CAT Quantitative Aptitude- 5 - Free MCQ Practice Test with solutions, Quant

Let P be the sum and r be the rate of interest (expressed as a decimal fraction).
We have,
P (1 + r)² - P = 0.40(6Pr)
P (1 + 2r + r² - 1) = 2.4Pr
2r + r² = 2.4r
r² = 0.4r
r = 0.4 i.e., 40%
Hence, option 3.

Fixed monthly cost = Rs. 40,000
First we need to compute the Total Revenue.
Revenue made by lending 10 DVDs in May = 31 x 10 x 300 = Rs. 93,000
Revenue made by lending 12 CDs in May = 31 x 12 x 200 = Rs. 74,400
But then the lending rate in May is 40% (it is gradually increasing by 5%)

= Rs. 37,200
Similarly, the lending rate in May is 56% (it is gradually increasing by 8%)

= Rs. 41,664
Total direct cost of a DVD = 50x31 x 10 = Rs. 15,500
Total direct cost of a CD = 30 x 31 x 12 = Rs. 11,160
Maintenance cost = Rs.5,000
The revenue earned from all these products = (Actual money made by lending DVDs and CDs) - (direct cost of DVDs and CDs) - (Maintenance)
= Rs. 37,200 + Rs. 41,664 - Rs. 15,500 - Rs. 11,160 - Rs. 5,000
= Rs. 47,204
Profit or loss = revenue - expenses
= Rs. 47 ,204 - Rs . 40,000
= Rs. 7,204
This is positive that means his revenue is more than expense.
We have a profit of Rs. 7,204.
Hence, option 2.

When the bucket reached the 5^th student, 5/3^rd solution in the bucket has been replaced by water 4 times.
Also, when the bucket reaches the 5th student 5/3^rd of the solution in the bucket has been spilled.
After 4 replacements, amount of milk in solution

When the bucket is passed to the 5^th student 5/3 litres of the solution is spilled.
1/6^th of the solution is spilled.
1/6^th of the milk is spilled.
As equal percentages of milk and water are spilled, the percentage of milk in the bucket is
4.8/10 x 100 = 48%
Percentage of milk in the bucket when it was passed to the 5^th student = 48%
Hence, option 3.

The initial worth of the ball was Rs. 50,000. The worth of silver doubles every 2 years.
Worth of the silver ball at the end of 6 years = Rs. (50,000 x 8) = Rs. 4,00,000
Robin invests Rs. 4,00,000 in a new platinum ball and the worth of a platinum ball is directly proportional to its volume.

r₂ = 8 cm

Sum of the ages of the family 10 years ago = 33 × 8 = 264 years
After another 4 years, sum of their ages = 264 + (4 × 8) = 296 years
As Mrs. Edwards then died at the age of 64 years and William was born, sum of their ages would be 232 years but the number of members would still be 8.
Similarly, after 3 more years, the sum of the ages of the family members would be 184 years (232 +( 3 × 8) - 72 + 0) and the number of members would still be 8.
Presently i.e. after 3 more years, the sum of the ages of the members would be 208 years and the number of family members would still be 8.
Thus, the present average age of the Edwards = 208/8 = 26 years

Let the speed of the bus be b km/hr
The speed of the train = 12b km/hr
Let the bus normally travel for 3x hours and let the train normally travel for 2x hours.
3xb + 2.4xb = 810
xb = 150
If the train travels for the number of hours that the bus travels,

5.6xb - 810 = 0.4b
5.6(150) -810 = 0.4b 
b = 75
The speed of the bus = 75 km/hr 
Answer: 75

Thus, f(x) + f(1 - x) =
Therefore,

Therefore, adding these, we have:

Only one solution is possible.

Answer: 1

Consider the equation, x³ - x - 1 = 0,
x³ = x + 1 ...(i)
On multiplying the equation by x and x² to get,
x⁴ = x² + x . . . (ii)
x⁵ = x³ + x² . ..(iii)
x⁵ = x³ + x² = (x + 1) + x² ... [Using (i)]
= (x + x²) + 1
= x⁴ + 1 ... [Using (ii)]

x⁵ - x⁴ = 1
 
Hence, option 4.

The value of the three digit number decreases by 99% when its digits are reversed, we can see that the number with reversed digits must be 0.01 times the original three digit number.
The number formed on reversal of the digits must be a single digit number
(The number thus formed would become more than three digits on multiplication by 100).
The last two digits of the original three digit number must be zero, and the first digit must be a nonzero digit.
There are 9 such numbers: 100, 200, 300,..., 900.

Answer: 9

We have 1 + 2x > 0

Also we have



Consider the given inequality






4(1 + 2x) < 49


Hence, option 3.

The number of ways in which three points can be selected out of the eight vertices of a cube is ⁸C₃ = 56
We can see that in most of these 56 ways, the three points selected form a right angled triangle. So, we look at possible selections of three points which would not give a right angled triangle.
We can see that if two of these three points are the endpoints of one of the edges of the cube, then irrespective of which of the remaining vertices is the third point, we always get a right angle. If two of the three points are endpoints of a body diagonal of the cube, then the third point will always be such that it will form an edge of the cube with one of the two points, so we again get a right angle.
So, the sides of the triangle formed by the three points should neither be edges nor the body diagonals of the cube, i.e. all three sides should be face diagonals.
To count the number of such triangles, consider the 'top' face ABCD of a cube ABCDWXYZ (the points corresponding to A, B, C and D on the bottom face are W, X, Y and Z respectively). It has two face diagonals, AC and BD. With AC as one of the sides, we can have two acute triangles, ΔACX and ΔACZ. With BD as one of the sides, we can have two acute triangles, ΔBDY and ΔBDW.
Similarly, WXYZ has two face diagonals, WY and XZ, from which we get four acute triangles, ΔBWY, ΔDWY, ΔAXZ and ΔCXZ. Thus, 8 triangles out of 56 are not right angled triangles.
56 - 8 = 48
Answer: 48

Let BC = x, BD = 4x
CD = BD - BC = 3x
Also AB² = BC x BD = 4x²
AB = 2x
Now , in right angled  ΔABD
BD² = AB² + AD² 
(4x)² = (2x)² + (24)²
16x² = 4x² + (24)² 
12x² = (24)²
x² = 48 = 16 x 3

 CD = 3x

Hence, option 3.

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - be - ca)
In the above identity if we replace c by -c, we get,
a³ + b³ + c³ - 3abc = (a + b - c)(a² + b² + c² - ab + bc + ca) ... (i)
By triangle inequality we know that
a + b> c or (a + b - c) > 0 ...(ii)
Also we can write,

Combining (ii) and (iii) and replacing in (i), we get,

Hence, option 4.

As shown in the graph, we get the intersection points (-4, 8), (4, 8), (0, 12) and (0, 4).
The enclosed region would be a square with each side of length of 4√2
The radius of the inscribed circle = 2√2
It's area = 8π
The radius of the circle circumscribing the square will be 4.
It’s area = 16π
The difference between these two areas = 16π - 8π = 8π
Hence, option 2.

i.xy = 4

∴ A line parallel to the y-axis can intersect the graph o f xy = 4 at only one point.
ii. y = tan(x)

∴ A line parallel to the y -axis can intersect the graph o f y = tan(x) at only one point.
iii. x² + y² = 4

∴ A line parallel to the y-axis can intersect the graph of x² + y² = 4 twice.
iv. x = y²

∴ A line parallel to the y -a x is can intersect the graph o f x = y² twice.

v. |x|+ |y| = 2

∴ A line parallel to the 7-axis can intersect the graph of |x| + |y| = 2 twice.
Hence, option 3.
Alternatively,
A line parallel to the y-axis intersects the graph of an equation twice if for a given value of x, there can be two values of y.
Consider the given equations:
i. xy = 4
∴ y = 4/x
For a given value of x, y can take only one value.
A line parallel to the y-axis intersects the graph of xy = 4 only once.
ii. y = tan x
For a given value of x, y can take only one value.
iii. x² + y² = 8 .
∴ y² = 8 - x²
Clearly, for a given value of x, y can take two values.
∴ A line parallel to the y-axis intersects the graph of x² + y² = 8 twice.
iv. x = y²
Again, here it is clear that for a given value of x, y can take two values.
v. |x| + [y| = 2
∴ |y| = 2 - |x|
For every value of |x|, y can take a positive and a negative value.
Thus we can see that a line parallel to the y-axis intersects the graphs of (iii), (iv) and (v) twice.
Hence, option 3.

Let people who:

like ice-cream only = Do not like chocolate = x
like chocolate only = Do not like ice-cream = y
like neither chocolate nor ice-cream = z
From conditions given in the question,
People who like both ice-cream and chocolate = 0.25y
x + 0.25y = 0.66(x + z) ...(i)
x = 0.9 (y + z) ... (ii)
Solving (i) and (ii),
y/z = 2/3
The ratio of people who like both to none = 0.25y : z = 1 : 6
Hence, option 2.

Three questions are attempted from the first five.
These can be selected in ⁵C₃ = 10 ways
Also the remaining 7 questions can be attempted from the next 10 in ¹⁰C₇ = 120 ways
Therefore the total number of ways in which questions can be selected in the exam, if three questions are attempted from the first five questions, is 10 x 120 = 1200

Case II:

Four questions are attempted from the first five.
These can be attempted in ⁵C₄ = 5 ways
Also the remaining 6 questions can be attempted from the next 10 in ¹⁰C₆ = 210 ways
Therefore the total number of ways in which questions can be selected in the exam, if four questions are attempted from the first five questions, is 5 x 210 = 1050

Case III:

All five questions are attempted from the first five.
These can be attempted in ⁵C₅ = 1 way
Also the remaining 5 questions can be attempted from the next 10 in ¹⁰C₅ = 252 ways
Therefore the total number of ways in which questions can be selected in the exam, if five questions are attempted from the first five questions is 1 x 252 = 252
Therefore the total number of ways of selecting questions in the exam is 1200+ 1050 + 252 = 2502
Answer: 2502

Suppose the potter places m pots in 1 row and makes a total of n rows, then the total number of pots he makes is mn, where: 120 < mn < 300

According to the question, mn = (m + 6) ( n - 10)
Solving m = 
Since m cannot be a non-integral value, n must be a multiple of 5.
Also, n > 10 because if n ≤ 10, then m is negative or 0 (at n = 10).
Putting n = 15, we get m = 3 and mn = 45 which is less than 120. Hence, it is incorrect.
Putting n = 20, we get m = 6 and mn = 120 which is equal to 120. Hence, it is incorrect.
Putting n = 25, we get m = 9 and mn = 225 which is between 120 and 300. Hence, it is correct.
Putting n = 30, we get m = 12 and mn = 360 which is beyond 300. Hence, it is incorrect.

Sum of the original set of 20 observations = 20 x X

Sum of the new set of 17 observations = 17 x X
Now, the new set is formed by discarding three of the elements of the original set. 20X- (12 +19 + 74) = 17X
3X= 105
X = 35 
Answer: 35

Alternatively,

As the mean of the original set does not change when the three numbers are discarded, hence, the mean of the original set is same as the mean of the three discarded numbers.
The mean of the three discarded numbers is (19 + 74 + 12)/3 = 35 
X= 35
Answer: 35

According to the diagram in the previous question,

∴ Number of members selected for only C = 5x = 15 x 5 = 75

Hence, option 4.

Let A require ‘a’ days, B require ‘b’ days and C require ‘c’ days to complete the work separately.
A and B together both can complete a work in 24 days

A, B and C all three can complete a work in 18 days.

Solving equation (i) and (ii), we get,
∴ c = 72 days
∴ C requires 72 days to complete the work separately.
So, B requires 72 days to complete the work separately.
Putting this value of b = 72 in equation (i), we get,
a = 36 days




∴ The number of days required to complete the work = 16 + 24 = 40 days
Alternatively,
Let the total amount of work to be done be the L.C.M. of 24 and 18 i.e. 72 units.

Let A, B and C complete a, b and c units of work per day respectively when working alone.
∴ a + b = 72/24 = 3 units/day
and a + b + c = 72/18 = 4 units/day
∴ c = 4 - 3 = 1 unit/day
∴  b =1 unit/day and a = 2 units/day
Now, A, B and C work together for the first 8 days.
In this time frame, they complete 8(a + b + c) = 8(1 + 1 + 2) = 32 units of work.
Then, B and C work together for the next 8 days.
So, in this period, they complete 8(b + c) = 8(1 + 1) = 16 units o f work.
Thus, by the time, C starts working alone, 32 + 16 = 48 units of work has been completed.
Now, C completes the remaining 24 units of work alone by completing 1 unit per day.
Thus, C takes 24 days to complete the remaining work alone.
So, total time taken = 8 + 8 + 24 = 40 days.
Answer: 40