Numerical Computation & Estimation Short Notes for GATE DA

GATE DA Numerical Computation & Estimation: 6 chapters, 12 previous year questions (80% of Quantitative Aptitude), 683 practice questions and one solved question from each chapter.

About Numerical Computation & Estimation Short Notes

Quick revision sheets for Numerical Computation & Estimation in GATE DA. Every chapter is condensed into key formulas, shortcuts and common traps so you can revise 6 chapters fast before the exam.

Numerical Computation & Estimation Weightage in GATE DA

Numerical Computation & Estimation accounts for 12 of 15 Quantitative Aptitude previous year questions in our bank (80%), about 4 per paper across 3 papers.

Numerical Computation & Estimation Chapter Matrix

Chapter	Topics	PYQs	Share of unit PYQs	Practice questions
Number Systems, Digit Problems and Divisibility	Digit Constraints and Number Formation	2	17%	103
Exponents, Logarithms and Infinite Series	Exponent and Logarithm Laws, Infinite Series Summation	3	25%	167
Percentages, Ratios and Financial Computation	Percentage Change in Geometry, Investment Returns and Weighted Percentages	2	17%	114
Descriptive Statistics and Data Normalization	Mean, Median and Mode Constraints, Standardization and Z-Score Normalization	2	17%	127
Combinatorial Counting and Bijections	Digit-Based Counting with Divisibility Constraints, Bijections and Self-Inverse Mappings	2	17%	112
Geometry and Mensuration	Circle Geometry and Chord-Angle Relations	1	8%	60

One Solved Question from Each Numerical Computation & Estimation Chapter

Question 1 · Number Systems, Digit Problems and Divisibility MCQ

Consider all $5$ -digit numbers formed using distinct digits from the set ${1, 2, 3, 4, 5, 6}$ . If these numbers are arranged in ascending order, what is the rank of the number $42513$ among those that are divisible by $3$ ?

A.
139
B.
371
C.
163
D.
115

Correct Answer:

Step-by-Step Solution

Key idea: This is a lexicographical ranking problem with a divisibility constraint. We must first identify which subsets of $5$ digits yield a sum divisible by $3$ , then count how many valid numbers precede $42513$ .

Step 1: Identify valid $5$ -digit subsets.

The sum of all $6$ digits is $1 + 2 + 3 + 4 + 5 + 6 = 21$ .

A $5$ -digit number is formed by omitting one digit $x$ . The sum of its digits is $21 - x$ .

For the sum to be divisible by $3$ , $21 - x$ must be a multiple of $3$ , which means $x$ must be a multiple of $3$ .

Thus, $x \in {3, 6}$ .

Case 1: Omit $3$ . Digits used: ${1, 2, 4, 5, 6}$ .

Case 2: Omit $6$ . Digits used: ${1, 2, 3, 4, 5}$ .

Step 2: Count valid numbers strictly less than $42513$ .

The target $42513$ uses digits ${1, 2, 3, 4, 5}$ , so it belongs to Case 2.

Numbers starting with $1$ : Can be from Case 1 or Case 2. Each has $4! = 24$ permutations. Total = $48$ .
Numbers starting with $2$ : Similarly, $24 + 24 = 48$ .
Numbers starting with $3$ : Case 1 doesn't contain $3$ . Case 2 has $4! = 24$ . Total = $24$ .
Numbers starting with $4$ and less than $42513$ :
From Case 1 ( ${1, 2, 4, 5, 6}$ ):
$41 xxx$ : $3! = 6$ .
$421 xx$ : $2! = 2$ .
$425 xx$ : Remaining digits ${1, 6}$ . Numbers are $42516, 42561$ . Both are $> 42513$ . So $0$ .

Total Case 1 = $6 + 2 = 8$ .

From Case 2 ( ${1, 2, 3, 4, 5}$ ):
$41 xxx$ : $3! = 6$ .
$421 xx$ : $2! = 2$ .
$423 xx$ : $2! = 2$ .
$425 xx$ : Remaining digits ${1, 3}$ . Numbers are $42513, 42531$ . Strictly less is $0$ .

Total Case 2 = $6 + 2 + 2 = 10$ .

Step 3: Calculate the rank.

Total numbers before $42513 = 48 + 48 + 24 + 8 + 10 = 138$ .

Rank = $138 + 1 = 139$ .

Answer: 139

Question 2 · Exponents, Logarithms and Infinite Series MCQ

An algorithm's time complexity is modeled by a series of operations. In the $k$ -th iteration ( $k \geq 1$ ), the theoretical number of operations is $k \cdot (0.5)^{k - 1}$ . However, due to a known initialization overhead, the actual number of operations in the first iteration ( $k = 1$ ) is fixed at 4, while for all $k \geq 2$ , it strictly follows the formula $k \cdot (0.5)^{k - 1}$ .

What is the total number of operations performed over infinitely many iterations?

A.
4
B.
5
C.
6
D.
7

Correct Answer:

Step-by-Step Solution

Key idea: This is an arithmetico-geometric series with a modified first term. We must calculate the theoretical infinite sum, subtract the theoretical first term, and add the actual first term.

Step 1: Identify the theoretical series.

$S_{t h eo} = \sum_{k = 1}^{\infty} k \cdot (0.5)^{k - 1}$ .

Step 2: Apply the standard arithmetico-geometric formula $\sum_{k = 1}^{\infty} k x^{k - 1} = \frac{1}{( 1 - x ) ^{2}}$ with $x = 0.5$ .

$S_{t h eo} = \frac{1}{( 1 - 0.5 ) ^{2}} = \frac{1}{0.25} = 4$ .

Step 3: Isolate the theoretical first term.

For $k = 1$ , the formula gives $1 \cdot (0.5)^{0} = 1$ .

Step 4: Adjust for the constraint.

The problem states the actual first iteration has 4 operations, not 1.

$S_{a c t u a l} = S_{t h eo} - (theoretical k = 1) + (actual k = 1)$

$S_{a c t u a l} = 4 - 1 + 4 = 7$ .

Answer: D

Question 3 · Percentages, Ratios and Financial Computation MCQ

The length of a rectangle is increased by $30%$ and its width is decreased by $20%$ . If the original area of the rectangle was $500 cm^{2}$ , what is the area of the new rectangle in $cm^{2}$ ?

A.
520
B.
550
C.
750
D.
250

Correct Answer:

Step-by-Step Solution

Key idea: Use multiplication factors for successive percentage changes instead of adding or subtracting percentages directly.

Step 1: The length multiplier is $1 + \frac{30}{100} = 1.3$ .

Step 2: The width multiplier is $1 - \frac{20}{100} = 0.8$ .

Step 3: The total area multiplier is the product of the individual multipliers: $1.3 \times 0.8 = 1.04$ .

Step 4: The new area is the original area multiplied by the total area multiplier: $500 \times 1.04 = 520 cm^{2}$ .

Answer: 520

Question 4 · Descriptive Statistics and Data Normalization MCQ

Five integers are picked from 0 to 20, with possible repetitions, such that their arithmetic mean is 14, median is 14, and they have a unique mode of 16. Ignoring permutations, the number of ways to pick these five integers is _____

A.
0
B.
1
C.
2
D.
3

Correct Answer:

Step-by-Step Solution

Key idea: This is an integer constraint problem where the median and mode fix the upper elements, and the mean restricts the sum of the remaining lower elements. The unique mode condition acts as a strict boundary filter.

Step 1: Let the sorted integers be $x_{1} \leq x_{2} \leq x_{3} \leq x_{4} \leq x_{5}$ .

Step 2: The median is 14, so $x_{3} = 14$ .

Step 3: The unique mode is 16. Since $x_{3} = 14$ , the value 16 can only occupy $x_{4}$ and $x_{5}$ . Thus, $x_{4} = 16$ and $x_{5} = 16$ . The frequency of 16 is exactly 2.

Step 4: The mean is 14, so the total sum is $5 \times 14 = 70$ .

Step 5: Calculate the sum of the remaining elements: $x_{1} + x_{2} = 70 - (14 + 16 + 16) = 24$ .

Step 6: Find valid pairs for $(x_{1}, x_{2})$ such that $0 \leq x_{1} \leq x_{2} \leq 14$ and $x_{1} + x_{2} = 24$ .

The possible pairs are:

(10, 14): Here, 14 appears twice ( $x_{2}, x_{3}$ ) and 16 appears twice ( $x_{4}, x_{5}$ ). This creates a bimodal distribution, violating the unique mode condition.
(11, 13): Here, 14 appears once, and 16 appears twice. The mode is uniquely 16. This is valid.
(12, 12): Here, 12 appears twice and 16 appears twice. This is bimodal, violating the unique mode condition.

Conclusion: There is exactly 1 valid way to pick the integers.

Answer: 1

Question 5 · Combinatorial Counting and Bijections MCQ

Let $N_{4}$ be the number of 4-digit positive integers formed using distinct digits from ${1, 2, 3, 4, 5, 6}$ that are divisible by 4. Let $N_{11}$ be the number of such integers that are divisible by 11.

A student assumes that the divisibility of a number by 4 and by 11 are independent events, and estimates the number of integers satisfying both conditions as $\frac{N _{4} \times N _{11}}{360}$ .

What is the actual number of 4-digit integers formed from ${1, 2, 3, 4, 5, 6}$ without repetition that satisfy both conditions?

A.
12
B.
14
C.
16
D.
18

Correct Answer:

Step-by-Step Solution

Key idea: This requires calculating two separate counts ( $N_{4}$ and $N_{11}$ ), recognizing the flaw in the independence assumption, and then finding the exact intersection by checking valid partitions.

Step 1: Calculate $N_{4}$ . Last two digits must be a multiple of 4.

Valid pairs from ${1..6}$ : 12, 16, 24, 32, 36, 52, 56, 64 (8 pairs).

For each pair, the first two digits are chosen from the remaining 4 digits: $4 \times 3 = 12$ ways.

$N_{4} = 8 \times 12 = 96$ .

Step 2: Calculate $N_{11}$ . Alternating sum $(d_{1} + d_{3}) - (d_{2} + d_{4}) = 0 ⟹ d_{1} + d_{3} = d_{2} + d_{4}$ .

We need to partition 4 chosen digits into two pairs with equal sums.

Total sum of 6 digits is 21 (odd). To get an even sum for 4 digits, we must exclude 2 digits with an odd sum.

There are 9 such pairs to exclude. Checking each, exactly 7 sets of 4 digits can be partitioned into equal sums:

${3, 4, 5, 6}$ , ${2, 3, 5, 6}$ , ${2, 3, 4, 5}$ , ${1, 3, 4, 6}$ , ${1, 2, 5, 6}$ , ${1, 2, 4, 5}$ , ${1, 2, 3, 4}$ .

For each set, there are 2 ways to assign the pairs to $(d_{1}, d_{3})$ and $(d_{2}, d_{4})$ , and $2! \times 2! = 4$ ways to arrange within pairs.

$N_{11} = 7 \times 2 \times 4 = 56$ .

Step 3: The student's estimate is $\frac{96 \times 56}{360} = 14.93$ , which is invalid since counts must be integers. We must find the actual intersection.

Step 4: Check the 7 valid sets for divisibility by 4 (last two digits div by 4).

${3, 4, 5, 6}$ : Pairs ${3, 6}, {4, 5}$ . Last two div by 4: 36. Rem ${4, 5}$ sum=9. Matches! (2 ways).
${2, 3, 5, 6}$ : Pairs ${2, 6}, {3, 5}$ . Last two div by 4: 32, 36, 52, 56. None match the pair sums. (0 ways).
${2, 3, 4, 5}$ : Pairs ${2, 5}, {3, 4}$ . Last two div by 4: 52. Rem ${3, 4}$ sum=7. Matches! (2 ways).
${1, 3, 4, 6}$ : Pairs ${1, 6}, {3, 4}$ . Last two div by 4: 16. Rem ${3, 4}$ sum=7. Matches! (2 ways).
${1, 2, 5, 6}$ : Pairs ${1, 6}, {2, 5}$ . Last two div by 4: 16, 52. Both match! (4 ways).
${1, 2, 4, 5}$ : Pairs ${1, 5}, {2, 4}$ . Last two div by 4: 24. Rem ${1, 5}$ sum=6. Matches! (2 ways).
${1, 2, 3, 4}$ : Pairs ${1, 4}, {2, 3}$ . Last two div by 4: 32. Rem ${1, 4}$ sum=5. Matches! (2 ways).

Total actual intersection = $2 + 0 + 2 + 2 + 4 + 2 + 2 = 14$ .

Answer: 14.

Question 6 · Geometry and Mensuration MCQ

Points $A$ , $B$ , and $C$ lie on a circle. A point $P$ on the circle is such that the chord $A B$ subtends an angle of $4 0^{\circ}$ at $P$ , and the chord $B C$ subtends an angle of $2 5^{\circ}$ at $P$ . Let $M$ and $m$ be the maximum and minimum possible values of the angle subtended by the chord $A C$ at $P$ . Find $M - m$ .

A.
50
B.
65
C.
80
D.
15

Correct Answer:

Step-by-Step Solution

Key idea: The angle subtended by a chord at a point on the circle depends on the relative order of the points. We must consider the casework for the rays $P A$ , $P B$ , $P C$ .

Step 1: The angle subtended by $A B$ at $P$ is $∠ A P B = 4 0^{\circ}$ . The angle subtended by $B C$ at $P$ is $∠ B P C = 2 5^{\circ}$ .

Step 2: The angle subtended by $A C$ at $P$ is $∠ A P C$ . Depending on the position of $B$ relative to $A$ and $C$ from $P$ 's perspective, the ray $P B$ can be between $P A$ and $P C$ , or outside.

Step 3: Case 1: $P B$ is between $P A$ and $P C$ . Then $∠ A P C = ∠ A P B + ∠ B P C = 4 0^{\circ} + 2 5^{\circ} = 6 5^{\circ}$ .

Step 4: Case 2: $P B$ is not between $P A$ and $P C$ . Then $∠ A P C = ∣∠ A P B - ∠ B P C ∣ = ∣4 0^{\circ} - 2 5^{\circ} ∣ = 1 5^{\circ}$ .

Step 5: The maximum value $M = 6 5^{\circ}$ and the minimum value $m = 1 5^{\circ}$ .

Step 6: $M - m = 65 - 15 = 50$ .

Answer: 50

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Numerical Computation & Estimation Short Notes for GATE DA

About Numerical Computation & Estimation Short Notes

Numerical Computation & Estimation Weightage in GATE DA

Numerical Computation & Estimation Chapter Matrix

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