SBI 2024 aptitude
Quantitative, reasoning, and verbal drills with solutions
This page collects SBI placement papers from 2024 with practice questions, worked solutions, and the exam pattern students reported that cycle. Use it when you want drive history: what the first round looked like, which topics repeated, and how to approach solutions. Work the sets below under a timer, then compare with newer material so your prep matches both established SBI patterns and recent shifts.
SBI 2024 aptitude
Quantitative, reasoning, and verbal drills with solutions
SBI 2024 coding
DSA practice aligned to SBI online assessments
SBI interview experience
Round structure and tips from student reports
SBI prep guide
Study plan and weekly schedule
Timed placement-style MCQs with score and explanations after you submit. Use it to check speed and accuracy before the real test.
Quantitative
If the price of a commodity increases by 25% and then decreases by 20%, what is the net change in price?
Correct answer: No change (0%)
Verbal
Choose the correct form: "If I _____ rich, I would travel the world."
Second conditional uses "were" for all subjects (subjunctive mood).
Reasoning
A is taller than B but shorter than C. D is shorter than B. Who is tallest?
C > A > B > D → C is tallest.
Verbal
Synonym of "Abundant":
Abundant means plentiful.
Quantitative
Find x: 3x + 7 = 22
3x = 15 → x = 5.
Reasoning
Find the next number: 6, 11, 21, 41, ?
×2−1 pattern → 81.
Quantitative
If A:B = 3:4 and B:C = 5:6, find A:B:C.
Correct answer: 15:20:24
Reasoning
A walks 9 km north, then 12 km east. Distance from start is:
Right triangle → 15 km.
Reasoning
Odd one out: Cat, Dog, Tiger, Car
Others are animals.
Quantitative
The next term in AP 7, 10, 13, 16, … is:
Common difference = 3 → next = 19.
Verbal
Antonym of "Benevolent":
Benevolent ↔ malevolent.
Quantitative
If the sum of the first 50 natural numbers is S, what is the value of S?
Sum of first n natural numbers = n(n + 1)/2 For n = 50: S = 50 × 51 / 2 = 1275
Quantitative
15 men can complete a work in 20 days. After 5 days, 5 men leave. How many more days will it take to complete the work?
Correct answer: 22.5 days
Reasoning
What is the value of x? Statement I: x² - 7x + 12 = 0 Statement II: x > 2
Statement I: x² - 7x + 12 = 0 (x - 3)(x - 4) = 0 x = 3 or x = 4 Statement II: x > 2 (both 3 and 4 satisfy this) Combining both: x = 3 or x = 4 (both positive and > 2)
Verbal
Fill: Neither the teacher nor the students _____ present.
With neither…nor, verb agrees with nearer subject (students) → were.
Your score
0/15(0%)
| Section | What shows up | Prep focus |
|---|---|---|
| Aptitude / logical | Quant, reasoning, sometimes verbal | Timed sectional accuracy |
| Coding / programming logic | Easy-medium DSA or output-style MCQs | Handle tricky inputs |
| Technical interview | OOPs, DBMS, OS, projects | Explain aloud |
| HR | Fit, location, intent | A few real examples ready |
First round: SBI PO / Clerk / Specialist Cadre Test
Skills emphasized: Aptitude, reasoning, GA, English
Languages: Specialist tech cadres differ
These are practice-style questions aligned to patterns students report for SBI drives around 2024. They are not leaked live papers. Work them timed, then read the solutions only after you have an answer.
Problem: If 30% of a number is 150, what is the number?
Solution: 0.3x = 150 → x = 150 / 0.3 = 500.
Answer: 500
Problem: The average of five numbers is 20. One number 30 is replaced by 10. What is the new average?
Solution: Old sum = 5 × 20 = 100. New sum = 100 − 30 + 10 = 80. New average = 80 / 5 = 16.
Answer: 16
Problem: Find the next term: 2, 6, 12, 20, 30, ?
Solution: Pattern: 1×2, 2×3, 3×4, 4×5, 5×6, 6×7. Next = 6 × 7 = 42.
Answer: 42
Problem: What is the probability of drawing an ace from a standard 52-card deck?
Solution: There are 4 aces in 52 cards. Probability = 4/52 = 1/13.
Answer: 1/13
Problem: A number is increased by 20% and then decreased by 20%. What is the net percentage change?
Solution: Start with 100 → 120 → 96. Net change = 4% decrease. Formula: successive +a then −a gives −(a²/100)% = −4%.
Answer: 4% decrease
Problem: If the price of an item rises by 25%, by what percent should consumption fall so that expenditure stays the same?
Solution: Required reduction = r/(100+r) × 100 with r = 25. = 25/125 × 100 = 20%.
Answer: 20%
Problem: An article is marked 40% above cost and sold after a 10% discount. Find the profit percent.
Solution: SP = CP × 1.4 × 0.9 = 1.26 CP. Profit = 26%.
Answer: 26%
Problem: An article sold at 10% loss would give 5% profit if sold for ₹60 more. Find the cost price.
Solution: 0.9P + 60 = 1.05P 60 = 0.15P P = 60 / 0.15 = ₹400.
Answer: ₹400
Problem: Find the odd one out: 3, 5, 7, 9, 11
Solution: 3, 5, 7, and 11 are prime. 9 = 3×3 is composite, so it is the odd one out.
Answer: 9
Problem: If CAT is coded as DBU, how is DOG coded in the same way?
Solution: Each letter moves +1 in the alphabet: C→D, A→B, T→U. D→E, O→P, G→H → EPH.
Answer: EPH
Problem: Pointing to a photograph, Ravi says, ‘She is the daughter of my mother’s only son.’ How is the girl related to Ravi?
Solution: Ravi’s mother’s only son is Ravi himself (assuming one son). The girl is therefore Ravi’s daughter.
Answer: Daughter
Problem: Statements: All engineers are graduates. Some graduates are managers. Conclusion: Some engineers are managers. Does it follow?
Solution: The ‘some graduates’ who are managers need not overlap with the engineers. The conclusion does not follow necessarily.
Answer: Does not follow
Problem: Move all zeros in an array to the end while keeping the relative order of non-zero elements.
Approach: Two pointers: write non-zeros toward the front, then fill the remainder with zeros. Or swap zeros as you scan.
Complexity: O(n) time, O(1) space
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Rotate an array to the right by k steps. Example: [1,2,3,4,5,6,7], k = 3 → [5,6,7,1,2,3,4].
Approach: Normalize k %= n. Reverse the whole array, reverse the first k elements, then reverse the rest. That yields the rotation in place.
Complexity: O(n) time, O(1) space
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Find the first non-repeating character in a string and return its index, or -1 if none exists.
Approach: Count frequencies in one pass (hash map or array of 26 for lowercase). Second pass returns the first index with count 1.
Complexity: O(n) time
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Design a stack that supports push, pop, top, and getMin in average O(1) time.
Approach: Keep a parallel min-stack (or store pairs). When pushing, also push the new minimum. When popping, pop both stacks.
Complexity: O(1) per operation amortized
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given an integer array, find the contiguous subarray with the largest sum and return that sum. Example: [-2,1,-3,4,-1,2,1,-5,4] → 6 (from [4,-1,2,1]).
Approach: Keep a running sum. If the running sum drops below 0, reset it to 0 before taking the next element (or track the best ending-here value). Track the global maximum as you scan once from left to right.
Complexity: O(n) time, O(1) extra space
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given a mutable character array representing a string, reverse it in place without allocating another array of the same size.
Approach: Use two pointers at the start and end. Swap characters, then move inward until the pointers meet. Watch empty and single-character inputs.
Complexity: O(n) time, O(1) extra space
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Write a function that returns true if n is prime and false otherwise. Handle n < 2 correctly.
Approach: Return false for n < 2. Trial-divide from 2 to floor(sqrt(n)). If any divisor divides n evenly, it is composite; otherwise prime.
Complexity: O(√n) time
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Problem: Given a string containing only ‘()[]’, decide whether the brackets are balanced and correctly nested.
Approach: Scan left to right with a stack. Push opening brackets. On a closing bracket, the stack top must be the matching opener. At the end the stack must be empty.
Complexity: O(n) time, O(n) space
SBI tip: Restate the problem, sketch a brute-force idea, then tighten it. Call out edge cases (empty input, single element, overflow) before you write code.
Students usually say the first round is time-tight - easy marks vanish if you sit too long on one hard question. For SBI, skim the paper in a couple of minutes, mark what you can finish cleanly, and protect accuracy. Languages people commonly use: Specialist tech cadres differ.
| Area | Why it matters at SBI |
|---|---|
| Aptitude | What usually helps you clear the first round |
| Core CS (OOPs / DBMS / OS) | Technical interview depth |
| India’s Largest Public Bank awareness | Helps in managerial / HR conversations |
| Communication | Explain your approach clearly; keep a few real examples ready for HR |
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