Quantative aptitude problems

1. si = p*n*i

         =1200*12*0.18

        =2592

2. 6200 = p*2/12*i----------------------------1

7400 = p*3/12*i

therefore,

6200+7400 = p*2/12*i+p*3/12*i

13600 = 1/6 pi +1/4 pi

            =10/24 pi

pi = 13600*24/10

    =32640(principal 32000 + interest 640)

3. 2*p*10/12*i = 3*p*n*i

or,5/3 *p*i = 3*p*i*n

or,9*n = 5

or,n=5/9=0.56=0.6.

1. The interest rate is 18% per annum as such per month it is 18/12% i.e. 1.5%. as the interest per month is 1.5% i = p*r*n/100;

1200 = p*1.5/100 implies p = 1200*100/1.5 =80,000

2.the question doesn't specify whether the interest is compound or simple

if simple interest

A = P+I = P + P*r*n/100

for 2 years

6200 = P + P*r*2/100

for 3 years

7400 = P + P*r*3/100

subtracting first from second we get P*r/100 = 1200

substituting it in the first equation we get 6200-2*(P*r/100)=P

i.e. P = 6200 - 2*(1200) = 6200 - 2400 = 3800

r = 1200*100/3800 = 31.5% (approx)

if compound interest

P(1+r)^2 = 6200; P(1+r)^3 = 7400 dividing second by first we get (1+r) = 7400/6200

implies r = 1200/6200 = 19.35%

P = 6200/(1.1935)^2 = 4352.22

3. Again the question doesn't specify whether the Interest is simple or compound.

Simple Interest:

P+P*r*10/100 = 2P implies r= 10% (cancel P on both sides)

to triple the amount  P+ P*10*n/100 = 3P implies n = 20 i.e. 20 years

Compound Interest

P(1+r)^10 = 2P implies r = 7.17% (take logarithms on both sides)

(1.0717)^n = 3 n = 15.85 years

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