Latest Elitmus Aptitude Question SOLUTION: Q. In a triangle A is the greatest angle ana A+7B=155 then what will be the range of c

Answer 30 < c < 85
Solution:
A+7B=155 ..........(1)
A+B+C=180 ..........(2)
from (1) & (2)
6B-25=C ........(3)
now try the least possible value of B and the max value of B and get the range of C.
for example:
clearly put B=1 you will get C=31 hence C should be greater than 30.
Now put B=10 , u will get C=85 but B+C=95 hence A=85 but it is given A is greatest.
So C should be less than 85. :)

Ans

79>c>31

Given

a + 7 b=155

and a is greatest

that means

a = 92 85 78 71 64 57 50 43

b = 63 70 77 84 91 98 105 112

after dividing b by 7

b = 9 10 11 12 12 14 15 16

then

when a=78 and b = 11 then range of c is 91

when a = 85 and b =10 then range of c is 85

Above two cases cant be possible then

when we take a=92 b=9 then c=79 which is acceptable...........

85>c>25
explaination : the triangle is isoceles , sum of the angles is 180

http://www.careerbless.com/qna/discuss.php?questionid=915

arnab can u explain in detail

any body written the exam this option is theire or no

I had a dobuet let it be 84 then the possible combinations be (85,11),(86,10),87,9).......... like which doesn't satisfies A+B=155,and A+B+C=180 at a time

25

and MIRZA what if you put B= 1/6

25

if you put c=25 you will get B=0 hence c>25 if you put c=85 you will get A=85 not a valid angle as A is largest of all hence c

A+B+C=180
A must be >=90
B+C

Answer: 25 < C < 85

Explanation

A + 7B = 155 ---(1)
A + B + C = 180 ---(2)
A>B ---(3)
A>C ---(4)

From (1)
B =
155
−
A
7
155−A7

Substituting the value of B in (2)
A +
(
155
−
A
)
7
(155−A)7 + C = 180
7A + 155 - A + 7C = 1260
6A + 7C = 1105
=> C =
(
1105
−
6
A
)
7
⋯
(
5
)
(1105−6A)7⋯(5)

Above equation represent C in terms of A. From this equation, we can see
a. C gets maximum value when A is lowest.
b. C gets minimum value when A is highest.

Find minimum and maximum values of A subject to A>B and A>C which when substituted in the above equation gives possible range of C.

From (1), A can be close to 155 when B is near zero.
i.e., the highest value A can take is close to 155 (with A < 155)

if A = 155, C =
(
1105
−
6
×
155
)
7
(1105−6×155)7 = 25
Therefore, C > 25 ---(6)

From (1) we have A + 7B = 155. Let B=A. The equation becomes 8A=155 => A = 19.375.
For A > 19.375, A will be greater than B ---(7)

From (2) we have, C =
(
1105
−
6
A
)
7
(1105−6A)7. Let C=A.
The equation becomes A =
(
1105
−
6
A
)
7
(1105−6A)7
=>13A = 1105
=> A = 85
i.e., For A > 85, A will be greater than C ---(8)

Since A is the largest, from (7) and (8),we can see that minimum value of A close to 85 (with A > 85)

if A = 85, C =
(
1105
−
6
×
85
)
7
(1105−6×85)7 = 85
Therefore, C < 85 ---(9)

Combining (6) and (9)
i.e., 25 < C < 85