P over B equals R over 100 => P/B = R/100 => P/5950 = 48000/100 => P = 5950 * 480 = 2856000 Or 59.50 * 48000 = 2856000
The answer depends on whether you are dealing with simple interest of compound interest. Suppose P = Principle R = Rate (in % per annum) T = Time (in years) I = Interest Then for simple interest: I = P*R*T/100 so that P = 100*I/(R*T) For compound interest P+I = P*(1+R/100)T so that P = I/[(1+R/100)T - 1]
p = 50q/100 = 1/2 q r = 40q/100 = 2/5 q p = (1/2)/(2/5) = (1/2)(5/2) = 5/4 r or 1 1/4 r Thus, p is 125% of r.
P{1-r/100}^n _p where p is the principal and r the rate and n the years
A = P*(1+R/100)T Where A = amount P = Principal R = Interest Rate (in percentage), and T = Time Since R and T are known, you can calculate (1+R/100)T = k, say. Then A = P*k so that P = A/k
P=B×RB=P÷RR=P÷B
6.5%Formula for finding Simple InterestSI [Interest] = (P×R×T)/100P [sum] = (SI×100)/(R×T)R [Rate/year] = (SI×100)/(P×T)T [Time] = (SI×100)/(P×R)whereS.I. = Simple Interest,P = Principal or Sum of amount,R = % Rate per annum,T = Time Span
Suppose you have a variable which grows at r% per period [year] and you want to find out how many periods, t, it will take before it grows, from a starting value of P to reach the value Q.The approximate method used by younger pupils is trial and error. This is because the y do not have the necessary mathematical knowledge.After t periods, the variable in question will have increased from P to Q whereQ = P*(1+r/100)tthen Q/P = (1+r/100)ttaking logarithms, log(Q/P) = log[(1+r/100)t] = t*log(1+r/100)and so finally, t = log(Q/P) / log(1+r/100).
P*r*t divided by 100
Converse: If p r then p q and q rContrapositive: If not p r then not (p q and q r) = If not p r then not p q or not q r Inverse: If not p q and q r then not p r = If not p q or not q r then not p r
p = r - c r - c = p r - c - r = p - r -(-c) = -(p) c = -p
Let c= original cost Let p= percent reduced by Let r= cost after reduction Equation: c=p*(100%-r)