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a, b and c are three interior angles of a triangle ABC. If a, b and c are three integers, then what is the number of possible values of a – b + c, which are less than zero ?
- A. 88
B. 90
C. 178
D. 179
E. 180
05 Dec 2018, 01:57
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EgmatQuantExpert
a, b and c are three interior angles of a triangle ABC. If a, b and c are three integers, then what is the number of possible values of a – b + c, which are less than zero ?
- A. 88
B. 90
C. 178
D. 179
E. 180
The sum of the angles of a triangle is 180.
Hence we can say that:
a+b+c=180
Now we need to find all possible cases when:
a-b+c<0 or
a+c<b
Thing to note: A and C has to be at least 1.
Hence,
A+C=180-b
Hence:
B>90
But B cannot be 180 or 179 as A and c have to be atleast 1.
Hence the maximum value of B is 178
Total possible values:
178-90=88
Hence answer should be A.
General Discussion
05 Dec 2018, 02:06
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EgmatQuantExpert
a, b and c are three interior angles of a triangle ABC. If a, b and c are three integers, then what is the number of possible values of a – b + c, which are less than zero ?
- A. 88
B. 90
C. 178
D. 179
E. 180
For a triangle the sum of three sides a+b+c=180
since here in we need to find "the number of possible values of a – b + c, which are less than zero"
or say a+c<b and (a+c)-b <0 ;
since c=180-a-b
a – b + c, substitute value of c that a-b+180-a-b
=> -2b=-180
=> b=90
so if b=90 , then a+c has to be 90 , and value of a,c can be any thing (89,1), (81,9) ... but since we need to find the number of possible values of a – b + c, which are less than zero ,value of b has to be 91 and the values of a & c can be anything from (88,1) , (1,88) total values such values which we would fetch us sum < 0 would be IMO 88 ; so IMO A is correct..
