Numbers and Percentages

General Aptitude carries 15 of GATE's 100 marks, the same in every paper. Its quantitative questions need no advanced mathematics, only careful, quick arithmetic. This lesson covers the number facts that appear most often.

Divisibility rules

HCF and LCM

• HCF (GCD): the largest number dividing all of them. LCM: the smallest number all of them divide.
• For two numbers: HCF × LCM = product of the numbers. HCF(84, 126) = 42, so LCM = 84 × 126 / 42 = 252.
• Word problems: "largest tile that fits exactly" → HCF; "when will the bells ring together again" → LCM.
• The smallest number that leaves remainder 3 when divided by 4, 6 or 10 is LCM(4, 6, 10) + 3 = 63.

Remainders and last digits

Work with remainders instead of huge numbers:

• (a × b) mod n = ((a mod n) × (b mod n)) mod n, and similarly for sums.
• Powers repeat in cycles: 2¹, 2², 2³ mod 7 are 2, 4, 1, then the cycle repeats with length 3. Since 100 = 3 × 33 + 1, 2¹⁰⁰ mod 7 = 2¹ mod 7 = 2.
• Last digits repeat with cycle length at most 4: 7¹, 7², 7³, 7⁴ end in 7, 9, 3, 1. Since 2026 = 4 × 506 + 2, 7²⁰²⁶ ends in 9.

Counting multiples

• Multiples of k from 1 to N: ⌊N / k⌋.
• Multiples of k between a and b: ⌊b / k⌋ − ⌊(a − 1) / k⌋. Three-digit multiples of 7: ⌊999/7⌋ − ⌊99/7⌋ = 142 − 14 = 128.
• "Divisible by 2 or 3" from 1 to 100, by inclusion–exclusion: 50 + 33 − 16 = 67.

from math import gcd, lcm

print("HCF(84, 126) =", gcd(84, 126), "| LCM(84, 126) =", lcm(84, 126))
print("2^100 mod 7 =", pow(2, 100, 7), "| last digit of 7^2026 =", pow(7, 2026, 10))
print("three-digit multiples of 7:", sum(1 for n in range(100, 1000) if n % 7 == 0))
print("1..100 divisible by 2 or 3:", sum(1 for n in range(1, 101) if n % 2 == 0 or n % 3 == 0))

HCF(84, 126) = 42 | LCM(84, 126) = 252
2^100 mod 7 = 2 | last digit of 7^2026 = 9
three-digit multiples of 7: 128
1..100 divisible by 2 or 3: 67

Percentages

• x% of y = x·y / 100.
• Percentage change = (new − old) / old × 100.
• Successive changes of a% and b% combine to a + b + ab/100 percent (not a + b). A 20% rise followed by a 20% fall gives 20 − 20 − 400/100 = −4%.
• If A is r% more than B, then B is r/(100 + r) × 100 % less than A. A price that rises 25% must fall 20% to return to the start (25/125 = 20%).

A worked example: a shop raises a price by 15% and then gives a 10% discount. The net change is 15 − 10 − 150/100 = +3.5%: the customer still pays more than the original price.

Practice questions