What is the value of \(0.1 \times 0.01 \times 1000\)?
Answer: A
Count the decimal places as you go.
\[ 0.1 \times 0.01 = 0.001 \]
\[ 0.001 \times 1000 = 1 \]
Equivalently \( 10^{-1} \times 10^{-2} \times 10^{3} = 10^{0} = 1 \).
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A student has to secure 40% marks to pass. He gets 178 marks and fails by 22 marks. The maximum marks are:
Answer: B
The student gets 178 marks and fails by 22 marks. This means the passing marks are 178 + 22 = 200.
Let the maximum marks be M.
We are given that the passing mark is 40% of the maximum marks.
So, 40% of M = 200.
\(0.40 \times M = 200\).
\(M = \frac{200}{0.40} = \frac{2000}{4} = 500\).
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What is the result of dividing 0.025 by 100?
Answer: D
Dividing by 100 moves the decimal point two places left.
\[ \frac{0.025}{100} = 0.00025 \]
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A number is first increased by 20% and then decreased by 10%. Find the net percentage change.
Answer: C
We can use the successive percentage change formula: \(A + B + \frac{AB}{100}\).
Here, A = +20 (increase) and B = -10 (decrease).
Net change = \(20 + (-10) + \frac{20 \times (-10)}{100}\)
= \(10 - \frac{200}{100} = 10 - 2 = 8\)%.
Since the result is positive, it is an 8% increase.
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Half of 1 percent, written as a decimal, is:
Answer: A
\[ 1\% = 0.01 \]
\[ \frac{1}{2} \times 0.01 = 0.005 \]
Half of one per cent is 0.5%, not 0.5 - the per cent sign is worth a factor of 100.
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Simplify: \(36 \div 2 \times 3 + 4 - 6\)
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: \(58 - 6 = 52\)
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The sum of all prime numbers between 60 and 80 is:
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.
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Which fraction is the largest among 2/3, 3/4, 4/5, and 5/6?
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.
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A number when divided by 361 gives a remainder of 47. What remainder does it give when divided by 19?
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.
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What is the approximate value of \(1002 \div 49.98 \times 99\)?
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, \(20 \times 100 = 2000\).
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