Free Quant Practice Test - CAT 

The number should be of the form $a^6$, where 6 is the LCM of 2 and 3.

The required numbers are $1^6,2^6$  and $3^6$.

Solving, we get (4x + 3)$\times$3+2 = 71 $⟹$ x=5.

The smallest number possible is 1002 (this is a 4k+2 number, it cannot be expressed as difference of squares) .

Therefore our equation becomes. $1\times 1+2\times 2+........N\times n=1002.$

N=13.93

i.e. when the ${14}^{th}$ tap is open the tank gets filled.

S= 90+80+80=250

X= 100

Using the Maxima of all shortcut
$\frac{S-X}{N-1}=\frac{150}{2}=75$

Suppose in the vessels A $\&$ B, one litre of liquid is present.

So, Amount of milk in vessel A =$\frac{4}{9}$        ;         Amount of milk in vessel B = $\frac{7}{9}$

The required mixture in vessel C should be $\frac{1}{2}$ milk.

So, let the ratio be x: 1. Applying alligation:

$\frac{\left(\frac{5}{18}\right)}{\left(\frac{1}{18}\right)}=x:1\to x=5.$

Use the reverse gear approach

Option (a) f(2) = 32 - 32 + 24 - 24 - 8 + 8 = 0

Option (b) f(-2) $\ne$ 0

Option (c) f(3) $\ne$ 0

Option (d) f(-3) $\ne$ 0

Adding the three equations we get,  $2(x+y+z)=9.2⟹x+y+z=4.6$.
Subtracting first eq. from the above eq. we get $\left\{y\right\}+\left[z\right]=0.8⟹\left\{y\right\}=0.8$ and $\left[z\right]=0$.
Subtracting second eq. from the above eq. $\left\{x\right\}+\left[y\right]=1.4⟹\left\{x\right\}=0.4$ and $\left[y\right]=1$.
Subtracting third eq. from the above eq. $\left[x\right]+\left\{z\right\}=2.4⟹\left[x\right]=2$ and $\left\{z\right\}=0.4$.
$\therefore x=2.4,y=1.8$ and $z=0.4$.
Hence, choice (c) is the right answer.

Now since tan 45 = 1,BC=DB=x
Also, tan 30 = $\frac{1}{\sqrt{3}}=\frac{x}{50+x}$
Thus x($\sqrt{3}$ - 1) = 50
or, x = 25($\sqrt{3}$ + 1)m

Let the equation be $A{x}^5+B{x}^4+C{x}^3+D{x}^4+Ex+F=0$
Sum of the product of the roots taken 1 at a time $=\frac{-B}{A}$
$\to =-\frac{B}{A}$

Sum of the product of the roots taken 2 at a time $=\frac{C}{A}$
$\to (2\times 3)+(2\times 4)+(2\times 5)+(2\times 6)+(3\times 4)+(3\times 5)+(3\times 6)+(4\times 5)+(4\times 6)+(5\times 6)=\frac{C}{A}$
$\to 155=\frac{C}{A}$

Sum of the product of the roots taken 3 at a time $=-\frac{D}{A}$
$\to (2\times 3\times 4)+(2\times 3\times 5)+(2\times 3\times 6)+(2\times 4\times 5)+(2\times 4\times 6)+(2\times 5\times 6)+(3\times 4\times 5)+(3\times 4\times 6)+(3\times 5\times 6)+(4\times 5\times 6)=-\frac{D}{A}$
$\to 580=-\frac{D}{A}$

Sum of the product of the roots taken 4 at a time $=\frac{E}{A}$
$\to (2\times 3\times 4\times 5)+(2\times 3\times 4\times 6)+(2\times 3\times 5\times 6)+(2\times 4\times 5\times 6)+(3\times 4\times 5\times 6)=1044$
$\to 1044=\frac{E}{A}$

Sum of the product of the roots taken 5 at a time$=-\frac{F}{A}$
$\to (2\times 3\times 4\times 5\times 6)=-\frac{F}{A}$
$\to 720=-\frac{F}{A}$

Take A=1. Then B= -20,C=155, D= -580, E=1044, F=-720

Let ST=TM=TU=x; UV=VW=y.

Then RS= 18-(2x+y); PW=10-y=PM=PQ.               

PR=RQ+PQ= 18-(2x+y)+ (10-y)=12; $\Rightarrow$ x+y=8. Now, TV=TU+UV=x+y. Hence TV=8.

Draw a square of side x and an arc of radius x.

 
All rectangles with diagonal less than or equal to x will lie within or on the quadrant of the circle. The unshaded region is the favorable area.
Hence, required probability = $\frac{x^2(1-\frac{\pi }{4})}{x^2}\times 100$ = 0.215 = 21.5% chance

$(8^2{)}^{2n}-(7^2{)}^{2n}$ is of the form $a^n-b^n$ where the power is always even because 2n is always even.

That means it is divisible by $8^2+7^2$ as well as $8^2-7^2$.

Check for n=1, it becomes $8^4-7^4$, which is divisible by $8^2+7^2$ and $8^2-7^2$.

Thus. the given number is divisible by both 15 and 113.

It can be easily observed that ${1470}^{367}$ has more number of zeroes as compared to

${1430}^{343}.$ hence, the rightmost non-zero integer will depend on the unit digit of only ${1430}^{343}$, which in turn will depend on the unit digit of $3^{343}$

343 can written as 4k+3

In the power cycle of 3 (3,9,7,1), the third digit is 7. Hence, the last non-zero integer in the expression is 7.

AM$=\frac{(2^{a_1}+4^{a_2}+8^{a_3})}{3}$ and GM $=(2^{a_1}\times 4^{a_2}\times 8^{a_3}{)}^{\left(\frac{1}{3}\right)}$

Using condition AM$\ge$GM, we get $\frac{(2^{a_1}+4^{a_2}+8^{a_3})}{3}\ge (2^{a_1}\times 4^{a_2}\times 8^{a_3}{)}^{\left(\frac{1}{3}\right)}$

$2^{a_1}+4^{a_2}+8^{a_3}\ge 3(2^{a_1}\times 2^{2a_2}\times 2^{3a_3}{)}^{\left(\frac{1}{3}\right)}$

$\ge 3(2^{a_1+2a_2+3a_3}{)}^{\left(\frac{1}{3}\right)}$

But $a_1+2a_2+3a_3=1$

Answer =$(27\times 2{)}^{\frac{1}{3}}$

Any power of 9 when divided by 6, gives a remainder 3. hence answer $=\frac{11\times 3}{6}$ ((As there are 11 terms)$=\frac{33}{6}$= remainder =3

We can use the last two digits technique to solve this question.

$6^{17}=3^{17}\times 2^{17}$
$3^{17}=(3^4{)}^4\times 3=(3^2\times 3^2{)}^4\times 3=(09\times 09{)}^4\times 3={81}^4\times 3=21\times 3=63$
$2^{17}=(2^{10}{)}^1\times 2^7=24\times ......28=.....72$
Answer is the tens digit of the product of 72 and 63 = ...36.

Method 1- Unitary method

Consider a 2x2 grid

 No. of different paths = (1+ 2x2 +1)

For a 3x3 = (1 + 3x3 + 3x3 + 1) ..

For a 4x4 =( 1+ 4x4 + 6x6 + 4x4 +1) ..

Thus, answer is 70

Method 2- Using SàD

Shortcut:- No of rows=4. number of columns=4.

i.e. using the one zero method, total number of paths= ${}^{4+4}{C}_4$= 70

For two players, we need one match. The first round, having 256 players will need $\frac{256}{2}$= 128 matches.

Now there are 128 players left. To ensure that all players play a match, we need $\frac{128}{2}$= 64 matches

Continuing in the same way

Total number of matches = 128+64+ 32+ 16+8+4+2+1= 255

OR

In 1 match 1 player is eliminated

For tournament to conclude there should be only 1 winner and hence 255 players need to lose.
Hence we need 255 matches to eliminate 255 players.

Assume a GP, 1, 2, 4 with common ratio =2

Assume an AP with common difference 1 = 1, 2, 3, 4

Here $a^{th}$,$b^{th}$ and $c^{th}$ term is 1,2 and 4

Look in the answer options where you get answer 2 after substitution

Take p=q=r=s=1.

(1+1)(1+1)(1+1)(1+1) = 16

There are 12 letters, in which 'S' appears 4 times and 3 vowels are A,E,U. 

Take all vowels (AUE) as a single object

So number of permutations when vowels occur together = $\frac{10!}{4!}$ $\times$ 3!
Arrangements of 12 letters take all at atime = $\frac{12!}{4!}$
Required arrangements when al vowels do not occur together
= $\frac{12!}{4!}$ - $\frac{10!}{4!}$ $\times$ 3!
= $\frac{10!}{4!}$ (12 $\times$ 11 -6)
= $\frac{10!}{4!}$ $\times$ 126

A = 126

To find the minimum number of spectators who are fans of all 5 teams, take the difference from 100 from each, and the difference from 100 again for the total

100-90=10

100-85=15

100-70=30

100-75=25

100-80=20

total = 100

100-100=0

Minimum no. of spectators who are all 5 teams' fans= 0

At n=1, we should get $\frac{3}{4}$ (this is the sum to 1 term of the series)

Option (a) gives $=\frac{2}{4}=\frac{1}{2}$               

Option (b) gives $\frac{3}{4}$

Option (c) gives $\frac{3}{8}$

Option(d) gives 8

Answer= option b

Number = Quotient $\times$ Divisor + Remainder

Given that:

N = rX+ 28 and N = s*10X+125

$\Rightarrow$ rX+28 = s*10X + 125

Solving the above two equations, we get X(r-10s) = 97.

Therefore X is a factor of 97. So possible values are 1 and 97.

Now X leaves a remainder of 28. Therefore its greater than 28. Thus X=97. 5X=485.

5X will also leave a remainder of 125 (since 5X is a factor of 10X and 5X>125).

Since the only condition we need to maintain is that the diagonals need to intersect at right angles, assume the simplest quadrilateral which satisfies this condition- a square. Now the question can be solved by simple graphical division.

Take the side of the square =p=q=2. Area = 4.

We need the area of the shaded region, which is half of that of the square= 2. Look in the answer options for 2 on substituting p=2 and q=2. Answer is option (a)

The diameter of the circle = diagonal of the square = 2$\sqrt{2}$

Substituting in answer options, we get answer as option(c)

The woman is twice as efficient as the child hence she gets paid twice the money= 10200. Therefore the man gets paid 18000 - (5100+10200)= 2700. The man did

$\frac{2700}{18,000}$  of the work.

$={\frac{9}{60}}^{th}$ in one day.Hence his efficiency is $\frac{\left(\frac{1}{5}\right)}{\left(\frac{9}{60}\right)}=75\%.$

This question can be solved using alligation. For alligation; remember to always consider the cost price only.

Selling Price (SP)

Cost Price (CP)

SP of rice 1 = 2.70/kg                           CP (rice 1) = $\frac{2.7}{0.9}$ = 3/kg

SP of rice 2 = 4.5/kg                             CP (rice 2) = $\frac{4.5}{1.125}$ = 4/kg

SP of mixture = 3.95/kg                      CP (mixture) = $\frac{3.95}{1.25}$ = 3.16/kg

Required Ratio = 84:16

S1 club members increase in an arithmetic progression.

Hence on June $7^{th}$ ,the number of members in club S1 is  n+6(10.5n)=64n.

S2 club members increase in an geometric progression.

Hence on June $7^{th}$  ,the number of members in club S2 is  $n{r}^6=64n....$(As given in the question).

Therefore r=2.  

Assumption

Assume n=2

Then

g(2) = g(1) +g(1) = 2g(1)= 2k

substitute n=2 in answer options to eliminate wrong options.

Hence, answer is option (b)

Assume the roots p and q = 1 and 2 respectively.

Sum of roots $=-(-\frac{2n}{m})=3$

Product of roots $=\frac{v}{m}=2$

Write all the terms is the form of variable m,

$\frac{2n}{m}=3\Rightarrow n=\frac{3m}{2}$

$\frac{v}{m}=2\Rightarrow v=2m$

$p^{\frac{1}{2}}{q}^2+p^3{q}^3+p^8{q}^3=20$

Substitute values in terms of m in answer options. Only option b gives 20

For product of three numbers to be positive it must have either 0 or 2 negative numbers.

Case 1: 0 negative numbers.

3 positive numbers can be selected out of 5 in ${}^5{C}_3$ ways = 10 ways.

Case 2: 2 negative numbers

2 negative numbers can be selected out of 4 in ${}^4{C}_2$ ways = 6 ways.

1 positive number can be selected out of 5 in ${}^5{C}_1$ ways = 5 ways.

Hence this combination can be selected in a total of $(6\times 5)=30$ ways.

Drop a perpendicular PM on the side DC.
Let PN be 'x' and side of square be 2a
$\Rightarrow$$\frac{1}{2}\times 2a\times$ (2a + x) = 8a${}^2$
$\Rightarrow$x = 6a
$\frac{\Delta PQR}{\Delta PDC}={\left(\frac{6a}{8a}\right)}^2=\frac{9}{16}\Rightarrow \Delta PQR=\frac{9}{16}\Delta PDC$
$\Delta PQR=\frac{9}{16}(2\times ABCD)$
Required Ratio = $\frac{9}{8}$
Hence, option (c)

The minimum value will occur when $p^2=\frac{625}{p^2}$

$p^4=625\Rightarrow p=5$

Putting the value of p in the equation, min value =$p^4=625\Rightarrow p=5\Rightarrow =25+\frac{625}{25}=25+25=50$