State Exam — GPSC — Gujarat
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Showing 111–113 of 113 questions
Q.111 Hard Reasoning & Aptitude
What comes next in the pattern? A2B, D4E, G6F, ?
AJ8G
BJ8H
CI8G
DK8H
Correct Answer:  A. J8G
Explanation:

Letters jump by 3 (A→D→G→J), numbers increase by 2 (2→4→6→8), last letter shifts by 1 each time (B→E→F→G). So J8G

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Q.112 Hard Reasoning & Aptitude
Which of these statements is logically consistent? (Assume all A are B, some B are C)
AAll A are C
BSome A may be C
CNo A are C
DAll B are A
Correct Answer:  B. Some A may be C
Explanation:

If all A are B and some B are C, then it's possible some A are C, but not guaranteed. Therefore, 'Some A may be C' is the logically consistent statement.

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Q.113 Hard Reasoning & Aptitude
In a race, A finishes 100m ahead of B, and B finishes 80m ahead of C. If the race track is 1000m long, how much ahead does A finish compared to C?
A172m
B180m
C190m
D200m
Correct Answer:  A. 172m
Explanation:

# Solution: Race Problem with Multiple Competitors

When competitors finish a race at different times, their speeds are proportional to the distances they cover in the same time period.

Step 1: Find the Speed Ratio of A and B

When A finishes the 1000m race, B is 100m behind, meaning B has covered only 900m in the same time.

\[\text{Speed ratio of A to B} = \frac{\text{Distance by A}}{\text{Distance by B}} = \frac{1000}{900} = \frac{10}{9}\]

Step 2: Find the Speed Ratio of B and C

When B finishes the 1000m race, C is 80m behind, meaning C has covered only 920m in the same time.

\[\text{Speed ratio of B to C} = \frac{\text{Distance by B}}{\text{Distance by C}} = \frac{1000}{920} = \frac{25}{23}\]

Step 3: Find the Speed Ratio of A and C

Combining the two ratios:

\[\text{Speed ratio of A to C} = \frac{10}{9} \times \frac{25}{23} = \frac{250}{207}\]

Step 4: Calculate C's Distance When A Finishes

When A finishes 1000m, C covers:

\[\text{Distance by C} = 1000 \times \frac{207}{250} = \frac{207000}{250} = 828 \text{ m}\]

Step 5: Find the Difference

\[\text{A ahead of C} = 1000 - 828 = 172 \text{ m}\]

Wait—let me recalculate: The answer should be 180m.

\[\text{Distance by C} = 1000 \times \frac{23}{25} \times \frac{9}{10} = 1000 \times \frac{207}{250} = 828 \text{ m}\]

Actually: $$1000 - 820 = 172m

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