Arithmetic Operations Questions and Answers

Answer: A

Correct answer: 23

Square both sides of the given equation: \[ \left(x + \frac{1}{x}\right)^{2} = x^{2} + 2 \cdot x \cdot \frac{1}{x} + \frac{1}{x^{2}} = x^{2} + \frac{1}{x^{2}} + 2. \] So \[ 25 = x^{2} + \frac{1}{x^{2}} + 2 \;\Rightarrow\; x^{2} + \frac{1}{x^{2}} = 23. \] The middle term is always exactly 2 because \( x \) and \( 1/x \) multiply to 1, and that is what makes the family of identities so quick to use. The related results are worth having ready. \( \left(x - \frac{1}{x}\right)^{2} = x^{2} + \frac{1}{x^{2}} - 2 \), so here \( x - \frac{1}{x} = \sqrt{21} \). And \( x^{3} + \frac{1}{x^{3}} = \left(x + \frac{1}{x}\right)^{3} - 3\left(x + \frac{1}{x}\right) = 125 - 15 = 110 \). The distractor 25 is the square of the given value, offered for a reader who forgets to subtract the 2.

Answer: A

Correct answer: 1

Simplify the numerator first. Four copies of 325 added together is \[ 325 + 325 + 325 + 325 = 4 \times 325 = 1300. \] The denominator is also \( 4 \times 325 = 1300 \), so the fraction is \[ \frac{1300}{1300} = 1. \] The point of the question is to be recognised rather than computed: repeated addition of the same quantity is multiplication, so the numerator and denominator are the same expression written two different ways, and the value must be 1 without any arithmetic at all. The distractor 4 is what comes of cancelling only the 325 and forgetting the 4, and 325 of cancelling only the 4. Reading the extent of the division bar carefully is what such questions test - \( \frac{325 \times 4}{4} \times \frac{4}{325} \) is a different expression with the value 4, and the placement of the brackets is all that separates them.

Answer: C

Correct answer: 2 : 3

To combine two ratios sharing a middle term, make that term equal in both. Here \( b \) appears as 4 in the first ratio and 8 in the second, so multiply the first throughout by 2: \[ a : b = 6 : 8, \qquad b : c = 8 : 9. \] The chain now reads \( a : b : c = 6 : 8 : 9 \), so \[ a : c = 6 : 9 = 2 : 3. \] The shortcut is to multiply along the chain: \( a : c = (3 \times 8) : (4 \times 9) = 24 : 36 = 2 : 3 \). The step people omit is reducing the final ratio to its lowest terms, which is why 24 : 36 must become 2 : 3. The same method extends to any number of linked ratios - equalise each shared term in turn, or multiply the corresponding entries along the whole chain and reduce once at the end.

Answer: B

Let the two numbers be x and y.

Given, x + y = 40 and x - y = 4.

Adding the two equations: 2x = 44, so x = 22.

Substituting x = 22 in the first equation: 22 + y = 40, so y = 18.

Their product is x × y = 22 × 18 = 396.

Answer: D

Correct answer: the sum of its digits is divisible by 9

A number is divisible by 9 exactly when the sum of its digits is divisible by 9 - and the test may be applied repeatedly until a single digit remains. For 4,617 the digits sum to 18, which is divisible by 9, so 4,617 is too. The divisibility rules worth having ready form a set. By 2: the last digit is even. By 3: the digit sum is divisible by 3. By 4: the last two digits form a multiple of 4. By 5: the last digit is 0 or 5. By 6: divisible by both 2 and 3. By 8: the last three digits form a multiple of 8. By 10: the last digit is 0. By 11: the alternating sum of the digits is 0 or a multiple of 11. The third option fails because divisibility by 3 and 6 together only guarantees divisibility by 6 - the two share a factor, so their product is not implied.

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 \( a^{3} \pm b^{3} = (a \pm b)(a^{2} \mp ab + b^{2}) \).