Aptitude - Simple Interest - Discussion

Please explain how do you get 7500 and 6400.

In this explanation how - 22x will come can any one explain me.

C let us assume that principle invested in a is x
& b be y
total amt invested is 13900
so x+y=13900 or y=13900-x rit-----1
foorming quad eq. of simple int.
x*14*2/100 + y*11*2/100 = 3508 i=prt/100
or 28x/100+22y/100=3508
28x+22y=350800
substitution of y frm 1
28x+22(13900-x)=350800 now solve.

Good thinking.

pin/100 + (13900-p)in/100=3508 where p is principle
i is interest rate and n is no.of years
p(28/100)+(13900-p)22/100=3508
28p+13900*22-22p=3508*100
28p-22p+305800=350800
6p=45ooo
p=7500
13900-7500=6400

Formula is simple interest(S.I)=P*T*R/100;

P=Principal amount or sum
T=Time
R=Interest rate

By using this formula we can solve this.

Can anyone tell me how to do this answer easily and fast without using pen ?

Amount invested is nothing but the principal.

Now principal is divided into two parts.

Therefore P1+P2 = 13900 Rs-------equation no.1.

Now,

S.I with rate 14%=S.I1=(P1*2*14)/100 -----as for two years N=2.

S.I with rate 11%=S.I2=(P2*2*11)/100 -----as for two years N=2.

Therefore total S.I=S.I1+S.I2 ---------equation 2.

But S.I = 3508 Rs.

Therefore Equation 2 becomes,

3508 = [ (P1*2*14)/100 ] + [(P2*2*11)/100 ].

3508 = [ P1*14 +P2*11 ] [2/100]---TAKE 2& 100 common.

(3508*100)/2=[ P1*14 +P2*11 ] -----equation 3.

Solve equation 1 & 3 simultaneously,

Hence P1 = 7500 ----for A scheme at rate 14%.

P2 = 6400 ----for B scheme at rate 11%.

Can someone please explain me the 2nd step 28X-22X = 350800 - (13900x22) how it came?

Assume the amount invested in scheme B be 6400.

Then the amount invested in scheme A is 13900-6400 = 7500.

14% in 7500 = 1050.

For two years (1050*2) = 2100.

11% in 6400 = 704.

For two years (704*2) = 1408.

Adding these two gives (2100 + 1408) = 3508.

Therefore we assumed correctly and the answer is 6400.