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03 Jun 2018, 11:53
Kudos
Bookmarks
E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
25%
(medium)
Question Stats:
83% (02:05) correct
17%
(02:44)
wrong
based on 122
sessions
History
Date
Time
Result
Not Attempted Yet
If 5a – 3b = –9 and 3a + b = 31, then a + b =
A) –7
B) –3
C) 6
D) 13
E) 19
A) –7
B) –3
C) 6
D) 13
E) 19
03 Jun 2018, 12:48
Kudos
Bookmarks
Bunuel
Elimination method:-
Multiply equation (1) with 3 and equation (2) with 5, then subtracting (2) from (1), we have
3*(5a – 3b = –9) --------(1)
5*(3a + b = 31) --------(2)
(-) (-) (-)
.......................................................
-14b = -182
or, b=\(\frac{-182}{-14}\)=13
from (2) we have, 3a=31-b
or, a=\(\frac{(31-b)}{3}\)
Now substituting the value of b, we have
a=\(\frac{(31-13)}{3}\)=\(\frac{18}{3}\)=6
So, we have a+b=6+13=19.
Hence, Option (E) is the answer.
03 Jun 2018, 20:01
Kudos
Bookmarks
Bunuel
Given:
5a – 3b = –9 ...(P)
3a + b = 31 ....(Q)
Multiply (Q) by 3 to get (R), then add (P) and (R)
5a – 3b = –9 ...(P)
9a + 3b = 93 ...(R)
-------------------
14a = 84
a = 6
Find b
From (Q): 3a + b = 31
(3)(6) + b = 31
b = (31-18) = 13
(a + b) = (6 + 13) = 19
Answer E
