1 Find the true discount on £ 8. d. Drawn. Discounted. Ans. 147 3 2 Mar. 6 at 6 months, June 11 at 6 p.c. £2 2 108% 202 10 0 Mar. 31 at 7 months May 8 at 6 p.c.£6 5 1,8791 SQUARE ROOT. 25109 A number multiplied by itself is said to be squared, or raised to the second power. The number or factor so multiplied is called the square root. Ex.-22-2×2=4. The extraction of any root is indicated by writing the index of the root within the sign✔ Ex. 24-22×2=2. 38=32×2×2=2.. 4/16=4/2×2×2×2=2. The first nine numbers are the respective square roots of 1, 4, 9, 16, 25, 36, 49, 64, 81. Here we first mark off the figures in pairs or periods, from the right or units, and then take the nearest square number to the first period, placing its square root in the answer. 2. Subtract and bring down the next period, double the figure of the root, and add a cypher, for the trial divisor. Having ascertained how many times the trial divisor will divide, set the number in the answer as the next figure of the root, adding it also to the trial divisor. Then divide and proceed as before. When the example contains a decimal the periods are marked from the right before the point, and from the left after the point. The Cube of a number is that number raised to the third power, thus the cube of 2=8 or 2×2×2=8. The cube root is the number which has been raised to the third power. Thus8= 2, or 3 2 × 2 × 2 = 2. The method of extracting the cube root, though similar to that for the square root, is rather more complicated, and ought to be thoroughly learnt. The reason for the rule will be best understood when the candidate has learnt Algebra. The explanation cannot be here given, for want of space, and would serve no practical purpose, as the rule is easily acquired by a little practice. Divide into periods of 3 figures each from the right, find the cube nearest to the first period, subtract it from it, and bring down the next period, making 35279. A little to the left set down the multiplication of the first figure of the root by 3, which is seen in the example to be 15. Multiply this by the root figure 5, making 75, to which affix two noughts, and one nought to the 15, making 150 and 7500 respectively. The 7500 is the trial divisor, which is found to divide 4 times into 35279. Place the 4 in the answer, add it to the 150 and multiply 154 by the root figure 4, adding the product 616 to the trial divisor, for the true divisor. Multiply the true divisor 8116 by the root figure 4, subtract and bring down the next period. 4 154 Add again the root figure 4 to on the left, making the sum of 162, or if preferable multiply the 4 in 154 by 3, adding the product 12 to 150; then square the root 4, and add the 16 to 616 ; to the 162 add a 0, and to 8748 8116 add two O as before for the new trial divisor, and proceed as before. making the sum of 8748 When the example contains a decimal, the periods are marked from the left of the point, as in the square root. Extract the cube roots of 315126110125. 360488070613. 8002.400240008. 327 and .244140625. Ans. 6805. Ans. 7117. Ans. 20.002. Ans. .75 and .625 SOME MISCELLANEOUS QUESTIONS, WITH THEIR SOLUTIONS. How many sovereigns, half-sovereigns, crowns, florins, shillings, sixpences, and threepences, and of each an equal number, are there in £79,, 8,, 9 ? Every collection of one of each of these coins amounts to £1,, 18,, 9; hence there will be as many coins of each kind as £79 8 9 contains £1,, 18 9; that is, there will be 41 A tea merchant mixes 27 lbs. of tea at 1s.,, 11d. a lb., 45 lbs at 28.,, 7d., and 12 lbs. at 3s.,, 6d. If these are cost prices, at what rate per lb. must he sell the mixture to gain 10 per cent. |