# Free Quant Practice Test - CAT

### Question 1

Ken is the best sugar cube retailer in the nation. Trevor, who loves sugar, is coming over to make an order. Ken knows Trevor cannot afford more than 127 sugar cubes, but might ask for any number of cubes less than or equal to that. Ken has to prepare n cups of cubes, where n is as small as possible, with which he can satisfy any order Trevor might make. How many cubes are in the cup with the most sugar?

22

64

81

124

#### SOLUTION

Solution :B

This question is entirely based on the powers of 2 less than 127.Thus the answer is 26=64 cubes in the cup with the most sugar.

How is that? Let us look at this important concept using the unitary approach:

Take the number 5.

In other words the question is - how many numbers would you need so that all the numbers from 1 to 5 are possible using addition.

The powers of 2 less than 5 are 20=1, 21=2 and 22=4.

1= 1

2= 2

3= 1+2

4=4

5=4+1

Thus, we see that with powers of 2 we can easily write all the numbers below the highest power of 2 taken.

In case we take up to 22, then we can write all values upto 23−1 or 7.

20=1

21=2

21+20=3

22=4

22+20=5

22+21=6

22+21+20=7

Hence with 3 numbers you can write upto 7.

Hence to add up to 127 you will have to consider the powers of 2 till 26.So the cups must contain 1, 2, 4, 8, 16, 32, and 64 cubes of sugar, with 64 being the most number of cubes in one cup.

### Question 2

Let 3a=4, 4b=5, 5c=6, 6d=7, 7e=8 and 8f=9. What is the value of the product abcdef?

1

2

√6

3

#### SOLUTION

Solution :B

8f=9→8=91f

7e=8→7e=91f→7 = 91ef

Continuing in the same way we get 3=91abcdef

⟹3=32abcdef

⟹2abcdef=1⇒abcdef=2

### Question 3

There are a certain number of chairs inside the hall. If the chairs are arranged 5 chairs per row, 7 chairs per row, and 8 chairs per row then 3, 2 and 5 chairs remain respectively. It is known that there are more than 600 chairs and less than 700 chairs inside the hall. Find the number of chairs remaining if the chairs are arranged 13 chairs per row.

2

3

4

5

#### SOLUTION

Solution :B

This question can be solved using Chinese Remainder theorem.

Suppose the no. of chairs is N. It is also known that 600<N<700.

Suppose the quotients when N is divided by 5, 7 and 8 and leaves the remainders 3, 2 and 5 be A, B

and C respectively.

Then, N = 5A+3 = 7B+2 = 8C+5.......................(1)

Solving first for

5A+3=7B+2

We get the first integral solution for (A,B) at (4,3).

The first solution for the above pair of equations = 23.

This increases in an Arithmetic Progression of 35 (LCM of 7 and 5)

Thus the general form of the above equation = 23+35k

Let us equate this to the last part of eqn (1).

8C+5=23+35K

The first integral solution for this occurs at C=11 and K=2

Thus the first number which when divided by 5, 7 and 8; leaves the remainders 3, 2 and 5 is 93.

This number is the first term of an AP with a common difference of 280 (LCM of 7, 5 and 8)

Thus, the number which will lie in the range of 600-700 is 653.

653 when divided by 13 leaves a remainder of 3; which is option (b).

### Question 4

Each corner of a square subtends an angle of 30∘ at the top of a tower 'h' meters high standing in the centre of the square. If 'a' is the length of the each side of square then the relation between h and a is

2a2= h2

2h2 = a2 ^{ }

3a2^{ } =2h2

2h2^{ } = 3a2

#### SOLUTION

Solution :C

We can go ahead and assume values for H and deduce the value of a, substitute it in the given equations and eliminate the wrong answer choices.

Let us assume the value of h= √3

As it is a 30-60-90 triangle we will have the ratio as 1:√3:2.

Hence the diagonal of the square will be 2 and the side 'a' will be √2

Now applying this in the answer options we see that only option (c)gives us RHS= LHS.

### Question 5

Frank and Joe are playing ping pong. For each game, there is a 30% chance that Frank wins and a 70% chance that Joe wins. During a match, they play games until someone wins a total of 21 games. What is the expected value of number of games played per match?

30

28

27

29

#### SOLUTION

Solution :A

Approach 1

The expected value of the ratio of Frank's to Joe's score is 3:7, so Frank is

expected to win 3 games for each of Joe's 7. Extending the same ratio, 3:7 = 9:21. Thus the expected number of games in a match is 30.

Approach 2- Reverse Gear Approach

The ratios imply that Frank wins 7 out of 10 matches and Joe wins 3 out of 10 matches. Total number of matches = 710xN + 310xN (N is the number of matches possible)Look for a multiple of 10 in the answer options. You can mark option (a) directly

### Question 6

The following list of numbers is given: 21, 40, 59, 78, 97, . . . , 4562. Find the number of integers in this list whose HCF with 240 is not more than 1.

40

41

64

78

#### SOLUTION

Solution :C

The Euler's number ϕ(n) for 240 is given as:

ϕ(240)=240(1−12)(1−13)(1−15)=6421, 40, 59, 78, 97, . . . , 4562 are in Arithmetic Progression with common difference as 19.

Number of terms =[(4562−21)19]+1=(454119)+1=239+1=240

Now, applying the interesting property of numbers:

"If a is prime to n, the number of terms of the Arithmetic Progression x, x+a, x+2a, . . . , x+(n-1)a which are prime to n is ϕ(n).

ϕ(n) = no. of positive integers smaller than n and co-prime to n.

Hence the number of positive integers in the above list whose HCF with 240 is not greater than 1

= The number of positive integers in the list which are prime to 240=ϕ(240) = 64.

### Question 7

Find the approximate value of x, for this equation to be satisfied: xxxxx−4=0. Given that (am)n=amn

1.41

1.92

1.54

1.732

#### SOLUTION

Solution :A

Go from answer options

If x =1.41 =√2 then we get

√2√2√2√2√2=4

Hence the equation is satisfied.

### Question 8

A parallelogram is divided into 20 regions of equal area by drawing line segments parallel to one of its diagonals. What is the ratio of the longest line segment to one the shortest one?

3:2

4:3

3:1

5:2

#### SOLUTION

Solution :C

The figure looks as above.ΔABE and ΔCDE are similar.

The area of each part is equal.

Thus the ratios of the areas= square of the ratios of the corresponding sides⇒ If area of ∆CDE = x then area of ΔABE =9x

∴CDAB=√9xx = 3:1

### Question 9

N is a five-digit number such that all its digits are non-zero and even, and there is only a single repetition. If the repeated digits are adjacent, and N is divisible by 4, how many possible values can N take?

36

48

44

52

#### SOLUTION

Solution :B

The digits of N can be 2, 4, 6 and 8. Since N is divisible by 4, the last two digits must be 24 or 28 or 44 or 48 or 64 or 68 or 84 or 88.

When the last two digits are the same, i.e., 44 or 88, there can be no more repetition; then, the first three digits can be chosen in 3! Ways. Thus, the number of possibilities = 3! x 2 = 12.

When the last two digits are distinct, we get the following cases:

i] Third digit is same as fourth digit.

Number of ways remaining two digits can be chosen = 2!

Ii] Third digit is same as second digit.

Number of ways = 2!

iii] Second digit is same as first digit.

Number of ways = 2!

Thus, when the number ends with 24, the number of possibilities = 2! + 2! + 2! = 6.

Since there are 6 cases where the last two digits are distinct, the total number of ways =6 × 6+ 12 = 36 + 12 = 48. Hence, [b].

### Question 10

Find the complete range of values of x for which (x2−7x+13)x<1.

(−∞,+∞)

(0,∞)

(−∞,4)

(−∞,0)∪(3,4)

#### SOLUTION

Solution :D

Check for values which are included in the range in 2 options and not there in two options i.e. substitute x=0, equation is not satisfied at x=0; thus option (a) and option (c) can be eliminated next, check for values which are there in option (b) and not there in option (d) or vice versa. Put x=1The equation is not satisfied at x=1, LHS= 6 and RHS=1. Thus the correct range is option (d)

### Question 11

The average age of n persons is 55.Two persons aged 41 and 47 respectively leave the group and three new persons aged between 60 and 65 join the group. If the average age of the group goes up by one year and if the initial number of people in the group is a multiple of 7, what is the number of persons now in the group?

42

43

44

45

#### SOLUTION

Solution :B

Let x,y,x be the ages of 3 persons which lies between 60 and 65

The equation governing the problem 55n−41−47+(x+y+z)n+1=56

Also given n

is a multiple of 7

55n- 88 +k = 56n + 56 where k=x+y+z, n+ 144=k

Now k ranges between 3 × 61 and 3× 64

183 ≤ n+44 ≤ 192

39 ≤ n ≤ 48Thus n =42, the latest number n + 1 =42+1=43

ShortcutGo from answer options. The initial number of people in the group is a multiple of 7. The answer options give the number of initial people + 1. Thus the correct answer is option (b).

### Question 12

Find the area of the biggest possible triangle whose base is 10 cm and perimeter 36 cm.

48 cm2

54 cm2

66 cm2

60 cm2

#### SOLUTION

Solution :D

Among all triangles having a specified base and specified perimeter, an isosceles triangle with the specified base has the largest area.

Hence the triangle is isosceles with perimeter 36 cm and base 10 cm.Each equal side of the triangle -

=36−102=13 cm

∴ Height of the triangle

=√132−(102)2.

=√169−25=12 cm

∴ Area of the triangle

12×10×12=60 cm2.

### Question 13

ABCD is a rectangle, with AP = PQ = QB and points R, S, T dividing the side CD in four equal parts. Find the area of the shaded region if the area of the rectangle is A sq. units..

A12

2A9

A8

A4

#### SOLUTION

Solution :C

We know that the area of all triangles between two parallel lines and having a common base is the same

Thus, the triangle can also be represented as follows

we can say that, area (RQS) = area (RR1S)

Using the concept of graphical division, we can divide the figure into 8 equal regions as follows.

The required region is A8 units.

### Question 14

What is the remainder when p - q is divided by 100, if p=299+499+699+899+.........+10099 and q=199+399+599+799+..........+9999?

50

10

20

30

#### SOLUTION

Solution :A

99=4k+3

Thus, if you take the individual differences (upto 10)

For (299+499+699+899+1099)−(199+399+599+799+999)⇒

Last digit of this expression = (8+4+6+2+0) -(1+7+5+3+9) = -5 = 5 (10-5)

However, since there are 10 such groups; (since 1099 or multiples of 1099 are exactly divisible by 100)

=5×10=50 (which will always end in zero)Thus the remainder when divided by 100 is the last two digits = 50 itself

### Question 15

The total amount of money spent on the purchase of pencils, rubbers and pens is Rs. 45. The cost of one pencil, one rubber and two pens is Rs. 2, Rs. 3 and Rs. 8 respectively. The total number of pencils bought is greater and less than the number of pens and the number of rubbers bought respectively. How many different combinations of the number of pencils, rubbers and pens bought are possible?

1

2

3

4

#### SOLUTION

Solution :C

Let x, y and z be the number of pencils, rubbers and pens bought. The cost of one pencil, one rubber and one pen is Rs.2, Rs.3 and Rs.4 respectively.

Now, 2x + 3y + 4z = 45.∴x=45−3y−4z2

Also,z<45−3y−4z2<y

Or, 6z+3y < 45 and 5y+4z >45.Possible values of y and z that satisfy the above equation is (7, 3); (9, 2) and (11, 1) in that order. Therefore, there are 3 different possible combinations.

### Question 16

Let three 3-digit integers abc, bca, cab when divided respectively by 10, 8 and 3 leave the same remainder 1. If 2<b<8, then a = ?

3

4

5

#### SOLUTION

Solution :A

abc leaves remainder of 1 when divided by 10⟹c=1.

bca leaves remainder of 1 on dividing by 8⟹4b+2c+a=8m+1 and similarly a+b+c = 3n+1.

On solving we get 3b = 8m -1 - 3n ⟹ m = 2, 5, 8,... or 3p-1 format.

Now, 4b+2+a = 8m+1 ⟹ 4b+a = 8m-1

For p=3, m=8, 4b+a = 63 means b > 8. So, p cannot be greater than 2.

For p=2, m=5, 4b+a = 39, if a = 3, 7 means b≥8 which is not possible.

So, p=1, m=2, 4b+a = 15, possible value of a = 3.

Hence, choice (a) is the right option.

### Question 17

A circle of radius 3 crosses the centre of a square of side length 2. Find the approximate positive difference between the areas of the non-overlapping portions of the figures

26

24

22

cannot be determined

#### SOLUTION

Solution :B

Let the area of the square= s, the area of the circle= c, and the area of the overlapping portion= x. The area of the circle not overlapped by the square is "c -x” and the area of the square not overlapped by the circle is "s - x”, so the difference between these two is (c -x) -(s -x) = c -s = 9π2 -4 .(approximately = 24.26). Hence the correct answer is option (b).

### Question 18

How many pairs of positive integers (a, b) are there such that a and b have no common factor greater than 1 and ab+14b9a is an integer?

1

2

3

4

#### SOLUTION

Solution :D

Let u=ab. Then the problem is equivalent to finding all possible rational numbers 'u' such that

u+149u=k, for some integer k.

This equation is equivalent to 9u2−9uk+14=0

9u2−9uk+14=0 whose solutions are,

u=9k±√81k2−50418

=k2±16√9k2−56

Hence u is rational if and only if:

√9k2−56 is rational, which is true if and only if 9k2−56 is a perfect square.

Suppose that 9k2−56=S2 for some positive integer S.

(3k+S)(3k-S)=56

The only factors of 56 are 1,2,4,7,8,14,28 and 56.

So, (3k-S) and (3k+S) is one of the ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no integral solutions.

The cases (2, 28) and (4, 14) yield k = 5 and k = 3 respectively.

If k=5, then u=13 or u=143.

If k=3, then u=23 or u=73.

Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3).

### Question 19

Let a, b, c, d and e be distinct integers such that (6 - a) (6 - b) (6 - c) (6 - d) (6 - e) = 45. What is a+b+c+d+e?

25

28

27

16

#### SOLUTION

Solution :A

If 45 is expressed as a product of five distinct integral factors, the unique way is:

45 = (+1) (-1) (+3) (-3) 5 (no factor has its absolute value greater than 5)

⇒ (6 - a) (6 - b) (6 - c) (6 - d) (6 - e) = (1) (-1) (3) (-3) 5

The corresponding values of a, b, c, d and e are 5, 7, 3, 9 and 1 and their sum is 5 + 7 + 3 + 9 + 1 = 25.

### Question 20

Sum of the first 30 terms of an arithmetic progression is 0. If the first term is -29, then find the sum of the 28th, 29th and 30th terms of this arithmetic progression.

81

84

-84

-81

#### SOLUTION

Solution :A

Let the common difference of the arithmetic progression be 'd'.

Sum of first 30 terms of the arithmetic progression

=302×[2(−29)+(30−1)d]

Hence, 15(−58+29d) = 0

Hence, d=2

Sum of 28th, 29th and 30th term of this arithmetic progression

= 3(-29) + (27 + 28 +29) × 2 = 81

### Question 21

A test has 50 questions. A student scores 1 mark for a correct answer, −13 for a wrong answer, and −16 for not attempting a question. If the net score of a student is 32, the number of questions answered wrongly by that student cannot be less than

6

3

2

5

#### SOLUTION

Solution :B

Let the number of correct answers be 'x', number of wrong answers be 'y' and number of questions not attempted be 'z'.

Thus,x+y+z=50 .............(i)

And x−y3−z6=32

The second equation can be written as,

6x-2y-z=192 ...................(ii)

Adding the two equations we get,

7x−y=242⇒x=2427+y7=242+y7

Since, x and y are both integers, y cannot be 1 or 2. The minimum value that y can have is 3.

Hence option (b)

### Question 22

There are several boxes with coins bearing a value (1 to n) each. The coins in different boxes are grouped as follows:- 1(first box), 2+3(second box), 4+5+6(third box), 7+8+9+10(fourth box), ..... and so on. The value of the coins in the 15th box is:

2098

1695

5675

1687.5

#### SOLUTION

Solution :B

123456

This equation is based on the n(n+1)2 pattern, where the last coin in the nth box will have the value n(n+1)2.

Thus the last coin in the 15th box will have a value =15×162=120

The first coin will have a value = 106

Thus the sum =152(106+120)=1695.As there are 15 coins in the 15th box and it is 15 numbers ... it has to be a multiple of 15. Only option (b) satisfies this condition.

### Question 23

Disha wants to paint her room brown. For that she uses a certain violet paint which contains 30% blue pigment and 70% red pigment by weight. She mixes that violet paint with a certain green paint which contains 50% blue pigment and 50% yellow pigment. When these paints are mixed to produce the brown paint which she uses, it contains 40% blue pigment. If the brown paint weighs 10 grams, then the red pigment contributes how many grams of that weight?

2.8 gms

3.5 gms

4gms

5.3 gms

#### SOLUTION

Solution :B

Use alligation,

Considering only the blue pigment in all the paints

Thus, 5 gms of each paint is used. Weight of red paint = 710×5 =3.5 gms. Option (b)

### Question 24

In a board meeting, 8 delegates from number 1 to 8 are sitting around a circular table. Number 4 and 5 always sit together and number 1 always sits next to number 4. If number 8 always sits exactly opposite number 3, in how many ways can the seating be done?

6!

6C2 × 3

4×3!

4! × 2

#### SOLUTION

Solution :C

Let's fix number 8 at a slot at the table, automatically, number 3 also gets fixed. Out of the remaining 6 delegates, 4 and 5 will be together and can be arranged in 2! Ways amongst each other. Also for each of these ways, number 1 can sit next to 4 in only one way.

Now 1,4,5 can be placed only in 2 ways(See below).

This question now becomes the arrangement of the remaining 3 people × 4= 3! × 4. Option (c)

### Question 25

S1,S2,S3....Sn is the sum of an infinite GP with first terms as 1,2,3....n and common ratios as 12,13....1n+1 then find S1+S2+S3....+Sn

12n2

12n(n+3)

n2

#### SOLUTION

Solution :C

S1=2

S2=3

S1+S2=5

Here n=2

Go from answer options

n(n+1)=2.3 =6

12n2=12×4=2

12n(n+3)=12(2)(5)=5

n=2

n2=4

### Question 26

What are the last two digits of 1325255?

25

35

75

#### SOLUTION

Solution :A

Any number ending with 25 raised to any power will always end in 25.

### Question 27

ABCD is a cyclic quadrilateral while triangle ABF and triangle CEF are similar triangles. Which of the following cannot be the value of angle BAC if ∠BEC = 30∘?

20

40

60

50

#### SOLUTION

Solution :A

Let ∠ACD = ∠BAC = x. Connect D to B such that the line intersects AC at O.

As the quadrilateral ABCD is cyclic, ∠AOD = 2 * ∠ACD. (Because O is the centre of the circle around the cyclic quadrilateral, and angle subtended at center is twice the angle subtended at a point on the circumference by the same arc AD, here.

So, ∠AOD = 2x

∠EBA = ∠BED = 30

∠BDC = ∠DBE + 30

∠AOD is exterior angle for triangle COD.

So, ∠AOD = 2x = 30 + ∠DBE + x

∠x = 30 + ∠DBE.So, ∠BAC can't be less than 30 degrees so 'a' is the answer.

### Question 28

What is the remainder when 120212 is divided by 1000?

120

1

0

296

#### SOLUTION

Solution :D

Write (1202)12=(1200+2)12. In this expansion there will be 13 terms.

From the binomial theorem, the (r+1)th term is given as tr+1=nCran−rbr

The 13th term is 212=4096. 12th term is 12C11×12001×211=12−×12−00×2048−=−−−200.

When we add all 13th and 12th term, we get 296 as the last three digits.

The previous terms (1 to 11) will not make a difference to the answer as there will be a minimum of 12002 involved in the other terms, therefore the terms will end with more than 3 zeroes.

### Question 29

There is a ship which is moving away from an iceberg 150 m high, takes 2 mins to change the angle of elevation of the top of the iceberg from 600 to 450. What is the speed of the ship?

2.46 kmph

1.3 kmph

1.9 kmph

1.7 kmph

#### SOLUTION

Solution :C

In a 45-45-90 triangle, sides are in the ratio 1:1:√2

Thus OB= 150

In a 30-60-90 triangle, sides are in the ration 1:√3:2

Thus OA= 150√3

Distance travelled in 2 mins = AB

= 150- 50√3 = 63.4 m

speed = (63.42)×(601000)= 1.9 kmph

### Question 30

b) f (0) = 2

What is the value of f(98)?

0

2

98

100

#### SOLUTION

Solution :D

2 = f(0) = f(-98 + 98)

= -98 + f(98)f(98) = 98 + 2 = 100. Hence option (d)

### Question 31

Find the remainder when 3001000 is divided by 19?

#### SOLUTION

Solution :This needs the application of basic remainder theorem, Euler's theorem and frequency method.

First find the remainder when 300 is divided by 19 = 15.

The problem now changes to 15100019.

From Euler's theorem, the Euler's number of 19 is 18 (for all prime numbers, Euler's number=N-1).1000=18×55+10

Problem now changes to 151019.

1519 gives remainder -4. So the remainder of (−4)1015 or (4)1015 needs to be found.

Now (4)215 gives a remainder of 1.

Going by the frequency method,

(4)1015 will give a remainder of (1)515, i.e 1.

### Question 32

An bullet can travel in the air at a speed of 1000 kmph. When walls of 10 cm thickness are placed in its path, the speed at which the bullet emerges out of the last wall in its path diminishes by a quantity which varies as the square root of the number of walls placed in its path. With four walls its emergent speed is 300 kmph. The greatest number of walls that can be placed in front of the bullet so that the bullet emerges out of the final wall in its path is :

#### SOLUTION

Solution :

Decrease in speed =1000-300 = 700 kmph.

700 = k√4⇒ k= 350

Now, for the speed to decrease to zero.

1000 = k√4=350√x, where x is the number of walls in the way which will cause the speed of the emergent bullet to become 0.

x=8.16

So,if 8 walls are placed in the bullet's way, there will be some residual emergent speed.If 9 walls are placed, it will not emerge out of the 9 th wall.

### Question 33

Suppose f(x) = (x+1)3. If g(x) is the function whose graph is reflection of the graph of f(x) with respect to the line x = 3 and h(x) is the reflection of g(x) with respect to the x-axis, then h(8) equals:

#### SOLUTION

Solution :Reflection of the functionf(x) = (x+1)3 about the line x=3 is going to be g(x) =(x - 7)3. g(x) is the translation of f(x) 4 units to the right of x=3. h(x) being the reflection of g(x) about the x-axis is going to be -(x - 7)3 So the value of h(x) at x=8 will be -(8 - 7)3^{ }= -1.