Arithmetic Operations Questions and Answers

Answer: A

Correct answer: 1

Apply the difference of two squares to the numerator: \[ (0.98)^{2} - (0.02)^{2} = (0.98 + 0.02)(0.98 - 0.02) = 1 \times 0.96 = 0.96. \] Dividing by 0.96 gives \[ \frac{0.96}{0.96} = 1. \] The numbers are chosen so that the sum is exactly 1 and the difference is exactly the denominator, which is the signal that the identity is intended. Squaring both terms and subtracting - \( 0.9604 - 0.0004 = 0.96 \) - gives the same answer but far more slowly and with more scope for a slip in the decimal places. The general form is worth recognising in any arrangement: \[ \frac{a^{2} - b^{2}}{a - b} = a + b, \qquad \frac{a^{2} - b^{2}}{a + b} = a - b. \] Spotting which factor the denominator matches is the whole of the work.

Answer: A

Correct answer: 0.152

This is the difference of two cubes, but here direct computation is quick enough. \[ 0.6^{3} = 0.216, \qquad 0.4^{3} = 0.064, \] so \[ 0.216 - 0.064 = 0.152. \] The identity gives the same result and is worth practising: \[ a^{3} - b^{3} = (a-b)(a^{2} + ab + b^{2}) = 0.2 \times (0.36 + 0.24 + 0.16) = 0.2 \times 0.76 = 0.152. \] The decimal places are what usually go wrong. Cubing a number with one decimal place gives three, so both 0.216 and 0.064 must carry three - and writing 0.64 for \( 0.4^{3} \) is the classic slip, which is why 0.512 and 0.128 are offered. A quick sanity check helps: the answer must be smaller than \( 0.6^{3} = 0.216 \) and positive, which alone rules out 0.512.

Answer: A

Correct answer: 0.09

This is the identity for the sum of two cubes: \[ a^{3} + b^{3} = (a+b)(a^{2} - ab + b^{2}). \] The numerator is \( a^{3} + b^{3} \) and the denominator is exactly the second factor, \( a^{2} - ab + b^{2} \), so the whole expression reduces to \[ \frac{(a+b)(a^{2}-ab+b^{2})}{a^{2}-ab+b^{2}} = a + b. \] With \( a = 0.05 \) and \( b = 0.04 \), the answer is \( 0.05 + 0.04 = 0.09 \). The whole point of the question is recognising the pattern rather than computing anything: multiplying the six decimal products out is slow and invites arithmetic slips. The companion form appears just as often: \[ \frac{a^{3} - b^{3}}{a^{2} + ab + b^{2}} = a - b. \] Note the sign in the middle term flips between the two, which is the detail to get right.

Answer: A

Let the three consecutive even numbers be \(x, x+2, x+4\).

Their sum is \(x + (x+2) + (x+4) = 144\).

\(3x + 6 = 144\).

\(3x = 138\).

\(x = 46\).

The numbers are 46, 48, and 50. The smallest number is 46.

Answer: B

Correct answer: 0.04

Write the number as a fraction of powers of ten: \[ 0.000064 = \frac{64}{1000000} = \frac{4^{3}}{100^{3}}. \] Taking the cube root of numerator and denominator separately, \[ \sqrt[3]{0.000064} = \frac{4}{100} = 0.04. \] The reliable method with decimals is to count places. A cube root divides the number of decimal places by three, so six decimal places become two. Since \( \sqrt[3]{64} = 4 \), the answer is 4 with two decimal places, which is 0.04. The same rule applied to a square root halves the decimal places: \( \sqrt{0.0064} = 0.08 \), because four places become two and \( \sqrt{64} = 8 \). Checking is quick: \( 0.04^{3} = 0.04 \times 0.04 \times 0.04 = 0.000064 \). Miscounting the places is the only real difficulty, and the distractors are all off by exactly one.

Answer: A

Correct answer: 3.96

Move the decimal point in pairs, since a square root halves the number of decimal places. \( 12.96 \) has two decimal places, so its root has one: \( \sqrt{12.96} = 3.6 \). \( 0.1296 \) has four, so its root has two: \( \sqrt{0.1296} = 0.36 \). Adding, \[ 3.6 + 0.36 = 3.96. \] The pattern is worth seeing as a whole. From \( \sqrt{1296} = 36 \) follows \( \sqrt{12.96} = 3.6 \), \( \sqrt{0.1296} = 0.36 \) and \( \sqrt{0.001296} = 0.036 \) - each step of two decimal places in the number moving the point one place in the root. The distractor 3.6 is what comes of computing only the first term, and 4.32 of adding the roots as though the decimal shifted the same amount in both. Checking is easy: \( 3.6^{2} = 12.96 \) and \( 0.36^{2} = 0.1296 \).

Answer: A

Correct answer: 9899

Rewrite the mixed number as a round figure minus a small fraction, which is what makes the arithmetic collapse: \[ 99\frac{98}{99} = 100 - \frac{1}{99}, \] since \( \frac{98}{99} = 1 - \frac{1}{99} \). Now multiply: \[ \left(100 - \frac{1}{99}\right) \times 99 = 9900 - 1 = 9899. \] The alternative is to convert to an improper fraction: \( \frac{99 \times 99 + 98}{99} = \frac{9801 + 98}{99} = \frac{9899}{99} \), and multiplying by 99 gives 9899 - the same answer by a longer route. The technique generalises to any mixed number just below a round figure. \( 999\frac{998}{999} \times 999 = 999000 - 1 = 998999 \), and \( 49\frac{48}{49} \times 49 = 2450 - 1 = 2449 \). The distractor 9900 is the product before subtracting the 1, and 9801 is \( 99^{2} \) alone.

Answer: C

The sum of the first \( n \) odd numbers is always \( n^{2} \).

\[ 1 + 3 + 5 + \cdots + 39 = 20^{2} = 400 \]

The pattern is easy to see: 1, 4, 9, 16 … each partial sum is a perfect square.

Answer: A

Align the decimal points and add.

\[ 0.600 + 0.660 + 0.066 + 6.606 = 7.932 \]

Padding each number to three decimal places first keeps the columns straight.

Answer: C

Correct answer: 111

If the HCF is 37, both numbers are multiples of it, so write them as \( 37a \) and \( 37b \) where \( a \) and \( b \) are co-prime - they can share no further factor, or the HCF would be larger. Their product is \[ 37a \times 37b = 1369ab = 4107 \;\Rightarrow\; ab = 3. \] The only co-prime pair with product 3 is \( 1 \) and \( 3 \), so the numbers are \( 37 \times 1 = 37 \) and \( 37 \times 3 = 111 \). The larger is 111. Checking: \( 37 \times 111 = 4107 \) and \( \text{HCF}(37, 111) = 37 \). The co-primality requirement is what makes the method work and is the step usually forgotten. Had \( ab \) been 12, for instance, the admissible pairs would be only \( 1 \times 12 \) and \( 3 \times 4 \) - not \( 2 \times 6 \), since 2 and 6 share a factor.