If a = 20, b = 35 and c = 5 then (4a + b) / c is
Answer: A
Substitute and work through the numerator first.
\[ 4a + b = 4(20) + 35 = 80 + 35 = 115 \]
\[ \frac{115}{c} = \frac{115}{5} = 23 \]
The whole numerator is divided by \( c \), not just the \( b \) - the 115 on offer is what you get by stopping one step early.
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By factorization of (17x - 34), answer must be
Answer: C
Both terms share a factor of 17.
\[ 17x - 34 = 17x - 17 \times 2 = 17(x - 2) \]
Checking: \( 17x - 34 \). Taking out \( x \) instead is not possible, since 34 has no factor of \( x \).
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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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Which of the following is the value of (x+ 1/x)²?
Answer: C
Apply \( (p+q)^{2} = p^{2} + 2pq + q^{2} \) with \( p = x \) and \( q = \tfrac{1}{x} \).
\[ \left(x + \frac{1}{x}\right)^{2} = x^{2} + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^{2}} \]
\[ = x^{2} + \frac{1}{x^{2}} + 2 \]
The middle term simplifies to the constant 2 because \( x \) cancels with \( \tfrac{1}{x} \) - which is why this identity is so useful for problems linking \( x + \tfrac{1}{x} \) to \( x^{2} + \tfrac{1}{x^{2}} \).
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Algebraic expression of word expression "2 times number a increased by 3" is
Answer: B
"2 times number \( a \)" is \( 2a \), and "increased by 3" adds 3 to it.
\[ 2a + 3 \]
The order matters: \( 3a + 2 \) would be three times the number increased by two.
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Simplify 15ax² ⁄ 5x
Answer: B
Divide the coefficients and subtract the exponents.
\[ \frac{15ax^{2}}{5x} = \frac{15}{5} \cdot a \cdot x^{2-1} = 3ax \]
The \( x \) in the denominator lowers the power by one; leaving it as \( x^{2} \) is the usual slip.
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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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Sum of 2a + b + 5c and -5a - 2b + 3c is
Answer: C
Add the like terms.
\[ a: \ 2 - 5 = -3 \]
\[ b: \ 1 - 2 = -1 \]
\[ c: \ 5 + 3 = 8 \]
\[ \therefore\ -3a - b + 8c \]
The \( c \) terms are both positive, so that coefficient grows while the other two turn negative.
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Factorization of p(4q + 3) +4(4q+3) leads to
Answer: C
Both terms share the bracket \( (4q + 3) \).
\[ p(4q+3) + 4(4q+3) = (p + 4)(4q + 3) \]
Checking: \( 4pq + 3p + 16q + 12 \), which is what the original expands to.
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By solving expression 4[3s - 2(6s - 8)], answer will be
Answer: C
Work from the inside out.
\[ 2(6s - 8) = 12s - 16 \]
\[ 3s - (12s - 16) = 3s - 12s + 16 = -9s + 16 \]
\[ 4(-9s + 16) = -36s + 64 \]
The minus in front of the inner bracket flips both its terms, turning \( -16 \) into \( +16 \).
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