CUET GAT 2026: Most Important Topics & Questions - Score Maximum

The GAT (General Aptitude Test) section of CUET 2026 is crucial for undergraduate admissions in central universities. This section features questions covering General Knowledge, Current Affairs, Logical Reasoning, Quantitative Aptitude, and General Mental Ability.

If you aim to maximize your score in the GAT, it is essential to focus on the right topics. We have compiled a set of high-frequency questions - complete with their answers - drawn from the most critical topics within the CUET GAT syllabus.

We have curated key questions from the most vital CUET GAT topics, including the Number System, Percentages, Profit & Loss, Time & Work, Speed ​​& Distance, Averages, Ratio & Proportion, Simple & Compound Interest, and Data Interpretation.

These questions are designed based on the NTA's examination pattern and trends observed in previous years. For smart and effective preparation, be sure to practice these questions. You can access the topic-wise questions provided below.

CUET GAT Important Topics with Questions

Quantitative Aptitude (Quant)

This section tests numerical ability and problem-solving skills. Master formulas and shortcuts to solve questions quickly.

Average

The Average (or औसत) of a set of numbers is the sum of all entries divided by the total number of entries. This means Sum of All = Average × Total Number of Entries. For example, if the average of 12 numbers is 46, their initial sum is 12 × 46 = 552. If four numbers (20, 8, 12, 40) are removed (sum = 80), the new sum becomes 552 - 80 = 472, and the new count is 12 - 4 = 8. The new average is then 472 / 8 = 59. (To multiply 12 by 46, consider 12 as 10 + 2: 46 × 10 = 460 and 46 × 2 = 92, totaling 552.)

Ratios and Proportions

A ratio shows a relationship between quantities and is never given in fractions. To find exact values from a ratio like 2:5, a constant (x) is used, making them 2x and 5x. The "constant x" is crucial for solving ratio problems involving exact values.

For fractional ratios, convert them to whole numbers by finding the Least Common Multiple (LCM) of denominators. For example, to convert 1/5 : 9/18 : 11/16, the LCM of 5, 18, 16 is 720. Multiplying each fraction by 720 gives 144 : 360 : 495. If the difference between the greatest (495x) and smallest (144x) numbers is 78, then 351x = 78, so x = 2/9. The largest number (495x) would then be 495 × (2/9) = 110.

Mean Proportion

For three numbers a, x, and b, if x is the mean proportion between a and b, then a/x = x/b, which implies x² = ab, and thus x = √ab. If 42 is the mean proportion between x and 294, then x / 42 = 42 / 294. Solving for x gives x = (42 × 42) / 294 = 6.

Highest Common Factor (HCF) and Least Common Multiple (LCM)

The HCF is the largest positive integer that divides two or more numbers without a remainder. It represents the common factor in all numbers. For example, HCF(16, 56) = 8.

A 100% guaranteed important formula for examinations is the relationship between HCF, LCM, and two numbers:

Product of Two Numbers = HCF × LCM

• If two numbers are in the ratio 8:7 and their HCF is 30, the numbers are 8 × 30 = 240 and 7 × 30 = 210. Their LCM can be found as (240 × 210) / 30 = 1680.

• If the HCF of two numbers is 18 and their LCM is 1296, and one number is 162, the other number (N2) is (18 × 1296) / 162 = 144.

• The LCM of 45, 63, and 81 is found by prime factorization: 45 = 3² × 5, 63 = 3² × 7, 81 = 3⁴. The LCM is 3⁴ × 5 × 7 = 2835.

• If the product of two numbers is 324 and their HCF is 3, their LCM is 324 / 3 = 108.

Ages

Age problems involve setting up equations for past, present, or future ages. If A is the present age:

• Age 'x' years from now = A + x

• Age 'x' years ago = A - x

For example, if A is 9 years older than B (A = B + 9), and in 10 years, A will be twice as old as B was 10 years ago (A + 10 = 2(B - 10)), then:

1. Substitute A = B + 9 into the second equation: (B + 9) + 10 = 2B - 20

2. Simplify: B + 19 = 2B - 20

3. Solve for B: 39 = B.

4. Then A = 39 + 9 = 48.
   The present age of A is 48 years.

A key concept is that the age difference between any two individuals remains constant throughout their lives. (If there's a 2-year age gap, it existed at birth, exists now, and will always exist.) If Gopal and Manoj have a 2-year difference, and after 18 years their age ratio is 21:22, the 1-unit difference in ratio corresponds to 2 years. So, 1 unit = 2 years. Their ages after 18 years are 42 and 44. Their present ages are 24 and 26. The sum of their ages 10 years ago would be (24-10) + (26-10) = 14 + 16 = 30 years.

Financial Calculations: Income, Expenditure, and Savings

The fundamental principle is Income = Expenditure + Savings. If Sudha's expenditure is 150% more than her savings, and 150% = 3/2, this means if Savings = 2 units, then Expenditure = 2 + 3 = 5 units. For instance, if Savings = ₹2000, Expenditure = ₹5000, and Income = ₹7000.

If her expenditure decreases by 7% (5000 - 350 = ₹4650) and savings increase by 21% (2000 + 420 = ₹2420), her new income is ₹4650 + ₹2420 = ₹7070. The percentage change in income is ((7070 - 7000) / 7000) * 100 = 1% increase.

Percentage Problems

If 20% of a number is added to 90, the result is the number itself. This implies that 90 represents the remaining 80% of the number (100% - 20%). Therefore, 80% of that number is 90. Similarly, if 20% of a number is added to 78, the result is the number itself, so 80% of that number is 78.

To find what percentage of B's marks A's marks were, if B got 50 and A got 75, calculate (75 / 50) * 100 = 150%.

Election Problem: Multi-Step Percentage Calculation

Question: Let Total Registered Voters = x. If 70% cast their vote, and of those, 60% voted for candidate X, then 40% voted for candidate Y. If Y received 8400 votes:

x * (70/100) * (40/100) = 8400

x * 0.7 * 0.4 = 8400

x * 0.28 = 8400

x = 8400 / 0.28 = 30,000 Total Registered Voters. (Consider a school with 100 students (total voters). If 90% vote, then 90 students vote. If 60% of *those who voted go to A, and 40% to B, then 40% of 90 students = 36 students voted for B. This shows how percentages are applied sequentially.*)

Costing and Pricing Concepts: MRP, SP, CP

• MRP (Marked Retail Price): Price displayed.

• SP (Selling Price): Actual selling price.

• CP (Cost Price): Price for the seller.

Discount is always applied on MRP, and Profit/Loss is always calculated on CP. (Imagine a pen: MRP is ₹50 (sticker price). You get a discount, so SP is ₹30 (what you pay). The shopkeeper bought it for CP ₹20 (his cost).).

If a shopkeeper marks goods at ₹1980 (MRP) and allows a 12% discount (3/25), SP is 22 units if MRP is 25 units (MRP:SP = 25:22). If he makes 32% profit (8/25) on CP, SP is 33 units if CP is 25 units (CP:SP = 25:33).

To combine these ratios, equalize SP (LCM of 22 and 33 is 66):

MRP:SP = 75:66 (multiplying by 3)

CP:SP = 50:66 (multiplying by 2)

Combined ratio: CP : SP : MRP = 50 : 66 : 75.

Since MRP = ₹1980 (75 units), 1 unit = 1980/75. CP = 50 units = (1980/75) * 50 = ₹1320.

If a trader marks goods 35% above CP (7/20), then CP:MRP = 20:27. He gives a 15% discount (3/20), so MRP:SP = 20:17.

Equalizing MRP (LCM of 27 and 20 is 540):

CP:MRP = 400:540 (multiplying by 20)

MRP:SP = 540:459 (multiplying by 27)

Combined ratio: CP : MRP : SP = 400 : 540 : 459.

Profit = SP - CP = 459 - 400 = ₹59. Profit percentage = (59 / 400) * 100 = 14.75%.

If Ravi bought a watch for ₹4800, spent ₹200 on repair, and sold it for ₹6000:

• Total CP = ₹4800 + ₹200 = ₹5000.

• SP = ₹6000.

• Profit = ₹1000.

• Profit % = (1000 / 5000) * 100 = 20%.

Discount Problem: "Buy N Get M Free" Offer

For a "Buy 16 Get 9 Free" offer, the discount is on the total number of items received. You pay for 16 but get 25 (16+9). The free items are 9. If each item is ₹1, the discount is ₹9 on a total MRP of ₹25. 

Net percentage discount = (9 / 25) * 100 = 36%.

Simple Interest Problem: Effect of Rate Change

If a sum was invested for 4 years, and a 3% higher interest rate would yield an additional ₹960, this additional amount comes from the 3% extra for 4 years, which is a total of 12% of the Principal. So, 12% of P = ₹960. P = (960 * 100) / 12 = ₹8000.

Compound Interest Calculation (Formula-Free Method)

Compound interest is successive percentage growth.

To calculate the amount of ₹3,200 at 10% compound interest for 2 years:

• Year 1: 10% of 3200 = 320. Amount = 3200 + 320 = ₹3520.

• Year 2: 10% of 3520 = 352. Amount = 3520 + 352 = ₹3872.
  Alternatively, use the successive percentage formula (a + b + ab/100) for 10% and 10%: 10 + 10 + (10*10)/100 = 21%.
  Compound Interest = 21% of 3200 = ₹672. Total Amount = 3200 + 672 = ₹3872.

Partnership Investment and Profit Sharing

Profit is distributed among partners in the same ratio as their investments.

If Anmol and Ayushi jointly invested ₹4,600, total profit was ₹7,500, and Anmol's profit was ₹2,700:

Anmol's Profit Ratio = 2700 / 7500 = 9/25.

This ratio is also Anmol's investment ratio to the total investment.

Anmol's Investment = (9/25) * 4600 = ₹1656.

Mixture and Alligation (Acid & Water Ratio)

In a 330-liter solution with acid:water ratio of 1:2, initial acid = 110 liters, water = 220 liters. If only acid is added to get a new ratio of 8:5, the water quantity remains constant (220 liters). In the new ratio, 5 parts represent water, so 5 units = 220 liters. 1 unit = 44 liters. Final acid = 8 units * 44 = 352 liters. Acid added = 352 - 110 = 242 liters.

Ages Ratio (Building Analogy)

Five years ago, ratio of Building 1:Building 2 ages was 7:2. Five years hence, the ratio will be 3:1. The time gap between "5 years ago" and "5 years hence" is 10 years.

Let ages 5 years ago be 7x and 2x. Ages 5 years hence are 7x+10 and 2x+10.

(7x+10) / (2x+10) = 3/1

7x + 10 = 6x + 30

x = 20.

Building 2's age 5 years ago = 2x = 40 years.

Present age of Building 2 = 40 + 5 = 45 years.

This can also be solved using the "Mentos Method" by cross-multiplying ratios and using the time gap.

Train and Tunnel Length

When a train crosses a tunnel, the total distance covered = Length of Train (L) + Length of Tunnel.

Speed = Distance / Time.

If a train crosses tunnels of 2708m in 88s and 2195m in 76s, and its speed is constant:

(L + 2708) / 88 = (L + 2195) / 76

Simplifying (dividing denominators by 4):

(L + 2708) / 22 = (L + 2195) / 19

19(L + 2708) = 22(L + 2195)

19L + 51452 = 22L + 48290

3L = 3162

L = 1054 meters.

Speed, Distance, Time (Returning Speed)

Distance = Speed * Time. If distance is constant: D = S1 * T1 = S2 * T2.

Neelu walks a distance in 6 hours and returns in 9 hours with a speed decrease of 2 km/hr.

Let original speed = S1, returning speed = S2.

S1 = D/6, S2 = D/9.

S1 - S2 = 2

(D/6) - (D/9) = 2

(3D - 2D) / 18 = 2

D/18 = 2 => D = 36 km.

Returning speed S2 = D/9 = 36/9 = 4 km/hr.

Work and Time (Men & Women)

Total Work = Efficiency * Number of Days.

5 Men (M) do work in 16 days. 9 Women (W) do same work in 8 days.

5M * 16 = 9W * 8

80M = 72W => 10M = 9W.

So, 1 Man = (9/10) Women.

Total work = 9W * 8 = 72 Woman-Days.

Workforce = 3 Women + 2 Men = 3W + 2(9/10)W = 3W + 9/5W = (15+9)/5W = 24/5 Women.

Days to complete = Total Work / Combined Workforce = 72 / (24/5) = 72 * (5/24) = 3 * 5 = 15 days.

Work and Time (Man-Days-Work Formula)

The formula is m1 * d1 / w1 = m2 * d2 / w2.

If 3 men dig a 39-feet well in 4 days (m1=3, d1=4, w1=39), how many men (m2) are needed to dig a 78-feet well (w2=78) in 12 days (d2=12)?

(3 * 4) / 39 = (m2 * 12) / 78

12 / 39 = (m2 * 12) / 78

(12 * 78) / 39 = m2 * 12

12 * 2 = m2 * 12

m2 = 2 men.

Geometry - External Angle of a Triangle

The exterior angle of a triangle is equal to the sum of its two opposite interior angles.

If exterior Angle BCD = 145 degrees and interior Angle ABC = 70 degrees, then the other opposite interior angle (Angle A) = 145 - 70 = 75 degrees.

Geometry - Isosceles Triangle Angles

In an isosceles triangle, if two sides are equal (AB = AC), then the angles opposite to these sides are also equal (Angle ABC = Angle ACB). The sum of interior angles in a triangle is 180 degrees.

If Angle BAC = 50 degrees and AB = AC, then Angle ABC = Angle ACB = (180 - 50) / 2 = 65 degrees.

The exterior angle Angle ACD can be found in two ways:

1. Exterior Angle Theorem: Angle ACD = Angle BAC + Angle ABC = 50 + 65 = 115 degrees.

2. Angles on a Straight Line: Angle ACB + Angle ACD = 180 degrees. So, 65 + Angle ACD = 180 => Angle ACD = 115 degrees.

Algebraic Identities (Solving for a+b given a-b and ab)

Given a - b = 6 and a * b = 135. Find a + b.

1. Use (a - b)² = a² + b² - 2ab:
   6² = a² + b² - 2(135)
   36 = a² + b² - 270
   a² + b² = 36 + 270 = 306.

2. Use (a + b)² = a² + b² + 2ab:
   (a + b)² = 306 + 2(135)
   (a + b)² = 306 + 270 = 576
   a + b = √576 = 24.

Divisibility Rule of 8

A number is divisible by 8 if its last three digits are divisible by 8.

For 28p0, the last three digits are 8p0. If p=1, 810 is not divisible by 8 (810/8 = 101.25). So, 1 cannot be the value of p.

Divisibility Rule of 11

A number is divisible by 11 if the difference between the sum of digits at odd places and the sum of digits at even places is 0 or a multiple of 11.

For 78p3945:

• Odd places (from right): 5 + 9 + p + 7 = 21 + p.

• Even places (from right): 4 + 3 + 8 = 15.

• Difference = (21 + p) - 15 = 6 + p.
  For divisibility by 11, 6 + p must be 11 (as p is a single digit).
  p = 11 - 6 = 5.

Divisibility by 6, 9, and 12

• Divisibility by 6: By 2 and 3.

• Divisibility by 9: Sum of digits by 9.

• Divisibility by 12: By 3 and 4.
  If a number is divisible by 6 (thus by 2 and 3) and by 9 (thus by 3), but not by 12, it must fail the divisibility by 4 test. For example, 126 is divisible by 6 (even, sum=9) and 9 (sum=9), but not by 12 because its last two digits (26) are not divisible by 4.

Reasoning

Tests logical thinking and pattern recognition. Regular practice improves speed and accuracy.

Counting Consecutive Odd Numbers

In a given sequence, we count sets of three consecutive odd numbers (Odd-Odd-Odd). The analysis confirms two such sequences.

Counting Letters in a Series

To find letters appearing after Q but before E in a specific series, a direct count is needed. After careful re-checking, there are three such letters.

Identifying Position of an Element

The position of 'L' from the left end in the series is the fifth position.

Counting Symbol Patterns

To count sequences of Symbol-Symbol-Number, direct scanning of the series reveals two such patterns.

Word Transformation and Vowel Counting

If each letter in a word is changed to its next letter, and we need to find how many transformed words have no vowels, the analysis indicates that only one such word exists.

Letter Series Pattern Identification (IXBG, EWHEVH, CUK)

For the series IXBG, EWHEVH, CUK:

1. First Letter: I(9) -> G(7) -> E(5) -> C(3). Pattern is -2. Next is A(1).

2. Second Letter: X(24) -> W(23) -> U(21). Pattern is -1. Next is T(20).

3. Third Letter: B(2) -> E(5) -> H(8) -> K(11). Pattern is +3. Next is N(14).
   The derived next term is ATN.

Letter Series with Consistent Subtraction (VMQ, PKO, RIM, PGK)

For the series VMQ, PKO, RIM, PGK, the pattern is a consistent subtraction of 2 from the numerical value of each letter.

Applying this to PGK:

P(16) - 2 = N(14)

G(7) - 2 = E(5)

K(11) - 2 = I(9)

The derived next term is NEI.

Number Series Pattern (Arithmetic/Multiplicative)

A series (X, Y, Z) follows the pattern: Y = (X * 3) + 7 and Z = Y + 4.

Example: For 11, 40, 44: (11 * 3) + 7 = 40, and 40 + 4 = 44. The key operations are * 3, then + 7, then + 4.

Identifying Analogous Letter Series Pattern (OH, QK, TV)

Given reference series OH, QK, TV:

1. First Letter: O(15) -> Q(17) (+2); Q(17) -> T(20) (+3). Pattern: +2, +3.

2. Second Letter: H(8) -> K(11) (+3); K(11) -> V(22) (+11). Pattern: +3, +11.
   Option C (RKTN and WY) matches this pattern:

3. R(18) -> T(20) (+2); T(20) -> W(23) (+3).

4. K(11) -> N(14) (+3); N(14) -> Y(25) (+11).
   Thus, Option C is the correct match.

Word Transformation by Letter Reordering

This involves reordering letters of a word based on a sequential pattern. While specific examples like ECHO and URGE were discussed, the precise reordering rule was not consistently clarified. The conclusion points to Option A as the correct answer related to a transformed word "TW".

General Knowledge

Covers history, economy, and science topics. Important for scoring easy marks in CUET GAT.

Delhi Sultanate: Razia Sultana

Razia Sultana was removed from the throne in 1240 AD, having reigned from 1236 AD - 1240 AD. She belonged to the Slave Dynasty (Mamluk Dynasty or Aibak Dynasty) and was the first and only female Muslim ruler of Medieval India. She was the daughter of Shamsuddin Iltutmish, who notably nominated her as his successor over his sons.

Indian Economy & Five Year Plans

The Eighth Five-Year Plan (1992 - 1997) was launched after the Structural Adjustment Policies, also known as LPG Reforms (Liberalization, Privatization, Globalization). These policies were introduced in 1991 to combat a severe balance of payments crisis and are associated with the Rao-Manmohan Model, named after Prime Minister P.V. Narasimha Rao and Finance Minister Dr. Manmohan Singh.

Science (Atmosphere Layers and Radio Transmission)

The Thermosphere is the atmospheric layer that aids radio transmission. It contains the Ionosphere, which has electrically charged particles that reflect radio waves back to Earth, enabling long-distance communication. The layers of Earth's atmosphere from lowest to highest are Troposphere, Stratosphere, Mesosphere, Thermosphere, and Exosphere. (Exosphere is the highest layer, like an "exit" from Earth. Troposphere is the lowest layer.)

Science (Human Brain Anatomy and Function)

The Medulla (Medulla Oblongata), part of the Hindbrain, controls involuntary actions such as blood pressure, heartbeat, and respiration. The Hindbrain also includes the Cerebellum (controls precision, posture, balance) and Pons (regulatory and reflective system, connecting brain regions).