Free Quant Practice Test - CAT 

xxyy = xx00 + yy = 11(x00 + y) = 11(x0y)

xxyy is a perfect square, so if one divisor is 11, then there should be one more 11 to get ${11}^2$.

As xxyy = 11(x0y), x0y should be divisible by 11, so x + y has to 11.

The time taken by the train between any two adjacent stations can have only two values. So, it must cross one of the intermediate stations exactly between that of the neighbouring stations. So, the train should either cross B at 1:30 pm or D at 4 pm

There are 13 couples: 13 males & 13 females In addition to it, there are 5 single males and 7 single females  Now, 5 single males can shake hands with 20 females, therefore 5*20 = 100 ways Also, each of 13 males (from the couples) can shake hands with 12 females (except his wife) + 7 single females. So 13 * 19 = 247 ways. Therefore, in total 347 ways

To get a set of 9 numbers, none of which are divisible by 10, we need to consider the lowest valued number in the set to have a value 1 above the divisor of 10. 0 is a divisor, one above that will be 1 and the lowest valued number in the set is n-4.

So, n-4=1. Now subsequent sets will occur in an AP with a common difference of 10.

1,11,21,31,41.....91 as $n\le 100$

No. of such numbers $=\frac{91-1}{10}+1=\frac{90}{10}+1=9+1=10$

X=100

One =A, Two =B, Tree=C, Four= D, Five =E

Atleast 1 = A+B+C+D+E = 99----------------------(1)

Atleast 2= B+C+D+E=49-----------------------------(2)

Atleast 3= C+D+E= 24--------------------------------(3)

Atleast 4= D+E=14------------------------------------(4)

Atleast 5= E=4-------------------------------------------(5)

4-5= D+E-E = D= 14-4=10 (Exactly 4 = 10)

3-4= C+D+E-(D+E) =C = 24-14=10 (Exactly 3 =10)

Total = 10+10=20

A quadratic equation can be written as

$x^2$ - (sum of roots)x + (product of roots) =0

Arithmetic mean(p) = $\frac{\text{Sum of roots}}{2}\Rightarrow$ sum of roots = 2p

Geometric mean (q) = $\sqrt{\text{(Product of roots)}}\Rightarrow$ product of roots = $q^2$

The required equation is $x^2-2px+q^2=0$    

Suppose we take x = 50-paisa coins, y = 25-paisa coins and z =10-paisa coins.

We have x + y + z = 85 and 50x + 25y + 10z = 1875, because x, y and z are all integers, z should be a multiple of 5.

Possible values are
$\begin{array}{ccc}x& y& z\\ 2& 63& 20\\ 5& 55& 25\\ 8& 47& 30\\ ....\\ ....\\ 23& 7& 55\end{array}$

So there are eight solutions.

You should notice here that an increase of 5 in z and 3 in x and decrease of 8 in y neither changes the number of coins nor the value.

Let a be the number of students who do not study at home and do not attend classes.
Let b be the number of students who do not study at home but attend classes.
Let c be the number of students who study at home and attend classes also.
Let d be the number of students who study at home but do not attend classes.
Given: $d+b=\frac{2}{3}(b+c+d)$ and $d+c+b=\frac{3}{4}(a+b+d)$
Also, b = $\frac{7d}{5}$ and c = $\frac{a}{2}$.
If $d+b=2x⟹b+c+d=3x$, and $a+b+d=4x$
$(a+b+d)+(c+b+d)-(b+d)=300$
$\Rightarrow x=60$
$\Rightarrow a+c=180$

Option (c) is the right answer.

Answer is option (b)

Area of inner rectangle = $\frac{1}{2}$ * (area of rhombus)

Sum of area of all the rhombuses = $\frac{1}{2}$ + $\frac{1}{8}$ + $\frac{1}{32}$ + ........$\infty$ = $\frac{\frac{1}{2}}{1-\frac{1}{4}}$ = $\frac{2}{3}$
$\therefore$ Required ratio = $\frac{\frac{4}{3}}{\frac{2}{3}}$ = 2:1

We must cut the longest edges, so the similar piece has dimensions $L\times 100\times k$ for some $1\le k<H$.
The shortest edge of this piece cannot be L, so it must be k. Thus $L\times 100\times H$ and $k\times L\times 100$ are similar.
Thus, $\frac{L}{k}=\frac{100}{L}=\frac{H}{100}$.
So, HL = ${100}^2$. ${100}^2$ = $2^4\times 5^4$
So ${100}^2$ has 25 factors, of which $\frac{(25-1)}{2}=12$ pairs are $<100$.

Revenue will be maximum when the cost = number of seats, as revenue = (cost of each seat)$\times$(number of seats) i.e. the value of both = 400

Method 1: Conventional Approach

Given : $a_{n+1}=a_n{a}_{n-1}^2$
$a_{n+1}{a}_n=(a_n{a}_{n-1}{)}^{2^1}$
$(a_n{a}_{n-1}{)}^2-(a_{n-1}{a}_{n-2}{)}^{2^2}$
$..$
$..$
$(a_2{a}_1{)}^{2^{n-1}}=(a_1{a}_0{)}^{2^n}$
Multiplying all equations, we get
$a_{n+1}{a}_n=(a_1{a}_0{)}^{2^n}=8^{2^n}$
Now $a_{10}{a}_9=8^{2^9},a_9{a}_8=8^{2^8}......$
$a_{10}=8^{2^9-2^8+2^7-2^6+2^5-2^4+2^3-2^2}\times a_2$ $(a_2={8.1}^2=8)$
$⟹a_{10}=8^{\left(\frac{2^{10}-1}{3}\right)}$
$⟹a_{10}=2^{(2^{10}-1)}$

Method 2: Shortcut: Double Substitution

This technique involves writing the answer choice in terms of variables. We need to find $a_{10}$.

Thus n=10. Now rewrite all the options in terms of n.

$a.2^{(2^{n+1}-1)}b.2^{(2^n-1)}c.2^{(2^{n-1}-1)}d.2^{(2^{n-2}-1)}$
At n=2 in the question, $a_2=a_1\times a_0^2=8\times 1=8$
Now put n=2 in each of the answer choices and check where you are getting 8.
$a.2^{(2^3-1)}\ne 8b.2^{(2^2-1)}=8c.2^2\ne 8d.2^0\ne 8$
Thus, options a, c, d can be eliminated. Answer is option (b).

Required Ratio = 1: 2

Graphs will intersect at only one point in the first quadrant

In the first quadrant, both f(x) and g(x) will be positive.
2x- 3 - 1 = x${}^2$ - 2x
x${}^2$ - 4x + 4 = 0
$\therefore$ (x- 2)${}^2$ = 0

x=2  at only one point.

To solve this problem quickly, you can come up with likely values forxthat would make the mean equal to the median.
Three values that would make the set symmetrical are 0, 10, and 20:
{0, 4, 8, 12, 16}
{4, 8, 10, 12, 16}
{4, 8, 12, 16, 20}

To verify if there are any other possibilities for x, we need to evaluate each range of "x”

(1) If x is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median:

$\frac{40+x}{5}=840+x=40x=0$

(2) If x is between 8 and 12, then the median is equal to x.

$\frac{40+x}{5}=x40+x=5x40=4xx=10$

(3) If x is greater than 12, then the median is equal to 12.

$\frac{40+x}{5}=1240+x=60$

The above series can be written as (2-1)*1! + (3-1)*2! + (4-1) 3! +...+(64-1)*63!

= 2! - 1! + 3! - 2! + 4! - 3! +.....+ 64! - 63!

= 64! - 1!

Therefore the remainder is (-1) or 63.

Method 1:  Using variables

Let the no of pipes be x.

Number of pipe days needed to fill tank A is $\frac{x}{2}+\frac{x}{4}$

This is double the amount taken to fill tank B....

The equations is,$\frac{\frac{x}{2}+\frac{x}{4}}{2}=(\frac{x}{4}+1)$

Thus, x=4.

Method 2: Using numbers (Assumption)

Assume answer to be option (b) - a middle answer option

Let A have a capacity of 300 litres

B will have a capacity 150 litres

Now 4 pipes fill A for half a day = effectively 2 pipes working for an entire day = 100 litres are filled by the first half

And 2  pipes fill A for another half = effectively 1 more pipe fills A for an entire day

This means that 3 pipes fill A completely= 300 litres

Thus, each pipe fills 100 litres

Let us verify it with B

Given that in the first day, 2 pipes fill B for half a day $\Rightarrow$ one pipe fills B for an entire day = 100 litres

Now 50 litres is left which Is filled by one pipe in half a day.

TOTAL = 150 litres for B

Thus, our assumption is correct

Total = 3+4+5=12

Conventional Approach :
Given inequality is $|x+3|>x^2-11x+30$
We consider 2 cases 
Case (i) : $x>-3$
$|x+3|=x+3$
$x+3>x^2-11x+30\Rightarrow x^2-12x+27<0$
$(x-9)(x-3)<0$
 x= (3,9)
Case (ii) ; $x<-3$
|x+3|= -x-3
$-x-3>x^2-11x+30$
$\Rightarrow x^2-10x+33<0$
$\Rightarrow x^2-10x+25+8<0=(x-5{)}^2+8<0$
$(x-5{)}^2+8$ is always a positive quantity, irrespective of the value of x. Hence we do not have any value for x in the interval. Hence  the valid interval for x is (3,9). 

Shortcut- Elimination Approach

Step 1: Choose a value for x which is present in 2 answer options and absent in the other 2. Eg) x=4 satisfies the range specified in option (a) and (d), but does not form a part of the range in option (b) and (c)

At x=4

LHS= |4+3|=7

RHS= 16 - 44+30 = 2. LHS$>$RHS. It is satisfied. This means that answer options, which do not include x=4 can be directly eliminated as they can NEVER be the right answer

Step 2- Choose a value for x which is there in one of the remaining 2 options and absent in the other

Eg) Take x=10. This option satisfies the range in option (a) but is absent in option (d)

At x=10

LHS= |10+3|=13

RHS= 100-110+30= 20

LHS$<$RHS. This violates the condition in the question. This means that option (a) can never be the right answer.

Thus, by elimination, answer is option (d) 

20 is equalized by 3 pairs of primes (37,3), (29,11) and (23,17).

17 is equalized by 3 pairs of primes (31,3), (29,5) and (23,11).

18 is equalized by 4 pairs of primes (31,5), (29,7), (23,13) and (19,17).

15 is equalized by 3 pairs of primes (23,7), (19,11) and (17,13).

Let assume the point be centroid of the triangle.
The perpedicular distance from centroid to side of an equilateral triangle = $\frac{\sqrt{3}p}{6}$
a+b+c =  $\frac{\sqrt{3}p}{2}$

$f\left(x\right)=1-f(1-x)\Rightarrow f\left(x\right)+f(1-x)=1$                                             $f\left(\frac{1}{2}\right)=1-f\left(\frac{1}{2}\right)\Rightarrow f\left(\frac{1}{2}\right)=\frac{1}{2}$

$f\left(\frac{99}{100}\right)=f(1-\frac{1}{100})=1-f\left(\frac{1}{100}\right)$

$f\left(\frac{98}{100}\right)=f(1-\frac{2}{100})=1-f\left(\frac{2}{100}\right)$

So,$f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+.....+f\left(\frac{98}{100}\right)+f\left(\frac{99}{100}\right)$

$=f\left(\frac{1}{100}\right)+f\left(\frac{2}{100}\right)+.....+1-f\left(\frac{2}{100}\right)+1-f\left(\frac{2}{100}\right)$

$\left[f\left(\frac{50}{100}\right)\right]=f\left(\frac{1}{2}\right)=\frac{1}{2}$

$=49+f\left(\frac{50}{100}\right)$

$=49+\frac{1}{2}=49.5$

Hence option (b).

Let us see the sequence of the numbers:

Number  Last term of the number

$11$
$23$
$36$
$410$
$\stackrel{-}{n}\frac{\stackrel{-}{n}(n+1)}{2}$

We have to find the value of n for the 2000th term.

If n = 62, the last term that ends with n is $(\frac{1}{2}\times 62\times 63)$ = 1953.

Therefore, the next 63 terms are 63. So the 2000th term is 63. So the remainder when divided by 4 is 3.

Assume no. of Xeta=1 (you can assume any value which satisfies $2^x-1$)

Then, x = 1.

If there is 1 Xeta, it will die after the $1^{st}$ second.

Put x=1 in the options and check which option gives answer as 1.
Answer will be option (a) - x seconds.

$2^{nd}$ second $⟹$ The remaining Xeta will die. Answer is (d).

PRO $\left(12\right)+11=2\times 3\times 5\times 7\times 11+11$
$=11\times (2\times 3\times 5\times 7+1)$
$=11\times (210+1)$
$=11\times 211$

Since 211 is prime, the largest prime factor = 211.

Using alligation, we can directly solve this question

Take either petrol or kerosene.

Petrol in cask 1 =$\frac{7}{10}$

Petrol in cask 2 =$\frac{3}{4}$

Petrol in mixture =$\frac{11}{15}$


Ratio to take the mixture $=\frac{1}{60}:\frac{1}{30}=1:2$

2 same, 1 different, 1 different = ${}^3{C}_2{\ast }^3{C}_2=9$

3 same 1 different = ${}^2{C}_2{\ast }^3{C}_2=6$

4 same = 1

Total = 16

$\frac{(12.4w+26)}{(w+5)}=12$

On solving, w=85

We need a number of the form ABC where $A+B+C\le 5$, and A,B,C are whole numbers.

Method 1 - Trial and Error - Manual approach

List out all the possibilities manually,

Case 1: When sum of digits=1

A+B+C=1

We can solve this using the similar to different approach where there is 1 zero and 1 one. Number of combinations ${=}^3{C}_2=3$

Case 2: When sum of digits=2

A+B+C=2

Number of possibilities = Answer based on 2 zeroes and 2 ones ${=}^4{C}_2=6$

Case 3: When sum of digits=3

A+B+C=3

Number of possibilities = Answer based on 3 zeroes and 2 ones ${=}^5{C}_2=10$

Case 4: When sum of digits=4

A+B+C=4

Number of possibilities = Answer based on 4 zeroes and 2 ones ${=}^6{C}_2=15$

Case 5: When sum of digits=5

A+B+C=5

Number of possibilities = Answer based on 5 zeroes and 2 ones ${=}^7{C}_2=21$

Total = 21+15+10+6+3 = 55

Method 2- Shortcut

We will use a dummy variable "D” to take the place of the $\le$ sign.

Then the equation becomes -

A+B+C+D=5 (Where D can take values 0,1,2,3,4,5)

$\begin{array}{cc}A+B+C& D\\ 5& 0\\ 4& 1\\ 3& 2\\ 2& 3\\ 1& 4\\ 0& 5\end{array}$

Using the same Similar to Different approach, we can get the answer based on 5 zeroes and 3 ones as ${}^8{C}_3=56$

We must remove the case where D=5, because in that case A+B+C=0, which is not possible. Thus, the answer is 56-1=55.

Take 2 and $\frac{1}{2}$ as the two positive numbers whose product is unity. Here n=2, only option (c) and (d) is satisfied. The other two can never be true.

Take the 4 positive numbers as 1,1,1 and 1. Here n=4, option (c) is not true. Hence, answer is option (d).