GMAT Question of the Day - Daily to your Mailbox; hard ones only

 It is currently 18 Dec 2018, 02:57

GMAT Club Daily Prep

Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.

Customized
for You

we will pick new questions that match your level based on your Timer History

Track

every week, we’ll send you an estimated GMAT score based on your performance

Practice
Pays

we will pick new questions that match your level based on your Timer History

Events & Promotions

Events & Promotions in December
PrevNext
SuMoTuWeThFrSa
2526272829301
2345678
9101112131415
16171819202122
23242526272829
303112345
Open Detailed Calendar
• Happy Christmas 20% Sale! Math Revolution All-In-One Products!

December 20, 2018

December 20, 2018

10:00 PM PST

11:00 PM PST

This is the most inexpensive and attractive price in the market. Get the course now!
• Key Strategies to Master GMAT SC

December 22, 2018

December 22, 2018

07:00 AM PST

09:00 AM PST

Attend this webinar to learn how to leverage Meaning and Logic to solve the most challenging Sentence Correction Questions.

A right angled triangle has its sides in Arithmetic progression and be

Author Message
TAGS:

Hide Tags

Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

Show Tags

Updated on: 01 Aug 2018, 09:43
2
8
00:00

Difficulty:

95% (hard)

Question Stats:

41% (02:04) correct 59% (17:31) wrong based on 85 sessions

HideShow timer Statistics

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

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.
Math Expert
Joined: 02 Aug 2009
Posts: 7112
Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

Show Tags

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 a-d,a,a+d
$$a^2-2ad+d^2+a^2=a^2+2ad+d^2......a^2-4ad=0$$
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 perimeter

All 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 9-1=8 as sides 15:20:25 will not be counted
_________________

1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html

GMAT online Tutor

Senior PS Moderator
Joined: 26 Feb 2016
Posts: 3327
Location: India
GPA: 3.12
A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

Show Tags

Updated on: 01 Aug 2018, 23:16
1
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 right-angled 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 right-angled triangle are 3x-4x-5x

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 right-angled triangles can be formed.
_________________

You've got what it takes, but it will take everything you've got

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.
Edited!
Intern
Joined: 12 Jul 2018
Posts: 3
Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

Show Tags

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 9-4=5 , 5 such triangle are possible
pls correct if i m wrong
Intern
Joined: 29 Jun 2018
Posts: 4
Location: India
GMAT 1: 720 Q49 V39
WE: Analyst (Investment Banking)
Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

Show Tags

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.

Director
Status: Learning stage
Joined: 01 Oct 2017
Posts: 931
WE: Supply Chain Management (Energy and Utilities)
A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

Show Tags

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.

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.
_________________

Regards,

PKN

Rise above the storm, you will find the sunshine

Director
Joined: 14 Dec 2017
Posts: 518
Location: India
Re: A right angled triangle has its sides in Arithmetic progression and be  [#permalink]

Show Tags

03 Aug 2018, 11:51
1
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.

Thanks,
GyM
_________________
Re: A right angled triangle has its sides in Arithmetic progression and be &nbs [#permalink] 03 Aug 2018, 11:51
Display posts from previous: Sort by

A right angled triangle has its sides in Arithmetic progression and be

 Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®.