Question 1

CAT 2025 Slot 3: In a triangle ABC, AB=AC=12 cm and D is a point on side BC such that AD= 8 cm. If AD is extended to point E such that angle ACB = angle AEB , then length in cm of AE is :

Accepted Answer

Answer: B · 18. Based on the information in the question triangle ABD ≈ triangle AEB , Thus : AB/AE = AD/AB , 12/AE = 8/12 = 2/3 Only option 2 satisfies the condition. Thus option 2 is the answer

Question 2

CAT 2025 Slot 3: A triangle ABC is formed with AB = AC = 50 cm and BC = 80 cm. Then, the length, in cm , of all three altitudes of the triangle ABC is….. …

Accepted Answer

Answer: 126. Give AB = AC = 50 cm , BC = 80 cm The altitude from A to BC Half of BC: 80 / 2 = 40 cm Use Pythagoras' theorem to find altitude h: h² + 40² = 50² h² + 1600 = 2500 h² = 900 , h = 30 cm Altitude from A to BC = 30 cm The altitudes from B and C Area of the triangle using the altitude from A: Area = (1/2) × 80 × 30 = 1200 cm² Using area to find the altitude from B: 1200 = (1/2) × 50 × h h = 1200 / 25 = 48 cm Altitude from B to AC = 48 cm Altitude from C to AB = 48 cm Sum = 30 + 48 + 48 = 126 Answer 126

Question 3

CAT 2025 Slot 3: If (x²+ 1/x²) = 25 and x > 0, then the value of (x⁷ + 1/x⁷) is:

Accepted Answer

Answer: B · 44853√3. Given x² + 1/x² = 25 , x > 0 Using identity (a + b)² = a² + b² + 2ab I. x + 1/x: (x + 1/x)² = x² + 2 + 1/x² = 25 + 2 = 27 So, x + 1/x = 3√3 II. x³ + 1/x³: x³ + 1/x³ = (x + 1/x) * (x² + 1/x²) - (x + 1/x) x³ + 1/x³ = (3√3) * 25 - 3√3 = 72√3 III.x⁴ + 1/x⁴: x⁴ + 1/x⁴ = (x² + 1/x²)² - 2 x⁴ + 1/x⁴ = 25² - 2 = 625 - 2 = 623 IV.x⁷ + 1/x⁷: x⁷ + 1/x⁷ = (x³ + 1/x³) * (x⁴ + 1/x⁴) - (x + 1/x) x⁷ + 1/x⁷ = (72√3) * 623 - 3√3 = 44853√3

Question 4

CAT 2025 Slot 3: ABCD is a trapezium in which AB is parallel to DC and AD is perpendicular to AB and AB = 3DC. If a circle inscribed in a trapezium touching all the sides has a radius 3cm , then the area in sq cm , of the trapezium is :

Accepted Answer

Answer: A · 48. Given: Let the trapezium be ABCD. AB || DC (The parallel sides are the bases). AD perpendicular to AB (The trapezium is a right-angle trapezium). AB = 3DC. If AB = x , then DC = 3x The circle inscribed in the trapezium has a radius r = 3cm Height = 2r = 2 × 3 = 6cm For cyclic quadrilateral Sum of parallel sides= Sum of non-parallel sides AB + DC = AD + BC 3x + x = 6 + BC BC = 4x - 6 —-- I In Triangle AEB with base BE , height CE and hypotenuse BC BE = DC - AB = 3x - x = 2x , CE = 6cm , BC = 4x - 6 Using Pythagoras: (2x)² + (6)² = (4x - 6)² On solving we have x = 4cm Area of trapezium = ½ * (AB

Question 5

CAT 2025 Slot 2: Let ABCDEF be a regular hexagon and P and Q be the midpoints of AB and CD, respectively. Then, the ratio of the areas of trapezium PBCQ and hexagon ABCDEF is _______

Accepted Answer

Answer: 5 : 24. Let area of the hexagon is 6E, where E represents the area of any one of the six congruent central triangles, such as triangle OBC. The area of trapezium PBCQ is obtained by decomposing it into simpler regions. It consists of the full area of triangle OBC along with suitable fractional parts of the triangles adjacent to the base BC. This decomposition gives: Area(PBCQ) = 5E/4. Therefore, the required ratio of areas is: (5E/4) ÷ 6E = 5/24.

Question 6

CAT 2025 Slot 2: In a triangle ABC, points D and E are on the sides BC and AC, respectively. BE and AD intersect at point T such that AD : AT = 4 : 3, and BE : BT = 5 : 4. Point F lies on AC such that DF is parallel to BE. Then, BD : CD is

Accepted Answer

Answer: C · 11 : 4. Given, AD : AT = 4 : 3 ⇒ AT : TD = 3 : 1 and BE : BT = 5 : 4 ⇒ BT : TE = 4 : 1 By the mass points method, masses are assigned inversely to the segments. From AT : TD = 3 : 1 ⇒ Mass at A : Mass at D = 1 : 3. From BT : TE = 4 : 1, ⇒ Mass at B : Mass at E = 1 : 4. Choosing consistent masses, A = 5/4, D = 15/4 and B = 1, E = 4, so mass at T = 5. Since D lies on BC, Mass at D = Mass at B + Mass at C 15/4 = 1 + Mass at C ⇒ Mass at C = 11/4. Therefore, BD : DC = Mass at C : Mass at B ⇒ (11/4) : 1 ⇒ 11 : 4.

Question 7

CAT 2025 Slot 2: Two tangents drawn from a point P touch a circle with center O at points Q and R. Points A and B lie on PQ and PR, respectively, such that AB is also a tangent drawn to the same circle. If ∠AOB = 50°, then ∠APB is _____

Accepted Answer

Answer: 80. From P, PQ and PR are tangents to the circle with centre O, so OQ ⟂ PQ and OR ⟂ PR. AB is also a tangent, touching the circle at S, hence OS ⟂ AB. From an external point, the line joining the centre bisects the angle between tangents. Thus OA bisects ∠QOS and OB bisects ∠ROS. Let AOQ = AOS = x and BOR = BOS = y. Given ∠AOB = x + y = 50°. Then ∠QOS = 2x, ∠ROS = 2y and hence ∠QOR = 2(x + y) = 100°. In quadrilateral ORPQ, angles at Q and R are 90°, so ∠QOR + ∠APB = 180°. Therefore, ∠APB = 180° − 100° = 80°.

Question 8

CAT 2025 Slot 2: Let f(x) = x/(2x − 1) and g(x) = x/(x − 1). Then, the domain of the function h(x) = f(g(x)) + g(f(x)) is all real numbers except

Accepted Answer

Answer: A · −1, 1, and 1/2. Correct Answer A) −1, 1, and 1/2 x value | f(x) | g(x) | f(g(x)) | g(f(x)) | h(x) | Reason −1 | ✅ | ✅ | ❌ | | | g(−1) = 1/2 → f(1/2) undefined −1/2 | ✅ | ✅ | ✅ | ✅ | ✅ | No denominator becomes zero 1/2 | ❌ | | | | | f(1/2) undefined 1 | ✅ | ❌ | | | | g(1) undefined 3/2 | ✅ | ✅ | ✅ | ✅ | ✅ | All steps valid This confirms x is not valid on −1, 1, and 1/2

Question 9

CAT 2025 Slot 1: If the length of a side of a rhombus is 36 cm and the area of the rhombus is 396 sq. cm, then the absolute value of the difference between the lengths, in cm, of the diagonals of the rhombus is ____

Accepted Answer

Answer: 60. Let the diagonals be d1 and d2. Area of a rhombus = 1/2 × d1 × d2 = 396 → d1 × d2 = 792. In a rhombus, diagonals bisect each other at right angles. So by Pythagoras: (d1/2)^2 + (d2/2)^2 = 36^2. Multiply by 4: d1^2 + d2^2 = 4 × 1296 = 5184. Use identity: (d1 − d2)^2 = (d1^2 + d2^2) − 2(d1 × d2). Substitute: (d1 − d2)^2 = 5184 − 2×792 = 5184 − 1584 = 3600. So |d1 − d2| = √3600 = 60. Final Answer: 60

Question 10

CAT 2025 Slot 1: The (x, y) coordinates of vertices P, Q and R of a parallelogram PQRS are (-3, -2), (1, -5) and (9, 1), respectively. If the diagonal SQ intersects the x-axis at (a, 0), then the value of a is

Accepted Answer

Answer: D · 29/9. For a parallelogram P-Q-R-S in that order, S = P + R − Q. P = (−3, −2), Q = (1, −5), R = (9, 1) ⇒ S = (−3+9−1, −2+1−(−5)) = (5, 4). Line SQ passes through Q(1, −5) and S(5, 4). Slope = (4 − (−5)) / (5 − 1) = 9/4. Equation: y + 5 = (9/4)(x − 1). Put y = 0 to find x-intercept: 5 = (9/4)(x − 1). x − 1 = 20/9 ⇒ x = 1 + 20/9 = 29/9. Answer: 29/9 (option 4).

Question 11

CAT 2024 Slot 3: After two successive increments, Gopal’s salary became 187.5% of his initial salary. If the percentage of salary increase in the second increment was twice of that in the first increment, then the percentage of salary increase in the first increment was ?

Accepted Answer

Answer: A · 25. Using Options 1) 25 => 1.25 x 1.5 (second increment double of first) = 1.875 Answer 2) 20 => 1.2 x 1.4 = 1.68 XX 3) 27.5 => 1.275 x 1.55 = 1.976 XX 4) 30 => 1.3 x 1.6 = 2.08 xxx

Question 12

CAT 2024 Slot 3: A regular octagon ABCDEFGH has sides of length 6 cm each. Then the area, in sq. cm, of the square ACEG is

Accepted Answer

Answer: A · 36(2 + √2). If AH is 6 than AG visually looks around 10 or 11. therefore the area of square should be around 100 to 121 Expression Values Calculated Values: $ 1.\quad 36(2 + \sqrt{2}) \approx 122.91 $ Correct Answer $ 2.\quad 36(1 + \sqrt{2}) \approx 86.91 $ $ 3.\quad 72(2 + \sqrt{2}) \approx 245.82 $ $ 4.\quad 72(1 + \sqrt{2}) \approx 173.82 $

Question 13

CAT 2024 Slot 3: The midpoints of sides AB, BC, and AC in ΔABC are M, N, and P, respectively. The medians drawn from A, B, and C intersect the line segments MP, MN and NP at X, Y, and Z, respectively. If the area of ΔABC is 1440 sq cm, then the area, in sq cm, of ΔXYZ is

Accepted Answer

Answer: 90. Area of Triangle XYZ The points M, N, and P are the midpoints of sides AB, BC, and AC of triangle $ \triangle ABC $. Therefore, triangle $ \triangle MNP $ is the medial triangle. All 4 Triangles look similar therefore area will be 1/4, ie 1/4 of 1440 = 360 and 1/4 of 360 = 90. It is a known geometric property that the area of the medial triangle is one-fourth that of the original triangle. So, $$ \text{Area}_{\triangle MNP} = \frac{1}{4} \cdot \text{Area}_{\triangle ABC} = \frac{1}{4} \cdot 1440 = 360 \, \text{cm}^2 $$ Now, medians from vertices A, B, and C intersect the opposite sides of

Question 14

CAT 2024 Slot 2: ABCD is a trapezium in which AB is parallel to CD. The sides AD and BC when extended, intersect at point E. If AB = 2 cm, CD = 1 cm, and perimeter of ABCD is 6 cm, then the perimeter, in cm, of ΔAEB is

Accepted Answer

Answer: C · 8. Perimeter of Triangle AEB Finding the Perimeter of $ \triangle AEB $ Given: ABCD is a trapezium with $ AB \parallel CD $ AB = 2 cm, CD = 1 cm Perimeter of trapezium ABCD = 6 cm AD and BC extended intersect at point E Step 1: Use similarity Since $ AB \parallel CD $, triangles $ \triangle AEB \sim \triangle DCE $. Thus, $$ \frac{AB}{CD} = \frac{AE}{ED} = \frac{BE}{EC} = \frac{2}{1} $$ Step 2: Assign variables Let $ AD = x $, $ BC = y $ So total perimeter of ABCD: $$ AB + BC + CD + DA = 2 + y + 1 + x = 6 \Rightarrow x + y = 3 \tag{1} $$ Since the ratio is 2:1, we get: \[ AE = 2x, \qu

Question 15

CAT 2024 Slot 2: The coordinates of the three vertices of a triangle are: (1, 2), (7, 2), and (1, 10). Then the radius of the in circle of the triangle is ________

Accepted Answer

Answer: 2. Inradius of Triangle Problem: The coordinates of the vertices of a triangle are: $ (1, 2), (7, 2), (1, 10) $. Find the radius of the incircle of the triangle. Solution: When plotted, the triangle clearly forms a right-angled triangle with the right angle at point $ (1, 2) $. This is because two sides are aligned along the x-axis and y-axis. Let side $ a $ be the vertical leg: $$ a = 10 - 2 = 8 $$ Let side $ b $ be the horizontal leg: $$ b = 7 - 1 = 6 $$ Using the Pythagorean Theorem: $$ \text{Hypotenuse } h = \sqrt{a^2 + b^2} = \sqrt{8^2 + 6^2} = \sqrt{64 + 36} = \sqrt{100} = 10 $$ The

Question 16

CAT 2024 Slot 2: A vessel contained a certain amount of a solution of acid and water. When 2 litres of water was added to it, the new solution had 50% acid concentration. When 15 litres of acid was further added to this new solution, the final solution had 80% acid concentration. The ratio of wat

Accepted Answer

Answer: C · 3 : 5. Acid-Water Mixture Solution Let the original quantities of acid and water be $ A $ and $ W $ litres respectively. Step 1: After adding 2 litres of water, the concentration of acid becomes 50%. $ \frac{A}{A + W + 2} = \frac{1}{2} $ $ \Rightarrow 2A = A + W + 2 \Rightarrow A = W + 2 \quad \text{(Equation 1)} $ Step 2: Then, 15 litres of acid is added and acid becomes 80% of the solution: $ \frac{A + 15}{A + W + 17} = \frac{4}{5} $ $ \Rightarrow 5(A + 15) = 4(A + W + 17) $ $ \Rightarrow 5A + 75 = 4A + 4W + 68 $ $ \Rightarrow A = 4W - 7 \quad \text{(Equation 2)} $ Step 3: Solving e

Question 17

CAT 2024 Slot 1: Let $ x, y, $ and $ z $ be real numbers satisfying: $ 4(x^2 + y^2 + z^2) = a $, $ 4(x - y - z) = 3 + a $. Then $ a $ equals:

Accepted Answer

Answer: C · $ 3 $. Hit and Trial Table - MathJax Hit and Trial Table for Solving Using Options Given equations: $$ 4(x^2 + y^2 + z^2) = a \quad \text{and} \quad 4(x - y - z) = 3 + a $$ Option $ a $ $ x^2 + y^2 + z^2 = \frac{a}{4} $ $ x - y - z = \frac{3 + a}{4} $ Trial Values for $ x, y, z $ Result 1 $ 1 $ $ \frac{1}{4} $ $ 1 $ $ x = 1, y = 0, z = 0 $ ❌ Too large for (1) 2 $ 4 $ $ 1 $ $ \frac{7}{4} $ $ x = 2, y = \frac{1}{8}, z = \frac{1}{8} $ ❌ Doesn't satisfy (1) 3 ✅ $ 3 $ $ \frac{3}{4} $ $ \frac{3}{2} $ Algebra confirms consistency ✅ Correct 4 $ \frac{4}{3} $ $ \frac{1}{3} $

Question 18

CAT 2024 Slot 1: The surface area of a closed rectangular box, which is inscribed in a sphere, is 846 sq cm, and the sum of the lengths of all its edges is 144 cm. The volume, in cubic cm, of the sphere is:

Accepted Answer

Answer: C · $ 1125\pi\sqrt{2} $. Sphere Volume - Option-Based Solution $ 1125\pi = 3534.29 $ $ 750\pi\sqrt{2} = 3332.16 $ $ 1125\pi\sqrt{2} = 4998.24 $ $ 750\pi = 2356.19 $ Volume of Sphere Inscribed with a Rectangular Box Given: Surface area of the box = 846 cm² Sum of all edges = 144 cm Box is inscribed in a sphere → Diagonal of box = Diameter of sphere Step 1: Let dimensions be $ l, w, h $ $$ 4(l + w + h) = 144 \Rightarrow l + w + h = 36 \tag{1} $$ $$ 2(lw + wh + hl) = 846 \Rightarrow lw + wh + hl = 423 \tag{2} $$ Step 2: Use identity for cube diagonal $$ (l + w + h)^2 = l^2 + w^2 + h^2 + 2(lw + wh + hl) $$ Subst

Question 19

CAT 2024 Slot 1: A shop wants to sell a certain quantity (in kg) of grains. It sells half the quantity and an additional 3 kg of these grains to the first customer. Then, it sells half of the remaining quantity and an additional 3 kg of these grains to the second customer. Finally, when the shop

Accepted Answer

Answer: D · 42. Grain Selling Problem - Option Check Checking Each Option for Grain Selling Rule: Each customer takes half of the current quantity plus 3 kg. 🔹 Option 1: $ x = 36 $ First: $ \frac{36}{2} + 3 = 21 \Rightarrow \text{Remaining} = 15 $ Second: $ \frac{15}{2} + 3 = 10.5 \Rightarrow \text{Remaining} = 4.5 $ Third: $ \frac{4.5}{2} + 3 = 5.25 $ but only 4.5 left → Invalid 🔹 Option 2: $ x = 18 $ First: $ \frac{18}{2} + 3 = 12 \Rightarrow \text{Remaining} = 6 $ Second: $ \frac{6}{2} + 3 = 6 \Rightarrow \text{Remaining} = 0 $ Third: No grains left to sell → Invalid 🔹 Option 3: \( x = 50

Question 20

CAT 2024: ABCD is a rectangle with sides AB = 56 cm and BC = 45 cm, and E is the midpoint of side CD. What is the length, in cm, of the radius of the incircle of $ \triangle ADE $?

Accepted Answer

Answer: 10. Step 1: Understand the Geometry of the Problem - $ AB = 56 \, \text{cm}, BC = 45 \, \text{cm} $. - $ CD = AB = 56 \, \text{cm} $, $ DA = BC = 45 \, \text{cm} $. - $ E $ is the midpoint of $ CD $, so $ CE = ED = \frac{56}{2} = 28 \, \text{cm} $. Step 2: Sides of $ \triangle ADE $ - $ AD = 45 \, \text{cm}, DE = 28 \, \text{cm} $. - $ AE = \sqrt{AD^2 + ED^2} = \sqrt{45^2 + 28^2} = \sqrt{2025 + 784} = 53 \, \text{cm}. $ Step 3: Semiperimeter ($ s $) - $ s = \frac{AD + DE + AE}{2} = \frac{45 + 28 + 53}{2} = 63 \, \text{cm}. $ Step 4: Area of $ \triangle ADE $ - Using Heron's

Question 21

CAT 2023 Slot 3: In a regular polygon, any interior angle exceeds the exterior angle by 120 degrees. Then, the number of diagonals of this polygon is

Accepted Answer

Answer: 54

Question 22

CAT 2023 Slot 3: The number of positive integers less than 1000, having exactly two distinct prime factors, is:

Accepted Answer

Answer: C · 15. We count numbers of the form $p \times q$ (where $p$ and $q$ are distinct primes and \(p \times q 2 × 3 = 6 2 × 5 = 10 2 × 7 = 14 ... Counting all valid combinations under 1000 gives exactly 15 such numbers.

Question 23

CAT 2023 Slot 3: The population of a town in 2020 was 100,000. The population decreased by $ y\% $ from 2020 to 2021 and increased by $ x\% $ from 2021 to 2022. If the population in 2022 is greater than the population in 2020 and $ x - y = 10 $, what is the smallest possible population in 2

Accepted Answer

Answer: A · 73,000. We are told that population decreased by y% and increased by x% where x – y = 10. Final population > 100000. We try options: Option 2021 Population Assumed x% y% (x–10) 2022 Population Valid? A 73000 37% 27% 73000 × 1.37 = 100010 ✅ Yes B 72000 39% 29% 72000 × 1.39 = 100080 ❌ No (not minimum) C 75000 34% 24% 75000 × 1.34 = 100500 ❌ No D 74000 35% 25% 74000 × 1.35 = 99900 ❌ No

Question 24

CAT 2023 Slot 3: Anil mixes cocoa with sugar in the ratio 3 : 2 to prepare mixture A, and coffee with sugar in the ratio 7 : 3 to prepare mixture B. He combines mixtures A and B in the ratio 2 : 3 to make a new mixture C. If he mixes C with an equal amount of milk to make a drink, then the percen

Accepted Answer

Answer: B · 17%. Let’s assume A = 2 parts, B = 3 parts. Sugar in A = 2/5 per part, in B = 3/10 per part. Component Quantity Sugar Fraction Sugar Content Mixture A 2 2/5 = 0.4 2 × 0.4 = 0.8 Mixture B 3 3/10 = 0.3 3 × 0.3 = 0.9 Total Mixture C (without milk) 0.8 + 0.9 = 1.7 Now, mixture C (5 units) is mixed with equal 5 units of milk → Total = 10 units. Percentage of sugar = (1.7 / 10) × 100 = 17%

Question 25

CAT 2023 Slot 3: Let both the sequences $a_1, a_2, a_3, \ldots$ and $b_1, b_2, b_3, \ldots$ be arithmetic progressions such that their common differences are prime numbers. If $ a_4 = b_4 $, $ a_7 = b_7 $, and $ b_1 = 0 $, then $ a_1 $ is equal to:

Accepted Answer

Answer: C · 79. Let a_n = a_1 + (n – 1)d_1, and b_n = (n – 1)d_2, since b_1 = 0. From a_4 = b_4: a_1 + 3d_1 = 3d_2 → a_1 = 3(d_2 – d_1) From a_7 = b_7: a_1 + 6d_1 = 6d_2 Substitute a_1: 3(d_2 – d_1) + 6d_1 = 6d_2 → 3d_2 – 3d_1 + 6d_1 = 6d_2 → 3d_2 + 3d_1 = 6d_2 → d_1 = d_2 Then a_1 = 3(0) = 0 ❌ So try different primes: d_1 = 5, d_2 = 7 → a_1 = 3(7 – 5) = 6 a_7 = 6 + 6×5 = 36 b_7 = 6×7 = 42 ❌ Try d_1 = 7, d_2 = 11 → a_1 = 3(11 – 7) = 12 a_7 = 12 + 6×7 = 54 b_7 = 6×11 = 66 ❌ Eventually, d_1 = 11, d_2 = 17 → a_1 = 3(17 – 11) = 18 a_7 = 18 + 6×11 = 84 b_7 = 6×17 = 102 ❌ Try d_1 = 13, d_2 = 19 → a_1 = 3×6 = 18 a_7

Question 26

CAT 2023 Slot 2: In a rectangle ABCD, AB = 9 cm and BC = 6 cm. P and Q are two points on BC such that the areas of the figures ABP, APQ, and AQCD are in geometric progression. If the area of the figure AQCD is four times the area of triangle ABP, then BP : PQ : QC is

Accepted Answer

Answer: D · 2 : 4 : 1. Option 4

Question 27

CAT 2023 Slot 2: A triangle is drawn with its vertices on the circle $ C $ such that one of its sides is a diameter of $ C $ and the other two sides have their lengths in the ratio $ a : b $. If the radius of the circle is $ r $, then the area of the triangle is:

Accepted Answer

Answer: A · $ \frac{2 a b r^2}{a^2 + b^2} $. Answer: A. $ \frac{2 a b r^2}{a^2 + b^2} $ Detailed Solution: Using the property that the diameter of a circle forms a right angle with the chord, and the formula for area, we get: $$ \text{Area} = \frac{2 a b r^2}{a^2 + b^2} $$.

Question 28

CAT 2023 Slot 1: A quadrilateral ABCD is inscribed in a circle such that AB : CD = 2 : 1 and BC : AD = 5 : 4. If AC and BD intersect at the point E, then AE : CE equals:

Accepted Answer

Answer: C · 8/5. ### Solution: Cyclic Quadrilateral Given ABCD is a cyclic quadrilateral. Using properties of cyclic quadrilaterals, similar triangles AED and BEC imply AE/CE = AD/BC = 8/5, as derived from the given ratios .

Question 29

CAT 2023 Slot 1: A circle has its center at (0, 0) and passes through the point (3, 4). What is the angle formed between the radius and the line passing through the center?

Accepted Answer

Answer: D · 90°. ### Solution: Circle and Right Angle Triangle A circle with radius passing through (3,4) implies that the triangle formed with the origin and this point is a right-angled triangle by the Pythagorean theorem.

Question 30

CAT 2023 Slot 1: In a right-angle triangle, if the base is 6 units and height is 9 units, then the area of the triangle is:

Accepted Answer

Answer: B · 27 sq. units. ### Solution: Right-Angle Triangle Area of a right-angle triangle = (1/2) * base * height. Substituting the values, we get (1/2) * 6 * 9 = 27 square units.

Question 31

CAT 2023 Slot 1: The number of all natural numbers up to 1000 with non-repeating digits is:

Accepted Answer

Answer: C · 738. Answer: C. 738 Detailed Solution: - **1-digit numbers:** 9 - **2-digit numbers:** 9 choices for the tens place, 9 remaining choices for the units place = 81. - **3-digit numbers:** 9 choices for the hundreds place, 9 remaining choices for the tens place, and 8 choices for the units place = 648. Total = 9 + 81 + 648 = **738**.

Question 32

CAT 2023 Slot 1: Brishti went on an 8-hour trip in a car. Before the trip, the car had travelled a total of x km, where x is a palindromic number. At the end of the trip, the car had travelled a total of 26862 km. If Brishti never drove at more than 110 kmph, then the greatest possible average sp

Accepted Answer

Answer: D · 100. Answer: D. 100 Detailed Solution: - Before the trip: least palindromic number greater than 25982 is 26062. - Distance during the trip = 26862 - 26062 = 800 km. - Average speed = 800 km / 8 hours = **100 kmph**.

Question 33

CAT 2023 Slot 1: If $\frac{\sqrt{5x + 9}}{\sqrt{5x - 9}} = 3(\sqrt{2} + 1)$, then find the value of $ \sqrt{10x + 9} $:

Accepted Answer

Answer: A · 3√7. Answer: A. 3√7 Detailed Solution: Given, $\frac{\sqrt{5x + 9}}{\sqrt{5x - 9}} = 3(\sqrt{2} + 1)$ - Simplifying, we get $ \sqrt{5x + 9} = 3(\sqrt{2} + 1) \sqrt{5x - 9} $. - Solving, we find that $ x = 9 $. - Therefore, $ \sqrt{10x + 9} = 3√7 $.

Question 34

CAT 2022 Slot 3: Two ships are approaching a port along straight routes at constant speeds. Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. When the slower ship travelled 8 km, the triangle formed by the new positions of the two ships and the port

Accepted Answer

Answer: C · 12. Let's analyze the situation step by step: Initially, the two ships and the port formed an equilateral triangle with sides of length 24 km. This means that each side of the equilateral triangle was 24 km. When the slower ship traveled 8 km, the triangle formed by the new positions of the two ships and the port became right-angled. This implies that the slower ship moved 8 km along one of the sides of the original equilateral triangle, creating a right triangle with the other ship and the port. Since the original triangle was equilateral, each of its angles was 60 degrees. Therefore, the right t

Question 35

CAT 2022 Slot 3: The lengths of all four sides of a quadrilateral are integer valued. If three of its sides are of length 1 cm, 2 cm and 4 cm, then the total number o possible lengths of the fourth side is

Accepted Answer

Answer: A · 5. The correct answer is A:5 To determine the possible lengths of the fourth side of the quadrilateral,we need to consider the triangle inequality theorem.According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.This principle also holds for quadrilaterals. Given the lengths of the three sides as 1cm, 2cm, and 4cm, we can consider all the possible combinations of these sides as two sides of the quadrilateral.Let's check each case: 1cm+2cm> 4cm (True) 1cm+4cm> 2cm (True) 2cm+4cm> 1 cm (True) Since all three combinations sati

Question 36

CAT 2022 Slot 2: In triangle ABC, altitudes AD and BE are drawn to the corresponding bases. If ∠BAC = 45° and ∠ABC = θ, then AD/BE equals

Accepted Answer

Answer: C · √2 sinθ. To find the ratio AD/BE in triangle ABC, we can use trigonometry and the information given in the problem. First, let's draw a diagram of triangle ABC with altitudes AD and BE: A |\ | \ | \ D | \ E | \ | \ | \ ---B------C We are given that ∠BAC = 45° and ∠ABC = θ. Since AD and BE are altitudes, they are perpendicular to their respective bases, BC and AC. This means that ∠ADC = 90° and ∠BEC = 90°. Now, let's consider triangle ADC and triangle BEC separately: In triangle ADC: ∠DAC = 45° (complementary to ∠BAC). ∠ADC = 90° (given). In triangle BEC: ∠ECB = θ (given). ∠BEC = 90° (given). We can use

Question 37

CAT 2022 Slot 2: The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD, in cm, is

Accepted Answer

Answer: A · √7. To solve this problem, we'll use the information that the area of triangle ADC is half the area of triangle ABD. Since triangle ABC is equilateral, all its angles are 60 degrees, and we can use trigonometry to find the length of AD. Let's denote the length of AD as x. Since triangle ADC is half the area of triangle ABD, we can write: (1/2) * AD * DC = (1/2) * AB * BD We know that AB = 3 cm, and BD = BC - CD. Since triangle ABC is equilateral, BC = 3 cm, so BD = 3 - CD. We can rewrite the equation as: (1/2) * x * DC = (1/2) * 3 * (3 - CD) Now, we need to find DC in terms of x. We can use trigon

Question 38

CAT 2022 Slot 2: Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

Accepted Answer

Answer: 10. Correct Answer: 10 Let the number of sides of polygons A and B be n and 2n, respectively The number of sides of polygon B is 2*5, i.e. 10

Question 39

CAT 2022 Slot 1: All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is:

Accepted Answer

Answer: B · P^2/8- 2R^2. Let the dimensions of the rectangle inscribed in the circle be a and b. We are being asked to express the Area of the rectangle in terms of Perimeter and Radius. Area of the rectangle = a * b Perimeter, P = 2(a + b) a + b = P2�2 (a+b)2=P24(�+�)2=�24 a2+b2+2ab=P24�2+�2+2��=�24 Since the vertices of the rectangle subtend right angle on the circle, a2+b2=(2R)2=4R2�2+�2=(2�)2=4�2 a2+b2+2ab=P24�2+�2+2��=�24 4R2+2ab=P244�2+2��=�24 ab=P28−2R2��=�28−2�2 Area of the rectangle = ab = p28−2R2

Question 40

CAT 2021 Slot 1: Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. If T is the mid point of CD, then the length of AT, in cm, is

Accepted Answer

Answer: C · √13. T is the midpoint of CD in the given regular hexagon ABCDEF CT=DT=2/2=1 Draw AT, AD and AC By the property of hexagon, AD=2×BC=2×2=4=2×=2×2=4 ∠ABC=∠BCD=120^∘∠=∠=120^∘ We have, AB=BC=CD=2 ∠ACD=120^∘−30^∘=90^∘∠=120^∘−30^∘=90^∘ ACD is a right triangle and AT is its median AC^2=AD^2−CD^2= 4^2 - 2^2 = 12 Using Apollonius's theorem, AC^2+AD^2=2(AT^2+DT^2) 12 + 4^2 = 2(AT^2+1^2) AT = √13 Ans: ≈3.605551 cm