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Manhattan Question Set # 10

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by richachampion » Sun Oct 09, 2016 6:02 pm
if b<c,d and c>0 which of the following cannot be true if b,c and d are integers?

a. bcd>0
b. b+cd<0
c. b-cd>0
d. b/cd<0
e. (b^3)cd<0

OA: C
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by regor60 » Wed Oct 12, 2016 10:34 am
As far as I can tell, that answer only works if the question is reworded to state that b<c<d.

As currently worded, it is unclear whether this is the case, so that d could be greater than b, but also less than 0, in which case all of the answers could be true.

Do number lines with different possible scenarios:

0_____b___c___d

0______b___d___c

b_______0___c___d

b_______0____d___c

b________d____0___c


And examine each question using above.
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by fiza gupta » Thu Oct 13, 2016 5:01 am
Given : b<c,d>0,c>0 and a,b,c are integers.

b<c -> so b can be either b>0 or b<0 and cd>0(because c and d are positive integers)

a. bcd>0
can be true if b>0

b. b+cd<0
can be true if b<-cd
example if b=-10, c=2, d=3
-10+(2)(3) = -4<0

c. b-cd>0
b > cd - can never be true because
given b<c (multiplying c with b(being integer) will further increase the value of c)
b < cd (always)

d. b/cd<0
can be true if b<0

e. (b^3)cd<0
can be true if b<0

so c is correct
Fiza Gupta
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by Matt@VeritasPrep » Fri Oct 14, 2016 12:52 am
If this
b<c,d
means b<c and b<d, then they're all possible. Examples:

A:: c = 3, d = 2, b = 1

B:: c = 5, d = 1, b = -10

C:: c = 2, d = -2, b = -3

D:: c = 3, d = 2, b = -1

E:: c = 3, d = 2, b = -1
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by Matt@VeritasPrep » Fri Oct 14, 2016 1:01 am
Since , and < are on the same key, though, I'm going to guess the question reads
b < c < d


Using this, we know that

c > 0

Since d > c, we also have d > 0.

Since c > 0 and d > 0, we must have c*d > 0.

If b - cd > 0, then b > cd. So b > 0.

But if b > 0 and b is an integer, then b cannot be greater than the product of two other positive integers greater than itself. (For instance, we can't have 1 > 2*3, 1 > 3*4, ..., 2 > 3*4, ..., etc.)

So b - cd > 0, or b > cd, is impossible.

(Note that this IS possible if 1 > b > 0. For instance, we could have b = 1/2, c = 2/3, and d = 7/10. It just won't work for integers.)
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by Matt@VeritasPrep » Fri Oct 14, 2016 1:03 am
To delve a little further into my last post, suppose we have

d > c > 0 > b.

Then b > cd is obviously impossible, since it implies negative > positive * positive. So we must have

d > c > b > 0. (If b = 0, then we'd have 0 > pos * pos, which again is nonsensical.)

Now let's work with this. If d > c > b > 0, and b > cd, then we also have c > cd. Since c > 0, dividing both sides by c gives 1 > d. Since d > c, we also have 1 > c.

So since c and d are positive, we have 1 > d > c > b > 0. So there aren't any integer solutions here.
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