Govt. Exams
Entrance Exams
2^a = 32 = 2^5, so a = 5. And 3^b = 81 = 3^4, so b = 4. Therefore a + b = 5 + 4 = 9
Let numbers be x and x+2. Then x(x+2) = 195. x² + 2x - 195 = 0. Using formula: x = 13. Numbers are 13 and 15. Sum = 28
Divisors of 16: 1,2,4,8,16. Product = 1×2×4×8×16 = 1024. Or use formula: if n has d divisors, product = n^(d/2). Here d=5, so product = 16^2.5 = 1024
Let number = x. Then x - x/3 = 30. So 2x/3 = 30, x = 45
Let number be x. x + 1/x = 2.5. Multiply by x: x² - 2.5x + 1 = 0. x = (2.5 ± √(6.25-4))/2 = (2.5 ± 1.5)/2. x = 2 or 0.5.
For divisibility by 11: alternate sum of digits. 12321: (1+3+1) - (2+2) = 5 - 4 = 1. Recheck: (1+3+1) - (2+2) = 5-4=1. Actually 1-2+3-2+1 = 1. Try: 1-2+3-2+1 = 1. Check: 12321/11 = 1120.09... Correct: (2+2) - (1+3+1) = 4-5 = -1, still divisible.
Let numbers be 3k and 5k where HCF(3k, 5k) = k = 4. Numbers are 12 and 20. Sum = 32.
# Solution: Finding the Difference Between Two Numbers
When two numbers have a known sum and product, we can use algebraic identities to find their difference without calculating the numbers individually.
Step 1: Set Up the Algebraic Identity
Let the two numbers be \(a\) and \(b\). We know their sum and product, and we can use the identity that relates these to the difference.
We use the identity: \[(a - b)^2 = (a + b)^2 - 4ab\]
Step 2: Calculate the Difference
Substitute the given values into the identity.
The difference between the two numbers is 4.
Answer: (B) 4
# Step-by-Step Explanation
Step 1: Find the Least Common Multiple (LCM)
To find numbers divisible by both 6 and 9, we need numbers divisible by their LCM.
- 6 = 2 × 3
- 9 = 3²
- LCM(6,9) = 2 × 3² = 18
Step 2: Identify the requirement
We need numbers between 1 and 100 that are divisible by 18.
Step 3: List all multiples of 18 in this range
- 18 × 1 = 18 ✓
- 18 × 2 = 36 ✓
- 18 × 3 = 54 ✓
- 18 × 4 = 72 ✓
- 18 × 5 = 90 ✓
- 18 × 6 = 108 ✗ (exceeds 100)
Step 4: Count the valid numbers
The multiples are: 18, 36, 54, 72, and 90 = 5 numbers
Conclusion:
Answer A: 5 is correct because numbers divisible by both 6 and 9 must be divisible by their LCM of 18. Only five multiples of 18 fall between 1 and 100. The other options overcounted by either miscalculating the LCM or incorrectly counting multiples.
Let numbers be x, x+2, x+4, x+6, x+8. Sum = 5x + 20 = 120, so x = 20. Largest = 20 + 8 = 28.
