angle C = 180 \implies (6x + 10) + (x + y) = 180 \implies 7x + y = 170 \) ...(i)
\( \angle B + \angle D = 180 \implies 5x + (3y - 10) = 180 \implies 5x + 3y = 190 \) ...(ii)
Solving these: \( x = 20, y = 30 \).
Angles: \( \angle A = 130^\circ, \angle B = 100^\circ, \angle C = 50^\circ, \angle D = 80^\circ \).