UPSC CSAT Quantitative Aptitude: Number System, HCF-LCM, Percentages, Ratio-Proportion

Must?Know

• Natural numbers start from 1; whole numbers include 0; integers include negative numbers, zero, and positives; rational numbers are expressible as p/q where q-0.
• The sum of first n natural numbers is n(n+1)/2; sum of first n odd numbers is n²; sum of first n even numbers is n(n+1).
• A prime number has exactly two distinct positive divisors: 1 and itself; 2 is the only even prime number; 1 is not a prime.
• Twin primes are pairs of primes differing by 2 (e.g., 11 and 13); co-prime numbers have HCF = 1 (e.g., 8 and 15).
• Every composite number can be expressed as a product of primes uniquely (Fundamental Theorem of Arithmetic); e.g., 60 = 2² × 3 × 5.
• For any two numbers a and b, a × b = HCF(a,b) × LCM(a,b); if a and b are co-prime, LCM = a × b.
• HCF of fractions = HCF of numerators / LCM of denominators; LCM of fractions = LCM of numerators / HCF of denominators.
• The smallest number divisible by a set of numbers is their LCM; the largest number dividing all is their HCF.
• When dividing a number by another, Dividend = Divisor × Quotient + Remainder; remainder is always less than divisor.
• If a number leaves the same remainder r when divided by a, b, c, then required number is LCM(a,b,c) × k + r.
• Percentage means per hundred; x% = x/100; 25% = 1/4, 33.33%-1/3, 66.66%-2/3, 75% = 3/4.
• Successive percentage changes: a% increase followed by b% increase results in net change = a + b + (ab/100)%.
• If A is r% more than B, then B is less than A by [r/(100+r)] × 100%; e.g., if A is 25% more than B, B is 20% less than A.
• Population after n years, if increasing at r% p.a., is P(1 + r/100)?; decreasing: P(1 – r/100)?.
• Ratio a:b represents a/b; if two ratios a:b and c:d are equal, then a/b = c/d-ad = bc (proportion).
• If a/b = c/d, then (a+b)/(a–b) = (c+d)/(c–d) (componendo and dividendo); applicable only if a-b and c-d.
• If a quantity is divided in the ratio m:n, parts are [m/(m+n)]×total and [n/(m+n)]×total.
• In direct proportion, a/b = constant; in inverse proportion, a × b = constant; e.g., speed and time for fixed distance.
• If A:B = 2:3 and B:C = 4:5, then A:B:C = 8:12:15 (by making B same via LCM of 3 and 4 = 12).
• To compare fractions, cross-multiply: 3/7 vs 4/9-3×9 = 27, 4×7 = 28-4/9 > 3/7.
• Decimal to fraction: terminating decimals (e.g., 0.25 = 1/4); recurring decimals (e.g., 0.333... = 1/3).
• A number ending in 0, 2, 4, 6, 8 is divisible by 2; sum of digits divisible by 3-divisible by 3; last two digits divisible by 4-divisible by 4.
• Divisibility by 11: difference between sum of digits at odd places and even places is 0 or divisible by 11.
• The number of trailing zeros in n! is determined by the number of times 5 appears in prime factorization: floor(n/5) + floor (n++ floor(n/25) + floor(n/125) + ...
• For two numbers, if HCF = H and ratio = a:b (in lowest terms), then numbers are Ha and Hb.

Difficulty Level

Intermediate – Questions often combine multiple concepts (e.g., ratio with percentage or LCM with remainders), requiring careful interpretation and calculation.

Common UPSC Traps

Trap: "LCM of fractions is calculated like integers" – Fact: LCM of fractions = LCM of numerators / HCF of denominators (not LCM of denominators).
Trap: "If A is 20% more than B, then B is 20% less than A" – Fact: B is less than A by (20/120)×100-16.67%, not 20%.
Trap: "1 is a prime number" – Fact: 1 has only one positive divisor; primes must have exactly two distinct divisors.
Trap: "HCF can be greater than the smaller number" – Fact: HCF of two numbers cannot exceed the smaller number; it is a common divisor.

Practice MCQs

Question: What is the least number which when divided by 6, 9, 12, and 18 leaves a remainder of 3 in each case?
A) 39 B) 63 C) 75 D) 111
Answer: A
Explanation: LCM of 6, 9, 12, 18 is 36; required number = 36k + 3; smallest such number is 36×1 + 3 = 39.
Why others fail: D (111) is 36×3 + 3, which is valid but not the least.

Question: Two numbers are in the ratio 3:5. If each is increased by 10, the ratio becomes 5:7. What is the sum of the numbers?
A) 24 B) 32 C) 40 D) 48
Answer: C
Explanation: Let numbers be 3x and 5x; (3x+10)/(5x+10) = 5/7-7(3x+10) = 5(5x+10)-x = 5; sum = 8x = 40.
Why others fail: B (32) may result from incorrect equation setup or solving error.

Question: A number is increased by 25% and then decreased by 25%. The net change is:
A) 6.25% decrease B) 5% decrease C) 0% change D) 6.25% increase
Answer: A
Explanation: Net change = 25 – 25 – (25×25)/100 = –6.25%; decrease of 6.25%.
Why others fail: C (0%) is tempting, assuming symmetry, but percentage changes are not symmetric.

Question: The HCF and LCM of two numbers are 12 and 240 respectively. If one number is 60, the other is:
A) 48 B) 50 C) 52 D) 54
Answer: A
Explanation: Product of numbers = HCF × LCM = 12 × 240 = 2880; other number = 2880 / 60 = 48.
Why others fail: B (50) may result from misusing formula or arithmetic error.

Question: What is the number of trailing zeros in 100!?
A) 20 B) 22 C) 24 D) 26
Answer: C
Explanation: Number of 5s = floor(100/5) + floor(100/25) + floor(100/125) = 20 + 4 + 0 = 24.
Why others fail: A (20) counts only multiples of 5, missing higher powers like 25, 50, 75, 100.

Last?Minute Revision

• Sum of first n natural numbers: n(n+1)/2
• Sum of first n odd numbers = n²
• 2 is the only even prime number
• 1 is neither prime nor composite
• HCF × LCM = product of two numbers
• LCM of fractions = LCM(numerators)/HCF(denominators)
• HCF of fractions = HCF(numerators)/LCM(denominators)
• 25% = 1/4, 20% = 1/5, 12.5% = 1/8
• 33.33% = 1/3, 66.66% = 2/3
• Net % change after a? and b? = a – b – ab/100
• If A:B = a:b, B:C = c:d-A:B:C = ac : bc : bd
• For same remainder r, number = LCM(divisors) × k + r
• Divisibility by 11: (sum odd digits – sum even digits) ÷ 11
• Trailing zeros in n! = sum of floor(n/5^k) for k=1,2,...
• Componendo and dividendo: if a/b = c/d? (a+b)/(a–b) = (c+d)/(c–d)
• Product of two numbers = HCF × LCM
• Co-prime numbers have HCF = 1
• Twin primes: difference = 2 (e.g., 17,19)
• Fundamental Theorem of Arithmetic: unique prime factorization
• 0.333... = 1/3, 0.1666... = 1/6
• Ratio a:b implies a/b
• Inverse proportion: a × b = constant
• Direct proportion: a/b = constant
• Population after n years at r% growth: P(1 + r/100)^n
• If A is r% more than B, B is [r/(100+r)]×100% less than A
• verify from standard source