Algebraic methods: Substitution method and Elimination method.

Adding gives (20x=60), so (x=3) and (y=\frac{9}{2}). Therefore (x+2y=12); do the final step separately.

From the first equation, (x=4y-14). Substitute carefully and verify the result in both equations.

Adding gives (6x=48), so (x=8). Substituting in the first equation gives (16-7y=5), so (y=\frac{11}{7}).

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.

Multiply both equations by (12). This gives (4x+3y=84) and (3x+4y=96), so adding gives (7x+7y=180).

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.

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.

Multiply the first equation by (4) to get (x+4y=36). Multiply the second by (2) and solve to get (y=8).

Putting (x=5) in the second equation gives (y=\frac{5}{3}). Then (5a+5=25), so (a=4).

Putting (y=3) in the second equation gives (x=4). Then (16+3k=34), so verify the parameter carefully.

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.

Putting (x=5) in the first equation gives (y=\frac{13}{2}). Then (5m-13=12), so (m=5).

Let boat speed be (b) and stream speed be (s), so (b+s=14), (b-s=10). Subtracting gives (2s=4), so (s=2).

Let correct answers be (c) and wrong answers be (w), so (c+w=30) and (5c-2w=108). Elimination gives (7c=168), so (c=24).

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.

Let the prices be (x) and (y), so (x+y=275) and (x-y=65). Subtracting gives (2y=210), so the cheaper ticket is (105) rupees.

Let (u=x+y) and (v=x-y). Solving (3u+4v=59), (5u-2v=37) gives (u=9), (v=8), so (x=\frac{17}{2}).

Let (u=\frac{1}{x}) and (v=\frac{1}{y}). Solve (3u+2v=13), (2u-v=3) carefully before choosing.

Let (u=x-2) and (v=y+1). Solving (3u+2v=31), (5u-2v=21) gives values to substitute back for (x+y).

Let (u=2x-y) and (v=x+y). Solve the two equations first, then convert back to (x) and (y).