IBPS RRB Clerk Prelims 2024 Reasoning Questions with Solutions

Explanation

Sorting all five words in alphabetical order from left to right gives: BAT DCE LOG SPY TAN

The word at the second position from the left end is DCE.

Explanation

The 3rd, 5th, 6th, and 11th letters of the word DETERIORATE are T, R, I, and E, respectively. Combining these four letters yields multiple valid English words, such as TIRE, TIER, and RITE. As per the given instructions, when more than one meaningful word can be formed, Z is the required response.

Explanation

The building consists of three floors with Flat X to the west of Flat Y. D lives on an odd-numbered floor, and since E lives immediately above D in a different flat, D must live on Floor 1 and E on Floor 2. The condition that as many floors are above D as below F places F on Floor 3. Since D and F live in different named flats, D is in Flat X and F is in Flat Y. A lives north-west of C, placing A on Floor 2, Flat X and C on Floor 1, Flat Y. Finally, B occupies Floor 3, Flat X. Thus, D lives on Floor 1, Flat X.

Explanation

Based on the solved floor-flat arrangement, Flat X is occupied by B on Floor 3, A on Floor 2, and D on Floor 1. Flat Y is occupied by F on Floor 3, E on Floor 2, and C on Floor 1. Comparing the pairs, C and E both live in Flat Y, placing them in the same named flat.

Explanation

B occupies Flat X on Floor 3. Moving two floors below Floor 3 leads to Floor 1. On Floor 1, C resides in Flat Y.

Explanation

Flat X is to the west of Flat Y, meaning Flat Y is the easternmost column. Since F lives in Flat Y of Floor 3, there is no flat further east of F, making the statement that no one lives to the east of F correct.

Explanation

Floor 1 is occupied by D and C, and Floor 3 is occupied by B and F, both of which are odd-numbered floors. A lives on Floor 2, which is an even-numbered floor. Thus, A is the odd one out.

Explanation

From the combined relation J ≤ K < L = M > N ≥ O, comparing J and M yields J ≤ K < M, which simplifies to J < M, verifying Conclusion II. Comparing K and N involves opposite inequality symbols (< and >), so no definite relationship can be established between K and N, making Conclusion I false.

Explanation

From the statement P = Q ≤ R < S = T ≥ U, evaluating Q and T gives Q ≤ R < S = T, which leads directly to Q < T, making Conclusion I true. Evaluating S and U gives S = T ≥ U, which directly yields S ≥ U, making Conclusion II true.

Explanation

Given X ≥ Y < Z = A ≤ B = C, the comparison between X and A contains conflicting directional signs (≥ and <), so Conclusion I cannot be established. Similarly, comparing X and B also involves conflicting signs, so Conclusion II cannot be established.