Each entry in the multiplication table above is an integer that is eit

h= b*c
h is not equal to zero so neither b nor c is zero.
but this doesn't tell about a.
not sufficient.

Statement-2
c=f
a*c= f
this expression can hold if c=f=0
a can have any value.
and If C=F= any value, a is always 1.
not sufficient

Statement 1+2
value of c is except zero.
and for these value, a is always 1.

C is correct answer.

First, the significance of multiplication table needs to be understood.
This is like a 3*3 matrix and results in the following products ,
a^2 = d
a*b = e
a*c = f
a*b = e
b^2 = g
b*c = h
c*a = f
c*b = h
c^2 = j

Now statement 1 says ,
h is not equal to 0.
That means neither b nor c equals zero as b*c = h;
But we cant tell anything about a.

Statement 2 says , c =f
a*c = f
a*c = c
c*(a-1) = 0
Now a can be equal to 1 or c can be equal to 0 .
So a can be equal to 1 or a may not be equal to 1 .
Not sufficient.

Together ,
c *(a-1) = 0
From statement h not equal to zero means c is not equal to zero.
If c is not equal to zero then (a-1) = 0 ( As we can divide the zero by c which is a non zero quantity )
Hence , a = 1
So option C is the answer.

Please give me KUDO s if you liked my explanation.
GMATNinja VeritasKarishma

Bunuel

Can you help me with this question?

1. I got confused by what the table is. The "x" on the top left corner: I read it as x and not multiplication.

2. Even if I go past point 1, I am still not sure on how to approach this question.

Hey TheNightKing ,
Please refer to my explanation above. Please let me know if you have any doubt.
Also, please give me kudos if you find it helpful.

Thank you Sir. Since you wrote each of the multiplications explicitly it became easier to see what's happening.

Hey sayan
I am not able to understand the login behind the multiplication results explanations

any letter is an integer positive, negative or zero

(1) insufic
bc=h≠0, so (b,c,h) are not 0
no info about a or d

(2) insufic
c=f, ac=f, ac=c, ac=c
c=0, then, a=anything
a=1, then, c=anything

(1/2) sufic
c≠0, thus,
a=1, then, c=anything≠0

Ans (C)

What is the value of a?

(1) h ≠ 0
b * c = h

This statement tells us that neither b nor c are zero. However, we can't determine the value of a. INSUFFICIENT.

(2) c = f

The chart tells us c * a = f

We can conclude a = 1 or c = 0
if c = 0, a = any value
if a = 1, c = any value

INSUFFICIENT.

(1&2) From statement 1 we know c ≠ 0. Therefore we can conclude a = 1. SUFFICIENT.

Answer is C.
_________________

why C please?

the table also shows c * c = j --> f * f = j ; f # j => f # 0 ==> c #0 ==> B is sufficient.

willnguyen2409, that is now how this multiplication table works. Only the first operation that you have mentioned is correct. Sayan has explained the table perfectly in the following manner -

This is like a 3*3 matrix and results in the following products ,
a^2 = d
a*b = e
a*c = f
a*b = e
b^2 = g
b*c = h
c*a = f
c*b = h
c^2 = j

The above calculations are the only ones being shown by the table. Let me know if this doesn't help and you need more clarity on why the answer is C.

still not sure about it. isn't statement 2 say c =f ==> f^2 = j

Yes, that is correct. But, what if C = 0, then A could literally B any value and hence that statement is not sufficient.

We need statement 1 to make sure that C is not equal to 0 as it says that b*c=h and h ≠ 0, which implies that neither C nor B is 0.

Hope that helps!

Given:
a*c = f

(1) h is not 0.

c*b = h --> c can not be 0.

a can take different values. Not sufficient.


(2): c = f

a can be 0 if c and f are both 0:
0*0 = 0

a can be 1:
1*1 = 1
1*10 = 10

a can be negative if c and f are both 0:
-1*0 = 0

a can take different values. Not sufficient.


(1) + (2):

If c can not be 0, a can not be negative.
If c can not be 0, a can not be 0.

a can only be 1:

a*c = f, c = f -->

1*1 = 1
1*10 = 10

Statement 1 is clearly insufficient individually
Statement 2 :
ac=f from table gives a=1 ; however a=0 also satisfies the condition
therefore insufficient
Now staement 1 and 2 together
bc=h ,=>b and c not equal to zero and bc = f , therefore f has to be non-zero
hence a=1 suff
Therefore IMO C

Great Qn; never get thrown off by the really hard looking problem. They’re never too difficult to solve. This is the evidence

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Thanks Basshead. Why is the need to test below two cases when c = 0, a=1 already? And how is a could be any value? Not sure what did I miss? Thanks for your time in advanced.

if c = 0, a = any value
if a = 1, c = any value

Lets take the statements directly,
1) h is non zero
and from table,
cb=h
Therefore, c and b are non zero from 1.
This doesn't help find value of a.
2) c=f
and from table,
f=ac
before you go cancelling the common value c and f. Please note that we can cancel it only if we know that c and f are non zero.
and if we cannot cancel then this statement also doesn't help us get value of a.

Combining, we can say in the eqn. 2, c and f are non zero from enq. 1. So, we can cancel the common value of c and f from statement 2 inference.
and we get a=1.
This is a single answer. hence combined statement is the answer. Ans-C.