BOROUGH SCHOOL EXAMINATION.

Ashburton Guardian, Volume V, Issue 1544, 6 July 1885, Page 2

BOROUGH SCHOOL EXAMINATION.

Ta ths Editor. Sir, —Will you kindly allow me to correct a statement which you made in your issue of the Ist instant. In that issue you published one of the questions put at the recent examination of the Ashburton Borough School. Your presumed answer thereto is very presumptive and bad on the face of it. The question put is :—“ What is the smallest sum of money with which I can purchase either pigs at 16s each, lambs at L2 each, calves at L 5 each, or heifers at L 8 6s each V' Now you presumed Ll 6 10s to be the smallest. I would like to ask how many pigs at 16s, or lambs at L2, or calves at L 5, you could purchase for L'6 10a. I am afraid the last pig, lamb or calf, as the case may be, would be a sort of freak of nature. The examiners, I take it, wished the pupils to find a sum with which a number of any of the above mentioned animals could be purchased and such a sum must bo common to each of the above mentioned amounts. This is obtained by a process called least common multiple, and instead of being the amount yon mentioned, it is .660, a totally different amount from Ll 6 10s. With your permission, Sir, I will show the working according to arithmetic. To begin with, we will reduce the price of each animal to;a common value, and as a shilling is the lowest value we will reduce the price of each animal to shillings Thus we have pigs at 16s each, lambs at 40s each, calves at 100 s each, and heifers at 165 s each. It now remains to find an amount which will be common to each of those amounts, i.e , an amount into which we can divide each without leaving a remainder, and the least amount with which the latter can be done will be the answer to our question. Now, we can find a number which will divide one or more of the above amounts, as, for example, 5 will divide 40, 100 and 165 without leaving a remainder. Again, it is patent that 16 is equal to 4x9x2, 40=5x4x2, 100=5 ] x4x6, 165=5x33. Now, it will be seen that 4x 2 is common to 16 and 40, 5x4 to 40 and 100, and 5 to 100 and 165, so that a number into which we can divide either 16, 40, 100 or 165 must contain the factors of each of those numbers Thus 4x2x2x6x5x33 contain each factor of those numbers, and the result of the latter numbers multiplied together will be the h-as* number common to all—otherwise the least common multiple According to arithmetic the working would stand thus—--4 j 16 40 100 ’65 2)4 10 'ia 165 5} 2 5 25 165 2 I 5 33 4x2xsx2xsx 33=1320/= £660. which is the least amount that will satisfy our question. I hope that the above explanation will prove satisfactory to everybody, and especially to Mr D. Thomas. 1 understand, Sir, that at the meeting held the' other night be challenged any one to do the sums written on the blackboard in the schoolroom, and in particular pointed out the above sum. I trust that he will now fulfil the practical part of his challenge, and kindly forward cheque for the amount thereof (LlO 10s, I believe) to yourself, and I will guarantee to see that the amount is distributed in a charitable and deserving manner* Apologising fur trespassing at such a length, I am, etc, Junius. Ashburton, 6th July, 1885. [Wo thank our correspondent for his full exposition of a very useful inle of arithmetic. We were acquainted with it. In referring to the question we wished to point out the ambiguity given to it b> the substitution of the word or for and. We shall be very pleased to act as tbe channel of communication between Mr Thomas and Junius, but the former gentleman’s challenge applied to all the questions set, one of which was without an answer.— Ed. Q.]