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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

\(3a-2b-2c=32 => 5a-2a-2b-2c=32 =>5a-2(a+b+c)=32\)

therefore \(a+b+c=\frac{5a}{2}-16\) ----------------------------(1)

Also \(2b+2c=3a-32\)

Now, \(\sqrt{3a}- \sqrt{2b + 2c} = 4\) substitute the value of \(2b+2c\)

\(\sqrt{3a}- \sqrt{3a-32} = 4\). square both sides to get

\(3a+3a-32-2\sqrt{3a(3a-32)}=16\). this can be written as

\(6a-48=2\sqrt{3a(3a-32)}\) \(=>\) \(3a-24=\sqrt{3a(3a-32)}\). Again square both sides to get

\(9a^2+24^2-2*3a*24=9a^2-96\). solve this to get \(a=12\)

Substitute the value of \(a\) in equation (1) or \(a+b+c=\frac{5*12}{2}-16=14\)

Option E

Same answer. The highlighted portion would be 9a^2 - 96a?
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Same answer. The highlighted portion would be 9a^2 - 96a?

Hi Richak91

Thanks for highlighting the typo error :-)
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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


Lets start solving this question.
Formula required : (x−y)(x+y)=(x2−y2)

\(\sqrt{3a}- \sqrt{2b + 2c} = 4\) ..............(i)

3a – 2b – 2c = 32
(\(\sqrt{3a}- \sqrt{2b + 2c}\))(\(\sqrt{3a}+ \sqrt{2b + 2c}\)) = 32
(4)(\(\sqrt{3a}+ \sqrt{2b + 2c}\)) = 32
(\(\sqrt{3a}+ \sqrt{2b + 2c}\)) = 8 ........................ (ii)

Adding (i) & (ii) , we get (\(2 \sqrt{3a}\)) = 12
a = 12

Subtracting (i) from (ii) , we get (\(2 \sqrt{2b + 2c}\)) = 4
b+c = 2

So, a+b+c = 12+2 = 14

Answer E
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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.
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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 ??
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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.
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rulingbear




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!

3a>32 already includes [m]\sqrt{3a} is be greater than 4.
So we should further work on 3a>32..

But your observation is appreciable and surely can be used here to solve the problem using plugging method..
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shashankism
rulingbear




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!

3a>32 already includes [m]\sqrt{3a} is be greater than 4.
So we should further work on 3a>32..

But your observation is appreciable and surely can be used here to solve the problem using plugging method..

There is no need to actually work further once you know 3a>32, a is most likely 12 and the answer can only be 14, given the options.
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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
Easy strategy would be to replace these numbers by variables to simplify the equation -

Let 3a = x and 2b + 2c = y, then

\(x - y = 32\)
\((\sqrt{x}- \sqrt{y})*(\sqrt{x}+ \sqrt{y}) = 32\) --- (1)
\(\sqrt{x}- \sqrt{y} = 4\) --- (2)

From (1) and (2)

\(\sqrt{x}+ \sqrt{y} = 8\) --- (3)

From (2) and (3)

\(2*\sqrt{x} = 12\)
\(\sqrt{x} = 6\)
x = 36
3*a = 36
a = 12 --- (4)

\(\sqrt{y} = 2\)
y = 4
2*b + 2*c = 4
b + c = 2 --- (5)

From (4) and (5)

a + b + c = 14

Answer: E
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