Algebraic methods: Substitution method and Elimination method.

Use elimination to solve the simultaneous equations and then evaluate the requested expression. Add 15x + 2y = 54 to 5x − 2y = 6. The y terms cancel, giving 20x = 60 and therefore x = 3. Substitute x = 3 into the first equation: 15(3) + 2y = 54, so 45 + 2y = 54 and 2y = 9. The question asks for x + 2y, not just y, so calculate 3 + 9 = 12. Hence option A is correct. Verification in the second equation gives 5(3) − 9 = 6. The other options differ from the exact value and cannot be obtained from the solved pair. The revised options are distinct and include the correct result.

The appropriate governing method is elimination because the coefficients of y are -7 and +7, which are opposites. Add the two equations term by term: (2x - 7y) + (4x + 7y) = 5 + 43. The y terms cancel, leaving 6x = 48, so x = 8. Substitute x = 8 into the first equation: 2(8) - 7y = 5. Thus 16 - 7y = 5, so 7y = 11 and y = 11/7. The ordered pair is therefore (8,11/7), which is option A. Checking in the second equation gives 4(8) + 7(11/7) = 32 + 11 = 43. The other listed pairs fail to satisfy both equations simultaneously, even if one equation may appear close for a particular pair.

The governing idea is to remove denominators before using elimination. The least common multiple of 3 and 4 is 12, so multiply both equations by 12. The first becomes 4x + 3y = 84, and the second becomes 3x + 4y = 96. Add these equations: 7x + 7y = 180. Factor the left side to get 7(x + y) = 180, and divide by 7: x + y = 180/7. Therefore option B is correct. Finding x and y separately is unnecessary because the requested sum appears immediately after addition. The integer choices 24, 26, and 28 are not equal to 180/7 and could arise from careless denominator clearing, incorrect addition, or rounding. Exact fraction arithmetic is required.

The governing concept is solving a linear system accurately after clearing decimal coefficients. Multiply 0.25x + y = 9 by 4 to obtain x + 4y = 36. Multiply x - 0.5y = 2 by 2 to obtain 2x - y = 4. From the second equation, x = 2 + y/2. Substitute this into the first: 2 + y/2 + 4y = 36. Since y/2 + 4y = 9y/2, we get 9y/2 = 34. Multiplying by 2 gives 9y = 68, hence y = 68/9. Therefore option C is correct. The decimal equations are equivalent to the multiplied equations, so no solution is lost. The values 6, 7, and 9 fail at least one original equation; rounding would also destroy the exact result.

Answer: 4, so option B is correct. A common solution means that the same pair of values (x, y) must satisfy both equations. We know x = 5, so first use the second equation to find y: 2x - 3y = 5. Substituting x = 5 gives 2(5) - 3y = 5, or 10 - 3y = 5. Therefore 3y = 5 and y = 5/3. Now substitute x = 5 and y = 5/3 into the first equation: ax + 3y = 25 becomes 5a + 3(5/3) = 25. Since 3(5/3) = 5, we get 5a + 5 = 25, so 5a = 20 and a = 4. Check: 4(5) + 3(5/3) = 20 + 5 = 25, and 2(5) - 3(5/3) = 10 - 5 = 5. If a = 3, the first expression is 20; if a = 5, it is 30; and if a = 6, it is 35. Thus none of A, C, or D works. Memory cue: a common solution must be substituted into both equations, not just one.

The key principle is that the stated solution must satisfy both equations. Begin with the first equation and use x = 5: 3(5) + 2y = 28. Thus 15 + 2y = 28, giving 2y = 13 and y = 13/2. Substitute x = 5 and y = 13/2 into the second equation: 5m - 2(13/2) = 12. The product 2(13/2) equals 13, so 5m - 13 = 12. Therefore 5m = 25 and m = 5. Hence option C is correct. A check gives 15 + 13 = 28 in the first equation and 5(5) - 13 = 12 in the second. Values 3, 4, and 6 would make the second left side 2, 7, and 17, respectively, rather than 12.

Let c be the number of correct answers and w be the number of wrong answers. Since all 30 questions are counted, c + w = 30. The marking rule gives the second linear equation 5c - 2w = 108, because correct answers add marks and wrong answers subtract marks. From the first equation, w = 30 - c. Substitution gives 5c - 2(30 - c) = 108, so 5c - 60 + 2c = 108, 7c = 168, and c = 24. Thus option C is correct. Verification is immediate: 24 correct answers give 24 × 5 = 120 marks, while 6 wrong answers deduct 6 × 2 = 12 marks, leaving 120 - 12 = 108. The other options do not produce the stated score.

Use elimination because multiplying the first equation by 2 creates -10y, which cancels the +10y in the second equation. Thus 2(6x - 5y = 8) gives 12x - 10y = 16. Add this to 9x + 10y = 83: 21x = 99, so x = 33/7. Substitute this into the first equation: 6(33/7) - 5y = 8. Since 6(33/7)=198/7 and 8=56/7, we obtain 5y=142/7 and y=142/35. Convert x to denominator 35: x=165/35. Therefore x+y=(165+142)/35=307/35, so option B is correct. The integer distractors 8, 9, and 10 are not the exact sum and do not satisfy the solved system.

The coefficients of y are +5 and -5, so addition eliminates y immediately. Add the equations: (14x + 5y) + (7x - 5y) = 77 + (-7), giving 21x = 70 and x = 10/3. Substitute this value into 14x + 5y = 77: 14(10/3) + 5y = 77. Since 140/3 is subtracted from 231/3, 5y = 91/3, so y = 91/15. The requested expression is y - x. Convert x to fifteenths: 10/3 = 50/15. Thus y - x = 91/15 - 50/15 = 41/15. Therefore option C is correct. The integer choices 5, 6, and 8 cannot equal this exact fractional value and result from incomplete or incorrect elimination.

Add the equations because the y coefficients are -7 and +7, so they cancel: (12x - 7y) + (4x + 7y) = 9 + 39. This gives 16x = 48 and therefore x = 3. Substitute x = 3 into the second equation: 4(3) + 7y = 39, so 12 + 7y = 39 and 7y = 27. Hence y = 27/7. Now evaluate the requested expression exactly: 2x - y = 2(3) - 27/7 = 6 - 27/7 = 42/7 - 27/7 = 15/7. Therefore option B is correct. The values 2, 4, and 5 may result from forgetting the coefficient 2, using the wrong sign, or rounding y, but none is the exact value.

This is a pair of linear equations solved by elimination. Let c be the price of one chair and t be the price of one table. The statements give 3c + 2t = 4900 and 2c + 3t = 5600. Multiply the first equation by 2 and the second by 3: 6c + 4t = 9800 and 6c + 9t = 16800. Subtracting eliminates c and gives 5t = 7000, so t = 1400. Therefore one table costs ₹1400, making option C correct. Substitution gives c = 700; then 3(700)+2(1400)=4900 and 2(700)+3(1400)=5600, confirming the result. The other options do not satisfy both original totals.