(a) The annual investment that should be made.

Question

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Answer 1

Question

Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Expert Solution

Answer to Problem 1P

The annual investment is $12158.5.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n−1] ...... (I)

Here, the compound amount in future is F, the interest rate is i and time period is n.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5.

Conclusion:

Thus, the annual investment to be made is $12158.5.

To determine

(b)

The investment per year by employee and employer.

Expert Solution

Answer to Problem 1P

The amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

3% of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A′=$65000×3%=$65000×3100=$1950

The amount to be invested by employer is $1950 per year.

Calculate the investment by employee.

Ai=$12158.5−$1950=$10208.5

The amount to be invested by employee is $10208.5 per year.

Conclusion:

Thus, the amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

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Answer 2

Question

Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Expert Solution

Answer to Problem 1P

The annual investment is $12158.5.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n−1] ...... (I)

Here, the compound amount in future is F, the interest rate is i and time period is n.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5.

Conclusion:

Thus, the annual investment to be made is $12158.5.

To determine

(b)

The investment per year by employee and employer.

Expert Solution

Answer to Problem 1P

The amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

3% of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A′=$65000×3%=$65000×3100=$1950

The amount to be invested by employer is $1950 per year.

Calculate the investment by employee.

Ai=$12158.5−$1950=$10208.5

The amount to be invested by employee is $10208.5 per year.

Conclusion:

Thus, the amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

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Answer 3

Question

Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Expert Solution

Answer to Problem 1P

The annual investment is $12158.5.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n−1] ...... (I)

Here, the compound amount in future is F, the interest rate is i and time period is n.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5.

Conclusion:

Thus, the annual investment to be made is $12158.5.

To determine

(b)

The investment per year by employee and employer.

Expert Solution

Answer to Problem 1P

The amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

3% of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A′=$65000×3%=$65000×3100=$1950

The amount to be invested by employer is $1950 per year.

Calculate the investment by employee.

Ai=$12158.5−$1950=$10208.5

The amount to be invested by employee is $10208.5 per year.

Conclusion:

Thus, the amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

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Answer 4

Question

Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Expert Solution

Answer to Problem 1P

The annual investment is $12158.5.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n−1] ...... (I)

Here, the compound amount in future is F, the interest rate is i and time period is n.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5.

Conclusion:

Thus, the annual investment to be made is $12158.5.

To determine

(b)

The investment per year by employee and employer.

Expert Solution

Answer to Problem 1P

The amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

3% of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A′=$65000×3%=$65000×3100=$1950

The amount to be invested by employer is $1950 per year.

Calculate the investment by employee.

Ai=$12158.5−$1950=$10208.5

The amount to be invested by employee is $10208.5 per year.

Conclusion:

Thus, the amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

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Answer 5

Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Expert Solution

Answer to Problem 1P

The annual investment is $12158.5.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n−1] ...... (I)

Here, the compound amount in future is F, the interest rate is i and time period is n.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5.

Conclusion:

Thus, the annual investment to be made is $12158.5.

To determine

(b)

The investment per year by employee and employer.

Expert Solution

Answer to Problem 1P

The amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

3% of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A′=$65000×3%=$65000×3100=$1950

The amount to be invested by employer is $1950 per year.

Calculate the investment by employee.

Ai=$12158.5−$1950=$10208.5

The amount to be invested by employee is $10208.5 per year.

Conclusion:

Thus, the amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Answer 6

Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Expert Solution

Answer to Problem 1P

The annual investment is $12158.5.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n−1] ...... (I)

Here, the compound amount in future is F, the interest rate is i and time period is n.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5.

Conclusion:

Thus, the annual investment to be made is $12158.5.

Answer 7

Chapter 10A, Problem 1P

To determine

(a)

The annual investment that should be made.

Answer 8

Expert Solution

Answer to Problem 1P

The annual investment is $12158.5.

Answer 9

Expert Solution

Answer 10

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

Concept used:

Write the expression for annual investment.

A=F[i(1+i)n−1] ...... (I)

Here, the compound amount in future is F, the interest rate is i and time period is n.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5.

Conclusion:

Thus, the annual investment to be made is $12158.5.

Answer 13

To determine

(b)

The investment per year by employee and employer.

Expert Solution

Answer to Problem 1P

The amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

3% of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A′=$65000×3%=$65000×3100=$1950

The amount to be invested by employer is $1950 per year.

Calculate the investment by employee.

Ai=$12158.5−$1950=$10208.5

The amount to be invested by employee is $10208.5 per year.

Conclusion:

Thus, the amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Answer 14

To determine

(b)

The investment per year by employee and employer.

Answer 15

Expert Solution

Answer to Problem 1P

The amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Answer 16

Expert Solution

Answer 17

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

The compound amount in future is $2000000.

The time period is 30 years.

The earning of engineer is $65000 per year.

3% of the salary is contributed each year to retirement savings.

Calculations:

Assume the interest rate is 10%.

Calculate the annual investment.

Substitute $2000000 for F, 10% for i and 30 for n in Equation (I).

A=$2000000[0.1(1+0.1)30−1]=$2000000[0.1(1.1)30−1]=$2000000(0.00607)=$12158.5

Calculate the investment by the employer.

A′=$65000×3%=$65000×3100=$1950

The amount to be invested by employer is $1950 per year.

Calculate the investment by employee.

Ai=$12158.5−$1950=$10208.5

The amount to be invested by employee is $10208.5 per year.

Conclusion:

Thus, the amount to be invested by the employer is $1950 per year.

The amount to be invested by the employee is $10208.5 per year.

Answer 20

Ch. 10A - Prob. 1P Ch. 10A - Prob. 2P Ch. 10A - Prob. 3P Ch. 10A - Prob. 4P Ch. 10A - Prob. 5P Ch. 10A - Prob. 6P Ch. 10A - Prob. 7P Ch. 10A - Prob. 8P Ch. 10A - Prob. 9P Ch. 10A - Prob. 10P

(a) The annual investment that should be made.

(a) The annual investment that should be made.

Answer to Problem 1P

Explanation of Solution

(b) The investment per year by employee and employer.

Answer to Problem 1P

Explanation of Solution

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Chapter 10A Solutions

(a)

The annual investment that should be made.

(b)

The investment per year by employee and employer.