What is the value of \((32 \times 12) + (17^2) - 95\)?
Answer: A
Follow the BODMAS rule. First, perform multiplication and find the square, then do addition and subtraction.
Step 1: \(32 \times 12 = 384\).
Step 2: \(17^2 = 289\).
Step 3: The expression becomes \(384 + 289 - 95\).
Step 4: \(384 + 289 = 673\).
Step 5: \(673 - 95 = 578\).
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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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Evaluate \(123 \times 5 + 123 \times 15\).
Answer: A
Using the distributive property \(a \times b + a \times c = a \times (b+c)\).
Here, a = 123, b = 5, c = 15.
The expression is \(123 \times (5 + 15) = 123 \times 20\).
\(123 \times 20 = 2460\).
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Find the value of \(5^3 + 6^3 + 7^3 + ... + 10^3\).
Answer: A
We use the formula for the sum of the cubes of the first n natural numbers: \(S_n = [\frac{n(n+1)}{2}]^2\).
The required sum is (Sum of cubes from 1 to 10) - (Sum of cubes from 1 to 4).
Sum of cubes from 1 to 10 = \([\frac{10(10+1)}{2}]^2 = (\frac{110}{2})^2 = 55^2 = 3025\).
Sum of cubes from 1 to 4 = \([\frac{4(4+1)}{2}]^2 = (\frac{20}{2})^2 = 10^2 = 100\).
The required sum = \(3025 - 100 = 2925\).
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What is the smallest number that must be added to 2203 to make it a perfect square?
Answer: C
We need to find the nearest perfect square greater than 2203.
We know \(40^2 = 1600\) and \(50^2 = 2500\). The number is between 40 and 50.
Let's try numbers ending in 3 or 7. \(47^2 = (50-3)^2 = 2500 - 300 + 9 = 2209\).
\(46^2 = (47-1)^2 = 2209 - 94 + 1 = 2116\).
The next perfect square after 2203 is 2209.
The number to be added is \(2209 - 2203 = 6\).
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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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Find the unit digit of the expression \(11^1 + 12^2 + 13^3 + 14^4 + 15^5 + 16^6\).
Answer: B
We only need to find the unit digit of each term and add them.
Unit digit of \(11^1\) is 1.
Unit digit of \(12^2\) is the unit digit of \(2^2\), which is 4.
Unit digit of \(13^3\) is the unit digit of \(3^3=27\), which is 7.
Unit digit of \(14^4\) is the unit digit of \(4^4=256\), which is 6.
Unit digit of \(15^5\) is 5.
Unit digit of \(16^6\) is 6.
Sum of unit digits = \(1+4+7+6+5+6 = 29\).
The unit digit of the final expression is the unit digit of 29, which is 9.
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Find the value of: \(1 - 2 + 3 - 4 + 5 - 6 + ... + 99 - 100\).
Answer: A
We can group the terms in pairs:
(1 - 2) + (3 - 4) + (5 - 6) + ... + (99 - 100).
Each pair sums to -1.
There are 100 terms in total, so there are 100/2 = 50 such pairs.
The total sum is 50 × (-1) = -50.
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What is the number which when multiplied by 13 is increased by 180?
Answer: B
Let the number be x.
According to the question, 13x = x + 180.
13x - x = 180.
12x = 180.
x = 180 / 12 = 15.
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If one-third of one-fourth of a number is 15, then three-tenth of that number is:
Answer: D
Let the number be x.
One-third of one-fourth of the number is \((\frac{1}{3}) \times (\frac{1}{4}) \times x = 15\).
\(\frac{x}{12} = 15\), so x = 180.
Now, we need to find three-tenth of the number:
\((\frac{3}{10}) \times 180 = 3 \times 18 = 54\).
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