shashankism
rulingbear
Bunuel
If 3a – 2b – 2c = 32 and \(\sqrt{3a}- \sqrt{2b + 2c} = 4\), what is the value of a + b + c ?
A. 3
B. 9
C. 10
D. 12
E. 14
I will use plugging in . Given the equations and the options, a is likely 12 and a+b+c is 14. For further checking, 2b+2c = 4 and \sqrt{2b + 2c} = 2, making a=b=1. Hence, a+b+c = 12+2(1) = 14.
I don't think plugging in is a good idea for this question.... How did u directly arrived at a = 12 ??
Good question. It needs keen observation
1. From the first equation, 3a is likely to be greater than 32
2. From the second equation, [m]\sqrt{3a} is likely to be greater than 4
Put this inference together and a could be 12, i.e 3*12= 36>32, and \sqrt{3*12} = 6>4
The same could be used to know that a= b= 1, as explained in my earlier post. Even without doing this you should know by now that the answer is E.
This is a good
must be true type of question, it could also make a decent data sufficiency question. Just observing keenly will obviously save a lot of time!
Another way to look at it is the form of the equation, take the first equation, 3a - 2a- 2b= 32. This can be rearranged as 3a - 2(a+b)= 32, therefore a must be an even number and a+b will also be even. Given this, a+b+c would be an even number of the x+2y. Only 10,12 and 14 are viable. Looking at the second equation will tell you that x/a would be 12 as 3a would form the only proper square.