GMAT Data Sufficiency Discussions

MBA GMAT
1361
stmt 1:

2*(3/2)* (4/3) >1

r=4/3 => 1/r
(9)*(4/3)*(1/4) >1

r = 1/4 => 1/r >1

not suff to ans.

stmt 2

(pqr)^2 >1

=> pqr >1

this again is stmt 1

so no suff.

ans .. E




Statement 2 is only pqr^2 and not (pqr) ^2 ...please correct me if I'm wrong.
1362
If P,Q,R,S are positive inters , and P/Q=R/S , Is R is divisible by 5?


i.p is divisible by 140
ii.Q= 7^x , where x is positive integer




I guess ans should be C since p,q,r,s are integers !

R = P/Q * S

St 1 : P is divisible by 140 ...nothing about Q or S ...Hence, not sufficient
St 2 : Q = 7^x ...again nothing about P or S ...not suff

Combined :
P = 2^2*5*7* m (where m is any integer)
Q = 7^x

Now, since R has to be an integer, (P*S)/Q has to be an integer ...or in other words, all 7's of Q have to be absorbed by P and S combined
and we still have 5 in the numerator ...Hence R has to be a multiple of 5 ..

Sufficient ..Ans C
1363
In the rectangular coordinate system, are the points (a, b) and (c, d) equidistant from the origin?



(1) a/b = c/d


(2) Square root of (a^2) + Square root of (b^2) = Square root of (c^2) + square root of (d^2).

nuttyvarun Says
I guess option B..!!..

akhanna Says
If option b is square root of (a^2 + b^2) = square root of (c^2 + d^2), then option b by itself is sufficient. But if the choice is given as in your post, then I think it should be option e.


I guess the Ans option should be C

St 1 : equal ratio tells us nothing ...not sufficient

St 2 : Sq Root of positive no is positive only
Hence, it translates to a+|b| = |c|+d

Same total does not necessarily mean equidistant points ..
5+3 = 4+4 , but(5,3) and (4,4) are not equidistant ..Not sufficient

Combined :
With same total and same ratio , i guess the co-ordinates are same ..
Have done this is ina hurry ...Would be happy if somebody could disprove this ..

I mean if i randomly select a= 10 and b =5
a+b = 15 ...now, for c+d = 15 and c/d = 2:1, there is no other possibility for (c,d) except 10, 5

IMO ,Ans C
1364
I guess the Ans option should be C

St 1 : equal ratio tells us nothing ...not sufficient

St 2 : Sq Root of positive no is positive only
Hence, it translates to a+b = c+d

Same total does not necessarily mean equidistant points ..
5+3 = 4+4 , but(5,3) and (4,4) are not equidistant ..Not sufficient

Combined :
With same total and same ratio , i guess the co-ordinates are same ..
Have done this is ina hurry ...Would be happy if somebody could disprove this ..

I mean if i randomly select a= 10 and b =5
a+b = 15 ...now, for c+d = 15 and c/d = 2:1, there is no other possibility for (c,d) except 10, 5

IMO ,Ans C


you are right.
a+b=c+d
a/b=c/d
apply compendo to the above-
a+b/d=c+d/d
b=d hence a=b
so coordinates are same.
1365

a=3
b=4

c=4
d=3

a/b ??????>?>?>?> c/d
but a^2 + b^2 = c^2 + d^2 = 25

trying another one;
a=-5
b=10

c=5
d=10
again a/b ??????>?>?>?> c/d
but a^2 + b^2 = c^2 + d^2 = 125

I guess answer should be option B.. your thoughts

I guess the Ans option should be C

St 1 : equal ratio tells us nothing ...not sufficient

St 2 : Sq Root of positive no is positive only
Hence, it translates to a+b = c+d

Same total does not necessarily mean equidistant points ..
5+3 = 4+4 , but(5,3) and (4,4) are not equidistant ..Not sufficient

Combined :
With same total and same ratio , i guess the co-ordinates are same ..
Have done this is ina hurry ...Would be happy if somebody could disprove this ..

I mean if i randomly select a= 10 and b =5
a+b = 15 ...now, for c+d = 15 and c/d = 2:1, there is no other possibility for (c,d) except 10, 5

IMO ,Ans C
1366
a=3
b=4

c=4
d=3

a/b
c/d
but a^2 + b^2 = c^2 + d^2 = 25

trying another one;
a=-5
b=10

c=5
d=10
again a/b
c/d
but a^2 + b^2 = c^2 + d^2 = 125

I guess answer should be option B.. your thoughts


Hey Varun, unfortunately u have chosen bad set of nos on both occassions ..

When we evaluate St 2 only, we are not bothered about the ratios ...So why the urge to take same absolute values .

2 more egs from my end ...
We need to maintain a+b=c+d

eg1 a=15 , b = 5 , c=10 , d=10
But, 15^2 + 5^2 is not equal to 10^2 + 10 ^2

Similarly , a=1, b=3 , c=2, c=2

1^2 +3^2 is not equal to 2^2+2^2

Infinite more egs ...
Hence, Ans C ..

Hope that helps !!
1367

hey.. I just noticed my mistake.. I took square root of the sum of squares.. and not of individual squares.. sorry for that confusion..

1368
Is 1/r>1?
(1) pqr>1
(2) pqr^2>1


Is ans the for the below DS - both statements are suff to ans the q


Is 1/r>1?
(1) pqr>1
(2) pqr^2>1

stmt 1:

2*(3/2)* (4/3) >1

r=4/3 => 1/r
(9)*(4/3)*(1/4) >1

r = 1/4 => 1/r >1

not suff to ans.

stmt 2

(pqr)^2 >1

=> pqr >1

this again is stmt 1

so no suff.

ans .. E

deepakraam Says
Statement 2 is only pqr^2 and not (pqr) ^2 ...please correct me if I'm wrong.


my reasoning for the problem :
For this sum it would not matter whether it is pqr^2 OR (pqr) ^2
Ans should be E in both cases ...

Let us take St 2 as pqr^2

Question : it translates to is 0
St 1 : Tells us nothing, if product of 3 nos is greater than 1, r may or maynot be greater than 1 ...not suff ...

St 2 : pqr^2>1
Now, r^2 is always non negative , hence pq is positive ..
but cannot say whether r is greater than 1 or not ...not suff ..

Combined :
pq is +ve, r is +ve and hence pqr is also +ve, however we cannot still say whether r is greater than 1 or not ...Not suff ...
Hence, Ans E ..

Can give few egs of p,q,r if reqd ...
1369

In what proportions should coffee of type A, type B and type C be mixed to get coffee which costs $8 for every 100 ounces?
1. The ratio in which coffee of type A and coffee of type C are mixed is 2:1.
2. Coffee of type A, B and C costs $4, $10 and $6 respectively for 100 ounces



In what ratio Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?
1. Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio of 2:3.
2. The amount of milk in 100 gallons of Solution 1 is 80 gallons more than that of water in the same solution. Further, 50 gallons of Solution 2 contain 10 gallons more milk than water.

1370
my reasoning for the problem :
For this sum it would not matter whether it is pqr^2 OR (pqr) ^2
Ans should be E in both cases ...

Let us take St 2 as pqr^2

Question : it translates to is 0
St 1 : Tells us nothing, if product of 3 nos is greater than 1, r may or maynot be greater than 1 ...not suff ...

St 2 : pqr^2>1
Now, r^2 is always non negative , hence pq is positive ..
but cannot say whether r is greater than 1 or not ...not suff ..

Combined :
pq is +ve, r is +ve and hence pqr is also +ve, however we cannot still say whether r is greater than 1 or not ...Not suff ...
Hence, Ans E ..

Can give few egs of p,q,r if reqd ...



------------------------------------------------------------------------
pq and r are positive , but individually r is 1 or positive and even more than 1 can not be explained . Hence insufficient.

Answer: Only E.


------------------------------------------------------------------------

1371
In what proportions should coffee of type A, type B and type C be mixed to get coffee which costs $8 for every 100 ounces?
1. The ratio in which coffee of type A and coffee of type C are mixed is 2:1.
2. Coffee of type A, B and C costs $4, $10 and $6 respectively for 100 ounces



In what ratio Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?
1. Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio of 2:3.
2. The amount of milk in 100 gallons of Solution 1 is 80 gallons more than that of water in the same solution. Further, 50 gallons of Solution 2 contain 10 gallons more milk than water.




Answer:

C.

:p
1372

In what proportions should coffee of type A, type B and type C be mixed to get coffee which costs $8 for every 100 ounces?
1. The ratio in which coffee of type A and coffee of type C are mixed is 2:1.
2. Coffee of type A, B and C costs $4, $10 and $6 respectively for 100 ounces


The answer should be c. This is the case of 3 equation and 3 variables.
Both the choices are required to answer the question.

In what ratio Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?
1. Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio of 2:3.
2. The amount of milk in 100 gallons of Solution 1 is 80 gallons more than that of water in the same solution. Further, 50 gallons of Solution 2 contain 10 gallons more milk than water.
Each of the choices can answer the question. so the answer is D.

1373

HI Guys,

I thin answer is E.

Assuming we need a ounces of coffee a,b ounces of cofee b and c ounces of coffee c we have the following

equation 1 tells us c=2a

equation 2 tells me

if i were to have a ,b and c ounces respectively cost me 8 dollrs then

(4a/100) + (10b/100) + (6c/100) = 8
or
4a + 10b + 6c =800 ..this does not give me a:b:c..

even if u use equatoin 1 which is c=2a i dont get a:b:c. So i think answer should be E.

Whats the real answer.

1374
In what proportions should coffee of type A, type B and type C be mixed to get coffee which costs $8 for every 100 ounces?
1. The ratio in which coffee of type A and coffee of type C are mixed is 2:1.
2. Coffee of type A, B and C costs $4, $10 and $6 respectively for 100 ounces


The answer should be c. This is the case of 3 equation and 3 variables.
Both the choices are required to answer the question.

let a,b,c are the required quantity in ounces.the equations formed will be-
a+b+c=100
4a+10b+6c=800
from statement1.a=2c
solving the three equations we will get distint values of a,b,c.
answer C.

In what ratio Solution 1 and Solution 2 be mixed to get a solution which contains water and milk in the ratio of 3:7?
1. Solution 1 contains water and milk in the ratio 1:9 and Solution 2 contains water and milk in the ratio of 2:3.
2. The amount of milk in 100 gallons of Solution 1 is 80 gallons more than that of water in the same solution. Further, 50 gallons of Solution 2 contain 10 gallons more milk than water.
Each of the choices can answer the question. so the answer is D.

second one is also correct.each one is suffecient
1375

ans for the milk and water prblm is "Each statement alone is suff to ans the q"

1376

Doubt with a q already posted here..:o

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Already Posted solution:

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
(10-n)/10 * n/9 * 2 = 7/15 ==> n=3
Hence, option D

But both bulbs are drawn simultaneously, hence order does not matter.
should it not be (10-n)/10 * n/9= 7/15 ==> Complex roots.

Hence only A is sufficient.??

Thanks

1377
Doubt with a q already posted here..:o

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Already Posted solution:

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
(10-n)/10 * n/9 * 2 = 7/15 ==> n=3
Hence, option D

But both bulbs are drawn simultaneously, hence order does not matter.
should it not be (10-n)/10 * n/9= 7/15 ==> Complex roots.

Hence only A is sufficient.??

Thanks


I think the ans is D

total light bulbs =10 and defective =n where n
from 1) nc2/10c2 = 1/15 so we can calc n. so 1) alone is suff

from 2)nc1*(10-n)c1/10c2=7/15 here we get n=3,7. but n
2) alone is suff.

Hence D


1378
I think the ans is D

total light bulbs =10 and defective =n where n
from 1) nc2/10c2 = 1/15 so we can calc n. so 1) alone is suff

from 2)nc1*(10-n)c1/10c2=7/15 here we get n=3,7. but n
2) alone is suff.

Hence D




But why will simple probability without combinations not work here??

Prob of 1 defective and 1 non defective = n/10 * (10-n)/9
(Order does not matter so I don't multiply by 2)

Can you tell y my reasoning is wrong.

Thanks again
1379
But why will simple probability without combinations not work here??

Prob of 1 defective and 1 non defective = n/10 * (10-n)/9
(Order does not matter so I don't multiply by 2)

Can you tell y my reasoning is wrong.

Thanks again

hey natzmyid

I used combinations to find the probability and in turn the value of n.
1380
Doubt with a q already posted here..:o

A box contains 10 light bulbs, fewer than half of which are defective. Two bulbs are to be drawn simultaneously from the box. If n of the bulbs in box are defective, what is the value of n?
(1) The probability that the two bulbs to be drawn will be defective is 1/15.
(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.

Already Posted solution:

(2) The probability that one of the bulbs to be drawn will be defective and the other will not be defective is 7/15.
(10-n)/10 * n/9 * 2 = 7/15 ==> n=3
Hence, option D

But both bulbs are drawn simultaneously, hence order does not matter.
should it not be (10-n)/10 * n/9= 7/15 ==> Complex roots.
Hence only A is sufficient.??
Thanks


Yes order does not matter i.e u cud either choose non defective first and then defective or other way round ..Hence u multiply by 2 ...

Refer bold statement - u invariably have chosen non defective first ..OR u have fixed the postions for each of the 2 outcomes ...

I mean , there are 2 poss such that one of the bulbs to be drawn will be defective and the other will not be defective:

1) Non defective & defective ( ur calculation) i.e (10-n)/10 * n/9
OR
2) Defective and non defective i.e n/10 *(10-n)/9.

Add the 2 eqns and equate to 7/15 ...
This is simply combinations
Correct way has been explained by siddharth ..

Hope that helps !!