# Free Quant Practice Test - CAT

### Question 1

The infinite sum 513+55132+555133+5555134+..... equals

5625

4621

6536

5526

5222

#### SOLUTION

Solution :C

513+55132+555133+5555134+.....=59(913+99132+999133+.............)

=59[10−113+100−1132+1000−1133..............]

=59[{(1013)1+(1013)2+(1013)3+..........}−{(113)+(113)2+(113)3+........}]Now solve this infinite GP to get 6536 Hence, option (c)

### Question 2

The set of all positive integers is the union of 2 disjoint subsets {f(1),f(2),f(3),...}& {g(1),g(2),g(3),...}, where f(1)<f(2)<f(3)<.....& g(1)<g(2)<g(3)<...... g(n)=f(f(n))+1 for n = 1,2,3,......

What is the value of g(1)?

3

2

1

Indeterminate

#### SOLUTION

Solution :B

f(x) is a set of all odd numbers and g(x) is a set of all even numbers. Thus, g(1) = f(f(1)) + 1 = 2.

### Question 3

In the figure, AB = OA(radius of the circle). If Y = kX. The value of k is

#### SOLUTION

Solution :The above diagram is self-explanatory. 180-4x+x+y=180

Hence y=3x, k=3.

### Question 4

Three distinct prime numbers less than 10 are taken and all the numbers are formed by arranging all the digits taken. Now the difference between the largest and the smallest number is 495. It is also given that the sum of the digits is more than 13. What is the product of the digits?

42

70

105

Cannot be determined

#### SOLUTION

Solution :B

Prime numbers less than 10 = 2, 3, 5, 7.

If the difference between the largest and the smallest number is ending in 5, the prime numbers in the end position have to be 7 and 2.

The smallest and largest numbers are of form 2_7 and 7_2

Since it is given that the sum of the digits is >13, x will be 5.

Verifying, 752-257 = 495. Answer is option (b).

### Question 5

ax + 2y = 30 and 3x + by = 15 are two equations in variables x and y. For what condition you cannot find the solution of these simultaneous equations.

ab ≠ 6

a2+b2+20=9a+3b.

ab = 6

#### SOLUTION

Solution :C

ax + 2y = 30.................................(1)

3x + by = 15.................................(2)Multiplying equation (2) by 2

6x +2by = 30........................(3)

Equating eqn (1) and (3)

ax +2y = 6x +2by

We cannot find the solution when a=6 and b=1, answer is option (c)

### Question 6

If S1,S2,S3,....Sn are the sums of infinite geometric series whose first terms are 1, 2, 3,...n and whose ratios are 12,13,14,1(n+1) respectively, then find the value of S21+S22+S23+....+S26

140

139

91

140

#### SOLUTION

Solution :B

S1=11−12=2

S2=21−13=3

Thus, the respective sums are 2,3,4,5,6,7

Sums of square of these numbers =[n(n+1)(2n+1)6]−1=139.

### Question 7

PQR is a right angled triangle with PQ=8, QR=6, QT as altitude. If a circle is drawn with QT as diameter, what is the area of the shaded region?

π−24

(π−2)(125)26

(π−2)(125)22

(π−4)(65)22

#### SOLUTION

Solution :C

Let PT=x, Then 82-x2 = 62 - (10-x)2; ⇒ x = 6.4

Thus QT = 82 - 6.42 = 4.8; ST=4.8√2;

Area of square STUQ =(4.8√2)2;

Area of circle: π (2.4)2;

Area of the shaded region: 12(π(2.4)2−(4.8√2)2)=(π−2)(125)22

### Question 8

In a country, all numbers are represented with the help of 3 alphabets, a, b & c. 15 is written as abc; 6 is written as bc; 60 is written as bcbc. How would 17 be written in that country?

abb

bab

baa

aba

#### SOLUTION

Solution :A

If only 3 digits are involved, the system has to be in base 3 with the digits 0,1,2. We can confirm the base using the examples provided.

Thus, a=1 ,b=2 and c=0 (you can reconfirm using the other examples as well)

17 in base 3 = = abb.

### Question 9

There is an octagonal die with 8 faces having values 1 to 8 on each face. It is thrown 4 times in a sequence. In how many ways can you get a sum which is less than or equal to 20?

1245

1395

1575

2865

#### SOLUTION

Solution :D

This is a question based on both upper limit and lower limit

Given that A+B+C+D≤20

This can be written as A+B+C+D+E=20

Where 1≤A,B,C,D≤8

Applying a lower limit of 1 to A,B,C,D

A+B+C+D+E=16

Number of ways = 20C4 = 4845

Giving an upper limit of 8 to A (this makes A=9)

A+B+C+D+E=8

Number of solutions = 12C4

Similarly, violating the conditions for A,B,C & D= 4*12C4

### Question 10

ABCABC is a six digit number where A,B,C are three consecutive natural numbers. Find the remainder when ABCABC is divided by 1001?

1

997

Cant be determined

None of these

#### SOLUTION

Solution :D

ABCABC = ABC×1000 + ABC = ABC×1001.

∴ Remainder when ABCABC is divided by 1001 is 0.

### Question 11

There are two women participating in a chess tournament. Every participant played two games with the other participants. The number of games that the men played between themselves proved to exceed by 66 the number of games that the men played with the women. How many games were played?

78

210

156

128

#### SOLUTION

Solution :C

Let the number of men = n

Number of games between men and women = 4n

Given that 2(nC2−2n)=66

Solving for n, we get n= 11

Thus total number of games played = 13C2=78×2=156

### Question 12

Integers a, b, c, d are such that −5≤a≤4, −2≤b≤6, −3≤c≤8 and −4≤d≤6. If X = ab + bc + cd + da, then find the least value of X?

-96

-72

0

-144

#### SOLUTION

Solution :A

There are only 2 possibilities for minimum value -

Thus, the answer is option (a).

### Question 13

Amar, Akbar and Anthony ran on a racetrack, with Amar finishing 160 m ahead of Akbar and 400 m ahead of Anthony. Akbar finished the race 300 m ahead of Anthony. The three of them ran the entire distance with their respective constant speeds. What was the length of the racetrack?

600m

800 m

1000 m

500 m

1200 m

#### SOLUTION

Solution :B

Using the concept of ratios of speed (speed directly proportional to distance), we can easily arrive at the total distance. The images below givea picture of the distances when Amar and Akbar finish the race

Length of the race track = 400 +x

The ratio of Akbar and Anthony's speed can be arrived at as

240+xx=400+x100+xNow go from answer options, answer will be option (b) 800 m

Verifying, x = 400

640400=800500. Thus, answer is option (b)

### Question 14

(2+x+x3)5=a0+a1x+.....+a15x15. find the value of a0+2a1+a2+2a3....+2a15_{ }

1024

1536

1586

2048

#### SOLUTION

Solution :B

At x=1

a0+a1+.....+a15=45.............(1)

At x= -1

a0−a1+......−a15=0.................(2)

(1)-(2)=2a1+2a3.......2a15=45.........(3)

([(1)+(2)]2)=a0+a2...=452........(4)

(3)+(4)=a0+2a1+a2+2a3....+2a15=45+[452]=1536.

### Question 15

Let f(x) = 24x+2 for x ϵ Real numbers. Find out the value of f(12001) + f(22001)+ ...... + f(20002001)

2000

2002

1050

1020

1000

#### SOLUTION

Solution :E

f(1 - x) = 241−x+2 = f(x) = 2∗4x4x+2 = f(x) = 4x4x+2

f(x) + f(1 - x) = 1So, the desired sum is 1000.

### Question 16

Sum to p terms of the series 12+(12+32)+(12+32+52)+.... is

13(p3+3p2−p)

13(p4+2p4)

16p(p+1)(2p2+2p−1)

None of these

#### SOLUTION

Solution :C

At p=2, sum = 1+10 =11

Put p=2 in answer options

Only Option c gives 11

### Question 17

Aamir would take 9 hours to build a chimney alone, and Saif would take 10 hours to build it alone. When they work together they talk a lot, and their combined output is decreased by 10 bricks per hour. Working together, they build the chimney in 5 hours. How many bricks are in the chimney?

500

900

950

1000

1900

#### SOLUTION

Solution :B

Using the conventional approach :Saif takes 10 hrs.

So total time they both would take without talking =10×919=9019 hrs.

But they take 5 hrs =9519 hrs.So time wasted is 519 hrs.

They work for 5 hrs and put 10 bricks less per hour, so they had to put in extra time for 5×10=50 bricks.

Now, extra time taken= 519 hrs.

so, in 519 hrs: 50 bricks

5 hrs: x

x=950 bricks.

But this includes the 50 bricks extra put in the time 519 hrs. So total number of bricks is 900.Using percentage approach:

Aamir takes 9 hours,⇒ he does 11.11% of the work every hour

Saif takes 10 hours ⇒ he does 10% of the work every hour

Together they finish 21.11% of the work

Given that they take 5 hours to finish the work after chatting, which means they are actually finishing 1005=20% of the work

The remaining 1.1% accounts for the 10 bricks decrease due to chatting

Thus, 100% corresponds to 900. Answer is option (b)

### Question 18

In a college, 35% of the students play at least two out of three games- Cricket, Hockey and Football. Football is played by 45% of the students. 15% of the people play both Cricket and Hockey, while 10% of the people play all the three games. If 7% of the students do not play any game, what percentage of the students play only Football?

10%

15%

25%

30%

Cannot be determined

#### SOLUTION

Solution :B

Using the venn diagram approach

II +III= 35

a+b=20

Hence percentage playing only Football = 45-10-20 =15%

### Question 19

Consider the sequence defined recursively by p1=a (any positive integer), and pn+1=−1pn+1, n=1,2,3 For which of the following values of n must

pn=a?

15

18

19

14

#### SOLUTION

Solution :C

p1=a; p2=−1a+1;p3=(−1−1a);p4=a.

Hence the pattern p1=p4=p7=p10=p13=p16=p19....=a

Hence, choice (c) is the right answer

### Question 20

In the figure the larger circle has dia 40 units and the smaller circle has dia 10 units. Both the circles touch each other at P and the smaller circle and ABCD touch each other at Q. PQ is the dia of the smaller circle. If ABCD is a square and Q is the mid point of AD, then its side is?

7√17

7+3√17

8+4√19

8√29

#### SOLUTION

Solution :C

From the above figure, Now BRO is a right angled triangle. Solving for x, we get

X= 22 - 4 √19

Side length of a square = 8 + 4 √19

### Question 21

In the given DPQR and D PST; PS=8, ST=7, QR=35 and QS=PT. Also, ∠PST = ∠PRQ. What is the value of PT?

3

4

2

5

#### SOLUTION

Solution :C

In the figure ∠PST and ∠PRQ are similar (A,A,A).

If SQ = x = PT and TR = y =,(xx+8)=735; x = 2

### Question 22

How many natural number solutions are there for the equation, A+B+C = 100?

99C2

99C3_{ }

101C2_{ }

99C99

#### SOLUTION

Solution :A

Natural number solutions means each of A,B and C can take a value ≥ 1

This is a Similar to Different question, with a lower limit condition. (Refer ebooklet for indepth explanation)

To take care of the condition, that we are dealing with natural numbers,

give A,B &C 1 initially

(A+1)+(B+1)+(C+1)=100

Thus, this reduces to A+B+C=97.

This question is now based on arrangement of 97 zeroes and 2 ones=99C2

The answer is 99C2

_{ }which is option (a)

### Question 23

A quadratic function attains its minimum value of 5 at x=1 and holds a value of 6 at x=0. Find the value of the function at x=5

18

21

25

45

#### SOLUTION

Solution :B

Standard form of a quadratic equation = f(x) = ax2+bx+c=0

Substitute x=0

Given that f(x)=c=6

Substitute x=1

a+b+c=5

a+b =-1................(1)

The minimum value of a function occurs at x=−(b2a), which is given as 1

Thus, 2a=-b

Substituting back in eqn (1) a=1 and b=-2

f(x)=x2−2x+6Thus, at x=5, f(x)=21. Option (b)

### Question 24

A company has to construct a flyover. It deployed 10 men who can finish the work 30 days each working 8 hrs a day. After five days of work due to heavy rain 50% of the flyover constructed so far collapsed. After the incident the management decided to stop the construction work during the period it rains. Also the Management deployed 5 more men and 10 women to complete the project faster. From the sixth day onwards it rained every alternative day for 2 hours each. Find the fraction of construction work left after the 17th day. Assume that all the men work with same efficiency, and all women workwith half of the efficiency of men. Also assume that it rains only during the working hours.

160

1112

2130

#### SOLUTION

Solution :D

Work completed in 5 days = (16). Since 50% of the bridge collapsed so total work completed after 5 days =

(112) Remaining work after 5 days = (1112)

Since women efficiency is half so 10 women will finish the work in 60 days i.e each day they will finish (160) part of work.

15 men will finish = (120) part of work every day. So from the sixth day onwards the work finished everyday =120+160=115. Now from sixth days onwards it rains alternatively for 2 working hours.

Hence total numbers of hours lost = 6×2 = 12 hrs = 1.5 days ( 8 hrs per day)

So till 17th day instead of remaining 12 days the work was done effectively on only 10.5 days.

So the work completed in 10.5 days =1015+130=2130.

Fraction of work left after 17th day = work left after 5 days - work done in 12 days =1112−2130=1360

Hence option D is correct.

Alternative approach: (using units of work done)

Total number of units of work to be done = 10×30×8 = 2400

In the first 5 days, 10×5×8 =400 units of word is done. However, 200 units of work is lost

Thus, remaining work is 200 units

From the 6th day 5 more men and 10 more women are joining in ⇒ 10 more men are joining in

Thus, on every day that it rains; number of units of work done =6×20=120

And on every day that it does not rain; number of units of work done =8×20=160

Total units of work done =120×6+160×6+200=1880

Remaining work = 2400-1880 = 540. Required ratio =(5402400)=(1360)

### Question 25

How many pairs of positive integers, x, y exist; such that x2+3y and y2+3x are both perfect squares?

0

1

2

#### SOLUTION

Solution :D

Since x and y are positive, we may write -

x2+3y=(x+a)2, and

y2+3x=(y+b)2where a, b are positive integers.

Expanding, we find that the squared terms cancel, leaving the linear simultaneous equations -

3y=2ax+a2

3x=2by+b2

Solving, we obtain -

x=2a2b+3b29−4ab

y=2b2a+3a29−4abSince a and b are positive, the numerator in each fraction will be positive. For the denominator to be positive, we must have ab = 1 or 2.

If (a,b) = (1,1), (1,2), (2,1), then, respectively, (x,y) = (1,1), (16,11), (11,16). Hence, these are the only solutions.

### Question 26

An elevator starts at the basement with 9 people (excluding the elevator operator) and discharges them all by the time it reaches the top floor, which is floor number 6. In now many ways could the operator have perceived the people leaving the elevator if all the people look the same? (there are totally 6 floors excluding the basement)

2002

1009

1576

2492

#### SOLUTION

Solution :A

This is a question based on distributing similar things to different groups (SàD)

i.e. the distribution of 9 similar things to 6 different groups

A+B+C+D+E+F=9

Solving, we get 14C5= 2002.

### Question 27

When a 4 digit number pqrs is multiplied by N, the 4 digit number repeats itself to give an 8 digit no. If the 4 digit number has all distinct digits then N is a multiple of:

11

33

77

73

#### SOLUTION

Solution :D

The number is of the form pqrs, which when multiplied by N, gives a number of the form pqrspqrs.

This is possible if N = 10001.

However, 10001=73×137. Thus N has to be a multiple of 73 or 137. Answer is option (d).

### Question 28

How many solutions are there for the equation A+B+3C = 10 if A,B,C are whole numbers?

22

25

24

14

#### SOLUTION

Solution :B

It is given that A,B and C can take values ≥0

Give values to C from 0 to 3

At C=0, the equation changes to A+B=10. Number of solutions will be based on arrangement of 10 zeroes and 1 one = 11C1= 11

At C=1, the equation changes to A+B=7. Number of solutions=8C1=8

At C=2, the equation changes to A+B=4. Number of solutions=5C1=5

At C=3, the equation changes to A+B=1. Number of solutions=2C1=1Total number of solutions = 11+8+5+1 = 25

### Question 29

What is the remainder when 77+777+7777+77777.........+777777777 is divided by 8?

-1

1

7

0

#### SOLUTION

Solution :D

78 remainder = -1

Thus, any odd power of 7, when divided by 8, will give a remainder of -1 itself.

(−1)7+(−1)77+(−1)777......8

= -1-1-1-...8 times = - 8

The remainder when -8 is divided by 8, is 0.

### Question 30

The numbers x,y,z are proportional to 2,3,5. The sum of x,y and z is 100. The number y is given by the equation y = ax - 10. Then 'a' is

#### SOLUTION

Solution :x : y: z: :2 : 3 : 5

∴x=2y3 and z=5y3

x+y+z=100

2y3+y+5y3=100

y= 30 & x=20

But y= ax-10

30 = 20a - 10

a=2

### Question 31

The smallest possible radius of a circle such that it is possible to place 6 points on the circumference with an integer distance between any two is x/π. The value of 'x' is

#### SOLUTION

Solution :The smallest radius means the smallest circumference.

Circumference = 2πr

With six points equidistant with distance between them to be integer, this integer has to be smallest i.e. 1

Circumference = 6 units

2πr = 6

r = 3π

So, x = 3

### Question 32

A cone has a circular base of radius 2 units and vertex of height 5 units directly above the centre of the circle. A cube is formed directly above the centre of the circle. The cube has four vertices in the base and four on the sloping sides. What is the approximate length of each side of the cube?

2.35 cm

4.15 cm

3.25 cm

2.22 cm

#### SOLUTION

Solution :D

We can solve this question just by using proportions. (because we have two similar triangles)let the side of the cube=x, the height of the cone =h=5 and the radius of the base=2

ΔADC ≈ΔAEF

rh=x2h−x

⇒2(5 - x) = 5x2

⇒(10 - 2x) = 2.5x

⇒ x = 2.22 units

### Question 33

A book shopkeeper gives a discount of 30% on a book and makes a profit of 20% . If the book shopkeeper had given a discount of Rs. 400, he would have incurred a loss of 20% . Find the marked price (in Rs) of the book.

750

600

820

900

#### SOLUTION

Solution :A

Let the marked price of the book be x.Selling Price (SP) after discount of 30%

=x(1−30100)=7x10

It is known that there is a profit of 20% of SP 7x10

∴CP=7x10(100100+20)=7x12.

New SP=x-400 and on this new SP there is a loss of 20%.

⇒7x12(1−20100)=x−400

⇒7x15=x−400⇒400=x−7x15

⇒9x15=400⇒x=750Shortcut- Using the reverse Gear Approach

Always start with a middle answer option

Assuming 750 as the right answer option.

If he gives a discount of 30% then the selling price will be 750 - 225=525, with which he is able to make a profit of 20%.

Thus, cost price will be =(5251.20)=437.5

Again if he had given a discount of 400 Rs. He would have sold the book at 750 - 400=350 he incurred a loss of 20% or (3500.8)=437.5 hence that is the correct option.

### Question 34

If x2−gx+3=0 and 2x2−6x+g=0 have a common root, then find the value(s) of the common root.

1

2

3

None of these

#### SOLUTION

Solution :C

If two equations have a common root then they can be found by equating them:

When we equate we get x2+x(g−6)+(g−3)=0

Now the Determinant of this equation has to be equal to 0

(g−6)2−4(1)(g−3)=0

We get the values of g= 4 or 12Re-substituting in the original equation we see that only g=4 will satisfy the equation

i.e You will get the roots to be 1,2 and 1,3 hence 1 is the common root.

x2−gx+3=0 is equation 1.

2x2−6x+g=0 is equation 2.