Simplify (3x ⁄ 2y)(x ⁄ 3y)
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 \( \tfrac{x^{2}}{6y^{2}} \).
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Simplify (1 ⁄ (x² - 1)) ÷ (1 ⁄ (x + 1))
Answer: A
Dividing by a fraction means multiplying by its reciprocal.
\[ \frac{1}{x^{2}-1} \div \frac{1}{x+1} = \frac{1}{x^{2}-1} \times (x+1) \]
Factorising the denominator:
\[ = \frac{x+1}{(x-1)(x+1)} = \frac{1}{x-1} \]
The \( (x+1) \) cancels, which is why the difference of squares has to be split before the simplification becomes visible.
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Simplify (1 ⁄ s) - (1 ⁄ t)
Answer: B
Put both fractions over the common denominator \( st \).
\[ \frac{1}{s} - \frac{1}{t} = \frac{t}{st} - \frac{s}{st} = \frac{t-s}{st} \]
The order matters: subtracting \( \tfrac{1}{t} \) leaves \( t \) first in the numerator, so \( \tfrac{s-t}{st} \) is the negative of the answer.
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Simplify (2 ⁄ tanA) + (4 ⁄ tanB)
Answer: A
The common denominator is \( \tan A \tan B \).
\[ \frac{2}{\tan A} + \frac{4}{\tan B} = \frac{2\tan B}{\tan A \tan B} + \frac{4\tan A}{\tan A \tan B} \]
\[ = \frac{2\tan B + 4\tan A}{\tan A \tan B} \]
Each numerator picks up the other denominator - swapping them is the error the second option represents.
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On solving 2p - 3q - 4r + 6r - 2q + p, answer will be
Answer: C
Collect like terms.
\[ 2p + p = 3p \]
\[ -3q - 2q = -5q \]
\[ -4r + 6r = 2r \]
\[ \therefore\ 3p - 5q + 2r \]
Only terms in the same variable may be combined; \( p \), \( q \) and \( r \) stay separate.
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Answer of factorization of expression 4z(3a + 2b - 4c) + (3a + 2b - 4c) is
Answer: A
Both terms share the bracket \( (3a + 2b - 4c) \), so take it out as a common factor.
\[ 4z(3a+2b-4c) + 1 \cdot (3a+2b-4c) = (4z + 1)(3a + 2b - 4c) \]
The second term carries an invisible coefficient of 1, which is what becomes the \( +1 \) in the first bracket - dropping it would leave \( 4z(3a+2b-4c) \) alone.
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By factorizing expression 2bx + 4by - 3ax -6ay, answer must be
Answer: B
Group the terms in pairs and factor each.
\[ 2bx + 4by - 3ax - 6ay = 2b(x + 2y) - 3a(x + 2y) \]
The bracket \( (x + 2y) \) is now common:
\[ = (2b - 3a)(x + 2y) \]
Taking out \( -3a \) rather than \( 3a \) from the second pair is what makes both brackets identical.
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If -4x + 5y is subtracted from 3x + 2y then answer will be
Answer: C
Subtracting means taking the first expression away from the second, so every sign in it flips.
\[ (3x + 2y) - (-4x + 5y) = 3x + 2y + 4x - 5y \]
\[ = 7x - 3y \]
The \( -4x \) becomes \( +4x \) under the subtraction - leaving it negative would give \( -x - 3y \).
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Answer of factorization of expression (3x - 2y)(a + b) + (4x - 3y)(a + b) is
Answer: D
Both terms share the factor \( (a + b) \), so take it out first:
\[ (3x - 2y)(a+b) + (4x - 3y)(a+b) = \left[(3x - 2y) + (4x - 3y)\right](a+b) \]
Collecting like terms inside the bracket:
\[ 3x + 4x = 7x, \qquad -2y - 3y = -5y \]
\[ \therefore\; (7x - 5y)(a + b) \]
The sign is the trap: both \( y \) terms are negative, so they add to \( -5y \) rather than cancelling.
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If Ana is x years old then 5 times Ana's age three years ago is
Answer: B
Three years ago Ana was \( x - 3 \) years old, and five times that is
\[ 5(x - 3) \]
The subtraction happens inside the bracket, before the multiplication: \( 5x - 3 \) would be five times her present age less three.
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