Practice Library
All MCQs
Browse exam-wise, subject-wise, and country-wise MCQs with explanations.
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Correct Answer: A. Making read-only reference files available to task nodes
Explanation:
Small lookup tables or configuration files can be localized with tasks.
This supports efficient map-side enrichment.
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Correct Answer: D. Too many distinct counters create coordination and memory overhead
Explanation:
Counters are aggregated through the framework and are not intended as arbitrary per-key storage.
A small, meaningful set is most effective.
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Correct Answer: C. A distributed metric aggregated across task attempts
Explanation:
Counters track events such as malformed records or processed rows.
They are useful for monitoring and data-quality checks.
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Correct Answer: B. Successful mapper outputs remain available on their local nodes for refetching
Explanation:
The new reducer attempt retrieves the same intermediate partitions again.
Lost mapper output may require only the affected mappers to be rerun.
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Correct Answer: A. It reruns the map task, possibly on another node
Explanation:
Map input remains in HDFS and can be read again.
The scheduler chooses an available location for the retry.
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Correct Answer: D. One execution instance of a map or reduce task
Explanation:
A task can have multiple attempts because of failures or speculation.
Only a successful committed attempt contributes final output.
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Correct Answer: C. When tasks write to external systems with non-idempotent side effects
Explanation:
Duplicate attempts can repeat external updates or messages.
Side effects must be idempotent or speculation should be controlled.
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Correct Answer: B. Running duplicate attempts of unusually slow tasks
Explanation:
The framework may launch another attempt on a different node.
The first successful attempt is accepted, reducing straggler impact.
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Correct Answer: A. Adding an extra value to a hot key to spread its records across partitions
Explanation:
Salting creates multiple temporary versions of a frequent key.
A later stage combines the partial results.
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Correct Answer: D. A key associated with an exceptionally large number of values
Explanation:
A hot key can overload one reducer under ordinary partitioning.
Salting or specialized aggregation may distribute its workload.
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Correct Answer: C. An uneven distribution of keys or records across tasks
Explanation:
Skew causes some tasks to process much more data than others.
Stragglers then delay the entire job.
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Correct Answer: B. All intermediate keys are sent to a single reducer and final output can be globally sorted by key
Explanation:
One reducer receives all partitions, creating a global key order.
This can become a serious scalability bottleneck.
