A new electric car company holds a limited-time sales event. On the fi

Question

A new electric car company holds a limited-time sales event. On the first day, 3 cars are sold. On each subsequent day of the event, 3 more cars are sold than on the previous day. For how many days does the event last?

(1)  If the event had lasted 2 more days , there would have been  84 more cars sold  on the last day of the event than on the first.

(2) On exactly 9 days during the event, the number of cars for the day is a multiple of 9.

Bunuel · Answer

This is a modified version of the following question: a-new-electric-car-company-holds-a-limited-time-sales-event-153087.html

chetan2u · Answer

A new electric car company holds a limited-time sales event. On the first day, 3 cars are sold. On each subsequent day of the event, 3 more cars are sold than on the previous day. For how many days does the event last?
 the first day is 3, next 6 then 9 and so on.. so it is arithmetic progression. In AP we can find the number of terms from the value of term

(1)  If the event had lasted 2 more days , there would have been  84 more cars sold  on the last day of the event than on the first.
 so on last day, the cars sold = 84+3-2*3=81
So 3+(n-1)*3=81......n-1=78/3=26...N=27.. 
Suff

(2) On exactly 9 days during the event, the number of cars for the day is a multiple of 9.
Since it is increasing by 3, every third term will be multiple of 9..
So total terms become 3*9 to 3*10-1, so from 27 to 29..
Three answers possible..
Insufficient

A

dabaobao · Answer

Given: 
Arithmetic Progression with a1 = 3 and d = 3

Wanted: 
n?

1) Last day sales: a1 + (n+2 - 1)d = a1 + (n+1)d

n+2 day sale - First day sale = 84

3 + (n+1)3 - 3 = 84
n+1 = 28
n = 27

Sufficient

2) Gives no new info. We already know from the question stem that all sales are going to be multiple of 3 since a1 = 3 and d = 3.

Not Sufficient

ANSWER: A

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A new electric car company holds a limited-time sales event. On the fi

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