Principal = Rs.(100 x 5400)/(12 x 3)= Rs. 15000.
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 ? 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
R=S[i/(1+i)^n-1]
Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.
Area of each tile =
Required number of tiles = = 814
We know that, After every 400 years, the same day occurs.
Thus, if 9th August 2016 is Saturday, before 400 years i.e., on 9th August 1616 has to be Saturday.
Age of the 15th student = [15 * 15 - (14 * 5 + 16 * 9)] = (225-214) = 11 years.
area of the room = 544 x 374 sq.cm
size of largest square tile = H.C.F of 544cm and 374 cm= 34 cm
area of 1 tile = 34x34 sq cm
no. of tiles required = (544 x 374) / (34 x 34) = 176
The population grew from 3600 to 4800 in 3 years. That is a growth of 1200 on 3600 during three year span.
Therefore, the rate of growth for three years has been constant.
The rate of growth during the next three years will also be the same.
Therefore, the population will grow from 4800 by = 1600
Hence, the population three years from now will be 4800 + 1600 = 6400
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