GMAT Problem Solving Discussions

MBA
1530
Happens all the times....

glad to be of help and you can actually thank me my clicking the 'Thank' button rather than send a message too :-).

All the Best...

Alternatively,

Let m = x1*y1*z1 and v = x2*y2*z2
Given f(m) = 9(fn) =>30x1y1z1 = 9(30x2y2z2)
Hence 9x1y1z1 = 9x2y2z2

The ques here implies finding the least diff b/w m and u
Least value of u whose pdt of digits is non-zero = 111
Hence 9(1.1.1) = x1y1z1
least value of m can be 119

hence diff b/w m and v= 119-111 = 8

Hence y not A
1531
Alternatively,

Let m = x1*y1*z1 and v = x2*y2*z2
Given f(m) = 9(fn) =>30x1y1z1 = 9(30x2y2z2)
Hence 9x1y1z1 = 9x2y2z2

The ques here implies finding the least diff b/w m and u
Least value of u whose pdt of digits is non-zero = 111
Hence 9(1.1.1) = x1y1z1
least value of m can be 119

hence diff b/w m and v= 119-111 = 8

Hence y not A

I'm sorry.
I read lster tht definition of f(n) is in terms of exponents and not multiplication of terms.

soory for the above post.
1532
Ans is D

For both a-b and a/b to be an even integer, both a and b have to be even
Also, since a/b is an even int, a>b and a is a multiple of b

(a+2)/2 always give an odd integer whereas (b+2)/2 will always be alwas be even.

Hence OA is D

Jishnu you cannot use a as 2 coz it doesn't satisfy the two conditions.
Morover a small correction for Shri reply: (b+2)/2 can be even or odd...and not necessarily even always...take a=8 and b=4...
(8+2)/2=5 ...true here
(4+2)/2=3 ... is a even.

Regards,
1533

Puys,

Please help me with the following problem.
Q. In a certain animal population, for each of the first 3
months of life, the probability that an animal will die
during that month is 1/10. For a group of 200 newborn
members of the population, approximately how many
would be expected to survive the first 3 months of
life?
(A) 140
(B) 146
(C) 152
(D) 162
(E) 170

I thought the answer was A....probability of newborn death during first 3 months will be 1/10 + 1/10 + 1/10 = 3/10. Hence, 3/10 of 200 are possible deaths.....i.e 60. Therefore number expected to survive is 200 - 60 = 140.
But the OA is B....Can someone please explain how?

Regards
MSD

1534
Puys,

Please help me with the following problem.
Q. In a certain animal population, for each of the first 3
months of life, the probability that an animal will die
during that month is 1/10. For a group of 200 newborn
members of the population, approximately how many
would be expected to survive the first 3 months of
life?
(A) 140
(B) 146
(C) 152
(D) 162
(E) 170

I thought the answer was A....probability of newborn death during first 3 months will be 1/10 + 1/10 + 1/10 = 3/10. Hence, 3/10 of 200 are possible deaths.....i.e 60. Therefore number expected to survive is 200 - 60 = 140.
But the OA is B....Can someone please explain how?

Regards
MSD

The probability of survival for 200 animals for 3 months
=9/10 * 9/10 *9/10 *200=145.8=146
1535

Hi,

Can you explain the answer in detail...

Regards
Madhav

1536

Following is a DS question

Q. Does x = y?

(1) x = y
(2) x^2 = y^2

I have a doubt...Should we consider x and y to be negative after removing the mod sign? Is it typical to assume any number as negative after the mod sign is removed?

MSD

1537
Following is a DS question

Q. Does x = y?

(1) x = y
(2) x^2 = y^2

I have a doubt...Should we consider x and y to be negative after removing the mod sign? Is it typical to assume any number as negative after the mod sign is removed?

MSD

Is the OA E?
1538

Need help with the following question. Can someone please explain in detail?

Team No. of games won
A 4
B 7
C 9
D 2
E 2
X ?

According to the incomplete table above, if each of
the 6 teams in the league played each of the other
teams exactly twice and there were no ties, how
many games did team X win? (Only 2 teams play in
a game.)

A. 4 B.5 C.6 D.8 E.10

Regards
MSD

1539
ashishjha100 Says
Is the OA E?


Yes. The OA is E.....I chose A because I assumed that after removing mod sign both x and y will be negative numbers.

Regards
MSD
1540
Need help with the following question. Can someone please explain in detail?

Team No. of games won
A 4
B 7
C 9
D 2
E 2
X ?

According to the incomplete table above, if each of
the 6 teams in the league played each of the other
teams exactly twice and there were no ties, how
many games did team X win? (Only 2 teams play in
a game.)

A. 4 B.5 C.6 D.8 E.10

Regards
MSD

Is the OA C..........
1541
Hi,

Can you explain the answer in detail...

Regards
Madhav

We need the no: of animals which will survive in (1st month) AND (2nd month) AND (3rd month)

since the probability for an animal to die is 1/10 then the probability for its survival is 1 - 1/10= 9/10

so required no: is =9/10 * 9/10 *9/10 *200=145.8=146
1542

How can 3 be even???

1543
ashishjha100 Says
Is the OA C..........


Yes. The OA is C......can you explain how?

Regards
MSD
1544
Need help with the following question. Can someone please explain in detail?

Team No. of games won
A 4
B 7
C 9
D 2
E 2
X ?

According to the incomplete table above, if each of
the 6 teams in the league played each of the other
teams exactly twice and there were no ties, how
many games did team X win? (Only 2 teams play in
a game.)

A. 4 B.5 C.6 D.8 E.10

Regards
MSD

No: of matches when 6 teams each play twice with each other = 2 * 6C2 = 30

so the total wins should be 30.
The table given in the question sums to 24

so wins by X = 30 - 24 = 6
1545
Need help with the following question. Can someone please explain in detail?

Team No. of games won
A 4
B 7
C 9
D 2
E 2
X ?

According to the incomplete table above, if each of
the 6 teams in the league played each of the other
teams exactly twice and there were no ties, how
many games did team X win? (Only 2 teams play in
a game.)

A. 4 B.5 C.6 D.8 E.10

Regards
MSD

Is the OA C?
1546

The other way of doing this would be

200 * 9/10 = 180 survived ( 1st Month )
180 * 9/10 = 162 survived ( 2nd Month )
162 * 9/10 = 145.8 ~ 146 survived ( 3rd Month )

Hope this helps the understanding.
Jishnu 😃

1547

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?

A. 1:2
B. 4:5
C. 1:1
D. 3:2
E. 5:3

The OA is A.

Can some one please solve this one with reasons!

1548
A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?

A. 1:2
B. 4:5
C. 1:1
D. 3:2
E. 5:3

The OA is A.

Can some one please solve this one with reasons!


Let revenue for 1996 as:
Cars -> a and Trucks ->b

So in 1997 --> revenue from cars : ( a - a * (11/100))
revenue from trucks : ( b + b * (7/100))

Cars + Trucks in 1996 -> a+b

Given that there is a 1% increase -->

(a+b) + (a+b) * (1/100) = ( a - a * (11/100)) + ( b + b * (7/100))

=> a+b = 7b-11a
=> 12a = 6b
=> a/b = 6/12 => 1:2

And so 'A' it is
1549

Car revenues of 96= x Truck revenues of 96= y

Car revenues of 97= 0,89x Truck revenues of 97 = 1,07 y

0,89x+1,07y= 1,01 (x+y)

After number crunching: y=2x or x/y= 1/2

A certain company that sells only cars and trucks reported that revenues from car sales in 1997 were down 11 percent from 1996 and revenues from truck sales in 1997 were up 7 percent from 1996. If total revenues from car sales and truck sales in 1997 were up 1 percent from 1996, what is the ratio of revenue from car sales in 1996 to revenue from truck sales in 1996?

A. 1:2
B. 4:5
C. 1:1
D. 3:2
E. 5:3

The OA is A.

Can some one please solve this one with reasons!