Critical Reasoning Questions for CAT [Download PDF]

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CAT Critical Reasoning Questions PDF
CAT Critical Reasoning Questions PDF

Critical Reasoning (CR) questions are asked as a part of Reading Comprehension questions in the CAT VARC section. These types of questions are generally challenging in the RCs. If you find these questions a bit tough, make sure you solve more RCs that involve CR based questions. Learn how to answer questions on Critical Reasoning concepts. You can check out these Critical Reasoning questions from the CAT Previous year’s papers. Practice a good number of questions in the CAT Critical Reasoning so that you can answer these questions with ease in the exam. In this post, we will look into some important Critical Reasoning Questions for CAT VARC. These are a good source for practice; If you want to practice these questions, you can download this Important CAT Critical Reasoning Questions PDF below, which is completely Free.

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Question 1: The value of $8\times6\div15\div4\times16+7\div12\times6-5\div8\times12$ is:

a) 12.6

b) 10

c) 8.8

d) 3.2

1) Answer (C)

Solution:

$8\times6\div15\div4\times16+7\div12\times6-5\div8\times12$
$=8\times\dfrac{6}{15\times4}\times16+\dfrac{7}{12}\times6-\dfrac{5}{8}\times12$

$=12.8+3.5-7.5=12.8-4=8.8$

Question 2: The value of $(3\dfrac{4}{9}+2\dfrac{3}{7}-1\dfrac{1}{5}-4\dfrac{2}{7})\div(5\dfrac{2}{9}+\dfrac{3}{5}\times4\dfrac{4}{9})\times(3\dfrac{3}{5}\times\dfrac{4}{9}+2\dfrac{2}{5}) = 2-p$. ‘p’ lies between:

a) 1.5 and 1.6

b) 1.8 and 1.9

c) 1.4 and 1.5

d) 1.7 and 1.8

2) Answer (B)

Solution:

$(3\dfrac{4}{9}+2\dfrac{3}{7}-1\dfrac{1}{5}-4\dfrac{2}{7})$

$=3+2-1-4+\dfrac{4}{9}+\dfrac{3}{7}-\dfrac{1}{5}-\dfrac{2}{7}$

$=\dfrac{140+135-90-63}{315}=\dfrac{122}{315}$

$(5\dfrac{2}{9}+\dfrac{3}{5}\times4\dfrac{4}{9})$

$=\dfrac{47}{9}+\dfrac{3}{5}\times\dfrac{40}{9}$

$=\dfrac{47}{9}+\dfrac{8}{3}$

$=\dfrac{71}{9}$

$(3\dfrac{3}{5}\times\dfrac{4}{9}+2\dfrac{2}{5})$

$=\dfrac{18}{5}\times\dfrac{4}{9}+\dfrac{12}{5}$

$=\dfrac{8}{5}+\dfrac{12}{5}=\dfrac{20}{5}=4$

$(3\dfrac{4}{9}+2\dfrac{3}{7}-1\dfrac{1}{5}-4\dfrac{2}{7})\div(5\dfrac{2}{9}+\dfrac{3}{5}\times4\dfrac{4}{9})\times(3\dfrac{3}{5}\times\dfrac{4}{9}+2\dfrac{2}{5})$

$=\dfrac{122}{315}\times\dfrac{9}{71}\times4 = 2-p$

$2-p = 0.1963$
$p = 2-0.1963 = 1.8037$ which lies between 1.8 and 1.9.

Question 3: If $7\times16807\div343\div49^{-3}\times2401^{-2} = 117649^x$, then the value of $46656^x$ is:

a) 6

b) $7^3$

c) 9

d) $8^6$

3) Answer (A)

Solution:

$7\times16807\div343\div49^{-3}\times2401^{-2}$
$=7\times7^5\div7^3\div(7^2)^{-3}\times(7^4)^{-2}$
$=7\times7^5\times7^{-3}\times7^6\times7^{-8}$
$=7^{1+5-3+6-8} = 7^1$

Given, $7^1 = 117649^x$
$7^1 = (7^6)^x$
$7^1=7^6x$
⇒ 6x = 1
x = $\dfrac{1}{6}$

Therefore, The value of $46656^x = 46656^\frac{1}{6} = 6$.

Question 4: The value of $2\div7\times21\times4\div8\div9\div12\times16\div6\times18\times15$ is:

a) 24.67

b) 23.33

c) 20

d) 28

4) Answer (C)

Solution:

$2\div7\times21\times4\div8\div9\div12\times16\div6\times18\times15$

$=\dfrac{2\times21\times4\times16\times18\times15}{7\times8\times9\times12\times6}$

$=20$

Question 5: Pipe A can fill a tank in x hours. Pipe B is 33.33% more efficient than A. The time taken for Pipe A to completely fill the tank is four hours less than the time taken for Pipe C to empty the completely filled tank. If all the three pipes are opened, the time taken to fill 66.67% of the tank is 3 hours 12 minutes, then find the value of x.

a) 8

b) 10

c) 12

d) 15

5) Answer (A)

Solution:

Let the total capacity of the tank be 3x units.

Efficiency of A = $\dfrac{3x}{x} = 3$ units/hour.

Efficiency of B = 33.33% more than that of A = $\dfrac{4}{3}\times3 = 4$ units/hour.

Time taken by Pipe C to empty the tank = x+4 hours.

Efficiency of C = $-\dfrac{3x}{x+4}$ units/hour. (Negative indicates draining)

Given, Time taken to fill 66.67% of tank is 3 hr 12 min = 3.2 hours.

66.67% of 3x = $\dfrac{2}{3}\times3x = 2x$ units.

$\dfrac{2x}{3+4-\dfrac{3x}{x+4}}=3.2$

$\dfrac{2x^2+8x}{4x+28}=3.2$

$\dfrac{x^2+4x}{2x+14}=3.2$

$x^2-2.4x-44.8=0$

$5x^2-12x-224=0$
$(5x+28)(x-8)=0$
$x=\dfrac{-28}{5}$ or $x=8$

Since the time cannot be in negative, the value of x is 8.

Question 6: Pipe P and a drain pipe R can together fill the tank in x hours. Another pipe Q and the drain pipe R can together fill the tank in 3x hours. Pipes P and Q can together fill the tank in 14.4 hours. If the efficiency of the pipes P and Q together is 10 times of the pipes Q and R together, then find the time taken by all the three pipes to fill the tank together.

a) 20.57 hours

b) 22.63 hours

c) 21.33 hours

d) 24.67 hours

6) Answer (A)

Solution:

Given, Pipe P and a drain pipe R can together fill the tank in x hours.
(P+R) → x hours
Another pipe Q and the drain pipe R can together fill the tank in 3x hours.
(Q+R) → 3x hours.
Pipes P and Q can together fill the tank in 14.4 hours.
Let the total capacity of the tank be 72x units. (LCM of x,3x,14.4)
Efficiency of P+R = 72 units/hour.
Efiiciency of Q+R = 24 units/hour.
Efficiency of P+Q = 5x units/hour.

Given, Efficiency of P+Q = 10 $\times$ Efficiency of Q+R
$5x = 10\times24$
$x = 48$
Then, Time taken by the pipes are,
P+R → 48 hours
Q+R → 144 hoursP+Q → 14.4 hours.

Then, The total capacity of the tank = LCM of 48,144 and 14.4 = 144 units.
Efficiencies of the pipes:
P+Q = 10 units per hour
Q+R = 1 unit per hour
P+R = 3 units per hour.

⇒ 2(P+Q+R) = 10+1+3 = 14
⇒ P+Q+R = 7.

Then, P, Q and R can together fill the tank in $\dfrac{144}{7} = 20.57$ hours.

Question 7: A sum of money of Rs.15000 is lent in two parts such that one part is lent at Simple Interest at 11.11% per annum for two years and the other part is lent at Compound Interest at 12.5% per annum compounded annually for two years. If the difference between the interests earned from both the schemes is Rs.507, then find the interest earned on the sum of money which was invested at simple interest is now invested at 8.33% per annum Compound interest compounded annually for 2 years.

a) Rs.1677.77

b) Rs.1432.67

c) Rs.1756

d) Rs.1237.5

7) Answer (D)

Solution:

Given, Principal = Rs.15000.

Let the sum of money invested at Simple interest be Rs.9x. (Since, 11.11% = $\dfrac{1}{9}$)

Interest earned in two years = $\dfrac{9x\times2}{9} = Rs.2x.$

Sum of money invested at Compound Interest = Rs.(15000-9x).

Amount earned in two years = $(15000-9x)\times\dfrac{9}{8}\times\dfrac{9}{8}$

Interest earned through Compound Interest = $(15000-9x)\times\dfrac{81}{64}-(15000-9x)$

Given, $(15000-9x)\times\dfrac{81}{64}-(15000-9x) – 2x = 507$

$\dfrac{151875}{8}-\dfrac{729x}{64}-15000+9x-2x=507$

$9x-2x-\dfrac{729x}{64}+\dfrac{151875}{8}-15000=507$

$\dfrac{31875}{8}-\dfrac{281x}{64}=507$

$\dfrac{281x}{64}=\dfrac{31875}{8}-507$

$\dfrac{281x}{64}=\dfrac{27819}{8}$

$281x=222552$

$x=792$.

Therefore, Sum of money invested at Simple Interest = 9x = Rs.7128.

Hence, Required Interest = $7128\times\dfrac{13}{12}\times\dfrac{13}{12} = Rs.8365.5$.

Hence, Interest earned = 8365.5 – 7128 = Rs.1237.5

Question 8: In approximately how many years will a certain sum of money become 21 times of itself at 7% per annum Compound interest compounded annually?

a) 45 years

b) 84 years

c) 68 years

d) 39 years

8) Answer (A)

Solution:

Let the sum of money be Rs.x
Let the time period be ‘T’ years.
Amount after T years = Rs.343x

$x\times\dfrac{107}{100}^T = 21x$

$\dfrac{107}{100}^T = 21$

$1.07^T = 21$

From the options,
A) $1.07^\text{45} \approx 21$
B) $1.07^\text{84} \approx 294$
C) $1.07^\text{68} \approx 100$
D) $1.07^\text{39} \approx 14$

Hence, Option A is the correct answer.

Question 9: A person invested a certain sum of money at 15% Simple interest per annum for three years. He then withdraws the whole amount and invested the whole amount in a scheme which offers 20% per annum for the first year and 10% per annum for the second year. He withdraws the whole amount which is equal to Rs.5742. What is the sum which he invested initially?

a) Rs.2500

b) Rs.3000

c) Rs.4000

d) Rs.4500

9) Answer (B)

Solution:

Let the initial investment be Rs.x.

Amount after 3 years = $\dfrac{x\times15\times3}{100}+x = Rs.1.45x$

Amount after next two year = $1.45x\times1.2\times1.1 = Rs.1.914x$.

Given, 1.914x = 5742

x = $\dfrac{5742}{1.914} = 3000$

Hence, Option B is the correct answer.

Question 10: Which of the following yields the highest interest of all when invested for 2 years?
I. Rs.24000 invested at 12.5% per annum Simple Interest.
II. Rs.19440 invested at 11.11% per annum Compound interest compounded annually.
III. Rs.28750 invested at 10% per annum compounded half-yearly.

a) Only I

b) Both II and III

c) Only III

d) Both I and III

10) Answer (C)

Solution:

I. Principal = Rs.24000
Rate of interest = 12.5%
Time period = 2 years.
Interest = $\dfrac{24000\times12.5\times2}{100} = Rs.6000$.

II. Principal = Rs.19440
Rate of interest = 11.11% = $\dfrac{1}{9}$
Amount after two years = $19440\times\dfrac{10}{9}\times\dfrac{10}{9} = Rs.24000$

Interest = Rs.24000-19440 = Rs.4560.

III. Principal = Rs.28750.
Rate of interest = 10% per annum compounded half-yearly.
Amount after two years = $28750\times1.05\times1.05\times1.05\times1.05 = Rs.34945.8$.
Interest = Rs.34945.8 – Rs.28750 = Rs.6195.8
Hence, Option C is the correct answer.

Question 11: Three years ago, the average age of P, Q and R was 21 and that of Q and R five years ago was 18 years. Find the presents age of P.

a) 25 years

b) 26 years

c) 27 years

d) 23 years

11) Answer (B)

Solution:

According to the question,
The sum of ages of P, Q and R three years ago be = 21 x 3 = 63 years
So present age = 63 + 9 = 72
Sum of ages of Q and R five years ago = 36 years
So present age = 36 + 10 = 46
So present age of A = 72 – 46 = 26 years

Question 12: If $x^3$ = xyz = 27, where x, y, z are positive integers, then find the value of z, for x = y = z $\neq$1

a) 3

b) 9

c) 6

d) Data insufficient

12) Answer (A)

Solution:

According to the question,
Prime factors of 27 = 3 x 3 x 3
$x^3$ = xyz = 27
$x^2$ = yz
$x^3$ = 27
x = 3
9 = yz,
Only possible way where yz = 9 is for y = 3 and z = 3
So z = 3

Question 13: If a = (6 + $4\sqrt{2}$), then find the value of $\sqrt{a} + \frac{1}{\sqrt{a}}$.

a) 1

b) ${3+\sqrt{2}}$

c) 4

d) $\frac{6+\sqrt{2}}{2}$

13) Answer (D)

Solution:

According to the question,
a = (6 + $4\sqrt{2}$)
a = $4 + 2 + 2 \times 2 \times \sqrt{2}$
a = $(2 + \sqrt{2})^2$
So,
$\sqrt{a} = (2 + \sqrt{2})$
So,
$\frac{1}{\sqrt{a}} = \frac{1}{(2 + \sqrt{2})}$
$\frac{1}{\sqrt{a}} = \frac{2 – \sqrt 2}{(2 + \sqrt 2)(2 – \sqrt 2)}$, solving this we get
$\frac{1}{\sqrt{a}} = \frac{2 – \sqrt{2}}{2}$

So,
$\sqrt{a} + \frac{1}{\sqrt{a}}$
$(2 + \sqrt{2}) + \frac{2 – \sqrt{2}}{2} = \frac{6+\sqrt{2}}{2}$

Question 14: If (x + $\frac{1}{x})^2 = 3$, then find the value of $x^{84} + x^{72} + x^{36} + x^{24} + x^6 + 1$.

a) 1

b) 4

c) 5

d) 6

14) Answer (B)

Solution:

According to the question,
(x + $\frac{1}{x})^2 = 3$
x + $\frac{1}{x} = \sqrt{3}$
Now, cubing both side we get,
(x + $\frac{1}{x})^3 = 3\sqrt{3}$
$x^3 + \frac{1}{x^3} + 3(x + \frac{1}{x}) = 3 \sqrt{3}$
$x^3 + \frac{1}{x^3} + 3(\sqrt{3}) = 3 \sqrt{3}$
$x^3 + \frac{1}{x^3} = 0$
Or, $x^6 + 1 = 0$, $x^6 = -1$

$x^{84} + x^{72} + x^{36} + x^{24} + x^6 + 1$
${x^6}^{14} + {x^6}^{12} + {x^6}^{6} + {x^6}^4+ x^6 + 1$
1 + 1 + 1 + 1+ 0 = 4

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