Number System, LOGICAL REASONING, VERBAL ABILITY Unit Test for CAT: 68 Questions with Solutions & Analysis

Attempt the Number System, LOGICAL REASONING, VERBAL ABILITY 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: 26 · Verbal Ability and Reading Comprehension: 24 · Data Interpretation and Logical Reasoning: 18

Free sample questions from Number System, LOGICAL REASONING, VERBAL ABILITY Unit Test

Question 1 · Verbal Ability and Reading Comprehension MCQ

There is a sentence that is missing in the paragraph below. Look at the paragraph and decide where (option 1, 2, 3, or 4) the following sentence would best fit.
Sentence: Many have had to leave their homes behind, with more than 1.3 million people being displaced due to the drought.
Passage: Somalia has been dealing with an enormous humanitarian catastrophe, driven by the longest and most severe drought the country has experienced in at least 40 years. ___(1)___. Five consecutive rainy seasons have failed, causing more than 8 million people - almost half of the country’s population - to experience acute food insecurity. ___(2)___. More than 43,000 people are believed to have lost their lives, with half of the lives lost likely being children under five. The damage the drought has caused is far-reaching. ___(3)___. Farmers have lost all their agricultural income, while pastoralists have lost more than 3 million livestock, impoverishing entire communities, and leaving them on the brink of famine. ___(4)___. Some, like the pastoralists, may never be able to go back as their livelihoods have been irreversibly wiped out.

A.
Option 4
B.
Option 2
C.
Option 3
D.
Option 1

Correct Answer:

Step-by-Step Solution

Key idea: This is a forward-link consequence insertion question, recognisable because the missing sentence introduces displaced people, and a later sentence refers to people who may never be able to go back.

Step 1: Read the missing sentence. It states that many people have had to leave their homes behind, with more than 1.3 million people being displaced due to the drought.

Step 2: Look for the strongest clue after the blanks. The sentence after Blank 4 says that some, like the pastoralists, may never be able to go back as their livelihoods have been irreversibly wiped out.

Step 3: Identify the reference. The phrase "go back" only makes logical sense if people have already been established as having left their homes. Therefore, the displacement sentence must precede this idea.

Step 4: Test Blank 4. The flow becomes: farmers and pastoralists suffer severe losses; many have to leave their homes; some may never be able to go back. This creates a clean Consequence $\to$ Displacement $\to$ Inability to return sequence.

Step 5: Reject other blanks. Blank 1 is too early because the paragraph has not yet explained the scale of the crisis. Blank 2 is possible as a general escalation, but it weakens the direct link between displacement and the final "go back" sentence. Blank 3 interrupts the transition from general damage to specific livelihood losses.

Answer: Option 4 (A).

Question 2 · Data Interpretation and Logical Reasoning NAT

Common Description: Instructions [25 - 29]
The figure below shows a network with three parallel roads represented by horizontal lines R-A, R-B, and R-C and another three parallel roads represented by vertical lines V1, V2, and V3. The figure also shows the distance (in km) between two adjacent intersections. Six ATMs are placed at six of the nine road intersections. Each ATM has a distinct integer cash requirement (in Rs. Lakhs), and the numbers at the end of each line in the figure indicate the total cash requirements of all ATMs placed on the corresponding road. For example, the total cash requirement of the ATM(s) placed on road R-A is Rs. 22 Lakhs.

The following additional information is known.
1. The ATMs with the minimum and maximum cash requirements of Rs. 7 Lakhs and Rs. 15 Lakhs are placed on the same road.
2. The road distance between the ATM with the second highest cash requirement and the ATM located at the intersection of R-C and V3 is 12 km. How many ATMs have cash requirements of Rs. 10 Lakhs or more?

Correct Answer:

Step-by-Step Solution

Key idea: Finding exact element sets from total sum constraints and integer partitioning. This is recognisable by a grid with row/column sums and a bounded set of distinct integers.

Step 1: Calculate the required excluded sum.

The grid contains 6 ATMs with distinct integer values. The total sum of all ATMs is the sum of the row totals: $22 + 20 + 20 = 62$ .

The problem states the minimum value is 7 and maximum is 15. The possible values are integers from 7 to 15 inclusive.

Sum of all integers from 7 to 15 = 99.

We need to select 6 distinct integers that sum to 62. This means we must EXCLUDE exactly 3 integers from the set ${7, 8, 9, 10, 11, 12, 13, 14, 15}$ that sum to $99 - 62 = 37$ .

Step 2: Find the excluded triple.

The excluded numbers cannot be 7 or 15 (as they are explicitly stated to be present).

We must pick 3 distinct numbers from ${8, 9, 10, 11, 12, 13, 14}$ that sum to 37.

The maximum possible sum of 3 numbers from this subset is $12 + 13 + 14 = 39$ .

To get 37, we must drop the sum by 2 from the maximum. The only valid distinct triple is ${10, 13, 14}$ .

(Note: ${11, 12, 14}$ sums to 37, but testing this set against the grid column constraints reveals it cannot form valid row pairs, leaving ${10, 13, 14}$ as the unique working exclusion).

Step 3: Identify the final ATM set.

Excluding ${10, 13, 14}$ leaves the set $S_{1} = {7, 8, 9, 11, 12, 15}$ .

Step 4: Count ATMs $\geq 10$ Lakhs.

The values in $S_{1}$ that are 10 or more are 11, 12, and 15.

There are exactly 3 such ATMs.

Answer: 3

Question 3 · Data Interpretation and Logical Reasoning MCQ

Common Description: Instructions [30 - 33]
The above is a schematic diagram of walkways (indicated by all the straight-lines) and lakes (3 of them, each in the shape of rectangles - shaded in the diagram) of a gated area. Different points on the walkway are indicated by letters (A through P) with distances being OP = 150 m, ON = MN = 300 m, ML = 400 m, EL = 200 m, DE = 400 m.

The following additional information about the facilities in the area is known.
1. The only entry/exit point is at C.
2. There are many residences within the gated area; all of them are located on the path AH and ML with four of them being at A, H, M, and L.
3. The post office is located at P and the bank is located at B. One person enters the gated area and decides to walk as much as possible before leaving the area without walking along any path more than once and always walking next to one of the lakes. Note that he may cross a point multiple times. How much distance (in m) will he walk within the gated area?

A.
2800
B.
3000
C.
3800
D.
3200

Correct Answer:

Step-by-Step Solution

Key idea: Eulerian path maximization on a geometrically restricted subgraph.

Step 1: Interpret the constraint. The person must walk next to a lake, meaning they can only traverse the perimeter edges of the 3 rectangular lakes.

Step 2: Identify valid walkways. To maximize distance without repeating any path, we must find the longest valid trail on the lake perimeters starting from the entry/exit point C.

Step 3: Calculate total distance. The optimal path traverses the 400 m walkways 4 times (e.g., DE, LM, KN, EF), the 300 m walkways 6 times (e.g., CD, KL, MN, KJ, GF, FC), and the 200 m walkways 2 times (FK, JG).

Total distance = $(4 \times 400) + (6 \times 300) + (2 \times 200) = 1600 + 1800 + 400 = 3800$ m.

Answer: C

Question 4 · Quantitative Ability MCQ

Let a, b, m and n be natural numbers such that a > 1 and b > 1. If $a^{m} b^{n} = 14 4^{145}$ , then the largest possible value of $n - m$ is

A.
580
B.
290
C.
289
D.
579

Correct Answer:

Step-by-Step Solution

Pattern: this is an extremal exponent allocation question — recognisable because two unknown bases raised to unknown powers are set equal to one fixed number's power ( $a^{m} b^{n} = 14 4^{145}$ ), and we must extremize a combination of the exponents ( $n - m$ ).

Step 1 — Break 144 into primes.

$144 = 16 \times 9 = 2^{4} \times 3^{2}$ .

So $14 4^{145} = 2^{4 \times 145} \times 3^{2 \times 145} = 2^{580} \times 3^{290}$ .

Step 2 — See what $a, b$ are allowed to be.

The right side only has the primes $2$ and $3$ . If $a$ or $b$ contained any other prime, that prime would show up in $a^{m} b^{n}$ , which it does not. So $a$ and $b$ must be built only from $2$ 's and $3$ 's.

Step 3 — Make $m$ as small as possible.

We want the largest $n - m$ . Since $m$ is a natural number, the smallest legal value is $m = 1$ (not $0$ — natural numbers here start at $1$ ). Use $m = 1$ .

Step 4 — Decide where the two exponents (580 and 290) go.

With $m = 1$ , the equation becomes $a \cdot b^{n} = 2^{580} 3^{290}$ . We want $n$ as large as possible. To pull a big exponent out as $n$ , use the smallest possible base for $b$ and let it absorb the larger of the two available exponents (580), while $a$ quietly absorbs the smaller leftover exponent (290).

So take $b = 2$ , requiring $b^{n} = 2^{580} \Rightarrow n = 580$ , and $a = 3^{290}$ (which is a valid natural number greater than 1).

Check: $a^{m} b^{n} = (3^{290})^{1} \times 2^{580} = 2^{580} 3^{290} = 14 4^{145}$ . ✓

Step 5 — Compute the answer.

$n - m = 580 - 1 = 579$ .

Why the reverse choice is a trap: if instead you let $a = 2^{580}$ (absorbing the bigger exponent into $a$ ) and $b = 3$ (absorbing the smaller exponent, 290, as $n$ ), you get $n = 290$ , so $n - m = 289$ — a smaller, wrong value. The 289 option exists specifically to catch students who assign the exponents without checking which assignment actually maximises $n$ . Always route the larger exponent into $b^{n}$ (via the smallest base) when you are trying to maximise $n$ .

Question 5 · Quantitative Ability MCQ

Let n and m be two positive integers such that there are exactly 41 integers greater than $8^{m}$ and less than $8^{n}$ , which can be expressed as powers of 2. Then, the smallest possible value of n + m is

A.
42
B.
44
C.
14
D.
16

Correct Answer:

Step-by-Step Solution

Key idea: This is a power-form counting question. The trick is to rewrite everything as a power of the same base, then count how many integer exponents lie strictly between two bounds.

Why this applies: The question compares $8^{m}$ and $8^{n}$ but asks about numbers expressible as "powers of 2" — so converting the base-8 boundaries into base-2 boundaries makes the counting direct.

Step 1 — Rewrite in base 2:

Since $8 = 2^{3}$ :

$8^{m} = 2^{3 m}, 8^{n} = 2^{3 n}$

Step 2 — Count powers of 2 strictly between these:

We want integers $2^{p}$ with

$2^{3 m} < 2^{p} < 2^{3 n} ⟺ 3 m < p < 3 n$

Since $p$ is an integer, $p$ ranges from $3 m + 1$ to $3 n - 1$ . The count of such integers is:

$(3 n - 1) - (3 m + 1) + 1 = 3 n - 3 m - 1 = 3 (n - m) - 1$

Step 3 — Set this equal to 41 (given):

$3 (n - m) - 1 = 41 ⟹ 3 (n - m) = 42 ⟹ n - m = 14$

Step 4 — Minimise $n + m$ subject to $n - m = 14$ :

Write $n = m + 14$ . Then

$n + m = 2 m + 14$

This is minimised by taking $m$ as small as possible. Since $m$ is a positive integer, the smallest allowed value is $m = 1$ , giving $n = 15$ .

$n + m = 2 (1) + 14 = 16$

Common trap: Forgetting the strict inequality (using $\leq$ instead of $<$ ) changes the count formula and gives a wrong equation for $n - m$ .

Answer: 16

Question 6 · Quantitative Ability NAT

For a 4-digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4-digit number satisfying the above conditions is

Correct Answer:

4195

Step-by-Step Solution

Pattern: this is a "digit place-value equations" question. You are told sums of overlapping groups of digits and one digit relationship, and asked for the maximum number — so the trigger words are "sum of digits in ... places" and "highest possible".

Step 1 — Name the digits.

Let the 4-digit number be $\overline{ab c d}$ with thousands digit $a$ , hundreds $b$ , tens $c$ , units $d$ .

Step 2 — Translate every sentence into an equation.

"sum of thousands, hundreds, tens digits is 14": $a + b + c = 14$ ... (i)
"sum of hundreds, tens, units digits is 15": $b + c + d = 15$ ... (ii)
"tens digit is 4 more than units digit": $c = d + 4$ ... (iii)

Step 3 — Combine (i) and (ii) to link $a$ and $d$ directly.

Subtracting (ii) from (i): $(a + b + c) - (b + c + d) = 14 - 15$ , so $a - d = - 1$ , i.e. $a = d - 1$ .

This is the key move — it removes $b, c$ and connects the two digits we actually need to control ( $a$ , the most valuable digit, and $d$ ).

Step 4 — Apply digit bounds.

$a \geq 1$ (leading digit can't be 0) $\Rightarrow d \geq 2$ .

From (iii), $c = d + 4 \leq 9 \Rightarrow d \leq 5$ .

So $d \in {2, 3, 4, 5}$ , giving $a = d - 1 \in {1, 2, 3, 4}$ respectively.

Step 5 — Maximize the number correctly (most significant digit first).

To get the highest 4-digit number, we must maximize $a$ first — not $d$ or $c$ in isolation. Since $a = d - 1$ increases with $d$ , the largest $a$ comes from the largest allowed $d$ , which is $d = 5$ , giving $a = 4$ .

Step 6 — Find the remaining digits for $d = 5$ .

$c = d + 4 = 9$ .

From (ii): $b = 15 - c - d = 15 - 9 - 5 = 1$ .

Check (i): $a + b + c = 4 + 1 + 9 = 14$ ✓.

Step 7 — State the number.

$\overline{ab c d} = 4, 1, 9, 5 \Rightarrow N = 4195$ .

Answer: 4195.

Question 7 · Quantitative Ability NAT

For any natural number $n$ let $a_{n}$ be the largest integer not exceeding $n$ . Then the value of $a_{1} + a_{2} + ..... + a_{50}$ is

Correct Answer:

217

Step-by-Step Solution

Pattern: this is a floor-function block-summation question — recognisable because we must sum $⌊ n ⌋$ (or a similar floor expression) across a range of $n$ , which is best done by grouping $n$ into blocks where the floor value stays constant.

Step 1 — Understand when $⌊ n ⌋ = k$ .

$⌊ n ⌋ = k$ exactly when $k^{2} \leq n \leq (k + 1)^{2} - 1$ , i.e. $n$ runs over $2 k + 1$ consecutive integers (from $k^{2}$ to $k^{2} + 2 k$ ).

Step 2 — Build the blocks up to n = 50.

$k = 1$ : $n = 1$ to $3$ (3 values) → contributes $1 \times 3 = 3$
$k = 2$ : $n = 4$ to $8$ (5 values) → contributes $2 \times 5 = 10$
$k = 3$ : $n = 9$ to $15$ (7 values) → contributes $3 \times 7 = 21$
$k = 4$ : $n = 16$ to $24$ (9 values) → contributes $4 \times 9 = 36$
$k = 5$ : $n = 25$ to $35$ (11 values) → contributes $5 \times 11 = 55$
$k = 6$ : $n = 36$ to $48$ (13 values) → contributes $6 \times 13 = 78$
$k = 7$ : $n = 49$ to $50$ — the block would normally run to $63$ , but our sum stops at $n = 50$ , so only $2$ values (49, 50) are included → contributes $7 \times 2 = 14$

Step 3 — Check the total count of n's covered.

$3 + 5 + 7 + 9 + 11 + 13 + 2 = 50$ ✓ (matches $n = 1$ to $50$ exactly).

Step 4 — Add up all contributions.

$3 + 10 + 21 + 36 + 55 + 78 + 14 = 217$

Trap: the final block ( $k = 7$ ) is a partial block — it does not run the full $2 k + 1 = 15$ values (which would go up to $n = 63$ ), only up to $n = 50$ . Forgetting to cut this last block short (and instead using the full 15 values) is the single most common mistake in this pattern.

So the answer is $217$ .

Question 8 · Quantitative Ability MCQ

If $m$ and $n$ are natural numbers such that $n > 1$ , and $m^{n} = 2^{25} \times 3^{40}$ , then $m - n$ equals

A.
209932
B.
209937
C.
209942
D.
209947

Correct Answer:

Step-by-Step Solution

This is a "perfect power" question — recognizable because you're told a number is expressible as $m^{n}$ for natural numbers with $n > 1$ , and asked to find $m$ and $n$ themselves.

Step 1: Look at the prime factorization already given.

$m^{n} = 2^{25} \times 3^{40}$

Step 2: For $m$ to be a natural number, $m^{n} = 2^{25} 3^{40}$ means $m$ itself must be of the form $2^{a} 3^{b}$ , and then

$m^{n} = 2^{an} \times 3^{bn} = 2^{25} \times 3^{40}$

So $an = 25$ and $bn = 40$ .

Step 3: This means $n$ must divide both 25 and 40 — $n$ is a common factor of the two exponents.

Divisors of 25: 1, 5, 25. Divisors of 40: 1, 2, 4, 5, 8, 10, 20, 40. The only common divisor greater than 1 (since $n > 1$ is required) is $n = 5$ .

Step 4: With $n = 5$ : $a = 25/5 = 5$ and $b = 40/5 = 8$ .

$m = 2^{5} \times 3^{8} = 32 \times 6561 = 209952$

Step 5: Compute $m - n$ .

$m - n = 209952 - 5 = 209947$

Trap to avoid: don't pick any divisor of 40 (like 10 or 20) without checking it also divides 25 exactly — $n$ must work for BOTH exponents simultaneously, since it's the same $n$ in $m^{n}$ .

Answer: m − n = 209947 (Option D).

Question 9 · Quantitative Ability NAT

A school has less than 5000 students and if the students are divided equally into teams of either 9 or 10 or 12 or 25 each, exactly 4 are always left out. However, if they are divided into teams of 11 each, no one is left out. The maximum number of teams of 12 each that can be formed out of the students in the school is

Correct Answer:

150

Step-by-Step Solution

Pattern: this is a simultaneous congruence (same-remainder) question — recognisable because the same remainder (4) shows up across several divisors (9,10,12,25), plus one extra clean-divisibility condition (11) to pin down the exact number.

Step 1 — Convert "remainder 4 with 9, 10, 12, 25" into one condition.

If $N$ leaves remainder $4$ when divided by each of $9, 10, 12, 25$ , then $N - 4$ is exactly divisible by all of them, i.e. by their LCM.

Step 2 — Compute the LCM.

$9 = 3^{2}, 10 = 2 \times 5, 12 = 2^{2} \times 3, 25 = 5^{2}$ .

$lcm = 2^{2} \times 3^{2} \times 5^{2} = 4 \times 9 \times 25 = 900$

Step 3 — Write the general form of N.

$N - 4 = 900 k ⟹ N = 900 k + 4, k = 1, 2, 3, \dots$

Step 4 — Apply the "less than 5000" limit.

$900 k + 4 < 5000 \Rightarrow k < 5.55 \dots$ , so $k \in {1, 2, 3, 4, 5}$ (since $900 \times 5 + 4 = 4504 < 5000$ but $900 \times 6 + 4 = 5404 > 5000$ ).

Step 5 — Apply the "divisible by 11 with no remainder" condition.

We need $N \equiv 0 (mod 11)$ . First reduce $900 mod 11$ : $900 = 81 \times 11 + 9$ , so $900 \equiv 9 (mod 11)$ .

So $N \equiv 9 k + 4 (mod 11)$ . Check each $k$ :

$k = 1 : 9 (1) + 4 = 13 \equiv 2$
$k = 2 : 9 (2) + 4 = 22 \equiv 0$ ✓
$k = 3 : 9 (3) + 4 = 31 \equiv 9$
$k = 4 : 9 (4) + 4 = 40 \equiv 7$
$k = 5 : 9 (5) + 4 = 49 \equiv 5$

Only $k = 2$ works.

Step 6 — Find N.

$N = 900 (2) + 4 = 1804$ .

Step 7 — Maximum number of teams of 12.

$1804 \div 12 = 150$ remainder $4$ (consistent with our very first condition!). So the maximum number of full teams of 12 is $⌊ 1804/12 ⌋ = 150$ .

Trap: a student who stops at " $N = 900 k + 4$ " without also checking every $k$ against the mod-11 condition might pick the wrong $k$ (e.g. assuming $k = 1$ works without checking), or forget to apply the "less than 5000" bound at all and miss that only one valid $N$ exists.

Question 10 · Quantitative Ability MSQ

Let $a$ , $b$ , $m$ and $n$ be natural numbers with $a > 1$ and $b > 1$ such that

$a^{m} b^{n} = 18 0^{10} .$

Which of the following can be the value of $n - m$ ?

A.
-19
B.
0
C.
19
D.
20

Correct Answer:

["A","B","C"]

Step-by-Step Solution

Key idea: this is an extremal exponent allocation question, but instead of asking only for the maximum, it asks which differences are possible. We need both bounds and constructions.

Step 1: Prime-factorize the fixed number.

$180 = 2^{2} 3^{2} 5$

Therefore:

$18 0^{10} = 2^{20} 3^{20} 5^{10}$

The largest prime exponent available is 20.

Step 2: Bound $n$ and $m$ .

Since $b > 1$ , $b$ contains at least one prime. If that prime appears in $b$ with exponent at least 1, then in $b^{n}$ its exponent is at least $n$ .

But no prime exponent in $18 0^{10}$ exceeds 20. Hence:

$n \leq 20$

Similarly:

$m \leq 20$

Also $m, n$ are natural numbers, so $m \geq 1$ and $n \geq 1$ .

Therefore:

$n - m \leq 20 - 1 = 19$

and:

$n - m \geq 1 - 20 = - 19$

Step 3: Show the endpoint values are possible.

For $n - m = 19$ , take:

$b = 2, n = 20, a = 3^{20} 5^{10}, m = 1$

Then:

$a^{1} b^{20} = 3^{20} 5^{10} 2^{20} = 18 0^{10}$

For $n - m = - 19$ , take:

$a = 2, m = 20, b = 3^{20} 5^{10}, n = 1$

Then:

$a^{20} b^{1} = 2^{20} 3^{20} 5^{10} = 18 0^{10}$

Step 4: Show 0 is possible.

We need $m = n$ . Since the exponents of $18 0^{10}$ are $20, 20, 10$ , we can use $m = n = 10$ :

$a = 4, b = 45, m = n = 10$

Then:

$4^{10} 4 5^{10} = (4 \times 45)^{10} = 18 0^{10}$

Step 5: Check 20.

$n - m = 20$ would require $n \geq 21$ because $m \geq 1$ , but we already proved $n \leq 20$ . So 20 is impossible.

Answer: A, B, C

Question 11 · Quantitative Ability MCQ

If $x$ , $y$ and $z$ are natural numbers satisfying

$4^{2 x} \cdot 2^{y} \cdot 9^{x} \cdot 2 7^{z} \cdot 5^{x + y + z} = 2^{2 z + 3} \cdot 3^{5 y + 1} \cdot 5^{3 x + y},$

which of the following ordering relations is correct?

A.
$x < y < z$
B.
$x < z < y$
C.
$y < x < z$
D.
$z < y < x$

Correct Answer:

Step-by-Step Solution

Key idea: this is a prime-exponent equality question. The equation contains products of powers, so the correct move is to rewrite everything using prime bases and then match exponents.

Step 1: Rewrite the left-hand side in prime bases.

$4^{2 x} = (2^{2})^{2 x} = 2^{4 x}$

So the power of 2 on the left is:

$4 x + y$

Also:

$9^{x} = (3^{2})^{x} = 3^{2 x}$

$2 7^{z} = (3^{3})^{z} = 3^{3 z}$

So the power of 3 on the left is:

$2 x + 3 z$

The power of 5 on the left is:

$x + y + z$

Step 2: Match the exponent of 5.

Right-hand side has $5^{3 x + y}$ , so:

$x + y + z = 3 x + y$

Cancel $y$ :

$z = 2 x$

Step 3: Match the exponent of 2.

Left: $4 x + y$ . Right: $2 z + 3$ .

$4 x + y = 2 z + 3$

Substitute $z = 2 x$ :

$4 x + y = 4 x + 3$

Therefore:

$y = 3$

Step 4: Match the exponent of 3.

Left: $2 x + 3 z$ . Right: $5 y + 1$ .

$2 x + 3 z = 5 y + 1$

Substitute $y = 3$ and $z = 2 x$ :

$2 x + 3 (2 x) = 15 + 1$

$8 x = 16$

$x = 2$

Then:

$z = 2 x = 4$

Step 5: Compare.

$x = 2, y = 3, z = 4$

Therefore:

$x < y < z$

Answer: A

Number System, LOGICAL REASONING, VERBAL ABILITY Unit Test for CAT: 68 Questions with Solutions & Analysis

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