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A right angled triangle has its sides in Arithmetic progression and be
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Updated on: 01 Aug 2018, 09:43
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A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible? A. 9 B. 8 C. 7 D. 6 E. 5 Edit: corrected answer options. Thank you pushpitkc @chetan2u
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Originally posted by PKN on 31 Jul 2018, 11:48.
Last edited by PKN on 01 Aug 2018, 09:43, edited 2 times in total.



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Re: A right angled triangle has its sides in Arithmetic progression and be
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31 Jul 2018, 17:48
PKN wrote: A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible? A. 8 B. 7 C. 6 D. 5 E. 4 Algebraically let's how many are there..Let sides be ad,a,a+d \(a^22ad+d^2+a^2=a^2+2ad+d^2......a^24ad=0\) So a=4d or multiple of 4.. So 3:4:5 or 6:8:10 and so on Otherwise also we know 3:4:5 is a right angled triangle.. Atmost one is multiple of 10, means it may not have any multiple of 10 or just one at the max. Least perimeter = 3+4+5=12 Next bigger perimeter will be 12*2 when sides are 6:8:10 So total possible perimeter will be the greatest multiple of 12 just less than 115 12*9=108, so 9 such perimeterAll 9 will have atmost one multiple of 10.. Ans 9 but not in choices But if you are asking for atmost one multiple of 5.. Then only Ans will be 91=8 as sides 15:20:25 will not be counted
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A right angled triangle has its sides in Arithmetic progression and be
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Updated on: 01 Aug 2018, 23:16
PKN wrote: A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible? A. 8 B. 7 C. 6 D. 5 E. 4 From the question stem, we can deduce the following information 1. The sides of the rightangled triangle are in an AP. 2. At most one of the sides is a multiple of 10 3. The perimeter is less than 115 We need a set of numbers which is both a Pythagorean triplet and in an AP One such set of numbers that is both in an AP and sides of a rightangled triangle are 3x4x5x The triangle possible are 3,4,5(when x = 1)  6,8,10(when x = 2)  9,12,15(when x = 3)  12,16,20(when x = 4)  15,20,25(when x = 5)  18,24,30(when x = 6)  21,28,35(when x = 7)  24,32,40(when x = 8)  27,36,45(when x = 9)  At x = 10, the perimeter of the triangle is 12x or 120(and this fails condition 3) Therefore, there are at least 9(Option A) possibilities where rightangled triangles can be formed.
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Originally posted by pushpitkc on 31 Jul 2018, 23:36.
Last edited by pushpitkc on 01 Aug 2018, 23:16, edited 2 times in total.
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Re: A right angled triangle has its sides in Arithmetic progression and be
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01 Aug 2018, 00:32
atmost 9 options are possible out of which option with d= 2,4,5,6,8 will have atleast one side in multiple of 10 so, one of its kind is allowed and hence 94=5 , 5 such triangle are possible pls correct if i m wrong



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Re: A right angled triangle has its sides in Arithmetic progression and be
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01 Aug 2018, 09:46
PKN wrote: A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible? A. 8 B. 7 C. 6 D. 5 E. 4 Let the sides be x, x+y, x+2y. Then, by Pythagorean theom. (x+2y)^2 = x^2 + (x+2y)^2 Simplifying the above equation we get 3*(y/x)^2 + 2*(y/x) 1 = 0, solving the quadratic equation we get y=2x/3 Thus, the three sides are x, 5x/3, 7x/3 and the perimeter is 5x. Now as the perimeter is less than 115 and the sides are integers both x and y will also be integers. Hence the possible values of the perimeter (5x) are 110, 105, 100......5. so the possible values for x are 22, 21, 20 ...... 1. However, as y=2x/3 and y is an integer x should be a factor of 3. Therefore, x can take the value of 3, 6, 9, 12, 15, 18, 21 i.e 7 possible values and hence 7 possible triangles. Please help me identify where i went wrong.



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A right angled triangle has its sides in Arithmetic progression and be
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01 Aug 2018, 10:13
prashant17p wrote: PKN wrote: A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible? A. 8 B. 7 C. 6 D. 5 E. 4 Let the sides be x, x+y, x+2y. Then, by Pythagorean theom. \((x+2y)^2\) = \(x^2\) + \((x+2y)^2\)Simplifying the above equation we get 3*(y/x)^2 + 2*(y/x) 1 = 0, solving the quadratic equation we get y=2x/3 Thus, the three sides are x, 5x/3, 7x/3 and the perimeter is 5x. Now as the perimeter is less than 115 and the sides are integers both x and y will also be integers. Hence the possible values of the perimeter (5x) are 110, 105, 100......5. so the possible values for x are 22, 21, 20 ...... 1. However, as y=2x/3 and y is an integer x should be a factor of 3. Therefore, x can take the value of 3, 6, 9, 12, 15, 18, 21 i.e 7 possible values and hence 7 possible triangles. Please help me identify where i went wrong. Hi prashant17p, Welcome to GMAT Club. Highlighted portion is wrong. It should be \((x+2y)^2\)=\(x^2\)+ \((x+y)^2\)Also, At most one of its sides is a multiple of 10 implies NOT MORE THAN ONE of the sides of the right angled triangle is a multiple of 10. You need to take care of this statement while determining possible combinations of 3 sides of the right angled triangles.
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Re: A right angled triangle has its sides in Arithmetic progression and be
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03 Aug 2018, 11:51
PKN wrote: A right angled triangle has its sides in Arithmetic progression and being integers. At most one of its sides is a multiple of 10. Its perimeter is less than 115. How many such triangles are possible? A. 9 B. 8 C. 7 D. 6 E. 5
Let the sides be \((a  d)\), \(a\), \((a + d)\), where d is the common difference for the sides to form an AP. Given the perimeter \((a  d) + a + (a + d) < 115\) Hence we have \(a < 38.3\), since a is an integer we get \(a = 1,2,......38\) Also, the given triangle is a right triangle, hence \((a d)^2 + a^2 = (a + d)^2\), the longest side as the hypotenuse. we get \(a = 4d\), hence \(a\) is a multiple of 4 & has a value between 1 & 38. we get 9 such values for \(a\) to satisfy above criterion. Answer A. Thanks, GyM
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