Aptitude - Simple Interest - Discussion

By using I and II;

S.I for 2 years = 750-690 = 60 as S.I is divided uniformly for each year, For 1 year = Rs. 30 and For 3 year = 90,
Principal = 690-90 = 600.

By using II and III
As Total Sum = Principal + Interest i.e 750 = P + I (after 5 years) ------> (1)
P = I *100 / R * T = I *100 / 5*5 = 4I,
Put value of P in eq. 1, 750 = 5I , I = 150 and P = 750 -150 = Rs. 600,

Similarly By using I and III.
P = I *100 / R*T = I*100/5*3 = 20/3(I),
690 = 20/3(I) + I , By solving this we get I = Rs.90 and P = 690 - 90 = Rs. 600.

Hence answer is E Any two of the three.

As Simple as Like that, I will explain here;

Given as Statement 1: P = 690 N= 3 yrs.
Given as Statement 2: P = 750 N= 5 yrs.
By Using these statements We get SI for 2 yrs is (750-690 = 60).

Therefore; SI for 1 yr is 30.

So, SI for 3 yrs will be 90 and that of for 5 yrs will be 150.
We know that Amount = Principal + SI.
Here for 3 yrs W.K.T SI is 90 and Amount is 690.
and therefore Principal is 690-90=600.
And by using statements 2 and 3 we can solve by the following method.
P + (P*5*5)\100 = 750.
100P + 25P = 750*100.
125P = 75000.
P = 600.

Using I and III;
Let S.I be 'x'.
Hence, P = 690 - x.
x = [(690- x) *5*3] / 100.
Solve for 'x' and find P.
Similarly, it works using II and III

Now, using I and II.
Let x be the interest rate.
Hence, P + 3Px = 690, simplify to P(1+3x) =690 -----> (a)
P + 5Px = 750, simplify to P(1+5x)=750 -----> (b).

Now dividing a by b, P gets canceled and solves for x.
x = 0.05 or 5%.
Now substitute x in the equation a or b to get P.

So we can use any of the three to solve.
Thanks.

For those who need help regarding how to divide equation I and II (as suggested by abcd).

750 = P+(P*R*5)/100 => 75000 = 100P + 5PR. (i).

690 = P+(P*R*3)/100 => 69000 = 100P + 3PR. (ii).

Subtracting (i) & (ii) , we get.

R = 3000/P.

Now substitute it in any one either (i) or (ii) , we get.

P = 600.

Thanks.

As Amount = principal+ SI.

Where si= prt/100.
This is the basic formulae that are applied here.

If suppose we take equationn 1 and 2 then,
P + P.3.r% = 690---(1)
P + P.5.r% = 750---(2)

Divide equation 1 by 2 to get r and then p,

And if we take eqn 1 and eqn 3 we will get;
"P=(100*si)/(r*t) ".

In very simple level I explain to 9 how to get answer using ll & similarly only from (ll) we get P+S.I = 750 & T = 5.

And from (lll) we get R = 5.

Now P+(P*5*5)\100 = 750.

100P+25P = 750*100.
125P = 75000.
P = 600.

We can find answer using I and II as follows:.

5PR% = 750 - P.

3PR% = 690 - P --- Dividing them.
---------------.
5 750-P.

= 3 690-P.

3450 - 5P = 2250 - 3P.
1200 = 2P.
P = 600.

Using I and III.

I = PNR%.

I. Amount after 3 years = 690.
II. Amount after 5 years = 750.

Difference = SI for 2 years = 60.

SI for 1 year = 30,
SI for 3 years = 90,
SI for 5 years = 150,

Thus, Principal = 600.

There wasn't any information given with r% in the question (statements 1, 2). So I don't think taking are in both the cases is correct. So option D should be correct clarify.

AMOUNT = S. I + P.

690 - P = 3PR -> (1).
750 - P = 5PR -> (2).

Divide Eqn 2 by 1;

P = 600.

Use P = 600 in section I & III.

So, the correct option is 'E'.