Arithmetic Operations Questions and Answers

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.

Answer: C

Correct answer: \( \frac{1}{5} \)

Write each factor as a single fraction: \[ 1 - \frac{1}{n} = \frac{n-1}{n}. \] The product becomes \[ \frac{2}{3} \times \frac{3}{4} \times \frac{4}{5} \times \cdots \times \frac{9}{10}. \] Every numerator cancels against the denominator of the previous factor - a telescoping product - leaving only the first numerator and the last denominator: \[ \frac{2}{10} = \frac{1}{5}. \] The general result is that \[ \prod_{n=k}^{m}\left(1 - \frac{1}{n}\right) = \frac{k-1}{m}, \] which here is \( 2/10 \). Recognising the cancellation is the whole of the work; multiplying the eight fractions out is slow and prone to error. The same structure appears with \( \left(1 + \frac{1}{n}\right) \) factors, which give \( \frac{m+1}{k} \) instead - the fraction inverts, so it is worth checking which form is presented.

Answer: C

Correct answer: 64

Find the largest perfect square not exceeding 2,000. Since \( 44^{2} = 1936 \) and \( 45^{2} = 2025 \), the largest is 1,936. The number to subtract is therefore \[ 2000 - 1936 = 64. \] The companion question - the least number to be added to make a perfect square - uses the next square up: \( 2025 - 2000 = 25 \). Reading which is asked decides the answer, and the two are always different. The method is to take the square root, discard the fractional part for subtraction or round up for addition, and square the result. Here \( \sqrt{2000} \approx 44.7 \), so 44 and 45 are the two squares to consider. The distractor 44 is the square root itself rather than the difference, which is the usual slip. Knowing the squares up to 50 by heart makes questions of this kind almost immediate.

Answer: C

Correct answer: 5

Express 32 as a power of 2: \[ 32 = 2 \times 2 \times 2 \times 2 \times 2 = 2^{5}. \] Since the bases are equal, the exponents must be equal, so \( x = 5 \). The rule being used is that \( a^{m} = a^{n} \) implies \( m = n \), provided \( a \) is positive and not 1. The powers of 2 are worth knowing by heart, since they appear constantly in both aptitude and computing questions: 2, 4, 8, 16, 32, 64, 128, 256, 512, 1024. The laws of indices that go with this are the other things to have ready: \[ a^{m} \times a^{n} = a^{m+n}, \quad \frac{a^{m}}{a^{n}} = a^{m-n}, \quad (a^{m})^{n} = a^{mn}, \quad a^{0} = 1, \quad a^{-n} = \frac{1}{a^{n}}. \] The distractor 16 is 32 halved, offered for a reader who confuses the exponent with the value.

Answer: A

Correct answer: 65

A number leaving the same remainder 5 on division by each divisor is 5 more than a common multiple of them all. The smallest such multiple is the LCM: \( 12 = 2^{2} \times 3 \), \( 15 = 3 \times 5 \), \( 20 = 2^{2} \times 5 \), so \[ \text{LCM} = 2^{2} \times 3 \times 5 = 60. \] The required number is therefore \[ 60 + 5 = 65. \] Checking: 65 divided by 12 gives remainder 5, by 15 gives 5, and by 20 gives 5. The general form is \( \text{LCM} \times n + r \), so the next such numbers are 125, 185 and so on. A variation asks for a number leaving different remainders whose difference from the divisor is constant - say remainders 7, 10 and 15 for divisors 12, 15 and 20, each 5 short. The answer is then \( \text{LCM} - 5 = 55 \), and recognising which form is being asked is the whole difficulty.

Answer: B

Correct answer: 0.13

A square root halves the number of decimal places, and \( \sqrt{169} = 13 \). Since 0.0169 has four decimal places, the root has two: \[ \sqrt{0.0169} = 0.13. \] Written as a fraction the reasoning is plain: \[ \sqrt{0.0169} = \sqrt{\frac{169}{10000}} = \frac{13}{100} = 0.13. \] Checking is immediate: \( 0.13 \times 0.13 = 0.0169 \). The rule requires an even number of decimal places for the root to terminate, which is why such questions always supply two, four or six. A cube root divides the places by three instead, so \( \sqrt[3]{0.000027} = 0.03 \). The perfect squares up to 30 are worth knowing by heart, since almost every question of this kind is built on one: 121, 144, 169, 196, 225, 256, 289, 324, 361, 400.