Free Quant Practice Test - CAT 

Conventional Method Join PO, OQ and OR

There are 3 triangles whose sum of areas = area of the equilateral triangle

Area of a triangle $=\frac{1}{2}\times$ base $\times$ height

Thus Area of $△POR=\frac{1}{2}\times PR\times OM$

Area of $△QOR=\frac{1}{2}\times QR\times ON$

Area of $△POQ=\frac{1}{2}\times PQ\times OL$

Sum = Area of $△POR+$Area of $△QOR+$Area of $△POQ=$ Area of $△PQR$

$=\frac{1}{2}\times PR\times OM+\frac{1}{2}\times QR\times ON+\frac{1}{2}\times PQ\times OL$

Area of an equilateral triangle $=\frac{\sqrt{3}}{4}\times a^2$ , where 'a' is the length of 1 side
$\frac{\sqrt{3}}{4}\times 5^2=\frac{1}{2}\times PR\times (OM+OL+ON)$ (Since, PR = QR = PQ)
$⟹25\frac{\sqrt{3}}{4}=\frac{1}{2}\times 5\times (OM+OL+OM)$
$⟹OM+ON+OL=5\frac{\sqrt{3}}{2}$

Shortcut:- Assumption

Note the most important word in the question-ANY. We can assume ANY point inside the triangle. We will consider a point infinitely close to P.

Thus, the diagram will look as follows.

The answer can now be obtained in a single step.
OL = 0, OM =0 and ON is the height of an equilateral triangle $=5\frac{\sqrt{3}}{2}$
Thus, $OL+OM+ON=5\frac{\sqrt{3}}{2}$

Assumption Technique

This is a question with variables. Thus, we can assume any value which satisfies the basic conditions for x,y and z

Assume any Geometric Progession, say  1        2          4          8 .

This implies that the 2 Geometric Means between 1 and 8 are

y=2 and z=4.

Now, Arithmetic Mean, x = \(\frac {8 + 1}{2} = 4.5 .Substitutethe values for x,y,z  in $\frac{y^3+z^3}{xyz}$

Answer = $\frac{2^3+4^3}{2\times 4\times 4.5}$ = 2.

Shortcut of a shortcut!

You can assume ANY Geometric Progression.

 Assume the most basic GP 1    1    1    1  and the most basic   AP                1             1              1,

then x=1, y=1 z=1. Answer on substitution in $\frac{y^3+z^3}{xyz}$=2, which will make the calculation even faster.

Conventional Approach

To write a number as a sum of two or more consecutive positive integers, we need to form an AP with "n” terms with a common difference 1

Sum to "n” terms of an AP= $\frac{n}{2}$(2a + (n - 1)d) where d=1 (as the numbers are consecutive) n(2a+(n-1))=200

thus, we need to write 200 as a product of 2 numbers, where one number is odd and the other is even

this can be done in 2 ways

200= 5$\times$40

200=25$\times$8

Thus 100 can be expressed in 2 ways as a sum of two or more consecutive positive integers

Shortcut:-

The answer is just one step. The number of ways of writing any number as a sum of two or more

consecutive positive integers= number of odd factors of that number-1

100 = 2${}^2$$\times$5${}^2$ Number of odd factors=3

Number of ways of expressing 100 as a sum of two or more consecutive positive integers= 3 - 1=2

Conventional Method

From the data given in the question

$S_3=16=\frac{a(r^3-1)}{r-1}$................................(1)
$S_6=16+128=144=\frac{a(r^6-1)}{r-1}=\frac{a(r^3-1)(r^3+1)}{r-1}$................................(2)
9 = (r${}^3$ + 1)
Hence r =2
Substituting in equation 1, we get
a = $\frac{16}{7}$
We know that, 
$S_n=\frac{a(r^n-1)}{r-1}$
Substituting the values of a and r found above, we get
$S_n=\frac{16}{7}(2^n-1)$
Thus, answer is option (b) 
Shortcut-
Substitution The answerto this question is in the question itself!
At n=3, the sum should be 16
Hence, in the correct answer option, at n=3, we should get 16
Option (a)= $\frac{16}{7}$(2${}^3$ + 1)$\ne$16
Option(b)=$\frac{16}{7}$(2${}^3$ - 1) = 16
Option (c)=$\frac{9}{7}$(3${}^3$ - 1)$\ne$16
Option (d)=$\frac{16}{7}$(3${}^3$ - 1)$\ne$16
Options (a), (c) and (d) can never be the answer. Answer is option (b) 

Shortcut of last two digits:

When the question is about finding the remainder when you divide a number by 100, what we actually need to find out is the last two digits, for which we have a direct shortcut. For an indepth explanation, please refer the demo tutorial

$2^{87}=(2^{10}{)}^8\times 2^7$

$(2^{10}{)}^{evevpower}willalwaysendin..76$

$2^7$ ends in ...28

Therefore 76x...28=28

The hexagon can be graphically divided as follows

Hence, there are totally 54 equilateral triangles out of which 24 are unshaded. Each of the triangles have a  side 2. Total unshaded area = $\frac{\sqrt{3}}{4}$ x 4 x 24 = 24$\sqrt{3}$

Let f(x) =x${}^9$ + 5x${}^8$ - x${}^3$ + 7x+3 = 0

We can solve this question easily using Descartes' Rule

According to Descartes' rule maximum number of positive real roots = number of sign

changes in f(x) = 2

Similarly, Maximum number of negative real roots = number of sign changes in f(-x)

Note: To find f(-x) replace "x” by "-x” in each instance

therefore f(-x) = - x${}^9$ + 5x${}^8$ + x${}^3$ - 7x+3

Maximum number of negative real roots = number of sign changes in f(-x) = 3

Zero cannot be a root because constant part is also involved in equation.

So maximum number of real roots = 2 + 3 = 5


DESCARTES' RULE (Points to Remember)
Maximum number of positive real roots = number of sign changes in f(x)
Maximum number of negative real roots = number of sign changes in f(-x)
If a constant term is present, do not consider 0 as a root or else, you need to consider it

In this case, the minimum of all is that part of the Venn diagram which has 5 overlapping regions. This can be found using a shortcut technique.

Here X=100,

Step 1- Taking the difference of each from X.

100-90=10

100-85=15

100-70=30

100-75=25

100-85=15

Total (Y) $\sum$ = 95% (Percentage of fans who do not support atleast 1 country)

Step 2- Take the difference of Y from 100, 100-95 = 5%

Minimum percentage of fans who support all 5 countries = 5%

The correct answer can be marked in 30 seconds if you know that this is an application of "number of terms in an Arithmetic Progression”. If we distribute 500 consecutive numbers with a common difference of 9, number of terms can only be $\frac{500}{9}$ = 55 or (55+1). You can directly mark option (c)

Why a common difference of 9? Because the first number with a property of its digit sum=5 is 05, the next term is 14 (1+4=5), the next is 23(2+3=5) and so on

The difference between any two consecutive terms with this property is 9.

Thus, the above question is nothing but finding the number of terms in an Arithmetic Progression with the terms as 5,14.... 500. 

Notice another interesting thing

When 5 is divided by 9,the remainder is 5

When 14 is divided by 9, the remainder is 5

When 23 is divided by 9, the remainder is 5

3 different ways of asking the same question

1) How many numbers less than or equal to 500 will leave a remainder of 5 when divided by 9?

2) Find the number of terms in the AP 5,14,23,32...........500 ?

3) If B = sum of digits of A, C = sum of digits of B, .........X = X itself. If X = 5 find the different possible values of A given that 0$<$A $<$500

This question can be visualized as a similar to different question in P$\&$C (refer e booklet for indepth explanation of concept)

Number of columns from A to B= 3 and rows= 2. Answer is based on 3 zeros and 2 ones =${}^5{C}_2=10$ ways.

Similarly, number of columns from B to C= 4 and number of rows= 2. Answer is based on 4 zeroes and 2 ones= ${}^6{C}_4$ =15. Total number of ways = 15$\times$10=150. Option (d)

Method:

Let the distance between home and office =d

Suppose he reaches the office on time, the time taken = x minutes

Case 1: When he reaches office 20 minutes late, time taken = x+20

Case 2: when he reaches office 10 minutes early, time taken = x-10

As the distance traveled is the same, the ratio of speed in case 1 to the speed in case 2 will be the inverse of the time taken in both cases.

To understand this better ,Let us call the horizontal distances as H and the vertical distances as V.

The person needs to first go to B.

Number of horizontal distances = Number of rows =2.

Number of vertical distances =Number of columns =3.

i.e.., HHVVV.

Now the question changes to the number of ways of arranging HHVVV=$\frac{5!}{3!2!}{=}^5{C}_2=10$ ways.

Similarly, from B to C, the number of distances =HHVVVV. Number of arrangements=$\frac{6!}{4!2!}=15$ ways .Total number of ways =$10\times 15=150.$

Ratio of speed in both cases = 4:6 = 2:3

Ratio of time in both cases = 3:2

Therefore,$\frac{x+20}{x-10}=\frac{3}{2}$

2x+40=30x-30$\to$ x=70 minutes.

Taking case 1: $\frac{4}{d}=\frac{60}{90}(90=x+20)d=\frac{360}{60}=6km.$

Shortcut- Constant Product Approach

Assume original speed= 4km/hr.

Percentage increase in speed from 4 kmph to 6 kmph= $50\%or\frac{1}{2}$

$\frac{1}{2}$ increase in speed will result in $\frac{1}{3}$ decrease in original time (from Constant Product Rule)

=30 minutes.(from given data). Thus, Original time= 90 minutes= 1.5 hours 

Answer is Distance = 4 $\times$ 1.5=6 km

In case, you want to break down the shortcut in steps, you should proceed like this:

There are 2 cases here,

Case 1 $\Rightarrow$ speed = 4 km/hr ,              20 imn late.

Case 2 $\Rightarrow$ speed = 6 km/hr ,              10 min early.

Step 1: Calculate the increase or decrease in the parameters speed /time.

To calculate increase / decrease we need to consider a case as original case and then calculate the increase or decrease over it.

We will take the first case as the original case,so we can easily observe that the increase in speed s $50\%$ or $\frac{1}{2}$ from case 1 to case 2 $(\times 100=50\%)$

Step 2: Inverse Proportionality.

Now based on inverse proportionality there will always be a or $33\%$ or decrease in time from case 1 and case 2.

Step 3: Calculating the actual time and distance.

Suppose in the case 1 which we have taken as original, time taken is 'x' mins.

So due,to $\frac{1}{2}$ increase in speed we have decrease in time of 'x' =30 mins.

(form 20 minutes late to 10 mins early there is a decrease of 30 minutes.)

x=90 minutes.

So in the case 1, he is taking 90 mins to travel at a speed of 4 km/hr.So,distance =$1.5\times 4=6$ km.


The explanation may be long, but when you actually apply it; you can solve time consuming questions in very little time,helping you save precious minutes which you can use in your weaker areas ! 

Conventional method

Replacing f(x) with y, the given function is

y= (4-(x-7)³)^1/5

To find the inverse function, we need to write an equation for x in terms of y

Raising both sides to the power 5, we get

y⁵ = 4 - (x-7)³

 (x-7)³ = 4- y⁵

Taking cube root on both sides

x-7 = (4 - y⁵)^1/3

x = 7 + (4 - y⁵)^1/3

We have an expression for x interms of y. We get f^-1(x) if we replace y with x in the above equation

Hence, f^-1(x) = 7 + (4 - x⁵)^1/3. Answer option (a)

Shortcut

What is the basic definition of an inverse function?

An Inverse function is one which can be expressed as f(y) = x when f(x) = y

Using, this very definition, we can use a common sense approach to arrive at the answer In no time at all

1) Substitute a suitable value for x (it can be any value). in this case, for our convenience we will take x=7. Then at x=7, f(x)= 4^1//5

2) Now use the definition of inverse function. Put x=4^1/5 in each of the answer options and see where you get f(x)=7.  You are just exchanging "x” and "y” values!

This is the concept of inverse function after all!

Only option (a) gives 7 at x=4^1/5

All other options can be eliminated as for x=4^1/5, we do not get 7. Answer is option (a)

Let x dl of pure milk be added. The concentration of pure milk is 100%

Total milk content in x dl of milk = x * 100% = x dl

Total milk content in 800 dl of 15% mixture = 800 * 15% = 120 dl

Therefore, total milk content in final mixture = x + 120

Total volume of mixture = 800 + x

Hence concentration of milk in final mixture = $\frac{x+120}{800+x}$

This is given to be 32%.

Hence $\frac{x+120}{800+x}$ = 32 %

Solving, we get x = 200 dl.

Shortcut- Using Alligation

There may be different ways to solve this question, but one of the fastest approaches is using the concept of Alligation. Pure milk has 100% concentration, and it is mixed with a solution with 15% milk to make it 32%. This information can be used to get the ratio of the mixture as follows

For every 4 parts of 15% milk, 1 parts of 100% milk needs to be added.

Hence to 800 dl of the 15% solution, $\frac{800}{4}$ = 200 dl of 100% milk needs to be added

Let us take some simple numbers to see if there is a pattern

If you take a number, say 6, In the first round it will be on, 2${}^{nd}$ it will be off, 3${}^{rd}$ it will be on, and 6${}^{th}$ it will be off. Hence its final state will be off.

Consider a number 4-In the first, second and fourth round it will be ON, OFF & finally ON

Consider a number 9. in the first, third and 9${}^{th}$ round it will change its state and its final state will be on

From this, you can deduce that the state of the bulb is determined by the property of the number; i.e. all numbers with odd number of factors will remain in an "on” state and all other numbers will be off. Conclusion- All the bulbs which are on are "perfect square” numbered bulbs

Thus, the bulbs which will be on are all in the positions of numbers which are perfect squares.

The question is thus, asking for the number of perfect squares till 1000= 31 (i.e. 1${}^2$,2${}^2$...........31${}^2$)

Conventional method

Volume of a cylinder, V = $\pi$r${}^2$h ( r = radius and h = height)

Given that r+h=6. Hence, h = 6-r

Hence, V = $\pi$r${}^2$(6-r)

V = 6$\pi$r${}^2$ - $\pi$r${}^3$

For maximizing volume, we need to differentiate V with respect to r and equate it to 0.

$\frac{dv}{dr}$ = 0

$\frac{dv}{dr}$ = 12pr - 3$\pi$r${}^2$ = 0

Hence, r = 4

R + h is given as 6, hence h = 2

V = $\pi$r${}^2$h = $\pi$$\times$4${}^2$$\times$2 = 32$\pi$

Alternate Method:

Volume of a cylinder = $\pi$r${}^2$h ( r =radius and h= height)

Given that r+h=6

To maximize $\pi$r${}^2$h, we need to maximize r${}^2$h which happens when $\frac{r}{2}$ = $\frac{h}{1}$,

$\Rightarrow$ r = 4 and h =2

Maximum volume = $\pi$$\times$4${}^2$$\times$2 = 32$\pi$

Points to remember

2. If ab=constant, the minimum value of a+b will be obtained at a=b

3. If a+b+c is a constant, then a${}^m$. b${}^n$ .c${}^p$ is maximum when $\frac{a}{m}$ = $\frac{b}{n}$ = $\frac{c}{p}$  (the above question is an example of this)

Numbers divisible by 2 = 50

Numbers divisible by 3 = 33

Numbers divisible by 5 = 20
Numbers divisible by 6 (LCM of 2 & 3) = 16
Numbers divisible by 10 (LCM of 2 & 5) = 10
Numbers divisible by 15 (LCM of 3 & 5) = 6
Numbers divisible by 30 (LCM of 2, 3 & 5) = 3

Numbers divisible by 2, 3 & 5 combined = 50 + 33 + 20 - (16 + 10 + 6) + 3 = 103 - 32 + 3 = 106 - 32 = 74
Numbers not divisible by 2, 3 & 5 = 100 - 74 = 26

To select a team of 15, you have to select almost 3 all rounders.
Therefore the required number of ways
                                     = ${}^7$C${}_5$ $\times$ ${}^7$C${}_6$ $\times$ ${}^1$C${}_1$ $\times$ ${}^5$C${}_3$ 
                                     =1470. 

Since rule says that = $\frac{a}{sinA}=\frac{b}{sinB}=\frac{c}{sinC}=k$
Therefore, a=k(sinA), b=k(sinB) and c=k(sinC)
We can write $\frac{cosA}{a}=\frac{cosB}{b}=\frac{cosC}{c}$ as
$\frac{cosA}{ksinA}=\frac{cosB}{ksinB}=\frac{cosC}{ksinC}$ = cot A=cot B=cot C 
$\Rightarrow$ A=B=C(Equilateral Triangle). 

In one hour pipe A can fill $\frac{100}{20}$ = 5% of the tank

In one hour pipe B can empty $\frac{100}{25}$ = 4% of the tank

(A+B)'s work = 5-4 = 1%

To fill a one fourth of empty tank, it will take = $\frac{25}{1}$ = 25 hours

Distance travelled by Karnataka express till 4 am = 9*60 = 540 kmph

      Distance remaining = (800-540)km = 260 km

      Relative speed = (60+90)kmph = 150 kmph

Time required to travel This distance = $\frac{260}{150}$ * 60 = 104 minutes = 1 hr 44 mins

Time at which train meets = 4 am + 1 hr 44 mins

                                      = 5.44 am.

The numbers can be arranged in ascending order min $<$ median $<$ max

We have min + max =13

$Mean=\frac{min+med+max}{3}=median-\frac{12+med}{3}=2$

On solving we get median =9.5.

'abc' has exactly 3 factors, so 'abc' should be square of prime number.
Let the square root of 'abc' be 'p' (where 'p' is prime).

'abcabc' = abc $\times$ 1001 = $p^2\times 13\times 11\times 7$

If 'p' is any prime number other than 13, 11 or 7, then no. of factors = 3 x 2 x 2 x 2 = 24 (no. will be of the form $p^2\times 13\times 11\times 7$)

If 'p' is any prime number among 13, 11 or 7, then no. of factors = 4 x 2 x 2 = 16 (no. will be of the form $7^3\times 13\times 11,ifp=7$)

$\therefore$ No. of factors can be 16 or 24.

Let the smaller number be 7x, then the larger number will be (7x+7) = 7(x+1)

Since x and (x+1) are co-prime, the LCM would be $7\times x\times (x+1)$.

Maximum value of x such that LCM is a three digit number is x=11.

So, smaller number = 77.

Put x=5, 2f(5)+3f(25)=36.

Put x=25, 2f(25)+3f(5)=676.

f(5) = 391.2 

Discount to Nilay =$1-0.8\times 0.4=68\%$
Discount to Nilabh =$1-0.8\times 0.6\times 0.5=76\%$

Let a,b,c be the number of units sold to Vikalp, Nilay, and Nilabh respectively.

So,  0.2a+0.68b+0.76c=0.56(a+b+c)

So, 5c=3.(3a-b),

So,c is multiple of 3.
If,c=3 then (a,b)= (2,1),(3,4), (4,7)
If, c=6 then (a,b)= (4,2), (5,5), (6,8)

Hence 6 solutions.

List all multiples of perfect squares (without repeating any number) and subtract this from 99.

4 - there are 24 multiples of 4 (4,8,12,.....96)

9 - there are 11 multiples, 2 common with 4 (36 and 72) so, add 9 multiples

16 - 0 new multiples

25 - 3 new multiples (25,50,75)

36 - 0 new ones

49 - 2 {49,98}

64 - 0

81 - 0

Total multiples of perfect squares are 38. There are 99 numbers in total. So, there are 61 numbers that are not multiples of perfect squares.

A v/s B:

A has higher $\%$ of alcohol than B, A has to give 200 ml of wine. Considering that A has $60\%$ of strength mixture, B will end up having 800 ml of wine that has $\frac{60\%\times 1+48\%\times 3}{4}=51\%$ of alcohol.

B v/s C:

B has higher $\%$ of alcohol than C, B has to give 200 ml of wine to C now. C will end up having wine that has,

3 sinx +4 cosx + r $\ge$ 0
3 sinx + 4 cosx $\ge$ -r
$5\times (\frac{3}{5}sinx+\frac{4}{5}cosx)\ge -r$
Let $\frac{3}{5}$ = cos A $\Rightarrow$ sin A $\Rightarrow \frac{4}{5}$
5(sin x cos A +sin Acos X)$\ge$ -r
5(sin(X+A)) $\ge$ -r
We have -1 $\le$ sin (angle) $\le$ 1
5 sin(X+A)$\ge$ -5
$r_{min}$ =5.

We know that

$\frac{(x+3y)}{4}=60$ or x+3y =240 and $x>y$

As $0<x<100$, let's start with x=99. When x= 99, y=47.

When x=96, y=48 and so on. x has to be greater than y. Now in the limiting case, when x=60, y=60.

So when x=63, y=59. So x can take the values of 63,66,69...... till 99, a total of 13 values. 

For this condition,
Series is 109,117,125,..........997
a= 109 ,d=8 , n=112
$S_n=\frac{n}{2}[2a+(n-1\left)d\right]$
$=\frac{112}{2}[2\times 109+(112-1\left)8\right]$
      =61936.

Since each box call hold five balls number of ways in which balls; could be distributed so that none is empty are (2. 2, 1) (3, 1, 1)
i.e., (${}^5$C${}_2$ . ${}^3$C${}_2$ . ${}^1$C${}_1$ + ${}^5$C${}_3$ . ${}^2$C${}_1$ . ${}^1$C${}_1$ ) $\times$ 3!
    =(30+20) $\times$ 6 = 300 

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 $\to$ x = 1.