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“A drill has to drill holes between 2mm and 25mm in diameter. Eight spindle speeds are required and the cutting speed is to be 20m/min. Find the eight speeds using Arithmetic Progression
7 years ago
Maths Question asked by Douglas
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2 Answers
Highest speed (1000x20)/(3.14x2) = 3185 r/m = l (last term of A.P)
Lowest speed (1000x20)/(3.14x25)= 255 r/m = a (first term of A.P)
a + (n -1)d = l where d is the difference
255 + 7d = 3185
d = (3185-255)/7 = 419 (nearest whole number)
So rounded speeds are 255,674,1093,1512,1931,2350,2769,3188
Lowest speed (1000x20)/(3.14x25)= 255 r/m = a (first term of A.P)
a + (n -1)d = l where d is the difference
255 + 7d = 3185
d = (3185-255)/7 = 419 (nearest whole number)
So rounded speeds are 255,674,1093,1512,1931,2350,2769,3188
There is something wrong with the question. The speed is given in terms of a single depth - 20m/min. Therefore, no progression can be calculated. There are two measures for the hole diameters but no progression is required.
Assuming there should be 8 drill bits ranging, in arithmetic progression, between 2 mm and 25 mm, the diameters are:
2 mm, 5 2/7 mm, 8 4/7 mm, 11 6/7 mm, 15 1/7 mm, 18 3/7, 21 5/7 mm and 25 mm (steps of 3 2/7 mm).
Assuming there should be 8 drill bits ranging, in arithmetic progression, between 2 mm and 25 mm, the diameters are:
2 mm, 5 2/7 mm, 8 4/7 mm, 11 6/7 mm, 15 1/7 mm, 18 3/7, 21 5/7 mm and 25 mm (steps of 3 2/7 mm).
The answer given is correct. I agree with the Anand that the question is not well asked. The drill bits turn through 20m every minute, so the 1000x20 is convert it into mm. C = Pi x d, so the operations 1000x20/pix2 and 1000x20/pix25 are to find the speeds in revolutions per minute. This question isn`t really a maths question, but an engineering question.
