When the 80-cent oil alone is taken, in what ratio to the 60-cent must it be used? When the $1.20 oil alone is taken, in what ratio to the 60-cent oil must it be used? When the $1.20 and 80-cent oils are mixed gallon for gallon, how much 60-cent oil must be added? When 1 gal. of the $1.20 oil and 3 of the 80-cent oil are taken, how much 60-cent oil must be added? If three-fourths of the mixture consist of the 60-cent oil, what per cent of each of the other two must be taken?

12. A solder composed of tin and lead, specific gravities 7.29 and 11.35, has a specific gravity of 10.44. What is the weight of each metal in a kilogram of solder?

AVERAGE OF PAYMENTS.

Ex. John Smith has given to William Jones notes as follows: $150, due May 15; $200, due June 30; $500, due July 31. He wishes to pay them all at one time. When shall they be considered due?

If all the notes were paid May 15, Smith would lose the use of $200 for 46 days, and of $500 for 67 days.

The use of $200 for 46 days is equal to the use of $200 × 46 for 1 day; and the use of $500 for 67 days is equal to the use of $500×67 for 1 day. Smith would, therefore, lose what is equivalent to the use of $9200 + $33,500 = $42,700 for 1 day, and is entitled to keep the $150 + $200 + $500 = $850 as many days after May 15 as are required for the use of $850 to equal the use of $42,700 for 1 day, or 43700 days 50.2 days. Hence, the equated time for paying all the notes is 50 days after May 15; that is, July 4.

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The work may be arranged as follows:

13. Find the equated time for the payment of $250 due in 3 mos., $400 due in 6 mos., $700 due in 8 mos.

14. Find the equated time for the payment of $300 due in 30 dys., $500 due in 60 dys., and $200 due in 90 dys.

15. Find the equated time for the payment of $325 due now, $200 due in 30 dys., $460 due in 60 dys., and $150 due in 90 dys.

16. Find the equated time for the payment of $240 due May 10, $420 due July 2, $310 due Sept. 14, and $600 due Oct. 1.

17. Find the equated time for the payment of $275 due June 21, $175 due July 16, $200 due Aug. 6, and $150 due Sept. 3.

18. Find the equated time for the payment of $112.30 due July 6, $115.25 due July 30, $232.15 due Sept. 4, and $102.36 due Oct. 1.

Ex. A owes B $200 due in 10 mos. If he pays $120 in 4 mos. when should he pay the balance?

By paying $120 in 4 months A loses the use of $120 for 6 months, which is equal to the use of $720 for 1 month. Therefore, he is entitled to keep the balance ($80) 720 months 9 months after its maturity.

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19. A owed B $2000 payable in 4 mos., but at the end of 1 mo. he paid him $500, at the end of 2 mos. $500, and at the end of 3 mos. $500. In how many months is the balance due?

20. A merchant bought, Feb. 11, 1881, a bill of goods amount

ing to $1700, on 4 months' credit; but he paid Mar. 22, $400, Apr. 20, $220, May 10, $300. When is the balance due?

Find the equated time of maturity of each of the following bills, and the amount due at settlement, including interest at 6% :

JAMES PRICE, to JOHN BATES, Dr.

21.

Find the equated time for payment, reckoning from July 15, which is the earliest date that any item becomes due, and find the interest on the whole bill from the equated time to Oct. 18. As this is not banking business, allow no days of grace.

23. Find the equated time of maturity of each side of the

NOTE. On the Dr. side is the statement of the goods sold by Bacon & Co. to Adams & Co.; and on the Cr. side is the statement of the cash paid by Adams & Co. to Bacon & Co.

Ex. Find the time for payment of the balance of an account, if the debit and credit sides, when equated, stand as follows:

If the account were settled at the later date, May 30, the $900 on the Dr. side would have been on interest 1 day, and this is equivalent to having the balance, $75, on interest 200 of 1 day = 12 days. ience, the balance should begin to draw interest 12 days before May 30; that is, May 18.

Ex. Find the time for payment of the balance of an account if the debit and credit sides, when equated, stand as follows:

Due Jan. 20, 1881, $850

Due Feb. 18, 1881, $950

Difference in equated time, 29 days.

Balance of account, $950 - $850 = $100.

If the account were settled at the later date, Feb. 18, the $850 would have been on interest 29 days, which is equivalent to having the balance, $100, on interest 58 of 29 days = 2464 days.

Hence, to increase the Dr. side by an equal amount of interest, the balance should remain unpaid 247 days; that is, the balance is due Oct. 23, 1881.

375. From the two preceding examples is derived the following

RULE FOR EQUATING ACCOUNTS.

Find the equated time for each side of the account.

Multiply the side of the account that falls due first by the number of days between the dates of the equated times of the two sides, and divide the product by the balance of the ac

count.

The quotient will give the number of days to the maturity of the balance, to be counted forward from the later date when the smaller side falls due first, and backward when the larger side falls due first.

Find the time for paying the balance in the following

NOTE. In finding the equated time of accounts it is customary to neglect cents if less than 50, and if 50 or more to consider them as $1. A fraction of a day in the result is rejected if less than ; if or more it is called a day.

376. When an account is settled by cash at any other date than that at which the balance becomes due, the interest is found on the balance for the interval between the day of settlement and the day the balance is due, and is added to, or deducted from, the balance, according as the settlement is made after or before the balance is due.

377. Another method is by computing the interest on each item, from its date to the day of settlement. (The time is reckoned in days.)

When the balance of account and the balance of interest fall on

opposite sides, the cash balance is their difference.