Free Quant Practice Test - CAT 

$510=2\times 3\times 5\times 17$.
Use the concept of Euler's theorem:
$510(1-\frac{1}{2})(1-\frac{1}{3})(1-\frac{1}{5})(1-\frac{1}{17})=128$
$Numberofpairs=\frac{128}{2}=64$.

Therefore, answer option (d) stands out as the correct answer choice. 

Using graphical division, we can divide the figure to look like the following: 

Looking at the figure, we can easily make out that the shaded region is made out of 16 triangles,
Each being 1/64${}^{th}$ the size of the larger square. (Since, area of each triangle =$\frac{1}{4}$ * $\frac{1}{16}$ is area of square) 
Therefore, area of shaded region = $(\frac{1}{4}\times \frac{1}{16}\times 16\times 16)$cm${}^2$
Thus, required area of shaded region = 4cm${}^2$.

Method 1:
Let the capacity of glass be 100 ml.

$\begin{array}{ccc}& RedWine& Water\\ Initial& 100& 0\\ Afterthefirstoperation& 80& 20\\ Afterthesecondoperation& 60& 40\\ Afterthethirdoperation& 40& 60\end{array}$

Method 2: Quantity of red wine left after 3 operations, when the glass originally contains $100\%$ wine from which $\frac{1}{5},\frac{1}{4}$ and $\frac{1}{3}$ of total wine are taken out each time and replaced is
$=(1-\frac{1}{5})(1-\frac{1}{4})(1-\frac{1}{3})=40\%$


 

This is because if we are writing a number N as a difference of squares, then -
$N=a^2-b^2⟹N=(a+b)(a-b)$. If (a+b, a-b) is a pair of one odd and one even number, 'a' and 'b' can't be natural numbers.
Hence, both the factors need to be either even or odd.
In the case of 80, the possible cases are $80=2\times 40;=4\times 20;=8\times 10.$
Thus, there are only three instances where both are even. Hence, the answer is option (b).

A Radius of semicircles ACB, AC and BC are 2.5, 2 and 1.5 respectively.
Area of Semi circles (ACB), (AC) and (BC) are $\frac{6.25\pi }{2}$, 2$\pi$ and $\frac{2.25\pi }{2}$ respectively. And area of
triangle ABC =$\frac{1}{2}$ $\times$ 4 $\times$ 3 = 6
Sum of the area of region A and B = (Area of semicircle (ACB) - Area of triangle (ABC)} = $(\frac{6.25\pi }{2}-6)$
Area of Shaded region = (Sum of the area of semicircles AC and BC - Sum of the area of region A and B)
Area of Shaded region = $(2\pi +\frac{2.25\pi }{2})$ - $(\frac{6.25\pi }{2}-6)$ = $\left(\frac{6.25\pi }{2}\right)$ - $(\frac{6.25\pi }{2}-6)$ = 6 

Alternate Solution:
As the options are far apart we can easily arrive at the answer by approximation. The shaded
region is approximately half of the areas of semicircles AC and BC = $\frac{1}{4}(4\pi +2.25\pi )$ = 6(approximately) 

Every clock shall display the correct time when it is late or fast by hours which are a multiple of 12 .

As per the question, the first clock loses 3 minutes in an hour. Therefore, to lose 12 hours, it will take $\frac{12\times 60}{3}=240$ hours =10 days.

Also, the second clock gains 2 minutes inan hour. Therefore, to gain 12 hours, it will take : $\frac{12\times 60}{2}=360$ hours = 15 days.

Therefore, after 30 days (LCM of 15 and 10), both the watches will show the correct time .

A red marble can be drawn from Box I afterthe two rounds of transfers in the following four ways:
Case 1: R$\to$R$\to$R
Probability = $(\frac{5}{11}\times \frac{5}{110}\times \frac{5}{11})$
Case 2: R$\to$B$\to$R
Probability = $(\frac{5}{11}\times \frac{5}{110}\times \frac{4}{11})$
Case 3: B$\to$B$\to$R
Probability = $(\frac{6}{11}\times \frac{6}{110}\times \frac{5}{11})$
Case 4: B$\to$R$\to$R
Probability = $(\frac{6}{11}\times \frac{6}{110}\times \frac{6}{11})$
Thus resulting probability = sum of probabilities of individual cases
=$(\frac{5}{11}\times \frac{5}{110}\times \frac{5}{11})$ +$(\frac{5}{11}\times \frac{5}{110}\times \frac{4}{11})$ +$(\frac{6}{11}\times \frac{6}{110}\times \frac{5}{11})$ +$(\frac{6}{11}\times \frac{6}{110}\times \frac{6}{11})$ = $\frac{621}{1210}$
Therefore, answer option (c) is the correct answer choice.

Approach 1:

You can directly mark the answer as 25 $\times$ x = 20 $\times$ 5 $\Rightarrow$ x=4
Proof:
Observe that
ac = 25 sq units ....(1)
bc= 20 sq units ... (2)
ad = 5 sq units ... (3)
Dividing (1) by (3), we get: $\frac{c}{d}$ = 5 units ... (4)
Dividing (1) by (2) we get $\frac{a}{b}$ =5/4 units ... (5)
​St Multiplying (4) and (5) we get:
Sit $\frac{ac}{bd}$ = $\frac{25}{4}$ sq units
Since its already given that ac = 25 sq units, therefore the missing area 'X' =4 square units.

Approach 2:
Use the simple number approach to break up the values in each cell as shown below 

thus x=4x1=4 

Method 1
There are four possible cases that satisfy the condition of forming 4-letter passwords with at least one symmetric letter.
Case 1: 1 symmetric letter + 3 unsymmetric letters
Case 2: 2 symmetric letters + 2 unsymmetric letters
Case 3: 3 symmetric letters + 1 unsymmetric letter
Case 4: All 4 symmetric letters
= {(${}^{11}$C${}_1$ $\times$ ${}^{15}$C${}_3$) + (${}^{11}$C${}_2$ $\times$ ${}^{15}$C${}_2$) + (${}^{11}$C${}_3$ $\times$ ${}^{15}$C${}_1$) + (${}^{11}$C${}_4$) }
=5005 + 5775 + 2475 + 330
=13585 $\times$ 4! = 326040
Method 2
Total number of cases- cases where there are no symmetric letters

${}^{26}$C${}_4$ -${}^{15}$C${}_4$ = 13585 $\times$ 4! = 326040

Method 1:
Assume that the distance between the towns is x km. The first time the cars pass each other, the first car (from town A) has gone 70 km and the other has gone
(x-70)km. These distance were covered in the same time, so the ratio of the speeds of the cars is $\frac{70}{x-70}.$
The next time they meet, the first car has travelled (x+40) km and the second car has travelled (2x-40) km.Thus the ratio of the speeds is also equal to $\frac{(x+40)}{2x-40}.$ We can equate the two ratios to get $\frac{70}{x-70}=\frac{(x+40)}{2x-40}$ , so 70(2x-40)=(x+40)(x-70). This gives $140x-2800=x^2-30x-2800$, so, x(x-170) =0. The towns are 170 km apart.

Method 2: Reverse Gear:
Go from answer options,
Since the time is constant in both cases, the distance ratios should also remain the same. Assuming that option (b) is correct

$\frac{70}{100}=\frac{210}{300}$

Hence,Distance Ratio 1 (first Meeting =Distance Ratio(Second Meeting) Our Assumption is correct. Answer os option (b)).
 

From the figure, it is evident the required area = $\frac{AreaofSquare-AreaofCircle}{4}\times$2
Therefore, required area = $\frac{\left[\right(2{)}^2-\pi \left(1{)}^2\right]}{4}\times 2=\frac{4-\pi }{2}$

Shortcut: 4-pi-2 method
This is the most straightforward example of the circle inside a square pattern.
Therefore, we know that Area of square: Area of Circle = 4:$\pi$
Thus,required area = $\frac{4-\pi }{2}$ 

From the data given in the question, we can draw a venn diagram as follows 

Since N=30, the sum of the shaded region is 10. Hence 9+x=10 & x=1. Answer is option (d) 
 

Method 1- Conventional Approach
f(x) = x${}^{15}$- 2013{x${}^{14}$ - x${}^{15}$ + x${}^{12}$ - ......... + x${}^2$ - x} or
f(x) = x${}^{15}$- 2013{-x + x${}^2$ - x${}^3$ + ........ - x${}^{13}$ + x${}^{14}$} 
f(x) = x${}^{15}$ - $\left(\frac{2013x(x^{14}-1)}{(x+1)}\right)$
Now, Put x = 2012
f(2012) = (2012)${}^{15}$ - $\left(\frac{2013.2012({2012}^{14}-1)}{(2012+1)}\right)$
f(2012) = (2012)${}^{15}$ - 2012(2012^{14} - 1) 
f(2012) = 2012.

Method 2- Assumption
Assume 2012 = n and 2013 = (n+ 1). Put x=1. Now replace the options in terms of "n" as
a) (n-1)           b) n               c) (n + 1)                 d) (n + 2) 
f(1)=1 - 2 + 2 - 2 + 2 - 2 + 2 - ......................-2 + 2 
f(1) = 1. Options (a),(c) & (d) can be eliminated as only option (b) gives us 1. 

From the diagram above, AC${}^1$=16
and BC${}^2$=9.
Also C${}^1$ - C${}^2$=25.
PC${}^1$ = 7 (16-9)
AB=PC${}^2$
C${}^1$ - P - C${}^2$ is a right angled triangle
($C^1-C^2{)}^2$ = PC${}^{12}$+PC${}^{22}$ 
Thus, PC${}^2$ =$\sqrt{625-49}$
= $\sqrt{576}$ = 24 

Given that surface area of the original cube = 24 units.
$\therefore$  Length of each side of the cube =$\sqrt{\frac{24}{6}}$= 2 units.
The radius of the biggest possible sphere that can be kept inside the cube of each side 2 units.
= $\frac{2}{2}$ = 1 unit.
Now, the diagonal of the second cube inscribed in the sphere = Diameter of sphere.
$\Rightarrow$$\sqrt{3}$ (length of each side of second cube) = 2 units.
$\Rightarrow$Length of each side of second cube =$\frac{2}{\sqrt{3}}$ unit
$\therefore$ surface area of the inner cube = 6 x${\left(\frac{2}{\sqrt{3}}\right)}^2$
= $6\times \frac{4}{3}=$ 8 units

For a ray to get back to its original orientation, it requires an even no. of operations. Since, all the rays get reversed after 42 operations and if there are k rays $\Rightarrow$ 2n+k = 42 $\Rightarrow$ k has to be even. Hence, choice (d) is the right answer

Hence option (c)

Option (a)
g(4) = 0 also f(0) = 0. So, g(4) = f(0)
Out of the given options, only (a) satisfies g(4) = f(0) when you put x = 0 Hence, the answer is
option (a) 

New Profit percentage = $120.50\%$

$\begin{array}{ccc}Tubelight& 10& 15\\ Bulb& 42& 33\\ & 52& 48\end{array}$

From this we can deduce that the cost of the tubelight is very close to our assumption. We can also directly mark option (d) as the correct answer without verifying because with a Rs.10 tubelight we are getting the new cost as Rs.48 ; thus the cost of the tubelight has to be slightly more than Rs.10 to get the cost of the package as Rs.50 !
 

(20 + x)(10y + p) = (10p + y)(10x + 2) $\Rightarrow$ px = 2y. For p = 5 or 7 $\Rightarrow$ y = p and x = 2, and for p = 3 $\Rightarrow$ y = 3, 6 or 9 and x = 2, 4 or 6 respectively.

The initial sum of all pages in the book $=\frac{200\times 201}{2}=20100.$
If we try to visualize the binding of the book, it would consist of 100 leaves, printed back to back so that the total number of pages in the book = 200.
If we pull out the middle sheet from the book, pages 99, 100, 101 and 102 would get pulled out.
Therefore, resulting sum of all the page numbers still remaining in the book = 20100 - (99+100+101+102) = 19698

So total paint used = 45 + 20 = 65litres