# Free Quant Practice Test - CAT

### Question 1

The number of positive integers which divide (25)! are -

213.33.52

28.32.52

211.32.5

#### SOLUTION

Solution :A

32!=231.314.57.74.112.132.17.19.23.29.31 (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)=213.33.52 factors.

Hence, choice (a) is the right answer.

### Question 2

A number when divided by 3,4,5,6,7 gives remainders 2,3,4,5,6 respectively. The smallest such 5 digit number is?

10019

10076

11079

10079

#### SOLUTION

Solution :D

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

2nd number = 419+420 = 839

3rd= 839+420 = 1259 and so on.

First 5 digit number = 10080-1 = 10079.

### Question 3

If three consecutive numbers are selected randomly from the first 100 natural numbers, then what is the probability that their product of the three numbers is divisible by 24?

3471

6376

6198

4387

#### SOLUTION

Solution :C

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 982 = 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 = 6198. Hence option (c)

### Question 4

The sum of n terms of an AP is 4pn(n+3).The sum of squares of these terms is:

643p2(n3+6n2+12)

323p2(2n3+9n2+13n)

643p2(4n3+12n2+2n)

1283p2(n3+6n2+12.)

#### SOLUTION

Solution :B

Suppose the AP is 1,2,3

At n=1, sum =1

4(p)(1)(4)=1 ⇒ p =116

Sum of squares = 12= 1 itself

Look for 1 in the answer options

Only at option b

323×1256×(24)=1Answer is option (b)

### Question 5

A shopkeeper keeps 100 marbles in two boxes, A and B, such that Box A contains only red and Box B contains only green marbles. Barring an equal number of marbles from each box, he sold all the others and earned Rs. 25 and Rs. 15 from Box A and Box B respectively. Later, he sold half of the remaining marbles. His total earning from all the transactions accumulated to Rs. 45. What is the cost of each marble - red or green.?

40 paise

25 paise

75 paise

50 paise

#### SOLUTION

Solution :D

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

⇒ 100e - 5 = 45 (from (iii)

100e = 50⇒ e = 0.5 = 50 Paise.

Hence, [d].

### Question 6

In a class,110 students like physics, 80 like chemistry, 120 like mathematics and 70 like English. 20 students like only physics, 15 like only chemistry, 30 like only mathematics and 15 like only English. What can be the maximum possible number of students in the class ?

180

210

380

230

#### SOLUTION

Solution :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.

### Question 7

Sum of the series

S=1+12(1+2)+13(1+2+3)+14(1+2+3+4)+

.....upto 19 terms is

110

98

104

116

118

#### SOLUTION

Solution :C

n th term is of the form 1n×∑n

The series is of the term 22+32+42+52.....202

Thus, the sum is 12×∑n−1=12×(20×212)−1=104.

### Question 8

ABCDEF is a regular hexagon, with O as the centre. There is another regular hexagon with 2 of the end points PQ inscribed in the above hexagon (again with centre O). Find the area of the unshaded region, given that FPQC lies on a straight line and FP=CQ=2 and FP = 13 FO.

20√3

24√3

18√3

28√3

#### SOLUTION

Solution :B

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 = √34 x 4 x 24 = 24√3

### Question 9

The product "n” of three positive integers is 6 times their sum. If one of the numbers is the sum of the other two, then the sum of all possible value(s) of "n” is?

336

252

240

192

#### SOLUTION

Solution :A

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.

### Question 10

What is the sum of all 5 digit nos that are formed by using 1,3, 5, 7, 9 taken one at a time?

5555500

66600

6666600

666600

#### SOLUTION

Solution :C

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

Here n=5

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

### Question 11

If three positive real numbers x, y, z satisfy y - x = z - y and xyz = 4, then what is the minimum possible value of y ?

21/3

22/3

21/4

23/4

#### SOLUTION

Solution :B

Condition implies x,y,z is in AP and xyz is const. Hence, y is min when x+y+z is min, that will happen when x=y=z= 22/3

### Question 12

The diameter of a circle X is given by ([1p2]+p2), which of the following best approximates the circumference of circle X?

(Given that [1p] + p = 3)

27

22

30

12

#### SOLUTION

Solution :B

⇒([1p]+p)2 = 9

⇒([1p2]+p2+2) = 9

⇒([1p2]+p2) = 7

Circumference = 2πr ≈ 22

### Question 13

If a ≠ b, and ab ≠ 0, then when a is replaced by 1a and b is replaced by 1b everywhere in the expression (a+b)(a−b), the resulting expression is equal to

−(a+b)(a−b)

(a+b)(a−b)

(a−b)(a+b)

(b−a)(a+b)

#### SOLUTION

Solution :A

(a+b)(a−b)=−(a+b)(a−b)

### Question 14

Let x be a prime number such that x≥23. Given that y= x!+1. The number of primes in the list y+1, y+2, y+3...y+x-1 is?

x-1

2

1

None of these

#### SOLUTION

Solution :D

For 1≤n≤x−1, y+n=x!+n+1, which is divisible by n+1 always.

Therefore, there is no prime number in this list.

### Question 15

The number of integral values of a for which (5a−1)<(a+1)2<7a−3

#### SOLUTION

Solution :First (5a−1)<(a+1)2⇒a2+1+2a>5a−1⇒(a−2)(a−1)>0⇒a<1;a>2

Second (a+1)2<7a−3⇒(a−4)(a−1)<0⇒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.

### Question 16

The sum of 4 consecutive two-digit odd numbers, when divided by 10 becomes a perfect square. Which of the following can possibly be one of these 4 numbers?

23

49

87

75

#### SOLUTION

Solution :C

The sum has to end with a zero. The numbers have to end with 7,9,1,3. Option (d) is out. Go from answer options, answer is option (c).

The numbers are 87, 89, 91, 93 which add up to 360. 36 is a perfect square.

### Question 17

Suppose n is an integer such that the sum of digits of n is 2; and 1010<n<1011. The number of different values of n is?

11

10

9

55

#### SOLUTION

Solution :A

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 10!9!×1! = 10 ways

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

### Question 18

In the cube shown below, the difference in length between line segment *QR* and line segment PQ is approximately what percentage of the distance from *P* to *R*?

10%

20%

30%

40%

50%

#### SOLUTION

Solution :C

Assume the side of the cube =5

Length of body diagonal = QR = 5√3

PQ= 5√2

QR-PQ ≈ 1.5

PR= 5

⇒Required percentage = 1505 = 30%

### Question 19

If m times the mth term of an AP is equal to n times its nth term, find the (m+n)th term of that AP. Given that m is not equal to n.

tm+tn

tm−tn

tm×tn

0

#### SOLUTION

Solution :D

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.

### Question 20

Ajay and Arun start running simultaneously from the diametrically opposite ends of a circular track towards each other at 15km/h and 25km/h respectively. After every 10 minutes their speed reduces to half of their current speeds. If the length of the circular track is 1500 m, how many times will Ajay and Arun meet on the track?

9

11

7

8

#### SOLUTION

Solution :B

40(1+12+14+18....)=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

1500×8+750=12750 , which is lesser than 13,333.33.

Hence they meet 9 times.

### Question 21

How many natural numbers are there such that their factorials are ending with 5 zeroes?

1

2

>2

None of these

#### SOLUTION

Solution :D

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=5×2).

10! is 1×2×3×4×(5)×6×7×8×9×(2×5). 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.

### Question 22

What is the minimum value of p and q in the equation x3−px2+qx−8=0 where p and q are positive real numbers and roots of the equation are real?

4,8

6,12

8,12

Cannot be determined

#### SOLUTION

Solution :B

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

x3-(sum of the roots) x2+(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

### Question 23

The mid points of an equilateral triangle are joined. Forming a second triangle, a third triangle is formed joining the midpoints of the second and the process continued infinitely. If the area of the first triangle is 24 units, then find the sum of areas of all triangles.

26

28

32

36

#### SOLUTION

Solution :C

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 14 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 =14 of the second triangle

Thus this forms an infinite Geometric progression where the common ratio is 14 and the initial term (Area of first triangle)=24

S=a1−r=241−14=24×43=32 units. Answer is option (c)

### Question 24

Consider the set S={2,3,4......2n+1}, where n is a positive integer larger than 2007. Define X as the average of odd integers in S and Y as the average of the even integers in S. What is the value of X-Y?

1

12n

2008

0

#### SOLUTION

Solution :A

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).

### Question 25

A group of students equal in number to the square root of half the whole class is selected for a project. Two more students were selected as captain and vice captain. Rest of the class which form 89 th of class were given leave. Find the total number of students in the class?

64

72

80

81

#### SOLUTION

Solution :B

Method 1:Conventional MethodThe 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=√x2

also, given that x−√x2−2=89x

Replace x by a2

Then the equation changes to a2−a√2−2=89a2

Solving, we get x=72

### Question 26

The population of a town was 3600 three years back. It is 4800 right now. What is the rate of growth of population, if it has been constant over the years and has been compounding annually?

15%

5%

12%

10%

#### SOLUTION

Solution :D

Final Value = Initial Value (1+r100)t

Thus, Rate =[(FVIV)1t−1]×100

⇒FVIV=4836=1.33

Going from answer options -

If 10 is the answer -

1.13=1.331≈1.33=FVIV

Therefore, rate is approximately 10%.

### Question 27

In how many different ways can the six faces of cube be painted by 6 different colours?

45

40

30

66

#### SOLUTION

Solution :C

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 = [(6×5×3!)]6 = 30

### Question 28

Find the no. of non negative integral solutions of: x1+x2+x3=15.

15

15C3

17!2!15!

#### SOLUTION

Solution :C

Non negative integral solutions means x1,x2,x3≥0

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

Applying 0's and 1's method:

Total no. of ways =17!2!15!. Hence option (c) is the correct answer.

### Question 29

Find the complete range of values of "p” for which (p2−7p+13)p<1?

(−∞,+∞)

0,∞

(−∞,4)

(−∞,0)∪(3,4)

#### SOLUTION

Solution :D

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).

### Question 30

What is the maximum value of the function f(x) = min (7 - 9x, 3x + 4) for 0 < x < 9?

234

785

194

383

#### SOLUTION

Solution :C

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 → x = 14

So, max f(x) = min (194 , 194) = 194

### Question 31

Find the remainder when 12333 is divided by 10.

3

9

7

1

#### SOLUTION

Solution :A

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)

12333=1234k+1=...3. Thus, the last digit of this number will be 3.

### Question 32

A plane MH370 is reported to have crashed somewhere in the rectangular region shown in fig. The probability that it crashed inside the lake (shown in the figure) is P110 . Find the value of 'P'. Type the answer in the blank below:

#### SOLUTION

Solution :Solution: Area of the entire region where the helicopter can crash = (5.5×10)km2 =55 km2

Area of the lake = (3×3.5) km2 =10.5 km2

Probability that helicopter crashed inside the lake = 10.555=21110.

So, P = 21

### Question 33

21 workers working 5hours a day for 30 days can build a platform of width 150 meters, length 20 meters and height 15 meter. How many days will 24 workers working 7 hours a day will take if they have to build a platform of width 720meter, length 15 meter and height 10 meters?

Type the answer in the blank below:

#### SOLUTION

Solution :Solution:

Equate the amount of work in terms of man - hours and volume of the platform.21 workers, 5 hours a day, 30 days = 150×20×15

24 workers, 7 hours a day, N days = 720×15×10

Taking the ratio ,21×5×3024×7×N=150×20×15720×15×10

N =45

### Question 34

'A' is the set of natural numbers less than 100. 'B' is the subset of 'A' such that 'B' contains two digit numbers from set 'A' which are divisible by exactly two numbers. 'C' is the subset of 'B' such that 'C' contains numbers whose sum of digits is divisible by 2. 'D' is the subset of 'C' such that 'D' contains numbers whose sum of digits is a perfect square or a perfect cube. Then find the median of the set 'D'?

Type the answer in the blank below:#### SOLUTION

Solution :A be the set of natural numbers less than 100.

∴ A = {1, 2, 3, 4, ... , 99}

Now, B is the subset of A such that B contains two digit numbers which are divisible by exactly two numbers from the set A .

∵ Only prime numbers are the numbers which are divisible by 1 and itself only.

∴ B = {11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47, 53, 59, 61, 67, 71, 73, 79, 83, 89, 97}

Now, C is the subset of B such that C contains numbers whose sum of digits is divisible by 2.

∴ C = {11, 13, 17, 19, 31, 37, 53, 59, 71, 73, 79, 97}

Now, D is the subset of C such that D contains numbers whose sum of the digits is a perfect square or a perfect cube.

∴ D = {13, 17, 31, 53, 71, 79, 97}

∴ Median of the set D = 53