CAT Practice Test - 3 
(Quantitative Ability )

  MOCK TEST - 3 - Quantitative Ability (QA Section)

  Quantitative Ability is one of the section of CAT exam along with Data Interpretation and Logical Reasoning and Verbal Ability and Reading comprehension. Practice tests or Mock tests help aspirants to prepare for the exams.

   Quantitative Ability is the ability of a person to make empirical enquiries through numerical data gathering and analysis by performing mathematical or statistical computations.

   Maximum Time for this section: 40 : 00 (Minutes : Seconds)

CAT Practice Test - 3 (Quantitative Ability )

Tips for practice test:
1. Attend all the  questions.
2. Take a Maximum time of 40 : 00 (Minutes : Seconds).
3. Complete two cycles, first go of the entire section in 29 : 00 (Minutes : Seconds) by solving at least 16 to 18 questions completely. Second go of the entire section in 11 : 00 (Minutes : Seconds) by attempting last 8 to 6 questions.

Details:

Total number of Questions: 66
Total number of MCQ's Questions: 50
Total Number of Non-MCQ's (TITA) Questions: 22

Total number Quantitative Ability (QA) Questions: 24
Total number Quantitative Ability (QA) MCQ's Questions: 16
Total number Quantitative Ability (QA) non-MCQ's (TITA) Questions: 08

Total number Data Interpretation and Logical Reasoning (DILR) Questions: 22
Total number Data Interpretation and Logical Reasoning (DILR) MCQ's Questions: 17
Total number Data Interpretation and Logical Reasoning (DILR) non-MCQ's (TITA) Questions: 05

Total number Verbal Ability & Reading Comprehension (VARC) Questions: 26
Total number Verbal Ability & Reading Comprehension (VARC) MCQ's Questions: 17
Total number Verbal Ability & Reading Comprehension (VARC) non-MCQ's (TITA) Questions: 09

Maximum Time: 40 : 00 (Minutes : Seconds) per section

Note-1: Use of simple calculator is allowed.
Note-2: Not allowed to change section once selected.

Quantitative Ability (QA)

Question No. 01: Not an MCQ (No negative marking)

 The angle (in degreesº) covered by minute hand in 1 hour and 22 minutes is ______.

The angle turned by minute hand in 60 minutes is 360º.
The angle turned in 1 hour and 22 minutes is
= 360º + (360º x 22/60)
= 360º + 132º= 492º

Question No. 02: MCQ (with negative marking)

 Find the percentage change required in the constant term of a quadratic equation ax² + bx + c = 0 in order to make another quadratic equation from this with a = 1, one root of the second quadratic equation is same as first equation but opposite in sign and the other root of the second equation is three times the second root of the first equation. 

Options:
Option (a): 200% increase
Option (b): 300% increase
Option (c): 400% decrease
Option (d): 100% increase

Let the equation be
x² - 2x + 1 = 0 ---- (1)

Roots for the above quadratic equation are 1, 1
and
x² - x - 2 = 0 ---- (2)

Roots for the above quadratic equation are -2, 3
Roots of both the equations satisfy the given conditions.
Hence, required percentage

400% decrease (= ((1 - (-3)) / 1) x 100 = 400%)

Question No. 03: MCQ (with negative marking)

 Directions for the Questions 03 and 04: These questions are based on the following information let x and y are real number, such that

  f(x, y) = |x + y|,
  F(f(x, y )) = - f(x, y) 
 and G(f(x, y)) = - F(f(x, y ))

Find the value of (6.G(f(7, 3)) + F(f(3, 2))) / (2.F(f(3, 4)) + f(1 , 2))

Options:
Option (a): 5
Option (b): -5
Option (c): 0
Option (d): -4

(2.G(f(1, 2)) + F(f(3, 4))) / (f(3 , 4) + F(f(2, 1)))
Consider 6.G(f(7, 3)) + F(f(3, 2))) = 60 - 5 = 55
Now consider (2.F(f(3, 4)) + f(1 , 2) = - 14 + 3 = - 11
55/-11 = -5
Option (b): -5

Question No. 04: MCQ (with negative marking)

 Directions for the Questions 03 and 04: Read the following information and answer the questions.
let a and b are real number, such that

    f(a, b) = |a + b|,  
    F(f(a, b)) = -f(a, b),
    G(f(a, b)) = - F(f(a, b ))

4. Which of the following expression gives -log₂ 16 . a² as a result?

Options:
Option (a): F (f(a, a)) G(f(a, a)) . 4
Option (b): - F (f(a, a)) G(f(a, a))
Option (c): F (f(a, a)) G(f(a, a))
Option (d): - F (f(a, a)) G(f(a, a)) / 4

F(f(a,-a)) = 0 and G(f(a,-a)) = 0
F(f(a,a)) = -2a and G(f(a,a)) = 2a
F(f(a,a)).G(f(a,a)) = -4a²
log₂16 = 4
F (f(a, a)) G(f(a, a)) = -log₂16 . a²

Option (c): F (f(a, a)) G(f(a, a))

Question No. 05: Not an MCQ (No negative marking)

 If the arithmetic mean for 20, 30, 35, 42, P, 56 is 40. find P.

arithmetic mean is
= (20 + 30 + 35 + 42 + P +56) / 6 = 40
183 + P = 240
P = 240 - 183 = 57

Question No. 06: MCQ (with negative marking)

 If p, q, and r are integers such that p² = rq² Then it follows

Options:
Option (a): p is an even number
Option (b): if r divides p, then r is a perfect square.
Option (c): r divides p
Option (d): q² divides p

if r divides p, then r is a perfect square.

Question No. 07: Not an MCQ (No negative marking)

 Find f((22)₄,(7)₈,(15)₁₀). If f(x, y, z) = xy + yz + zx, where x, y, and z are in decimal numbers.

(22)₄ = 4¹ x 2 + 4° x 2 = 4 x 2 + 1 x 2 = 8 + 2 = (10)₁₀
(7)₈ = (7)₁₀
(15)₁₀ = (15)₁₀
Now the numbers are in common base of 10. f(x, y, z) = xy + yz + zx
= f(10, 7, 15) = (10 x 7) + (7 x 15) + (10 x 15)
= 70 + 105 + 150 = 325

Question No. 08: MCQ (with negative marking)

 A publisher prints 5,000 copies of 'Fortune Teller' at a cost of ₹ 3,000.00/-. He gave 500 copies free to different philanthropic institutions. He allowed a discount of 25% on the published price and gave one copy free for every 25 copies bought at a time. He was able to sell all the copies in this manner. If the published price is ₹ 4.25/-, then what is his overall gain or loss percentage in the whole transaction?

Options:
Option (a): 370%
Option (b): 350%
Option (c): 357%
Option (d): 344%

Cost = ₹ 3,000.00/-
Published Price ₹ 4.25/-
SP = 75/100 x 4.25 = ₹ 3.1875/-
Number of free copies = (5000/25) = 200 + 500 = 700
SP = 4300 x 3.1875 = ₹13,706.25
Percentage Gain = approx 357%

Question No. 09: MCQ (with negative marking)

 If the average salary of 17 employees in a company is A. The salary of the employees when arranged in either an ascending or a descending order was found to be in arithmetic progression. Five employees with 3rd, 7th, 9th, 11th, 15th highest salary left the job. The average of the remainder of the class was B. Then

Options:
Option (a): A = B
Option (b): A > B
Option (c): A < B
Option (d): Data Insufficient

Option (a): A = B
Given: 17 employees.
Its been told that if the salary of the employees is arranged in either an ascending or a descending order was found to be in arithmetic progression.
3rd highest is 2 places away from first, so is 15th from last, Thus surplus generated by the 3rd will be same as deficit incurred due to the 15th. 
Similarly, 7th is 6 places away from first, so is 11th from last, Thus surplus generated by the 7th will be same as deficit incurred due to the 11th.
and 9th which is the middle from first and last. All of the above left the job. Hence A =B.

Question No. 10: MCQ (with negative marking)

 2 x Infinite sum of the series 1/2 + 1/4 + 1/6 + 1/8 ... = ?

Options:
Option (a): approx. 0.57722
Option (b): 1/2
Option (c): 2
Option (d): 2.577

Option (a): approx. 0.57722
We know that, Infinite sum of the series 1/2 + 1/4 + 1/6 + 1/8 ... = γ/2 
where γ≈0.57721566490153286… is the Euler-Mascheroni constant.
thus 2 x γ/2 = γ or γ≈0.57721566490153286 = approx. 0.57722

Question No. 11: Not an MCQ (No negative marking)

 If 10 coins are tossed together, what is the probability of getting exactly 2 tails.

The event of getting exactly 2 tails will be the combination of 2 tails and 8 heads.
The number of arrangements with these combinations is 10!/(2! x 8!) = 3628800/(2x40320) = 45
Therefore the probability is 45/2¹⁰ = 45/1024 = 0.044

Question No. 12: MCQ (with negative marking)

 P works twice as fast as Q, whereas P and Q together can work three times as fast as R. If P, Q and R work together on a job, in what ratio should they share the earning?

Options:
Option (a): 2 : 1 : 1
Option (b): 4 : 2 : 1
Option (c): 4 : 3 : 2
Option (d): 4 : 2 : 3

Ratio of efficiency, P = 2Q --- (1)
and P + Q = 3R --- (2),
From (1) and (2), Q = R
P = 2Q = 2R
Hence, P : Q : R = 2 : 1 : 1

Question No. 13: MCQ (with negative marking)

 Three cockroaches J, M and R decided to take a race from one corner of the room to the diagonally opposite corner of the room. J can fly, M can walk and R can walk only along the edges. If all of them rack the destination at the same time by taking the smallest paths, what is the ratio of the speeds of J, M and R. the room is cube of 2m³?

Options:
Option (a): √3 + 1 : 3 : √3
Option (b): √3 : 5 : 3
Option (c): √3 : √2 : 3
Option (d): √6 : √(2(1 + √2)) : 6

Room is 2 x 2 x 2 m³
Now diagonal of the cube = √6 m
Distance covered by J = √6 m
Distance covered by M = √(2 + 2√2) = √(2(1 + √2)) m
Distance covered by R = 2 + 2 + 2 = 6 m
Ratio =√6 : √(2(1 +√2)) : 6

Question No. 14: Not an MCQ (No negative marking)

 A wall of measurements 20m x 2m x 4m is to be constructed with bricks of dimensions 10 cm x 10 cm x 8 cm. Find the number of bricks required to construct the wall.

Volume of the wall = 2000 x 200 x 400 cc
Volume of the brick = 10 x 10 x 8 cc
Number of bricks = (2000 x 200 x 400) / (10 x 10 x 8)
= 2,00,000.

Question No. 15: MCQ (with negative marking)

If a, b, c and d are non - zero real numbers such that

(a² + b² + c²)(b² + c² + d²) ≤ (ab + bc + cd)², 

If a, b, c are in Arithmetic Progression then which of the following is true

Options:
Option (a): 2b = a + c.
Option (b): b² = ac
Option (c):  b = 2ac/a + c
Option (d): Insufficient Data.

Option (b): b² = ac

Question No. 16: MCQ (with negative marking)

Directions for questions 16 and 17: Read the passage below and solve the equations based on it.

   [x] = Greatest integer less than or equal to x.

  {x} = Smallest integer more than or equal to x.

Given that 'x' is not an integer, then what is the value of [x] - {x}?

Options:
Option (a): -1
Option (b): 1
Option (c): 0
Option (d): Depends upon the value of x.

-1

Question No. 17: MCQ (with negative marking)

Directions for questions 16 and 17: Read the passage below and solve the equations based on it.

   [x] = Greatest integer less than or equal to x.

   {x} = Smallest integer more than or equal to x.

Given that 'x' is not an integer, then [x] + {x} is?

Options:
Option (a): An even integer.
Option (b): An odd integer.
Option (c): Positive number.
Option (d): Non-Positive number.

An odd integer.

Question No. 18: Not an MCQ (No negative marking)

A man can row 7 km in an hour in a lake. But in a river the speed of water current is 3 km/hr and takes 4 hours for him to go to a new place and return, find the distance from the starting point to the new place.

Let the distance be x km.
Upstream speed = 7 - 3 = 4 kmph.
Downstream speed = 7 + 3 = 10 kmph.
Total time = x/4 + x/10 = 4 hours.
x/4 + x/10 = 14x/40 = 4
=> 14x = 160 => x = 80/7 = 11³/₇ or approx 11.43

Question No. 19: MCQ (with negative marking)

The probability that students pass in Maths, Science and Social Studies are m, c and s respectively. Of these subjects, a student has a 50% chance of passing in at least one, a 50% chance of passing in at least two, and a 40% chance of passing in exactly two subjects. Which of the following relations are true?

Options:
Option (a): m + c + s =¹⁹/₂₀
Option (b): m + c + s =¹¹/₁₀
Option (c): mcs = ¹/₂₀
Option (d): mcs = ¹/₄

Option (b): m + c + s =¹¹/₁₀
P(m ∪ c ∪ s) = 1/2
P(m ∩ c) + P(c ∩ s) + P(m ∩ s) − 2c(m ∩ c ∩ s)=1/2
P(m ∩ c) + P(c ∩ s) + P(m ∩ s) − 3c(m ∩ c ∩ s)= 1/3
⇒ m + c + s − ms − mc −cs + mcs = 1/2 ------ (i)
⇒ ms + cs + cm − 2mcs = 1/2 ------ (ii)
⇒ ms + cs + cm − 3mcs = 2/5 ------ (iii)
mcs = 1/2 - 2/5 = 1 / 10 from (ii) and (iii)
m + c + s = 1/2 + 1/2 + 1/10 = ¹¹/₁₀

Question No. 20: MCQ (with negative marking)

A ball with diameter 15 cm fell in to a hole filled with water in such a way that the top of the ball is 5 cm above the ground level. What is the circumference in centimeters of the circle formed by the contact of the water surface with the ball?
 
Options:
Option (a): 10√2π
Option (b): 50 π
Option (c): 5√2π
Option (d): 10 π

10√2π

Question No. 21: Not an MCQ (No negative marking)

In a solution of 50 liters of milk and honey in which milk forms 66%. How much honey must be added to this solution to make it a solution in which milk forms 60%?

We will apply allegation rule here taking 66% concentration solution (of milk and water) mixed with a 0% concentration solution (pure water to give) 60% concentration solution.
Applying alligation rule, we get the ratio of initial solution and water as
60 : 6 = 10 : 1
Since there were 50 lts of the solution initially, water to be added is
50/10 = 5 lts

Question No. 22: MCQ (with negative marking)

Directions for questions 22 and 23: Read the following and solve these questions based on it:

People purchased shares of brands A, B, and C from stock market for $ 1,000

a. People who purchased brand A's share had a loss of $ 658
b. People who purchased brand A's and B's shares had a loss of $ 166
c. People who purchased brand B's share had a loss of $ 372
d. People who purchased brand B's and C's shares had a loss of $ 434
e. People who purchased brand C's share had a loss of $ 590
f.  People who purchased brand C's and A's shares had a loss of $ 126

Find the loss (assuming no profit) suffered by people who purchased shares of

22. brand C but not brand A

Options:
Option (a): 312
Option (b): 226
Option (c): 266
Option (d): 206

Option (d): 206
Apply the formula:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) +  n(A ∩ B ∩ C)
n(A ∪ B ∪ C) = a + b + c + d + e + f + g = 1000
n(A) = a + b + d + e = 658
n(B) = b + d + c + f = 372
n(C) = d + e + f + g = 590
n(A ∩ B) = b + d = 166
n(B ∩ C) = d + e = 434
n(C ∩ A) = d + f = 126
1000 = 658 + 372 + 590 - 166 - 434 - 162 +  n(A ∩ B ∩ C)
n(A ∩ B ∩ C) = 106

Mobirise Website Builder

Venn - diagram

From the equations above:
b + c + 126 = 372 => b + c = 246,
434 + f + g = 590 => f + g = 156,
a= 1000 - 246 - 434 - 156 = 164
b = 658 - 434 - 164 = 60
c = 186
d = 106
e = 38
f = 372 - 166 - 186 = 20
g = 136
From Venn-Diagram,
the loss (assuming no profit) suffered by people who purchased shares of brand C but not brand A 
= 186 + 20 = 206

Question No. 23: MCQ (with negative marking)

Directions for questions 22 and 23: Read the following and solve these questions based on it:

People purchased shares of brands A, B, and C from stock market for $ 1,000

a. People who purchased brand A's share had a loss of $ 658
b. People who purchased brand A's and B's shares had a loss of $ 166
c. People who purchased brand B's share had a loss of $ 372
d. People who purchased brand B's and C's shares had a loss of $ 434
e. People who purchased brand C's share had a loss of $ 590
f. People who purchased brand C's and A's shares had a loss of $ 126

Find the total loss (assuming no profit) suffered by people who purchased shares of

23. brands A and B not brand C.

Options:
Option (a): 560
Option (b): 710
Option (c): 620
Option (d): 338

Option (d): 338
Apply the formula:
n(A ∪ B ∪ C) = n(A) + n(B) + n(C) - n(A ∩ B) - n(B ∩ C) - n(C ∩ A) + n(A ∩ B ∩ C)
n(A ∪ B ∪ C) = a + b + c + d + e + f + g = 1000
n(A) = a + b + d + e = 658
n(B) = b + d + c + f = 372
n(C) = d + e + f + g = 590
n(A ∩ B) = b + d = 166
n(B ∩ C) = d + e = 434
n(C ∩ A) = d + f = 126
1000 = 658 + 372 + 590 - 166 - 434 - 162 + n(A ∩ B ∩ C)
n(A ∩ B ∩ C) = 106

Mobirise Website Builder

Venn - diagram

From the equations above:
b + c + 126 = 372 => b + c = 246,
434 + f + g = 590 => f + g = 156,
a= 1000 - 246 - 434 - 156 = 164
b = 658 - 434 - 164 = 60
c = 186
d = 106
e = 38
f = 372 - 166 - 186 = 20
g = 136
From Venn-Diagram,
the loss (assuming no profit) suffered by people who purchased shares of brands A or B but not brand C
= a + e + g = 164 + 38 + 136 = 338

Question No. 24: Not an MCQ (No negative marking)

If the income of David and Rahim are in the ratio of 3 : 5 and their expenditures in the ratio of 5 : 7. Find the ratio of their savings, given that Rahim saves one fourth of his income.

We take appropriate constants of proportionality in the two cases.
Income of David is 3x
Income of Rahim is 5x
Expenditure of David is 5y
Expenditure of Rahim is 7y
Savings of David = 3x - 5y
Savings of Rahim = 5x - 7y
We have been given that Rahim saves a fourth of his income.
5x - 7y = ¹/₄(5x)
=> 15x = 28y => x/y = 28/15
=> x = 28K and y = 15K
=> (3x - 5y)/(5x - 7y)
=> ((3 x 28) - (5 x 15))/((5 x 28) - (7 x 15))
=> 9k / 35k
=> Therefore, Ratio of their savings is 9 : 35

CAT Practice Test - 3: Data Interpretation & Logical Reasoning

CAT Practice Test - 3: Verbal Ability & Reading Comprehension

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