A certain roller coaster ride has between 80 and 110 people waiting in

Question

A certain roller coaster ride has between 80 and 110 people waiting in line to board. If riders are let on only in groups of 5 there will be 2 riders that do not get on. If the riders are let on only in groups of 6 all riders will be able to get on the roller coaster. How many riders are in the line?

A. 82
B. 87
C. 90
D. 102
E. 110

A. 82
B. 87
C. 90
D. 102
E. 110

Vinayak Shenoy · Answer

IMO D

let 'x' be the number of riders. Given
x=5k+2 and x=6q also x is between 80-110
so multiples of 5= 80, 85, 90, 95, 100, 105, 110
multiples of 6= 84, 90, 96, 102, 108
Condition: If riders are let on only in groups of 5 there will be 2 riders that do not get on. If the riders are let on only in groups of 6 all riders will be able to get on the roller coaster.
only value satisfying this condition within a group of 80-110 riders is 102.

Abhishek009 · Answer

The number must leave a remainder of 2 when divided by 5 and must be completely divisibly by 6

Among the given options only (D) satisfies...

Another way... 
 
Since the number is completly divisible by 6 , sum of the digits must be divisible by 2 and the digit must be even...

Only (D) satisfies...

One more way -

So, the units digit of the number must be 2 or 7 , check the options, only (A) & (D) satisfies... Now use any of the above ways to eliminate the other option...

GSBae · Answer

Solved in 33 seconds.

If we let X = the number of people on the roller coaster, we know that X is somewhere between 80 - 110.

If riders are let on only in groups of 5 there will be 2 riders that do not get on.  This tells us that X = 2 (mod 5). Therefore, our only possible choices for X are (82, 87, 92, 97, 102, 107).

If the riders are let on only in groups of 6 all riders will be able to get on the roller coaster.   This tells us that X = 0 (mod 6). We can check our available options and see which one is evenly divisible by 6.

82 = 60 + 22 = +4 mod 6.

Noticing this, let's just see how much each additional option increases:

87 = (82  + 5  = 4 + 5 = 9, not divisible by 6

92 = 9 + 5 = 14, not divisible by 6

97 = 92 + 5 = 14 + 5 = 19, not divisible by 6

102 = 97 + 6 = 19 + 5 = 24,  DIVISIBLE BY 6 .

Therefore, 102 = 0 (mod 6) and = 2 (mod 5).

Answer: D

AnisMURR · Answer

N= 5q+2

N= 6K

let s take 102!

102= 5*20+2
102=6*17

So Answer D

wayne10rooney · Answer

The number of riders must be i. multiple of 6 and ii. multiple of 5 with remainder 2 i.e. 5x+2

there are only 3 options satisfying the condition ii
 i.e. 82 87 102

to satisfy condition i number should be multiple of 2 and 3 
only number satisfying this condition is 102

Hence D

Alternate:- When riders are let on only in groups of 6 all riders will be able to get on the roller coaster hence the total number of riders must be multiple of 6.
Only number from the option is 102

Badari · Answer

Total number of riders = 5p + 2 or 6q.

The first number that fits in is 12. Numbers following will be 12 + multiple of 30 ( LCM of 5 and 6).

Therefore, the total number of riders will be 30k+12.

102 is the only possibilities as the number of riders is between 80 to 110.

Among the given options 102 is the only answer.

deenesh2309 · Answer

Here we are looking for number which is inbetween 80 and 110 and leaves 2 when divided by 5 and completely divisible by 6.
Odd option (B) can be ruled out straight. Validate rest of 4 options. Only D matches.

GMATinsight · Answer

"If riders are let on only in groups of 5 there will be 2 riders that do not get on"
i.e. No. of Riders = 5a+2

If the riders are let on only in groups of 6 all riders will be able to get on the roller coaster
i.e. No. of Riders = 6b

A multiple of 6 between 80 and 110 may be 84, 90, 96, 102, 108

Among them a no. which divided by 5 leaves remainder 2 = 102

Hence, No. of Riders = 102

Answer: Option D

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A certain roller coaster ride has between 80 and 110 people waiting in

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