Mathematics and Logical Reasoning Quiz

Assuming base e: let log(x) = t, so the equation becomes 2t - 5t + 6 = 0, giving t² - 5t + 6 = 0. This factors as (t-2)(t-3) = 0, so t = 2 or t = 3. Therefore x = e² or x = e³.

Let bananas on day 1 be x. Each day he eats 6 more: x, x+6, x+12, x+18, x+24. Sum = 5x + 60 = 100, so 5x = 40 and x = 8.

We need to minimize 3x + 4y given x²y³ = 6. Using AM-GM inequality or Lagrange multipliers, we can rewrite as (3x/2) + (3x/2) + (4y/3) + (4y/3) + (4y/3) ≥ 5×[(3x/2)²(4y/3)³]^(1/5) = 5×[(9x²/4)(64y³/27)]^(1/5) = 5×[(576x²y³/108)]^(1/5) = 5×[(16×6/3)]^(1/5) = 10. Equality occurs when all terms are equal.

Total ways to choose any subset from 12 balls = 2^12 = 4096. Subtract selections with no green (2^7 = 128) and no blue (2^8 = 256). Add back selections with neither (2^3 = 8) since counted twice. 4096 - 128 - 256 + 8 = 3720.

The pattern is 2^n + n for the nth term starting from n=1: 2^1+1=3, 2^2+2=6, 2^3+3=11, 2^4+4=20. The fifth term is 2^5+5=32+5=37. Alternatively, differences double-plus-one: +3, +5, +9, +17... where each difference follows 2^k+1.

For 5 consecutive numbers with average n, the numbers are n-2, n-1, n, n+1, n+2, summing to 5n. Adding the next two numbers (n+3, n+4) gives sum 7n+7 over 7 numbers, so new average is n+1.

Let w be wickets before. Then (12.4w + 26)/(w + 5) = 12.0, giving 12.4w + 26 = 12w + 60, so 0.4w = 34 and w = 85. The decrease by 0.4 in average means 85 wickets were taken before the last match.

For 5 team members, if 4 aged 3 years (total +12), the new member must be 15 years younger than the old member to keep the average the same as 3 years ago (compensating for all 5 members aging 3 years = 15 years total).

From March 9 to May 21, 2001: March has 31 days so 31-9 = 22 days from March 9-31, plus 30 days in April, plus 21 days in May = 73 days. SI = (18200 × 7.5 × 73)/(100 × 365) = 273.

For money to double at 4% compound interest: 2 = (1.04)^n. Testing n=25: 1.04^25 ≈ 2.66 > 2, so 25 years works. This uses the compound interest formula A = P(1+r)^n.

If SI = 6000 after 10 years and principal triples in 5 years, then in 5 years the interest equals 2×principal (since tripling means gain of 2P). So rate = (2P/P)×(100/5) = 40%. Then SI after 10 years = (P × 40 × 10)/100 = 4P. Working backwards: if 4P = 12000, then original SI of 6000 means this checks out.

Assuming 'S.I on o' is a typo for 'S.I on b': if b = ar (SI on a) and c = br (SI on b) at same rate r and time, then c = ar^2. This gives b^2 = (ar)^2 = a(ar^2) = ac.

Assuming 'S.I on o' is a typo for 'S.I on b': if b = ar (SI on a) and c = br (SI on b) at same rate r and time, then c = ar^2 and b^2 = a(ar^2) = ac. This is a duplicate of question 94628.

After 20% discount: value = 80% or 0.8. After 10% discount: 0.8 × 0.9 = 0.72. After 25% discount: 0.72 × 0.75 = 0.54. Final value is 54% of original, so equivalent single discount = 100% - 54% = 46%.

After successive discounts of 20%, 10%, and 25%, the final price is 0.8 × 0.9 × 0.75 = 0.54 of original, meaning the customer pays 54% and the equivalent single discount is 46%. This is a duplicate of question 94630.

If a:b = c:d = k, then a = bk and c = dk. Substituting: (pa+qc)/(pb+qd) = (pbk+qdk)/(pb+qd) = k(pb+qd)/(pb+qd) = k = a/b = c/d. Since a:b = c:d by given condition, the expression equals both ratios.

If a:b = c:d = k (say), then a = bk and c = dk. Substituting in (pa+qc)/(pb+qd) gives (pbk+qdk)/(pb+qd) = k(pb+qd)/(pb+qd) = k = a/b = c/d. The expression simplifies to the original ratio, so both a:b and c:d are equivalent answers. Option D correctly captures this.