1) A person got a job with salary * 1,80,000 per year. He was given 10.000
increment per year. Complete the following activity to find the number of
years (n), when his salary will be * 2,50,000.
Answers
Step-by-step explanation:
In 8 years his annual salary will be Rupees 250000.
Solution :
Proof : There is an increment of Rupees 10,000 in salary per year.
This is an A.P. with d = 10,000.
Here, a=180000,tn=2,50,000,n=?
tn=a+(n−1)d ….(Formula)
∴250000=180000+(n−1)×10000
∴25=18+n−1 ....(Dividing both the sides by 10000)
∴25=17+n
∴n=25−17
∴n=8
Given:
The salary of a person=Rs.1,80,000 per year
Increment=Rs.10,000 per year
To find:
The number of years after which the salary becomes Rs.2,50,000
Solution:
The salary becomes Rs.2,50,000 after 8 years.
We can find the number of years by following the given steps-
We know that the person is getting an increment of a fixed amount every year.
Let the number of years be n.
The salary in the initial year=Rs.1,80,000
We know that with an equal increase each year, the salary every year forms an arithmetic progression.
The common difference, d= Rs.10,000
The first term of the sequence, a=Rs.1,80,000
The last term of the sequence, l= Rs.2,50,000
Number of years=n
We know that the last term of an A.P., l= a +(n-1)d
On putting the values, we get
2,50,000=1,80,000+(n-1)×10,000
2,50,000-1,80,000=10,000(n-1)
70,000=10,000(n-1)
70,000/10,000=n-1
7=n-1
n=7+1
n=8
The number of years, n=8
Therefore, the salary becomes Rs.2,50,000 after 8 years.