If a and b are nonnegative integers and 4a + 3b = 32, how many values

Question

If a and b are nonnegative integers and 4a + 3b = 32, how many values of b are there?

A. 1
B. 2
C. 3
D. 4
E. 5

BrentGMATPrepNow · Accepted Answer

If we focus solely on the value of a we can see that the GREATEST value of a is 8, since any a-value greater than 8 will make it impossible to satisfy the equation (4a + 3b = 32) without requiring b to be a negative value.

So, we need only check the values of a from 0 to 8. 
a =  0 : we get 4( 0 ) + 3b = 32. This means 3b = 32, so b = 32/3 NOT AN INTEGER VALUE
a =  1 : we get 4( 1 ) + 3b = 32. This means 3b = 28, so b = 28/3 NOT AN INTEGER VALUE
a =  2 : we get 4( 2 ) + 3b = 32. This means 3b = 24, so b = 24/3 = 8   AN INTEGER VALUE  
a =  3 : we get 4( 3 ) + 3b = 32. This means 3b = 20, so b = 20/3 NOT AN INTEGER VALUE
a =  4 : we get 4( 4 ) + 3b = 32. This means 3b = 16, so b = 16/3 NOT AN INTEGER VALUE
a =  5 : we get 4( 5 ) + 3b = 32. This means 3b = 12, so b = 12/3 = 4   AN INTEGER VALUE  
a =  6 : we get 4( 6 ) + 3b = 32. This means 3b = 8, so b = 8/3 NOT AN INTEGER VALUE
a =  7 : we get 4( 7 ) + 3b = 32. This means 3b = 4, so b = 4/3 NOT AN INTEGER VALUE
a =  8 : we get 4( 8 ) + 3b = 32. This means 3b = 0, so b = 0/3 = 0   AN INTEGER VALUE

Answer: C

Cheers,
Brent

theLucifer · Answer

4a +3b = 32 
3b = 32- 4a 
3b = 4*8 -4a
3b = 4* (8-a)
Since we have find the integer value of b, 8-a should be divisible by 3.
So Ans = 2 (b)

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Abhishek009 · Answer

The following combination of ( a , b ) are available -

1. a = 2, b = 8
2. a = 5, b = 3
3. a = 8, b = 0
  
Hence, correct answer will be (C) 3

mcrimmin · Answer

B cannot be 3. If B is any odd value A will not be an integer.

The good values for B are 0. 4 and 8. Answer is still C.

RMD007 · Answer

Alternate Approach:

4a + 3b = 32.
For a moment, if we ignore 4a, we know that maximum possible range of values of b will be 11(including 0) (3*10= 30). and we know that b cannot be 10.
Now, 4a will always be even and to arrive at sum 32, an even number, 3b should be even number. So b should be even number.
So possible values of b - (0,2,4,6,8).

Plugging the possible set we can figure out that b would take 0, 4, 8. --> C

ScottTargetTestPrep · Answer

Simplifying, we have:

3b = 32 - 4a

3b = 4(8 - a)

b = 4(8 - a)/3

Note that b must be an integer, so (8 - a) must be evenly divisible by 3. Since a can be 2, 5, or 8, b has three possible values.

Answer: C

satya2029 · Answer

a and b are non negative integers.
4a+3b=32
4a is always even, 32 is even. 3b must be even. For this, b must be even. 4a always comes in the multiple of 4, and 32 is also a multiple of 4. So, 3b must be multiple of 4. 
b=0, 4,8
and if we increase the value of b, value of a will tend towards negative integer( a and b are non negative integers)
C:)

bhargava05 · Answer

4a + 3b = 32

the first value for which this is true is when a= 8 and b =0

from now add the value of 4 to b and substract 3 form 8
we get

a b
8 0
5 4
2 8

so non negative integers for b are 0 4 8 which is 3

bumpbot · Answer

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If a and b are nonnegative integers and 4a + 3b = 32, how many values

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If a and b are nonnegative integers and 4a + 3b = 32, how many values

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