HP 35s User Manual

							 
 
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HP 35s Loan down payments 
 
 
 
 
Loan down payments 
 
The Time Value of Money on the HP 35s 
 
Practice solving loan down payment problems 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
						

							 
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HP 35s Loan down payments 
 
hp calculators - 2 - HP 35s Loan down payments - Version 1.0 
Loan down payments 
 
Down payments are often made on loans to lower the required payment. Other reasons for down payments can be to 
ensure the loan applicant has an equity interest in the loan collateral, which would make the loan applicant less likely to 
abandon the property, since the property would be worth more than the loan balance. Down payments are also required 
to ensure an investment in the property has been made by the loan applicant, thereby reducing the risk to the lender that 
the loan will be abandoned.  
The process to be used is to input the payment the applicant can afford and determine the equivalent Present Value 
(PV). The difference between this PV and the actual loan amount will be the required down payment.  
The Time Value of Money on the HP 35s 
 
To solve time value of money problems on the HP 35s, the formula below is entered into the flexible equation solver built 
into the calculator. This equation expresses the standard relationship between the variables in the time value of money 
formula. The formula uses these variables: N is the number of compounding periods; I is the periodic interest rate as a 
percentage (for example, if the annual interest rate is 15% and there are 12 payments per year, the periodic interest 
rate, i, is 15÷12=1.25%); B is the initial balance of loan or savings account; P is the periodic payment; F is the future 
value of a savings account or balance of a loan. 
 
Equation: P x 100 x ( 1 - ( 1 + I ! 100 )^ -N)  ! I + F x ( 1 + I ! 100 ) ^ -N + B 
 
To enter this equation into the calculator, press the following keys on the HP 35s: 
 
!#$%&&$4%4%()*%&&+,-
./+*)(0$4%()*%&&+,.-
/(12 
 
To verify proper entry of the equation, press 
 
34 
 
and hold down the 4-key. This will display the equation’s checksum and length. The values displayed should be a 
checksum of CEFA and a length of 41. 
 
To solve for the different variables within this equation, the 56 button is used. This key is the right shift of the 
! key. 
 
Notes for using the SOLVE function with this equation: 
 
1) If your first calculation using this formula is to solve for the interest rate I, press %57)-before beginning. 
2) Press !. If the time value of money equation is not at the top of the list, press 8or 9 to scroll through the 
list until the equation is displayed. 
3) Determine the variable for which you wish to solve and press: 
a) 56/ to calculate the number of compounding periods. 
b) 56) to calculate the periodic interest rate. Note: this will need to be multiplied by the number of 
compounding periods per year to get the annual rate. If the compounding is monthly, multiply by 12. 
c) 561 to calculate the initial balance (or Present Value) of a loan or savings account. 
d) 56# to calculate the periodic payment.   
						

							 
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HP 35s Loan down payments 
 
hp calculators - 3 - HP 35s Loan down payments - Version 1.0 
e) 560 to calculate the future value of a loan or savings account. 
4) When prompted, enter a value for each of the variables in the equation as you are prompted and press :. The 
solver will display the variables’ existing value. If this is to be kept, do not enter any value but press : to 
continue. If the value is to be changed, enter the changed value and press :. If a variable had a value in a 
previous calculation but is not involved in this calculation (as might happen to the variable P (payment) when solving 
a compound interest problem right after solving an annuity problem), enter a zero for the value and press :. 
5) After you press : for the last time, the value of the unknown variable will be calculated and displayed. 
6) To do another calculation with the same or changed values, go back to step 2 above. 
 
The SOLVE feature will work effectively without any initial guesses being supplied for the unknown variable with the 
exception noted above about the variable I in this equation. This equation follows the standard convention that money in 
is considered positive and money out is negative. 
 
The practice problems below illustrate using this equation to solve a variety of loan down payment problems. 
 
Practice solving loan down payment problems 
 
Example 1: Leigh Anne wants to buy a car and can afford a payment of $400 a month. If the car costs $25,000  
 and Leigh Anne can get a 72 month loan at 6.9%, compounded monthly, how much must she give as a  
 down payment to lower her payment to $400 a month? 
 
Solution: First, enter the time value of money equation into the HP 35s solver as described earlier in this document. 
 Press !-and press 8or 9 to scroll through the equation list until the time value of 
 money equation is displayed. Then compute the present value of a loan for 72 months of $400 per month  
 at Leigh Annes interest rate. To do this, press:- 
-561  
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution.  
 The value of 0.0000 is displayed below if this is the first time the time value of money equation has been  
 solved on the HP 35s calculator. If any previous equations have used a variable used in the time value of  
 money equation, they may already have been assigned a value that would be displayed on your HP 35s  
 display. Follow the keystrokes shown below and the solution should be found as described. 
 
 Figure 1 
--
-In either RPN or algebraic mode, press: ;&&: 
 
 Figure 2 
 
-In RPN mode, press: 2%?*: 
 In algebraic mode, press: *%?2:-
   
						

							 
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HP 35s Loan down payments 
 
hp calculators - 4 - HP 35s Loan down payments - Version 1.0 
 Figure 3 
 
-In either RPN or algebraic mode, press: @?:-
 
 Figure 4 
 
-In either RPN or algebraic mode, press: &: 
 
 Figure 5  
 With a payment of $400 per month, Leigh Anne can afford a loan amount of $23,527.99. To buy the  
 car costing $25,000, Leigh Anne must make a down payment of the difference. 
 
 In RPN mode, press: ?A&&&( 
 In algebraic mode, press: (?A&&&2-
 
 Figure 6  
Answer: To lower her monthly payment to $400, Leigh Anne needs to make a $1,472.01 down payment.  
 
Example 2: Jane is looking to buy a house and can afford a payment of $1,200 a month. If the house costs $270,000  
 and Jane can get a 30 year loan at 5.4%, compounded monthly, how much must Jane give as a down  
 payment to lower her payment to $1,400 a month? 
 
Solution: First, enter the time value of money equation into the HP 35s solver as described earlier in this document. 
 Press !-and press 8or 9 to scroll through the equation list until the time value of money  
 equation is displayed. Then compute the present value of a loan for 72 months of $400 per month  
 at Janes interest rate. To do this, press: 
-
-561 
 
 The HP 35s SOLVER displays the first variable encountered in the equation as it begins its solution.  
 The displays for these prompts are not shown in this example. Follow the keystrokes shown below and  
 the solution should be found as described.   
						

							 
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HP 35s Loan down payments 
 
hp calculators - 5 - HP 35s Loan down payments - Version 1.0 
 In RPN mode, press: %;&&:-- (Enters P) 
--A=;2%?*: (Enters I) 
--B&2%?$:  (Enters N) 
--&:-   (Enters F) 
 
 In algebraic mode, press: %;&&:-- (Enters P) 
--A=;*%?2: (Enters I) 
--B&$%?2:  (Enters N) 
--&:-   (Enters F) 
 
 Figure 7 
 
 With a payment of $1,400 per month, Jane can afford a loan amount of $249,318.47. To buy the  
 house costing $270,000, Jane must make a down payment of the difference. 
 
 In RPN mode, press: ?@&&&&( 
 In algebraic mode, press: (?@&&&&2-
 
 Figure 8 

Answer: To lower her monthly payment to $1,400, Jane needs to make a $20,681.53 down payment.   
						

							 
 
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HP 35s  Average sales prices 
 
 
 
 
Averages and standard deviations 
 
Practice finding average sale prices and  
standard deviations 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
 
   
						

							 
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HP 35s Averages sales prices 
 
hp calculators - 2 - HP 35s Averages sales prices - Version 1.0 
Averages and standard deviations 
 
The average is defined as the sum of all data points divided by the number of data points included. It is a measure of 
central tendency and is the most commonly used. A standard deviation is a measure of dispersion around a central 
value. To compute the standard deviation, the sum of the squared differences between each individual data point and 
the average of all the data points is taken and then divided by the number of data points included (or, in the case of 
sample data, the number of data points included minus one). The square root of this value is then taken to obtain the 
standard deviation. The property of the standard deviation is such that when the underlying data is normally distributed, 
approximately 68% of all values will lie within one standard deviation on either side of the mean and approximately 95% 
of all values will lie within two standard deviations on either side of the mean. This has application to many fields, 
particularly when trying to decide if an observed value is unusual by being significantly different from the mean. 
 
On the HP 35s, values are entered into the statistical / summation registers by keying in the number (or pair of numbers) 
desired and pressing !. This process is repeated for all numbers or pair of numbers.  When entering a pair of 
numbers in RPN or algebraic mode, key the Y value, press , then key the X value and press !.  
 
To view the mean, press #$. To view the standard deviation, press %&. When either of these is pressed, the 
HP 35s displays a menu of possible values. Items on this menu are viewed by pressing the  or ( cursor keys.  
 
To use a value displayed on the menu, press the  button and the value will be copied for further use. This is 
illustrated in the problems below. 
 
Practice finding average sale prices and standard deviations 
 
Example 1: The sales price of the last 10 homes sold in the Parkdale community were: $198,000; $185,000;  
 $205,200;$225,300; $206,700; $201,850; $200,000; $189,000; $192,100; $200,400. What is the 
 average of these sales prices and what is the sample standard deviation? Would a sales price of  
 $240,000 be considered unusual in the same community? 
 
Solution: Be sure to clear the statistics / summation memories before starting the problem. 
 
 %)*+
+
 The keystrokes are the same whether in RPN or algebraic mode: 
 
 ,-.///!,.0///!1/01//!+
+1102//!1/34//!1/,.0/!+
+1/////!,.-///!,-1,//!+
+1//*//! 
  
 To find the average, press: #$. Figure 1 displays the menu shown. 
 
 Figure 1 
 
 To find the sample standard deviation, press: %&. Figure 2 displays the menu shown. +
   
						

							 
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HP 35s Averages sales prices 
 
hp calculators - 3 - HP 35s Averages sales prices - Version 1.0 
 Figure 2 
 
 To find the value two standard deviations above and below the average, press the following: 
 
 In RPN mode: 
 
 %&15#$67%879 
 
 In algebraic mode: 
 
 %&516#$ (computes the above value) 
 #$915%& (computes the below value) 
 
 Figure 3 
 
Answer: The average sales price is $200,355 and the sample standard deviation is $11,189. Within two  
 standard deviations on either side of this average, in this case between $177,977 and $222,733, 95%  
 of all home sales prices should fall. If a home were to sell for $240,000 in this area, it would be an  
 unusual event. Figure 3 indicates the display in RPN mode. 
 
Example 2: The sales price of the last 7 homes sold in the real estate office’s zip code were: $245,000; $265,000;  
 $187,000; $188,000; $203,000; $241,900; $222,000. What is the average of these sales prices and what  
 is the sample standard deviation?  
  
Solution: Be sure to clear the statistics / summation memories before starting the problem. 
 
 %)*+
 
 The keystrokes are the same whether in RPN or algebraic mode: 
 
+1*0///!130///!,.4///!+
+,..///!1/2///!1*,-//!+
+111///! 
 
 To find the average sales price, press: #$. Figure 4 displays the menu shown. 
 
 Figure 4  
 To find the sample standard deviation, press: %&. Figure 5 displays the menu shown. +
   
						

							 
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HP 35s Averages sales prices 
 
hp calculators - 4 - HP 35s Averages sales prices - Version 1.0 
 Figure 5 
 
Answer: The average sales price was $221,700 and the standard deviation was $30,318.81 
 
Example 3: Julie has bought gas this week while showing houses at four gasoline stations as follows: 15 gallons at  
 $1.56 per gallon, 7 gallons at $1.64 per gallon, 10 gallons at $1.70 per gallon and 17 gallons at $1.58  
 per gallon. What is the average price of the gasoline purchased? 
 
Solution: The HP 35s has a weighted average mean calculation built-in that will solve this problem easily. Be sure  
 to clear the statistics / summation memories before starting the problem. 
 
 %)*+
 
 In RPN or algebraic mode, press: 
 
 ,0,:03!4,:3*!,/,:4! 
 ,4,:0.!+
+
+To find the weighted average price of gasoline purchased, press: #$((. Figure 6 displays  
 the menu shown. 
 
 Figure 6 
 
Answer: The average price per gallon Julie has paid this week while showing houses is slightly less than $1.61.