Algebraic methods: Substitution method and Elimination method.

Correct answer: C. 5

From 4x − y = 11, we get y = 4x − 11. Substituting this into the second equation gives 2x + 3(4x − 11) = 23, so 14x = 56 and x = 4. Therefore, y = 4(4) − 11 = 5.

Correct answer: B. 9

Adding the equations eliminates y and gives 10x = 40, so x = 4. Substituting into 3x − 2y = 2 gives 12 − 2y = 2, hence y = 5. Therefore, x + y = 4 + 5 = 9.

Correct answer: A. 1

Adding the equations eliminates y and gives 11x = 44, so x = 4. Substituting into 2x + 5y = 23 gives 8 + 5y = 23, so y = 3. Thus, x − y = 4 − 3 = 1.

Correct answer: C. 5

Subtract the second equation from the first: (6x + 7y) − (6x − 2y) = 55 − 10, so 9y = 45 and y = 5. The equal coefficients of x make subtraction a direct way to eliminate x.

Correct answer: D. \(\frac{172}{13}\)

The equations are \(3x+4y=26\) and \(5x-2y=22\). Multiplying the second equation by 2 gives \(10x-4y=44\). Adding it to the first equation eliminates \(y\): \(13x=70\), so \(x=\frac{70}{13}\). Substituting this into the second equation gives \(y=\frac{32}{13}\). Therefore, \(2x+y=2\left(\frac{70}{13}\right)+\frac{32}{13}=\frac{172}{13}\). Hence, option D is correct. Exam tip: Substitute the calculated values back into both original equations to verify the solution.

Correct answer: B. \((x,y)=(4,6)\)

Adding the two equations gives \((11x+4y)+(7x-4y)=68+4\), so \(18x=72\). Hence, \(x=4\). Substituting \(x=4\) into \(7x-4y=4\) gives \(28-4y=4\), and therefore \(y=6\). Thus, the correct solution is \((4,6)\). In option A, substituting \(x=3\) does not satisfy the second equation. Exam tip: When the coefficients of one variable are opposites, add the equations to eliminate that variable directly.

Correct answer: B. 18

Substitute x = 3y − 2 into 2x + y = 31: 2(3y − 2) + y = 31, so 7y = 35 and y = 5. Then x = 3(5) − 2 = 13, so x + y = 13 + 5 = 18.

Correct answer: C. 10

Adding the two equations eliminates y because 5y and -5y cancel: 12x=36, so x=3. Substituting x=3 into 4x+5y=7 gives 12+5y=7, hence y=-1. Therefore, 3x-y=3(3)-(-1)=10. In an exam, be careful when subtracting a negative value, as the sign changes; this is the most common error in this question.

Correct answer: C. 25

Let the greater number be x and the smaller number be y. Then x + y = 41 and x − y = 9. Adding these equations gives 2x = 50, so the greater number is x = 25.

Correct answer: A. 3/8

Let the numerator be x, so the denominator is x + 5. Then (x + 3)/(x + 6) = 2/3, giving 3x + 9 = 2x + 12 and x = 3. The denominator is 8, so the original fraction is 3/8.

Correct answer: B. (61/11) rupees

Let the price of one pencil be \(p\) rupees and that of one eraser be \(e\) rupees. The equations are \(3p+2e=31\) and \(2p+5e=47\). Multiplying the first equation by 5 and the second by 2 gives \(15p+10e=155\) and \(4p+10e=94\). Subtracting, \(11p=61\), so \(p=\frac{61}{11}\) rupees. Therefore, option B is correct. The value 6 rupees in option C is close, but it does not satisfy the equations exactly. Exam tip: In the elimination method, make the coefficients of one variable equal before adding or subtracting the equations.

Correct answer: A. 12:1

Subtract the first equation from the second: 3x = 24, so x = 8. Substituting into 2x + 3y = 18 gives y = 2/3. Thus x:y = 8:2/3 = 24:2 = 12:1.

Correct answer: B. 37/7

From the first equation, x = 4y − 14. Substituting into the second gives 3(4y − 14) + 2y = 32, so 14y = 74 and y = 37/7.

Correct answer: A. 94

Let the tens digit be x and the units digit be y. Then x + y = 13. The original number is 10x + y and the reversed number is 10y + x. Using the given condition, (10x + y) − (10y + x) = 45, so 9(x − y) = 45 and x − y = 5. Solving x + y = 13 and x − y = 5 gives x = 9 and y = 4. Therefore, the original number is 94. Option 85 has a digit sum of 13, but reversing it reduces the number by only 27, so it is incorrect. Exam tip: represent a two-digit number as 10x + y and its reversal as 10y + x.

Correct answer: B. \(\frac{570}{29}\)

Multiplying both equations by 10 gives \(2x-5y=10\) and \(5x+2y=110\). Multiply the first equation by 2 and the second by 5 to obtain \(4x-10y=20\) and \(25x+10y=550\). Adding them gives \(29x=570\), so \(x=\frac{570}{29}\). Although \(20\) is a nearby value, it does not satisfy the system. Exam tip: For linear equations containing fractions, first multiply by the LCM to remove the fractions.

Correct answer: C. 6

Multiplying both equations by 10 gives \(3x+2y=31\) and \(6x-2y=23\). Adding these equations eliminates \(y\): \(9x=54\), so \(x=6\). Therefore, option C is correct. In an exam, first clear the decimals and then eliminate the variable whose coefficients are opposites.

Correct answer: C. 6

Using y = 3 in 4x − 2y = 10 gives 4x − 6 = 10, so x = 4. Substituting x = 4 and y = 3 into the first equation gives 16 + 3k = 34, hence k = 6.

Correct answer: C. 5

In the given solution, y=2. Substituting this into the second equation, 3x-y=7, gives 3x-2=7, so x=3. Substituting x=3 and y=2 into the first equation, px+y=17, gives 3p+2=17; hence 3p=15 and p=5. Therefore, option C is correct. Exam tip: First use the given value to find the other variable, and then substitute both values into the equation containing the parameter.

Correct answer: B. (2) km/h

The direct answer is option B, 2 km/h. Downstream speed equals boat speed plus stream speed, while upstream speed equals boat speed minus stream speed. Let the still-water boat speed be b and stream speed be s. Downstream speed is 42/3=14 km/h, so b+s=14. Upstream speed is 30/3=10 km/h, so b-s=10. Subtract the second equation from the first: (b+s)-(b-s)=14-10, giving 2s=4 and therefore s=2 km/h. Option A, 1, would make the difference between downstream and upstream speeds only 2, not 4. Option B is correct. Option C, 3, would give a speed difference of 6, and option D, 4, would give a difference of 8; neither matches the data. The boat speed would be b=12 km/h, confirming downstream 14 and upstream 10. Memory cue: stream speed is half the difference of downstream and upstream speeds.

Correct answer: B. 24 years

Let Ram’s and Shyam’s present ages be \(r\) and \(s\) years, respectively. From the first condition, \(r-s=6\). After four years, their total age will be \((r+4)+(s+4)=50\), giving \(r+s=42\). Adding the two equations, \(2r=48\), so \(r=24\) years. Therefore, option B is correct. Exam tip: in age-sum problems, add the elapsed years to both people’s ages before forming the equation.

Correct answer: C. ₹105

Let the cheaper and costlier ticket prices be c and e. Then c + e = 275 and e − c = 65. Subtracting the second equation from the first gives 2c = 210, so the cheaper ticket costs ₹105.

Correct answer: C. 19/7

Let u = 1/x and v = 1/y. The equations become 3u + 2v = 13 and 2u − v = 3. From the second equation, v = 2u − 3. Substituting gives 3u + 2(2u − 3) = 13, so 7u = 19 and 1/x = u = 19/7.

Correct answer: D. 53/4

Set u = x − 2 and v = y + 1. Adding 3u + 2v = 31 and 5u − 2v = 21 gives 8u = 52, so u = 13/2. Then 2v = 31 − 3(13/2) = 23/2, so v = 23/4. Thus x = u + 2 = 17/2 and y = v − 1 = 19/4, giving x + y = 53/4.

Correct answer: D. 103/17

Multiply the second equation by 3 to get 15x − 9y = 42. Adding this to 2x + 9y = 61 gives 17x = 103, so x = 103/17.

Correct answer: D. 50/7

Multiply the first equation by 2 to obtain 4x + 10y = 62. Adding the second equation eliminates y and gives 7x = 50, so x = 50/7.