An application needs to make use of the SQS service for sending and receiving messages. The application takes 60 seconds to process a message. Assuming that a queue has been created with the default settings, which one of the following must be implemented?

Answer options:

A.Call the ChangeMessageVisibility API and increase the timeout.
B.Call the DeleteMessage API to increase the timeout.
C.Call the ChangeMessageVisibility API and decrease the timeout. Call the DeleteMessage API to delete the message.
D.Call the DeleteMessage API to delete the message from the queue first. Call the ChangeMessageVisibility API and increase the timeout.

Answer – A
The AWS Documentation mentions the following.
Changes the visibility timeout of a specified message in a queue to a new value. The default visibility timeout for a message is 30 seconds. The minimum is 0 seconds. The maximum is 12 hours.
For example, you have a message with a visibility timeout of 5 minutes. After 3 minutes, you call ChangeMessageVisibility with a timeout of 10 minutes. You can continue to call ChangeMessageVisibility to extend the visibility timeout to the maximum allowed time. If you try to extend the visibility timeout beyond the maximum, your request is rejected.
https://docs.aws.amazon.com/AWSSimpleQueueService/latest/SQSDeveloperGuide/sqs-dg.pdf
Options B and D are incorrect since you first need to call ChangeMessageVisibility API.
Option C is incorrect since the ChangeMessageVisibility API should be used to increase the timeout.
For more information on the visibility timeout, please refer to the below URL-
https://docs.aws.amazon.com/AWSSimpleQueueService/latest/APIReference/API_ChangeMessageVisibility.html