Boost your speed and accuracy on any competitive exam. Discover the top tricks for solving Numerical Ability & Quantitative Aptitude MCQs. Start practicing today!
Tips & Tricks Concepts
Multiplication by 11: For a 2-digit number ab, the result is a | a+b | b.
Finding Squares (Ending in 5): The square of n5 is n*(n+1) followed by 25.
Unit Digit Method: To check a multiplication, just multiply the last digits of the numbers.
Approximation: If options are far apart, round off numbers before calculating.
Interchangeability: x% of y is the same as y% of x.
Fraction Conversion: Memorize fraction-to-percentage values (e.g., 1/8 = 12.5%) for fast calculation.
Successive Change: For two successive percentage increases x and y, the net increase is (x + y + (xy/100))%.
Successive Discounts: For two successive discounts x% and y%, the net discount is (x + y - (xy/100))%.
Combining Ratios: If A:B = m:n and B:C = p:q, then A:B:C = mp : np : nq.
Relative Speed (Opposite): When objects move in opposite directions, their speeds add up.
Relative Speed (Same): When objects move in the same direction, their speeds are subtracted.
Average Speed (Two Equal Distances): Average Speed = (2 * s1 * s2) / (s1 + s2).
LCM Method: Assume total work = LCM of the individual days to avoid fractions.
Efficiency Concept: Work = Time × Efficiency.
CI Rule of 72: The years it takes for money to double is approximately 72 / (interest rate).
CI vs SI (2 Years): The difference between CI and SI for 2 years is P(R/100)².
Observation over Calculation: Scan the graph/table for the answer before doing any complex math.
Fibonacci Sequence: A sequence where each number is the sum of the two preceding ones, starting from 0 and 1. Example: 0, 1, 1, 2, 3, 5, 8, 13...
Fundamental Logarithm Rules 📜
The Product Rule: logb(MN) = logb(M) + logb(N). Use this to combine separate log terms into one or to break a complex log down.
The Quotient Rule: logb(M/N) = logb(M) - logb(N). Simplifies expressions involving division inside a logarithm.
The Power Rule: logb(Mk) = k ⋅ logb(M). This is the most useful rule for solving equations as it helps bring down exponents.
Change of Base & Key Identities
Change of Base Formula: logb(M) = logc(M) / logc(b). Essential when you need to simplify an expression with multiple bases.
Log of 1: logb(1) = 0. The logarithm of 1 to any base is always 0.
Log of the Base: logb(b) = 1. The logarithm of a number to the same base is always 1.
Core Concepts for Solving MCQs
The Log-to-Exponential Relationship 🔄: logb(x) = y is the same as by = x. If you're stuck on a log equation, immediately rewrite it in its exponential form.
Equating Logarithms: If logb(M) = logb(N), then M = N. Use the other rules to get a single log on each side, then drop the logs and solve.
The Multiplication Principle (The Slot Method) 🎰
Concept: If a task has multiple steps, the total number of ways to complete it is the product of the ways to do each step. Think of filling slots (_ _ _) and multiplying the options for each slot.
Basic Permutation Formula (Order Matters!)
Rule: The number of ways to arrange r items from a set of n distinct items is P(n, r) = n! / (n-r)!
MCQ Tip: Look for keywords like "arrange," "order," "rank," or specific roles (e.g., President, VP). If order is important, it's a permutation.
Permutations with Repetition (Non-Distinct Items)
Rule: The number of distinct arrangements of n items with repetitions is n! / (r1! ⋅ r2! ...), where r is the count of each repeated item.
MCQ Tip: This is most common for arranging letters in a word with repeated letters, like "MISSISSIPPI."
Circular Permutations 🔄
Rule: The number of ways to arrange n distinct items in a circle is (n-1)!
MCQ Tip: Use for seating arrangements around a table. If items can be flipped (like a necklace), divide by 2.
Strategic Tips for Solving MCQs
Identify Keywords: "Arrange/order" points to permutations. "Choose/select" often points to combinations.
Use Casework: For problems with multiple conditions (e.g., "starts with a vowel"), break the problem into smaller cases, solve each, and add the results.
The Complement Rule (The "Not" Rule): Sometimes it's easier to find the total arrangements and subtract the ones you don't want. Useful for questions with phrases like "at least one" or "never together."
Test Your Knowledge
1. The Fibonacci sequence is the sequence of integers: [ ISRO CSE 2007 ]
The Fibonacci sequence is a special series of numbers where each number is the sum of the two that come before it. It starts with 0 and 1.
0 + 1 = 1
1 + 1 = 2
1 + 2 = 3
2 + 3 = 5
...and so on.
Trick to Remember
Just pick a number in the sequence and see if it's the sum of the two before it. For example, in option B, let's pick 8. The two numbers before it are 3 and 5. Does 3 + 5 = 8? Yes! This simple check confirms it's a Fibonacci sequence.
2. The exponent of 11 in the prime factorization of 300! is: [ GATE IT 2008 ]
Answer & Explanation
Correct Answer: C) 29
To find the highest power of a prime number (p) in a factorial (n!), we use Legendre's Formula. We repeatedly divide n by p and its powers, and sum up the integer parts of the results.
Here, n = 300 and p = 11.
Step 1: Divide 300 by 11 → ⌊300 / 11⌋ = 27
Step 2: Divide 300 by 11² (121) → ⌊300 / 121⌋ = 2
Step 3: Divide 300 by 11³ (1331) → ⌊300 / 1331⌋ = 0. We stop here.
Total Exponent = 27 + 2 = 29.
Trick to Remember
Think of it as "successive division." Keep dividing the main number (300) by the prime (11), and add up all the whole number results you get. It's a quick way to find the power of any prime in a large factorial without actual calculation.
3. 25 persons are in a room. 15 of them play hockey, 17 of them play football and 10 of them play both hockey and football. Then the number of persons playing neither hockey nor football is: [ GATE CSE 2010 ]
Answer & Explanation
Correct Answer: D) 3
This is a classic set theory problem. We first find the number of people who play at least one sport using the formula: |A ∪ B| = |A| + |B| - |A ∩ B|.
Number who play at least one sport = (Plays Hockey) + (Plays Football) - (Plays Both)
= 15 + 17 - 10
= 32 - 10 = 22
Now, we subtract this from the total number of persons to find those who play neither.
Number playing neither = Total Persons - Number playing at least one sport
= 25 - 22 = 3
Trick to Remember (Venn Diagram Method)
Imagine two overlapping circles. Start from the middle: the overlapping section is "Both," which is 10. The "Hockey only" circle is 15 - 10 = 5. The "Football only" circle is 17 - 10 = 7. Add up everyone in the circles: 5 + 7 + 10 = 22. The people outside the circles are Total - 22, which is 25 - 22 = 3.
4. If log(P) = (1/2)log(Q) = (1/3)log(R), then which of the following options is TRUE?
Answer & Explanation
Correct Answer: B) Q2 = PR
This problem uses the power rule of logarithms, which states that n * log(x) = log(xn).
First, apply this rule to the given equation:
Given: log(P) = (1/2)log(Q) = (1/3)log(R)
This becomes: log(P1) = log(Q1/2) = log(R1/3)
Since the logs are equal, their arguments must be equal:
P = Q1/2 → Squaring both sides gives Q = P2
P = R1/3 → Cubing both sides gives R = P3
Now, test the options by substituting Q and R with their P equivalents. Let's test option B:
Is Q2 = PR?
Substitute: (P2)2 = P * (P3)
Simplify: P4 = P4
The equation holds true, so option B is the correct answer.
Trick to Remember
Set the entire expression to a constant 'k'. This gives you: log(P)=k, log(Q)=2k, log(R)=3k. This shows a simple power relationship: Q is related to P by a square, and R is related to P by a cube (Q=P2, R=P3). You can quickly substitute these simple power relationships into the options to find the correct one.
5. Given the digits 2, 2, 3, 3, 3, 4, 4, 4, 4, how many distinct 4-digit numbers greater than 3000 can be formed? [ GATE CSE 2010 ]
Answer & Explanation
Correct Answer: B) 51
This is a permutation with repetition problem. Since the number must be greater than 3000, the first digit can only be 3 or 4. We solve this using two cases.
Case 1: The first digit is 3.
The remaining 3 digits must be formed from the set {2, 2, 3, 3, 4, 4, 4, 4}.
Three same digits: {4, 4, 4}. Arrangements: 3!/3! = 1.
All different digits: {2, 3, 4}. Arrangements: 3! = 6.
Total for Case 2 = 2 + (6 + 6 + 6) + 6 = 26.
Grand Total
Total distinct numbers = (Total from Case 1) + (Total from Case 2) = 25 + 26 = 51.
Trick to Remember: Case Work is Key
For permutation problems with conditions (like "greater than 3000") and repeated items, the safest strategy is Case Work. Break the problem down based on the conditional digit (here, the first digit being 3 or 4). Then, for each case, systematically find all possible combinations for the remaining slots and calculate their permutations. This structured approach prevents errors and ensures you don't miss any possibilities.
6. 5 skilled workers can build a wall in 20 days; 8 semi-skilled workers can build a wall in 25 days; 10 unskilled workers can build a wall in 30 days. If a team has 2 skilled, 6 semi-skilled and 5 unskilled workers, how long will it take to build the wall? [ GATE CSE 2010 ]< /p>
Answer & Explanation
Correct Answer: D) 15 days
This is a work-rate problem. The key is to find the work rate of an individual for each skill level and then combine the rates for the new team.
Step 1: Find the time for ONE worker of each type.
We calculate the "man-days" required for the job for each category.
New Team's total efficiency = (2 * 6) + (6 * 3) + (5 * 2) = 12 + 18 + 10 = 40 units/day.
Time to complete = Total Work / Total Efficiency = 600 / 40 = 15 days.
7. Hari(H), Gita(G), Irfan(I) and Saira(S) are siblings. All were born on 1st January. The age difference between any two successive siblings is less than three years. Given the following facts:
i. Hari's age + Gita's age > Irfan's age + Saira's age
ii. The age difference between Gita and Saira is one year. However, Gita is not the oldest and Saira is not the youngest.
iii. There are no twins.
In what order they were born (oldest first)? [ GATE CSE 2010 ]
Answer & Explanation
Correct Answer: B) SGHI
The most efficient way to solve this is to test each option against the given facts. The most restrictive fact is rule (ii).
Fact (ii) Analysis: Gita (G) and Saira (S) have an age difference of 1 year, so they must be born one after another. Also, G cannot be the oldest, and S cannot be the youngest.
Testing the Options:
A) HSIG (H > S > I > G): Fails. Gita and Saira are not born one after another. They are separated by Irfan.
C) IGSH (I > G > S > H): Fails. Let's check rule (i): H's age + G's age > I's age + S's age. This becomes (Youngest + 2nd Oldest) > (Oldest + 3rd Oldest). This is impossible because the oldest person's age will always be significantly higher.
D) IHSG (I > H > S > G): Fails. Let's check rule (i): H's age + G's age > I's age + S's age. This becomes (2nd Oldest + Youngest) > (Oldest + 3rd Oldest). This is also impossible.
B) SGHI (S > G > H > I): This is the only option left, let's verify it.
G & S are successive with a 1-year difference: Yes (S is oldest, G is 2nd). This is possible if S is 1 year older than G.
G is not oldest, S is not youngest: Yes. G is 2nd oldest and S is oldest. This condition is met.
No twins: Yes.
Successive age difference is less than 3 years: We can create a valid age set. For example, let their ages be S=14, G=13, H=11, I=10. All successive differences (1, 2, 1) are less than 3.
H + G > I + S: Using the sample ages: 11 + 13 > 10 + 14 => 24 > 24. This is not strictly greater. Let's try another set of ages allowed by the rules: S=14, G=13, H=12, I=10. The age gaps are (1, 1
8. If 137 + 276 = 435, how much is 731 + 672? [ GATE CSE 2010 ]
Answer & Explanation
Correct Answer: C) 1623
This is a logical reasoning problem, not a standard arithmetic one. The pattern is that the numbers are being added in base 8 (octal) instead of base 10.
Let's verify the example: 137 + 276 = 435 in base 8
Right column: 7 + 6 = 13. In base 8, 13 is (1 * 8) + 5. So, we write down 5 and carry over 1.
Middle column: 3 + 7 + 1 (carry) = 11. In base 8, 11 is (1 * 8) + 3. So, we write down 3 and carry over 1.
Left column: 1 + 2 + 1 (carry) = 4. We write down 4.
The result is indeed 435 in base 8.
Now, apply the same logic to 731 + 672:
Right column: 1 + 2 = 3. We write down 3.
Middle column: 3 + 7 = 10. In base 8, 10 is (1 * 8) + 2. We write down 2 and carry over 1.
Left column: 7 + 6 + 1 (carry) = 14. In base 8, 14 is (1 * 8) + 6. We write down 6 and carry over 1.
Since we have a final carried 1, the result is 1623.
Trick to Remember: Check the Base
When you see a math problem where the answer is wrong (e.g., 5 + 5 = 12), it's often a hint that the numbers are in a different base. The first place to check is the rightmost digit. In the example `137 + 276 = 435`, look at `7 + 6 = 13`. The result ends in a 5. Ask yourself, "In what base does 13 end in 5?" The answer is base 8, because 13 = 1×8 + 5. This quickly reveals the hidden pattern.
9. The cost function for a product in a firm is given by 5q2, where q is the amount of production. The firm can sell the product at a market price of Rs. 50 per unit. The number of units to be produced by the firm such that the profit is maximized is: [ GATE CSE 2012 ]
Answer & Explanation
Correct Answer: A) 5
To solve this, we need to find the profit function and then find the value of 'q' that maximizes it using calculus.
To find the maximum profit, we take the first derivative of the profit function with respect to q and set it to zero.
P'(q) = d/dq (50q - 5q2)
P'(q) = 50 - (2 * 5)q
P'(q) = 50 - 10q
Now, set the derivative to 0 to find the maximum:
50 - 10q = 0
10q = 50
q = 5
The firm should produce 5 units to maximize its profit.
Trick to Remember: The Marginal Rule (MR = MC)
A faster economic shortcut is to remember that profit is always maximized when **Marginal Revenue (MR) equals Marginal Cost (MC)**.
Marginal Revenue (MR) is the derivative of the Revenue function: R'(q) = 50.
Marginal Cost (MC) is the derivative of the Cost function: C'(q) = 10q.
Set them equal: MR = MC → 50 = 10q → q = 5. This method is often quicker than writing out the full profit function.
10. A container originally contains 10 litres of pure spirit. From this container, 1 litre of spirit is replaced with 1 litre of water. Subsequently, 1 litre of the mixture is again replaced with 1 litre of water and this process is repeated one more time. How much spirit is now left in the container? [ GATE CSE 2011 ]
Answer & Explanation
Correct Answer: D) 7.29 litres
This is a problem of repeated dilution. Each time the process is repeated, the amount of spirit is reduced by a certain fraction. The total volume of the liquid in the container remains constant at 10 litres.
Step-by-Step Calculation:
Initial Amount of Spirit: 10 litres.
After 1st Replacement: 1 litre of spirit is removed and replaced with water. The fraction of spirit remaining is (10-1)/10 = 9/10.
Amount of spirit left = 10 * (9/10) = 9 litres.
After 2nd Replacement: Now, 1 litre of the mixture is removed. The amount of spirit removed is 1 * (9/10). The fraction of spirit remaining from the previous step is again 9/10.
Amount of spirit left = 9 * (9/10) = 8.1 litres.
After 3rd Replacement: The process is repeated one more time.
Amount of spirit left = 8.1 * (9/10) = 7.29 litres.
So, 7.29 litres of spirit is left in the container.
Trick to Remember: The Dilution Formula
For repeated replacement problems, you can use a direct formula to save time:
Final Amount = Initial Amount × (1 - (Amount Replaced / Total Volume))n
Where 'n' is the number of times the operation is performed.
11. A transporter receives the same number of orders each day. Currently, he has some pending orders (backlog) to be shipped. If he uses 7 trucks, then at the end of the 4th day he can clear all the orders. Alternatively, if he uses only 3 trucks, then all the orders are cleared at the end of the 10th day. What is the minimum number of trucks required so that there will be no pending order at the end of the 5th day? [ GATE CSE 2011 ]
Answer & Explanation
Correct Answer: C) 6
This is a work-rate problem that can be solved with algebra. Let's define our variables:
B = Initial backlog of orders.
N = New orders received per day.
T = Orders one truck can ship per day (truck capacity).
Step 1: Formulate Equations from the Scenarios
The total orders cleared equals the initial backlog plus the new orders received over the period.
Scenario 1: 7 trucks for 4 days.
Total orders = B + 4N. Work done = 7 trucks * 4 days * T = 28T.
Equation (i): B + 4N = 28T
Scenario 2: 3 trucks for 10 days.
Total orders = B + 10N. Work done = 3 trucks * 10 days * T = 30T.
Equation (ii): B + 10N = 30T
Step 2: Solve the System of Equations
Subtract Equation (i) from Equation (ii) to eliminate B:
(B + 10N) - (B + 4N) = 30T - 28T
6N = 2T → 3N = T
This tells us that one truck's daily capacity is equal to three days' worth of new orders. Now find B by substituting T=3N into Equation (i):
B + 4N = 28 * (3N) = 84N
B = 84N - 4N = 80N
Step 3: Answer the Main Question
Let 'x' be the minimum number of trucks needed to clear all orders in 5 days.
Total Orders = Backlog (B) + 5 days of new orders (5N)
Work Done = x trucks * 5 days * T
Equation: B + 5N = 5xT
Substitute B=80N and T=3N into this equation:
80N + 5N = 5x * (3N)
85N = 15xN
85 = 15x
x = 85 / 15 = 17 / 3 ≈ 5.67
Since you cannot have a fraction of a truck, we must round up to the next whole number. Therefore, the minimum number of trucks required is 6.
Trick to Remember: Think in "Truck-Days"
A "truck-day" is the amount of work one truck does in one day.
• Scenario 1: 7 trucks × 4 days = 28 truck-days of work.
• Scenario 2: 3 trucks × 10 days = 30 truck-days of work.
The difference of 2 truck-days (30 - 28) was needed to handle the extra new orders received between day 4 and day 10 (which is 6 days).
So, 2 truck-days = 6 days of new orders. This means 1 truck-day = 3 days of new orders. This is a much faster way to find the core relationship (T=3N) and solve from there.
12. The variable cost (V) of manufacturing a product varies according to the equation V = 4q, where q is the quantity produced. The fixed cost (F) of production of the same product reduces with q according to the equation F = 100/q. How many units should be produced to minimize the total cost (V+F)? [ GATE CSE 2011 ]
Answer & Explanation
Correct Answer: A) 5
To find the minimum cost, we need to create a total cost function and then use calculus to find the quantity 'q' that minimizes it.
Step 1: Create the Total Cost Function
The total cost (C) is the sum of the variable cost (V) and the fixed cost (F).
Total Cost C(q) = V + F
C(q) = 4q + 100/q
Step 2: Find the Derivative of the Cost Function
To minimize the function, we find its first derivative with respect to q. It's easier to rewrite the function first:
C(q) = 4q + 100q-1
Using the power rule, the derivative C'(q) is:
C'(q) = 4 + (-1 * 100)q-2
C'(q) = 4 - 100/q2
Step 3: Set the Derivative to Zero and Solve for q
The minimum or maximum of a function occurs where its derivative is zero.
4 - 100/q2 = 0
4 = 100/q2
4q2 = 100
q2 = 25
q = 5
(We take the positive root since quantity cannot be negative). The total cost is minimized when 5 units are produced.
Trick to Remember: The AM-GM Shortcut
For cost functions in the form ax + b/x, the minimum value always occurs when the two terms are equal.
Simply set the variable cost equal to the fixed cost and solve:
Variable Cost = Fixed Cost
4q = 100/q
4q2 = 100
q2 = 25 → q = 5
This shortcut is much faster than using calculus for this specific type of problem.
13. P, Q, R and S are four types of dangerous microbes recently found in a human habitat. The area of each circle with its diameter printed in brackets represents the growth of a single microbe surviving the human immunity system within 24 hours of entering the body. The danger to human beings varies proportionately with the toxicity, potency and growth attributed to a microbe shown in the figure below:
A pharmaceutical company is contemplating the development of a vaccine against the most dangerous microbe. Which microbe should the company target in its first attempt? [ GATE CSE 2011 ]
Answer & Explanation
Correct Answer: D) S
As per the question, the danger of a microbe is directly proportional to its potency and growth, and inversely proportional to its toxicity.
Step 1: Formulate the Danger Equation
We can express the relationship as a formula for the Level of Danger (D):
Danger (D) ∝ Growth (G)
Danger (D) ∝ Potency (P)
Danger (D) ∝ 1 / Toxicity (T)
Combining these, we get the formula: D = (K × G × P) / T, where K is a constant of proportionality which we can take as 1 for comparison.
Step 2: Calculate the Danger Level for Microbe S
From the figure, for microbe S:
Potency (P) = 0.8
Toxicity (T) = 200
Growth (G) is the area of the circle. The value in the bracket (10) represents the diameter, so the radius is 5.
Area = π × r2 = π × 52 = 25π.
(Note: The provided answer image seems to use the diameter directly in the area calculation, which is unconventional. However, even with that method or the correct one, S remains the most dangerous).
Step 3: Compare Danger Levels
If we calculate the danger score for all four microbes, we find that microbe S has the highest value.
DP = (π×102 × 0.5) / 500 = π/10 ≈ 0.314
DQ = (π×152 × 0.4) / 300 = 0.3π ≈ 0.942
DR = (π×52 × 0.9) / 400 ≈ 0.056π ≈ 0.176
DS = (π×202 × 0.8) / 200 = 1.6π ≈ 5.026
Wait, re-reading the figure shows the diameters are P(20), Q(30), R(10), S(40). Let's re-calculate with correct diameters.
DP = (π×102 × 0.5) / 500 = π/10 = 0.1π
DQ = (π×152 × 0.4) / 300 = 0.3π
DR = (π×52 × 0.9) / 400 = 0.05625π
DS = (π×202 × 0.8) / 200 = 1.6π
Clearly, DS has the highest danger level. Therefore, the company should target microbe S first.
Trick to Remember: The "Danger Score" Method
For any proportionality problem, create a simple "score" formula. Put all the factors that make something 'more' of a quality (like danger) in the numerator, and all the factors that make it 'less' in the denominator.
Danger Score = (Growth × Potency) / Toxicity
Then, calculate this score for each option. The one with the highest score is your answer. This turns a word problem into a simple calculation race.
14. What will be the maximum sum of 44, 42, 40, ... ? [ GATE CSE 2013 ]
Answer & Explanation
Correct Answer: C) 506
This is an Arithmetic Progression (AP) where the terms are decreasing. The sum will be maximum if we only add the positive terms. As soon as we add a negative term, the sum will start to decrease.
Step 1: Identify the AP parameters
First term (a) = 44
Common difference (d) = 42 - 44 = -2
Step 2: Find the number of non-negative terms
We need to find the last term that is greater than or equal to zero. Let's find 'n' for the term an ≥ 0.
Formula for nth term: an = a + (n-1)d
44 + (n-1)(-2) ≥ 0
44 - 2n + 2 ≥ 0
46 ≥ 2n
23 ≥ n
So, there are 23 non-negative terms. The 23rd term will be the last one to add for the maximum sum.
Step 3: Calculate the sum of these 23 terms
We use the formula for the sum of an AP: Sn = n/2 * (2a + (n-1)d).
S23 = 23/2 * (2 * 44 + (23-1)(-2))
S23 = 23/2 * (88 + 22 * -2)
S23 = 23/2 * (88 - 44)
S23 = 23/2 * 44
S23 = 23 * 22 = 506
Trick to Remember: Sum of Positive Terms
For a decreasing AP, the maximum sum is always the sum of all its non-negative terms. To find the number of these terms quickly, you can use the formula: n = floor(|a/d|) + 1, but it can be tricky. A safer and quick way is to find the last term (l). We found the 23rd term is the last non-negative one. Let's calculate it: a23 = 44 + (22)(-2) = 0. Then use the simpler sum formula Sn = n/2 * (a + l).
S23 = 23/2 * (44 + 0) = 23 * 22 = 506. This is often much faster.
15. Given the sequence of terms, AD CG FK JP, the next term is: [ GATE CSE 2012 ]
Answer & Explanation
Correct Answer: D) PW
The key to solving this is to analyze the pattern of the first letters and the second letters of each pair separately.
Step 1: Analyze the First Letter of Each Pair
The series is A, C, F, J, ...
From A to C, we skip 1 letter (B). This is a +2 step.
From C to F, we skip 2 letters (D, E). This is a +3 step.
From F to J, we skip 3 letters (G, H, I). This is a +4 step.
The pattern is an increasing step: +2, +3, +4, ... The next step must be +5.
From J, a +5 step (skipping K, L, M, N, O) brings us to P.
Step 2: Analyze the Second Letter of Each Pair
The series is D, G, K, P, ...
From D to G, we skip 2 letters (E, F). This is a +3 step.
From G to K, we skip 3 letters (H, I, J). This is a +4 step.
From K to P, we skip 4 letters (L, M, N, O). This is a +5 step.
The pattern is an increasing step: +3, +4, +5, ... The next step must be +6.
From P, a +6 step (skipping Q, R, S, T, U, V) brings us to W.
Step 3: Combine the Results
The first letter is P and the second letter is W. Therefore, the next term in the sequence is PW.
Trick to Remember: Divide and Conquer
For letter series with pairs or groups of letters, don't try to solve the whole group at once. Use the "Divide and Conquer" strategy: split the problem into simpler individual series. Analyze the pattern of all the first letters, then all the second letters, and so on. This makes complex patterns much easier to spot.
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