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Snowflake ARA-C01 Exam With Confidence Using Practice Dumps

Exam Code:
ARA-C01
Exam Name:
SnowPro Advanced: Architect Certification Exam
Vendor:
Questions:
182
Last Updated:
May 11, 2026
Exam Status:
Stable
Snowflake ARA-C01

ARA-C01: SnowPro Advanced: Architect Exam 2025 Study Guide Pdf and Test Engine

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SnowPro Advanced: Architect Certification Exam Questions and Answers

Question 1

A new table and streams are created with the following commands:

CREATE OR REPLACE TABLE LETTERS (ID INT, LETTER STRING) ;

CREATE OR REPLACE STREAM STREAM_1 ON TABLE LETTERS;

CREATE OR REPLACE STREAM STREAM_2 ON TABLE LETTERS APPEND_ONLY = TRUE;

The following operations are processed on the newly created table:

INSERT INTO LETTERS VALUES (1, 'A');

INSERT INTO LETTERS VALUES (2, 'B');

INSERT INTO LETTERS VALUES (3, 'C');

TRUNCATE TABLE LETTERS;

INSERT INTO LETTERS VALUES (4, 'D');

INSERT INTO LETTERS VALUES (5, 'E');

INSERT INTO LETTERS VALUES (6, 'F');

DELETE FROM LETTERS WHERE ID = 6;

What would be the output of the following SQL commands, in order?

SELECT COUNT (*) FROM STREAM_1;

SELECT COUNT (*) FROM STREAM_2;

Options:

A.

2 & 6

B.

2 & 3

C.

4 & 3

D.

4 & 6

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Question 2

An Architect has a table called leader_follower that contains a single column named JSON. The table has one row with the following structure:

{

"activities": [

{ "activityNumber": 1, "winner": 5 },

{ "activityNumber": 2, "winner": 4 }

],

"follower": {

"name": { "default": "Matt" },

"number": 4

},

"leader": {

"name": { "default": "Adam" },

"number": 5

}

}

Which query will produce the following results?

ACTIVITY_NUMBER

WINNER_NAME

1

Adam

2

Matt

Options:

A.

SELECT lf.json:activities.activityNumber AS activity_number,

IFF(

lf.json:activities.activityNumber = lf.json:leader.number,

lf.json:leader.name.default,

lf.json:follower.name.default

)::VARCHAR

FROM leader_follower lf;

B.

SELECT

C.

value:activityNumber AS activity_number,

IFF(

D.

value:winner = lf.json:leader.number,

lf.json:leader.name.default,

lf.json:follower.name.default

)::VARCHAR AS winner_name

FROM leader_follower lf,

LATERAL FLATTEN(input => json:activities) p;

E.

SELECT

F.

value:activityNumber AS activity_number,

IFF(

G.

value:winner = lf.json:leader.number,

lf.json:leader,

lf.json:follower

)::VARCHAR AS winner_name

FROM leader_follower lf,

LATERAL FLATTEN(input => json:activities) p;

Question 3

How does a standard virtual warehouse policy work in Snowflake?

Options:

A.

It conserves credits by keeping running clusters fully loaded rather than starting additional clusters.

B.

It starts only if the system estimates that there is a query load that will keep the cluster busy for at least 6 minutes.

C.

It starts only f the system estimates that there is a query load that will keep the cluster busy for at least 2 minutes.

D.

It prevents or minimizes queuing by starting additional clusters instead of conserving credits.