Mixed-topic aptitude

This chapter consolidates fundamental concepts across logic, algebra, number theory, and calculus, presenting them within a mixed-topic aptitude framework. Proficiency in these integrated areas is crucial for success, as they form the bedrock for advanced problem-solving frequently encountered in the CMI examination.

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Chapter Contents

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| Topic |

|---|-------| | 1 | Logic plus sets and functions | | 2 | Algebra plus geometry | | 3 | Number theory plus combinatorics | | 4 | Calculus plus function reasoning |

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We begin with Logic plus sets and functions.

Part 1: Logic plus sets and functions

Logic plus Sets and Functions

Overview

Set theory and functions are natural homes for logical reasoning. Many aptitude problems in this area are not computational; they test whether you can understand statements precisely, negate them correctly, and connect logical structure to set inclusion and function properties. This topic trains that exact reasoning. ---

Learning Objectives

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Core Logical Tools

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Sets and Logic

So subset proofs are really implication proofs. ---

De Morgan's Laws

These are among the most important set identities. ---

Functions and Logic

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Negating Function Properties

This is very important in logic-based function questions. ---

Minimal Worked Examples

Example 1 Show that if $A\subseteq B$, then $\qquad A\cap C \subseteq B\cap C$ Take any $x\in A\cap C$. Then:

• $x\in A$

• $x\in C$

Since $A\subseteq B$, from $x\in A$ we get $x\in B$. So $x\in B$ and $x\in C$, hence $\qquad x\in B\cap C$ Therefore $\qquad A\cap C \subseteq B\cap C$ --- Example 2 Is the function $\qquad f(x)=x^2$ from $\mathbb{R}$ to $\mathbb{R}$ injective? No, because $\qquad f(1)=1=f(-1)$ but $\qquad 1\ne -1$ So the function is not injective. ---

Quantifiers

Quantifier order matters very much. ---

Standard Problem Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $A\subseteq B$, then which of the following is always true?" options=["$A\cap C \subseteq B\cap C$","$B\subseteq A$","$A\cup C = B\cup C$","$A^c \subseteq B^c$"] answer="A" hint="Translate subset into implication." solution="If $A\subseteq B$, then every element of $A$ is in $B$. Therefore every element of $A\cap C$ is in $B\cap C$, so $\qquad A\cap C \subseteq B\cap C$. Hence the correct option is $\boxed{A}$." ::: :::question type="NAT" question="How many of the following functions from $\mathbb{R}$ to $\mathbb{R}$ are injective: $f(x)=x$, $g(x)=x^2$, $h(x)=x^3$?" answer="2" hint="Test equal outputs carefully." solution="The function $f(x)=x$ is injective. The function $g(x)=x^2$ is not injective because $\qquad g(1)=g(-1)$. The function $h(x)=x^3$ is injective because equal outputs imply equal inputs. Hence exactly $\boxed{2}$ of them are injective." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["The contrapositive of $P\Rightarrow Q$ is $\neg Q \Rightarrow \neg P$","If a function is bijective, then it is injective","$(A\cup B)^c = A^c \cap B^c$","The converse of a true implication is always true"] answer="A,B,C" hint="Recall logical equivalence and De Morgan's law." solution="1. True.

True, because bijective means injective and surjective.

True by De Morgan's law.

False. A true implication need not have a true converse.

Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Show that if $A\subseteq B$, then $A\cap C \subseteq B\cap C$." answer="Take an arbitrary element of $A\cap C$ and prove it lies in $B\cap C$." hint="Use element-chasing." solution="Let $x\in A\cap C$. Then $\qquad x\in A \quad \text{and} \quad x\in C$ Since $A\subseteq B$, from $x\in A$ we get $\qquad x\in B$ Therefore $\qquad x\in B \quad \text{and} \quad x\in C$ Hence $\qquad x\in B\cap C$ So every element of $A\cap C$ belongs to $B\cap C$, which proves $\qquad A\cap C \subseteq B\cap C$." ::: ---

Summary

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Part 2: Algebra plus geometry

Algebra plus Geometry

Overview

Many strong aptitude problems in geometry are not solved by pure diagram chasing alone. They become easy only after translating geometric information into algebra: lengths into equations, slopes into relations, areas into formulas, and symmetry into coordinates. This topic trains exactly that bridge between geometry and algebra. ---

Learning Objectives

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Core Idea

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Main Coordinate Tools

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Area as an Algebra Tool

This is one of the cleanest algebraic tests in geometry. ---

Similarity and Ratios

This is often the hidden algebra behind a geometry ratio problem. ---

Minimal Worked Examples

Example 1 Find the point on the $x$-axis equidistant from $(2,3)$ and $(8,1)$. Let the point be $(x,0)$. Then equidistance gives $\qquad (x-2)^2 + 3^2 = (x-8)^2 + 1^2$ So $\qquad (x-2)^2 + 9 = (x-8)^2 + 1$ Expanding: $\qquad x^2 - 4x + 4 + 9 = x^2 - 16x + 64 + 1$ $\qquad -4x + 13 = -16x + 65$ $\qquad 12x = 52$ $\qquad x = \dfrac{13}{3}$ So the point is $\qquad \boxed{\left( \dfrac{13}{3}, 0 \right)}$ --- Example 2 Show that the points $(0,0)$, $(2,2)$, and $(5,5)$ are collinear. Their slopes are $\qquad \dfrac{2-0}{2-0} = 1$ and $\qquad \dfrac{5-2}{5-2} = 1$ Since the slopes are equal, the points are collinear. ---

Standard Problem Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="The point on the $x$-axis equidistant from $(1,2)$ and $(5,2)$ is" options=["$(2,0)$","$(3,0)$","$(4,0)$","$(5,0)$"] answer="B" hint="Points equidistant from two points lie on the perpendicular bisector." solution="Since the two points have the same $y$-coordinate, the midpoint of the segment joining them is $\qquad \left( \dfrac{1+5}{2}, 2 \right) = (3,2)$ Its perpendicular bisector is the vertical line $x=3$, which meets the $x$-axis at $(3,0)$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="Find the area of the triangle with vertices $(0,0)$, $(4,0)$, and $(0,3)$." answer="6" hint="Use base and height." solution="Take the base as the segment from $(0,0)$ to $(4,0)$, so base $=4$. The height from $(0,3)$ to the $x$-axis is $3$. Therefore the area is $\qquad \dfrac{1}{2}\cdot 4 \cdot 3 = 6$ Hence the answer is $\boxed{6}$." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If two non-vertical lines are parallel, their slopes are equal","If two non-vertical lines are perpendicular, the product of slopes is $-1$","Three points are collinear iff the area of the triangle formed by them is $0$","The midpoint of $(x_1,y_1)$ and $(x_2,y_2)$ is $(x_1+x_2,y_1+y_2)$"] answer="A,B,C" hint="Recall the standard coordinate formulas." solution="1. True.

True.

True.

False, because the midpoint is

$\qquad \left( \dfrac{x_1+x_2}{2}, \dfrac{y_1+y_2}{2} \right)$ Hence the correct answer is $\boxed{A,B,C}$." ::: :::question type="SUB" question="Show that the points $(1,1)$, $(3,5)$, and $(5,9)$ are collinear." answer="Their slopes are equal." hint="Compare the slopes of consecutive pairs." solution="Slope of the line through $(1,1)$ and $(3,5)$ is $\qquad \dfrac{5-1}{3-1} = \dfrac{4}{2} = 2$ Slope of the line through $(3,5)$ and $(5,9)$ is $\qquad \dfrac{9-5}{5-3} = \dfrac{4}{2} = 2$ Since the two slopes are equal, all three points lie on the same straight line. Hence the points are collinear." ::: ---

Summary

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Part 3: Number theory plus combinatorics

Number Theory Plus Combinatorics

Overview

Many aptitude-style problems combine counting with divisibility, parity, remainders, digit conditions, or residue classes. These questions are not purely number theory and not purely combinatorics; the real skill is to use arithmetic structure to reduce the counting work. ---

Learning Objectives

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Core Idea

A typical pattern is: $\qquad \text{count valid objects} = \text{count all objects satisfying an arithmetic rule}$ ---

Most Useful Arithmetic Tests

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Counting by Cases

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Residue-Class Thinking

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Digit-Based Problems

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Minimal Worked Examples

Example 1 How many $3$-digit numbers formed using the digits $1,2,3,4,5$ without repetition are even? For the number to be even, the last digit must be $2$ or $4$.

• choose last digit: $2$ choices

• choose first digit from the remaining $4$ digits: $4$ choices

• choose middle digit from the remaining $3$ digits: $3$ choices

So the count is $\qquad 2\cdot 4\cdot 3 = 24$ --- Example 2 How many $2$-element subsets of $\{1,2,3,4,5,6\}$ have even sum? An even sum comes from:

• even + even

• odd + odd

There are $3$ even numbers and $3$ odd numbers. So the count is $\qquad \binom{3}{2}+\binom{3}{2}=3+3=6$ ---

Common Strategies

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Common Mistakes

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Practice Questions

:::question type="MCQ" question="How many $2$-digit numbers are divisible by $5$?" options=["$9$","$18$","$10$","$20$"] answer="B" hint="A two-digit number divisible by $5$ must end in $0$ or $5$." solution="The tens digit can be any of $1,2,\dots,9$, so there are $9$ choices. The units digit must be $0$ or $5$, so there are $2$ choices. Hence the total number is $\qquad 9\cdot 2 = 18$. Therefore the correct option is $\boxed{B}$." ::: :::question type="NAT" question="How many $2$-element subsets of $\{1,2,3,4,5,6\}$ have odd sum?" answer="9" hint="An odd sum needs one odd and one even number." solution="The set has $3$ odd numbers and $3$ even numbers. To get odd sum, choose one odd and one even. Hence the number of such subsets is $\qquad 3\cdot 3 = 9$. Therefore the answer is $\boxed{9}$." ::: :::question type="MSQ" question="Which of the following are always true?" options=["An odd number plus an odd number is even","A number divisible by $3$ must have even digit sum","A number divisible by $5$ must end in $0$ or $5$","Two numbers with the same remainder mod $4$ differ by a multiple of $4$"] answer="A,C,D" hint="Check each using basic residue facts." solution="1. True. Odd + odd = even. 2. False. Divisibility by $3$ depends on digit sum being divisible by $3$, not being even. 3. True. 4. True by definition of congruence modulo $4$. Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="How many $3$-digit numbers formed from the digits $1,2,3,4,5$ without repetition are divisible by $5$?" answer="$24$" hint="For divisibility by $5$, fix the last digit first." solution="A number formed from the digits $1,2,3,4,5$ without repetition is divisible by $5$ only if its last digit is $5$, since $0$ is unavailable. So the last digit is fixed. Then the first digit can be chosen in $4$ ways, and the middle digit can be chosen in $3$ ways. Therefore the total number is $\qquad 4\cdot 3 = 12$. Hence the answer is $\boxed{12}$." ::: ---

Summary

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Part 4: Calculus plus function reasoning

Calculus plus Function Reasoning

Overview

Many harder function questions are not about computing derivatives mechanically. They ask what the derivative tells us about injectivity, monotonicity, number of roots, extrema, or shape. This topic combines function reasoning with calculus so that derivative information becomes a logical tool rather than just a computational rule. ---

Learning Objectives

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Core Idea

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Monotonicity

So derivative information often gives one-one behaviour immediately. ---

Critical Points

These are the main candidates for local maxima, local minima, or stationary non-extreme points. ---

Sign Chart Reasoning

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Number of Roots

This is a powerful reasoning tool in mixed aptitude questions. ---

Minimal Worked Examples

Example 1 Consider $\qquad f(x)=x^3-3x$ Then $\qquad f'(x)=3x^2-3=3(x^2-1)$ So critical points are $\qquad x=\pm 1$ Now:

• for $x<-1$, $f'(x)>0$

• for $-1<x<1$, $f'(x)<0$

• for $x>1$, $f'(x)>0$

Hence:

• local maximum at $x=-1$

• local minimum at $x=1$

--- Example 2 Let $\qquad f(x)=x^3+x$ Then $\qquad f'(x)=3x^2+1>0$ for all real $x$. Therefore $f$ is strictly increasing on $\mathbb{R}$, so it is injective and the equation $\qquad x^3+x=c$ has exactly one real solution for every real $c$. ---

Range and Extrema via Calculus

This is standard and essential in reasoning-based problems. ---

Common Patterns

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Common Mistakes

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CMI Strategy

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Practice Questions

:::question type="MCQ" question="If $f'(x)>0$ for all real $x$, then $f$ is" options=["constant","strictly increasing","strictly decreasing","periodic"] answer="B" hint="Derivative sign controls monotonicity." solution="If $f'(x)>0$ for all real $x$, then the function is strictly increasing on $\mathbb{R}$. Hence the correct option is $\boxed{B}$." ::: :::question type="NAT" question="For $f(x)=x^3+x$, how many real solutions does the equation $f(x)=0$ have?" answer="1" hint="Study monotonicity or factor directly." solution="We have $\qquad f(x)=x^3+x=x(x^2+1)$ Since $x^2+1>0$ for all real $x$, the only real solution is $\qquad x=0$ So the equation has exactly $\boxed{1}$ real solution." ::: :::question type="MSQ" question="Which of the following statements are true?" options=["If $f'(x)>0$ on an interval, then $f$ is increasing there","If $f'(a)=0$, then $a$ must be a local maximum","A strictly increasing function is injective","To find extrema on a closed interval, endpoints must also be checked"] answer="A,C,D" hint="Differentiate between stationary points and extrema." solution="1. True.

False. A zero derivative does not always give a local maximum.

True.

True.

Hence the correct answer is $\boxed{A,C,D}$." ::: :::question type="SUB" question="Show that the function $f(x)=x^3+x$ is one-one on $\mathbb{R}$." answer="Its derivative is always positive." hint="Use the derivative test for monotonicity." solution="Differentiate: $\qquad f'(x)=3x^2+1$ Since $\qquad 3x^2+1>0$ for all real $x$, the function $f$ is strictly increasing on $\mathbb{R}$. A strictly increasing function is injective. Therefore $f(x)=x^3+x$ is one-one on $\mathbb{R}$." ::: ---

Summary

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Chapter Summary

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Chapter Review Questions

:::question type="MCQ" question="A rectangular sheet of paper has dimensions 8 cm by 5 cm. A square of side length $x$ cm is cut from each corner, and the sides are folded up to form an open-top box. For what value of $x$ is the volume of the box maximized?" options=["1/2", "1", "3/2", "2"] answer="1" hint="Formulate the volume as a function of $x$ and use calculus to find its maximum. Remember to consider the domain of $x$." solution="Let the dimensions of the rectangular sheet be $L=8$ and $W=5$. When a square of side $x$ is cut from each corner, the base of the box will have dimensions $(L-2x)$ by $(W-2x)$, and the height will be $x$.
The volume of the box $V(x)$ is given by:

Substituting $L=8$ and $W=5$:




The domain for $x$ is determined by the physical constraints: $x > 0$, $8-2x > 0 \implies x < 4$, and $5-2x > 0 \implies x < 2.5$. Thus, $0 < x < 2.5$.

To maximize the volume, we find the derivative $V'(x)$ and set it to zero:

Set $V'(x) = 0$:

Divide by 4:

Factor the quadratic equation:

This gives two possible values for $x$: $x = 10/3$ or $x = 1$.
We must check which of these values lies within the valid domain $(0, 2.5)$.
$x = 10/3 \approx 3.33$, which is outside the domain $(0, 2.5)$.
$x = 1$, which is inside the domain $(0, 2.5)$.

To confirm that $x=1$ corresponds to a maximum, we can use the second derivative test:

At $x=1$:

Since $V''(1) < 0$, $x=1$ corresponds to a local maximum.
Thus, the volume of the box is maximized when $x=1$ cm."
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:::question type="NAT" question="How many positive integers less than 100 are not divisible by 2, 3, or 5?" answer="26" hint="Use the Principle of Inclusion-Exclusion. First, find the number of integers less than 100 that are divisible by 2, 3, or 5." solution="Let $N = 99$ be the total number of positive integers less than 100 (i.e., from 1 to 99).
We want to find the number of integers that are not divisible by 2, 3, or 5. This is equivalent to $N - |\text{divisible by 2 or 3 or 5}|$.
Let $A_2$ be the set of integers divisible by 2.
Let $A_3$ be the set of integers divisible by 3.
Let $A_5$ be the set of integers divisible by 5.

Using the Principle of Inclusion-Exclusion:
$|A_2| = \lfloor 99/2 \rfloor = 49$
$|A_3| = \lfloor 99/3 \rfloor = 33$
$|A_5| = \lfloor 99/5 \rfloor = 19$

$|A_2 \cap A_3| = \lfloor 99/\operatorname{lcm}(2,3) \rfloor = \lfloor 99/6 \rfloor = 16$
$|A_2 \cap A_5| = \lfloor 99/\operatorname{lcm}(2,5) \rfloor = \lfloor 99/10 \rfloor = 9$
$|A_3 \cap A_5| = \lfloor 99/\operatorname{lcm}(3,5) \rfloor = \lfloor 99/15 \rfloor = 6$

$|A_2 \cap A_3 \cap A_5| = \lfloor 99/\operatorname{lcm}(2,3,5) \rfloor = \lfloor 99/30 \rfloor = 3$

The number of integers divisible by 2, 3, or 5 is:
$|A_2 \cup A_3 \cup A_5| = (|A_2| + |A_3| + |A_5|) - (|A_2 \cap A_3| + |A_2 \cap A_5| + |A_3 \cap A_5|) + |A_2 \cap A_3 \cap A_5|$
$|A_2 \cup A_3 \cup A_5| = (49 + 33 + 19) - (16 + 9 + 6) + 3$
$|A_2 \cup A_3 \cup A_5| = 101 - 31 + 3$
$|A_2 \cup A_3 \cup A_5| = 73$

The number of positive integers less than 100 that are not divisible by 2, 3, or 5 is:
$N - |A_2 \cup A_3 \cup A_5| = 99 - 73 = 26$."
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:::question type="MCQ" question="Let $f: \mathbb{Z} \to \mathbb{Z}$ be a function defined by $f(x) = x^2 - 4x + 3$. Let $A = \{x \in \mathbb{Z} \mid f(x) \text{ is even} \}$. Let $B = \{x \in \mathbb{Z} \mid x \text{ is odd} \}$. Which of the following statements is true?" options=["$A \subset B$", "$B \subset A$", "$A \cap B = \emptyset$", "$A = B$"] answer="$A = B$" hint="Analyze the parity of $f(x)$ based on the parity of $x$. Recall that an integer is even if it is divisible by 2, and odd otherwise." solution="We need to determine the conditions under which $f(x) = x^2 - 4x + 3$ is even.
We can analyze the parity of $f(x)$ by considering $f(x) \pmod 2$.
$f(x) \equiv x^2 - 4x + 3 \pmod 2$
Since $4x$ is always even, $4x \equiv 0 \pmod 2$.
Since $3$ is odd, $3 \equiv 1 \pmod 2$.
So, $f(x) \equiv x^2 - 0 + 1 \pmod 2$
$f(x) \equiv x^2 + 1 \pmod 2$

For $f(x)$ to be even, $f(x) \equiv 0 \pmod 2$.
So, $x^2 + 1 \equiv 0 \pmod 2$, which implies $x^2 \equiv -1 \pmod 2$.
Since $-1 \equiv 1 \pmod 2$, we have $x^2 \equiv 1 \pmod 2$.

Now, let's consider the parity of $x$:
Case 1: $x$ is even.
If $x$ is even, then $x^2$ is even. So $x^2 \equiv 0 \pmod 2$.
In this case, $f(x) \equiv 0 + 1 \equiv 1 \pmod 2$, which means $f(x)$ is odd.

Case 2: $x$ is odd.
If $x$ is odd, then $x^2$ is odd. So $x^2 \equiv 1 \pmod 2$.
In this case, $f(x) \equiv 1 + 1 \equiv 2 \equiv 0 \pmod 2$, which means $f(x)$ is even.

Therefore, $f(x)$ is even if and only if $x$ is odd.
The set $A$ is defined as $A = \{x \in \mathbb{Z} \mid f(x) \text{ is even} \}$. From our analysis, $A = \{x \in \mathbb{Z} \mid x \text{ is odd} \}$.
The set $B$ is defined as $B = \{x \in \mathbb{Z} \mid x \text{ is odd} \}$.
Since both sets contain exactly the same elements, $A = B$.

Thus, the statement "$A = B$" is true."
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What's Next?