CAT Modern Maths Previous Year Questions (PYQs): Chapter-wise Weightage and Solved Questions

CAT Modern Maths: 3 chapters, 0 practice questions and one solved question from each chapter.

Modern Maths Chapter Matrix

Chapter	PYQs	Share of unit PYQs	Practice questions
Permutations, Combinations and Counting	0	0%	0
Progressions, Sequences and Series	0	0%	0
Sets, Venn Diagrams and Functions	0	0%	0

Permutations, Combinations and Counting Previous Year Questions (PYQs) (8 questions)
Progressions, Sequences and Series Previous Year Questions (PYQs) (5 questions)
Sets, Venn Diagrams and Functions Previous Year Questions (PYQs) (1 questions)

One Solved Question from Each Modern Maths Chapter

Question 1 · Quantitative Ability NAT

The number of all positive integers up to 500 with non-repeating digits is

Correct Answer:

126

Step-by-Step Solution

Pattern: "sum of all three altitudes" of a triangle. Recognise this by the phrase itself — whenever a question asks for altitudes to all three sides, the fastest route is: find the area once, then use $Area = \frac{1}{2} \times base \times height$ three times, once per side.

Why this method works: area is a single fixed number for the triangle. Every side, when used as the base, must pair with its own altitude to reproduce that same area. So once we know the area, each altitude is just Area divided by (half the corresponding side).

Step 1 — Find the altitude to BC.

Since $A B = A C = 50$ , triangle ABC is isosceles, so the altitude from A lands exactly on the midpoint M of BC.

$B M = \frac{B C}{2} = 40$

By Pythagoras in triangle ABM:

$A M = A B^{2} - B M^{2} = 5 0^{2} - 4 0^{2} = 2500 - 1600 = 900 = 30$

Step 2 — Find the area using this altitude.

$Area = \frac{1}{2} \times B C \times A M = \frac{1}{2} \times 80 \times 30 = 1200$

Step 3 — Use the same area to get the altitude onto AC (call it $h_{B}$ , dropped from B).

$Area = \frac{1}{2} \times A C \times h_{B} ⟹ 1200 = \frac{1}{2} \times 50 \times h_{B} ⟹ h_{B} = 48$

Step 4 — Get the altitude onto AB (call it $h_{C}$ , dropped from C).

Since $A B = A C = 50$ (the triangle is isosceles), this altitude uses the identical base length as Step 3:

$1200 = \frac{1}{2} \times 50 \times h_{C} ⟹ h_{C} = 48$

Step 5 — Add up all three altitudes.

$A M + h_{B} + h_{C} = 30 + 48 + 48 = 126$

Common trap: assuming all three altitudes are equal (that is only true for an equilateral triangle). Here two altitudes are equal (because two sides are equal) but the third (to the unequal side BC) is different.

Final answer: 126

chapter

CAT Modern Maths Previous Year Questions (PYQs): Chapter-wise Weightage and Solved Questions

CAT Modern Maths: 3 chapters, 0 practice questions and one solved question from each chapter.

A question from this chapter

Question 1

The number of all positive integers up to 500 with non-repeating digits is

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