# data[].attributes.body
1) 3 weeks

2) 5

# data[].attributes.body
(1) 3 weeks

(2) $5

(3) Bank B , $4 difference

# data[].attributes.body
CH1) 3weeks

CH2) 5

CH3) bank B gives 4$ more

# data[].attributes.body
(1)26600

(2)interestEarned = 46

Total amount = 400 + 46 = 446 , total amount

# data[].attributes.body
1-26600

2-446

# data[].attributes.body
Verbal A* :

r = 2.13

# data[].attributes.body
1-2400

b-2343.3

2-2.6%

# data[].attributes.body
(1)85

(2)1.9

# data[].attributes.body
(1)9080

# data[].attributes.body
1)3

2)5

3)4 $ more

# data[].attributes.body
9080.13

# data[].attributes.body
Verbal B :

(1)807 trees

# data[].attributes.body
9500 (1 - 0.9/100)^5 = 9080

# data[].attributes.body
500x10x1.7 / 100 = 85 dollars of interest earned

500 (1+r/100)^6 = 559.78

r= ( 6root(559.78/500) - 1 ) x 100 = 1.9%

# data[].attributes.body
verbal a*

1400(1+r/100)^4 = 1523.15

(4root(1523.15/1300) -1 ) x 100 = 2.13 %

# data[].attributes.body
1- 30000x(1-2/100)^6 = 26600 lions

2- 400x2.3x5 / 100 = 46 dollars of interest

400+46 = 446 dollars

# data[].attributes.body
Spatial B/A/A* :

(1)simple interest = 400

compound interest = 343

(2)r = 2.6

# data[].attributes.body
C3 - Bank B gives more

(200x6/100 - 200X4/100) = 4 dollar more

# data[].attributes.body
(1)35.5

# data[].attributes.body
spatial b/a/a*

a. i-

2000x2.5x8/100 = 400

2000+400 = 2400 dollars

2000(1+2.5/100)^8 = 2436.81 dollars

b- 1328.54+4540 = 4540 (1+r/100)^10

r = (10root(5868.58/4540)) - 1) x 100 = 2.6 %

# data[].attributes.body
1) 26600

2) 446

# data[].attributes.body
446

# data[].attributes.body
1 26609 lions

2 446 dollars

# data[].attributes.body
1) 30000*(0.98)^6= 26600

2) 400*(1+2.3*5/100) A= 400(1.115) = 446

# data[].attributes.body
1: 26,600

2: 446

# data[].attributes.body
verbal b

980(1-1.75/100)^11 = 807 trees

# data[].attributes.body
A1) exponential decay = a (1-r)^n

=30000 (1- 2/100)^6

=26575 = 26600 lions

A2) simple interest = PRT/100

=400*2.3*5/100

= $46

Total Amount = 400+46 = $446

# data[].attributes.body
q1) 26,600 , 300000(1-2/100)^6

q2) 446$

# data[].attributes.body
1 500x0.017x10

=85

2 559.78/500=(1+ r/100)6

=1.9

# data[].attributes.body
verbal a

a- 3500(1+19.5 /100)^2 = 4998.0875 ~ 5000

b- 3500(1+19.5/2 /100)^2x2 = 5077.92 ~ 5080 dollars

# data[].attributes.body
A) 2000*2.5*8/100 = 400

2000+400 = $2400

compound = 2000(1+2.5/100)^8 = $2436

# data[].attributes.body
Simple interest investment = $2400

Compound interest investment = $2343.32

Rate: 2.6 %

# data[].attributes.body
R=2.13

# data[].attributes.body
1)500x10x1.7 / 100 = 85 dollars

2)500 (1+r/100)^6 = 559.78

r= ( 6root(559.78/500) - 1 ) x 100 =

1.9%

# data[].attributes.body
verbal b 807 trees

# data[].attributes.body
A1) S.I = PRT/100

=2000*2.5*8/100

=$400

T.A= 2000+400=$2400

C.I= P(1+r/100)^n

=2000(1+2/100)^8

=$2343

A2) R = (n root of A/P - 1) *100

R= ((10 root of 1328.54/4540) - 1) *100