Math Word Problems and Arithmetic

A 20% profit on cost price gives a gain of Rs 160, so cost price = 160 / 0.20 = Rs 800 and selling price = Rs 960. Since 25% discount is applied on the marked price, selling price is 75% of marked price, so marked price = 960 / 0.75 = Rs 1280.

If the principal becomes 3 times in 10 years under simple interest, the interest earned is 2 times the principal. Rate = (2P x 100) / (P x 10) = 20%, so the rate of interest is 20%.

Under simple interest, if the principal doubles in 4 years, the interest earned equals the principal in 4 years. To become 3 times, the interest must equal twice the principal, which takes 2 x 4 = 8 years.

If money doubles in 3 years under compound interest, then doubling again takes another 3 years, giving 4 times the original in 6 years total. Mathematically, if (1+r)^3 = 2, then (1+r)^6 = 4, confirming 6 years.

For a number to be divisible by 80, it must be divisible by both 5 and 16. Divisibility by 5 requires the last digit to be 0 or 5. Since xy is a two-digit number at the end, we need 65300 + 100x + 10y ≡ 0 (mod 80). This simplifies to 60 + 10x + y ≡ 0 (mod 80). Testing values, when x=6 and y=0, we get 60 + 60 + 0 = 120, which is divisible by 80. Therefore x+y=6+0=6.

We need N-3 to be divisible by 21, 28, 36, and 45. Therefore N-3 must equal the LCM of these numbers. LCM(21,28,36,45) = LCM(3×7, 2²×7, 2²×3², 3²×5) = 2² × 3² × 5 × 7 = 4 × 9 × 5 × 7 = 1260. So N = 1260 + 3 = 1263.

From 1.5x = 0.04y, we get y = 1.5x/0.04 = 37.5x. Substituting in (y-x)/(y+x) = (37.5x-x)/(37.5x+x) = 36.5x/38.5x = 365/385 = 73/77 after dividing numerator and denominator by 5.

Let the number be N. One-seventh exceeds one-eleventh by 100: N/7 - N/11 = 100. This gives (11N - 7N)/77 = 100, so 4N/77 = 100. Therefore N = 100 × 77/4 = 25 × 77 = 1925.

Let Rita's age be 3k and mother's age be 8k. The difference is 8k - 3k = 35, so 5k = 35, giving k = 7. Rita is 21 and mother is 56. After 4 years: Rita = 25, mother = 60. The ratio becomes 25:60 = 5:12 after dividing by 5.

Using allegation method: Gold is 19 parts, copper is 9 parts, alloy is 15 parts. Gold contributes 4 parts above 15 (19-15), copper contributes 6 parts below 15 (15-9). The ratio is the inverse: gold:copper = 6:4 = 3:2. Alternatively: 19g + 9c = 15(g+c), so 19g + 9c = 15g + 15c, giving 4g = 6c, so g:c = 6:4 = 3:2.

Fill tap rate: 1/16 tank per minute. Empty tap rate: 1/8 tank per minute. Net rate when both open: 1/16 - 1/8 = 1/16 - 2/16 = -1/16 tank per minute (negative means emptying). Tank is half full (1/2). Time to empty: (1/2) ÷ (1/16) = (1/2) × 16 = 8 minutes.

In 729 ml with milk:water = 7:2, milk = (7/9)*729 = 567 ml, water = 162 ml. Let x ml water be added to make ratio 1:1. Then 567/(162+x) = 1/1, so 567 = 162+x, giving x = 405 ml.