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For all numbers a and b, the operation # is defined by a # b = 2a + 2b

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Bunuel
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For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  15 Oct 2019, 09:03
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A
B
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E

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5% (low)

Question Stats:

94% (00:29) correct 6% (00:51) wrong based on 173 sessions

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For all numbers a and b, the operation # is defined by a # b = 2a + 2b + 1. What is the value of 6 # 1 ?

A. 6
B. 7
C. 13
D. 14
E. 15
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  15 Oct 2019, 09:31
Bunuel wrote:
For all numbers a and b, the operation # is defined by a # b = 2a + 2b + 1. What is the value of 6 # 1 ?

A. 6
B. 7
C. 13
D. 14
E. 15


Here a=6 and b=1

Now 6#1=2*6+2*1+1=12+2+1=15

Ans. (E)
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  15 Oct 2019, 10:03
Bunuel wrote:
For all numbers a and b, the operation # is defined by a # b = 2a + 2b + 1. What is the value of 6 # 1 ?

A. 6
B. 7
C. 13
D. 14
E. 15

\(6 # 1\) , here \(a = 6\) & \(b = 1\)

So, \(a # b = 2*6 + 2*1 + 1\)

Or, \(a # b = 12 + 2 + 1\)

Or, \(a # b = 15\), Thus Answer must be (E)
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  15 Oct 2019, 10:29
Bunuel wrote:
For all numbers a and b, the operation # is defined by a # b = 2a + 2b + 1. What is the value of 6 # 1 ?

A. 6
B. 7
C. 13
D. 14
E. 15


Given; For all numbers a and b, the operation # is defined by a # b = 2a + 2b + 1.

Asked: What is the value of 6 # 1 ?

6 # 1 = 2*6 + 2*1 + 1 = 12 + 2 + 1 = 15

IMO E
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  15 Oct 2019, 10:46
For all numbers a and b, the operation # is defined by a # b = 2a + 2b + 1. What is the value of 6 # 1 ?

A. 6
B. 7
C. 13
D. 14
E. 15--> correct

Answer:
a # b = 2a + 2b + 1
6 # 1 = 2*6+2*1+1, as a=6 & b=1
6#1=15
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  15 Oct 2019, 11:42
6#1= 2*6 + 2*1 +1 = 12+2+1 =15

Hence E
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  16 Oct 2019, 19:19
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Bunuel wrote:
For all numbers a and b, the operation # is defined by a # b = 2a + 2b + 1. What is the value of 6 # 1 ?

A. 6
B. 7
C. 13
D. 14
E. 15


6 # 1 = 2(6) +2(1) + 1 = 12 + 2 + 1 = 15

Answer: E
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EgmatQuantExpert
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  22 Oct 2019, 22:12
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Solution


Given:
    • a # b = 2a + 2b + 1

To find:
    • 6 # 1

Approach and Working Out:
    • 6 # 1 = 2*6 + 2*1 + 1 = 12 + 2 + 1 = 15

Hence, the correct answer is Option E.

Answer: E
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  05 Oct 2020, 23:36
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Very obvious that when we get such an easy question, we simply substitute the values and get the answer.

We can try an alternative approach. Now Even * Even or Even * Odd is always Even. Therefore 2a and 2b are Even.

Also, Even + Even = Even and Even + Odd = Odd. So 2a + 2b + 1 = even + even + odd = odd


Among the answer options A and D get eliminated. Now a = 6. 2a + 1 is 13 with taking he value of 2b.

The answer has to be greater than 13. Options B and C get eliminated.



Option E

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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink] New post  06 Oct 2020, 01:03
a # b = 2a + 2b + 1

=> 6 # 1 = 2(6) + 2(1) + 1 = 15

Answer E
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Re: For all numbers a and b, the operation # is defined by a # b = 2a + 2b [#permalink]
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