Summative Ex 2 Repaired)

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    Summative Exercise 2 (Project 2)Question 1

    RM500 was deposited every month for 10 years at an interest rate of 6% compounded

    monthly.

    a)

    Find the amount at the end of 10 years and the interest earned.b) Find the interest rate compounded annually that is equivalent to 6% compoundedmonthly as given.

    (a.)

    #

    (b.) 1+ r = 1+ r = - 1

    =0.06678

    =6.168% #

    Question 2Maimunah received RM7, 000 from her father. She invest the money in bank that offers 5.0%

    interest compound semi-annually. The accumulated amount at the end of 4 years will be

    enough to pay for the down payment of a new car that cost RM60,000.

    a) What is the value of the down payment?b) To pay the balance, she plans to obtain a loan from a finance company and agree

    to pay equal monthly instalments for seven years. The finance company charges

    interest at 4.4% compounded monthly. Find the monthly instalment.

    (a.)

    P=RM7000

    i= n=42 #

    (b.)

    Balance = RM 60,000 RM 8,528.82

    = RM 51471.18

    A=RM51,471.18

    i=

    n=7x12=84

    continue...... at the back

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    51471.18=R(72.1828894)

    R=RM713.07 #

    Question 3

    1. Rasid bought a house for RM185, 000. He paid a 10% down payment and the balancewas borrowed from a bank that charged the interest at a rate of 8% compounded

    monthly. The loan is to be repaid in monthly instalments for 10 years. Find

    a) The monthly payment,b) The total interest charged.

    (a.)

    Down payment 10%

    =RM185, 000x10%

    =RM18, 500

    Balance = RM 185, 000- RM18, 500

    =RM166, 500

    A=RM166, 500

    i=

    n=10x12

    R=RM 2020.10 #

    (b.)

    Each month =RM 2,020.10

    Interest charge ,

    =(120x RM 2020.10) RM166, 500

    =RM 75,912.00 #

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    Question 4

    Jasper deposited Rm456 at the end of every 6 months for 8 years in a saving account that paid

    interest at a rate of 6.8% compounded semi-annually. Find

    a) The amount in the account just after her last deposit,b) The total interest earned.

    (a.)

    R=RM456 I= = 3.4%

    n=8 x 2= 16

    =RM 9487.13 #

    (b) = RM 9487.13-(16xRM456)

    = RM 2191.13 #

    Question 5

    Noraini borrowed RM80, 000 from a bank for 5 years at 7% compounded monthly.

    a) Find the monthly paymentb) If she has not paid her first four monthly payments, how much should she paid on

    the fifth payment to settle all the outstanding arrears?

    c) Immediately after paying the 19th monthly payment, she wanted to settle thebalance of the debt with a single payment. How much should this payment be?

    (a)

    A=RM 80 000 I=

    n=5x12=60

    R = RM 1584.10 #

    (b)

    Outstanding arrears

    S= = =RM 8013.45 #

    (c)Outstanding loan A = =RM 5, 7616.64 #

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    Question 6

    For a loan of RM 36000 the monthly payment is RM 872.50 for 4 years.if the loan is to be

    settle immediately after the 28th

    payment calculate the outstanding principle balance (OPB )

    by the rule of 78.

    R = RM 872.50

    n = 4

    Outstanding balance = ( 872.50 x 48 ) 36,000

    = RM 41,880 RM 36,000

    = RM 5,880

    Question 7

    Ahmad took a loan of RM 40,000 at 6.5% compounded monthly and had to pay equal

    monthly payment for three years. Calculate

    a) The monthly paymentb) The total interest

    (a) A = RM 40,000 = 6.5%n = 3 yearsA=R

    40000 = R = RM 1,225.96 #

    (b)Total interest

    I = S nR

    I = ( 1225.96 x 36 ) 40,000

    = 44,134.56 40,000

    = RM 4,134.56 #

    Question 8

    RM 150 was invested at the end of every month for 40 month in an account that paid 5%

    interest compounded monthly. After 40 month no more payment was made. Find the amount

    in the account at the end of 50 month.

    R = RM 150 n 1 = 40 months n 2 = 10 months i = 5% compounded monthly

    S=R S = P ( 1 + i ) nS = 6,514.25 ( 1 +

    )10

    = 150 = RM 6,790.82 #= RM 6,514.25

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    Question 9

    Mustafa deposited RM R every month into an account that give an interest of 8%

    compounded monthly for 2 years. Immediately after 2 years he withdrew all the money which

    amounted to RM 40000 from the account. Find the monthly deposit R.

    R = R n = 2 years i = 8% compounded monthly S = RM 40000

    S=R 40,000 = R 40,000 = R ( 25.93 )

    R = RM 1,542.43 #

    Question 10

    Indahsah save RM 200 at the end of each month into an account that give an interest of 12%

    compounded monthly for 10 years.a) Find the amount save in the account at the end of 10 yearsb) At the end of 10 years she made monthly withdrawals for the next 5 yeras.how much

    was the monthly withdrawal?(a)

    R = RM200 i = 12% compounded monthly

    n = 10 years

    S=R = 200 = RM 46,007.74 #

    (b)

    A=R 46 007.74 = R 46 007.74 = R ( 44.96 )

    R = RM 1,023.42 #

    Question 11

    Find the total interest earned if RM900 is deposited every three months for five years at an

    interest rate of 6% compounded quarterly.

    #

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    Question 12

    The cash price of a bus is RM 120 000. MMM Company bought the bus through an

    instalment plan of seven years and the interest rate charged was 8% compounded monthly.

    a) Find the monthly paymentb) After paying for 12 months, the company decided to settle all the outstanding balance.

    Find the amount the company had to pay on the 13 th payment and the total interest

    charged.

    (a)

    70.35 #

    (b)

    TOTAL INTEREST #

    Question 13

    Megat intends to buy an annuity scheme that offers an interest rate of 6% compounded

    monthly. He expects to receive RM 500 every month for the next five years starting a month

    from the date of purchase. How much would Megat have to pay for the scheme?

    #

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    Question 14

    Find the amount to be invested quarterly at 10% compounded every three months to

    accumulate RM 60000 in three years.

    Question 15A loan of RM 12000 was to be settled by making monthly payments for 4 years. If the interest

    rate was 9% compounded monthly, calculate the monthly payments.

    Question 16

    A retailer bought 100 pairs og shoes at RM 25 per pair. He wanted to have 15% net profit

    based on cost. The operating expenses were 8% based on cost. Find the selling price of each

    pair shoes and the total gross profit for the 100 pairs shoes

    #Gross profit =

    = =#

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    Question 17

    A retailer bought table for RM 500. Operating expenses incurred in selling the table were

    25% net profit based on cost. If the retailer wanted to make 12% net profit based on the cost

    a) The retail price,b) The gross profit,c) The net profit,d) The breakeven price,

    The maximum mark down per cent that can be offered without making any lost.if the table

    was sold at RM 600, find the net profit/loss.

    (a) Retail price

    #

    (b) Gross profit Gross profit =

    = =

    #

    (c) Net profit #

    (d) Breakeven price

    (e)

    Question 18

    The list price of an items is RM 1200 and the trade discounts given was 10%, 5% and 3%. Find

    a) The amount of discountb)A single discount that is equivalent to the chain discount rate

    a) NP=L-(1-r1)(1-r2)(1-r3)=1200(1-0.1)(1-0.05)(1-0.03)

    = RM 995.22

    Amount discount

    =1200-995.22

    =RM 995.22 #

    b) r=1-(1-r1)(1-r2)(1-r3)=1-(1-10%)(1-5%)(1-3%)

    =17.065%

    Markdown percent

    Proft and lost

    #

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    Question 19

    An invoice dated 11 May 2004 for RM 2650 inclusive of RM300 transportation charge was paid on 31

    may 2004. If the trade discount was 20% and the cash discount terms were 5/10, 4/30, n/60, find

    a) The trade discount offered

    b) The cash discount if entitled

    c) The amount paid

    a) TD=LP-NP=2350-1880

    = RM 470 #

    b) Cash discount=4%(2350)

    =75.2

    c) The amount paid =L(1-r1)(1-c)=2350(1-20%)(1-4%)+300

    =1804.8 + 300

    = RM 2104.8 #

    Questions 20A pen costing RM290 was marked up 30% based on the retail price. Find the retail price of the pen

    .

    R= ? C= RM 290 M=0.30(R)

    R= C+M

    R=290+0.3R

    R-0.3R=290

    0.7R=290

    R = RM 414.28

    Question 21

    A retailer bought an item for RM 700.The operating expenses were RM 100 and the gross

    profit was 50% on cost.a) Find the net profit and the retail price

    b) If the retailer decided to sell the item at the breakeven price find the markdownpercent

    a) Gross profit = NP + OE50% ( 700 ) = NP + 100

    350 100 = NP

    NP = RM 250

    Retail price = C + NP + OE

    = 700 + 250 + 100

    = RM 1,050 #

    b) Breakeven price = C + OE= 700 + 100= 800

    RP BEP

    = 1050 800= 250

    Markdown = x 100

    = 23.81%

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    Question 22

    On 23 January 2004 a retailer purchased some items at RM 4 400 and the transportation cost

    was RM 125.He was offered a trade discount of 25% and cash discount terms of

    5/10,2/30,n/60.Calculate the payment if he paid the invoice on 18 February 2004.

    23 January 2004 + 18 February 2004 = 26 days

    Cash discount = 2%

    LP = 4 400 125= 4 275

    TD = 25%

    Amount paid = L ( 1- r1 )(1- c) + 125

    = 4275 ( 1- 25%)(1-2%)+125

    = 3142.125 + 125

    = RM 3 267.125

    Question 23

    A retailer bought a set of bedroom furniture at RM 1500.The operating expenses were 20%

    based on cost price. The dealer wished to obtain 45% net profit based on cost.

    a) Find the seiling priceb) Find the breakeven pricec) Find the maximum markdown per cent that could be offered without incurring any

    loss if he set the seiling price at RM 2000, what was the net profit.

    a) R = C + NP + OE= 1500 + 45%( 1500) + 20%( 1500)= 1500 + 675 + 300

    = RM 2,475

    b) BEP = C + OE= 1500 + 20%(1500)= RM 1,800

    c) Maximum markdown per cent= R BEP

    = 2475 1800

    = 675

    = x 100

    = 27.27%

    d) NP = R BEP= 2000 1800

    = RM 200

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    Question 24

    A stall sells nasi lemakfor R 1.50 per pack at a mark-up of 30% based on cost. If the stall can

    sell 450 packs a day with operating expenses of RM 20 daily find

    a) The cost per packb) The breakeven price per packc) The net profit per daya) R = C + M

    1.50 = C + 30%C

    1.50 = 130%C

    C = RM 1.15 #

    b) BEP = C + OE= 1.15 + (

    )= RM 1.19 #

    c) NP = RP BEP= 1.50(450) 1.19(450)= RM 139.50 #

    Question 25

    A Manufacturer produces water filters at a cost of RM600 per unit.He sells the water filter at

    40% markup on cost. Encik Farid is a dealer.He purchased four units of water filters from the

    manufacturer. He sells the filters with net profit of 55% on cost and the operating expensesamount to 10% of the cost.

    a) What is the purchase price of the filter sold to farid?b) What is the selling price of the filter at Farids Shop?c) Find the total net profit on selling the four filters?

    (a)

    R= ? C=RM 600 M=40%(600)

    R= C+M

    R= 600+40%(600)

    R= RM 840 #

    (b)

    R= selling price C=RM 840NP=55%(840)

    OE=10%(840)

    R=C+NP+OE

    R=840+55%(840)+10%(840)

    R=RM 1386 #

    (c)

    NP=? RP= RM 5544 BEP= 3360+10%(3380)

    NP=RP-BEP

    NP=5544-3360+10%(3380)

    NP=RM 1,848 #

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    Question 26

    A dealer bought a computer for RM 3,600. Operating expenses were estimated to be 5% of

    the retail price. If the dealer wanted a 15% net profit based on the retail price, find

    a) The retail priceb) The net profitc) The mark-up

    (a)

    R= C+NP+OE

    R= 3,600+15%(R)+5%(R)

    3,600 = 1R-0.15R+0.05R

    3,600 = 0.8R

    R=

    R= RM4,500 #

    (b)

    NP = RP-BE

    = 4,500- ( 3,600+5%(4,500))

    = RM 675 #

    (C)

    4,500(15%) + 4,500(5%)

    = RM 900

    Question 27

    A trader bought a set of ornament that was listed at RM5,000 with trade cash of 20% and

    10%. If he sold the at the net profit of 30% based on cost and the operating expenses were

    5% on cost, find

    a) The gross profitb) The seiling pricec) The breakeven priced) The maximum markdown that could be given without incurring any loss.

    a)

    = L (1-r) (1-r)= 5000(1-20%) (1-10%)

    = RM 3,600

    GP=NP+OE= 3,600(30%) + 3,600(5%)

    = RM 1260 #

    b)R= C+NP+OE

    R= 3,600+3,600 (30%)+3,600(5%)R= RM 4,860 #

    (d)

    RM 4,860 - 3,780

    = RM 1,080

    (c)

    BE= C + OE

    BE= 3,600 + 3,600(5%)

    BE= RM 3,780