Arithmetic Operations Questions and Answers

Answer: B

Using BODMAS, we perform division and multiplication from left to right, then addition and subtraction.

Step 1: \(36 \div 2 = 18\)

Step 2: \(18 \times 3 = 54\)

Step 3: \(54 + 4 = 58\)

Step 4: \(58 - 6 = 52\)

Answer: A

The primes between 60 and 80 are

\[ 61,\ 67,\ 71,\ 73,\ 79 \]

\[ 61 + 67 + 71 + 73 + 79 = 351 \]

63, 65 and 69 are excluded as multiples of 3 or 5, and 77 is \( 7 \times 11 \) - the one most often mistaken for a prime.

Answer: D

Each fraction is one less than 1 by \( \tfrac{1}{n} \), so the largest denominator gives the largest fraction.

\[ \frac{2}{3} = 0.667, \quad \frac{3}{4} = 0.750, \quad \frac{4}{5} = 0.800, \quad \frac{5}{6} = 0.833 \]

So \( \tfrac{5}{6} \) is the largest. In general \( \tfrac{n}{n+1} \) increases with \( n \), approaching 1.

Answer: B

Correct answer: 9

Since \( 361 = 19^{2} \), any multiple of 361 is also a multiple of 19. Write the number as \( N = 361k + 47 \). The first term is divisible by 19, so the remainder on dividing \( N \) by 19 is the remainder of 47 alone: \[ 47 = 19 \times 2 + 9, \] giving a remainder of 9. The reasoning generalises: if a number leaves remainder \( r \) on division by \( d \), and \( e \) divides \( d \), then the remainder on division by \( e \) is simply \( r \bmod e \). The condition that \( e \) divides \( d \) is essential. Without it nothing can be concluded - a number leaving remainder 47 on division by 361 could leave any remainder at all on division by, say, 7. The error to avoid is giving 47 itself, or dividing 361 rather than 47.

Answer: B

For approximation, we can round the numbers to the nearest convenient values.

1002 ≈ 1000

49.98 ≈ 50

99 ≈ 100

The expression becomes \(1000 \div 50 \times 100\).

Following order of operations: \(1000 \div 50 = 20\).

Then, \(20 \times 100 = 2000\).

Answer: A

Convert the recurring decimals to fractions.

\[ 0.\overline{5} = \frac{5}{9}, \qquad 0.\overline{4} = \frac{4}{9} \]

\[ \frac{5}{9} + \frac{4}{9} = \frac{9}{9} = 1 \]

Adding the decimals directly gives \( 0.999\ldots \), which is exactly 1.

Answer: A

\[ S_{n} = \frac{n(n+1)}{2} \]

\[ S_{45} = \frac{45 \times 46}{2} = 45 \times 23 = 1035 \]

Answer: B

Let the two-digit number be 10x + y, where x is the ten's digit and y is the unit's digit.

The reversed number is 10y + x.

Their sum is \((10x + y) + (10y + x) = 11x + 11y\).

Sum = 11(x + y).

Since the sum is a multiple of 11, it will always be divisible by 11.

Answer: B

This is a telescoping series. Each term can be split using partial fractions: \(1/(n(n+1)) = 1/n - 1/(n+1)\).

The series becomes: \((\frac{1}{2} - \frac{1}{3}) + (\frac{1}{3} - \frac{1}{4}) + (\frac{1}{4} - \frac{1}{5}) + ... + (\frac{1}{9} - \frac{1}{10})\).

All the intermediate terms cancel out.

We are left with \(\frac{1}{2} - \frac{1}{10} = \frac{5-1}{10} = \frac{4}{10} = \frac{2}{5}\).

Answer: C

Correct answer: \( \frac{4}{5} \)

Each denominator is a product of consecutive integers: \( 2 = 1 \times 2 \), \( 6 = 2 \times 3 \), \( 12 = 3 \times 4 \), \( 20 = 4 \times 5 \). Every term can therefore be split as \[ \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}, \] giving \[ \left(1 - \frac{1}{2}\right) + \left(\frac{1}{2} - \frac{1}{3}\right) + \left(\frac{1}{3} - \frac{1}{4}\right) + \left(\frac{1}{4} - \frac{1}{5}\right). \] Everything between the ends cancels - a telescoping series - leaving \[ 1 - \frac{1}{5} = \frac{4}{5}. \] The general result is that the sum of the first \( n \) such terms is \( \dfrac{n}{n+1} \), and with \( n = 4 \) that is \( 4/5 \) directly. Adding the fractions by finding a common denominator of 60 gives \( 30 + 10 + 5 + 3 = 48 \), so \( 48/60 = 4/5 \) - correct, but slower.