Quantity I. A= p [1+`R/100` ]t ⇒ 3P = P [1+`R/100` ]4 ⇒ 3 = [1+ `R/100` ]4 …………(i) A= p [1+`R/100` ]t ⇒ 9P = P [1+`R/100` ]t ⇒ 32 = [1+ `R/100` ]t …………….(ii) From (i) and (ii) [1+`R/100` ]4 × 2 = [1+`R/100` ]t t = 8 years Quantity II. Since the amount was Rs 100. It becomes Rs 300 in 20 years. Rs 300 – Rs 100 = Rs 200 interest In 20 years, interest is Rs 200 In 1 year, the interest is Rs`200/20` Rate of interest = 10% per annum. = 100 x 50 / 100 x 10 = 5 years Hence Quantity I > Quantity II
If x = 15, find x5 - 16x4 + 16x3 - 16x2 + 16x - 16 = ?
If a = 55, then what is the value of
if a2 + b2 + c2 = 2(4a -5b -6c)-77 , then a-b-c = ?
If x² + px + q = 0 has roots 4 and -3, find the values of p and q.
For a =-4 and b = 5, value of a² – b² is:
If (x + y) = 6 and (x2 + y2) = 20, then the value of (x – y)2 is equal to:
((99.9 - 20.9)² + (99.9 + 20.9)² )/(99.9 x 99.9 + 20.9 x 20.9) = ?
...(p + q) = 8 and (p2 + q2 - 6) = 28. If p < q, then determine the value of (p/q).