NCERT Solutions Class 9 Mathematics Ganita Manjari 2 Chapter 13 Two Variables One Line Exercise 13.5

Exercise Set 13.5

Question 1. Form a pair of linear equations for each of the following problems and find their solutions.
(i) The sum of two integers is +5 and their difference is –21. Find the two numbers.
(ii) The difference between two numbers is 26 and one number is three times the other. Find the numbers.
(iii) The coach of a cricket team buys 7 bats and 6 balls for ₹8880. Later, she buys 3 bats and 5 balls for ₹4000. Find the cost of each bat and each ball.
(iv) The taxi charges in a city consist of a fixed charge together with the charge for the distance covered for every km. For a distance of 10 km, the total amount paid is ₹155 and for a journey of 15 km, the total amount paid is ₹220. What is the fixed charge and the charge per km? How much does a person have to pay for travelling a distance of 25 km?
(v) A fraction becomes equal to \( \frac{9}{11} \) if 2 is added to both the numerator and the denominator. If 3 is added to both the numerator and the denominator, it becomes equal to \( \frac{5}{6} \). Find the fraction.
(vi) If we add 1 to the numerator and subtract 1 from the denominator, a fraction reduces to 1. It becomes equal to \( \frac{1}{2} \) if we add 1 only to the denominator. What is the fraction?
(vii) Five years ago, Nuri was thrice as old as Sonu. Ten years later, Nuri will be twice as old as Sonu. How old are Nuri and Sonu?
(viii) The sum of the digits of a two-digit number is 9. Also, nine times this number is twice the number obtained by reversing the order of the digits. Find the number.
(ix) Meena went to a bank to withdraw ₹2000. She asked the cashier to give her ₹50 and ₹100 notes only. Meena got 25 notes in all. Find how many notes of ₹50 and ₹100 did she receive?
(x) A lending library has a fixed charge for the first three days and an additional charge for each day thereafter. Saritha paid ₹27 for a book kept for seven days, while Susy paid ₹21 for the book she kept for five days. Find the fixed charge and the charge for each extra day.
Answer:
(i) Let the integers be x and y: x + y = 5 ... (1), x − y = −21 ... (2).
Adding: 2x = −16, x = −8. From (1): y = 5 + 8 = 13.
Check: −8 + 13 = 5 ✓; −8 − 13 = −21 ✓. The integers are −8 and 13.
(ii) Larger number x, smaller y: x − y = 26 ... (1), x = 3y ... (2).
Substituting: 3y − y = 26, 2y = 26, y = 13, x = 39.
Check: 39 − 13 = 26 ✓; 39 = 3 × 13 ✓. The numbers are 39 and 13. (If negative numbers are allowed, −39 and −13 also work.)
(iii) Bat ₹x, ball ₹y: 7x + 6y = 8880 ... (1), 3x + 5y = 4000 ... (2).
(1) × 5: 35x + 30y = 44400; (2) × 6: 18x + 30y = 24000. Subtracting: 17x = 20400, x = 1200.
In (2): 3600 + 5y = 4000, 5y = 400, y = 80.
Check: 8400 + 480 = 8880 ✓; 3600 + 400 = 4000 ✓. Each bat costs ₹1200 and each ball ₹80.
(iv) Fixed charge ₹x, per-km charge ₹y: x + 10y = 155 ... (1), x + 15y = 220 ... (2).
(2) − (1): 5y = 65, y = 13. From (1): x = 155 − 130 = 25.
For 25 km: 25 + 25 × 13 = 25 + 325 = 350.
Fixed charge ₹25, charge per km ₹13; a 25 km trip costs ₹350.
(v) Let the fraction be \( \frac{x}{y} \).
\( \frac{x+2}{y+2} \) = \( \frac{9}{11} \) → 11x + 22 = 9y + 18 → 11x − 9y = −4 ... (1).
\( \frac{x+3}{y+3} \) = \( \frac{5}{6} \) → 6x + 18 = 5y + 15 → 6x − 5y = −3 ... (2).
(1) × 5: 55x − 45y = −20; (2) × 9: 54x − 45y = −27. Subtracting: x = 7. Then 77 − 9y = −4, 9y = 81, y = 9.
Check: \( \frac{9}{11} \) ✓ and \( \frac{10}{12} \) = \( \frac{5}{6} \) ✓. The fraction is \( \frac{7}{9} \).
(vi) Let the fraction be \( \frac{x}{y} \).
\( \frac{x+1}{y-1} \) = 1 → x + 1 = y − 1 → x − y = −2 ... (1).
\( \frac{x}{y+1} \) = \( \frac{1}{2} \) → 2x = y + 1 → 2x − y = 1 ... (2).
(2) − (1): x = 3. Then y = x + 2 = 5.
Check: \( \frac{4}{4} \) = 1 ✓; \( \frac{3}{6} \) = \( \frac{1}{2} \) ✓. The fraction is \( \frac{3}{5} \).
(vii) Nuri's age x years, Sonu's y years (now).
Five years ago: x − 5 = 3(y − 5) → x − 3y = −10 ... (1).
Ten years on: x + 10 = 2(y + 10) → x − 2y = 10 ... (2).
(2) − (1): y = 20. Then x = 10 + 40 = 50.
Check: 45 = 3 × 15 ✓; 60 = 2 × 30 ✓. Nuri is 50 years and Sonu is 20 years old.
(viii) Tens digit x, units digit y: x + y = 9 ... (1).
9(10x + y) = 2(10y + x) → 90x + 9y = 20y + 2x → 88x − 11y = 0 → 8x − y = 0 ... (2).
Adding (1) and (2): 9x = 9, x = 1, y = 8. Number = 18.
Check: 1 + 8 = 9 ✓; 9 × 18 = 162 = 2 × 81 ✓. The number is 18.
(ix) ₹50 notes: x, ₹100 notes: y. x + y = 25 ... (1); 50x + 100y = 2000, i.e. x + 2y = 40 ... (2).
(2) − (1): y = 15, so x = 10.
Check: 25 notes ✓; 500 + 1500 = 2000 ✓. Meena got 10 notes of ₹50 and 15 notes of ₹100.
(x) Fixed charge (first 3 days) ₹x, each extra day ₹y. 7 days = 3 + 4 extra; 5 days = 3 + 2 extra.
x + 4y = 27 ... (1), x + 2y = 21 ... (2). (1) − (2): 2y = 6, y = 3; x = 21 − 6 = 15.
Check: 15 + 12 = 27 ✓; 15 + 6 = 21 ✓. Fixed charge ₹15, extra charge ₹3 per day.

Teacher's Note:
1. In (x), only the days after the first three are charged extra: 7 days means 4 extra days, not 7.
2. In (v) and (vi), cross-multiply first to turn each condition into a linear equation.
3. Always check the answer against the original words, not just the equations.

Question 2. Form a pair of linear equations and find their common solutions graphically. 10 students of Grade 9 took part in a Mathematics quiz. If the number of girls is 4 more than the number of boys, find the number of boys and girls who took part in the quiz.
Answer:
Boys x, girls y. Total: x + y = 10 ... (1). Girls exceed boys by 4: y = x + 4, i.e. y − x = 4 ... (2).
Points for x + y = 10: (0, 10), (5, 5), (10, 0).
Points for y = x + 4: (0, 4), (−4, 0), (6, 10).
Plot the points and draw both lines.

The lines meet at (3, 7). Check: 3 + 7 = 10 ✓; 7 − 3 = 4 ✓.
3 boys and 7 girls took part in the quiz.

Question 3. Use the ratios \( \frac{a_1}{a_2} \), \( \frac{b_1}{b_2} \), \( \frac{c_1}{c_2} \) to determine whether the lines representing the following pairs of linear equations intersect at a point, are parallel or are coincident.
(i) 5x − 4y + 8 = 0; 7x + 6y − 9 = 0
(ii) 9x + 3y + 12 = 0; 18x + 6y + 24 = 0
(iii) 6x − 3y + 10 = 0; 2x − y + 9 = 0
Answer:
(i) \( \frac{a_1}{a_2} \) = \( \frac{5}{7} \), \( \frac{b_1}{b_2} \) = \( \frac{-4}{6} \) = \( -\frac{2}{3} \).
Since \( \frac{a_1}{a_2} \) ≠ \( \frac{b_1}{b_2} \), the lines intersect at a point (unique solution).
(ii) \( \frac{a_1}{a_2} \) = \( \frac{9}{18} \) = \( \frac{1}{2} \), \( \frac{b_1}{b_2} \) = \( \frac{3}{6} \) = \( \frac{1}{2} \), \( \frac{c_1}{c_2} \) = \( \frac{12}{24} \) = \( \frac{1}{2} \).
All three ratios are equal, so the lines are coincident (infinitely many solutions).
(iii) \( \frac{a_1}{a_2} \) = \( \frac{6}{2} \) = 3, \( \frac{b_1}{b_2} \) = \( \frac{-3}{-1} \) = 3, \( \frac{c_1}{c_2} \) = \( \frac{10}{9} \).
Since \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \), the lines are parallel (no solution).

Question 4. Which of the following pairs of linear equations have solutions? If they have solutions, find them graphically.
(i) x + y = 5, 2x + 2y = 10
(ii) x − y = 8, 3x − 3y = 16
(iii) 2x + y − 6 = 0, 4x − 2y − 4 = 0
(iv) 2x − 2y − 2 = 0, 4x − 4y − 5 = 0
Answer:
(i) Ratios: \( \frac{1}{2} \), \( \frac{1}{2} \), \( \frac{-5}{-10} \) = \( \frac{1}{2} \). All equal → coincident lines → infinitely many solutions.
Both equations give the points (0, 5), (2, 3), (5, 0), so they draw the same line.

Every point of the line is a solution: (x, 5 − x) for any x, e.g. (0, 5), (2, 3), (5, 0).
(ii) Ratios: \( \frac{1}{3} \), \( \frac{-1}{-3} \) = \( \frac{1}{3} \), \( \frac{-8}{-16} \) = \( \frac{1}{2} \). First two equal, third different → parallel lines → no solution.
x − y = 8 passes through (0, −8) and (8, 0); 3x − 3y = 16 through (0, \( -\frac{16}{3} \)) and (\( \frac{16}{3} \), 0).

(iii) Ratios: \( \frac{2}{4} \) = \( \frac{1}{2} \), \( \frac{1}{-2} \) = \( -\frac{1}{2} \). Unequal → the lines intersect → unique solution.
2x + y = 6 passes through (0, 6) and (3, 0); 4x − 2y = 4 through (0, −2) and (1, 0).

The lines meet at (2, 2). Check: 4 + 2 − 6 = 0 ✓; 8 − 4 − 4 = 0 ✓. Solution: x = 2, y = 2.
(iv) Ratios: \( \frac{2}{4} \) = \( \frac{1}{2} \), \( \frac{-2}{-4} \) = \( \frac{1}{2} \), \( \frac{-2}{-5} \) = \( \frac{2}{5} \). First two equal, third different → parallel → no solution.
2x − 2y = 2 passes through (0, −1) and (1, 0); 4x − 4y = 5 through (0, \( -\frac{5}{4} \)) and (\( \frac{5}{4} \), 0).

Summary: pairs (i) and (iii) have solutions ((i) infinitely many, (iii) the single solution (2, 2)); pairs (ii) and (iv) have none.

Teacher's Note:
1. Use the ratio test first; it tells you what the graph should look like before you draw it.
2. In (iv) the two lines are very close together, so draw them carefully on a large scale to see that they never meet.

Question 5. Half the perimeter of a rectangular garden, whose length is 4 m more than its width, is 36 m. Find the dimensions of the garden.
Answer:
Length l m, width w m. Half the perimeter = l + w, so l + w = 36 ... (1). Also l − w = 4 ... (2).
Adding: 2l = 40, l = 20. From (1): w = 16.
Check: 20 − 16 = 4 ✓; 20 + 16 = 36 ✓.
The garden is 20 m long and 16 m wide.

Question 6. Given the linear equation 2x + 3y − 8 = 0, write another linear equation in two variables so that the graphs of the pair formed represent (i) intersecting lines (ii) parallel lines (iii) coincident lines.
Answer:
Here a₁ = 2, b₁ = 3, c₁ = −8.
(i) Intersecting: need \( \frac{a_1}{a_2} \) ≠ \( \frac{b_1}{b_2} \). Take 3x − 2y − 7 = 0: \( \frac{2}{3} \) ≠ \( \frac{3}{-2} \). (They meet at (\( \frac{37}{13} \), \( \frac{10}{13} \)).)
(ii) Parallel: need \( \frac{a_1}{a_2} \) = \( \frac{b_1}{b_2} \) ≠ \( \frac{c_1}{c_2} \). Take 4x + 6y − 9 = 0: \( \frac{2}{4} \) = \( \frac{3}{6} \) = \( \frac{1}{2} \), but \( \frac{-8}{-9} \) = \( \frac{8}{9} \) ≠ \( \frac{1}{2} \).
(iii) Coincident: need all three ratios equal. Double the given equation: 4x + 6y − 16 = 0: \( \frac{2}{4} \) = \( \frac{3}{6} \) = \( \frac{-8}{-16} \) = \( \frac{1}{2} \).
Many other answers are possible.

Teacher's Note:
1. Parallel: scale a and b by the same factor but not c. Coincident: scale all three.
2. For intersecting lines, almost any equation works as long as a/b is different.

Question 7. Here is a problem that was posed by Mahāvīrāchārya in Gaṇita sāra saṅgraha (c. 850 CE). The price of 9 citrons and 7 fragrant wood-apples taken together is 107; and the price of 7 citrons and 9 fragrant wood-apples taken together is 101. O mathematician, tell me quickly the price of each citron and of each fragrant wood-apple.
(Hint: The given problem can be modelled as 9x + 7y = 107, 7x + 9y = 101. Can you figure out a way of solving these equations without directly using elimination or the substitution method? What is special in this pair of equations? The x and y coefficients are interchanged in the 2 equations. What will happen if we add the pair of equations? What will happen if we subtract the pair of equations? Can the resulting equations be solved to find the values of x and y?)
Answer:
Citron price x, wood-apple price y: 9x + 7y = 107 ... (1), 7x + 9y = 101 ... (2).
Special feature: the coefficients of x and y are swapped between the two equations.
Add: 16x + 16y = 208 → x + y = 13 ... (3).
Subtract (2) from (1): 2x − 2y = 6 → x − y = 3 ... (4).
These are very easy to solve. (3) + (4): 2x = 16, x = 8. Then y = 13 − 8 = 5.
Check: 72 + 35 = 107 ✓; 56 + 45 = 101 ✓.
Each citron costs 8 and each fragrant wood-apple costs 5.

Teacher's Note:
1. When coefficients are interchanged, adding and subtracting gives x + y and x − y directly.
2. This trick saves the large multipliers that elimination would need.

Question 8. 5 pencils and 7 pens together cost ₹50, whereas 7 pencils and 5 pens together cost ₹46. Find the cost of each pencil and pen.
Answer:
Pencil ₹x, pen ₹y: 5x + 7y = 50 ... (1), 7x + 5y = 46 ... (2). The coefficients are interchanged, so add and subtract.
Adding: 12x + 12y = 96 → x + y = 8 ... (3).
(2) − (1): 2x − 2y = −4 → x − y = −2 ... (4).
(3) + (4): 2x = 6, x = 3. Then y = 5.
Check: 15 + 35 = 50 ✓; 21 + 25 = 46 ✓.
A pencil costs ₹3 and a pen costs ₹5.

Question 9. Find the height of the stool.

Answer:
Let the stool be h cm high and the cat c cm tall.
Figure 1: the cat sits on the stool and reaches 85 cm: h + c = 85 ... (1).
Figure 2: the cat stands on the floor and the stool top is 25 cm above the cat's head: h − c = 25 ... (2).
Adding: 2h = 110, h = 55. Then c = 85 − 55 = 30.
Check: 55 + 30 = 85 ✓; 55 − 30 = 25 ✓.
The stool is 55 cm high (and the cat is 30 cm tall).