Question

Sarah buys a car costing ₹120500 its deprecated is a value by 5.5 % in the first 8% in the second year and 6% in the third year is the car worthy after 3 year

cost of car is 120500
discount of first year =5.5%
$cp \: of \: car \: \: \: 120500 \times 94.5 \div 100$
is equal to 113872.5

discount of second year = 8%

$second \: year \: cp \: of \: car \: \: 113872.5 \times 92 \div 100$
is equal to, 104762.5

discount of third year = 6℅

$third \: year \: cp \: of \: car \: \: 104462.7 \times 94 \div 100$
is equal to, 98,476.938



answer is 98,476.938​

Answer 1

Step-by-step explanation:

By using the equation of motion

v=u+at

By substituting the given values in the above equation

15=6+105t

Then the time is0.0857sec

The distance covered is given as

S=ut+21at2

S=6×0.0857+21×105×(0.0857)2

S=0.514+0.385

The distance covered is0.899m