Number Systems, Percentages & Ratio-Proportions
“Divisibility rules, remainder theorems, successive percentages, profit & loss, and mixture alligation.”
1.Core Quantitative Principles for GATE
§ 1.1.General Aptitude contributes 15 marks out of 100 in every GATE paper (5 questions of 1 mark each and 5 questions of 2 marks each). Scoring 13+ marks is the single highest-ROI strategy for securing a top-500 rank.
Successive Percentage Changes: When a quantity is modified by +x% followed by +y%, the net multiplier is (1 + x/100)(1 + y/100) = 1 + (x + y + xy/100)/100. For discounts (-d1% and -d2%), Net Discount = d1 + d2 - (d1 * d2) / 100.
Rule of Alligation: When two ingredients of prices C (cheaper) and D (dearer) are mixed to produce a mean price M, the ratio of weights is (Quantity of C) / (Quantity of D) = (D - M) / (M - C).
Modular Arithmetic & Remainders: Euler's Totient Theorem: If gcd(a, m) = 1, then a^{phi(m)} = 1 (mod m). For prime p, Fermat's Little Theorem states a^{p-1} = 1 (mod p).
If petrol price rises by 25% (+1/4), consumption must fall by (1 / (4 + 1)) = 1/5 = 20% to keep expenditure unchanged. If price rises by 1/n, consumption drops by 1/(n+1). If price falls by 1/n, consumption can rise by 1/(n-1).
2. Worked Problem: Repeated Dilution & Mixture Ratio
Do not subtract linear amounts; each removal takes away a mixture containing both wine and water.
Problem [Original Practice Problem · Modeled on GATE Pattern]:
A cask initially contains 80 liters of pure chemical solvent. 16 liters of the solvent are drawn off and replaced with water.
This operation is repeated two more times (total 3 operations).
What is the final quantity of chemical solvent remaining in the cask?
Step 1: Identify container capacity C, withdrawal amount x, and iterations n:
C = 80 liters, x = 16 liters, n = 3.
Step 2: Compute fractional dilution factor per operation:
Fraction remaining = (1 - x / C) = (1 - 16 / 80) = (1 - 1/5) = 4/5 = 0.8.
Step 3: Apply the repeated dilution formula:
Solvent remaining = C (1 - x / C)^n = 80 (4/5)^3
Solvent remaining = 80 (64 / 125) = (80 / 125) 64 = (16 / 25) * 64
Solvent remaining = 1024 / 25 = 40.96 liters.
Conclusion: 40.96 liters of chemical solvent remain, and (80 - 40.96) = 39.04 liters of water are present.
Primary Academic References & Exam Sources
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