Free Quant Practice Test - CAT 

Double substitution:
Step 1- Rewrite the answer options in terms of n (here, n=20)
$a\left)\frac{3^n-1}{2}b\right)\frac{3^{n+1}-1}{2}c)2^{n+1}-1d)3^n$
Let the single organism be called $A_0$. At t=2, the number of new microorganisms present would look as follows:

Thus, at t=2, there are 9 microorganisms alive. Put t=2 in the answer options, and check where you are not getting 9. Eliminate those options.
Only option (d) gives 9. Answer is option (d).

To find the tens place, we need to find the remainder when this number is divided by 100.
$\frac{7^{281}}{100}\mid r$$=\frac{\left(\right(7^4{)}^{70}\times 7^1)}{100}\mid r$$=\frac{\left(\right(..49\times ..49{)}^{70}\times 7^1)}{100}\mid r=\frac{\left(\right(..01{)}^{70}\times 07)}{100}\mid r=..07$
Similarly, $\frac{3^{264}}{100}\mid r$$=\frac{(3^4{)}^{66}}{100}\mid r$$=\frac{(3^2\times 3^2{)}^{66}}{100}\mid r$$=\frac{(09\times 09{)}^{66}}{100}\mid r$$=\frac{{81}^{66}}{100}\mid r=..81$
(66 is of the form 5n+1, and the tens digit of ${81}^n$ repeats after a cycle of 5)

Answer is the tens digit of $07\times 81=..6.$

Now, AO=4 & AOX=45${}^{\circ }$
$\Rightarrow$ AX = XO = 2$\sqrt{2}$
Thus, XY= 4 - 2$\sqrt{2}$ and the radius of the circle = $\frac{4-2\sqrt{2}}{2}$ = 2 - $\sqrt{2}$

Hence, option (d).

When he sells the mixture at Rs. 30/kg, he makes a profit of $20\%$ . This means the cost price of that mixture is $\frac{30}{1.2}$ = Rs. 25/kg.

Combine the variety 1 and variety 3 first to get a mixture with price Rs. 25/kg

Applying Alligation:

So, variety 1 $\&$ variety 3 are required in the ratio 1 : 1.

Now, combine variety 2 and variety 3 to get a mixture with price Rs. 25/kg

 Variety 1: Variety 3 = 1: 1

 Variety 2: Variety 3 = 5: 1

So, the required ratio in which all three have to be mixed = 1: 5: 2

If third variety is 2 kgs then second variety has to be 5 kg. Hence option (a)   

Number of those who opted for cooking which is more than the number of those who opted for exactly two of the three hobbies which in turn is more than those who opted for all three hobbies.

To maximize the number of girls who attend cooking classes, we only need to compare the girls who attend exactly two and who attend exactly three as all these will come under the individual subject.

So, assign 49 girls to all the three subjects and 51 girls in the exactly two. Also, 51 should be divided as equally as possible:

So, number of girls attending cooking class = 49 + 17 + 16 = 82. 

We know that number of girls attending stitching classes is maximum. Now to assign the minimum value to this, we need to minimize the number of girls attending exactly two and exactly three classes, and then distribute the rest of the numbers as equally as possible.

So, number girls attending stitching class = 32 + 2 + 1 = 35. Hence option (d)

To determine the maximum number of girls who opt for only Music Class, we need to maximize the number of girls attending Stitching as well as Music, and we will have to minimize all the other values. Thus,

So, number girls attending only music class = 48.Hence option (a)

$T_1=2^{88}$
$T_2=2^{88}-2^{87}=2^{87}(2-1)=2^{87}$
$T_3=2^{88}-2^{87}-2^{86}=2^{86}(2^2-2^1-1)=2^{86}$
Thus, the pattern is
$T_n=2^{89-n}$
Answer is $T_{34}=2^{55}$

Assumption

Assume a=0, b=0

g(0)=0+g(g(0))=0

Next assume a=1 and b=0

g(1)=1+g(g(0))=1+0=1

continue in the same way, assume a=2, b=0

g(2)=2+g(g(0))=2

Hence g(x)= x itself. Hence g(102)=102

The sum of n consecutive odd numbers is given as $n^2$ . Hence ${100}^2$ is divisible by 2, 4 and 8. Answer is option (a).

(Sum of n consecutive odd numbers $=n^2$ is derived from unitary method as follows : 1 + 3 + 5 + 7 + 9 +...... is a series (AP) with a common difference of 2. Therefore, the 100th term would be 99$\times$2+1= 199. Therefore the sum = $\frac{(1+199)100}{2}=10000$ or $n^2$.

$3B+5G=33.33\%..............\left(i\right)$

Time taken by 3 women $=\frac{100}{3G}$        ;      Time taken by 2 men $=\frac{100}{2B}$

$\frac{100}{3G}-\frac{100}{2B}=5\to \frac{33.33}{G}-\frac{50}{B}=5.................\left(ii\right)$

Option (a): B=2G $\to$  11G = 33.33$\%$ $\to$  G = 3.03 , B = 6.06. Substitute in (ii),  11 - (approx 8) = approx 3.

Option (b): B=2.5G $\to$  12.5G = 33.33$\%$ $\to$  G = $\frac{8}{3}$ , B=$\frac{20}{3}$, Substitute in (ii), 12.5 - 7.5 = 5  which is the difference given in the question.Hence option (b).

$3^4{5}^6{7}^3=3^4{35}^3{5}^3$

The question states that the given expression will be written as ${35}^{a}b$ where a and b are both positive integers, i.e. a can take values 1,2 or 3. For each of these values, b will have three different values. Hence answer is option (c).

Average speed of Savita over the complete journey = $\frac{180}{10}$ = 18 kmph.

This means auto took 2/5 of the total time and bus took 3/5 of the total time. So, the ratio of the time taken (bus: auto) = 3: 2.
So, ratio of distance covered = $3\times 20:2\times 15$ = 60 : 30 = 2 : 1.
Hence option (d) 

Thus 1-1=0. Remainder when ${10}^{25}-7$ is divided by 3=0.

The image can be redrawn as follows, by joining all the points

Thus, there are 24 unit squares. The number of unit squares shaded = 6. Required ratio = $\frac{1}{4}$ 

P(A) = $\frac{1}{3}$
P(A') = $\frac{2}{3}$
P(Bird Killed) = 1 - P(None of 3 shots hit)
$1-$ $\frac{2}{3}$ $\times$ $\frac{2}{3}$ $\times$ $\frac{2}{3}$ = $\frac{19}{27}$

$\angle$ABE = $\angle$BEC (alternate angles)

$\angle$AED = $\angle$EAB

$\angle$AED = $\angle$BEC

$\angle$EAB = $\angle$EBA

As BE bisects $\angle$ABC

$\angle$EBC = $\angle$ABE

From (1) and (2) as the two angles of the triangle ABE are equal to two angles of triangle BEC, the two triangles are similar

So, $\frac{BE}{CE}=\frac{AB}{BE}$

BE${}^2$ = (CE + ED)CE

Substituting BE = 12cm and CE = 9cm, we get ED = 7cm

Since Jay does $(\frac{1}{30}{)}^{th}$ of the work in one day, total work done by Jay in 18 days

 $=\frac{18}{30}=\frac{3}{5}$

Remaining $(\frac{2}{5}{)}^{th}$ work was done by Anup

Since Anup does $(\frac{1}{15}{)}^{th}$ of the work = 2/5/(1/15) $\frac{\left(\frac{2}{5}\right)}{\left(\frac{1}{15}\right)}$ = 6 days.

So, Anup joined Jay after (18−6) = 12 days.

There is no constraint given, other than that ABC is an equilateral triangle and AX∥BY∥CZ.

We can therefore assume that X,Y and Z are points on the side AC itself. If we assume the equilateral triangle to have a side=2, and the Y is the mid point of AC, then AY=YB=1

Area of the triangle

Look in the answer options and eliminate those options where you do not get $\sqrt{3}$ on substituting a=1 and b=1

$\frac{\sqrt{3}}{4}{a}^2=\sqrt{3}$

Answer is option (b)

There are 4 rows and 5 columns. Using the one-zero method, number of paths will be ${}^9$C${}_4$ = 126 ways

No. of 4-digit numbers in base 7 = 6$\times$7$\times$7$\times$7 = 2058.

At n=1,

Option a $=\frac{1}{\mathrm{3.2.3.4}}....$This can never be the answer

Option b $=\frac{1}{\mathrm{6.3.4.5}}....$This can also never be the answer

Option C $=\frac{1}{\mathrm{4.1.2.3}}....$ This can be the answer

Since an option "none of these" is there , do another check

At n=2, we should get $\frac{1}{24}+\frac{1}{120}$

Put n=2, in option c$=\frac{1}{\mathrm{4.3.2.4}}=\frac{1}{96}\ne \frac{11}{120}$  

Hence answer = option d.

Always be careful when there is an option none of these. 

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

x = 6, x = 2.

Alternatively :

Assume k=1; the solution becomes simpler.
Value is currently 64 and it becomes ${\frac{5}{8}}^{th}$ of that which is 40 =36+4 (This is the only possible breakup with 2 square numbers) hence, answer is option (c).

$\frac{{{27}^{16}}^3}{4}=\frac{{27}^{evennumber}}{4}=\frac{(-1{)}^{evennumber}}{4}=1$

Now the question changes to finding the unit digit of $7^{4n+1}$. From the power cycle of 7, unit digit will be 7. Hence, the remainder is 7.

Thus, the maximum number of boxes which can be added will very close to and slightly less than 8. Answer is 7 boxes

b + c = 150 - 50 = 100.

a + d = 50 (those who do not play football)

a + b = 90, d = 20

$\Rightarrow$a + 20 = 50 ⇒a = 30

b = 90 - 30 = 60, c = 100 - 60 = 40.

Option (d)
Side length of Beta cube = 2 cm
Volume of Beta cube = (2)${}^3$ = 8 cm${}^3$ .........-(1)
Diagonal length of Beta cube = side length of Alpha cube = 2$\sqrt{3}$ cm
Side length of Gamma cube = diagonal length of Alpha cube = (2 $\sqrt{3}$)$\sqrt{3}$ = 6 cm.
Diameter of sphere inside Gamma cube = Length of edge = 6 cm.
Radius of sphere =$\frac{6}{2}$ = 3 cm
Volume of sphere S
= $\left(\frac{4}{3}\right)\times \pi \times (3{)}^3$ = 36$\pi$.........(2)
Required ratio = S : q = 36$\pi$ : 8 = 9$\pi$ : 2 

Answer = sum of all 3. option- d