# Free Quant Practice Test - CAT

### Question 1

x! ends with n zeroes. (x+1)! ends with n+2 zeroes. 24≤x≤626. How many values are possible for x?

25

22

23

20

#### SOLUTION

Solution :D

If 24! ends in n zeroes, 25! will end in (n+2) zeroes.

Generalising this to all multiples of 25, they satisfy this condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.

However, if 124! ends in n zeroes, 125! will end in (n+3) zeroes.

Generalising this to all multiples of 125, they do not satisfy the condition: if (x)! ends with n zeroes then (x+1)! ends with (n+2) zeroes.

Infact, they end with (n+3) or (n+4) zeroes.

Therefore, the number of favourable cases between 24 to 626 can be represented as:

(Number of multiples of 25 - Number of multiples of 125)25 - 5 = 20.

### Question 2

If A > 11, B > 21, C > 30, D > 40 and E > 50, then how many positive integral solutions exist for A + B + C + D + E = 2012?

2012C4

2012C5

1859C4

#### SOLUTION

Solution :D

A + B + C + D + E = 2012 (A>11, B>21, C>30, D>40, E>50)

This is a lower limit Similar->Different question.

So, we assign the minimum values that can be taken by each of A, B, C, D and E and reframe the equation:

A + B + C + D + E = 1855 (Assigning 12 to A, 22 to B, 31 to C, 41 to D and 51 to E)

Using the 1-0 method:Required Answer = 1859C4.Option (d)

### Question 3

The figure below is a regular octagon. What fraction of its area is shaded?

13

14

15

38

#### SOLUTION

Solution :B

The diagram shows how the octagon can be divided into 4 congruent rectangles and 8 congruent triangles. Let R represent the area of a rectangle and let T represent the area of a triangle. Then the ratio of the shaded region to the area of the entire octagon is R+2T4R+8T=14

### Question 4

On every birthday of my life, my mother has seen to it that my cake contains my age in candles. Starting on my fourth birthday, I have always blown out all my candles. Before that age, I averaged a 50% total blowout rate. So far, I have blown out exactly 900 candles. How old am I

43

42

44

45

#### SOLUTION

Solution :B

n(n+1)2−1+2+32=900:n(n+1)2=903;

n(n+1) = 1806, n = 42.

### Question 5

In an election, BJP received 60% of the votes and Congress received the rest. If BJP won by 24 votes, how many people voted?

120

60

72

100

#### SOLUTION

Solution :A

If BJP received 60% of the votes this implies that Congress received 40% of the total number of votes. The difference between them, 20% , represents 24 votes. Therefore, the total number of votes cast was 5 × 24 =120.

### Question 6

Find the sum of given series Sn = 4 + 44 + 444 + 4444 + ...................n terms

49{109(10(n−1)−1)−n}

49{109(10n−1)−n}

49{10(10n−1)−n}

49{(10n−1)−n}

#### SOLUTION

Solution :B

Reverse Gear Approach

Assume n= 1, then S1 =4

Now put n=1in options.Eliminate those options , where you are not getting 4.

Only option (b) will give 4.

### Question 7

In a 4000 meter race around a circular stadium having a circumference of 1000 meters, the fastest runner and the slowest runner reach the same point at the end of the 5th minute, for the first time after the start of the race. All the runners have the same staring point and each runner maintains a uniform speed throughout the race. If the fastest runner runs at twice the speed of the slowest runner, what is the time taken by the fastest runner to finish the race?

20 min

15 min

10 min

5 min

#### SOLUTION

Solution :C

The ratio of the speeds of the fastest and the slowest runners is 2 : 1. Hence they should meet at only one point on the circumference i.e. the starting point (As the difference in the ratio in reduced form is 1). For the two of them to meet for the first time, the faster should have completed one complete round over the slower one. Since the two of them meet for the first time after 5 min, the faster one should have completed 2 rounds (i.e. 2000 m) and the slower one should have completed 1 round. (i.e. 1000 m) in this time. Thus, the faster one would complete the race (i.e. 4000 m) in 10 min.

### Question 8

The integers 1,2,3,4,......n are written on a blackboard. The following operation is then repeated until we will get only one number: In each repetition, any two numbers say a and b, currently on the blackboard are erased and a new number a + b is written. But one number was missed by mistake. The sum obtained was 1525. Which number was missed?

15

25

20

10

#### SOLUTION

Solution :A

n(n+1)2=1525; n(n+1)=3050; n=55 (approx.)

55×562=1540; missed number = 1540-1525 = 15.

### Question 9

What is the remainder when (1)7+(71)77+(771)777+(7771)7777+(77771)77777+(777771)777777 is divided by 50?

46

6

56

5

#### SOLUTION

Solution :B

To find the remainder when an expression is divided by 50, we need to find the last two digits of that number.

Last two digits of the given expression are 1 + 91 + 91 + 91 + 91 + 91 = 456 (Ten's digit of powers of 71 follow a cycle of 10).

∴45650, remainder=6.

### Question 10

A positive whole number M less than 100 is represented in base 2 notation, base 3 notation, and base 5 notation. It is found that in all three cases the last digit is 1, while in exactly two out of the three cases the leading digit is 1. Then M equals:

31

63

75

91

#### SOLUTION

Solution :D

Since the last digit in base 2, 3 and 5 is 1, the number should be such that on dividing by either 2, 3 or 5 we should get a remainder 1. The smallest such number is 31. The next set of numbers are 61, 91.

Among these only 31 and 91 are a part of the answer choices.

Among these, (31)10=(11111)2=(1011)3=(111)5. Thus, all three forms have leading digit 1.Hence the answer is 91.

### Question 11

At a new index of 12 shares, the shares of HBL, Niposys and Brilliance have a weightage of 7%, 13% and 1% respectively. What is the increase in the price of the other shares, if the rise in these three is 9%, 10% and 4%, when the index rises by 6%?

4%

3.4%

6%

None of these

#### SOLUTION

Solution :C

Lets say the value of the index = 100

HBL = 7

Niposys = 13

Brilliance = 1

Others = 100 - 21 = 79

Increased index value = 106

Increased individual values

HBL = 1.09 × 7=7.63

Niposys = 1.1 × 13 =14.3

Brilliance = 1.04 × 1 = 1.04

Total = 22.97Others = 106 - 22.97 = 83.03

Increase =83.03−7979×100=5.1% (approx)

### Question 12

ΔABC has a circle centred on vertex C that passes through points A and E. If ∠ABC ≈ ∠DEC, AC = 1 and CD = x, what is the distance between E and B?

2x

1−xx

1 - x

1 + x

#### SOLUTION

Solution :B

Since ∠DEC ≅ ∠ABC, it follows that ΔDEC ≈ ΔABC.

So, 1x=BE+11

⇒x(BE+1)=1

⇒(BE+1)=1x

⇒BE=1−xx

### Question 13

The number of positive integers n in the range 12<n<40 such that the product (n-1)(n-2)...3.2.1 is not divisible by n is:

5

7

13

14

#### SOLUTION

Solution :B

The question is asking you for all prime numbers between 12 and 40. There are 7 prime numbers.

### Question 14

A mixture of 125 litres of milk and water contains 20% water. What amount of water needs to be added to this milk-water mixture in order to increase the percentage of water to 25% of the new mixture?

7 L

5.66 L

8.33 L

None of these

#### SOLUTION

Solution :C

We need to find out how much of a solution of 100% water needs to be added to a solution containing 20% water to attain a dilution of 25%.

This can be found out as follows:Ratio = 75:5 = 15:1

i.e. for 15 parts of a 20% water solution, one part of 100% water solution needs to be added. Therefore, for a solution of 125 litres, (12515) litres of water needs to be added.

### Question 15

In the figure, EAF is a common tangent to the circles at the point A. Chords AC and BC of the smaller circles are produced to meet the larger circle at G and D respectively . Which of the following must be true?

1. ∠ADG = ∠EAG

2.∠ABD = ∠AG D

3. ∠BAE = ∠ADB

1 only

2 only

1 and 3 only

2 and 3 only

#### SOLUTION

Solution :A

From the tangent secant theorem, ∠ADG=∠EAG. Other equalities don't hold. Option(a)

### Question 16

If an=xn+1xn, n being a natural number, then which of the following is equal to an+3?

a1.an+2−an+1

a1.an+2−an+1+an

a1.an+2−an+1−an

a1.an+2+an+1−an

#### SOLUTION

Solution :A

Approach :1Using the options

a1.an+2exists in all options.

We will calculate it first.

a1.an+2 exists in all options.

We will calculate it first.

a1.an+2=(x+1x)(xn+2+1xn+2)

=xn+3+1xn+3+xn+1+1xn+1

⇒a1.an+2=an+3+an+1

⇒an+3=a1an+2−an+1

Approach 2:Assumption

Since the question is of the form "Variable to Variable”, assume any value for "n” and "x”

Assuming n=1 & x=1 . Then a1=2. Similarly a2=2 , a3=2 & a4=2

Eliminating answer options

Option (a) =2

Option (b) ≠ 2Option (c)≠ 2 & option (d)≠ 2 . Answer is option (a)

### Question 17

In a certain examination paper, there are n questions. For j = 1,2 ...n, there are 2n−j students who answered j or more questions wrongly. If the total number of wrong answers is 4095, then the value of n is:

12

11

10

9

#### SOLUTION

Solution :A

Let us say there are only 3 questions.

Thus there are 23−1=4 students who have answered 1 or more questions wrongly, 23−2=2 students who have answered 2 or more questions wrongly and 23−3=1 student who must have answered all 3 wrongly.

Thus total number of wrong answers =4+2+1=7=23−1=2n−1.In our question, the total number of wrong answers =4095=212−1. Thus n = 12.

### Question 18

Two circles centered at points A and B, respectively, intersect at points E and F as shown. These circles intersect segment AB at points C and D. If AC=1 and CD = DB = 2, determine EF.

2.4

4.8

2√7

#### SOLUTION

Solution :B

In triangle AFB, 12×AB×FP=12×AF×FB

⇒FP=125

Hence, EF= 245 = 4.8.

### Question 19

What are the last two digits of 384447?

54

34

44

64

#### SOLUTION

Solution :D

(3844)47=(...44)47=(11×4)47=(11)47×(2)94=..71×(2)90×24=..71×..24×16

(∵(210)odd number always ends in 24)The last two digits of the above product = 64.

### Question 20

A graph may be defined as a set of points connected by lines called edges. Every edge connects a pair of points. Thus, a triangle is a graph with 3 edges and 3 points. The degree of a point is the number of edges connected to it. For example, a triangle is a graph with three points of degree 2 each. Consider a graph with 12 points. It is possible to reach any point from any point through a sequence of edges. The number of edges, e, in the graph must satisfy the condition

11≤e≤66

10≤e≤66

11≤e≤65

0≤e≤11

#### SOLUTION

Solution :A

the question is basically asking us to find out the minimum number and maximum number of lines that can be drawn using 12 points.

We can draw a minimum of 11 lines. We can visualize this as one central point and all lines connecting to this point i.e. one point has a degree of 11 and the other 11 have a degree of 1.

There can be a maximum of 12C2 lines that can be drawn= 66

### Question 21

Find the greatest number of five digits, which is exactly divisible by 7, 10,14, 15, 21 and 28.

99840

99900

99960

99990

#### SOLUTION

Solution :C

The number should be exactly divisible by 14(7,2),15 (3, 5), 21 (3, 7), 28 (4, 7).

Hence, it is enough to check the divisibility for 3, 4, 5 and 7.Only (c) is divisible by all.

### Question 22

The rightmost non-zero digit of the number 302720 is?

#### SOLUTION

Solution :302720=32720×102720

The rightmost non-zero digit of 302720 will be the digit in the unit's place of 32720.

3's power cycle is 3, 9, 7, 1 and cyclicity is 4.

2720 = 680 × 4

∴The digit in the unit's place of 32720 is 1. ∴The rightmost non-zero digit of 302720 is 1.

### Question 23

When you reverse the digits of the number 13, the number increases by 18. How many other two digit numbers increase by 18 when their digits are reversed?

#### SOLUTION

Solution :Let 10x + y be a two digit number, where x and y are positive single digit integers and x>0.

Its reverse = 10y + x

Now, 10y + x - 10x - y = 18

∴ 9(y - x) = 18 ∴ y - x = 2

Thus y and x can be (1, 3), (2, 4), (3, 5), (4, 6), (5, 7), (6, 8) and (7, 9)

∴ Other than 13, there are 6 such numbers.

### Question 24

In how many ways can 7 identical red balls, 6 identical green balls, and 1 yellow ball be put in 14 distinct boxes such that each box contains only one ball?

#### SOLUTION

Solution :First, the only yellow ball can be placed in one of the 14 boxes in 14C1 = 14 ways.

Out of the remaining 13 boxes, we need to choose 6 places for the 6 identical green balls. This can be done in 13C6_{ }ways.Therefore, the total number of ways = 14 × 13C6

_{ }= 24024

### Question 25

Find the highest power of 5625 that can exactly divide 565!.

34

35

32

30

#### SOLUTION

Solution :A

5625=54×32.

∴ Highest power of 5625 in 565! will depend on the highest power of 54.

So, highest power of 54 in 565! is = Integral value of 113+22+44 = 34.

### Question 26

An arbelos is the region bordered by three mutually tangent circles with co-linear centers. See shaded region in the figure. If the area of this region is 42π and the radii of the two smaller circles have a ratio of 2 : 3. What is the radius of the largest circle(x)?

2√21

5√7

7√6

35

#### SOLUTION

Solution :B

Let the radii of the two smaller circles be 2

yand 3yrespectively. This makes the radius of the larger circle 5y. We are told that the area is 42π, so we have

42π = 12(π(5y)2−π(3y)2−π(2y)2),

y = √7.

But x = 5y; So, x = 5√7.

### Question 27

A train met with an accident 80 km away from old Delhi railway station. It travels at 56th of the usual speed and reaches the Kanpur central 1 hour 12 min late. Had the accident taken place 120 km further. It would have been only 1 hour late. What is the distance between Old Delhi Railway station and Kanpur Central?

800

720

700

None of these

#### SOLUTION

Solution :B

Since the speed is decreased by 1/6. So, the time will be increased by 15, which is equal to 1 hour 12 minutes. It means the normal time required for this remaining part of journey is 5 × 72 min = 360 min = 6 hour

Now, accident is supposed to happen 120 km further.

Since, the speed is decreased by 16. So, the time will be increased by 15, which is equal to 1 hour. Hence normal time required for this remaining part of journey is 5 × 1 = 5 hour.

Thus , it is clear that when the trains runs 120 km of its normal speed it takes 1 hour less, which implies that in 1 hour the train can run 120km with its normal speed. Thus the normal speed of train is 120km/h.

Since the train requires 6 hours at its normal speed of 120 km/h for the rest part of journey. So,

AK = 120 × 6 = 720.

Thus,

The total distance = Distance travelled before accident + Distance travelled after accident

= 80 + 720 = 800.

Option (a).

### Question 28

What digit does 20122012 end in?

2

6

4

8

#### SOLUTION

Solution :B

Since the number in the unit's place in 2012 is 2, we can find the last digit of the expression by considering only the power to which it is raised.

Now, we can write the following: 20122012=___24k. Therefore the last digit = 6.

### Question 29

All the page numbers from a book are added, beginning at page 1. However, one page number was added twice by mistake. The sum obtained was 805. Which page number was added twice?

20

22

29

25

#### SOLUTION

Solution :D

−n(n+1)2=805;n(n+1)=1610, most appropriate value of n= 39.

39×402=780; Therefore,the page number that was counted twice is : 805 -780 =25.

### Question 30

A 3x3x3 cube has three square holes, each with a 1 by 1 cross-section running from the centre of each face to the centre of the opposite face. The total surface area (in square units) of the resulting solid is:

24

48

72

78

#### SOLUTION

Solution :C

The surface area of the solid consists of 6 faces of dimension 3 × 3 each with a 1 × 1 square hole.

The walls of the 6 holes can each be unfolded to form 1 × 4 rectangles.

Thus the surface area is 6(3×3-1×1)+6×4 = 72

### Question 31

The area of a triangle whose vertices are (0,0), (12, 9) and (14, 6) is

#### SOLUTION

Solution :The simplest way to find this is to put a rectangle around the given triangle and then subtract off the areas of the right triangles that surround it.

(See Figure)

The area of the entire rectangle is 126, but after subtracting the areas of the 3 right triangles, whose areas are 54, 3, and 42, we are left with an area of 27.

### Question 32

In a class 40% of the students enrolled for Math and 70% enrolled for Economics. If 15% of the students enrolled for both Math and Economics, what % of the students of the class did not enroll for either of the two subjects?

5%

15%

0%

25%

#### SOLUTION

Solution :A

We know that (A ∪ B) = A + B - (A ∩ B), where (A ∪ B) represents the set of people who have enrolled for at least one of the two subjects Math or Economics and (A n B) represents the set of people who have enrolled for both the subjects Math and Economics.

Note: -

(A ∪ B) = A + B - (A ∩ B) ⇒ (A ∪ B) = 40 + 70 - 15 = 95%

That is 95% of the students have enrolled for at least one of the two subjects Math or Economics. Therefore, the remaining 5% did not enroll for either of the two subjects.

Approach 2- Using S-X Technique

S = 40% + 70% = 110%, X = ? and II = 15%

S - X = II, ⇒ 110% - X = 15% ⇒ X = 95%. Therefore, the remaining 5% did not enrol for either of the two subjects.

### Question 33

How many 5 digit numbers are divisible by 3 and contain the digit 6?

7499

8776

12503

None of these

#### SOLUTION

Solution :D

Consider first the number of 5-digit numbers divisible by 3. The smallest is 10002=3334×3, the largest is 99999=33333×3, so there are 30000 such numbers.

Now consider the numbers that do not contain the digit 6. There are 8 choices for the 1st digit, 9 choices for each of the 2nd, 3rd and 4th digits.

If the total of the first 4 digits is a multiple of 3, then the last digit must be 0, 3 or 9.

If it gives a remainder 1 when divided by 3, then the last digit must be 2, 5 or 8.

If it gives a remainder 2 when divided by 3, then the last digit must be 1, 4 or 7.

So in all cases there are 3 choices for the last digit. Hence the total number is 8×93×3=17496.

So the total no. of 5 digit numbers which are divisible by 3 and contains a 6 is 12504.

### Question 34

What is the first non-zero integer from the right in 29602960−13902908?

8

1

9

None of these

#### SOLUTION

Solution :C

We need not consider 29602960, as it will end in more number of zeroes.

Last non-zero digit of 13902908 = Unit digit of 1392908=1.Answer is (.......00 - .......1) = 9.