If Mike's class had 10 more people on it, it would have twice as many

Let, M = no. of students in Mike's class
D = no. of students in David's class

GIven: M + 10 = 2 D
Is M < D ?

(1) David's class has more than 5 people on it.
D > 5

Let D =6
M + 10 = 12 => M = 2 . So M < D

Let D = 20
M + 10 = 40 => M = 30. So M > D

Not Sufficient

(2) Mike's class has fewer than 10 people on it.
M < 10

Let M = 8
8 + 10 = 2 D => D = 9. M < D

Let M = 2
2 + 10 = 2 D => D = 6. M < D

Sufficient.

Answer B

given: m+10 = 2d

St1: d>5
(m+10)/2 >5 => m >0 and d>5
taking values d as 6 =>
if m is 2 then m<d
but if m is 7 then, m > d => N.Suff.

St2: m<10 => No info of d => Not Suff.

Combining both the sts. => 0<m<10 and d>5
Taking same values i.e. d as 6 =>
if m is 2 then m<d
but if m is 7 then, m > d => Not Suff. => ans . should be E

Waiting for OA.

The answer is: (1)David's class has more than 5 people on it.

Let's assume the number of people in Mike class is 1(since it is the least number of people we can use), 1+10=11.

Mike class would have twice that of David's class means David's class will be 1/2 of 11 which is 5.5,and it's greater than 5

Posted from my mobile device

Given: M+10=2D
To find: Is M<D?

(1) David's class has more than 5 people on it.
D>5

Let's test numbers.
Let D=6, then M=2...........M<D...YES
Let D=10, then M=10...........M=D...NO
Let D=11, then M=12...........M<D...NO

INSUFFICIENT!

(2) Mike's class has fewer than 10 people on it.
M<10
M+10=2D.....If you notice this equation, M+10 has to be a multiple of 2, isn't? So let's test M with 2,4,6 & 8.

Let M=2, then D=6...........M<D...YES
Let M=4, then D=7...........M<D...YES
Let M=6, then D=8...........M<D...YES
Let M=8, then D=9...........M<D...YES

SUFFICIENT!

IMP OPTION B

IMO-B

Let People in Mike Class = M , David Class = D
A/Q, M+10=2D

M=2D-10

(1) David's class has more than 5 people on it.
D>5
Therefore, D= { 6,7,8,9,10,11,.......}
Corresponding M= (2,4,6,8,10,12......}

Clearly , for D<10 , M<D
& D>10, M>D

Not sufficient

(2) Mike's class has fewer than 10 people on it.
M<10
Also , M>0 , so D>5
Now, M=2D-10
Therefore, D= {6,7,8,9}
Corresponding M= (2,4,6,8}
Clearly, D>M

Sufficient

This question has a nicely laid C trap.

Let #mike's class = M and #david's class = D
So,
M+10 = 2D

The question asks if D > M

S1:
D>5
We have no restriction on M, so if M=6 then D=8... This gives us D>M
BUT, if M=10 then D=10.....This gives M=D
INSUFF

S2:
M<10
Try any value of M 2,4,6,8 you will always have D>M
Therefore, SUFF
[B]

If Mike's class had 10 more people on it, it would have twice as many people as David's class. Are there fewer people on Mike's class than on David's?

(1) David's class has more than 5 people on it.

(2) Mike's class has fewer than 10 people on it.

given
x+10=2y
need to find whether x<y

#1
David's class has more than 5 people on it.
y>5 ;
wont be sufficient when y=10 we get x=10
#2
Mike's class has fewer than 10 people on it.
sufficient always as x will be even integer and y will be more than x
IMO B

Given: If Mike's class had 10 more people on it, it would have twice as many people as David's class.

Asked: Are there fewer people on Mike's class than on David's?

Let people in Mike's class be m and people in David's class be d.
m+10 = 2d. (1)

Asked: Is m<d?
m+10=2d
For m<d; 2d-10<d =>d<10 => 2d <20 => m+10<20 => m<10

(1) David's class has more than 5 people on it.
d>5
NOT SUFFICIENT

(2) Mike's class has fewer than 10 people on it.
m<10
SUFFICIENT

IMO B

is m<d?
m+10=2d… m=2d-10
is 2d-10<d… d<10?

(1) David's class has more than 5 people on it: d>5… is d<10? insufic;
(2) Mike's class has fewer than 10 people on it: m<10… 2d-10<10… d<10; sufic.

Answer (B).

Let M be the number of students in Mike's class and D be the number of students in David's class. The question stem gives us the equation 2D = M + 10.

We are being asked if M<D.

2D = M + 10 => 2D - M = 10. Since D and M cannot be negative integers, we have following cases

D = 5 , M = 0;
D = 6 , M = 2;
D = 7 , M = 4;
D = 8 , M = 6;
D = 9 , M = 8;
D = 10 , M = 10;
D = 11 , M = 12;
so on and so forth...

As we see that when M hits 10, D = M and then the value of D gets lower than M

Lets see the given conditions now.

(1) David's class has more than 5 people on it.

D can be any one of the cases mentioned above. Cant say.

2)Mike's class has fewer than 10 people on it.

If M < 10 then D > M. Thus option B gives the answer to the question.

OPTION B

Let number of people in Mike's class be x and number of people in David's class be y.
Then from the question stem, we can form the following equation:
x+10=2y.......... (1)

We are to determine if Mike's class has fewer people than that of David. i.e. is x<y?

1. David's class has more than 5 people in it.
This implies y>5.
If y=6, x+10=2*6 hence x=2.in this case x<y, hence Mike's class has fewer people than David's class.

However, if x=100,
x+10=2*100 hence x=190. In this case x>y, implying Mike's class has more people than David's class.
Statement 1 on its own is not sufficient.

2. Mike's class has fewer than 10 people in it.
Implying x<10.
For x to satisfy (1), then x must be an even number. So possible values of x are {8,6,4,2}.
When x=8, 8+10=2y meaning y=9.
x<y hence we can answer yes.
When x=2, 2+10=2y meaning y=6
x<y. Meaning Mike's class has fewer people than David's class.

Hence statement 2 on its own is sufficient.

The answer is therefore B.

Posted from my mobile device

david's class = D people D>=0
Mike's class = P people --> P >=0
Given:
P+10 = 2D
D=5+P/2 --> D>=5 and P must be even.
Question is -> P<D
p<5+p/2
p/2<5
p<10?

i)David's class has more than 5 people on it. not sufficient
ii)Mike's class has fewer than 10 people on it.-> P<10 --> Suffiecient.

My ans is B.

Stem: m + 10 =2d

S1. if d is 6, 7, 8, M would be 2, 4, 6 respectively. Hence m < d. but if d is 10, M = 10. Hence Not sufficient


s2. if m equals 9, 8, .....2, d equals 9.5, 9 and 6 . respectively. Hence M < D. Suff

Posted from my mobile device