Entrance Exams
Govt. Exams
Even numbers from 2 to 50: 2, 4, 6, ..., 50.
This is an AP with first term = 2, last term = 50, common difference = 2.
Number of terms = 25.
Sum = 25(2+50)/2 = 25 × 26 = 650
Let number be n.
Then n(n+1) = 342.
Solving: n² + n - 342 = 0.
Using quadratic formula or testing: 17 × 18 = 306 (no), 18 × 19 = 342 (yes).
So n = 18.
Wait, checking: 17 × 18 = 306, 18 × 19 = 342.
Answer is B.
Odd numbers between 10 and 30: 11, 13, 15, 17, 19, 21, 23, 25, 27, 29.
Count = 10
Let numbers be 5x and 7x.
Then 5x + 7x = 120, so 12x = 120, x = 10.
Larger number = 7 × 10 = 70
9+8+7+6+5+4+3 = 42.
Then 4+2 = 6.
Wait, let me recalculate: Sum = 42, which reduces to 4+2=6.
The answer should be D.
Actually 9+8+7+6+5+4+3=42, 4+2=6.
Using dividend = divisor × quotient + remainder: Number = 11 × 9 + 5 = 99 + 5 = 104
This question asks us to find the remainder when 527 is divided by 15 using the division algorithm.
We need to divide 527 by 15 and find what's left over.
Determine how many times 15 goes into 527 completely.
Subtract the product from the original number to find the remainder.
The remainder when 527 is divided by 15 is 2, so the answer is (A).
This question asks us to find the average value of the numbers 1 through 15.
The first 15 natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15.
Use the formula for sum of first n natural numbers: \[\text{Sum} = \frac{n(n+1)}{2}\]
Average is the sum divided by the count of numbers.
The average of the first 15 natural numbers is 8.
This question asks us to find an unknown number based on a sequence of arithmetic operations performed on it.
Let the unknown number be x. According to the problem, when x is multiplied by 8 and then 15 is subtracted, the result is 49.
Add 15 to both sides of the equation to move the constant to the right side.
Divide both sides by 8 to find the value of x.
The number is 8, which corresponds to answer choice (B).
Largest 3-digit number = 999, Smallest 3-digit number = 100.
Difference = 999 - 100 = 899