A tree increases annually by 1/8 th of its height. What will be its he

1.	A tree increases annually by 1/8 th of its height. What will be its height after 2 years, if its height today is 64 cm?1). 72 cm.2). 74 cm.3). 75 cm.4). 81 cm.
Answer» Given, present height of the tree = 64 cm The height of the tree increases by 1/8 th every year. After 1^st year Height of tree = 64 + 1/8 of 64 ⇒ Height of tree = 64 + 8 = 72 cm After 2^nd year Height of tree = 72 + 1/8 of 72 ⇒ Height of tree 72 + 9 = 81 cm Another approach, it can be considered a CI problem: PRINCIPAL (initial height) = 64 cm, rate = 1/8 × 100 % = 12.5 %, t= 2 years, A= final height For CI: $(A = P{\left( {1 + \frac{r}{{100}}} \right)^t})$ Where, A is the length at the end of time t, P is the height at the starting of the time, t is time in years. r is growth rate in percent ⇒ Final height = $(64{\left( {1 + \frac{{12.5}}{{100}}} \right)^2})$ ⇒ Final height = 64 × 1.125² ⇒ Final height = 64 × 1.265625 = 81 cm.

A tree increases annually by 1/8 th of its height. What will be its height after 2 years, if its height today is 64 cm?1). 72 cm.2). 74 cm.3). 75 cm.4). 81 cm.

Answer»

Given, present height of the tree = 64 cm

The height of the tree increases by 1/8 th every year.

After 1^st year

Height of tree = 64 + 1/8 of 64

⇒ Height of tree = 64 + 8 = 72 cm

After 2^nd year

Height of tree = 72 + 1/8 of 72

⇒ Height of tree 72 + 9 = 81 cm

Another approach, it can be considered a CI problem:

PRINCIPAL (initial height) = 64 cm, rate = 1/8 × 100 % = 12.5 %, t= 2 years, A= final height

For CI:

$(A = P{\left( {1 + \frac{r}{{100}}} \right)^t})$

Where,

A is the length at the end of time t,

P is the height at the starting of the time,

t is time in years.

r is growth rate in percent

⇒ Final height = $(64{\left( {1 + \frac{{12.5}}{{100}}} \right)^2})$

⇒ Final height = 64 × 1.125²

⇒ Final height = 64 × 1.265625 = 81 cm.

Discussion

No Comment Found

Related InterviewSolutions

1). 250002). 200003). 300004). 15000
An equal sum is invested in two different schemes. One scheme gives simple interest and the other gives compound interest (annual compounding). The sum of interest obtained after 2 years from both the schemes is Rs. 2961. If both schemes have 23% per annum interest rate, then what is the first year interest (in Rs.) of simple interest scheme?1). 5002). 11003). 7004). 900
Find the annual installment which will discharge a debt of Rs. 63840 in 3 years at 8% p.a. compound interest –1). Rs. 23309.702).3). Rs. 29407.904). Rs. 29804.50
Find the simple interest on Rs. 50 for 6 months at the rate of 10 paise per rupee per month.1). Rs. 352). Rs. 403). Rs. 254). Rs. 30
A sum of money invested at simple interest becomes 13/10 of itself in 2 years and 6 months. What is the rate (in percentage) of interest per annum?1). 102). 153). 124). 18
If the simple interest on a sum of money for 2 years at 5% per annum is Rs. 50, what is the compound interest on the same sum at the same rate and for the same time?1). Rs. 51.252). Rs. 52.453). Rs. 54.254). Rs. 60.25
1). Rs. 597.8122). Rs.553.53). Rs. 557.8124). Rs. 567.212
If Rs. 125000 amounts to Rs. 140608 at 8% compound interest, interest being compounded semi-annually in certain time period. If the same amount is invested at same rate of interest and time, the amount of simple interest is?1). Rs. 1400002). Rs. 1450003). Rs. 1500004). Rs. 200000
The time required for a sum of money to amount to five times itself at 16% simple interest p.a. will be1). 10 years2). 15 years3). 30 years4). 25 years
A certain sum becomes 'k' times in 6 years at compound interest. In 24 years it will become how many times?1). 4k2). k43). k34). 3k