How many numbers between 200 and 400 are divisible by 3, 4, and 5?
Answer: B
A number divisible by 3, 4, and 5 must be divisible by their LCM.
LCM(3, 4, 5) = \(3 \times 4 \times 5 = 60\).
We need to find the multiples of 60 between 200 and 400.
The multiples are: \(60 \times 4 = 240\), \(60 \times 5 = 300\), \(60 \times 6 = 360\).
There are 3 such numbers.
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On dividing a certain number by 357, the remainder is 39. On dividing the same number by 17, what will be the remainder?
Answer: C
Let the number be N. According to the division algorithm, N = Divisor × Quotient + Remainder.
N = 357k + 39, where k is the quotient.
We need to find the remainder when N is divided by 17.
N = (17 × 21)k + (17 × 2 + 5) = 17 × 21k + 17 × 2 + 5.
N = 17(21k + 2) + 5.
So, when the number is divided by 17, the remainder is 5.
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Which of the following fractions is the largest? 7/8, 13/16, 31/40, 63/80
Answer: A
To compare fractions, convert them to decimals or find a common denominator.
Converting to decimals:
\(7/8 = 0.875\)
\(13/16 = 0.8125\)
\(31/40 = 0.775\)
\(63/80 = 0.7875\)
Clearly, 0.875 is the largest, so 7/8 is the largest fraction.
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The sum of the digits of a two-digit number is 8. If the digits are reversed, the new number is 36 less than the original number. Find the number.
Answer: B
Let the ten's digit be x and the unit's digit be y.
The number is 10x + y. The reversed number is 10y + x.
Given, x + y = 8 (Equation 1).
Given, (10x + y) - (10y + x) = 36.
9x - 9y = 36, which simplifies to x - y = 4 (Equation 2).
Adding Equation 1 and Equation 2: 2x = 12, so x = 6.
From x + y = 8, we get 6 + y = 8, so y = 2.
The number is 10(6) + 2 = 62.
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What is the value of \(\sqrt{6+\sqrt{6+\sqrt{6+...}}}\) ?
Answer: B
Let \(x = \sqrt{6+\sqrt{6+\sqrt{6+...}}}\).
Then, we can write \(x = \sqrt{6+x}\).
Squaring both sides, we get \(x^2 = 6+x\).
\(x^2 - x - 6 = 0\).
Factoring the quadratic equation: \((x-3)(x+2) = 0\).
Since the value inside the square root is positive, x must be positive. So, x = 3.
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A number when divided by 6 leaves a remainder 3. When the square of the same number is divided by 6, the remainder is:
Answer: D
Let the number be N. Given, N = 6k + 3 for some integer k.
The square of the number is \(N^2 = (6k+3)^2 = (6k)^2 + 2(6k)(3) + 3^2\).
\(N^2 = 36k^2 + 36k + 9\).
\(N^2 = 36k^2 + 36k + 6 + 3 = 6(6k^2 + 6k + 1) + 3\).
This is in the form 6m + 3. So, when \(N^2\) is divided by 6, the remainder is 3.
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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
\[ 13x = x + 180 \quad\Longrightarrow\quad 12x = 180 \]
\[ x = 15 \]
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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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What is the value of \(587 \times 999\)?
Answer: A
Write 999 as \( 1000 - 1 \) to avoid the long multiplication.
\[ 587 \times 999 = 587 \times 1000 - 587 = 587{,}000 - 587 \]
\[ = 586{,}413 \]
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