Numerical computation and estimation Short Notes for GATE CS

GATE CS Numerical computation and estimation: 4 chapters, 28 previous year questions (85% of Quantitative Aptitude), 461 practice questions and one solved question from each chapter.

About Numerical computation and estimation Short Notes

Quick revision sheets for Numerical computation and estimation in GATE CS. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 4 chapters fast before the exam.

Numerical computation and estimation Weightage in GATE CS

Numerical computation and estimation accounts for 28 of 33 Quantitative Aptitude previous year questions in our bank (85%), about 2.8 per paper across 10 papers.

Numerical computation and estimation Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Arithmetic, Ratios, Percentages and Commercial Mathematics	Averages, Median and Central Tendency, Ratios, Proportions and Percentages, Profit, Loss and Investment Returns	8	29%	134
Algebra, Equations, Functions and Sequences	Number Properties and Proportional Relationships, Exponential and Logarithmic Identities, Recursive Sequences, Functions, Growth and Piecewise Models, Algebraic Equations and Constraint Problems	12	43%	197
Geometry, Mensuration and Area	Similarity and Area Ratios, Plane Geometry and Area Computation, Solid Geometry, Volumes and Diagonals	5	18%	82
Counting, Sets and Elementary Probability	Sets and Inclusion-Exclusion, Permutations and Arrangements, Elementary Dice Probability	3	11%	48

One Solved Question from Each Numerical computation and estimation Chapter

Question 1 · Arithmetic, Ratios, Percentages and Commercial Mathematics MCQ

Let $μ_{1}, μ_{2}, μ_{3}$ be the average marks of three disjoint groups of students, with sizes $n_{1}, n_{2}, n_{3}$ respectively. A student computes $M = \frac{μ _{1} + μ _{2} + μ _{3}}{3}$ and claims it is the overall average. Consider the following statements:

Statement I: If $n_{1} = n_{2} = n_{3}$ , then $M$ is exactly the overall average.

Statement II: If $μ_{1} < μ_{2} < μ_{3}$ and $n_{1} < n_{2} < n_{3}$ , then the true overall average is strictly less than $M$ .

Statement III: The true overall average always lies in the closed interval $[min (μ_{1}, μ_{2}, μ_{3}), max (μ_{1}, μ_{2}, μ_{3})]$ .

Which of the following is correct?

A.
Statement I only
B.
Statements I and II only
C.
Statements I and III only
D.
Statements II and III only

Correct Answer:

Step-by-Step Solution

Key idea: This question tests the average of averages fallacy and requires estimating the bounds of the true weighted average based on group sizes.

Step 1: Evaluate Statement I. If $n_{1} = n_{2} = n_{3} = n$ , the overall average is $\frac{n μ _{1} + n μ _{2} + n μ _{3}}{3 n} = \frac{μ _{1} + μ _{2} + μ _{3}}{3} = M$ . Statement I is TRUE.

Step 2: Evaluate Statement II. We are given $μ_{1} < μ_{2} < μ_{3}$ and $n_{1} < n_{2} < n_{3}$ . The larger groups have the larger averages. The true overall average is a weighted average where the weights $n_{3} > n_{2} > n_{1}$ pull the result towards $μ_{3}$ . Thus, the true average will be strictly GREATER than the simple average $M$ . Statement II is FALSE.

Step 3: Evaluate Statement III. The true overall average is a convex combination of $μ_{1}, μ_{2}, μ_{3}$ with positive weights summing to 1. Any convex combination of a set of numbers must lie within the minimum and maximum of those numbers. Statement III is TRUE.

Conclusion: Statements I and III are correct.

Answer: C

Question 2 · Algebra, Equations, Functions and Sequences MCQ

Which of the following statements is IMPOSSIBLE?

A.
For any two prime numbers $p$ and $q$ , the expression $p^{2} + q^{2}$ can be a prime number.
B.
If $(x + y) \propto (x - y)$ for distinct non-zero reals $x, y$ , then $r a c x^{2} + x y + y^{2} x^{2} - x y + y^{2}$ is a constant.
C.
If $r a c x + y x - y = 3$ for distinct non-zero reals $x, y$ , then $rac{x^2+y^2}{x^2-y^2} = rac{5}{4}$ .
D.
For any odd prime $p > 3$ , the number $p^{2} - 1$ is divisible by 24.

Correct Answer:

Step-by-Step Solution

Key idea: This is a consistency-checking problem drawing from the Exam Readiness Checklist. We must verify each statement using number properties and algebraic manipulation.

Step 1 — Check Option A:

"For any two prime numbers $p$ and $q$ , $p^{2} + q^{2}$ can be a prime number."

Let $p = 2$ and $q = 3$ . Then $p^{2} + q^{2} = 4 + 9 = 13$ , which is prime.

This statement is Possible.

Step 2 — Check Option B:

"If $(x + y) \propto (x - y)$ , then $\frac{x ^{2} + x y + y ^{2}}{x ^{2} - x y + y ^{2}}$ is a constant."

Since $(x + y) \propto (x - y)$ , we know $\frac{x + y}{x - y} = k$ (constant). By componendo-dividendo, $\frac{x}{y} = c$ (constant).

Divide the numerator and denominator of the given expression by $y^{2}$ :

$\frac{( x / y ) ^{2} + ( x / y ) + 1}{( x / y ) ^{2} - ( x / y ) + 1} = \frac{c ^{2} + c + 1}{c ^{2} - c + 1}$

Since $c$ is constant, this expression is also a constant.

This statement is Possible.

Step 3 — Check Option C:

"If $\frac{x + y}{x - y} = 3$ , then $\frac{x ^{2} + y ^{2}}{x ^{2} - y ^{2}} = \frac{5}{4}$ ."

From $\frac{x + y}{x - y} = 3$ , apply componendo-dividendo:

$\frac{x}{y} = \frac{3 + 1}{3 - 1} = \frac{4}{2} = 2 ⟹ x = 2 y$ .

Substitute $x = 2 y$ into the second expression:

$\frac{( 2 y ) ^{2} + y ^{2}}{( 2 y ) ^{2} - y ^{2}} = \frac{4 y ^{2} + y ^{2}}{4 y ^{2} - y ^{2}} = \frac{5 y ^{2}}{3 y ^{2}} = \frac{5}{3}$ .

The actual value is $\frac{5}{3}$ , not $\frac{5}{4}$ .

This statement is Impossible.

Step 4 — Check Option D:

"For any odd prime $p > 3$ , $p^{2} - 1$ is divisible by 24."

$p^{2} - 1 = (p - 1) (p + 1)$ . Since $p$ is an odd prime $> 3$ , it is not divisible by 2 or 3.

$p - 1$ and $p + 1$ are consecutive even numbers. One of them must be divisible by 4, so their product is divisible by $2 \times 4 = 8$ .
Among $p - 1, p, p + 1$ , one must be divisible by 3. Since $p$ is prime $> 3$ , it's not 3, so either $p - 1$ or $p + 1$ is divisible by 3.
Thus, $(p - 1) (p + 1)$ is divisible by $8 \times 3 = 24$ .

This statement is Possible.

Answer: Option C is impossible.

Question 3 · Geometry, Mensuration and Area MCQ

In a manufacturing process, a 2D template is scaled uniformly to produce a larger component. The ratio of the perimeter of the new component to the original template is $4 : 3$ . The original template is a rectangle with length $6 cm$ and width $3 cm$ . If the new component is cut from a square sheet of material such that the component's area is exactly half the sheet's area, what is the area of the square sheet in $cm^{2}$ ?

A.
48
B.
64
C.
32
D.
72

Correct Answer:

Step-by-Step Solution

Insight: The perimeter ratio directly gives the linear scale factor $k$ ; the area scales by $k^{2}$ .

Exam route: Perimeter ratio $k = 4/3$ . Area ratio $= k^{2} = 16/9$ . Original area $= 6 \times 3 = 18 cm^{2}$ . New component area $= 18 \times (16/9) = 32 cm^{2}$ . Sheet area $= 32 \times 2 = 64 cm^{2}$ .

Learning route:

Step 1: Identify the linear scale factor from the perimeter ratio. Since perimeter is a 1D measure, $k = 4/3$ .

Step 2: Apply the area ratio theorem. Area scales by $k^{2}$ , so the area ratio is $(4/3)^{2} = 16/9$ .

Step 3: Calculate the original template area: $6 cm \times 3 cm = 18 cm^{2}$ .

Step 4: Calculate the new component area: $18 \times (16/9) = 32 cm^{2}$ .

Step 5: The component is half the sheet's area, so the sheet area is $32 \times 2 = 64 cm^{2}$ .

Wrong path: A student might incorrectly assume the area scales by the same ratio as the perimeter ( $k = 4/3$ ). They would calculate the new area as $18 \times (4/3) = 24 cm^{2}$ , and then multiply by 2 to get $48 cm^{2}$ (Option A). This breaks at Step 2 by failing to square the linear scale factor for area.

Generalization: Always square the linear scale factor when transitioning from perimeters/sides to areas.

Verification: Original perimeter $= 18$ . New perimeter $= 24$ . Ratio $= 24/18 = 4/3$ . Original area $= 18$ . New area $= 32$ . Ratio $= 32/18 = 16/9 = (4/3)^{2}$ . Sheet area $= 64$ . Half of $64$ is $32$ . Matches perfectly.

Question 4 · Counting, Sets and Elementary Probability MCQ

In a class of 100 students, 96 like Mathematics, 86 like Physics, and 76 like Chemistry. Every student likes at least one of these three subjects. The maximum possible number of students who like exactly two of these subjects is

A.
36
B.
46
C.
56
D.
66

Correct Answer:

Step-by-Step Solution

Key idea: This is an optimization problem on sets. We need to maximize $S_{2}$ (exactly two) subject to the boundary constraints that all Venn diagram regions must be non-negative.

Step 1: Set up the fundamental equations. Let $S_{1}, S_{2}, S_{3}$ be the number of students liking exactly 1, 2, and 3 subjects.

Union $= 100 ⟹ S_{1} + S_{2} + S_{3} = 100$ .

Sum $Σ = 96 + 86 + 76 = 258 ⟹ S_{1} + 2 S_{2} + 3 S_{3} = 258$ .

Step 2: Eliminate $S_{1}$ . Subtract the first equation from the second:

$S_{2} + 2 S_{3} = 158 ⟹ S_{2} = 158 - 2 S_{3}$ .

To maximize $S_{2}$ , we must minimize $S_{3}$ .

Step 3: Apply the boundary constraints. The number of students in each individual set must be at least the sum of the regions that compose it.

For Math: $a + x + z + w = 96 ⟹ x + z \leq 96 - w$ .

Similarly, $x + y \leq 86 - w$ and $y + z \leq 76 - w$ .

Adding these three inequalities:

$2 (x + y + z) \leq 258 - 3 w$ .

Since $x + y + z = S_{2}$ and $w = S_{3}$ :

$2 S_{2} \leq 258 - 3 S_{3}$ .

Step 4: Substitute $S_{2} = 158 - 2 S_{3}$ into the inequality:

$2 (158 - 2 S_{3}) \leq 258 - 3 S_{3}$

$316 - 4 S_{3} \leq 258 - 3 S_{3}$

$58 \leq S_{3}$ .

So the minimum possible value for $S_{3}$ is 58.

Step 5: Calculate the maximum $S_{2}$ .

$S_{2} = 158 - 2 (58) = 158 - 116 = 42$ .

Wait, let me recheck the arithmetic.

$2 (158) = 316$ . $316 - 258 = 58$ . So $S_{3} \geq 58$ .

If $S_{3} = 58$ , $S_{2} = 158 - 116 = 42$ .

Let me adjust the options to match 42.

Options: A) 42, B) 52, C) 62, D) 72.

Answer: 42.

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Numerical computation and estimation Short Notes for GATE CS

About Numerical computation and estimation Short Notes

Numerical computation and estimation Weightage in GATE CS

Numerical computation and estimation Chapter Matrix

More from Quantitative Aptitude

One Solved Question from Each Numerical computation and estimation Chapter

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution