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Test: CAT Quantitative Aptitude- 4 - CAT MCQ


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22 Questions MCQ Test Quantitative Aptitude (Quant) - Test: CAT Quantitative Aptitude- 4

Test: CAT Quantitative Aptitude- 4 for CAT 2024 is part of Quantitative Aptitude (Quant) preparation. The Test: CAT Quantitative Aptitude- 4 questions and answers have been prepared according to the CAT exam syllabus.The Test: CAT Quantitative Aptitude- 4 MCQs are made for CAT 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: CAT Quantitative Aptitude- 4 below.
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Test: CAT Quantitative Aptitude- 4 - Question 1

If the sum of the roots of the quadratic equations mx2 + (2m - 1)x + 4 and (3m + 1)x2 - 6x + (2m - 3) are equal, then find the sum of values of m. 

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 1

The sum of the roots of a quadratic equation ax2 + bx + c is - b / a
⇒  -(2m - 1)/m = 6/(3m + 1)
⇒ 6m2 - m -1 = -6m
⇒ 6m2 + 5m -1 = 0
⇒ m = 1/6 or -1
Thus, the sum of the values of m: 1/6 -1 = -5/6

Test: CAT Quantitative Aptitude- 4 - Question 2

A right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm. find the volume of the solid so formed.

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 2


When a right triangle with sides 6 cm, 8 cm and 10 cm is revolved about the side 8 cm, then solid formed is a cone whose height, h = 8 cm.
The radius of the cone, r = 6 cm.
Slant height of the cone, l = 10 cm.
∴ Volume of the cone

Hence, the volume and surface area of the cone are 301.7 cm3.

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*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 4 - Question 3

The largest integer n, such that n + 10 divides n3 + 100, is


Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 3

As, we know that :
n3 + 1000 = (n + 10) (n2 - 10n + 100)
So,
n3 + 100 = (n + 10) (n2 - 10n + 100) - 900
Therefore, 900 must be divided by (n + 10)
Since largest divisor of 900 is 900, So n + 10 is divisible by 900 i.e. largest value of n is 890.

Test: CAT Quantitative Aptitude- 4 - Question 4

How many subsets of S = (1, 2, 3 , . . . , 400) have the product of their elements an even number?

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 4

For a given set containing N elements, the number of non-zero subsets = 2N - 1
Number of non-zero subsets in S = 2400 - 1

The product of the elements of a subset will be odd when all the elements of the subset are odd.

The odd elements in S are Sodd = (1, 3, 5 , ..., 399)
Number of non-zero subsets in Sodd = 2200-1

Number of subsets where product of its elements is even = (2400 - 1) -(2200-1) = 2400 - 2200

Hence, option 1.

Test: CAT Quantitative Aptitude- 4 - Question 5

Find the maximum area of the isosceles trapezium (in units2) whose unequal sides are 4 units and 6 units. Given that it is inscribed in a circle whose radius is 2√3

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 5

The maximum area of trapezium will be obtained when the two sides are on the opposite sides of the diameter.

The height of the trapezium = OE + OF
OE = √OC2 - √EC2 = √3
OF = √OB2 - √FB2 = 2√2

∴ the area of the trapezium = (1 / 2) x (2√2 + √3) x (6 + 4) = 5(2√2 + √3)

Hence, option 2.

Test: CAT Quantitative Aptitude- 4 - Question 6

The average speed of Mumbai metro train which travels from Ghatkopar to Versova is 30 km/hr and the distance between the two stations is 32 km. Which function represents the distance 'd(t)' remaining in a trip from Ghatkopar to Versova after a certain time 't' from the start? 

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 6

We know that,
Distance remaining = Total distance - Distance travelled
⇒ d(t) = 32 - 30t

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 4 - Question 7

Find the value of M in the figure given below,

 


Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 7

⇒ 3M + 9 + M + 5 + 140= 180
⇒ 4M = 26
⇒ M = 6.5°

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 4 - Question 8

6 equally spaced girls are standing along the circumference of a circle of radius 10 m and facing the center. What is the shortest distance between any two girls facing each other?


Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 8

If we draw a figure according to the information provided in the question then we can note that:
In the question it is good that each girl is facing the center of the circle.

Radius of circle =10m, Diameter of circle =20m

We need to find the shortest distance between any two girls faces each other.

We can see from the image that oppositely standing girls will only face each other so the shortest distance between any two girls facing each other will be equal to the diameter of the circle ie 20 m.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 4 - Question 9

Akhsay travels on a straight route joining P and Q in a way such that after each of the 20th, 40th and 60th minute from his starting point P, he increases his speed by 10 kmph and after each of the 70th, 80th, 90th, 100th and the 110th minute from the start, he decreases his speed by 10 kmph. If Akshay began his journey at a speed of 50 kmph and took exactly 2 hours to reach Q, the distance from P to Q (in km) is _______.


Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 9

The distances travelled for various time intervals are:


Hence, total distance in 2 hour = 
= 60 + 55 = 115 km

Test: CAT Quantitative Aptitude- 4 - Question 10

A shopkeeper gives a discount of 10% on the marked price and sells it for Rs. 540 to earn a profit of Rs. 90. Find the profit percentage if the product is marked Rs. 200 above its cost price for the same absolute value of discount.

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 10

Case 1:
Let the marked price be m.
The selling price = 0.9m
⇒ 0.9m = 540 
⇒ m = 600
Cost price = 540 -90 = Rs. 450

Case 2:
Discount = Rs. 60, Marked price = Rs. 650
Selling price = 650 - 60 = Rs. 590
Profit = 590 - 450 = Rs. 140
Profit percentage = (140/450) x 100 = 31.11%

Test: CAT Quantitative Aptitude- 4 - Question 11

A thief is running on a circular track of radius 5 m at 1.5 m / s . A policeman whose speed is twice that of the thief arrived at the starting point of the track 4 seconds after the thief. The policeman can either run along the circular track in the direction of the thief or go to the centre and then go to any point on the circle from the centre. Find the minimum distance covered by the thief after the policeman starts chasing him.

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 11

When the policeman arrives, the thief has traveled 6 meters.
So, the time taken by the policeman to catch the thief along the circular track = 6/1.5 = 4 seconds.

The total distance covered by the policeman = 12 meters.
Thus, the minimum distance is when the policeman takes the route via the center i.e. he travels a length of 2 x 5 = 10 meters.

The minimum time taken = 10/3 seconds.

The minimum distance covered by the thief = 1.5 x (10/3) = 5 meters

Hence, option 3.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 4 - Question 12

If x = 44 x 56 x 65 x 710 x 88, then how many factors of x are of the form 6 A2?


Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 12

 x = 44 x 56 x 65 x 710 x 88
⇒ x = 237 x 35 x 56 x 710 = (2 x 3) (236 x 34 x 56 x 710)

Power of 2 can take values of 0, 2 , 4 , . . . , 36 = 19
Power of 3 can take values of 0, 2,4 = 3
Power of 5 can take values of 0, 2,4, 6 = 4
Power of 7 can take values of 0, 2,4, 6, 8, 10 = 6

Total number of factors = 19 x 3 x 4 x 6 = 1368 

Test: CAT Quantitative Aptitude- 4 - Question 13

x is a positive integer with value at most equal to 110. How many values of x are possible if x is not a factor of (x - 1)!?

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 13

If x is a prime number greater than 1, then x is not a factor of (x - 1)!

There are 29 primes up to 110. 4 is also one such number.

Hence, option 4.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 4 - Question 14

The speed of a boat in still water is 24 kmph. If it can travel 20 km downstream in the same time as it can travel 12 km upstream, then it is ______ times as fast as the stream. Key in the number.


Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 14

Let the speed of the stream be v kmph.
Speed of boat in still water is 24 kmph.
Hence, the speed of boat upstream is (24 – v) kmph and the speed of boat down stream is (24 + v) kmph.

Putting v = 6 kmph:
Thus, the speed of boat in still water is 4 times as fast as the speed of the stream.

Test: CAT Quantitative Aptitude- 4 - Question 15

Find the minimum value of the quadratic expression 3x2 + 11x - 23.

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 15

The minimum value of a quadratic expression ax2 + bx + c (when a > 0) occurs at x = -b / 2a and its minimum value is -(b2 - 4ac) / 4a

The minimum value of the equation is = [4(3)(-23) - (11)2]/12 = -397 / 12

Hence, option 4.

Test: CAT Quantitative Aptitude- 4 - Question 16

50 square stone slabs of equal size were needed to cover, a floor area of 72 sq. m. The length of each stone slab is :

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 16

Area of each slab = 72 / 50 m= 1.44 m2
Length of each slab =√1.44 =1.2m =120cm

Test: CAT Quantitative Aptitude- 4 - Question 17

37% of the total number of people in a city read newspaper ABC and exceed the number of people who read XYZ by 8000. If 27% of the total population do not read any newspaper then find the population of the city if there are only two newspapers in the city. [Assume that no one reads more than one newspaper].

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 17

Let the total population of the town be x.
The number of people who read XYZ = 0.37x - 8000
0.37x+0.37x - 8000 + 0.27x = x
⇒ x = 8,00,000

Hence, option 1.

Test: CAT Quantitative Aptitude- 4 - Question 18

If x2 + x - 2 < 0; then which of the following statement is true?

1. x - 1 > 0
2. x - 1 < 0
3. x + 2 > 0
4. x - 2 < 0

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 18

x2 + x - 2 < 0
⇒ (x - 1) (x + 2) < 0
⇒ (x - 1) < 0 and (x + 2) > 0 or (x - 1) > 0 and (x + 2) < 0
⇒ x < 1, x > -2 or x > 1, x < -2 which is not possible.

∴ -2 < x < 1
⇒ (x - 1) < 0 and (x + 2) > 0 

Hence, option 3.

Test: CAT Quantitative Aptitude- 4 - Question 19

At 12:00 AM both the hour and a minute hand were overlapping each other. How many more minutes does the minute hand travel than the hour hand in the next 54 minutes?

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 19

When the minute hand travels 60 minutes, the hour hand travels 5 minutes.

The minute hand travels 55 minutes more than the hour hand in an hour.
In 54 minutes, the Minute hand travels = (54 x 55/ 60) = 49.5

Hence, option 3.

Test: CAT Quantitative Aptitude- 4 - Question 20

A and B start working together on a project and both have the same efficiency in the beginning. However the efficiency of A decreases to 0.6 times the usual after working for 2 hours. If the project can be finished in 120 man hours, then find the minimum number of days required to finish the work if both do an equal manhours of work each day and the sum of the total number of hours of work by both each day is 12. [ Initial efficiency of A = Efficiency of B = 1 man hour].

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 20

Let the number of hours worked by B be x.
The hours worked by A = (12 - x)
The manhours of work finished by A = (12 - x - 2) x 0.6 + 2

∵ (12 - x - 2) x 0.6 + 2 = x
⇒ x = 5

The total manhours of work done by them in a day = 2 * x = 2 * 5 = 10

Thus, the number of days required to finish the work = 120/10 =12 days

Hence, option 2.

Test: CAT Quantitative Aptitude- 4 - Question 21

The average number of apples per carton for three cartons of apple is 90. The average price per carton is Rs.1500. The average price of apples in the two cartons which do not contain the highest number of apples is 20 and 25 respectively. The price of the carton with the highest number of apples is 1900. If the highest difference in the number of apples in any two cartons is 100, then find the lowest number of apples in a carton.

Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 21

Let the number of apples in the carton with the least number of apples be x and the second least number of apples be y.
 ∵ x +y + x + 100 = 270
⇒ 2x +y = 170...(i)

Also, the total amount spent on the two cartons with x and y apples = 4500 - 1900 = 2600
⇒ 20x + 25y = 2600
⇒ 4x + 5y = 520.. .(ii)

From (i) and (ii)
x = 55,y = 60

The lowest number of apples in a cartoon = 55.

Hence, option 3.

*Answer can only contain numeric values
Test: CAT Quantitative Aptitude- 4 - Question 22

A saree is one-fourth by length green, two-fifth by length red and the remaining 3850 cm by length is black. What is the length of the saree in meters?


Detailed Solution for Test: CAT Quantitative Aptitude- 4 - Question 22

Let the length of the saree be ‘x’ cm.

x - (1/4) x - (2/5) x = 3850
⇒ (7/20) x = 3850
⇒ x = 11000 cm = 110 m

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