MATH1091 Assignment 1 Tadikonda PDF

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MATH 1091: Business Mathematics A-Student’s Name: Srisowmya TadikondaCalculator make and model: Normal scientific calculator by SHARPAssignment 1 Total marks: 1001 the sentences below. (5 marks) (a)Compensation for lending money that is based on one fixed sum is called: Interest(b)The formula to fin...

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MATH 1091: Business Mathematics

A-1

Student’s Name: Srisowmya Tadikonda Calculator make and model: Normal scientific calculator by SHARP

Assignment 1

Total marks: 100

1.Compl et et hes ent enc esbe l ow. ( 5mar ks ) ( a)Compens at i onf orl e ndi ngmoneyt hati sbas edononefixeds um i sc al l ed: I nt e r es t

( b)Thef or mul at ofindt hemat ur i t yval ueofas i mpl ei nt er es tl oani s : Thef or mul ai s S=P( 1+r t )

( c )At r ans f er abl epa pert hatc ommi t st hes i gne rt opayt hea mounts howni s c al l ed: Pr omi s s or yNot ei sanot et ha ti sat r ans f er abl epape rt ha tc ommi t st hes i gner t opa yt heamounts hown

( d)Toc al c ul at et hel e galduedat eofaCanadi anpr omi s s or ynot eyoumus t : Onec oul dc al c ul a t et hel e galdueda t eofaCanadi anpr omi s s or ynot eby a ddi ngt hr eeda yst ot hema t ur i t yda t e

( e)A l oanar r angementc ar r yi ngafluc t uat i ngr at eofi nt er es t , whi c hc anbe t er mi nat e datanyt i mebye i t herbor r owerorl ender , i sc al l ed: Al oana r r a nge mentc ar r yi ngafluc t ua t i ngr a t eofi nt er e s ti sc al l edde mand l oan

2. Fi ndt heexac tt i mebet weenMar c h1 7andOc t ober10.( 1mar k) Month

Total Amount of Days

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Assignment 1

March 16th

15 days

April

15 + 30 = 45 days

May

45 + 31 = 76 days

June

76 + 30 = 106 days

July

106 + 31 = 137 days

August

137 + 31 = 168 days

September

168 + 30 = 198 days

October 9th 198 + 9 = 207 days  There is a total of 207 days between March 16th and October 9th 3. Whati st hema t ur i t ydat eofa90daydebti nc ur r edonSept ember14 ?( 1mar k) Dat e

Da ys

Sept4th

26da ys

Oc t obe r

57da ys

November

87da ys

De c e mber

90da ys

3rd Sept4th–247da ys 246da ys+90da ys=337da ys 337da ys=De c ember3 r d =>The r e f or e , Dec e mber3rdi st hema t ur i t yda t e

4. Whati st hel ega ldueda t eofa1 20da ypr omi s s or ynot ei s s uedonApr i l6? ( 1mar k) April 16th = 106 days  106 + 120 = 226 days  226 + 3 = 229 days. 229 days from April 6th is August 17th. => Therefore, the legal due date would be August 17th.

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MATH 1091: Business Mathematics

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5. Cal c ul at et hes i mpl ei nt e r es tpa yabl eona6mont hl oanof$60 , 000i ft hei nt er es t r at ei s16. 5 %. ( 1mar k) P=6 0, 000

r=16. 5% =0. 1 65 16. 5*0. 010=0. 16 5

t=6/12

I=Pr t I=60,000( 0 . 165)( 6/12) I=6 0,000( 0. 0825) I=$4950 =>Thes i mpl ei nt e r e s tpa yabl ei s$ 4950.

6. Cal c ul a t et hei nt er es tpa ya bl eona60daypr omi s s or ynot ef or$ 4500i ft her at ei s 12% andt henot ewass i gnedonJ ul y2. ( 2mar ks ) P=4 500

r=12% =0. 12 12* 0. 01 0=0. 12

t=60+2=6 2 62 /365

I=450 0( 0. 12 )( 6 2/365) I=4500( 0. 0203835 6…) I=$ 91. 73 =>Thei nt e r e s tpa yabl ei s$ 91. 73

7. Whatpr i nc i palwi l le ar n$6 00i nt er es tat14% i n7mont hs ?( 2mar ks ) I=600

r=1 4% =0. 1 4 14*0. 010=0. 14

t =7 /12

I

=Pr t P=I /r t P=60 0/( 0. 14( 7/12) ) P=600/( 0. 081 666666…) P=$ 7, 346. 94 =>Ther e f or e , t hepr i nc i pali s$7 , 3 46. 94

8. How l ongwi l li tt a ke$1, 200t oea r n$120at10%?( 2mar ks ) P = 1,200

r = 10% = 0.10  10. 0.010 = 0.10

I = 120

I = Prt  t = I/Pr  t = 120 / (1,200(0.10))  t = 120 /  t = 1 x 12 = 12 months  12 months is a year (1 ½ years). So, 1 * 365 = 365 days. => Therefore, it will take a year to earn.

9. Whatr at eofi nt er es tc oul dear n$ 250onapr i nc i palof$3, 000i n300da ys ?( 2mar ks ) P = 3000

I = 250

t = 300/365

I = Prt  r = I/Pt  r = 250 / (3000(300/365))  r = 250 / (2465. 75342465753)  r = 0.101388  10.139% => Therefore, rate of interest is 10.139%

10. Whatwi l lbet heamountpa yabl eont hel ega lduedat eofa6mont hpr omi s s or y not ef or$5, 000s i gnedonAugus t22wi t hi nt e r es ta t13%?( 3mar ks )

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Assignment 1

August 22 + 6 months = February 22 August 22 = 234 days  365 – 254 = 131 days February 25th = 56 days – 1 day = 55 days + 1 day (for January 1st) = 56 days 131 days + 56 days = 185 days r = 13% = 0.13  13 * 0.010 = 0.13

P = 5000

t = 185/365

S = P(1+rt)  S = 5000(1+(0.13) (185/365))  S = 5000(1.06589)  = $5, 329.45.

therefore, S => Amount

payable is $5, 329.45.

11. Whati st hec os toffinanc i ngademandnot es i gnedonJ une6f or$2, 50 0i ft he i nt er es tr at ewas12. 5 % unt i lt hefir s tpar t paymentof$ 1750onSept ember1 4, af t e r whi c hi twentupt o13% unt i lt henot ewaspai di nf ul lonNovember3?( 4mar ks ) r = 12.5% = 0.125  12.5 *0.010 = 0.125

P = 2,500 th

t = June 6 – September 14th: 100/365 I = 2, 500(0.125) (100/365)  I = 2, 500 (0.03424658) = $85.62 September 14th partial payment deduction: $2, 500 - $1, 750 = $750 P = 750

r = 0.13

t = September 14th – November 3rd: 50/365

I = 750(0.13) (50/365)  I = 750(0.01780822) = $13.36 Total cost of financing = $85.62 + $13.36 = $98.98  Ther e f or e , t hec os toffina nc i ngf ort hi sde ma ndnot ei s$9 8. 98.

12. Us et hedec l i ni ngbal a nc emet hodt oc a l c ul at et hefinalpaymentneededonJ ul y7 t oc l earouta14. 5% de btof$5, 50 0i nc ur r edonMa r c h20i ft hef ol l owi ngpar t i al pa yment swer emade:$1, 500onMa y4and$2, 000onJ une6.( 5mar ks ) Dates March 20th – May 4th:

Work related to dates P = 5, 500

r = 14.5% = 0.145

 14.5 * 0.010 = 0.145 t = 45/365 I = 5, 500(0.145) (45/365) 

I=

5, 500(0.017876712) = $98.32 May 4th payment less interest =

$1,

500 - $98.32 = $1401.68 Unpaid balance = $5, 500 -

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$1,

MATH 1091: Business Mathematics

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401.68 = $4, 098.32 P = 4, 098.32 r =14.5% = 0.145

May 4th – June 6th:

 14.5 * 0.010 = 0.145

t=

33/365 I = 4, 098.32(0.145) (33/365) 

I=

4, 098.32(0.013109589) = $53.73 June 6th payment less interest = $2, 000 $53.73 = $1, 946.27 Unpaid balance = $4, 098.32 – 1, 946.27 = $2152.05 P = 2,152.05

June 6th – July 7th:

r = 14.5% = 0.145

 14.5 * 0.010 = 0.145 t = 31/365 I = 2, 152.05(0.145) (31/365) 

I

= 2, 152.05(0.012315068) = $26.50 Final payment = $2,152.05 + $26.50 = $2, 178.55  Ther e f or e , t hefinalpa yme nti s$2, 17 8. 55.

13. Whatwast hec os toft hel oandes c r i bedi nPr obl em 12?( 2mar ks ) I1 = 5,500(0.145) (45/365) = $98.32 I2 = $4,098.32(0.145) (33/365) = $53.73 I3 = $2,152.05(0.145) (31/365) = $26.50  Total Interest = $98.32 + $53.73 + $26.50 = $178.55

14. Api ec eofl andi sexpe c t edt ohaveaval ueof$1 25, 0 00i n4year s . I fmoneyi swor t h 12% s i mpl ei nt er es t , whati saf ai rpr i c et opa yf ort hel a ndt oday?I gnor el oc al t ax es . ( 3mar ks ) S = 125,000

r = 12% = 0.12  12 * 0.010 = 0.12

t=4

= P(1+rt)  P = S/(1+rt) P = 125,000(1 + (0.12)(4))  P = 125,000/(1 + (0.48))  P = 125,000/(1.48) = $84, 459 => Therefore, the fair price for the land today is $84, 46

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S

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Assignment 1

15. J ohnl entaf r i end$5000at10% a ndr ec ei ved$500i ni nt er es t . How l ongdi d J ohn’ sf r i endha veus eoft hemoney?( 3mar ks ) P = 5000

I = 500

r = 10% = 0.10  10 * 0.010 = 0.10

I = Prt  t = I/Pr t = 500 / (5000(0.10)  500 / 500 = 1 x 365 = 365 days => Therefore, John’s friend has use of that money for a year.

16. Af t er2year s ,your ec ei veanofferof$1 10, 00 0f ort hepr ope r t ypur c has edi n Pr obl em 14. Woul dyouac c eptt heoffer ?As s umet hatyouhavepai d$500ayearf or l andt ax esf oreac hyearyouha vehel dt hepr oper t y. Forf ul lmar ks , makes ur eyou s how yourwor k.( 4mar ks ) Property tax = 2 * 500 = 1000

P = $84, 459 + 1000 = 85, 459

r = 0.12

t=2

S = P(1+rt)  S = 85, 459 (1 + (0.12) (2))  85, 459(1.24) = $105, 969. 16  Therefore, the amount paid for everything is $105, 969. 16 and I am being offered $110, 00. So, I would accept the offer because it is over the amount I have paid… it would break even. However, I would also wait for other offers because I may get a better offer which will give me a bigger profit. Overall, I would accept the offer because I am getting a slight profit any which way.

17. Cal c ul at et hepr es entval ueofade btt hatwi l lamountt o$151 8. 33i n4mont hs ’ t i me ,i fs i mpl ei nt er es ti sc ha r ge dat12%.( 2mar ks ) S =1518.33 r = 12% = 0.12  12 * 0.010 = 0.12

t = 4/12

S = P(1+rt)  P = S/(1+rt) P = 1518.33/(1 + (0.12)(4/12))  P = 1518.33 /(1.04) = $1459.93 18. I fmone yi swor t h1 5%, whati st heval ueonJ une14ofanoni nt er e s t –bear i ng pr omi s s or ynot eofwhi c ht hef ac eval uei s$10, 00 0andwhi c hma t ur e son Dec embe r12?( 3mar ks ) S = 10, 000

r = 15% = 0.15  15 * 0.010 = 0.15

t = December 12th – June 14th = 181/365

P = S/(1+rt)  P = 10, 000/(1 + (0.15)(181/365))  P = 10,000 /(1.074383562) = $9,307.66

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MATH 1091: Business Mathematics

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19. Whatwoul dbet heval ueofa$14 , 000de btaf t e r6mont hsi fmoneyi swor t h17%? ( 2mar ks ) S = 14,000

r = 17% = 0.17  17 * 0.010 = 0.17

t = 6/12

P = S/(1 + rt)  P = 14,000/(1 + (0.17)(6/12))  P = 14,000 /(1.085) = $12,903. 23 20. I fmone yi swor t h1 6%, c al c ul at et hes i ngl eequi val entpayment9mont hsf r om now oft wodebt s :$30 0due2mont hsagoand$ 500duei n1yea r . Us e9mont hs f r om now ast hef oc aldat e. ( 4mar ks ) P1 = 300 2

P = 500

r = 16% = 0.16  16 * 0.010 = 0.16

t = 9 + 2 = 11/12

r = 16% = 0.16  16 * 0.010 = 0.16

t = 12 – 9 = 3/12 Discount

SP = [P(1+rt)] + [P/(1+rt)]  [300(1 + (0.16) (11/12)] + [500/ (1 + (0.16) (3/12)] SP = [300(1.1467)] + [500/ (1.04)]  344.01 + 480.77 = $824.78  Therefore, the single equivalent payment in 9 months from now is $824.78.

21. Fi ndt hes i zeoft woequalpayment s —oneduenow andt heot heri n8mont hs — t hatar et or epl ac et hr eeor i gi nalde bt sof$2000eac hdue6, 4,and2mont hsa go. Us ear at ef ormoneyof12% andnow ast hef oc aldat e.( 5mar ks ) P1 = 2000

r = 12% = 0.12  12 * 0.010 = 0.12

P2 = 2000

r = 0.12

t = 4/12

P3 = 2000

r = 0.12

t = 6/12

New Debt2 = x

t = 2/12

New Debt1 = x

r = 0.12

t = 8/12

[2000(1+(0.12) (2/12))] + [2000(1+(0.12) (4/12))] + [2000(1+(0.12) (6/12))] = x + [x/(1+(0.12)(8/12))] [2000(1.02)] + [2000(1.04)] + [2000(1.06)] = (1)x + [(1)x/ (1.08)] 2040 + 2080 + 2120 = 1x + 0.92592593  6, 240 = 1.92592593x  6, 240/1.92592593 = $3, 239.99 => Therefore, $3, 239.99 is the size of two equal payments.

22. Twodebt s —$500due1mont ha goand$700duei n3mont hs —ar et obe r epl ac edbyt hr eenew payment s: oneof$20 0duenow, oneof$4 00duei n4 mont hs , andt her ema i nderi n8mont hs ’t i me. Fi ndt hes i z eoft hefinal pa ymenti fmoneyi swor t h1 4%. Us enow ast hef oc aldat e.( 5mar ks )

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Assignment 1

OD 1= 500

r = 14% = 0.14  14 * 0.010 = 0.14

OD 2= 700

r = 0.14

t = 3/12  ND 2 = 400 ND 3 = x

r = 0.14

t = 1/12  ND 1 = 200 r = 0.14

t = 4/12

t = 8/12

[500(1+(0.14) (1/12))] + [700/ (1+(0.14) (3/12))] = 200 + [400/ (1+(0.14) (4/12))] + [x/ (1+(0.14) (8/12))] [500(1.011666667)] + [700/ (1.035)] = 200 + [400/ (1.046666667)] + [x/ (1.093333333)] 505.83 + 676.33 = 200 + 382.17 + [(1)x/ (1.09333)]  1182.16 = 582.17 + 0.91463415x 1182.16 – 582.17 = 0.91463415x  599.99/0.91463415 = $655.99 => Therefore, the final payment is $655.99.

23. Fi nd( a)t hepr oc eedsand( b)t hes i mpl edi s c ountofa3mont h,noni nt er es t bear i ngnot es i gne donJ ul y10wi t haf ac eva l ueof$15 , 000i fdi s c ount i ngoc c ur sat 13% s i mpl ei nt er es tonAugus t24.( 4mar ks ) Legal due date = October 13th a) S = 15, 000

r = 0.13

t = October 13th – August 24th = 50/365

P = 15, 000/ (1 + (0.13) (50/365)  P = 15, 000/ (1.01780822) = $14, 737.55  Therefore, the proceeds will be $14, 737.55 if paid on August 24th. b) Simple discount = 15, 000 – 14, 737.55 = $262.45 => Therefore, the simple discount

24. Fi ndt hepr oc eedsf r om a$8, 0 00not et hatma t ur esonMar c h5i fi ti sdi s c ount edon J anua r y2 0a t13%. As s umenogr a c e.( 3mar ks ) S = 8, 000

r = 13% = 0.13  13 * 0.010 = 0.13

t = January 20th – March 5th = 44/365

P = 8, 000/ (1 + (0.13) (44/365)  P = 8, 000/ 1.015671233 = $7, 876.56  Therefore, the proceeds will be $7, 876.56 if paid on Jan 20th. 25. Cal c ul at et hes i mpl edi s c ountona12%, 3mont hpr omi s s or ynot ef or$1 000t hat wass i gnedonSept ember6anddi s c ount e dat13% s i mpl ei nt e r es tonOc t ober10. ( 4mar ks )

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MATH 1091: Business Mathematics

Maturity value:

P = 1000

A-9

r = 0.12

t = September 6th – December 9th = 94/365

S = 1000(1 + (0.12) (94/365))  S = 1000(1.03090411) = $1, 030. 90 Proceeds:

S = 1, 030. 90

r = 0.13

t = October 10th – December 9th = 60/365

P = 1, 030. 90/ (1 + (0.13) (60/365))  P = 1, 030. 90/ (1.02136986) = $1, 009. 33 Simple discount = 1, 030. 90 – 1, 009.33 = $21.57 => Therefore, simple discount is $21.57. 26. I fa1 20da y , $ 15, 000not eear ni ng14% s i mpl ei nt er es ti sdi s c ount edat17%,60da ys bef or ei t sl egalduedat e, whatar et hepr oc ee ds ?( 4mar ks ) Maturity Value: P = 15,000 r = 14% = 0.14  14 * 0.010 = 0.14 t = 120 + 3 days = 123/365 S = 15, 000(1 + (0.14) (123/365))  S = 15, 000(1.04717808) = $15, 707.67 Proceeds: S = 15, 707.67

r = 17% = 0.17  17 * 0.010 = 0.17

t = 60/465

P = S(1+(rt))

P = 15, 707.67/ (1 + (0.17) (60/365))  P = 15, 707.67 / (1.02794521) = $15, 280.65  Therefore, the proceeds are $15, 280.65. 27. Suppos et hatyoudec i det odi s c ounta$14 0, 000not eyouar ehol di ngbys el l i ngi t t oabuye ronMa r c h1 5,whi c hwoul dbe4mont hsbef or ei ti sdue . I ft hedi s c ount r at ei s17. 5 %, wha twi l lt hepr oc eedsoft hes al ebe ?( 3mar ks ) Legal due date = July 18th because it is 4 months from March 15 with 3 days grace period. S = 140, 000 r = 17.5% = 0.175  17.5 * 0.010 = 0.175 t = July 18th – Mar 15th = 125/365 P = 140, 000/ (1 + (0.175) (125/365))  P = 140, 000/ (1.05993151) = $132, 084.00  Therefore, the proceeds of the sale will be $132, 084.00. 28. Me t c hos i nTr ac t orpur c has e dapat entonapi ec eofequi pmentf or $ 1. 5mi l l i onandagr eedt opa yt hef ul lamounti n15year s ,pl usi nt er e s t . Af t er8year s ,Met c hos i nTr ac t ori sdoi ngs owel lt ha ti tde c i dest or epa yt he f ul lpat entc os t .I fmoneyi swor t h6% s i mpl ei nt er es t , how muc hs houl d Me t c hos i nTr ac t oroffert hec ompanyi tpur c has edt hepat entf r om t os et t l e t heout s t andi ngde bt ?( 4mar ks ) Maturity Value:

P = 1,500,000

r = 6% = 0.06  6 * 0.010 = 0.06

years S = 1,500,000(1 + (0.06) (15))  S = 1,500,000(1.9) = $2,850,000 Remaining Debt:

S = 2,850,000

r = 0.06

t = 15 – 8 = 7 years

P = 2,850,000/ (1 + (0.06) (7))  P = 2,850,000/ (1.42) = $1,912,751.68

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t = 15

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Assignment 1

 Therefore, the settlement of the remaining debt should be $1,912,751.68 29. Youar et hi nki ngaboutbuyi nganew c ar . Thes al e sas s oc i a t et el l syout ha tyouc an dr i veawa ywi t ht hec art oda yi fyoupay$1 3, 000now and$15 , 000i n2year s . I fyou know t hatt hec ompanyc har ges26% i nt e r es tonout s t andi ngbal anc es , what s houl dt hec as hpr i c ebef ort hec art oda y?( 5mar ks ) P 1 = 13, 000

P 2 = 15, 000

r = 26% = 0.26  26 * 0.010 = 0.26

t=2

New debt = 13, 000 + [15, 000/ (1 + (0.26) (2))]  13, 000 + [15, 000/ (1.52)]  13, 000 + 9, 868.42 = 22, 868.42  Therefore, the new debt is $22, 868.42

30. Thec a rs a l esas s oc i at ei nPr obl em 29s ayst hatt her ei sas pec i alont oday. I fyou pa yc as h,y ouc anbuyt hec a rf or$2 4, 000,as avi ngsc l ai medt obeof$4, 000 .Woul d youac c eptt heoffer ?Whyorwhynot ?( 6mar ks ) No, I would not accept the offer because it is not $4000 saving… New debt = 13, 000 + [15, 000/ (1 + (0.26) (2))]  13, 000 + [15, 000/ (1.52)]  13, 000 + 9, 868.42 = 22, 868.42. So, 24, 000 – 4000 should be 20, 000 but 24, 000 - 22, 868. 42 = 1, 131.58. So, I would be paying about $1, 131.58 extra which is not beneficial for me hence why I would not accept the offer given by the sales associate.

31. Ac oupl ewant st os etas i dee noughmoneyf ort he i rdaught ert oa t t end c ol l ege. Thec os tofc ol l egei sexpec t edt obe$ 20, 000ayear . How muc hmoney woul dneedt obepl a c edi na nac c ountt oda yt hatpays10% s i mpl ei nt er e s t f ort hei nt er es tf r om t heac c ountt obeenought opa yt heannualc os tof a t t endi ngc ol l ege ?( 5mar ks ) I = 12, 000

r = 10% = 0.10  10 * 0.010 = 0.10

t=1

12, 000 = P (0.10) (1)  12, 000 = 0.10P  12, 000/0.10 = $120, 000 P = $120, 000  Therefore, the couple should save $120, 000 annually in order to save enough and pay annual cost for their child to go to college

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