CBSE Class 10 Mathematics Pair of Linear Equations in Two Variables Assignment Set 30

Very Short Answer Type Questions (1 Mark Each)

Question 1. Do the following pairs of linear equations have no solution? Justify your answer. (i) 2x + 4y = 3 and 12y + 6x = 6 (ii) x = 2y and y = 2x (iii) 3x + y − 3 = 0 and 2x + (2/3)y = 2.
Answer:
(i) Write as 2x + 4y − 3 = 0 and 6x + 12y − 6 = 0: \( \tfrac{2}{6} = \tfrac{4}{12} = \tfrac{1}{3} \ne \tfrac{-3}{-6} = \tfrac{1}{2} \). The lines are parallel: no solution.
(ii) x − 2y = 0 and 2x − y = 0: \( \tfrac{1}{2} \ne \tfrac{-2}{-1} \). Unique solution (0, 0): not a no-solution pair.
(iii) 3x + y − 3 = 0 and 2x + (2/3)y − 2 = 0: \( \tfrac{3}{2} = \tfrac{1}{2/3} = \tfrac{-3}{-2} = \tfrac{3}{2} \). The lines coincide: infinitely many solutions, not a no-solution pair.
Only pair (i) has no solution.
In simple words: Compare the ratios a1/a2, b1/b2, c1/c2.

Teacher's Note:
1. No solution needs a1/a2 = b1/b2 ≠ c1/c2.
2. Coincident lines have all three ratios equal.

Question 2. Do the following equations represent a pair of coincident lines? Justify. (i) 3x + (1/7)y = 3 and 7x + 3y = 7 (ii) −2x − 3y = 1 and 6y + 4x = −2 (iii) x/2 + y + 2/5 = 0 and 4x + 8y + 5/16 = 0.
Answer:
(i) \( \tfrac{3}{7} \) vs \( \tfrac{1/7}{3} = \tfrac{1}{21} \): the ratios differ, so the lines are not coincident (they intersect).
(ii) Write as 2x + 3y + 1 = 0 and 4x + 6y + 2 = 0: \( \tfrac{2}{4} = \tfrac{3}{6} = \tfrac{1}{2} \). Coincident.
(iii) \( \tfrac{1/2}{4} = \tfrac{1}{8} \), \( \tfrac{1}{8} \), but \( \tfrac{2/5}{5/16} = \tfrac{32}{25} \ne \tfrac{1}{8} \). The lines are parallel, not coincident.
In simple words: Equal ratios of all three coefficients mean coincident lines.

Teacher's Note:
1. Check a1/a2 = b1/b2 = c1/c2.
2. Only (ii) satisfies it.

Question 3. Are the following pairs of linear equations consistent? Justify. (i) −3x − 4y = 12 and 4y + 3x = 12 (ii) (3/5)x − y = 1/2 and (1/5)x − 3y = 1/6 (iii) 2ax + by = a and 4ax + 2by − 2a = 0, a, b ≠ 0 (iv) x + 3y = 11 and 2(2x + 6y) = 22.
Answer:
(i) \( \tfrac{-3}{3} = -1 \), \( \tfrac{-4}{4} = -1 \), \( \tfrac{-12}{-12} = 1 \): parallel, inconsistent.
(ii) \( \tfrac{3/5}{1/5} = 3 \), \( \tfrac{-1}{-3} = \tfrac{1}{3} \): different ratios, intersecting lines, consistent.
(iii) 2ax + by − a = 0 and 4ax + 2by − 2a = 0: \( \tfrac{2a}{4a} = \tfrac{b}{2b} = \tfrac{-a}{-2a} = \tfrac{1}{2} \): coincident, consistent (infinitely many solutions).
(iv) x + 3y − 11 = 0 and 4x + 12y − 22 = 0: \( \tfrac{1}{4} = \tfrac{3}{12} = \tfrac{1}{4} \ne \tfrac{-11}{-22} = \tfrac{1}{2} \): parallel, inconsistent.
In simple words: Consistent means at least one solution.

Teacher's Note:
1. Intersecting or coincident lines are consistent.
2. Parallel lines are inconsistent.

Question 4. For all real values of c, the pair of equations x − 2y = 8 and 5x − 10y = c have a unique solution. Justify whether it is true or false.
Answer:
\( \tfrac{1}{5} = \tfrac{-2}{-10} = \tfrac{1}{5} \). The first two ratios are always equal, so the lines are parallel (c ≠ 40) or coincident (c = 40) and the pair never has a unique solution.
The statement is false.
In simple words: Check a1/a2 against b1/b2.

Teacher's Note:
1. The ratio a1/a2 = b1/b2 for every c.
2. So uniqueness is impossible.

Short Answer Type Questions (2 Marks Each)

Question 5. For what value of k do the equations 3x − y + 8 = 0 and 6x − ky = −16 represent coincident lines?
Answer:
\( \tfrac{3}{6} = \tfrac{-1}{-k} = \tfrac{8}{16} \). So \( \tfrac{1}{k} = \tfrac{1}{2} \) and k = 2.
In simple words: All three ratios equal.

Teacher's Note:
1. 8/16 = 1/2 fixes k.
2. Check: 6x − 2y = −16 is 2 × the first.

Question 6. If the lines given by 3x + 2ky = 2 and 2x + 5y + 1 = 0 are parallel, then find the value of k.
Answer:
\( \tfrac{3}{2} = \tfrac{2k}{5} \ne \tfrac{2}{1} \). So \( 2k = \tfrac{15}{2} \) and k = 15/4 (and 3/2 ≠ 2, so the lines are indeed parallel).
In simple words: Parallel: a1/a2 = b1/b2 ≠ c1/c2.

Teacher's Note:
1. Rewrite 2x + 5y + 1 = 0 with c = 1.
2. Check the c-ratio is different.

Question 7. Find the value of c for which the pair of equations cx − y = 2 and 6x − 2y = 3 will have infinitely many solutions.
Answer:
We need \( \tfrac{c}{6} = \tfrac{-1}{-2} = \tfrac{-2}{-3} \). But \( \tfrac{1}{2} \ne \tfrac{2}{3} \), so no value of c gives infinitely many solutions. No such c exists.
In simple words: Check all three ratios.

Teacher's Note:
1. If two ratios already disagree, no value works.
2. The question has no valid c.

Question 8. For which values of k will the pair of equations kx + 3y = k − 3 and 12x + ky = k have no solution?
Answer:
\( \tfrac{k}{12} = \tfrac{3}{k} \ne \tfrac{k-3}{k} \). \( k^2 = 36 \), so k = ±6. For k = 6: \( \tfrac{3}{6} = \tfrac{1}{2} = \tfrac{k-3}{k} = \tfrac{3}{6} \): the ratios are equal, so infinitely many solutions. For k = −6: \( \tfrac{3}{-6} = -\tfrac{1}{2} \) and \( \tfrac{-9}{-6} = \tfrac{3}{2} \): different, so no solution.
The pair has no solution for k = −6.
In simple words: Solve a1/a2 = b1/b2 first, then test c.

Teacher's Note:
1. k = 6 gives coincident lines.
2. k = −6 gives parallel lines.

Question 9. If 2x + y = 23 and 4x − y = 19, find the values of 5y − 2x and \( \dfrac{y}{x} - 2 \).
Answer:
Adding: 6x = 42, so x = 7; then y = 23 − 14 = 9.
\( 5y - 2x = 45 - 14 = 31 \) and \( \dfrac{y}{x} - 2 = \dfrac{9}{7} - 2 = -\dfrac{5}{7} \).
In simple words: Add the equations to eliminate y.

Teacher's Note:
1. Solve for x and y before evaluating.
2. Check: 4(7) − 9 = 19.

Teacher's Note:
1. Then solve the pair.
2. Check: 1 + 12 = 13 and 3 + 4 = 7.

Long Answer Type Questions (3 Marks Each)

Question 11. In a competitive examination, one mark is awarded for each correct answer while 1/2 mark is deducted for every wrong answer. Jayanti answered 120 questions and got 90 marks. How many questions did she answer correctly?
Answer:
Let x be the number answered correctly and y the number answered wrongly. \( x + y = 120 \) and \( x - \tfrac{1}{2}y = 90 \).
From the first, y = 120 − x. Then \( x - \tfrac{120 - x}{2} = 90 \), so \( 2x - 120 + x = 180 \), \( 3x = 300 \) and x = 100.
Jayanti answered 100 questions correctly.
In simple words: Form the marks equation carefully.

Teacher's Note:
1. Wrong answers = 120 − 100 = 20.
2. Marks = 100 − 10 = 90.

Question 12. If x + 1 is a factor of 2x³ + ax² + 2bx + 1, find the values of a and b, given that 2a − 3b = 4.
Answer:
Since x + 1 is a factor, the value at x = −1 is zero: \( -2 + a - 2b + 1 = 0 \), so \( a - 2b = 1 \) ... (i). Given \( 2a - 3b = 4 \) ... (ii).
From (i) \( a = 1 + 2b \); in (ii): \( 2 + 4b - 3b = 4 \), so b = 2 and a = 5. a = 5, b = 2.
In simple words: Use p(−1) = 0.

Teacher's Note:
1. This uses the factor theorem.
2. Then solve the pair.

Teacher's Note:
1. Use ½ × base × height.
2. The x-axis triangle has height 4, the y-axis one has height 1.

Question 14. A motor boat can travel 30 km upstream and 28 km downstream in 7 hours. It can travel 21 km upstream and return in 5 hours. Find the speed of the boat in still water and the speed of the stream.
Answer:
Let the speed in still water be x km/h and the stream y km/h. \( \dfrac{30}{x - y} + \dfrac{28}{x + y} = 7 \) and \( \dfrac{21}{x - y} + \dfrac{21}{x + y} = 5 \).
Put \( u = \dfrac{1}{x - y} \), \( v = \dfrac{1}{x + y} \): \( 30u + 28v = 7 \) and \( u + v = \dfrac{5}{21} \). Then \( 2u = 7 - \dfrac{20}{3} = \dfrac{1}{3} \), \( u = \dfrac{1}{6} \), \( v = \dfrac{1}{14} \).
So x − y = 6 and x + y = 14: speed in still water 10 km/h, speed of the stream 4 km/h.
In simple words: Reduce to linear equations in 1/(x − y) and 1/(x + y).

Teacher's Note:
1. Upstream speed = x − y, downstream = x + y.
2. Check: 30/6 + 28/14 = 7.

Question 15. A two-digit number is obtained by either multiplying the sum of the digits by 8 and then subtracting 5 or by multiplying the difference of the digits by 16 and then adding 3. Find the number.
Answer:
Let the tens digit be a and the units digit b. \( 10a + b = 8(a + b) - 5 \Rightarrow 2a - 7b = -5 \) ... (i) and \( 10a + b = 16(a - b) + 3 \Rightarrow 6a - 17b = -3 \) ... (ii).
From (i) \( a = \dfrac{7b - 5}{2} \); in (ii): \( 3(7b - 5) - 17b = -3 \), \( 4b = 12 \), b = 3 and a = 8. The number is 83 (check: 8 × 11 − 5 = 83 and 16 × 5 + 3 = 83).
In simple words: Write the number as 10a + b.

Teacher's Note:
1. Form two equations, one for each description.
2. Check both conditions.

Question 16. A railway half ticket costs half the full fare, but the reservation charges are the same on a half ticket as on a full ticket. One reserved first class ticket from station A to B costs Rs. 2530. Also, one reserved first class ticket and one reserved first class half ticket from A to B cost Rs. 3810. Find the full first class fare from A to B and the reservation charges for a ticket.
Answer:
Let the full fare be Rs. x and the reservation charge Rs. y. \( x + y = 2530 \) and \( (x + y) + (\tfrac{x}{2} + y) = 3810 \).
The second gives \( \tfrac{x}{2} + y = 1280 \). Subtracting from the first: \( \tfrac{x}{2} = 1250 \), x = 2500 and y = 30.
Full first class fare = Rs. 2500; reservation charge = Rs. 30.
In simple words: Half ticket fare is x/2, reservation is the same y.

Teacher's Note:
1. Form two equations.
2. Check: 2500 + 30 = 2530.

Very Long Answer Type Questions (4 Marks Each)

Teacher's Note:
1. The region is a trapezium.
2. Area = ½ (sum of parallel sides) × distance.

Question 18. A shopkeeper sells a saree at 8% profit and a sweater at 10% discount, thereby getting a sum of Rs. 1008. If she had sold the saree at 10% profit and the sweater at 8% discount, she would have got Rs. 1028. Find the cost price of the saree and the list price (price before discount) of the sweater.
Answer:
Let the cost price of the saree be Rs. x and the list price of the sweater Rs. y. \( 1.08x + 0.90y = 1008 \) and \( 1.10x + 0.92y = 1028 \).
Multiply by 100: \( 108x + 90y = 100800 \) ... (i), \( 110x + 92y = 102800 \) ... (ii). Solving, x = 600 and y = 400.
Cost price of the saree = Rs. 600; list price of the sweater = Rs. 400.
In simple words: Profit 8% means selling price 108% of cost; discount 10% means 90% of list.

Teacher's Note:
1. Form two equations and solve.
2. Check: 648 + 360 = 1008.

Question 19. Susan invested certain amounts 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 amounts 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 she invest Rs. x in A (8%) and Rs. y in B (9%). \( 0.08x + 0.09y = 1860 \) and \( 0.09x + 0.08y = 1880 \).
Adding: \( 0.17(x + y) = 3740 \), so x + y = 22000. Subtracting: \( 0.01(x - y) = 20 \), so x − y = 2000. Hence x = 12000 and y = 10000.
She invested Rs. 12000 in scheme A and Rs. 10000 in scheme B.
In simple words: Interchanging means 9% on x and 8% on y.

Teacher's Note:
1. Add and subtract the two equations.
2. Check: 960 + 900 = 1860.

Question 20. Vijay 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 Re. 1 per banana, and got a total of Rs. 400. If he had sold the first lot at the rate of Re. 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 the lots have x and y bananas. \( \tfrac{2}{3}x + y = 400 \) ... (i) and \( x + \tfrac{4}{5}y = 460 \) ... (ii).
From (i) \( y = 400 - \tfrac{2x}{3} \). In (ii): \( x + 320 - \tfrac{8x}{15} = 460 \), so \( \tfrac{7x}{15} = 140 \), x = 300 and y = 200.
Total number of bananas = 300 + 200 = 500.
In simple words: Rs. 2 for 3 bananas means Rs. 2/3 each.

Teacher's Note:
1. Rs. 4 for 5 bananas means Rs. 4/5 each.
2. Check: 200 + 200 = 400 and 300 + 160 = 460.