IPMAT Indore 2023 · Quantitative Ability · Question 38

Step 1 — count the sample space. Three-digit numbers run from 100 to 999:

999-100+1=900

Step 2 — count the multiples of 3. From 102 to 999:

\frac{999-102}{3}+1=300

Step 3 — count the multiples of 4. From 100 to 996:

\frac{996-100}{4}+1=225

Step 4 — count the multiples of 12 (divisible by both). From 108 to 996:

\frac{996-108}{12}+1=75

Step 5 — inclusion–exclusion.

n\left( 3\text{or}4 \right)=300+225-75=450

Step 6 — take the complement.

n\left( \text{neither}\right)=900-450=450 \Rightarrow P=\frac{450}{900}=\frac{1}{2}

Answer: Option 1 — \frac{1}{2}

Shortcut. Use density instead of counting. Among any 12 consecutive integers, the fraction divisible by neither 3 nor 4 is

\left( 1-\frac{1}{3}\right)\left( 1-\frac{1}{4}\right)=\frac{2}{3}\times \frac{3}{4}=\frac{1}{2}

because 3 and 4 are coprime, which makes the two divisibility events independent.

The range 100–999 contains exactly 900=75\times 12 numbers — a whole number of blocks — so the density is exact and no boundary correction is needed:

P=\frac{1}{2}

Check the block-length condition before using this. The multiply-the-densities shortcut is exact only when the range is a whole multiple of the lcm of the divisors; otherwise it is an approximation and the ends must be counted by hand.