Question #8994e

3 Answers
Jan 13, 2018

Answer:

#=K68.25#
#=K3.25#

Explanation:

This can be solved using the formula:
#I="Pi"n#
where:
#I="Interest"#
#P="Principal"#
#i="interest rate"#
#n="no. of years"#

Plug in data provided to find the Interest #(I)# after after 3 years .
#I="Pin"#
#I=K650xx(0.035)/(cancel(yr))xx3cancel(yr)#
#I=K68.25#

Thus, the interest #(I)# per year is #K22.75#

If the interest rate is increased by 3.75% after the first year; the Interest #(I)# earned for the remaining 2 years can be computed as;
#I="Pin"#
#I=K650xx(0.0375)/(cancel(yr))xx2cancel(yr)#
#I=K48.75#

Thus, the total Interest earned before and after the increase can be computed as shown;
#I_(t)=I_(fy)+I_(s&ty)#
#I_(t)=K22.75+K48.75#
#I_(t)=K71.50#

Therefore, an increase of the Interest earned is
#=I_(3.50&3.75%)-I_(3.50%)#
#=K71.50-L68.25#
#=K3.25#

Jan 13, 2018

Answer:

(a) Interest earned after 3 years @ 3.5% Interest = #color(purple)(K 68.25)#

(b) Increase in Interest due to rise in interest to 3.75% after one year = #color(green)(71.5 - 68.25 = K 3.25)#

Explanation:

#I = (P N R) / 100#

Where I = Simple Interest earned, P = Principal, N = No. of years & R = Rate of Interest in percentage.

(a) Given #P = K 560, N = 3 years, R = 3.5%#

#I = ( 650 * 3 * 3.5) / 100 =color(red)( K 68.25)#

(b) New Rate of Interest #R_1 = 3.75%# & #N_1 = 2 years#

Interest earned #I_1 = (650 * 2 * 3.75) / 100 = color(blue)(K 48.75)#

Old rate of interest #R = 3.5%, Period N_1 = 1 year#

#I_o = (650 * 3.5 * 1) / 100 = color (blue)(22.75)#

Total Interest = 48.75 + 22.75 = K 71.5#

Increase in interest with the revised interest rate after 1 year #color(green)(I_i = 71.5 - 68.25 = K 3.25)#

Jan 13, 2018

Answer:

#"Part a: "K68.25#

#"Part b: "K3.25#

Explanation:

#color(blue)("Preamble")#

The symbol % cab be thought of as representing a fraction but one that is worth #1/100#. So we have:

#3.5% ->3.5xx% ->3.5xx1/100=3.5/100 #

A direct comparison is the the metric measure of centimetres. The part word centi signifies #1/100#. So we have #1/100xx1" metre"#

It is usual to translate the word of (as in #1/100# of 1 metre) to mean multiply. So #3.5%# of 'something' is stating #3.5%xx"'something'"#

The wording 'per anum'#-> #'per' means for each of 1 and 'anum' means 'year'. So 'peranum' is stating 'for 1 year'
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Answering the question")#

#color(brown)(ul("For EACH year we have:"))" simple interest"#

#3.5%" of "K650 color(white)("d") -> color(white)("d") 3.5/100xxK650 = color(cadetblue)(K22.75) color(red)(larr"Checked")#

#3.75%" of "K650 color(white)("d") ->color(white)("d") 3.75/100xxK650= color(sienna)(K24.375) color(red)(larr"Checked")#

..............................................................
#color(brown)("Question part a")" 3 years at 3.5%"#

#"3 years at 3.5% is: "3xxcolor(cadetblue)(K22.75) = K68.25 color(red)(larr" Corrected"#

.............................................................
#color(brown)("Question part b")#
Interest starts at #3.5%# but at the end of the first year it is changed to #3.75%#

1 year at #color(white)(.)3.5% color(white)("d") = color(white)("d") 1xx color(cadetblue)(K22.750)color(white)(d)=color(cadetblue)(Kcolor(white)(.)22.75)#
2 years at #3.75%=color(white)("d")2xxcolor(sienna)(K24.375)=ul(Kcolor(white)(".")48.75larr" Add")#
#color(white)("ddddddddddddddddddddddddddddd")Kcolor(white)(".")71.50#

So the increase in interest is:

#Kcolor(white)(".")71.50#
#ul(Kcolor(white)(".")68.25 larr" Subtract")#
#Kcolor(white)(".")color(white)("d")3.25color(red)(larr" Corrected"#