**Group:**- GRE - Group for prepration for GRE General Test
**Topic:**- GRE Math Question
**Posted by:**- Nishel R.

1. Question :

A light bulb blinks at 12 noon. It then blinks after 4 seconds, then after 8 seconds, then after 12 seconds and so on. How many times shall it have already blinked until it blinks at12.08pm?

Correct Answer: 15

Explanation:

The seconds after which the light bulb blinks are 0, 4, 8, 12 …These form an AP with common difference 4 and first term 0.

Let the number of times it blinks in 8 minutes be n.

Sum of n terms = 8*60 = 480

Sum of n terms in an AP where a is the first term and d is the common difference is given by

Sn = (n/2)[2a+(n-1)d]

(n/2)[2*0+4(n-1)] = 480

2n^2-2n-480=0

n^2-n-240=0

n^2-16n+15n-240=0

n(n-16)+15(n-16)=0

n=16

The bulb had already blinked 15 times before it blinked at 12.08 pm

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