# Free Quant Practice Test - CAT

If xxyy is a perfect square then x + y =?

A.

9

B.

10

C.

11

D.

12

#### SOLUTION

Solution : C

xxyy = xx00 + yy = 11(x00 + y) = 11(x0y)

xxyy is a perfect square, so if one divisor is 11, then there should be one more 11 to get 112.

As xxyy = 11(x0y), x0y should be divisible by 11, so x + y has to 11.

A train passes without stopping, through stations A, B, C, D and E (in order). The distance between any two stations is same. It is also known that the train runs only on two constant speeds changing its speed only at stations. If it crosses A, C and E at 12 pm, 3 pm and 5 pm respectively, then it can be inferred that the train crosses.

A.

B at 1:30pm

B.

D at 4pm

C.

D at 4:30pm

D.

Either (a) or (b)

E.

B at 1pm

#### SOLUTION

Solution : D

The time taken by the train between any two adjacent stations can have only two values. So, it must cross one of the intermediate stations exactly between that of the neighbouring stations. So, the train should either cross B at 1:30 pm or D at 4 pm

There are 13 couples, 5 single males and 7 single females in party. Every male shakes hand with every female once. But none shakes hand with his wife. How many handshakes in total took place in the party?

A.

247

B.

347

C.

360

D.

191

#### SOLUTION

Solution : B

There are 13 couples: 13 males & 13 females In addition to it, there are 5 single males and 7 single females  Now, 5 single males can shake hands with 20 females, therefore 5*20 = 100 ways Also, each of 13 males (from the couples) can shake hands with 12 females (except his wife) + 7 single females. So 13 * 19 = 247 ways. Therefore, in total 347 ways

How many natural number sets of the form Sn={n4,n3,n2,n1,n,n+1,n+2,n+3,n+4} (n is a natural number 100) and the set does not contain 10 or any integral multiple of 10?

A.

8

B.

9

C.

10

D.

15

#### SOLUTION

Solution : C

To get a set of 9 numbers, none of which are divisible by 10, we need to consider the lowest valued number in the set to have a value 1 above the divisor of 10. 0 is a divisor, one above that will be 1 and the lowest valued number in the set is n-4.

So, n-4=1. Now subsequent sets will occur in an AP with a common difference of 10.

1,11,21,31,41.....91 as n100

No. of such numbers =91110+1=9010+1=9+1=10

In a college that has 100 arts students, there are 5 optional subjects among History, Music, Dance, Debate and Photography to choose from. 99% of them choose at least one subject, 49% choose at least 2 subjects, 24% choose at least 3 subjects, 14% choose at least 4 subjects and 4% choose all the subjects. What is the sum of number of students who choose exactly three subjects and who choose exactly 4 subjects?

A.

50

B.

10

C.

15

D.

20

#### SOLUTION

Solution : D

X=100

One =A, Two =B, Tree=C, Four= D, Five =E

Atleast 1 = A+B+C+D+E = 99----------------------(1)

Atleast 2= B+C+D+E=49-----------------------------(2)

Atleast 3= C+D+E= 24--------------------------------(3)

Atleast 4= D+E=14------------------------------------(4)

Atleast 5= E=4-------------------------------------------(5)

4-5= D+E-E = D= 14-4=10 (Exactly 4 = 10)

3-4= C+D+E-(D+E) =C = 24-14=10 (Exactly 3 =10)

Total = 10+10=20

The quadratic equation in x such that the arithmetic mean of the roots is p and the geometric mean is q is

A.

x22qx+p2=0

B.

x2+2px+q=0

C. x2+2pxq2=0
D.

None of these

#### SOLUTION

Solution : D

A quadratic equation can be written as

x2 - (sum of roots)x + (product of roots) =0

Arithmetic mean(p) = Sum of roots2 sum of roots = 2p

Geometric mean (q) = (Product of roots) product of roots = q2

The required equation is x22px+q2=0

In how many ways can Rs. 18.75 be paid in exactly 85 coins consisting of 50-paisa, 25-paisa and 10-paisa coins?

A.

1

B.

2

C.

4

D.

8

#### SOLUTION

Solution : D

Suppose we take x = 50-paisa coins, y = 25-paisa coins and z =10-paisa coins.

We have x + y + z = 85 and 50x + 25y + 10z = 1875, because x, y and z are all integers, z should be a multiple of 5.

Possible values are
x  y  z2  63  205  55  258  47  30  ....  ....23  7  55

So there are eight solutions.

You should notice here that an increase of 5 in z and 3 in x and decrease of 8 in y neither changes the number of coins nor the value.

In a certain class of 300 students, the number of students who either do not study at home or do not attend classes is a third more than of those who either study at home or attend classes. The number of students who do not study at home but attend classes is two fifths more than those who study at home but do not attend classes, while the number of students who study at home as well as attend classes is half of those who neither study at home nor attend classes. If the number of students who only study at home or only attend classes is a third less than those who do either, then how many students who either do neither or do both?

A.

120

B.

150

C.

180

D.

210

#### SOLUTION

Solution : C

Let a be the number of students who do not study at home and do not attend classes.
Let b be the number of students who do not study at home but attend classes.
Let c be the number of students who study at home and attend classes also.
Let d be the number of students who study at home but do not attend classes.
Given: d+b=23(b+c+d) and d+c+b=34(a+b+d)
Also, b = 7d5 and c = a2.
If d+b=2xb+c+d=3x, and a+b+d=4x
(a+b+d)+(c+b+d)(b+d)=300
x=60
a+c=180

Option (c) is the right answer.

ABC is an equilateral triangle and AX,BY and CZ are three parallel lines. Find the area of triangle ABC

A.

a2+b2ab3

B.

a2+b2+ab3

C.

a2+b23

D.

a2b23

#### SOLUTION

Solution : B

There is no constraint given, other than that ABC is an equilateral triangle and AX || BY || CZ.
We can therefore assume that X,Y and Z are points on the side AC itself. If we assume the equilateral triangle to have a side=2, and the Y is the mid point of AC, then AY=YB=1
Area of the triangle = 34 a2 = 3
Look in the answer options and eliminate those options where you do not get √3 on substituting a=1 and b=1.

A quadrilateral is obtained by joining the mid-points of a rectangle. Further, a rectangle is obtained by joining the mid points of quadrilateral obtained above. Again another quadrilateral is obtained by joining the mid points of the second rectangle. This process is repeated for an infinite number of times. Find the ratio of the sum of areas of all the rectangles to the sum of the areas of all the quadrilaterals.

A.

2:1

B.

3:1

C.

4:1

D.

2:3

#### SOLUTION

Solution : A

The quadrilateral formed by joining the mid-points of a rectangle is a rhombus.
Area of rhombus = 12 * area of rectangle

Area of inner rectangle = 12 * (area of rhombus)

Let the area of the biggest rectangle be 1 square unit.
Therefore, Sum of area of all the rectangles obtained = 1 + 14 + 116 + ... +  = 1(114) = 43 square units

Sum of area of all the rhombuses = 1218132 + ........ = 12114 = 23
Required ratio = 4323 = 2:1

A rectangular block L×100×H, with L100H, where L and H are integers, is cut into two non-empty parts by a plane parallel to one of the faces, so that one of the parts is similar to the original. How many possibilities are there for (L, H)?

A.

10

B.

12

C.

20

D.

24

#### SOLUTION

Solution : B

We must cut the longest edges, so the similar piece has dimensions L×100×k for some 1k<H.
The shortest edge of this piece cannot be L, so it must be k. Thus L×100×H and k×L×100 are similar.
Thus, Lk=100L=H100.
So, HL = 1002. 1002 = 24×54
So 1002 has 25 factors, of which (251)2=12 pairs are <100.

A triangle ABC has 2 points marked on the side BC, 5 points marked on the side CA and 3 points marked on side AB. None of these marked points is coincident with the vertices of the triangle ABC. All possible triangles are constructed taking any three of these points and the points A, B, C as the vertices. How many new triangles have atleast one vertex common with the triangle ABC?

A.

156

B.

128

C.

107

D.

127

#### SOLUTION

Solution : D

Total points including the vertices = 13
Out of these 13 points,
5 points on side AB ( including B and A) are collinear
7 points on side CA (including C and A) are collinear
4 points on side BC (including B and C) are collinear
thus, total triangles possible 13C3 - 5C3 - 7C3 - 4C3 = 237
Out of these,
we should exclude those triangles which do not share the
vertices A or B or C
Such triangles =
133C3 - 52C3 - 72C3 - 0 = 109
We should also exclude the triangle ABC itself.
So, possible new triangles = 237 - 109 - 1 = 127.

AVR cinema hall has 500 seats. Cost of each ticket is Rs.300 per seat. At this price, the owner will be able to fill the entire hall. If the price of the ticket goes up by Re.1, the number of seats sold goes down by 1. At what price of the ticket, will he maximize his revenue?

A.

450

B.

350

C.

425

D.

None of these

#### SOLUTION

Solution : D

Revenue will be maximum when the cost = number of seats, as revenue = (cost of each seat)×(number of seats) i.e. the value of both = 400

For all positive integers n, and n1,
let an+1=ana2n1 and a0=1, a1=8 then a10=?

A.

2(2111)

B.

2(2101)

C.

2(291)

D.

2(281)

#### SOLUTION

Solution : B

Method 1: Conventional Approach

Given : an+1=ana2n1
an+1an=(anan1)21
(anan1)2(an1an2)22
.             .
.             .
(a2a1)2n1=(a1a0)2n
Multiplying all equations, we get
an+1an=(a1a0)2n=82n
Now a10a9=829, a9a8=828......
a10=82928+2726+2524+2322×a2 (a2=8.12=8)
a10=821013
a10=2(2101)

Method 2: Shortcut: Double Substitution

This technique involves writing the answer choice in terms of variables. We need to find a10.

Thus n=10. Now rewrite all the options in terms of n.

a. 2(2n+11)                b. 2(2n1)                c. 2(2n11)             d. 2(2n21)
At n=2 in the question, a2=a1×a20=8×1=8
Now put n=2 in each of the answer choices and check where you are getting 8.
a. 2(231)8              b. 2(221)=8             c. 228                 d. 208
Thus, options a, c, d can be eliminated. Answer is option (b).

AEB = 90 and BG is the bisector of ABC. DE is parallel to side BC. Find AF: FC.

A.

1: 2

B.

2: 3

C.

1: 4

D.

1: 3

#### SOLUTION

Solution : A

Let EBC = x
DE is parallel to BC, so DEB = x
BG is the angle bisector so DBE = x
=> BD = DE

AED = 90 - x, BAE = 90 - x
D is the midpoint. DE is parallel to BC
F is the midpoint of AC

Required Ratio = 1: 2

f(x) = | 2x -3 | - 1 ; g(x) = | x2 -2x | ; at how many points these two

graphs will intersect in the first quadrant?

A.

0

B.

1

C.

2

D.

3

#### SOLUTION

Solution : A

Graphs will intersect at only one point in the first quadrant

In the first quadrant, both f(x) and g(x) will be positive.
2x- 3 - 1 = x2 - 2x
x2 - 4x + 4 = 0
(x- 2)2 = 0

x=2  at only one point.

A set S of real values consists of the following 5 elements, {4, 8, 12, 16, x} in any order. For how many values of x does the mean of set S equal the median of set S?

A.

0

B.

1

C.

2

D.

3

#### SOLUTION

Solution : D

To solve this problem quickly, you can come up with likely values forxthat would make the mean equal to the median.
Three values that would make the set symmetrical are 0, 10, and 20:
{0, 4, 8, 12, 16}
{4, 8, 10, 12, 16}
{4, 8, 12, 16, 20}

To verify if there are any other possibilities for x, we need to evaluate each range of "x”

(1) If
is less than or equal to 8, then the median is equal to 8. We now set the mean equal to the median:

40+x5=8        40+x=40        x=0

(2) If x is between 8 and 12, then the median is equal to x.

40+x5=x        40+x=5x        40=4x       x=10

(3) If x is greater than 12, then the median is equal to 12.

40+x5=12        40+x=60

If (1*1! + 2*2! + 3*3! +...+63*63!) is divided by 64, find the remainder.

A.

0

B.

15

C.

63

D. Cannot be determined

#### SOLUTION

Solution : C

The above series can be written as (2-1)*1! + (3-1)*2! + (4-1) 3! +...+(64-1)*63!

= 2! - 1! + 3! - 2! + 4! - 3! +.....+ 64! - 63!

= 64! - 1!

Therefore the remainder is (-1) or 63.

There are certain number of pipes with equal efficiencies and two tanks A and B. Capacity of A is double that of B. These pipes fill tank A for half a day. Then half of them fill A and rest fill B for rest of the day. By the end of the day, tank A is completely filled. And it takes a single pipe half of next day to fill the remaining of tank B. How many pipes are there in total?

A.

3

B.

4

C.

5

D.

6

#### SOLUTION

Solution : B

Method 1:  Using variables

Let the no of pipes be x.

Number of pipe days needed to fill tank A is x2+x4

This is double the amount taken to fill tank B....

The equations is,x2+x42=(x4+1)

Thus, x=4.

Method 2: Using numbers (Assumption)

Let A have a capacity of 300 litres

B will have a capacity 150 litres

Now 4 pipes fill A for half a day = effectively 2 pipes working for an entire day = 100 litres are filled by the first half

And 2  pipes fill A for another half = effectively 1 more pipe fills A for an entire day

This means that 3 pipes fill A completely= 300 litres

Thus, each pipe fills 100 litres

Let us verify it with B

Given that in the first day, 2 pipes fill B for half a day one pipe fills B for an entire day = 100 litres

Now 50 litres is left which Is filled by one pipe in half a day.

TOTAL = 150 litres for B

Thus, our assumption is correct

There are 5 boxes, each containing 5 cubes of different colours-red, yellow, green, blue and black. I want to draw a different coloured cube from each box. If from the first and the second box, I drew red and green cubes respectively, then what is the maximum number of trials required at the remaining boxes so as to draw different coloured cubes?

A.

15

B.

12

C.

60

D.

120

#### SOLUTION

Solution : B

Number of trials needed at the 3rd box to ensure a colour other than green and red is picked is 3
Similarly, number of trials needed at the 4th box to ensure a new colour is
picked at the maximum=4
And number of trials in the last box to pick a different colour other than
that in the other 4 at a maximum=5.

Total = 3+4+5=12

Find the complete range of values of x that satisfies |x+3|>x211x+30.

A.

(,3)(3,)

B.

(-9, 3)

C.

(,3)(9,)

D.

(3, 9)

#### SOLUTION

Solution : D

Conventional Approach :
Given inequality is |x+3|>x211x+30
We consider 2 cases
Case (i) : x>3
|x+3|=x+3
x+3>x211x+30x212x+27<0
(x9)(x3)<0
x= (3,9)
Case (ii) ; x<3
|x+3|= -x-3
x3>x211x+30
x210x+33<0
x210x+25+8<0=(x5)2+8<0
(x5)2+8 is always a positive quantity, irrespective of the value of x. Hence we do not have any value for x in the interval. Hence  the valid interval for x is (3,9).

Shortcut- Elimination Approach

Step 1: Choose a value for x which is present in 2 answer options and absent in the other 2. Eg) x=4 satisfies the range specified in option (a) and (d), but does not form a part of the range in option (b) and (c)

At x=4

LHS= |4+3|=7

RHS= 16 - 44+30 = 2. LHS>RHS. It is satisfied. This means that answer options, which do not include x=4 can be directly eliminated as they can NEVER be the right answer

Step 2- Choose a value for x which is there in one of the remaining 2 options and absent in the other

Eg) Take x=10. This option satisfies the range in option (a) but is absent in option (d)

At x=10

LHS= |10+3|=13

RHS= 100-110+30= 20

LHS<RHS. This violates the condition in the question. This means that option (a) can never be the right answer.

Thus, by elimination, answer is option (d)

Two different primes are said to equalize an integer if they are the same distance from the integer on either sides of the number line. Which integer among the following is the most highly equalized?

A.

20

B.

18

C.

17

D.

15

#### SOLUTION

Solution : B

20 is equalized by 3 pairs of primes (37,3), (29,11) and (23,17).

17 is equalized by 3 pairs of primes (31,3), (29,5) and (23,11).

18 is equalized by 4 pairs of primes (31,5), (29,7), (23,13) and (19,17).

15 is equalized by 3 pairs of primes (23,7), (19,11) and (17,13).

If a,b,c are the lengths of the perpendiculars dropped from a point inside on an equilateral triangle of side "p” on its sides, then find a+b+c

A.

p2

B.

P

C.

(32)p

D.

3p

#### SOLUTION

Solution : C

Let assume the point be centroid of the triangle.
The perpedicular distance from centroid to side of an equilateral triangle = 3p6
a+b+c =  3p2

If f(x) = 1 - f(1-x) and then f(1100)+f(2100)+.....+f(98100)+f(99100).

A.

50

B.

49.5

C.

49

D.

98

#### SOLUTION

Solution : B

f(x)=1f(1x)f(x)+f(1x)=1                                             f(12)=1f(12)f(12)=12

f(99100)=f(11100)=1f(1100)

f(98100)=f(12100)=1f(2100)

So,f(1100)+f(2100)+.....+f(98100)+f(99100)

=f(1100)+f(2100)+.....+1f(2100)+1f(2100)

[f(50100)]=f(12)=12

=49+f(50100)

=49+12=49.5

Hence option (b).

Consider the non-decreasing sequence of positive integers:

1, 2, 2, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 5.... in which nth positive number appears n times. Find the remainder when the 2000th term is divided by 4.

A.

0

B.

1

C.

2

D.

3

#### SOLUTION

Solution : D

Let us see the sequence of the numbers:

Number  Last term of the number

1                  1
2                  3
3                  6
4                  10
n                  n(n+1)2

We have to find the value of n for the 2000th term.

If n = 62, the last term that ends with n is (12×62×63) = 1953.

Therefore, the next 63 terms are 63. So the 2000th term is 63. So the remainder when divided by 4 is 3.

Answer the questions based on the information given below.
A microorganism Xeta combines with another microorganism of its kind to form a single microorganism unit. If it does not find another Xeta every second, it dies. There are 2x1 Xeta in the incubation flask initially. Xeta will find another Xeta provided it is there in the same chamber.

When will all the Xeta be dead?
A.

x seconds

B.

x-2 seconds

C.

x-1 seconds

D.

2x seconds

#### SOLUTION

Solution : A

Assume no. of Xeta=1 (you can assume any value which satisfies 2x1)

Then, x = 1.

If there is 1 Xeta, it will die after the 1st second.

Put x=1 in the options and check which option gives answer as 1.
Answer will be option (a) - x seconds.

Answer the questions based on the information given below.
A microorganism Xeta combines with another microorganism of its kind to form a single microorganism unit. If it does not find another Xeta every second, it dies. There are 2x1 Xeta in the incubation flask initially. Xeta will find another Xeta provided it is there in the same chamber.

The time taken by a Xeta added initially to die, increases by 1 second. How many more Xetas need to be added in the incubation flask to ensure that this happens?
A.

3

B.

2x+2

C.

2(x+1)

D.

1

#### SOLUTION

Solution : D

Now we need to increase the time to die by 1 second. Assume values as previously done. The Xeta need to die after x+1 = 1+1 = 2 seconds.
Option (d) = 1+1 = 2 Xeta
1st second 2 Xeta will become one.

2nd second The remaining Xeta will die. Answer is (d).

If n is a positive integer greater than 1, then PRO(n) represents the product of all the prime numbers less than or equal to n. The largest prime factor of PRO(12)+11 is:

A.

2

B.

11

C.

13

D.

None of these

#### SOLUTION

Solution : D

PRO (12)+11=2×3×5×7×11+11
=11×(2×3×5×7+1)
=11×(210+1)
=11×211

Since 211 is prime, the largest prime factor = 211.

Two casks contain mixtures of petrol (P) and kerosene (K). in the first cask P:K=7:3 and in the second cask P:K= 3:1. in what ratio should the mixtures from the two casks be taken to give a mixture in which P:K is 11:4?

A.

2:3

B.

2:2

C.

1:2

D.

1:3

#### SOLUTION

Solution : B

Using alligation, we can directly solve this question

Take either petrol or kerosene.

Petrol in mixture =1115

Ratio to take the mixture =160:130=1:2

In how many ways can 4 balls be selected from a box containing 1 red, 2 blue, 3 white and 4 black balls such that at least two selected balls are of the same colour? It is known that all the balls of a colour are identical.
___

#### SOLUTION

Solution :

2 same, 1 different, 1 different = 3C23C2=9

3 same 1 different = 2C23C2=6

4 same = 1

Total = 16

A man, whose bowling average is 12.4, takes 5 wickets for 26 runs and thereby decreases his average by 0.4. The no. of wickets taken by him before his last match?
___

#### SOLUTION

Solution :

(12.4w+26)(w+5)=12

On solving, w=85

How many numbers are there between 1 and 1000 with their sum of digits  5?

A.

56

B.

55

C.

105

D.

75

#### SOLUTION

Solution : B

We need a number of the form ABC where A+B+C5, and A,B,C are whole numbers.

Method 1 - Trial and Error - Manual approach

List out all the possibilities manually,

Case 1: When sum of digits=1

A+B+C=1

We can solve this using the similar to different approach where there is 1 zero and 1 one. Number of combinations =3C2=3

Case 2: When sum of digits=2

A+B+C=2

Number of possibilities = Answer based on 2 zeroes and 2 ones =4C2=6

Case 3: When sum of digits=3

A+B+C=3

Number of possibilities = Answer based on 3 zeroes and 2 ones =5C2=10

Case 4: When sum of digits=4

A+B+C=4

Number of possibilities = Answer based on 4 zeroes and 2 ones =6C2=15

Case 5: When sum of digits=5

A+B+C=5

Number of possibilities = Answer based on 5 zeroes and 2 ones =7C2=21

Total = 21+15+10+6+3 = 55

Method 2- Shortcut

We will use a dummy variable "D” to take the place of the sign.

Then the equation becomes -

A+B+C+D=5 (Where D can take values 0,1,2,3,4,5)

A+B+CD504132231405

Using the same Similar to Different approach, we can get the answer based on 5 zeroes and 3 ones as 8C3=56

We must remove the case where D=5, because in that case A+B+C=0, which is not possible. Thus, the answer is 56-1=55.

Which of the following will be the sum of n positive numbers, given that their product is unity?

A.

a positive integer

B.

divisible by n

C.

equal to n+1n

D.

never less than n

#### SOLUTION

Solution : D

Take 2 and 12 as the two positive numbers whose product is unity. Here n=2, only option (c) and (d) is satisfied. The other two can never be true.

Take the 4 positive numbers as 1,1,1 and 1. Here n=4, option (c) is not true. Hence, answer is option (d).