Algebra, Geometry, DATA INTERPRETATION, READING COMPREHENSION Unit Test for CAT: 68 Questions with Solutions & Analysis

Attempt the Algebra, Geometry, DATA INTERPRETATION, READING COMPREHENSION Unit Test for CAT: 68 exam-level questions, detailed solutions and performance analysis. First questions free.

Paper breakdown

68 questions · 204 marks · 103.84 minutes. Quantitative Ability: 27 · Data Interpretation and Logical Reasoning: 22 · Verbal Ability and Reading Comprehension: 19

Free sample questions from Algebra, Geometry, DATA INTERPRETATION, READING COMPREHENSION Unit Test

Question 1 · Verbal Ability and Reading Comprehension MCQ

Common Description: The passage below is accompanied by four questions. Based on the passage, choose the best answer for each question.
RESIDENTS of Lozère, a hilly department in southern France, recite complaints familiar to many rural corners of Europe. In remote hamlets and villages, with names such as Le Bacon and Le Bacon Vieux, mayors grumble about a lack of local schools, jobs, or phone and internet connections. Farmers of grazing animals add another concern: the return of wolves. Eradicated from France last century, the predators are gradually creeping back to more forests and hillsides. “The wolf must be taken in hand,” said an aspiring parliamentarian, Francis Palombi, when pressed by voters in an election campaign early this summer. Tourists enjoy visiting a wolf park in Lozère, but farmers fret over their livestock and their livelihoods. . .
As early as the ninth century, the royal office of the Luparii—wolf-catchers—was created in France to tackle the predators. Those official hunters (and others) completed their job in the 1930s, when the last wolf disappeared from the mainland. Active hunting and improved technology such as rifles in the 19th century, plus the use of poison such as strychnine later on, caused the population collapse. But in the early 1990s the animals reappeared. They crossed the Alps from Italy, upsetting sheep farmers on the French side of the border. Wolves have since spread to areas such as Lozère, delighting environmentalists, who see the predators’ presence as a sign of wider ecological health. Farmers, who say the wolves cause the deaths of thousands of sheep and other grazing animals, are less cheerful. They grumble that green activists and politically correct urban types have allowed the return of an old enemy.
Various factors explain the changes of the past few decades. Rural depopulation is part of the story. In Lozère, for example, farming and a once-flourishing mining industry supported a population of over 140,000 residents in the mid-19th century. Today the department has fewer than 80,000 people, many in its towns. As humans withdraw, forests are expanding. In France, between 1990 and 2015, forest cover increased by an average of 102,000 hectares each year, as more fields were given over to trees. Now, nearly one-third of mainland France is covered by woodland of some sort. The decline of hunting as a sport also means more forests fall quiet. In the mid-to-late 20th century over 2m hunters regularly spent winter weekends tramping in woodland, seeking boars, birds and other prey. Today the Fédération Nationale des Chasseurs, the national body, claims 1.1m people hold hunting licences, though the number of active hunters is probably lower. The mostly protected status of the wolf in Europe—hunting them is now forbidden, other than when occasional culls are sanctioned by the state —plus the efforts of NGOs to track and count the animals, also contribute to the recovery of wolf populations.
As the lupine population of Europe spreads westwards, with occasional reports of wolves seen closer to urban areas, expect to hear of more clashes between farmers and those who celebrate the predators’ return. Farmers’ losses are real, but are not the only economic story. Tourist venues, such as parks where wolves are kept and the animals’ spread is discussed, also generate income and jobs in rural areas. Which one of the following has NOT contributed to the growing wolf population in Lozère?

A.
A decline in the rural population of Lozère.
B.
An increase in woodlands and forest cover in Lozère.
C.
The shutting down of the royal office of the Luparii.
D.
The granting of a protected status to wolves in Europe.

Correct Answer:

Step-by-Step Solution

Key idea: This is a 'Causal Verification / EXCEPT' question, recognisable by the capitalised NOT asking which factor did not contribute to a specific outcome.

Why it applies: The question requires verifying three true causes from the text and isolating the one false or irrelevant statement regarding the recent wolf population growth.

Step 1: Verify Option A (rural decline). Paragraph 3 states "Rural depopulation is part of the story... As humans withdraw, forests are expanding," which aided recovery. (True)

Step 2: Verify Option B (woodlands). Paragraph 3 notes forest cover increased, providing more habitat. (True)

Step 3: Verify Option D (protected status). Paragraph 3 explicitly states "The mostly protected status of the wolf... contribute to the recovery." (True)

Step 4: Verify Option C (Luparii). Paragraph 2 states the Luparii were shut down in the 1930s, which caused the wolves to disappear. This historical eradication effort did not cause the 1990s population recovery. (False)

Trap: Picking Option C because Luparii is a unique term found in the text. Students often select options containing specific text keywords without verifying the timeline or causal direction.

Answer: C

Question 2 · Verbal Ability and Reading Comprehension MCQ

Common Description: Instructions [9 - 12]
The passage below is accompanied by a set of questions. Choose the best answer to each question.
I have elaborated . . . a framework for analyzing the contradictory pulls on [Indian] nationalist ideology in its struggle against the dominance of colonialism and the resolution it offered to those contradictions. Briefly, this resolution was built around a separation of the domain of culture into two spheres—the material and the spiritual. It was in the material sphere that the claims of Western civilization were the most powerful. Science, technology, rational forms of economic organization, modern methods of statecraft—these had given the European countries the strength to subjugate the non-European people . . . To overcome this domination, the colonized people had to learn those superior techniques of organizing material life and incorporate them within their own cultures. . . . But this could not mean the imitation of the West in every aspect of life, for then the very distinction between the West and the East would vanish—the self-identity of national culture would itself be threatened. . . .
The discourse of nationalism shows that the material/spiritual distinction was condensed into an analogous, but ideologically far more powerful, dichotomy: that between the outer and the inner. . . . Applying the inner/outer distinction to the matter of concrete day-to-day living separates the social space into ghar and bāhir, the home and the world. The world is the external, the domain of the material; the home represents one’s inner spiritual self, one’s true identity. The world is a treacherous terrain of the pursuit of material interests, where practical considerations reign supreme. It is also typically the domain of the male. The home in its essence must remain unaffected by the profane activities of the material world—and woman is its representation. And so one gets an identification of social roles by gender to correspond with the separation of the social space into ghar and bāhir. . . .
The colonial situation, and the ideological response of nationalism to the critique of Indian tradition, introduced an entirely new substance to [these dichotomies] and effected their transformation. The material/spiritual dichotomy, to which the terms world and home corresponded, had acquired . . . a very special significance in the nationalist mind. The world was where the European power had challenged the non-European people and, by virtue of its superior material culture, had subjugated them. But, the nationalists asserted, it had failed to colonize the inner, essential, identity of the East which lay in its distinctive, and superior, spiritual culture. . . . [I]n the entire phase of the national struggle, the crucial need was to protect, preserve and strengthen the inner core of the national culture, its spiritual essence. . . .
Once we match this new meaning of the home/world dichotomy with the identification of social roles by gender, we get the ideological framework within which nationalism answered the women’s question. It would be a grave error to see in this, as liberals are apt to in their despair at the many marks of social conservatism in nationalist practice, a total rejection of the West. Quite the contrary: the nationalist paradigm in fact supplied an ideological principle of selection. Which one of the following explains the “contradictory pulls” on Indian nationalism?

A.
Despite its spiritual superiority, Indian nationalism had to fight against colonial domination.
B.
Despite its fight against colonial domination, Indian nationalism had to borrow from the coloniser in the spiritual sphere.
C.
Despite its scientific and technological inferiority, Indian nationalism had to fight against colonial domination.
D.
Despite its fight against colonial domination, Indian nationalism had to borrow from the coloniser in the material sphere.

Correct Answer:

Step-by-Step Solution

Key idea: This is a 'detail synthesis' question, recognizable because it asks for the meaning of a specific phrase ("contradictory pulls") based on the passage's central argument.

Step 1: Locate the phrase "contradictory pulls" in the first paragraph. The passage states the resolution was built around separating culture into material and spiritual spheres.

Step 2: Identify the pull in the material sphere. The text says, "To overcome this domination, the colonized people had to learn those superior techniques of organizing material life..." (i.e., borrow from the colonizer).

Step 3: Identify the counter-pull. The text adds, "But this could not mean the imitation of the West in every aspect of life... the self-identity of national culture would itself be threatened." Thus, they had to borrow materially but resist spiritually.

Step 4: Match with options. Option D accurately captures this tension: needing to adopt Western material techniques to fight domination, while trying to avoid total imitation.

Trap: Option B reverses the spheres (borrowing in the spiritual sphere), which the passage explicitly rejects to protect self-identity.

Answer: D

Question 3 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [30 - 34]
There are only four neighbourhoods in a city - Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1.
2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively. What BEST can be concluded about the number of new cases in Levmisto on Day 3?

A.
Either 2 or 3
B.
Exactly 2
C.
Exactly 3
D.
Either 0 or 1

Correct Answer:

Step-by-Step Solution

Key idea: This is a constrained pandemic-data puzzle. To determine Levmisto's Day 3 value, we build on the forced daily totals and grid deductions from the previous step.

Step 1: Recall forced daily totals.

$D_{1} = 5, D_{2} = 8, D_{3} = 9, D_{4} = 10, D_{5} = 11$ .

Step 2: Recall forced values for Levmisto (L) and Tyhrmisto (T).

From Day 1 (total 5, all $\geq 1$ ): L1=1, T1=1, P1=1, K1=2.

From Day 4 (total 10, P4=1): L4=3, T4=3, K4=3.

From Day 5 (total 11): L5=3, T5=3, K5=3, P5=2.

Step 3: Use Levmisto's grand total.

Total L = 12. We know L1=1, L4=3, L5=3.

So, $L_{2} + L_{3} = 12 - (1 + 3 + 3) = 5$ .

Step 4: Force Day 2.

Day 2 total is 8. Kitmisto is the only place with 3 cases. So K2 = 3.

This means L2 and T2 cannot be 3. The max for L2 and T2 is 2.

We also know P2 = 1 (from P's multiset and P1=1, P4=1, P5=2).

So, $L_{2} + T_{2} + P_{2} + K_{2} = 8 ⟹ L_{2} + T_{2} + 1 + 3 = 8 ⟹ L_{2} + T_{2} = 5$ .

Since neither can be 3, the only way to sum to 5 is $L_{2} = 2$ and $T_{2} = 2$ .

Step 5: Solve for L3.

We know $L_{2} + L_{3} = 5$ . Since $L_{2} = 2$ , $L_{3}$ must be exactly 3.

Answer: C

Question 4 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [30 - 33]
The table given below shows the amount, in grams, of carbohydrate, protein, fat and all other nutrients, per 100 grams of nutrients in seven food grains. The first column shows the food grain category and the second column its codename. The table has some missing values.

Food grain Category	Codename of the food grain	Carbohydrate	Protein	Fat	Other nutrients
Cereal	C1			0	12
Cereal	C2			3	10
Millet	M1	62	10
Millet	M2			7	16
Millet	M3	56		12
Pseudo-cereal	P1	66			10
Pseudo-cereal	P2		14		8

The following additional facts are known.
1. Both the pseudo-cereals had higher amounts of carbohydrate as well as higher amounts of protein than any millet.
2. Both the cereals had higher amounts of carbohydrate than any pseudo-cereal.
3. All the missing values of carbohydrate amounts (in grams) for all the food grains are non-zero multiples of 5.
4. All the missing values of protein, fat and other nutrients amounts (in grams) for all the food grains are non-zero multiples of 4.
5. P1 contained double the amount of protein that M3 contains. What is the median of the number of grams of protein in 100 grams of nutrients among these food grains?

Correct Answer:

Step-by-Step Solution

Key idea: This is a nutrition-and-composition table question. Each food grain's components are given per 100 grams, so every row must sum to 100. We must reconstruct all seven protein values to find the median.

Step 1: Use row sums for the cereals.

C1: Carb + Prot + 0 + 12 = 100 $⟹$ Carb + Prot = 88.

Carb is a non-zero multiple of 5. Prot is a non-zero multiple of 4.

Cereals have higher carb than any pseudo-cereal. P1 carb is 66, so C1 carb $\geq 70$ .

If Carb = 70, Prot = 18 (not mult of 4).

If Carb = 75, Prot = 13 (not mult of 4).

If Carb = 80, Prot = 8 (valid).

If Carb = 85, Prot = 3 (not mult of 4).

So C1: Carb = 80, Prot = 8.

C2: Carb + Prot + 3 + 10 = 100 $⟹$ Carb + Prot = 87.

If Carb = 70, Prot = 17 (no).

If Carb = 75, Prot = 12 (valid).

If Carb = 80, Prot = 7 (no).

So C2: Carb = 75, Prot = 12.

Step 2: Use pseudo-cereal and millet protein clues.

M3: 56 + Prot + 12 + Other = 100 $⟹$ Prot + Other = 32.

P1: 66 + Prot + Fat + 10 = 100 $⟹$ Prot + Fat = 24.

P1 protein = 2 $\times$ M3 protein.

Since P1 protein < 24, M3 protein < 12.

Also, pseudo-cereal protein > ANY millet protein. M1 protein is 10. So P1 protein > 10.

Thus, 2 $\times$ M3 protein > 10 $⟹$ M3 protein $\geq 6$ .

M3 protein must be a multiple of 4. The only multiple of 4 between 6 and 11 is 8.

So M3 protein = 8. (Then Other = 24).

Step 3: Find remaining proteins.

P1 protein = 2 $\times$ 8 = 16.

M2: Carb + Prot + 7 + 16 = 100 $⟹$ Carb + Prot = 77.

Carb is mult of 5, Prot is mult of 4.

If Carb = 65, Prot = 12 (valid).

If Carb = 70, Prot = 7 (no).

So M2 protein = 12.

P2 protein is given as 14.

Step 4: List and find the median.

Protein values: C1=8, C2=12, M1=10, M2=12, M3=8, P1=16, P2=14.

Sorted: 8, 8, 10, 12, 12, 14, 16.

The median (4th value) is 12.

Answer: 12

Question 5 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [30 - 34]
There are only four neighbourhoods in a city - Levmisto, Tyhrmisto, Pesmisto and Kitmisto. During the onset of a pandemic, the number of new cases of a disease in each of these neighbourhoods was recorded over a period of five days. On each day, the number of new cases recorded in any of the neighbourhoods was either 0, 1, 2 or 3.
The following facts are also known:
1. There was at least one new case in every neighbourhood on Day 1.
2. On each of the five days, there were more new cases in Kitmisto than in Pesmisto.
3. The number of new cases in the city in a day kept increasing during the five-day period. The number of new cases on Day 3 was exactly one more than that on Day 2.
4. The maximum number of new cases in a day in Pesmisto was 2, and this happened only once during the five-day period.
5. Kitmisto is the only place to have 3 new cases on Day 2.
6. The total numbers of new cases in Levmisto, Tyhrmisto, Pesmisto and Kitmisto over the five-day period were 12, 12, 5 and 14 respectively. Which of the two statements below is/are necessarily false?
Statement A: There were 2 new cases in Tyhrmisto on Day 3.
Statement B: There were no new cases in Pesmisto on Day 2.

A.
Statement A only
B.
Neither Statement A nor Statement B
C.
Statement B only
D.
Both Statement A and Statement B

Correct Answer:

Step-by-Step Solution

Key idea: This is a necessity-analysis pandemic puzzle, recognisable because the question asks which statements must be false. We need to determine whether each statement can ever be true under all the constraints. The safest route is to force the complete table.

Step 1: Force the daily totals.

Let daily totals be $D_{1}, D_{2}, D_{3}, D_{4}, D_{5}$ .

They are strictly increasing, and $D_{3} = D_{2} + 1$ .

Overall totals are:

Levmisto = 12, Tyhrmisto = 12, Pesmisto = 5, Kitmisto = 14.

Grand total = $43$ .

So:

$D_{1} + 2 D_{2} + D_{4} + D_{5} = 42$ .

Pesmisto can never exceed 2, so the maximum possible city total in one day is $3 + 3 + 3 + 2 = 11$ .

Thus $D_{5} \leq 11$ .

If $D_{2} \geq 9$ , then $D_{3} \geq 10$ , $D_{4} \geq 11$ , and $D_{5} \geq 12$ , impossible.

Therefore $D_{2} \leq 8$ .

Using maximum possible values $D_{4} = 10$ , $D_{5} = 11$ , and $D_{1} \leq D_{2} - 1$ :

$42 \leq (D_{2} - 1) + 2 D_{2} + 10 + 11 = 3 D_{2} + 20 ⟹ D_{2} \geq 8$ .

Therefore:

$D_{2} = 8$ , $D_{3} = 9$ , $D_{4} = 10$ , $D_{5} = 11$ , and $D_{1} = 5$ .

Step 2: Force Pesmisto.

Pesmisto total is 5. Its maximum is 2, and this happens only once.

So its five values must be one 2, three 1s, and one 0.

Day 5 total is 11, the maximum possible. Therefore Day 5 must be:

L=3, T=3, K=3, P=2. So P5=2.

Day 4 total is 10. Since P cannot have another 2, P4 must be 1, and the others must be 3.

So P4=1.

Step 3: Evaluate Statement B.

Day 2 has K as the only place with 3 cases, so K2=3 and L2, T2 are not 3.

Day 2 total is 8:

$L 2 + T 2 + P 2 + 3 = 8 ⟹ L 2 + T 2 + P 2 = 5$ .

If P2=0, then L2+T2=5. Since L2, T2 $\leq 2$ (as K is the only 3), the max sum is $2 + 2 = 4 < 5$ . Impossible.

Thus P2 cannot be 0.

Statement B ("no new cases in Pesmisto on Day 2") is necessarily FALSE.

Step 4: Evaluate Statement A.

We need to find P3. P total is 5. P5=2, P4=1. So P1+P2+P3 = 2.

Since Day 1 has at least 1 case in every neighbourhood, P1 $\geq 1$ .

If P2=2, then P1+P3=0 $⟹$ P1=0, contradicting P1 $\geq 1$ .

So P2 must be 1.

Then P1+P3 = 1. Since P1 $\geq 1$ , we must have P1=1 and P3=0.

So P3 = 0.

Day 3 total is 9. $L 3 + T 3 + P 3 + K 3 = 9 ⟹ L 3 + T 3 + K 3 = 9$ .

Since K3 $\leq 3$ , $L 3 + T 3 \geq 6$ .

The maximum for L3 and T3 is 3. So L3=3, T3=3.

Then K3 = 3.

So on Day 3, T3 is exactly 3.

Statement A ("2 new cases in Tyhrmisto on Day 3") is necessarily FALSE.

Both statements are necessarily false.

Answer: D

Question 6 · Quantitative Ability MCQ

For some real numbers a and b, the system of equations $x + y = 4$ and $(a + 5) x + (b^{2} - 15) y = 8 b$ has infinitely many solutions for x and y. Then, the maximum possible value of ab is

A.
33
B.
25
C.
15
D.
55

Correct Answer:

Step-by-Step Solution

This is an infinite-solutions linear-system question, recognisable because the question directly says "infinitely many solutions" for a pair of linear equations — that phrase always means the two equations must represent the SAME line, so their coefficient ratios must all be equal.

Step 1: Write both equations in standard form.

$x + y = 4 (a + 5) x + (b^{2} - 15) y = 8 b$

Step 2: For infinitely many solutions, the ratio of x-coefficients, y-coefficients, and constants must all be equal:

$\frac{a + 5}{1} = \frac{b ^{2} - 15}{1} = \frac{8 b}{4}$

Step 3: Simplify the third ratio: $\frac{8 b}{4} = 2 b$ . So we get two equations:

$a + 5 = 2 b and b^{2} - 15 = 2 b$

Step 4: Solve the quadratic in b.

$b^{2} - 2 b - 15 = 0 \Rightarrow (b - 5) (b + 3) = 0 \Rightarrow b = 5 or b = - 3$

Step 5: Find the corresponding a for each b using $a = 2 b - 5$ .

If $b = 5$ : $a = 2 (5) - 5 = 5$ , so $ab = 5 \times 5 = 25$ .
If $b = - 3$ : $a = 2 (- 3) - 5 = - 11$ , so $ab = (- 11) \times (- 3) = 33$ .

Step 6: Compare the two possible values of ab: 25 and 33. The maximum is 33.

Answer: 33.

Trap: Students often compute only one root of the quadratic in b and miss the second case, or forget to compare both ab values before picking the maximum.

Question 7 · Quantitative Ability MCQ

Let $△ A B C$ be an isosceles triangle such that AB and AC are of equal length. AD is the altitude from A on BC and BE is the altitude from B on AC. If AD and BE intersect at O such that $∠ A O B = 10 5^{\circ}$ , then $\frac{B E}{A D}$ equals

A.
$sin 1 5^{\circ}$
B.
$cos 1 5^{\circ}$
C.
$2 cos 1 5^{\circ}$
D.
$2 sin 1 5^{\circ}$

Correct Answer:

Step-by-Step Solution

This is the "Isosceles Triangle with Two Altitudes" (orthocenter angle) pattern: two altitudes of a triangle meet at the orthocenter, and the angle between them relates directly to the triangle's third angle.

Step 1 — Recognize the trigger.

AD and BE are altitudes, so their intersection point O is the orthocenter of $△ A B C$ . Whenever a question gives the angle between two altitudes, use the orthocenter angle rule:

$∠ A O B = 18 0^{\circ} - ∠ C$

Step 2 — Find angle C.

$10 5^{\circ} = 18 0^{\circ} - ∠ C \Rightarrow ∠ C = 7 5^{\circ}$

Since $A B = A C$ , the base angles are equal: $∠ B = ∠ C = 7 5^{\circ}$ , so $∠ A = 18 0^{\circ} - 15 0^{\circ} = 3 0^{\circ}$ .

Step 3 — Express BE and AD in terms of side AB.

In right triangle $A B D$ (right angle at $D$ ), $∠ A B D = ∠ B$ , so

$A D = A B sin B$

In right triangle $A B E$ (right angle at $E$ ), $∠ B A E = ∠ A$ , so

$B E = A B sin A$

Step 4 — Form the ratio.

$\frac{B E}{A D} = \frac{A B sin A}{A B sin B} = \frac{sin 3 0 ^{\circ}}{sin 7 5 ^{\circ}} = \frac{sin 3 0 ^{\circ}}{cos 1 5 ^{\circ}}$

Step 5 — Simplify using the double-angle identity $sin 3 0^{\circ} = 2 sin 1 5^{\circ} cos 1 5^{\circ}$ :

$\frac{sin 3 0 ^{\circ}}{cos 1 5 ^{\circ}} = \frac{2 sin 1 5 ^{\circ} cos 1 5 ^{\circ}}{cos 1 5 ^{\circ}} = 2 sin 1 5^{\circ}$

Trap avoided: it's easy to invert the ratio (compute AD/BE instead of BE/AD) or to stop at $sin 3 0^{\circ} / cos 1 5^{\circ}$ without simplifying to match the given option form.

Answer: $\frac{B E}{A D} = 2 sin 1 5^{\circ}$ .

Question 8 · Quantitative Ability MCQ

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 became right-angled. When the faster ship reaches the port, the distance, in km, between the other ship and the port will be

A.
4
B.
12
C.
8
D.
6

Correct Answer:

Step-by-Step Solution

This is the "Moving Ships and a Hidden Right Triangle" pattern: a moving triangle (formed by two ships and a fixed port) starts equilateral and later becomes right-angled, and the right-angle condition tells you exactly how far each ship has moved.

Step 1 — Recognize the trigger.

The port and two ships start as an equilateral triangle of side 24. Each ship travels straight toward the port, so the angle at the port between the two ships' paths always stays 60° (it never changes, since both ships move along their original straight lines toward the port).

Step 2 — Set up the new triangle after the slower ship moves 8 km.

Let the slower ship's new distance from the port be $24 - 8 = 16$ km. Let the faster ship's new distance from the port be $x$ km (unknown for now). The angle at the port between them is still $6 0^{\circ}$ .

Step 3 — Use the right-angle condition.

We're told the new triangle (port, slower ship, faster ship) is right-angled. Since the port's angle is fixed at 60° (not 90°), the right angle must be at one of the ships. Testing the right angle at the faster ship's new position gives a valid case: with angle at port = 60°, angle at faster ship = 90°, the remaining angle (at the slower ship) is 30°.

Step 4 — Apply the Law of Sines in this right triangle.

Side opposite the 90° angle (at the faster ship) is the port-to-slower-ship distance (16 km). Side opposite the 30° angle (at the slower ship) is the port-to-faster-ship distance ( $x$ ):

$\frac{16}{sin 9 0 ^{\circ}} = \frac{x}{sin 3 0 ^{\circ}} \Rightarrow x = 16 \times \frac{sin 3 0 ^{\circ}}{sin 9 0 ^{\circ}} = 16 \times \frac{1}{2} = 8$

Step 5 — Interpret this result.

The faster ship's new distance from port is 8 km, meaning it has already traveled $24 - 8 = 16$ km — twice as far as the slower ship's 8 km — in the same elapsed time. This confirms it truly is the faster ship (speed ratio 2:1).

Step 6 — Find how much further time the faster ship needs to reach the port.

The faster ship has 8 km left to cover. Since it covered 16 km in time $t$ , its speed is $16/ t$ , so the extra time to cover the remaining 8 km is:

$extra time = \frac{8}{16/ t} = \frac{t}{2}$

Step 7 — Find how far the slower ship travels in this extra time.

The slower ship's speed is $8/ t$ (it covered 8 km in time $t$ ). In the extra time $t /2$ , it covers:

$\frac{8}{t} \times \frac{t}{2} = 4 km$

Step 8 — Find the slower ship's total distance from the port.

Total distance traveled by the slower ship = $8 + 4 = 12$ km. Remaining distance from the port:

$24 - 12 = 12 km$

Trap avoided: it's easy to stop at Step 5 and report the faster ship's remaining distance (8 km) as the final answer — but the question asks for the slower ship's distance at the later moment when the faster ship reaches the port, which requires the extra time-projection in Steps 6–8.

Answer: 12 km.

Question 9 · Quantitative Ability NAT

The midpoints of sides AB, BC, and AC in $△ A B C$ 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 $△ A B C$ is 1440 sq cm, then the area, in sq cm, of $△ X Y Z$ is

Correct Answer:

Step-by-Step Solution

This is the PYQ pattern of finding the area of triangle XYZ formed inside the midpoint

triangle by the medians. You can recognise the pattern because M, N, P are midpoints of the

sides (forming the midpoint triangle MNP), and the three medians of the original triangle

cross the SIDES of this midpoint triangle at X, Y, Z. The cleanest way to handle this is

coordinates, since the ratio between the areas is fixed no matter the triangle's shape.

Step 1: Choose convenient coordinates.

Since area ratios do not change under this kind of coordinate placement (affine

transformations preserve area ratios), pick a simple triangle: A = (0,0), B = (2,0),

C = (0,2). This triangle has area = (1/2)22 = 2.

Step 2: Find the midpoints M, N, P.

M = midpoint of AB = (1, 0)

N = midpoint of BC = (1, 1)

P = midpoint of AC = (0, 1)

Step 3: Find X, the intersection of the median from A with segment MP.

The median from A goes to the midpoint of BC, which is N = (1,1). This median is the line

from (0,0) to (1,1), i.e. y = x.

Segment MP goes from M(1,0) to P(0,1), which lies on the line x + y = 1.

Solving y = x and x + y = 1 together: x + x = 1, so x = 0.5, y = 0.5.

X = (0.5, 0.5)

Step 4: Find Y, the intersection of the median from B with segment MN.

The median from B goes to the midpoint of AC, which is P = (0,1). This median goes from

B(2,0) to P(0,1).

Segment MN goes from M(1,0) to N(1,1), which lies on the vertical line x = 1.

Parametrising the median as (2 - 2t, t), setting the x-coordinate to 1 gives

2 - 2t = 1, so t = 0.5, giving y = 0.5.

Y = (1, 0.5)

Step 5: Find Z, the intersection of the median from C with segment NP.

The median from C goes to the midpoint of AB, which is M = (1,0). This median goes from

C(0,2) to M(1,0).

Segment NP goes from N(1,1) to P(0,1), which lies on the horizontal line y = 1.

Parametrising the median as (s, 2 - 2s), setting the y-coordinate to 1 gives

2 - 2s = 1, so s = 0.5.

Z = (0.5, 1)

Step 6: Compute the area of triangle XYZ using the coordinates found.

X = (0.5, 0.5), Y = (1, 0.5), Z = (0.5, 1).

Area = (1/2) * |x_X(y_Y - y_Z) + x_Y(y_Z - y_X) + x_Z(y_X - y_Y)|

= (1/2) |0.5(0.5 - 1) + 1(1 - 0.5) + 0.5(0.5 - 0.5)|

= (1/2) |0.5(-0.5) + 1*(0.5) + 0|

= (1/2) * |-0.25 + 0.5|

= (1/2) * 0.25 = 0.125

Step 7: Find the ratio of areas and apply it to the real triangle.

Ratio of area XYZ to area ABC = 0.125 / 2 = 1/16.

Since this ratio holds for any triangle (not just this coordinate example), apply it to the

given area:

Area of XYZ = 1440 * (1/16) = 90

Answer: the area of triangle XYZ is 90 sq cm.

Common trap: assuming X, Y, Z are the centroid or some other single special point, rather

than three distinct intersection points on the three different sides of the midpoint

triangle, leads to a wrong or undefined calculation. Working with actual coordinates avoids

this confusion completely.

Question 10 · Quantitative Ability NAT

In a circle of radius $13$ cm, two parallel chords have lengths $24$ cm and $10$ cm. Their four endpoints are joined to form a trapezium whose parallel sides are the two chords. What is the minimum possible area, in square cm, of this trapezium?

Correct Answer:

119

Step-by-Step Solution

Key idea: this is a parallel-chords trapezium question with a minimum-area twist. The hidden issue is that the two chords may lie on the same side of the centre or on opposite sides.

Step 1: Find the distance of each chord from the centre.

For a chord of length $L$ in a circle of radius $r$ ,

$d = r^{2} - (\frac{L}{2})^{2} .$

For the chord of length $24$ :

$d_{1} = 1 3^{2} - 1 2^{2} = 169 - 144 = 5.$

For the chord of length $10$ :

$d_{2} = 1 3^{2} - 5^{2} = 169 - 25 = 12.$

Step 2: Consider possible separations between the parallel chords.

If the chords are on the same side of the centre, their distance apart is

$12 - 5 = 7.$

If they are on opposite sides of the centre, their distance apart is

$12 + 5 = 17.$

Step 3: Choose the minimum height.

The minimum possible height of the trapezium is therefore $7$ .

Step 4: Compute the trapezium area.

The parallel sides are $24$ and $10$ , so

$Area = \frac{1}{2} (24 + 10) \times 7 = 17 \times 7 = 119.$

Answer: 119.

Question 11 · Quantitative Ability NAT

Real numbers $a$ , $b$ , and $c$ satisfy:

$3 a + 2 b - c = 10$

$a + 4 b + 2 c = 25$

What is the value of $5 a + 10 b + 3 c$ ?

Correct Answer:

Step-by-Step Solution

Key idea: This is a target expression via linear combination question. Recognisable because 3 variables, 2 equations, and a specific linear target — individual variables cannot be uniquely solved, but the target might be a fixed combination.

Step 1: Assume the target is a linear combination of given equations.

Let $5 a + 10 b + 3 c = x (3 a + 2 b - c) + y (a + 4 b + 2 c)$ .

Step 2: Match coefficients for each variable.

For $a$ : $3 x + y = 5$ ...(i)

For $b$ : $2 x + 4 y = 10 ⟹ x + 2 y = 5$ ...(ii)

For $c$ : $- x + 2 y = 3$ ...(iii)

Step 3: Solve for $x$ and $y$ using (i) and (ii).

From (ii): $x = 5 - 2 y$ .

Substitute into (i): $3 (5 - 2 y) + y = 5 ⟹ 15 - 6 y + y = 5 ⟹ - 5 y = - 10 ⟹ y = 2$ .

Then $x = 5 - 2 (2) = 1$ .

Step 4: Verify with equation (iii) to ensure consistency.

LHS: $- x + 2 y = - 1 + 2 (2) = 3$ . Matches RHS. ✓

The target IS a valid linear combination.

Step 5: Compute the target value.

Target = $x (Eq1) + y (Eq2) = 1 (10) + 2 (25) = 10 + 50 = 60$ .

Wait — my manual calculation gave 60, but I wrote answer as 45 above. Let me recheck.

Eq1: 10, Eq2: 25. x=1, y=2. 110 + 225 = 60.

I will correct the answer to 60 in the final output.

Answer: 60

Algebra, Geometry, DATA INTERPRETATION, READING COMPREHENSION Unit Test for CAT: 68 Questions with Solutions & Analysis

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