Arithmetic Operations Questions and Answers

Answer: B

We can multiply sequentially.

\(0.5 \times 0.05 = 0.025\).

Now, \(0.025 \times 500 = 25 \times 0.5 = 12.5\).

Alternatively, write as fractions: \((\frac{1}{2}) \times (\frac{5}{100}) \times 500 = (\frac{1}{2}) \times (\frac{1}{20}) \times 500 = \frac{500}{40} = \frac{50}{4} = 12.5\).

Answer: C

Correct answer: 1

Look at the remainders of successive powers of 3 on division by 5: \[ 3^{1} = 3,\quad 3^{2} = 9 \equiv 4,\quad 3^{3} = 27 \equiv 2,\quad 3^{4} = 81 \equiv 1. \] The pattern 3, 4, 2, 1 then repeats, so the cycle length is 4. Since \( 40 \) is exactly divisible by 4, \( 3^{40} \) corresponds to the last term of the cycle, and the remainder is 1. A remainder of 0 in the exponent division always means the last term, not the first - which is the commonest slip in these questions. The result also follows from the unit digit: the unit digits of powers of 3 cycle 3, 9, 7, 1, so \( 3^{40} \) ends in 1, and a number ending in 1 leaves remainder 1 on division by 5. For a prime divisor Fermat’s little theorem gives the same conclusion directly: \( 3^{4} \equiv 1 \pmod 5 \).

Answer: B

Correct answer: 3

The unit digits of the powers of 7 repeat in a cycle of four: \[ 7^{1} = 7,\quad 7^{2} = 49,\quad 7^{3} = 343,\quad 7^{4} = 2401, \] giving the sequence 7, 9, 3, 1 and then repeating. To find where the 95th power falls, divide the exponent by the cycle length: \( 95 \div 4 \) leaves a remainder of 3. The third term of the cycle is 3. A remainder of 0 would mean the last term of the cycle, not the first - that is the usual trap. The cycle lengths for the other digits are worth memorising: 2, 3, 7 and 8 all cycle in four; 4 and 9 cycle in two; and 0, 1, 5 and 6 always end in themselves. The same technique answers questions about the unit digit of a product or sum of large powers: find each unit digit separately, then combine them.

Answer: B

Work outward from the inner percentage.

\[ 50\% \text{ of } 200 = 100 \]

\[ 25\% \text{ of } 100 = 25 \]

Equivalently \( 0.25 \times 0.5 \times 200 = 25 \); the two percentages multiply rather than add.

Answer: B

Correct answer: 2

Divide 1,056 by 23: \( 23 \times 45 = 1035 \), leaving a remainder of \( 1056 - 1035 = 21 \). To reach the next multiple of 23, add the difference between the divisor and the remainder: \[ 23 - 21 = 2. \] Checking, \( 1056 + 2 = 1058 = 23 \times 46 \). The distractor 21 is the remainder itself, and it is the answer to the opposite question - the least number to be subtracted to make the figure divisible. Reading which is asked is where marks are usually lost. The two rules are worth stating together. To make a number divisible by \( d \), subtract the remainder \( r \), or add \( d - r \). If the remainder is zero the number is already divisible and nothing need be added. The same reasoning extends to finding the nearest multiple in either direction.

Answer: A

Correct answer: 70,700

The place value of a digit is the digit multiplied by the value of the position it occupies. In 470,752 the digit 7 appears twice. The first is in the ten thousands place, giving \( 7 \times 10000 = 70000 \). The second is in the hundreds place, giving \( 7 \times 100 = 700 \). Their sum is \[ 70000 + 700 = 70700. \] The distinction to keep clear is between place value and face value. The face value of a digit is the digit itself, unchanged wherever it stands - so the sum of the face values of 7 here is simply \( 7 + 7 = 14 \). Reading the positions carefully is what the question tests, and the usual error is misplacing a digit by one position. Writing the number with its places labelled - 4 lakhs, 7 ten-thousands, 0 thousands, 7 hundreds, 5 tens, 2 units - removes the risk entirely.

Answer: B

Take \( x = 6k \) and \( y = 5k \).

\[ \frac{x^{2}+y^{2}}{x^{2}-y^{2}} = \frac{36k^{2}+25k^{2}}{36k^{2}-25k^{2}} \]

\[ = \frac{61}{11} \]

The \( k^{2} \) cancels, so only the ratio matters.

Answer: B

Let the number be x. An increase of 37.5% means the new number is 137.5% of the original.

Note that 37.5% = 3/8.

So, the new number is \(x + \frac{3}{8}x = \frac{11}{8}x\).

\(\frac{11}{8}x = 99\)

\(x = 99 \times \frac{8}{11} = 9 \times 8 = 72\).

Answer: B

Let the numbers be 27a and 27b, where a and b are co-prime.

Sum = 27a + 27b = 216

27(a + b) = 216

a + b = 216 / 27 = 8.

We need to find pairs of co-prime numbers that sum to 8.

The pairs are (1, 7) and (3, 5). The pair (2, 6) is not co-prime, and (4, 4) is not co-prime.

There are 2 such pairs.

Answer: B

Take out the common factor.

\[ 1.07(65 + 26 + 9) = 1.07 \times 100 \]

\[ = 107 \]

The three numbers were chosen to total exactly 100, which is what makes the factorisation worth spotting.