Collection of 12 DS questions

(1) (x/3) = 2k+1 where k is an integer
x = 6k + 3
x is an odd integer; so x/4 may or may not have 3 reminder. NSF..

(2) x = 5m where m, a positive integer, could be 1, 2, 3, 4, 5, 6, 7 and so on..
If m = 1 or 2, 4, 5, reminder is not 0.
If m = 3, reminder is 3 or 7. NSF..

From 1 and 2: 5m = 6k + 3
If m = 3, k = 2 and x = 15. Then reminder is 3.
If m = 9, k = 7 and x = 45. Then reminder is 1.
If m = 15, k = 12 and x = 75. Then reminder is 3.

E..

(1) x = 12k+5 ........ k is an integer that could be 1 or 2 or so on........NSF
(2) x = 18m+11 ........ m is an integer that could be 1 or 2 or so on......NSF

1 and 2:
12k + 5 = 18m + 11...............where k>m.
12k = 18m + 6
2k = 3m + 1

If k = 2, m = 1. x = 29 and reminder is 5.
If k = 5, m = 3. x = 60 and r = 4
If k = 8, m = 5................. r = 0. NSF..

Thats E.

When x/3 is divided by 2, the remainder is 1, should be "translated" x/3=2k+1 --> x=6k+3 and not x=6k+6

6. How many perfect squares are less than the integer d?
(1) 23 < d < 33
(2) 27 < d < 37

1) For this condition d = 25 is perfect square of 5. Can we say there are 9, 16... only 2 perfect squares are < d ...? so Suff..
2) For this condition d = 36 is perfect square of 6. Can we say there are 9, 16 and 25.. only 3 perfect squares are < d ...? Suff..

D.

Is this explanation correct ?

9. Is x^2 equal to xy?
(1) x^2 - y^2 = (x+5)(y-5)
(2) x=y

1) x^2 - y^2 = x^2 - 25
=> - y^2 = -25...=> y^2 = 25....y = 5.....Insuff...

2) x = y....Suff..

B...Is there some trap ?

I may be wrong but this is what I could make out from the question stem
option1 - for this we have 23 < d < 33 where d can be 24 upto 32 so we can have 1,4,9,16 and 25 depending on value of d. If d = 24 then perfect squares should be 1,4,9,16. If d = 26 then we can have 1,4,9,16 and 25. Hence insuff
option 2 -for this we have 27 < d < 37 where d can be 28 upto 36. So perfect squares less than d will be 1,4,9,16 and 25. even if d =36 we need to find squares less than 36 so it will have only 5 perfect squares. hence suff

will go with B (I may be totally wrong with my understanding)

statement 1:
==========
x must be divisible by 3 but not by 2.So x must be odd multiple of 3 like 9,15,21,27..etc. Nt suff

statement 2:
==========
x is perfectly divisible by 5. x can be 5,10,15 etc.When we take 5 the rem is 1 and for 10 we get 2 and for 15 we get 3.Nt suff

combining both we x = 15,45,75....Still nt suff

So I will go with option E

Prime factors of 6 are 2 & 3; and prime factors of 9 are 3 & 3


1) 30 has prime factors of 2,5, & 3
6 has prime factors of 2 & 3

Therefore X must have one 5, one or zero 2s, and one or zero 3s.

Possibilities of X are 30 (2 x 5 x 3); 15 (5 x 3), & 10 (2 x 5)

We know LCM has to have one 5, two 3s and one 2. So LCM is 5 x 3 x 3 x 2 = 90

2) 45 has prime factors of 5, 3, 3
9 has prime factors of 3 & 3

X must have one 5 and could have two 3s, one 3, or zero 3s

Possiblities are 45 (5 x 3 x 3); 15 (5 x 3); and 5

We know LCM has to have two 3s, one 2, and one 5.

So LCM is 3 x 3 x 2 x 5 = 90

Answer is D... I think

1) Sufficient ... you can figure out the 100th prime # (from 1 to p+1)
2) Sufficient... you can count how many prime #s are between 1 & 3912 and that is P

D

we know each basket contains atleast 1 orange
1. we can have either 20 baskets having 1 orange each and when number of baskets are halved we will have 10 baskets with 2oranges each
we can also have 10 baskets with 2 oranges each and number of baskets are halved we will have 5baskets with 4oranges each
hence insuff

2. lets consider that we have 10 baskets having 2 oranges each. if the baskets are doubled then we will have 20 baksets each having 1 orange
but option says that on doubling the criteria that each basket has one orange will not suffice. So number of baskets are more than 10. and 20 is the only number which fits the criteria that each baket has atleast one orange.We will have 20 baskets with 1 orange each
Hence suff

will go with B

highlighted part is incorrect. However the answer is correct
the equation in 1 is x^2 - y^2 = (x+5)(y-5) which is equal to
x^2 - y^2 = xy -5x +5y -25 ----eqn1
for x^2 = xy value of x and y needs to be same.
On taking values as +/-5 for both x and y we can show the statement to be insuff

I'll try the ones that haven't been attempted -

4. What is the remainder when the positive integer x is divided by 8?
(1) When x is divided by 12, the remainder is 5.
(2) When x is divided by 18, the remainder is 11.
: E
if x = 29, it satisfies both (1) and (2) and the remainder when divided by 8 is 5.
if x = 65, it also satisfies both (1) and (2) but the remainder when divided by 8 is 1.
So both are Insuff.

7. The integers m and p are such that 2 is less than m and m is less than p. Also, m is not a factor of p. If r is the remainder when p is divided by m, is r > 1.
(1) The greatest common factor of m and p is 2.
(2) The least common multiple of m and p is 30.
: A
(A) implies that both m and p are even since they have 2 as a common factor. If thats the case, and given that m is not a factor of p means that the remainder will never be 0 or 1 so r > 1. So (A) is suff

(B) says that LCM(m,p) = 30. If m=3 and p=10, r =1; but if m = 6, p = 10, r = 4. So (B) is Insuff

8. A scientist is studying bacteria whose cell population doubles at constant intervals, at which times each cell in the population divides simultaneously. Four hours from now, immediately after the population doubles, the scientist will destroy the entire sample. How many cells will the population contain when the bacteria is destroyed?
(1) Since the population divided two hours ago, the population has quadrupled, increasing by 3,750 cells.
(2) The population will double to 40,000 cells with one hour remaining until the scientist destroys the sample.
: A - Need know at what intervals the population doubles.
A says that in 2 hours the population quadrupled so every hour it doubles. Also, say population was x when it got divided 2 hours ago, then at 2 hours population is 4x so increase is 3x = 3750. Solve for x and can find out what the population will be 4 hours from now. Hence Suff.

B only says that the population reached 40000 cells with one hour remaining and says nothing about the rate. Depending on the rate, population can multiply to different levels. Hence not suff.

Thanks for excellent questions Bunuel. It's truly appreciated. Since I really need practice on DS I tried to do them all. Unquestionably, I'm going to be wrong on some of them - so I look forward to seeing your answers.



1.
A. (x/3)/2=k+1 <=> x=6k+3 <=> x= 9,15,21,...
Plug in numbers: 9/4 gives remainder of 1, 15/4 gives remainder of 3 =>
Insufficient
B. x/5 is an integer. Since x>0 that means that x=5k <=> x=5,10,15...
Plug in numbers: 5/4 gives remainder of 1, 10/4 gives remainder of 2 =>
Insufficient
C. Plug in numbers by finding mutual possible values for x, e.g.: 15, 45.
15/4 gives remainder of 3. 45/4 gives remainder of 1. => Insufficient

So answer is E.

2. I have no idea what a gross revenue ratio is. Could someone tell me?


3. Lets formalize the information.The trucker paid a total of 155.76 including taxes of 10%, thus excluding taxes he paid 155.76/1.1

A. We are informed that the trucker paid $0.118 pr. gallon in taxes. Formalizing this
information we get that:
p*1.1 - p = $0.118 <=> p(1.1-1)= $0.118 <=> p= $0.118/0.1 = 1.18. To exclude
taxes just divide by 1.18/1.1. Sufficient
B. Formalize the information. We know the trucker paid a total of 155.76/1.1 excluding
taxes. To get the price pr. gallon

exluding taxes just divide by with 20. Sufficient

Hence answer is D.


4.

A. x/12 = k+5 <=> x=12k+60 <=> x=72, 84, 96,....
Plug in numbers: 72/8 gives a remainder of 0, 84/8 gives a remainder of 4. Insufficient

B. x/18 = k+11 <=> x=18k+11 <=> x=29,47,65
Plug in numbers: 29/8 gives a remainder of 5, 47/8 gives a remainder of 1. Insufficient
C. Since A will always be even while B will always be uneven and there are no mutual values.
Insufficient

Hence answer is E.


5.Remember the formula time*speed=distance <=> t*S=d

A. We are only given relative information on two of the three drivers, hence Insufficient
B. We are told that Marsha drives 9*50 = 450 < (1/3)*1500, so Marsha cannot have driven the longest distance.
That leaves Al and Pablo and therefore this statement is Insufficient
c. We know from (2) that Marsha cannot be the answer. The distance left is 1500-450 = 1050 miles. Moreover, we know
from (1) that Al drove one hour longer than Pablo, 5 miles pr. hour slower. By formalizing information we have
one equation in two unknowns which is Insufficient

Answer is E.


6.

A. Since there is a perfect square between 23 and 33 (25) the number of perfect squares cannot be determined.
Insufficient
B. Since there is a perfect square between 27 and 37 (26) the number of perfect squares cannot be determined.
Insufficient
C. Combining (1) and (2) we know that 27 < d < 33. There are no perfect squares in this interval and thus there
are 5 perfect squares less than d (1, 4, 9, 16, 25). Sufficient

So answer is C.


7. Formalize the information: 2 < m < p. p/m is not an integer.

A. GCF of m and p is 2, implying they are even numbers.
We know that m is not a factor of P. Plug in numbers:
6/4 gives remainder of 2, 6/10 gives remainder of 4. Insufficient
B. LCM of m and p is 30. Break down 30 into its prime factors: 2,3,5.
Since m < p and it is not a factor of p, it can be either 3, 5 implying
that p can be 10 and 6. Plug in numbers:
10/3 gives a remainder of 1, 6/5 gives a remainder of 1. Thus the remainder r=1
and therefore r is not >1. Sufficient

The answer is B.


8. We know that the population doubles at certain intervals and that it doubles (from an unknown quantity)
at exactly 4 hours before being destroyed.

A. We're told that the population has quadrupled in two hours. That means that the population has doubled 3 times
in two hours and therefore it will double 6 times in 4 hours. Furthermore we are told that 4x - x = 3750,
where x is population 2 hours ago <=> x=3750/3 = 1250. Thus the total population will be 1250*2^(3+6).
Sufficient
B. We cannot infer how frequent the population doubles using this information. Insufficient

Answer is A.


9. Is x^2 equal to xy?
(1) x^2 - y^2 = (x+5)(y-5)
(2) x=y

9. Rewrite the expression X^2=xy <=> xx=xy <=> x=y, thus the question reduces to whether x=y.

A. Rewrite (1): (x+y)(x-y) = (x+5)(y-5). This equation is solved for x=y=5, but not for
x=y=0. Thus the information is insuficcient.
B. Since the question reduced to whether x=y and we are told that x=y this is clearly Sufficient

So answer is B.


10. Since there are 20 oranges to be distributed and all baskets contain one orange, there can be a maximum of 20 baskets.
Furthermore since they are evenly distributed only these combinations (orange X baskets) remain: , 10*2, 5*4, 4*5, 2*10,

and 1*20.

A. This gives no additional information and is clearly Insufficient
B. There can be either 20, 10, 5, or 4 baskets. Since there are 20 oranges there number of baskets can be doubled without

problems for 10, 5 and 4 baskets. Thus there must be 20 baskets. Sufficient

So answer is B.


11.Since P is a prime greater than 2 it must be odd.

A. Since p+1 is even and >2 and thus is not a prime number, p must be the 99. (since we are excluding 2)
prime number (have no idea what that is though). Sufficient
B. Again this can be counted straightforward. Find out how many primes there are in the interval and subtract 1
(because of 2). Sufficient

Answer is D.


12. The LCM of 6 and 9 is are both factors of 3, but e.g. 9 is not a factor of 2, so the LCM of 6 and 9 is 18.

A. Since the LCM of x and 6 is 30, x must be a factor of 5. Thus the LCM is 18*5 = 90. Sufficient
B. Since the LCM of x and 9 is 45, x must be a factor of 5. Thus the LCM is 18*5 = 90. Sufficient

So answer is D.

Statement (1) is saying that there are 100 primes in the range between 1 and P+1 is basically telling us that P is the 100th prime . We can determine the single numerical value of P. So it's sufficient.

So D is the answer (as statement 2 is also sufficient).

10. A number of oranges are to be distributed evenly among a number of baskets. Each basket will contain at least one orange. If there are 20 oranges to be distributed, what is the number of oranges per basket?
(1) If the number of baskets were halved and all other conditions remained the same, there would be twice as many oranges in every remaining basket.
(2) If the number of baskets were doubled, it would no longer be possible to place at least one orange in every basket.

I know it's dangerous to assume, but I interpreted the term "distributed evenly" as basically saying that each basket has to have the same number of oranges.

Statement 1

If so, then based on Statement 1 we can have the following:

# of oranges per basket / # of baskets
10 / 2
2 / 10

INSUFFICIENT

Statement 2

For Statement 2, the same rations would meet the requirement.

INSUFFICIENT

Taken together, they are still insufficient... so

Answer: E?

I think this one was the toughest. I've already given the solution for this one previously, so here it is:

Before considering the statements let's look at the stem:
A. Population doubles at constant intervals, but we don't know that intervals.
B. Experiment will end in 4 hours from now.
C. We don't know when bacteria divided last time, how many minutes ago.

(1) Population divided 2 hours ago and increased by 3750 cells. Note that this statement is talking that bacteria quadrupled during 2 hours before NOW. So, starting point 2 hours ago, end of experiment 4 hours from now. Total 6 hours.
This statement gives ONLY the following info:
A. population of bacteria TWO hours ago - 1250.
B. population of bacteria now - 5000.

But we still don't know the interval of division. It can be 45 min, meaning that bacteria divided second time 30 min ago OR it can be 1 hour, meaning that bacteria just divided. Not sufficient.

(2) An hour before the end of experiment bacteria will double 40.000. Clearly insufficient.

(1)+(2) We can conclude that in 5 hours (2 hours before now+3 hours from now) population of bacteria will increase from 1250 to 40.000, will divide 5 times, so interval is 1 hour. The population will contain 40.000*2=80.000 cells when the bacteria is destroyed. Sufficient.

Answer: C.

The point is that from (1) we can not say what the interval of division is, hence it's not sufficient.
Please, tell me if you find this explanation not convincing and I'll try to answer your doubts.

thanx bunuel.... i missed the pt

Hi Bunuel... could you please explain the marked text in red? How did you arrive that the bacteria would divide 5 times..?

Thanks,
JT