The 800 Gmat Blog

Question:

Parasitic wasps lay their eggs directly into the eggs of various host insects in exactly the right numbers forany suitable size of host egg. If they laid too many eggs in a host egg, the developing wasp larvae would compete with each other to the death for nutrients and space. If too few eggs were laid, portions of the host egg would decay, killing the wasp larvae.
Which of the following conclusions can properly be drawn from the information above?
(A) The size of the smallest host egg that a wasp could theoretically parasitize can be determined from the
wasp’s egg-laying behavior.
(B) Host insects lack any effective defenses against the form of predation practiced by parasitic wasps.
(C) Parasitic wasps learn from experience how many eggs to lay into the eggs of different host species.
(D) Failure to lay enough eggs would lead to the death of the developing wasp larvae more quickly than
would laying too many eggs.
(E) Parasitic wasps use visual clues to calculate the size of a host egg.

Solution:

A:The answer is A
B: New information, from the passage there is no mention of anything to do with defenses
C: Could be true, but we don’t know for sure whether they learn from experience or they just know how to do it (do bees learn to sting from experience)
D: This is half true and half false, which makes it wrong
E: Again no mention of this anywhere in the stimulant so it is wrong

Leaving us with only A. because the stimulant says that wasps lay their eggs in exactly the right numbers, for us to know what this number is we would have to find out from the wasps behaviors

Question:

If x(x - 5)(x + 2) = 0, is x negative?
1) x² – 7x ≠ 0
(2) x² –2x –15 ≠ 0

Spolier:QA:C

Solution:

x(x - 5)(x + 2) = 0 => x= 0 or 5 or -2
(1) x^2 – 7x ≠ 0 => x(x-7) ≠ 0
=> x ≠ 0 & x ≠ 7.
So x=5 or -2 INSUFFICIENT
(2) x^2 –2x –15 ≠ 0
=> (x-5)(x+3)≠ 0
=> x ≠ 5 & x ≠ -3
Therefore x=0 or -2 INSUFFICIENT
Combining,
x ≠ 0, x ≠ 5 => x=-2 SUFFICIENT

The correct answer is C.

Question:

Approved April 24, 1800, the act of Congress that made provision for the removal of the government of the United States to the new federal city, Washington, D.C., also established the Library of Congress.

A. Approved April 24, 1800, the act of Congress that made provision for the removal of the government of the United States to the new federal city, Washington, D.C., also established
B. The act of Congress, which was approved April 24, 1800, making provision for the removal of the government of the United States to the new federal city, Washington, D.C., also established
C. The act of Congress approved April 24, 1800, which made provision for the removal of the government of the United States to the new federal city, Washington, D.C., and established
D. Approved April 24, 1800, making provision for the removal of the government of the United States to the new federal city, Washington, D.C., the act of Congress also established
E. Approved April 24, 1800, the act of Congress made provision for the removal of the government of the United States to the new federal city, Washington, D.C., also establishing

Spoiler QA:A

Solution:

A. Although the sentence is long it is correct. The act of congress that made… also established. The sentence maintains parallelism.
B. Incorrect. Which can not modify a clause.
C. Incorrect. Which can not modify a clause
D. Approved April 24, 1800, making provision for the removal of the government of the United States to the new federal city, Washington, D.C., the act of Congress also established
Incorrect. The sentence is awkwardly constructed.
E. Although the sentence is correct grammatically correct it changes the meaning of the sentence. The act of congress made…, also establishing . The way the sentence is constructed it makes it look as if the act of Congress did blah blah blah somehow also created the Library of Congress. By removing the that, the meaning of the sentence is changed.
The correct answer is A.

Question:

Famed for his masterful use of irony, many of Guy de Maupassant’s short stories have become classics due to the author slowly revealing at the end of each piece a tragic twist of fate.

A. Famed for his masterful use of irony, many of Guy de Maupassant’s short stories have become classics due to the author slowly revealing at the end of each piece a tragic twist of fate.

B. Many of Guy de Maupassant’s short stories have become classics because of how he famously and masterfully uses irony, evident in the slow revelation of a tragic twist of fate at the end of each piece.

C. Famed for using irony in a masterful way, many of Guy de Maupassant’s short stories have become classics because of the author slowly revealing a tragic twist of fate at the end of each piece.

D. Many of Guy de Maupassant’s short stories have become classics because of the author’s famed and masterful use of irony, evidenced in the slow revelation of a tragic twist of fate at the end of each piece.

E. Many of Guy de Maupassant’s short stories have become classics because he slowly revealed a tragic twist of fate at the end of each piece, demonstrating his famed and masterful use of irony.

Spoiler QA:D

Solution:

A. This contains a misplaced modifier. Modifier at start modifies “many of…” not Guy de Maupassant.

B. Because we are referring to Maupassant’s short stories we can not later use the subject pronoun he. He - references possessive “Guy de Maupassant’s” not Guy de Maupassant himself.It’s fine to use a possessive such as his.

C. This contains a misplaced modifier. Modifier at start modifies “many of…” not Guy de Maupassant.

D. This is the correct answer. It correct the possessive error by introducing a noun.

E. Because we are referring to Maupassant’s short stories we can not later use the subject pronoun he. He references possessive “Guy de Maupassant’s” not Guy de Maupassant himself. It’s fine to use a possessive such as his.

The correct answer is D.

Question:

Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.

(A) 648
(B) 2/9
(C) 1/55
(D) 864
(E) 1152
SPOILER: OA:D

Solution:

First I found how many total ways there are to arrange the girls and boys.
GB GB GB GB

4B 3B 2B 1B

G4 G3 G2 G1

Four girls can sit in the first seat. If one girl sits down, 3 girls can sit in the second available seat. so forth . . .
There are 4 x 3 x 2 x 1 ways to arrange the girls
There are 4 x 3 x 2 x 1 ways to arrange the boys
We can swap the orders of each girl-boy pair like this:
BG BG BG BG

So we multiply by 2
So the total is
4!4!2

Now we subtract out the cases where John and Susan sit together.
Think of JS and a single unit.

JS BG BG BG

If JS is in the first slot, we can arrange the rest of the girls and boys in
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
We can not swap them otherwise the alternate order is broken

Now we move JS to the second slot.
BG JS BG BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the third slot
BG BG JS BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the fourth slot
BG BG BG JS
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

So in total we have:
4 x 3!3! = 144 ways

We can also arrange the group with a girl first:
SJ GB GB GB
So in total we have:
4 x 3!3! = 144 ways

So the total ways S and J can sit together is:
3!3!4 + 3!3!4 = 288 ways

The answer would be:
4!4!2 - 3!3!8 = 864

The correct answer is D.

Question:

There are 11 marbles in a jar, six are red, three are blue, and 2 are white. If two marbles are to be drawn from the jar at random, without replacement, what is the probability that two marbles selected both are white?

(A) 1/11
(B) 2/9
(C) 1/55
(D) 2/55
(E) 4/121
SPOILER: OA:C

Solution:

There 11 total marbles, 2 of which are white.

The probability of choosing the first marble to be white is: 2/11

Since we are not replacing the marble, we have only 10 marbles left , 1 of which is white.

The probability of choosing the second marble to be white is: 1/10

So the probability of both marbles being white is 2/11 x 1/10 = 1/55

The correct answer is C.

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Question:

Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.

(A) 648
(B) 2/9
(C) 1/55
(D) 864
(E) 1152
SPOILER: OA:D

Solution:

First I found how many total ways there are to arrange the girls and boys.
GB GB GB GB

4B 3B 2B 1B

G4 G3 G2 G1

Four girls can sit in the first seat. If one girl sits down, 3 girls can sit in the second available seat. so forth . . .
There are 4 x 3 x 2 x 1 ways to arrange the girls
There are 4 x 3 x 2 x 1 ways to arrange the boys
We can swap the orders of each girl-boy pair like this:
BG BG BG BG

So we multiply by 2
So the total is
4!4!2

Now we subtract out the cases where John and Susan sit together.
Think of JS and a single unit.

JS BG BG BG

If JS is in the first slot, we can arrange the rest of the girls and boys in
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
We can not swap them otherwise the alternate order is broken

Now we move JS to the second slot.
BG JS BG BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the third slot
BG BG JS BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the fourth slot
BG BG BG JS
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

So in total we have:
4 x 3!3! = 144 ways

We can also arrange the group with a girl first:
SJ GB GB GB
So in total we have:
4 x 3!3! = 144 ways

So the total ways S and J can sit together is:
3!3!4 + 3!3!4 = 288 ways

The answer would be:
4!4!2 - 3!3!8 = 864

The correct answer is D.

Question:

Find the numbers of ways in which 4 boys and 4 girls can be seated alternatively in a row and there is a boy named John and a girl named Susan amongst the group who cannot be put in adjacent seats.

(A) 648
(B) 2/9
(C) 1/55
(D) 864
(E) 1152
SPOILER: OA:D

Solution:

First I found how many total ways there are to arrange the girls and boys.
GB GB GB GB

4B 3B 2B 1B

G4 G3 G2 G1

Four girls can sit in the first seat. If one girl sits down, 3 girls can sit in the second available seat. so forth . . .
There are 4 x 3 x 2 x 1 ways to arrange the girls
There are 4 x 3 x 2 x 1 ways to arrange the boys
We can swap the orders of each girl-boy pair like this:
BG BG BG BG

So we multiply by 2
So the total is
4!4!2

Now we subtract out the cases where John and Susan sit together.
Think of JS and a single unit.

JS BG BG BG

If JS is in the first slot, we can arrange the rest of the girls and boys in
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways
We can not swap them otherwise the alternate order is broken

Now we move JS to the second slot.
BG JS BG BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the third slot
BG BG JS BG
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

Now we move JS to the fourth slot
BG BG BG JS
Girls:
3 x 2 x 1
Boys:
3 x 2 x 1
or
3!3! = 36 ways

So in total we have:
4 x 3!3! = 144 ways

We can also arrange the group with a girl first:
SJ GB GB GB
So in total we have:
4 x 3!3! = 144 ways

So the total ways S and J can sit together is:
3!3!4 + 3!3!4 = 288 ways

The answer would be:
4!4!2 - 3!3!8 = 864

The correct answer is D.