What number should be subtracted from both terms of the fraction 15/19 to make it 3/4?
Answer: B
Let the number to be subtracted be x.
The new fraction is \(\frac{15-x}{19-x}\).
We are given that this equals 3/4.
\(\frac{15-x}{19-x} = \frac{3}{4}\)
Cross-multiply: \(4(15-x) = 3(19-x)\).
\(60 - 4x = 57 - 3x\).
\(60 - 57 = 4x - 3x\).
\(3 = x\).
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What is the value of the expression 101 + 102 + 103 + ... + 200?
Answer: B
The sum of an arithmetic progression is given by the formula \(S_n = \frac{n}{2}(a_1 + a_n)\), where \(n\) is the number of terms, \(a_1\) is the first term, and \(a_n\) is the last term.
Number of terms (n) = (Last Term - First Term) + 1 = (200 - 101) + 1 = 100.
First term (\(a_1\)) = 101.
Last term (\(a_n\)) = 200.
Sum = \(\frac{100}{2} \times (101 + 200) = 50 \times 301 = 15050\).
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Simplify: \(1 \div [1 \div \{1 \div (1 \div 1)\}]\).
Answer: A
Working from the innermost bracket outwards:
Step 1: \((1 \div 1) = 1\).
Step 2: \(\{1 \div 1\} = 1\).
Step 3: \([1 \div 1] = 1\).
Step 4: \(1 \div 1 = 1\).
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Find the sum of first 30 natural numbers.
Answer: B
The sum of the first n natural numbers is given by the formula \(S_n = \frac{n(n+1)}{2}\).
Here, n = 30.
Sum = \(\frac{30(30+1)}{2} = \frac{30 \times 31}{2} = 15 \times 31 = 465\).
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Which of the following numbers is divisible by 9?
Answer: B
A number is divisible by 9 if the sum of its digits is a multiple of 9.
A) 7+5+3+2+0 = 17 (Not divisible by 9)
B) 7+5+3+2+1 = 18 (Divisible by 9, since 18 = 9 × 2)
C) 7+5+3+2+2 = 19 (Not divisible by 9)
D) 7+5+3+2+3 = 20 (Not divisible by 9)
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A boy was asked to multiply a number by 25. He instead multiplied the number by 52 and got the answer 324 more than the correct answer. What was the number?
Answer: A
Let the number be x.
Correct multiplication = 25x.
Incorrect multiplication = 52x.
Given, 52x - 25x = 324.
27x = 324.
x = \(324 \div 27 = 12\).
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If a number is decreased by 4 and divided by 6, the result is 8. What would be the result if 2 is subtracted from the number and then it is divided by 5?
Answer: B
Let the number be x.
According to the first condition: \(\frac{x-4}{6} = 8\).
\(x-4 = 48\), so \(x = 52\).
Now, according to the second condition, subtract 2 from the number: \(52 - 2 = 50\).
Then divide by 5: \(50 \div 5 = 10\).
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What is the result of \(3.5 \times (60 \div 2.5)\)?
Answer: A
Using BODMAS, we solve the bracket first.
\(60 \div 2.5 = 60 \div (\frac{5}{2}) = 60 \times \frac{2}{5} = 12 \times 2 = 24\).
Now, multiply by 3.5: \(3.5 \times 24 = (\frac{7}{2}) \times 24 = 7 \times 12 = 84\).
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If \(1^2 + 2^2 + 3^2 + ... + 10^2 = 385\), then the value of \(2^2 + 4^2 + 6^2 + ... + 20^2\) is:
Answer: C
Let the required sum be S.
\(S = 2^2 + 4^2 + 6^2 + ... + 20^2\).
\(S = (2 \times 1)^2 + (2 \times 2)^2 + (2 \times 3)^2 + ... + (2 \times 10)^2\).
\(S = 2^2(1^2) + 2^2(2^2) + 2^2(3^2) + ... + 2^2(10^2)\).
\(S = 2^2 (1^2 + 2^2 + 3^2 + ... + 10^2)\).
\(S = 4 \times (\text{given sum})\).
\(S = 4 \times 385 = 1540\).
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What is the remainder when \(2^{31}\) is divided by 5?
Answer: C
We can use the concept of cyclicity of remainders.
\(2^1 \div 5\) gives remainder 2.
\(2^2 \div 5\) gives remainder 4.
\(2^3 \div 5\) gives remainder 3.
\(2^4 \div 5\) gives remainder 1.
\(2^5 \div 5\) gives remainder 2.
The cycle of remainders is (2, 4, 3, 1), which has a length of 4.
To find the remainder of \(2^{31} \div 5\), we divide the power 31 by the cyclicity 4.
\(31 \div 4\) gives a remainder of 3.
The remainder will be the 3rd term in the cycle, which is 3.
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