Reena took a loan of Rs. 1200 with simple interest for as many years as the rate of interest. If she paid Rs. 432 as interest at the end of the loan period, what was the rate of interest?

Correct Answer: 824.32

Correct Answer: Rs.24.64

Explanation:

For 1st year S.I =C.I.

 

Thus, Rs.16 is the S.I. on S.I. for 1 year, which at 8% is thus Rs.200

 

i.e S.I on the principal for 1 year is Rs.200

 

Principle =  R s . 100 * 200 8 * 1 = Rs.2500

 

Amount for 2 years, compounded half-yearly

  R s . 2500 * 1 + 4 100 4 = R s . 2924 . 4

 

C.I = Rs.424.64

 

Also,  S . I = R s . 2500 * 8 * 2 100 = R s . 400

 

Hence, [(C.I) - (S.I)] = Rs. (424.64 - 400) = Rs.24.64

Correct Answer: 4460

Explanation:

Let the sum be Rs.P.then
P(1+R/100)^3=6690?(i) and P(1+R/100)^6=10035?(ii)
On dividing,we get (1+R/100)^3=10025/6690=3/2.
Substituting this value in (i),we get:
P*(3/2)=6690 or P=(6690*2/3)=4460
Hence,the sum is rs.4460.

Correct Answer: 2 years

Correct Answer: 20000

Explanation:

Let sum=Rs.x

 

C.I. when compounded half yearly =  x 1 + 10 100 4 - x = 4641 10000

 

C.I. when compounded annually = x 20 100 2 - x = 11 25

  4641 10000 x - 11 25 x = 482

 

=> x=20000

Correct Answer: 3430

Explanation:

Let each installment be Rs.x. Then, 

(P.W. of Rs.x due 1 year hence) + (P.W of Rs.x due 2 years hence) + (P.W of Rs. X due 3 years hence) = 7620.

 

x 1 + 50 3 x 100 + x 1 + 50 3 x 100 2 + x 1 + 50 3 x 100 3 = 7620

 

  6 x 7 + 36 x 49 + 216 x 343 = 7620

 

  294 x + 252 x + 216 x = 7620 x 343

 

 => x = 3430

 

 Amount of each installment = Rs.3430

Correct Answer: I only or II and III only

Explanation:

I . P * R * 5 100 = P ? R = 20

  I I . P 1 + R 100 2 - P - P * R * 2 100 = 400 = > p R 2 = 4000000

  I I I . P * R * 1 100 = 2000 = > P R = 200000

  P R 2 P R = 4000000 200000 ? R = 20

 

Thus I only or (II and III) give answer.

 

 Correct answer is (D)

Correct Answer: Rs.8000

Explanation:

Compound Interest on P at 10% for 2 years when interest is compounded half-yearly

 

= P 1 + R 2 100 2 T - P = P 1 + 1 20 4 - P = P 21 20 4 - P

 

Simple Interest on P at 10% for 2 years =  P R T 100 = P × 10 × 2 100 = P 5

 

Given that difference between compound interest and simple interest = 124.05

  P * 21 20 4 - P - P 5 = 124 . 05

 

=> P 21 20 4 - 1 - 1 5 = 124 . 05

 P=8000

Correct Answer: 1261

Explanation:

Clearly, Rate = 5% p.a., Time = 3 years, S.I.= Rs. 1200. . .
So principal=RS [100*1200]/3*5=RS 8000
Amount = Rs. 8000 x [1 +5/100]^3 - = Rs. 9261.
.. C.I. = Rs. (9261 - 8000) = Rs. 1261.

Correct Answer: 5400

Explanation:

Interest for 1 year is the same whether it's simple interest or the compound interest.

Now interest of third year = 8575 - 7350 = 1225; means principal for this interest is 7350 if comound interest is takenIf 7350 is the principal interest = 1225

if 100 is the principal interest =(1225/7350) * 100 = 50/3 %

When a thing increases for two successive times the overall increase on initial amount = a + b + (a * b)/100

Therefore overall interest for two years = 50/3 + 50/3 + [(50/3)*(50/3)]/100 = 325/9 %

Therefore amount after 2 years = 100 + 325/9 = 1225/9

If 1225/9 is the amount principal =100

if 7350 is the amount principal =(900/1225)*7350 =5400

So sum = 5400; Rate = 50/3 %