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Q:

A chartered bank offers a five-year Escalator Guaranteed Investment Certificate.In successive years it pays annual interest rates of 4%, 4.5%, 5%, 5.5%, and 6%, respectively, compounded at the end of each year. The bank also offers regular five-year GICs paying a fixed rate of 5% compounded annually. Calculate and compare the maturity values of $1000 invested in each type of GIC. (Note that 5% is the average of the five successive one-year rates paid on the Escalator GIC.)

A) 1276.28	B) 1234
C) 1278	D) 1256

Answer & Explanation Answer: A) 1276.28

Explanation:

FV = $1000(1.04)(1.045)(1.05)(1.055)(1.06) = $1276.14

the maturity value of the regular GIC is

FV = $ 1000 x ${(1.05)}^{5}$ = $1276.28

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Q:

Kramer borrowed $4000 from George at an interest rate of 7% compounded semiannually. The loan is to be repaid by three payments. The first payment, $1000, is due two years after the date of the loan. The second and third payments are due three and five years, respectively, after the initial loan. Calculate the amounts of the second and third payments if the second payment is to be twice the size of the third payment.

A) 1389	B) 1359
C) 1379	D) 1339.33

Answer & Explanation Answer: D) 1339.33

Explanation:

Given:j=7% compounded semiannually making m=2 and i = j/m= 7%/2 = 3.5%
Let x represent the third payment. Then the second payment must be 2x.
PV1,PV2, andPV3 represent the present values of the first, second, and third payments.

Since the sum of the present values of all payments equals the original loan, then
PV1 + PV2 +PV3 =$4000 -------(1)

PV1 =FV/(1 + i)^n =$1000/(1.035)^4= $871.44

At first, we may be stumped as to how to proceed for
PV2 and PV3. Let’s think about the third payment of x dollars. We can compute the present value of just $1 from the x dollars

pv=1/(1.035)^10=0.7089188

PV2 =2x * 0.7089188 = 1.6270013x
PV3 =x * 0.7089188=0.7089188x
Now substitute these values into equation ➀ and solve for x.
$871.442 + 1.6270013x + 0.7089188x =$4000

2.3359201x =$3128.558

x=$1339.326
Kramer’s second payment will be 2($1339.326) =$2678.65, and the third payment will be $1339.33

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Q:

Other things being equal, would an investor prefer an interest rate of 10.5% compounded monthly or 11% compounded annually for a two-year investment?

A) 1232	B) 1243
C) 1254	D) 1262

Answer & Explanation Answer: A) 1232

Explanation:

i=j/m

FV= PV(1+ i)^n

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Q:

What annual payment will discharge a debt of Rs.7620 due in 3years at $16 \frac{2}{3}$ % per annum interest?

A) 5430	B) 4430
C) 3430	D) 2430

Answer & Explanation Answer: C) 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.

$[\frac{x}{(1 + \frac{50}{3 x 100})}] + [\frac{x}{1 + {(\frac{50}{3 x 100})}^{2}}] + [\frac{x}{1 + {(\frac{50}{3 x 100})}^{3}}] = 7620$

$\frac{6 x}{7} + \frac{36 x}{49} + \frac{216 x}{343} = 7620$

$294 x + 252 x + 216 x = 7620 x 343$

=> x = 3430

Amount of each installment = Rs.3430

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Q:

Divide Rs. 1301 between A and B, so that the amount of A after 7 years is equal to the amount of B after 9 years, the interest being compounded at 4% per annum.

A) Rs.625	B) Rs.626
C) Rs.286	D) Rs.627

Answer & Explanation Answer: A) Rs.625

Explanation:

Let the two parts be Rs. x and Rs. (1301 - x).

$x {(1 + \frac{4}{100})}^{7} = (1301 - x) {(1 + \frac{4}{100})}^{9}$

$\frac{x}{(1301 - x)} = {(1 + \frac{4}{100})}^{2} = (\frac{26}{25} * \frac{25}{26})$

=> 625x=676(1301-x)

1301x=676 x 1301x=676.
So,the parts are rs.676 and rs.(1301-676)i.e rs.676 and rs.625

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Q:

The difference between the compound interest and simple interest on a certain sum at 10% per annum for 2 years is Rs. 631. Find the sum.

A) 60100	B) 61100
C) 62100	D) 63100

Answer & Explanation Answer: D) 63100

Explanation:

Let the sum be Rs. x. Then,
C.I. = x ( 1 + ( 10 /100 ))^2 - x = 21x / 100 ,
S.I. = (( x * 10 * 2) / 100) = x / 5
(C.I) - (S.I) = ((21x / 100 ) - (x / 5 )) = x / 100
( x / 100 ) = 632 * x = 63100.
Hence, the sum is Rs.63,100.

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Q:

In what time will Rs. 1000 become Rs. 1331 at 10% per annum compounded annually?

A) 1years	B) 2years
C) 3years	D) 4years

Answer & Explanation Answer: C) 3years

Explanation:

Principal = Rs. 1000; Amount = Rs. 1331; Rate = 10% p.a. Let the time be n years. Then,
[ 1000 (1+ (10/100))^n ] = 1331 or (11/10)^n = (1331/1000) = (11/10)^3
n = 3 years

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Q:

If the simple interest on a sum of money at 5% per annum for 3 years is Rs. 1200, find the compound interest on the same sum for the same period at the same rate.

A) 1261	B) 1271
C) 1281	D) 1291

Answer & Explanation Answer: A) 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.

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