Arithmetic Operations Questions and Answers

Answer: D

Correct answer: 61

When a divisor leaves the same remainder for several numbers, it must divide exactly the differences between them - because the equal remainders cancel on subtraction. The differences are \[ 251 - 190 = 61, \qquad 373 - 251 = 122, \qquad 373 - 190 = 183. \] The required number is the HCF of these: 61 divides 122 and 183 exactly, so \[ \text{HCF}(61, 122, 183) = 61. \] Checking, each of 190, 251 and 373 leaves a remainder of 7 on division by 61. The technique is general and worth remembering in both directions. Where the remainders are equal but unknown, take the HCF of the differences. Where the remainders are given and different, subtract each remainder from its number first and then take the HCF of the results. Taking the HCF of the original numbers rather than the differences is the usual error.

Answer: B

Correct answer: 60

For any two numbers the product of the HCF and the LCM equals the product of the numbers themselves: \[ \text{HCF} \times \text{LCM} = a \times b. \] Substituting, \[ 12 \times 180 = 36 \times b \;\Rightarrow\; b = \frac{2160}{36} = 60. \] Checking: \( \text{HCF}(36, 60) = 12 \) and \( \text{LCM}(36, 60) = 180 \), so both conditions hold. The relation holds only for two numbers, and applying it to three is a common error. Two further checks are worth knowing. The HCF must always divide the LCM exactly - here \( 180/12 = 15 \) - so a question whose figures fail that test is inconsistent. And the HCF must divide both numbers while the LCM must be a multiple of both, which alone rules out 45 and 54 here since neither is divisible by 12.

Answer: B

Multiplying by 16 increases the number by 225, so the increase is fifteen times the number.

\[ 16x = x + 225 \quad\Longrightarrow\quad 15x = 225 \]

\[ x = 15 \]

Answer: B

Correct answer: 450

Factorise into primes: \[ 3600 = 2^{4} \times 3^{2} \times 5^{2}. \] A perfect cube requires every exponent to be a multiple of 3. Here the exponents are 4, 2 and 2. Dividing must reduce each to the nearest lower multiple of three - so \( 2^{4} \) becomes \( 2^{3} \), and \( 3^{2} \) and \( 5^{2} \) must go entirely, since their next lower multiple is \( 3^{0} \) and \( 5^{0} \). The divisor is therefore \[ 2^{1} \times 3^{2} \times 5^{2} = 2 \times 9 \times 25 = 450. \] Checking: \( 3600 \div 450 = 8 = 2^{3} \), a perfect cube. The companion question - the smallest number to multiply by - raises each exponent to the next multiple of three instead, giving \( 2^{2} \times 3^{1} \times 5^{1} = 60 \), and \( 3600 \times 60 = 216000 = 60^{3} \). The same method with multiples of two answers the equivalent questions about perfect squares.

Answer: A

Correct answer: 7400

Use the difference of two squares, \( a^{2} - b^{2} = (a+b)(a-b) \): \[ 87^{2} - 13^{2} = (87+13)(87-13) = 100 \times 74 = 7400. \] The identity turns two squarings and a subtraction into one multiplication, and it is worth looking for whenever a question presents \( a \times a - b \times b \). Examiners choose the numbers so that \( a+b \) is a round figure - here 100 - which is the signal that the identity is intended. The direct route confirms it: \( 7569 - 169 = 7400 \), but it takes appreciably longer under time pressure. The companion identities are worth having ready in the same way: \( (a+b)^{2} = a^{2} + 2ab + b^{2} \), \( (a-b)^{2} = a^{2} - 2ab + b^{2} \), and \( a^{3} \pm b^{3} = (a \pm b)(a^{2} \mp ab + b^{2}) \).

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.