Free Quant Practice Test - CAT 

$\frac{5}{13}+\frac{55}{{13}^2}+\frac{555}{{13}^3}+\frac{5555}{{13}^4}+.....=\frac{5}{9}(\frac{9}{13}+\frac{99}{{13}^2}+\frac{999}{{13}^3}+.............)$

$=\frac{5}{9}[\frac{10-1}{13}+\frac{100-1}{{13}^2}+\frac{1000-1}{{13}^3}..............]$

$=\frac{5}{9}[\{{\left(\frac{10}{13}\right)}^1+{\left(\frac{10}{13}\right)}^2+{\left(\frac{10}{13}\right)}^3+..........\}-\{\left(\frac{1}{13}\right)+{\left(\frac{1}{13}\right)}^2+{\left(\frac{1}{13}\right)}^3+........\}]$

Now solve this infinite GP to get $\frac{65}{36}$ Hence, option (c)

The above diagram is self-explanatory. 180-4x+x+y=180

Hence y=3x, k=3. 

Prime numbers less than 10 = 2, 3, 5, 7.

If the difference between the largest and the smallest number is ending in 5, the prime numbers in the end position have to be 7 and 2.

The smallest and largest numbers are of form 2_7 and 7_2

Since it is given that the sum of the digits is >13, x will be 5.

Verifying, 752-257 = 495. Answer is option (b).

ax + 2y = 30.................................(1)
3x + by = 15.................................(2)

Multiplying equation (2) by 2

6x +2by = 30........................(3)

Equating eqn (1) and (3)

ax +2y = 6x +2by
We cannot find the solution when a=6 and b=1, answer is option (c)

$S_1=\frac{1}{1-\frac{1}{2}}=2$

$S_2=\frac{2}{1-\frac{1}{3}}=3$

Thus, the respective sums are 2,3,4,5,6,7

Sums of square of these numbers $=\left[\frac{n(n+1)(2n+1)}{6}\right]-1=139.$

Let PT=x, Then 8${}^2$-x${}^2$ = 6${}^2$ - (10-x)${}^2$; $\Rightarrow$ x = 6.4

Thus QT = 8${}^2$ - 6.4${}^2$ = 4.8; ST=$\frac{4.8}{\sqrt{2}}$; 

Area of square STUQ =${\left(\frac{4.8}{\sqrt{2}}\right)}^2$;

Area of circle: $\pi$ (2.4)${}^2$;

Area of the shaded region: $\frac{1}{2}$$\left(\pi \right(2.4{)}^2-{\left(\frac{4.8}{\sqrt{2}}\right)}^2)=\frac{(\pi -2){\left(\frac{12}{5}\right)}^2}{2}$

If only 3 digits are involved, the system has to be in base 3 with the digits 0,1,2. We can confirm the base using the examples provided.

Thus, a=1 ,b=2 and c=0 (you can reconfirm using the other examples as well)

17 in base 3 = = abb.

This is a question based on both upper limit and lower limit

Given that A+B+C+D$\le$20

This can be written as A+B+C+D+E=20

Where 1$\le$A,B,C,D$\le$8

Applying a lower limit of 1 to A,B,C,D

A+B+C+D+E=16

$Numberofways={}^{20}{C}_4=4845$

Giving an upper limit of 8 to A (this makes A=9)

A+B+C+D+E=8

Number of solutions = ${}^{12}$C${}_4$

Similarly, violating the conditions for A,B,C & D= 4*${}^{12}$C${}_4$

ABCABC = ABC$\times$1000 + ABC = ABC$\times$1001.

$\therefore$ Remainder when ABCABC is divided by 1001 is 0.

Let the number of men = n

Number of games between men and women = 4n

Given that $2{(}^n{C}_2-2n)=66$

Solving for n, we get n= 11

Thus total number of games played = ${}^{13}{C}_2=78\times 2=156$ 

There are only 2 possibilities for minimum value -

Thus, the answer is option (a).

a picture of the distances when Amar and Akbar finish the race

Length of the race track = 400 +x

The ratio of Akbar and Anthony's speed can be arrived at as

$\frac{240+x}{x}=\frac{400+x}{100+x}$

Now go from answer options, answer will be option (b) 800 m

Verifying, x = 400

$\frac{640}{400}=\frac{800}{500}$. Thus, answer is option (b)

At x=1
$a_0+a_1+.....+a_{15}=4^5.............\left(1\right)$

At x= -1
$a_0-a_1+......-a_{15}=\mathrm{0.................}\left(2\right)$

(1)-(2)$=2a_1+2a_3.......2a_{15}=4^5.........\left(3\right)$

$\left(\frac{\left[\right(1)+(2\left)\right]}{2}\right)=a_0+a_2...=\frac{4^5}{2}........\left(4\right)$

(3)+(4)$=a_0+2a_1+a_2+2a_3....+2a_{15}=4^5+\left[\frac{4^5}{2}\right]=1536.$

f(1 - x) = $\frac{2}{4^{1-x}+2}$ = f(x) = $\frac{2\ast 4^x}{4^{x+2}}$ = f(x) = $\frac{4^x}{4^{x+2}}$
f(x) + f(1 - x) = 1

So, the desired sum is 1000.

At p=2, sum = 1+10 =11

Put p=2 in answer options

Only Option c gives 11

Saif takes 10 hrs.
So total time they both would take without talking $=\frac{10\times 9}{19}=\frac{90}{19}$ hrs.
But they take 5 hrs $=\frac{95}{19}$ hrs.

So time wasted is  $\frac{5}{19}$ hrs.

They work for 5 hrs and put 10 bricks less per hour, so they had to put in extra time for $5\times 10=50$ bricks.
Now, extra time taken= $\frac{5}{19}$  hrs.
so, in  $\frac{5}{19}$  hrs: 50 bricks
5 hrs: x
x=950 bricks.
But this includes the 50 bricks extra put in the time  $\frac{5}{19}$  hrs. So total number of bricks is 900.

 Aamir takes 9 hours,$\Rightarrow$  he does $11.11\%$ of the work every hour

Saif takes 10 hours $\Rightarrow$  he does 10% of the work every hour

Together they finish  $21.11\%$ of the work

Given that they take 5 hours to finish the work after chatting, which means they are actually finishing  $\frac{100}{5}=20\%$ of the work

The remaining $1.1\%$ accounts for the 10 bricks decrease due to chatting

Thus, $100\%$  corresponds to 900. Answer is option (b)

Using the venn diagram approach

II +III= 35

a+b=20

Hence percentage playing only Football = 45-10-20 =15%

$p_1=a;p_2=\frac{-1}{a+1};p_3=(-1-\frac{1}{a});p_4=a.$

Hence the pattern $p_1=p_4=p_7=p_{10}=p_{13}=p_{16}=p_{19}....=a$
Hence, choice (c) is the right answer

From the above figure,  Now BRO is a right angled triangle. Solving for x, we get

X=  22 - 4 $\sqrt{19}$

Side length of a square = 8 + 4 $\sqrt{19}$

In the figure $\angle$PST and $\angle$PRQ  are similar (A,A,A).
If SQ = x = PT and TR = y =,$\left(\frac{x}{x+8}\right)=\frac{7}{35}$; x = 2
 

Natural number solutions means each of A,B and C can take a value ≥ 1

This is a Similar to Different question, with a lower limit condition. (Refer ebooklet for indepth explanation)

To take care of the condition, that we are dealing with natural numbers, 

give A,B &C 1 initially

(A+1)+(B+1)+(C+1)=100

Thus, this reduces to A+B+C=97.

This question is now based on arrangement of 97 zeroes and 2 ones=${}^{99}$C${}_2$

The answer is ${}^{99}$C${}_2$ which is option (a)

Standard form of a quadratic equation = f(x) = $a{x}^2$+bx+c=0

Substitute x=0                 
Given that f(x)=c=6
Substitute x=1                 
a+b+c=5                        
a+b =-1................(1)
The minimum value of a function occurs at x=$-\left(\frac{b}{2a}\right)$, which is given as 1
Thus, 2a=-b
Substituting back in eqn (1) a=1 and b=-2
$f\left(x\right)=x^2-2x+6$

Thus, at x=5, f(x)=21. Option (b)

Work completed in 5 days = $\left(\frac{1}{6}\right)$. Since $50\%$  of the bridge collapsed so total work completed after 5 days =

$\left(\frac{1}{12}\right)$ Remaining work after 5 days = $\left(\frac{11}{12}\right)$

Since women efficiency is half so 10 women will finish the work in 60 days i.e each day  they will finish $\left(\frac{1}{60}\right)$ part of work.

15 men will finish = $\left(\frac{1}{20}\right)$ part of work every day. So from the sixth day onwards the work finished everyday $=\frac{1}{20}+\frac{1}{60}=\frac{1}{15}.$ Now from sixth days onwards it rains alternatively for 2 working hours.

Hence total numbers of hours lost =  $6\times 2$  = 12 hrs = 1.5 days ( 8 hrs per day)

So till 17th day instead of remaining 12 days the work was done effectively on only 10.5 days.

So the work completed in 10.5 days $=\frac{10}{15}+\frac{1}{30}=\frac{21}{30}.$

Fraction of work left after 17th day = work left after 5 days - work done in 12 days $=\frac{11}{12}-\frac{21}{30}=\frac{13}{60}$

Hence option D is correct.

Alternative approach: (using units of work done)

Total number of units of work to be done = $10\times 30\times 8$ = 2400

In the first 5 days, $10\times 5\times 8$ =400 units of word is done. However, 200 units of work is lost

Thus, remaining work is 200 units

From the $6^{th}$ day 5 more men and 10 more women are joining in $\Rightarrow$ 10 more men are joining in

Thus, on every day that it rains; number of units of work done $=6\times 20=120$

And on every day that it does not rain; number of units of work done $=8\times 20=160$

Total units of work done  $=120\times 6+160\times 6+200=1880$

Remaining work = 2400-1880 = 540. Required ratio =$\left(\frac{540}{2400}\right)=\left(\frac{13}{60}\right)$

Since x and y are positive, we may write -

$x^2+3y=(x+a{)}^2$, and
$y^2+3x=(y+b{)}^2$

where a, b are positive integers.

Expanding, we find that the squared terms cancel, leaving the linear simultaneous equations -
$3y=2ax+a^2$
$3x=2by+b^2$

Solving, we obtain -
$x=\frac{2a^{2}b+3b^2}{9-4ab}$
$y=\frac{2b^{2}a+3a^2}{9-4ab}$

Since a and b are positive, the numerator in each fraction will be positive. For the denominator to be positive, we must have ab = 1 or 2.

If (a,b) = (1,1), (1,2), (2,1), then, respectively, (x,y) = (1,1), (16,11), (11,16). Hence, these are the only solutions.

This is a question based on distributing similar things to different groups (SàD)

i.e.  the distribution of 9 similar things to 6 different groups

A+B+C+D+E+F=9

Solving, we get ${}^{14}$C${}_5$= 2002.

The number is of the form pqrs, which when multiplied by N, gives a number of the form pqrspqrs.

This is possible if N = 10001.

However, $10001=73\times 137$. Thus N has to be a multiple of 73 or 137. Answer is option (d).

Total number of solutions = 11+8+5+1 = 25

$\frac{7}{8}$ remainder = -1

Thus, any odd power of 7, when divided by 8, will give a remainder of -1 itself.

$\frac{(-1{)}^7+(-1{)}^{77}+(-1{)}^{777}......}{8}$

= -1-1-1-...8 times = - 8

The remainder when -8 is divided by 8, is 0.

The smallest radius means the smallest circumference.

Circumference = 2$\pi$r

With six points equidistant with distance between them to be integer, this integer has to be smallest i.e. 1

Circumference = 6 units

2$\pi$r = 6

r = $\frac{3}{\pi }$

So, x = 3

let the side of the cube=x, the height of the cone =h=5 and the radius of the base=2

$\Delta$ADC $\approx$$\Delta$AEF
$\frac{r}{h}=\frac{\frac{x}{2}}{h-x}$
$\Rightarrow$2(5 - x) = 5$\frac{x}{2}$
$\Rightarrow$(10 - 2x) = 2.5x
$\Rightarrow$ x = 2.22 units
 

Selling Price (SP) after discount of $30\%$

$=x(1-\frac{30}{100})=\frac{7x}{10}$

It is known that there is a profit of $20\%$ of SP $\frac{7x}{10}$

$\therefore CP=\frac{7x}{10}\left(\frac{100}{100+20}\right)=\frac{7x}{12}.$

New SP=x-400 and on this new SP there is a loss of $20\%$.

$\Rightarrow \frac{7x}{12}(1-\frac{20}{100})=x-400$

$\Rightarrow \frac{7x}{15}=x-400\Rightarrow 400=x-\frac{7x}{15}$

$\Rightarrow \frac{9x}{15}=400\Rightarrow x=750$

Shortcut- Using the reverse Gear Approach

Always start with a middle answer option

Assuming 750 as the right answer option.

If he gives a discount of 30% then the selling price will be 750 - 225=525, with which he is able to make a profit of $20\%$.

Thus, cost price will be  $=\left(\frac{525}{1.20}\right)=437.5$

Again if he had given a discount of 400 Rs. He would have sold the book at 750 - 400=350 he incurred a loss of $20\%$ or $\left(\frac{350}{0.8}\right)$=437.5 hence that is the correct option.

If two equations have a common root then they can be found by equating them:
When we equate we get $x^2+x(g-6)+(g-3)=0$
Now the Determinant of this equation has to be equal to 0
$(g-6{)}^2-4(1\left)\right(g-3)=0$
We get the values of g= 4 or 12

Re-substituting in the original equation we see that only g=4 will satisfy the equation

i.e You will get the roots to be 1,2 and 1,3 hence 1 is the common root.

$x^2-gx+3=0$ is equation 1.
$2x^2-6x+g=0$ is equation 2.