MAT1304, Finite Mathematics

Number:

MAT1304, Finite Mathematics

383.PV=$100, n=10, interest rate =7%

FV=PV[1+r]^{n}

=100*[1+0.07]^{10}

=$196.70

384.A= P [1+r/100]^{n}

A= 2000 [1+2.4/12/100]^{12}

=$2048.53

After5 payments of 168.84, the total amount paid =168. 84*5= $844.20

Thebalance remaining = $2048.53 – $844.20

=$1204.33

385.P = 59000 Annual interest = 621.27, rate = 12%

=59000 (1+12/100)^{1}

=66080/12 = 5507

=5507- 621.27

=4885.73

386

387.1500 now vs 2000 compounded quatery at a rate of 6.5% for 5 years

2000= P (1+0.01625)^{20}

P= 2000/1.38042

=1448.83. This implies that, the 1500 now is the better offer compredto 2000 in 5yrs

388. = 117,000 (1+6.5/100)^{15}

^{ }=117000*2.5718

=

389.A= P[1+r]^{n}

=$ 40000*(1+ 4/100)^{8}

=$ 54,722.88

390

391

392.28 years, 260000 on retirement, 6% semia annually

A=P [1+r/100]^{n}

^{ }260000= P [1+ (6/2)/100]^{28*2}

P= 260000/5.2346

=49669.51

393.D

394.A

395

396.

397. S_{5}=(6 [1-3^{5}])/ [1-3] = 726 (C)

399. a = 1856, r = ¼, find the 4^{th}term? a_{n}= a_{1}*r^{(n-1)}

a_{4th}= 1856* ¼ ^{(4-1)}

=1856/64

=29

400. n =2, A = 15000, r = 3%….quatery

15000= P* 1.0075^{8}

P= 15000/1.061599

=$ 14129.63

401.

402.$5100 at 4.6% monthly to reach 13477.83

13477.83=5100[1+4.6/100]^{n}

1.046^{n}=13477.86/5100 = 2.642712

nLog1.046 = Log 2.642712

n= Log 2.642712 / Log 1.046

n= 21yrs, 8 months

403.A= P [1+r/100]^{n}

^{ }=21200 = P [1+7/100]^{7}

P= 21200/ [1.07^{7}]

=13202.30

404.

405.6000 at 4% compounded quartery for 4yrs

A= 6000 [1+4/4/100]^{4*4}

=6000 * 1.1726

=7035.6

Interest= 7035.6 – 6000

=1035.6

406.P = 1200, r = 5%, t = 4yrs (quartery = *4)

A= 1200 [1+ 5/4/100] ^{4*4}

=1200 * 1.0125^{16}

=1463.88

407.

408.A= 3900(1 + (0.04 × 0.74)) = 4015.44

A= $ 4,015.44

References

Lial,M. L., Greenwell, R. N., & Ritchey, N. P. (2012). Finitemathematics (10th ed.). Boston, MA: Pearson Education.