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

Question

Competition Mode Question

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.

unraveled · Answer

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.
Let Mike’s class has M people and David’s has D. As per question
M + 10 = 2D
Statement 1) D > 5
Testing some numbers for D gives M as follows:
D = 6 M = 2; M < D Yes
D = 9 M = 8; M < D Yes
D = 15 M = 20; M < D No
INSUFFICIENT

Statement 2) M < 10
Taking cue from Statement 1 we test only for M = 9
Thus for M = 9, 2D = 19 which is not possible since D has to be an positive integer.
Therefore M can take only even values for all of which M < D.

PS: To check whether M < or > than D we have to find the value of M = D. Its M = D = 10 beyond which M > D and for D < 10 M < D always. Also note that M will always be even since M + 10 equates to 2D which is even.

SUFFICIENT.

Answer (B)

freedom128 · Answer

No of people Mike's class = M No of people David's class = D M+10 =2D -> M = 2D - 10 Q. M5 If D=6, then M=2*6-10=2. MD (No) NOT SUFFICIENT (2) M<10 If M=2, then D=(2+10)/2=6. M

Sayon · Answer

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

Ilishar · Answer

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.

Nerd100 · Answer

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

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EncounterGMAT · Answer

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

MayankSingh · Answer

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

Karmesh · Answer

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

Archit3110 · Answer

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

Kinshook · Answer

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

exc4libur · Answer

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).

prashant212 · Answer

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

eakabuah · Answer

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 x5. If y=6, x+10=2*6 hence x=2.in this case xy, 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

scorp17 · Answer

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 P<10 --> Suffiecient. My ans is B.

Aderonke01 · Answer

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

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egmat · Answer

Setup:
Let M = Mike's class, D = David's class

From the question stem: M + 10 = 2D → M = 2D - 10

Target: Is M < D?

Key Insight: Substitute M = 2D - 10 into the target question!

Is 2D - 10 < D?
Is D < 10?

So the real question becomes: Is D < 10?

---

Statement 1: David's class has more than 5 people (D > 5)

Case 1: D = 6 → D < 10 → YES
Case 2: D = 12 → D > 10 → NO

Two different answers possible → Not Sufficient

---

Statement 2: Mike's class has fewer than 10 people (M < 10)

M < 10
2D - 10 < 10
2D < 20
D < 10

This gives us a definite YES → Sufficient

---

Principle: In Yes/No DS questions, translate everything in terms of one variable. If a statement lets you definitively answer YES or NO (not both), it's sufficient.

Answer: B

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

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GMAT Data Sufficiency (DS) Questions

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

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