Free Quant Practice Test - CAT 

$32!=2^{31}{.3}^{14}.5^7{.7}^4{.11}^2{.13}^2.17.19.23.29.31$ (on prime factorization)

So it has $(31+1)(14+1)(7+1)(4+1)(2+1)(2+1)(1+1)(1+1)(1+1)(1+1)(1+1)=2^{13}.3^3{.5}^2$ factors.

Hence, choice (a) is the right answer.

Take the LCM of the numbers 3,4,5,6,7. First number to satisfy the given conditions will be LCM-1 = 420-1 = 419

Numbers will fall in an AP with a common difference = LCM

$2^{nd}$ number = 419+420 = 839

$3^{rd}$= 839+420 = 1259 and so on.

First 5 digit number = 10080-1 = 10079.

No. of ways of selecting 'r' consecutive things from 'n' consecutive things = n - r + 1

So, you can select 3 consecutive numbers in 100 - 3 + 1 = 98 ways.

Now, out of 98 selections: you can get either of the two combinations:

i) even - odd - even: If out of numbers chosen two are even and one is odd, then certainly the product of these numbers will be divisible by 24. No. of ways we can select even-odd-even is $\frac{98}{2}$ = 49

ii) odd - even - odd: The product of two odd numbers and one even number will be divisible by 12 if the even number is  a multiple of 8. We have 12 even multiples of 8 within 100. Hence we can have 12 combinations of odd- even - odd.

So, total number of favourable cases: 49 + 12 = 61

So, probability = $\frac{61}{98}$. Hence option (c)

Suppose the AP is 1,2,3

At n=1, sum =1

4(p)(1)(4)=1 $\Rightarrow$ p =$\frac{1}{16}$

Sum of squares = $1^2$= 1 itself

Look for 1 in the answer options

Only at option b

$\frac{32}{3}\times \frac{1}{256}\times \left(24\right)=1$

Answer is option (b)

Let the number of red and green marbles be a and b respectively.

Let the selling price of each marble be Rs.e.

Let the equal number of marbles not sold the first time = c.

Then, e(a - c) = 25                  ...         (i)

             e(b - c) = 15               ...         (ii)

The number of marbles that remained that remained with the shopkeeper = 2c

e(a - c) + e(b - c) + ec = 45

ec = 5                                               ...(iii) (from (i) and (ii))

Also, (100 - c)e - 45

$\Rightarrow$ 100e - 5 = 45                  (from (iii)

100e = 50

$\Rightarrow$ e = 0.5 = 50 Paise.

Hence, [d].

S = 110+80+120+70 = 380
We need to maximize X.

I = 80.

S = I + 2II + 3III + 4IV = 380.

To maximize X we need to minimize the overlapped regions. Assume III= 0, IV = 0.
So that 2II = 380-80 = 300.  II = 150.
X max = 80 + 150 = 230.

n th term is of the form $\frac{1}{n}\times \sum n$

The series is of the term $\frac{2}{2}+\frac{3}{2}+\frac{4}{2}+\frac{5}{2}.....\frac{20}{2}$

Thus, the sum is $\frac{1}{2}\times \sum n-1=\frac{1}{2}\times \left(\frac{20\times 21}{2}\right)-1=104.$

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 the numbers be x,y, x+y.

Therefore xy(x+y) = 12(x+y). Hence, xy= 12.

So, (x,y) = (1,12), (2,6) or (3,4).

Hence "n" = 156, 96 or 84.

Sum "s" (of all possible values of n) = 336.

The sum is given by (n-1)!$\times$(sum of all the digits) $\times$ (1111...n times)

Here n=5

Sum = 4! *(25)*(11111) = 6666600 option (c)

$\Rightarrow$${(\left[\frac{1}{p}\right]+p)}^2$ = 9
$\Rightarrow$$(\left[\frac{1}{p^2}\right]+p^2+2)$ = 9
$\Rightarrow$$(\left[\frac{1}{p^2}\right]+p^2)$ = 7
Circumference = 2$\pi$r $\approx$ 22

$\frac{(a+b)}{(a-b)}=-\frac{(a+b)}{(a-b)}$

For $1\le n\le x-1,y+n=x!+n+1$, which is divisible by n+1 always.

Therefore, there is no prime number in this list.

First $(5a-1)<(a+1{)}^2\Rightarrow a^2+1+2a>5a-1\Rightarrow (a-2\left)\right(a-1)>0\Rightarrow a<1;a>2$

Second $(a+1)2<7a-3\Rightarrow (a-4)(a-1)<0\Rightarrow 1<a<4$

So in this range its only a = 3 which satisfies the inequality. Therefore, no.of values of a satisfying the inequality is 1.

It is given that the number has 11 digits which add up to 2

When the first digit is 1,

The number can be formed from 9 zeroes and 2 ones which can be arranged in $\frac{10!}{9!}\times$1! = 10 ways

When the first digit is 2, we can have 1 more number. Total cases = 11. Answer is option (a)

Assume the side of the cube =5

Length of body diagonal = QR = 5$\sqrt{3}$

PQ= 5$\sqrt{2}$

QR-PQ $\approx$ 1.5

PR= 5

$\Rightarrow$Required percentage = $\frac{150}{5}$ = 30%

m(a + (m-1)d) = n(a + (n-1)d)
am + (m-1)dm = an + (n-1)nd
a(m-n) + d[m(m-1) - n(n-1)] = 0
(m-n)(a + d(m+n-1)) = 0
m is not equal to n.
a + d(m+n-1) = 0
(m+n) term is zero.

$40(1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}....)=13.33$

They cover a distance of 750 m for the first time they meet, and subsequently cover a distance of 1500 m each time they meet, hence

$1500\times 8+750=12750$ , which is lesser than 13,333.33.

Hence they meet 9 times.

The number of zeros in the end of a factorial depends on the number of 10s; i.e. effectively, on the number of 5s (since, $10=5\times 2$).

10! is $1\times 2\times 3\times 4\times \left(5\right)\times 6\times 7\times 8\times 9\times (2\times 5)$. The number of zeros at the end of 10!=2

(This is because the highest power of 5 till 10! is 2 (as there are 2 fives))

Continuing like this, 10!-14!, the highest power of 5 will be 2. The next 5 will be obtained at 15 = (5*3).

Therefore, from 15! To 19! - The highest power of 5 will be 3.

Similarly, from 20!-24! - Highest Power = 4

In 25, we are getting one extra five, as 25=5*5. Therefore, 25! to 29!, we will get the highest power of 5 as 6.
The answer to the question is, therefore, 0. There are no natural numbers whose factorials end with 5 zeroes.

A cubic equation can be expressed in terms of its roots as

$x^3$-(sum of the roots) $x^2$+(sum of the product of the roots taken 2 at a time) x-(product of the roots)=0

thus,

p= sum of roots taken one at a time

q= sum of roots taken two at a time

Product of roots =8 (given in question)

 'if the product of 2 or more numbers is constant, the sum will be minimum when they are equal'

i.e.If abc=constant, the minimum value of a+b+c will be obtained at a=b=c

Thus, For the sum of roots (p) to be minimum, 8 should be the product of 3 numbers that are equal to each other (difference between them should be equal). This number is 2. The roots are thus 2,2,2

P= 2+2+2= 6

And q= 2(2)+2(2)+2(2)= 12

Answer= option b

If you join the mid-points of a triangle, you form 4 smaller equilateral triangles as follows

Thus, the shaded triangle (second triangle in the question) has an area exactly $\frac{1}{4}$ of the area of the outer triangle. This process will continue and the next triangle which will be formed from the midpoints of the second triangle will have an area =$\frac{1}{4}$ of the second triangle

Thus this forms an infinite Geometric progression where the common ratio is $\frac{1}{4}$ and the initial term (Area of first triangle)=24

$S=\frac{a}{1-r}=\frac{24}{1-\frac{1}{4}}=\frac{24\times 4}{3}=32$ units. Answer is option (c)

The question is independent of n, which is shown below.

Take n=1. Then S= {2,3}. X=3 and Y=2. X-Y=1

Also, the number of odd and even numbers in the set are equal. If we take two consecutive numbers at a time, their difference will be 1.
The sum of all such differences, when divided by the no. of odd or even numbers will yield the result as 1.

Hence you can directly mark the answer as option (a). 

Method 1: Conventional Method

The traditional way to solve this question is by using equations

Let the number of students in the class=x

Then, if "y” is the number of students who are selected for the project, then $y=\sqrt{\frac{x}{2}}$

also, given that  $x-\sqrt{\frac{x}{2}}-2=\frac{8}{9}x$

Replace x by $a^2$

Then the equation changes to  $a^2-\frac{a}{\sqrt{2}}-2=\frac{8}{9}{a}^2$

Solving, we get x=72

Final Value =  Initial Value ${(1+\frac{r}{100})}^t$
Thus, Rate $=[{\left(\frac{FV}{IV}\right)}^{\frac{1}{t}}-1]\times 100$
$\Rightarrow \frac{FV}{IV}=\frac{48}{36}=1.33$

Going from answer options -
If 10 is the answer -
${1.1}^3=1.331\approx 1.33=\frac{FV}{IV}$
Therefore, rate is approximately 10%.

Mark a particular face X. (Later on, we will have to divide whatever answer we get by 6 because all 6 faces are identical at this point, and we have chosen one out of 6 randomly).

Put a colour on face X. There are 6 ways to do this.

Put another colour on the face opposite X. There are 5 ways to do this.

Arrange the remaining 4 colours in the remaining four places - there are 3! ways to do this (circular permutation of 4 colours = (4-1)!

Total = $\frac{\left[\right(6\times 5\times 3!\left)\right]}{6}$ = 30

Non negative integral solutions means $x_1,x_2,x_3\ge 0$

This is also a "similar to different” type of question.

Applying 0's and 1's method:

Total no. of ways $=\frac{17!}{2!15!}.$ Hence option (c) is the correct answer.

Use substitution method,

Put p= 5. Then LHS$>$RHS. Thus, the inequality is not satisfied for this value.

Answer can never be option (a) or (b)

Put p=2. Then LHS>RHS. Thus, the inequality is not satisfied for this value as well. Answer is option (d).

7- 9x is a decreasing function, while 3x +4 is an increasing function. The max of f(x) occurs
at the intersection of 7 - 9x and 3x + 4.
7- 9x = 3x + 4 $\to$ x = $\frac{1}{4}$
So, max f(x) = min $(\frac{19}{4},\frac{19}{4})$ = $\frac{19}{4}$

For remainder when a number is divided by 10, we need to check only the last digit.

We can write 33=4k+1. The first digit in the cycle of 3 = 3 (the cycle of 3 is 3,9,7,1)

${123}^{33}={123}^{4k+1}=...3$. Thus,  the last digit of this number will be 3.

Solution:  Area of the entire region where the helicopter can crash = (5.5$\times$10)km${}^2$ =55 km${}^2$

Area of the lake = (3$\times$3.5) km${}^2$  =10.5 km${}^2$

Probability that helicopter crashed inside the lake = $\frac{10.5}{55}=\frac{21}{110}$. 
So, P = 21

Solution: 
Equate the amount of work in terms of man - hours and volume of the platform.

21 workers, 5 hours a day, 30 days = $150\times 20\times 15$

24 workers, 7 hours a day, N days = $720\times 15\times 10$

Taking the ratio ,$\frac{21\times 5\times 30}{24\times 7\times N}=\frac{150\times 20\times 15}{720\times 15\times 10}$

N =45