Algebra Questions and Answers

Answer: B

Factorise each expression:

The L.C.M. takes the highest power of each factor: 2 cubed, x squared, (2x-3y) cubed and (4x^2+6xy+9y^2). So the L.C.M. is the expression in option B. Options with lower powers of 2 or of (2x-3y) miss part of (4x-6y) cubed.

The numerical factor 2 to the power 3 comes from (4x - 6y) cubed, because 4x - 6y = 2(2x - 3y), and the power of x comes from x squared in the first expression.

Answer: D

The number of 25-paise coins is 124, so the answer is None of these.

Let \(x\) be the 20-paise coins and \(y\) the 25-paise coins. In paise:

\[ \begin{aligned} x + y &= 324 \\ 20x + 25y &= 7100 \\ 20 \times 324 + 5y &= 7100 \\ 5y &= 620, \quad y = 124 \end{aligned} \]

Check: 200 coins of 20 paise make Rs. 40 and 124 of 25 paise make Rs. 31, Rs. 71 in all. 120, 200 and 144 do not satisfy both conditions (200 is the number of 20-paise coins).

Answer: B

The number of cows is 7.

Let there be \(c\) cows and \(h\) hens. Legs: \(4c + 2h\); heads: \(c + h\).

\[ \begin{aligned} 4c + 2h &= 2(c + h) + 14 \\ 2c &= 14, \quad c = 7 \end{aligned} \]

Every animal accounts for 2 legs in 'twice the heads', so the extra 14 legs are the cows' second pair of legs, 2 per cow. The number of hens does not matter. 5, 10 and 12 do not give 14 extra legs.

Answer: B

The value is 7.

Substitute \(x = 2\):

\[ \begin{aligned} 3x^{2} - 6x + 7 &= 3(2)^{2} - 6(2) + 7 \\ &= 12 - 12 + 7 = 7 \end{aligned} \]

The option printed '322 - 62 + 7' is the substitution step \(3 \cdot 2^{2} - 6 \cdot 2 + 7\) with its powers and multiplication signs lost, not a value. -7 comes from a sign error, and 0 from dropping the constant term.

Answer: C

The values are \(x = 3\) and \(y = 4\), so the answer is 3, 4.

From the first equation \(x = 15 - 3y\). Substitute into the second:

\[ \begin{aligned} 5(15 - 3y) - 2y &= 7 \\ 75 - 17y &= 7 \\ y &= 4,\ x = 15 - 12 = 3 \end{aligned} \]

Check: \(3 + 12 = 15\) and \(15 - 8 = 7\). The option 4, 3 has the right numbers in the wrong order, since the question asks for x and y respectively. 5, 3 fails the first equation (\(5 + 9 = 14\)), and -3, 6 fails the second (\(-15 - 12 = -27\)).

Answer: D

The value is 2.

Square the given equation twice:

\[ \begin{aligned} x^2 + \frac{1}{x^2} &= 2^2 - 2 = 2 \\ x^4 + \frac{1}{x^4} &= 2^2 - 2 = 2 \end{aligned} \]

In fact \(x + \frac{1}{x} = 2\) has only one solution, \(x = 1\), and then every power gives \(1 + 1 = 2\). 34 is the answer to the common version where \(x + \frac{1}{x} = 2\sqrt{2}\) or \(x^2 + \frac{1}{x^2} = 6\), and 4 is \(\left(x + \frac{1}{x}\right)^2\), the result of squaring once without subtracting the 2.

Answer: B

The value of \(\frac{x^2}{y^2}\) is 1.

Collect the terms of the given equation:

\[ \begin{aligned} x + 2y &= 2x + y \\ 2y - y &= 2x - x \\ y &= x \end{aligned} \]

Since x and y are equal, \(\frac{x^2}{y^2} = 1\) (taking \(y \neq 0\), so that the fraction is defined). The value 0 would need x = 0, and 2 or 4 would need x to be a multiple of y, which the equation does not allow.

Answer: D

There are 20 coins of 25 paise.

If there are k coins of 20 paise, there are 2k coins of 25 paise and, since that is half the number of 50 paise coins, 4k coins of 50 paise. Work in rupees:

\[ \begin{aligned} 0.5(4k) + 0.25(2k) + 0.2k &= 27 \\ 2k + 0.5k + 0.2k &= 27 \\ 2.7k = 27,\ k &= 10 \end{aligned} \]

So there are 10 coins of 20 paise, \(2k = 20\) of 25 paise and 40 of 50 paise. Check: 20 + 5 + 2 = 27 rupees. The option 10 is the number of 20 paise coins, not 25 paise coins.

Answer: B

The fares are Rs. 13 to B and Rs. 17 to C.

Let the fares be \(b\) and \(c\):

\[ \begin{aligned} 2b + 3c &= 77 \\ 3b + 2c &= 73 \\ 5b + 5c &= 150 \Rightarrow b + c = 30 \\ c - b &= 4 \end{aligned} \]

Subtracting the second equation from the first gives \(c - b = 4\), so \(c = 17\) and \(b = 13\). Check: \(2(13) + 3(17) = 77\) and \(3(13) + 2(17) = 73\). Rs. 17 and Rs. 13 swaps the two fares.

Adding and subtracting the two equations is the quickest way to solve a symmetric pair like this one.

Answer: D

Multiply numerators and denominators separately, then cancel.

\[ \frac{3x}{2y} \cdot \frac{x}{3y} = \frac{3x^{2}}{6y^{2}} = \frac{x^{2}}{2y^{2}} \]

The 3 in the numerator cancels one factor of the 6 - leaving it in gives \( \tfrac{x^{2}}{6y^{2}} \).