Caselets – NABARD Grade A Quantitative Aptitude
(For the NABARD Grade A exam – Quantitative Aptitude – 2026)
Caselets are short, real‑world “story” problems that test a combination of arithmetic, algebra, geometry, probability, rate‑time‑work, and other quantitative skills.
They are usually 2–4 lines long, involve multiple steps, and require a logical, systematic approach.
1. Simple Explanation
| What | Why it matters | How to read it |
|---|---|---|
| Definition | “Caselet” = case + let → a small case or story problem. | Look for the problem statement (the “story”) and the question (the exact thing you must find). |
| Structure | 1. Context – sets up a real‑world situation (e.g., farm, factory, travel). 2. Data – numbers, percentages, ratios, etc. 3. Question – what is asked (value, ratio, probability). |
Separate the context from the data. Use the data to build equations or compute directly. |
| Skill Mix | Usually requires two or more quantitative concepts. | Identify all concepts you might need (e.g., “rate‑time‑work” + “ratio”). |
Goal: Turn the information into a concise set of equations or formulas, solve, and round to required precision.
2. Key Points to Remember
| # | Concept | Quick Tips | Common Pitfalls |
|---|---|---|---|
| 1 | Reading the question | Highlight the exact quantity sought. | Assuming you need to find the wrong variable. |
| 2 | Identifying variables | Assign letters to unknowns (e.g., x, y). |
Mixing up variables when there are multiple unknowns. |
| 3 | Translating text to math | Convert words → equations: • “X is 30 % more than Y” → X = Y + 0.3Y = 1.3Y• “Z takes 5 days to finish work, B takes 7 days” → 1/Z + 1/B = 1/(Z+B) |
Mis‑interpreting “more than” vs. “less than”. |
| 4 | Rate‑Time‑Work | Use rate = work/time. For combined work: rate1 + rate2 = combined rate. |
Forgetting to convert to same time unit. |
| 5 | Ratio & Proportion | If A:B = m:n, then A = mk, B = nk. |
Wrong conversion of percentages to fractions. |
| 6 | Probability | Count favourable / total outcomes. Include “given that” conditions. | Ignoring conditional events. |
| 7 | Time‑Distance‑Speed | speed = distance / time; distance = speed × time. |
Mixing up units (km vs. m). |
| 8 | Algebraic Manipulation | Keep equations tidy; isolate one variable at a time. | Over‑complicating; solving for the wrong variable. |
| 9 | Checking | Plug answer back into original statement; verify units. | Skipping the sanity check. |
| 10 | Rounding | Follow exam instructions: usually to 1 decimal or nearest integer. | Rounding too early (loses precision). |
3. Worked Example (Full Caselet)
Caselet Problem
“A farmer has 300 kg of a mixture of rice and wheat. He sells 40 % of the mixture at ₹8 per kg and the remaining mixture at ₹10 per kg. The total revenue is ₹2 600. Find the amount of rice in the mixture.”
Step‑by‑Step Solution
| Step | Explanation | Equation |
|---|---|---|
| 1. Define Variables | Let R = kg of rice, W = kg of wheat. |
R + W = 300 |
| 2. Use Revenue Conditions | 40 % of mixture sold at ₹8/kg → 120 kg sold at ₹8/kg. 6 %? Wait 40 % of 300 = 120 kg. |
Revenue1 = 120 × 8 = 960 |
| 3. Remaining Mixture | 300 − 120 = 180 kg sold at ₹10/kg. | Revenue2 = 180 × 10 = 1800 |
| 4. Total Revenue Check | 960 + 1800 = 2760 (but problem says 2600). → Inconsistent? Let's adjust: maybe 40 % sold at ₹8, rest at ₹10 leads to 2760, not 2600. So maybe the mixture has different composition. Re‑think: We need to consider the composition of the 120 kg that were sold at ₹8 may not be all rice/wheat. Actually we should treat revenue as weighted average. | |
| 5. Weighted Average Approach | Let p be proportion of rice in the 120 kg sold at ₹8. Rice price? Suppose rice and wheat sold at same price? The problem statement ambiguous. Let's reinterpret: The farmer sells the entire mixture at two rates: 40 % of the total mixture at ₹8/kg and the remaining at ₹10/kg. The revenue is 2600. We need to find how many kg of rice are in the 300 kg mixture. Since rice and wheat are sold at the same rates (no differentiation), the composition of rice vs wheat does not affect revenue. Thus the problem seems unsolvable as stated. Maybe the correct problem is: “The farmer sells 40 % of the rice at ₹8/kg and the rest of the rice at ₹10/kg.” Let's change to that scenario. |
|
| 6. Corrected Problem | “A farmer has 300 kg of a mixture of rice and wheat. He sells 40 % of the rice at ₹8 per kg and the remaining rice at ₹10 per kg. The total revenue is ₹2 600. Find the amount of rice in the mixture.” | ** |