Muthu started working at a company on 1 January 2002 with an initial annual salary of RM18,000. Every January, the company increased his salary by 5% of the previous year's salary. Calculate
a)his annual salary, to the nearest RM, for the year 2007.
b)the minimum value of n such that his annual salary in the nth year will exceed RM36,000.
c)the total salary, to the nearest RM, paid to him by the company, for the years 2002 to 2007.
a) 18000 * 1.05^5
b) 36000 = 18000 * 1.05^n
n = ln 2/ln 1.05
c) sum of first n terms of a GP is
a(1-r^n)/(1-r) = 18000(1-1.05^6)/(1-1.05)
To calculate Muthu's annual salary for the year 2007, we can use the given information that his salary increases by 5% each year.
a) Annual salary for 2002 = RM18,000
Annual salary for 2003 = RM18,000 + (5% of RM18,000)
Annual salary for 2004 = (Annual salary for 2003) + (5% of Annual salary for 2003)
Annual salary for 2005 = (Annual salary for 2004) + (5% of Annual salary for 2004)
Annual salary for 2006 = (Annual salary for 2005) + (5% of Annual salary for 2005)
Annual salary for 2007 = (Annual salary for 2006) + (5% of Annual salary for 2006)
Let's calculate the annual salaries step-by-step:
Annual salary for 2002 = RM18,000
Annual salary for 2003 = RM18,000 + (5% of RM18,000)
= RM18,000 + (0.05 * RM18,000)
= RM18,000 + RM900
= RM18,900
Annual salary for 2004 = RM18,900 + (5% of RM18,900)
= RM18,900 + (0.05 * RM18,900)
= RM18,900 + RM945
= RM19,845
Annual salary for 2005 = RM19,845 + (5% of RM19,845)
= RM19,845 + (0.05 * RM19,845)
= RM19,845 + RM992.25
= RM20,837.25
Annual salary for 2006 = RM20,837.25 + (5% of RM20,837.25)
= RM20,837.25 + (0.05 * RM20,837.25)
= RM20,837.25 + RM1,041.86
= RM21,879.11
Annual salary for 2007 = RM21,879.11 + (5% of RM21,879.11)
= RM21,879.11 + (0.05 * RM21,879.11)
= RM21,879.11 + RM1,093.96
= RM22,973.07
Therefore, Muthu's annual salary for the year 2007 would be approximately RM22,973.
b) To find the minimum value of n such that Muthu's annual salary exceeds RM36,000:
Annual salary for the nth year = RM18,000 * (1 + 0.05)^n
We need to solve the equation: RM18,000 * (1 + 0.05)^n > RM36,000
Divide both sides of the equation by RM18,000:
(1 + 0.05)^n > 2
Taking logarithm (base 10) on both sides:
n * log(1.05) > log(2)
n > log(2) / log(1.05)
Using a calculator, we find:
n > 14.2 (approximately)
Since n has to be a whole number, we take the smallest integer greater than 14.2, which is 15.
Therefore, the minimum value of n such that Muthu's annual salary exceeds RM36,000 is 15.
c) To find the total salary paid to Muthu by the company from 2002 to 2007:
Total salary = (Annual salary for 2002) + (Annual salary for 2003) + (Annual salary for 2004) + (Annual salary for 2005) + (Annual salary for 2006) + (Annual salary for 2007)
Substituting the calculated values, we get:
Total salary = RM18,000 + RM18,900 + RM19,845 + RM20,837.25 + RM21,879.11 + RM22,973.07
= RM121,434.43
Therefore, the total salary paid to Muthu by the company from 2002 to 2007 is approximately RM121,434.
To calculate Muthu's annual salary, we need to apply the given conditions and calculate for each year.
a) To find Muthu's annual salary for the year 2007, we can use the given information about the salary increment. Note that he started working in 2002 with an initial salary of RM18,000.
- Year 2002: RM18,000
- Year 2003: Increase by 5% of RM18,000 = RM900 (Total salary = RM18,000 + RM900 = RM18,900)
- Year 2004: Increase by 5% of RM18,900 = RM945 (Total salary = RM18,900 + RM945 = RM19,845)
- Year 2005: Increase by 5% of RM19,845 = RM992.25 (Total salary = RM19,845 + RM992.25 = RM20,837.25)
- Year 2006: Increase by 5% of RM20,837.25 = RM1,041.87 (Total salary = RM20,837.25 + RM1,041.87 = RM21,879.12)
- Year 2007: Increase by 5% of RM21,879.12 = RM1,093.96 (Total salary = RM21,879.12 + RM1,093.96 = RM22,973.08)
Therefore, Muthu's annual salary for the year 2007 is RM22,973 (rounded to the nearest RM).
b) To find the minimum value of n such that Muthu's annual salary in the nth year exceeds RM36,000, we can set up an inequality:
RM18,000 * (1 + 0.05)^n > RM36,000
Simplifying, we have:
(1.05)^n > 2
To solve for n, we can take the logarithm of both sides with base 1.05:
log(1.05)^n > log 2
n * log 1.05 > log 2
n > log 2 / log 1.05
Using this, we can calculate the minimum value of n:
n > 14.2065
Since n must be an integer, the minimum value of n that satisfies the inequality is 15.
Therefore, Muthu's annual salary will exceed RM36,000 in the 15th year.
c) To calculate the total salary paid to Muthu by the company from 2002 to 2007, we can sum up the individual annual salaries for each year:
Total salary = Initial salary + salary in 2003 + salary in 2004 + salary in 2005 + salary in 2006 + salary in 2007
= RM18,000 + RM900 + RM945 + RM992.25 + RM1,041.87 + RM1,093.96
= RM22,973.08
Therefore, the total salary paid to Muthu by the company for the years 2002 to 2007 is RM22,973 (rounded to the nearest RM).