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HTML Code Creator MOCK TEST - 4 - 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)
Total number of Questions: 71
Total number of MCQ's Questions: 52
Total Number of Non-MCQ's (TITA) Questions: 19
Total number Quantitative Ability (QA) Questions: 24
Total number Quantitative Ability (QA) MCQ's Questions: 15
Total number Quantitative Ability (QA) non-MCQ's (TITA) Questions: 9
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: 25
Total number Verbal Ability & Reading Comprehension (VARC) MCQ's Questions: 20
Total number Verbal Ability & Reading Comprehension (VARC) non-MCQ's (TITA) Questions: 05
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.
While calculating the average weight of 42 students, teacher took 24 as divisor. Which gives the average as 49 kg. Find the correct average.
Options:
Option (a): 28 kg
Option (b): 27 kg
Option (c): 29 kg
Option (d): 30 kg
24 x 49 = 1176
1176 / 42 = 28
Option (a): 28 kg
Roof of a Godown makes an angle 45° with the horizontal, Front wall is of height 16m and the back wall is of height 24m, then the length of the roof is
Options:
Option (a): approx. 5.66 m.
Option (b): approx. 11.31 m.
Option (c): approx. 6.66 m
Option (d): approx. 4.66 m
length of the wire = sin 45° x (24 - 16)
= 0.707 x 8 = 5.66
Option (a): approx. 5.66 m.
If the H.C.F. of two numbers is 24 and factors of their L.C.M. are 17 and 18. The larger of the two numbers is:
Options:
Option (a): 432.
Option (b): 402.
Option (c): 430.
Option (d): 400.
Largest is 24 x 18 = 432
Option (a): 432.
A man spends ₹ 200 in January 2022 to purchase lottery tickets and wins ₹ 1,000, encouraged by this he starts purchasing lottery tickets worth ₹ 50 more than the previous month for the entire year. What is his annual spending on lottery tickets for the year 2022?
Options:
Option (a): 5750
Option (b): 5700
Option (c): 5650
Option (d): 5500
AP =(n/2)(2a+(n-1)d)
n = 12
a = 200
d = 50
AP = (12/2)((2 x 200) + (12 - 1)50)
= 5700
Option (b): 5700
The Oxford University Press compiled a 2200 page dictionary on the computers. Just before the dictionary went for printing, it was discovered that there was no page numbers. How many times should a typist press keys from 0 to 9 on the keyboard so as to number the dictionary from 1 to 2200?
single digit numbers: 1 to 9: 9 times
double digit numbers 10 to 99: 90 x 2 = 180 times
three digit numbers 100 to 999: 900 x 3 = 2700 times
four digit numbers 1000 to 2200: 1201 x 4 = 4804 times
total = 9 + 180 + 2700 + 4804 = 7693
If x is the first term, y the nth term and p be the product of n terms of a Geometric Progression, then the value of p⁴ will be
Options:
Option a: (xy)ⁿ ⁻ ¹
Option b: (xy)²ⁿ
Option c: (xy)¹ ⁻ ⁿ
Option d: None of the above.
Option b: (xy)²ⁿ
A man goes to a computer hardware shop with money just enough to buy a Blu-Ray player for his PC. He gets an unexpected discount of 10% on Blu-Ray player which enables him to purchase USB-External Hard-Disk worth ₹ 2,000.00/- in addition to the Blu-Ray player. What price does the man pay for the Blu-Ray player?
10% of Blu-Ray player is ₹ 2,000.00/-.
So he has to pay ((2000 x 100)/10)
= ₹ 20,000.00/-
The weight of a student is directly proportional to the height, considering this average age of the students in a class of 75 is 16. A 173 cm tall student has a weight of 60 kg. The average weight of the class is:
Options:
Option a: 60 kg
Option b: 64 kg
Option c: 66 kg
Option d: Data insufficient.
Option b: 64 kg
Weight is directly proportional to height and height is directly proportional to age.
A 173 cm tall student has a weight of 64 kg will be of 16 years.
There are two clocks. One of them loses 2 minutes in 12 hours and another one gains 6 minutes in 36 hours. Both are set right at 12 a.m. on January 1st 2000 (New Year). What is the next occurrence of both showing correct time
Options:
Option a: 12 a.m., March 31st 2000
Option b: 12 a.m., March 30th 2000
Option c: 12 noon, March 30th 2000
Option d: 12 noon, March 30th 2000
Option a: 12 a.m., March 31st 2000
One gains 6 minutes in 36 hours. That means it gains 2 minutes every 12 hours.
Another loses 2 minutes in 12 hours, That is, difference = 4 minutes every 12 hours.
They will be on the same time when the difference between them is 12 hours.
1 minute = 12/4 = 3 hours
Therefore, 3 x 12 x 60 = 2160 or 90 days.
=> January 31 days + February 29 days + March 30 days = total 90 days.
=> 12 a.m., March 31st 2000
If √(x + 3) + √(x + 7) = 10, find the value of [ x - ⁷⁹/₄ ].
√(x + 3) +√(x + 7) = 10
Squaring both sides, we get
x + 3 + x + 7 + 2(√(x + 3))(√(x + 7)) = 100
2x + 10 + 2(√(x + 3))(√(x + 7)) = 100
2(x + 5 + (√(x + 3))(√(x + 7))) = 100
x - 45 = - (√(x + 3))(√(x + 7))
Squaring both sides again, and solving we get
x =¹⁶⁷/₄
Thus,¹⁶⁷/₄ - ⁷⁹/₄ = 22.
Sum of all the real roots of the equation |x - 2|² + |x - 2| - 2 = 0 is
Options:
Option a: -4
Option b: 0
Option c: 4
Option d: 2
Taking (x - 2) > 0 we get x = 0, 3. So, x = 3
Taking (x - 2) < 0 we get x =1, 4. So, x = 1
Hence, sum of roots = 4.
Person A Speaks truth in 90% of the cases and Person B in 65% of the cases. In what percentage of cases are they likely to contradict each other in stating the same fact
Options:
Option
a: 42%
Option b: 0
Option c: 40%
Option d: 38%
Contradiction can occur if A speaks truth and B lies, B speaks truth and A lies.
So, the required probability is
= 0.9 x 0.35 + 0.65 x 0.1
= 0.38 or 38%
Kumar started a business with ₹ 25,000.00/- and after 6 months Satish joined him with ₹ 80,000.00/-. Kumar received ₹ 65,801.00/- including 10% of profits as commission for managing the business. What amount did Satish received?
Ratio of shares of profits is
(25,000 x 12) : (80,000 x 6) = 5 : 8
Let the total profit be P, as Kumar receives 10% of this as commission, the remaining 90% of P is shared in the ratio of 5 : 8.
Hence Kumar's receipts will be ⁵/₁₃ of 90% of total profit plus commission.
0.1 P + ⁵/₁₃(0.9 P) = 65,801
Upon solving this, we get P = ₹ 1,47,485.00/-,
Thus, Satish receives 147485 - 65801 = ₹ 81,684.00/-
In ▲PQR, PQ = 53, and PR = 70. A circle with center P and radius PQ intersect QR at points R and X. Moreover RX and QX have integer lengths. What is QR?
Options:
Option a: 11
Option b: 41
Option c: 61
Option d: 81
Option b: 41
Let x represent RX, and let y represent QX. QR = x + y.
Since the circle goes through Q and X, PQ = PX = 53.
Stewart's theorem yields a relation between the side lengths and a cevian length of a triangle.
Now use Stewart's Theorem.
xy(x + y) + 53² (x + y) = 70²y + 53²x
x²y + xy² +53²x + 53²y = 70²y + 53²x
x² + xy + 53² = 70²
(Since y cannot be equal to 0, dividing both sides of the equation by y is allowed.)
Or, x(x + y) = (70 + 53) (70 - 53)
Or, x(x + y) = 2091
The prime factors of 2091 are, 3 , 17, 41. Obviously, x < x + y.
Therefore, x is 17 and x + y is 41
If the number of square units in the area of a square is exactly half the number of units in its perimeter, then the diagonal of the square is equal to
Options:
Option a: 4
Option b: 2√2
Option c: 3√2
Option d: √2
Option b: 2√2
Let side of square = a,
then perimeter = 4a,
Area of the square = a²
According to the question area of a square is equal half its perimeter,
2 x a² = 4 x a
=> a = 2
Diagonal of a square = √2 x a
= 2√2
A small right angled triangle is obtained when piece of cardboard in the shape of a right angled triangle is cut along a line that is parallel to the hypothenuse with a 25% reduction in the length of the hypothenuse of the triangle. If the area of the smaller triangle is 40 sq units, find the area of the original triangle.
Options:
Option a: 71.11 sq units
Option b: 72.21 sq units
Option c: 73.13 sq units
Option d: 74.14 sq units
Let the length hypothenuse of the smaller triangle be 75 units, Thus bigger triangle has a length of 100 units.
Area of the original triangle = (40 x 100 x100) / ( 75 x 75 ) = 71.11 sq units
A set of natural numbers Sn is partitioned into subsets as: S₁={1}, S₂={2,3}, S₃={4,5,6}, S₄={7,8,9,10}, likewise:
Find the sum of the subset S₄₀.
Set S₁ ends with the sum of all natural numbers upto 1, Set S₂ ends with the sum of all natural numbers upto 2 and so on. Thus set S₃₉ would end with the sum of all natural numbers upto 39. That is, 1+2+3+4+.....+39 = 780. There for sum of all natural numbers in the set et S₃₉ will be the sum between 781 and 820,
That is: 781 + 782 + 783 + ... + 820
= 40/2 [(2 x 781) + ((40-1) x 1)] = 32020
formula used (s / 2)[(a x 2)+((n - 1)d)],
Where,
s: Number of elements in the series,
d: Difference between two consecutive numbers,
a: first number in the series.
A series in which any term is the sum of the preceding two terms is called a Fibonacci series. The first two terms are given initially and together they determine the entire series. If the sum of the squares of the ninth and the eighth terms of a Fibonacci series is 1597 then what is the 14th term of that series?
Options:
Option a: 233
Option b: 610
Option c: 377
Option d: 144
Option c: 377
Fibonacci Series:
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377.......
Find the number of even factors for the number 2700.
2700 = 27 x 100
= 3 x 3 x 3 x 2 x 2 x 5 x 5
= 2² x 3³ x 5²
= 2 x (3 + 1) x (2 + 1) = 24
Find the number of non-negative integer solutions to the system of equations
a + b + c + d + e = 25
and a + b + c = 10 is
Options:
Option a: 1022
Option b: 1056
Option c: 1046
Option d: 1066
Given a + b + c + d + e = 25 ---- (1)
a + b + c = 10 ------ (2)
Therefore d + e = 15 ------ (3)
Number of non-negative integral solutions of equation (2) is
=³⁺¹⁰⁻¹C₁₀ = ¹²C₁₀
Number of non-negative integral solutions of equation (3) is
= ²⁺¹⁵⁻¹C₁₅ = ¹⁶C₁₅
Therefore, Required number = ¹²C₁₀ x ¹⁶C₁₅ = 1056
There is a 50 liters solution of milk and water in which milk forms 60%. How much milk should be added in order for this solution to contain 70% milk?
We have 50 liters of water and milk mixture
60% of it is milk.
Then the amount of milk in 50 liters is 50*.6 = 30 liters .
Let x liters of milk is added to it.
Hence the amount of mixture is (50 + x) liters.
And the amount of milk in it is (30 + x) liters .
Hence, (30 + x)/(50 + x) = .7
Solving this x = approx. 16.67
Thus, approx. 16.67 liters of milk should be added to the solution.
Directions for questions 22 and 23: Read the following and solve the questions based on it
In the entrance examination of IIMs, there were 200 questions, each carried the same marks. The correct answers get 2 marks and there is 100% negative marks. A total of 70 candidates took the exam and it was found that the average of marks obtained by these 70 candidates was 240. The candidates were not required to attempt all the questions. None of the candidates got more number of incorrect answers that correct answers.
22. Which of the following is always true?
Options:
Option a: None of the applicants had a net score less than 240.
Option b: The net scores of the individuals would always be an even integer.
Option c: The net scores of the individuals would always be a positive integer.
Option d: The marks of the topper could be exactly 24% more than the average of the entire group.
Right answer = +2
Wrong answer = -2
it means student can only get even marks.
The net scores of the individuals would always be an even integer.
Directions for questions 22 and 23: Read the following and solve the questions based on it
In the entrance examination of IIMs, there were 200 questions, each carried the same marks. The correct answers get 2 marks and there is 100% negative marks. A total of 70 candidates took the exam and it was found that the average of marks obtained by these 70 candidates was 240. The candidates were not required to attempt all the questions. None of the candidates got more number of incorrect answers that correct answers.
23. If the topper had a net score of 390 and the eleventh ranker had a net score of 360 and the net score of the top eleven rankers were distinct, then what would be the average net score of the 59 students who got ranks from the fifth to the last?
Topper got 390 and eleventh ranker got 360. So Second to tenth rankers got 387, 384, 381, 378, 375, 372, 369, 366, 363.
Now, let x be the average of last 59 students.
Then,
390 + 387 + 384 + 381 + 378 + 375 + 372 + 369 + 366 + 363 + 360 + 59x .
x = 70 x 240
4125 + 59x = 16800 => 59x = 12675 => x = 214.8305
David and Rahim are friends whose income are in the ratio of 3 : 5 and their expenditures in the ratio of 5 : 7. Calculate 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
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