# Free Quant Practice Test - CAT

### Question 1

A synthetically developed microorganism multiplies continuously by giving birth to two new cells once in every minute. At time t=0, a single such new micro-organism is placed in a petridish. If each microorganism lives for exactly 25 minutes, then find the total number of live microorganisms in the dish at the end of the 20th minute?

320−12

321−12

221−1

320

#### SOLUTION

Solution :D

Double substitution:

Step 1- Rewrite the answer options in terms of n (here, n=20)

a) 3n−12 b)3n+1−12 c)2n+1−1 d)3n

Let the single organism be called A0. At t=2, the number of new microorganisms present would look as follows:

Thus, at t=2, there are 9 microorganisms alive. Put t=2 in the answer options, and check where you are not getting 9. Eliminate those options.

Only option (d) gives 9. Answer is option (d).

### Question 2

Find the tens place of the number 7281×3264.

2

4

6

8

#### SOLUTION

Solution :C

To find the tens place, we need to find the remainder when this number is divided by 100.

7281100∣r=((74)70×71)100∣r=((..49×..49)70×71)100∣r=((..01)70×07)100∣r=..07

Similarly, 3264100∣r=(34)66100∣r=(32×32)66100∣r=(09×09)66100∣r=8166100∣r=..81

(66 is of the form 5n+1, and the tens digit of 81n repeats after a cycle of 5)Answer is the tens digit of 07×81=..6.

### Question 3

If a, b, c are three consecutive natural numbers, then the expression (a + b - c) (a + c - b) (b + c - a) is:

positive

negative

non - negative

#### SOLUTION

Solution :D

Take any three consecutive natural numbers as: n-1, n, n + 1

Now add any two and subtract third from the sum.

n - 1 + n - (n + 1) = n - 2; n + n + 1 - (n - 1) = n + 2; n - 1 + n + 1 - n = n

So you will get either n - 2 or n + 2 or n. As you are considering natural numbers, smallest natural number is 1 ⟹ n -1 = 1 or n = 2. (So that n - 1 is a natural number)

So n - 2, n, n + 2 all three will be non - negative. So the given expression is also non - negative.

### Question 4

In the following figure, arc AB is a quadrant of the circle with centre at point O and radius 4 units. Further, OC = OD, OC:OA = 3:4. A circle C is drawn such that it touches the quadrant at the point Y and the mid-point of AB at X. What is the radius(in units) of the circle C?

2 - √2

√2 - 1

4 - 2√3

√3 - 1

#### SOLUTION

Solution :A

Now, AO=4 & AOX=45∘

⇒ AX = XO = 2√2

Thus, XY= 4 - 2√2 and the radius of the circle = 4−2√22 = 2 - √2

### Question 5

g(x) is defined as the sum of all the divisors of a natural number n.

If g(x) is less than 2n then it is placed in category A.

If g(x) is greater than 2n it is placed in category B.

Of the following numbers, which pair consists of numbers one of which can be placed in A and another in B?

32,64

6,64

7,28

14,84

#### SOLUTION

Solution :D

g(32)=1+2+4+8+16+32=63<2×32 (Category A)

g(64)=1+2+4+8+16+32+64=127<2×64 (Category A)

g(6)=1+2+3+6=12=2×6 (neither A nor B)

g(7)=1+7=8<2×7 (Category A)

g(28)=1+2+4+7+14+28=56=2×28 (Neither A nor B)

g(14)=1+2+7+14=24<2×14 (Category A)

g(84)=1+2+3+4+6+7+12+14+21+28+42+84=224>2×84 (Category B)Hence, option (d).

### Question 6

A merchant mixes three varieties of rice costing Rs.20/kg, Rs.24/kg and Rs.30/kg and sells the mixture at a profit of 20% at Rs.30/kg. How many kgs of the second variety will be in the mixture if 2 kgs of the third variety is there in the mixture?

5 kg

12 kg

15kg

8 kg

#### SOLUTION

Solution :A

When he sells the mixture at Rs. 30/kg, he makes a profit of 20% . This means the cost price of that mixture is 301.2 = Rs. 25/kg.

Combine the variety 1 and variety 3 first to get a mixture with price Rs. 25/kg

Applying Alligation:

So, variety 1 & variety 3 are required in the ratio 1 : 1.

Now, combine variety 2 and variety 3 to get a mixture with price Rs. 25/kg

Variety 1: Variety 3 = 1: 1

Variety 2: Variety 3 = 5: 1

So, the required ratio in which all three have to be mixed = 1: 5: 2

If third variety is 2 kgs then second variety has to be 5 kg. Hence option (a)

### Question 7

What is the maximum number of girls who opted for Cooking Classes?

75

82

83

81

#### SOLUTION

Solution :B

Number of those who opted for cooking which is more than the number of those who opted for exactly two of the three hobbies which in turn is more than those who opted for all three hobbies.

To maximize the number of girls who attend cooking classes, we only need to compare the girls who attend exactly two and who attend exactly three as all these will come under the individual subject.

So, assign 49 girls to all the three subjects and 51 girls in the exactly two. Also, 51 should be divided as equally as possible:

So, number of girls attending cooking class = 49 + 17 + 16 = 82.

### Question 8

In a group of 100 girls, each has to opt for one or more of the three hobby classes: Music, Cooking and Stitching. The number of students who opted for Stitching is more than the number of those who opted for Music which is more than the number of those who opted for Cooking which is more than the number of those who opted for exactly two of the three hobbies, which in turn is more than those who opted for all three hobbies. It is known that at least one student opted for all the three hobbies.

Determine the minimum number of girls who attend Stitching Classes.

27

36

38

35

#### SOLUTION

Solution :D

We know that number of girls attending stitching classes is maximum. Now to assign the minimum value to this, we need to minimize the number of girls attending exactly two and exactly three classes, and then distribute the rest of the numbers as equally as possible.

So, number girls attending stitching class = 32 + 2 + 1 = 35. Hence option (d)

### Question 9

In a group of 100 girls, each has to opt for one or more of the three hobby classes: Music, Cooking and Stitching. The number of students who opted for Stitching is more than the number of those who opted for Music which is more than the number of those who opted for Cooking which is more than the number of those who opted for exactly two of the three hobbies, which in turn is more than those who opted for all three hobbies. It is known that at least one student opted for all the three hobbies.

What is the maximum number of girls who opt for only Music Classes?

48

49

50

51

#### SOLUTION

Solution :A

To determine the maximum number of girls who opt for only Music Class, we need to maximize the number of girls attending Stitching as well as Music, and we will have to minimize all the other values. Thus,

So, number girls attending only music class = 48.Hence option (a)

### Question 10

Let Tn be defined as the sum to n terms of the series T. Find T34

T=288−287−286−285.......

254

255

256

234−1

#### SOLUTION

Solution :B

T1=288

T2=288−287=287(2−1)=287

T3=288−287−286=286(22−21−1)=286

Thus, the pattern is

Tn=289−n

Answer is T34=255

### Question 11

Given that a real value function g, is such that g(a+b)=a+g(g(b)) & g(0)=0. Find g(102)?

0

1

99

102

#### SOLUTION

Solution :D

Assumption

Assume a=0, b=0

g(0)=0+g(g(0))=0

Next assume a=1 and b=0

g(1)=1+g(g(0))=1+0=1

continue in the same way, assume a=2, b=0

g(2)=2+g(g(0))=2

Hence g(x)= x itself. Hence g(102)=102

### Question 12

The sum of the first 100 odd numbers is divisible by:

2, 4 and 8

2 and 4

2 only

None of these

#### SOLUTION

Solution :A

The sum of n consecutive odd numbers is given as n2 . Hence 1002 is divisible by 2, 4 and 8. Answer is option (a).

(Sum of n consecutive odd numbers =n2 is derived from unitary method as follows : 1 + 3 + 5 + 7 + 9 +...... is a series (AP) with a common difference of 2. Therefore, the 100th term would be 99×2+1= 199. Therefore the sum = (1+199)1002=10000 or n2.

### Question 13

Three boys and five girls together can finish a job in 3 days. Working on the same job 3 girls take 5 days more than the time required by 2 men. What is the ratio of efficiency of a boy to a girl?

2: 1

5: 2

3: 2

4: 1

#### SOLUTION

Solution :B

3B+5G=33.33%..............(i)

Time taken by 3 women =1003G ; Time taken by 2 men =1002B

1003G−1002B=5 →33.33G−50B=5.................(ii)

Option (a):B=2G → 11G = 33.33% → G = 3.03 , B = 6.06. Substitute in (ii), 11 - (approx 8) = approx 3.

Option (b):B=2.5G → 12.5G = 33.33% → G = 83 , B=203, Substitute in (ii), 12.5 - 7.5 = 5 which is the difference given in the question.Hence option (b).

### Question 14

If 35ab=345673 where a and b are both positive integers, then how many different values can b take?

1

2

3

>3

#### SOLUTION

Solution :C

345673=3435353

The question states that the given expression will be written as 35ab where a and b are both positive integers, i.e. a can take values 1,2 or 3. For each of these values, b will have three different values. Hence answer is option (c).

### Question 15

Savita covered 180 km distance in 10 hours. The first part of her journey she covered by bus, then she hired an auto. The speed of bus and auto is 20 kmph and 15 kmph respectively. The ratio of distances covered by bus and auto is:

1: 1

2: 3

3: 2

2: 1

#### SOLUTION

Solution :D

Average speed of Savita over the complete journey = 18010 = 18 kmph.

This means auto took 2/5 of the total time and bus took 3/5 of the total time. So, the ratio of the time taken (bus: auto) = 3: 2.

So, ratio of distance covered = 3×20:2×15 = 60 : 30 = 2 : 1.

Hence option (d)

### Question 16

What is the remainder when 1025−7 is divided by 3?

1

2

3

0

#### SOLUTION

Solution :D

10x when divided by 3, gives a remainder of 1 always. Also 73 gives a remainder of 1.Thus 1-1=0. Remainder when 1025−7 is divided by 3=0.

### Question 17

In the figure shown below, all the vertical lines are parallel to each other & are equally spaced. All the horizontal lines are parallel to each other and are equally spaced. What fraction of the area of ABCD, is shaded?

12

13

14

23

#### SOLUTION

Solution :C

The image can be redrawn as follows, by joining all the points

Thus, there are 24 unit squares. The number of unit squares shaded = 6. Required ratio = 14

### Question 18

The sum of p consecutive positive integers is 45. What is the value of p?

(1) p is even.

(2) p<9.

2

4

6

Cannot be determined

#### SOLUTION

Solution :D

Sum of first ten 10 consecutive natural numbers is 45 (p=10)

Also, the sum of 14+15+16 = 45 [453= (15)] (p=3)

Sum of 22+23 = 45 [452=22.5] (p=2)

Sum of 5, 6, 7, 8, 9, 10 = 45 [456=7.5] (p=6)

Thus, using both statements also, the answer cannot be determined.

Alternate Method:

p2(2a+p−1)=45

Now, possible values of p are 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90.

Out of these, two values of p can be considered - 2 and 6.

Thus, the answer cannot be determined.

### Question 19

Arjun can kill a bird once in 3 shots. On the assumption that he fires 3 shots, the probability that the bird is killed is x27. Find the value of x.

#### SOLUTION

Solution :P(A) = 13

P(A') = 23

P(Bird Killed) = 1 - P(None of 3 shots hit)

1− 23 × 23 × 23 = 1927

### Question 20

ABCD is a parallelogram. E is a point on CD such that BE bisects ∠ABC. BE = 12 cm, CE = 19 cm, ∠BEC = ∠AED. DE =

#### SOLUTION

Solution :∠ABE = ∠BEC (alternate angles)

∠AED = ∠EAB

∠AED = ∠BEC

∠EAB = ∠EBA

As BE bisects ∠ABC

∠EBC = ∠ABE

From (1) and (2) as the two angles of the triangle ABE are equal to two angles of triangle BEC, the two triangles are similar

So, BECE=ABBE

BE2 = (CE + ED)CE

Substituting BE = 12cm and CE = 9cm, we get ED = 7cm

### Question 21

Jay and Anup can do a job, each working alone in 30 and 15 days respectively. Jay started the work and after a few days, Anup joined him. They completed the work in 18 days from the start. After how many days, did Anup join A?

#### SOLUTION

Solution :Since Jay does (130)th of the work in one day, total work done by Jay in 18 days

=1830=35Remaining (25)th work was done by Anup

Since Anup does (115)th of the work = 2/5/(1/15) (25)(115) = 6 days.

So, Anup joined Jay after (18−6) = 12 days.

### Question 22

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

a2+b2−ab√3

a2+b2+ab√3

a2+b2√3

a2−b2√3

#### 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

Look in the answer options and eliminate those options where you do not get √3 on substituting a=1 and b=1

√34a2=√3

Answer is option (b)

### Question 23

A grid is set up as shown using 5 horizontal and 6 vertical lines. What are the no. of ways one can go from point P to point Q walking along the grids but not moving upwards or left or retracing a grid?

21

35

56

None of these

#### SOLUTION

Solution :D

There are 4 rows and 5 columns. Using the one-zero method, number of paths will be 9C4 = 126 ways

### Question 24

How many 4-digit positive integral numbers are there in base 7, if you are counting the numbers in the same base system?

2058

5666

6000

None of these

#### SOLUTION

Solution :A

No. of 4-digit numbers in base 7 = 6×7×7×7 = 2058.

### Question 25

Find the sum to n terms of the series 11.2.3.4+12.3.4.5+13.4.5.6+.......

13(n+1)(n+2)(n+3)

16(n+2)(n+3)(n+4)

14(n+1)(n)(n+2)

None of these

#### SOLUTION

Solution :D

At n=1,

Option a =13.2.3.4....This can never be the answer

Option b =16.3.4.5....This can also never be the answer

Option C =14.1.2.3.... This can be the answer

Since an option "none of these" is there , do another check

At n=2, we should get 124+1120

Put n=2, in option c=14.3.2.4=196≠11120

Hence answer = option d.

Always be careful when there is an option none of these.

### Question 26

There are 13 couples, 5 single men and 7 single women in party. Every man shakes hand with every woman once. But none shakes hand with his wife. How many handshakes in total took place in the party?

247

347

360

191

#### SOLUTION

Solution :B

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

### Question 27

The value of a gold bar is directly proportional to the square of its weight. If a gold bar weighing 8 kg breaks into 2 pieces, its total value decreases by 38 times. Find the weights of the two pieces.

4 kg, 4 kg

5 kg, 3 kg

6 kg, 2 kg

4.5 kg, 3.5 kg

#### SOLUTION

Solution :C

Value (V) = K ×82 = 64 K

Let weights be x kg and (8 - x) kg.

V1 = Kx2 and V2 = K(8−x)2

New value = K[x2+(8−x)2]

Given K[x2+(8−x)2]=58(64k)

x2+64+x2−16x=40

2x2−16x+24=0

x2−8x+12=0⇒(x−6)(x−2)=0x = 6, x = 2.

Alternatively :

Assume k=1; the solution becomes simpler.

Value is currently 64 and it becomes 58th of that which is 40 =36+4 (This is the only possible breakup with 2 square numbers) hence, answer is option (c).

### Question 28

Find the remainder when 727163 is divided by 10.

7

9

1

3

#### SOLUTION

Solution :A

We need to find out the unit digit of the given number.271634=27even number4=(−1)even number4=1

Now the question changes to finding the unit digit of 74n+1. From the power cycle of 7, unit digit will be 7. Hence, the remainder is 7.

### Question 29

A fishing motor boat sails at a speed of 50 kmph without any iron boxes on it. For every iron box of added to it, the speed of engine reduces by 10% of the earlier speed. At most how many boxes can be added so that the ship can cover a distance of 225 km in a maximum of 10 hrs? (Assume that the only additions which impede the performance of the boat are iron boxes and that all iron boxes have a weight of 7 kgs)

5

6

7

8

#### SOLUTION

Solution :C

Minimum speed at which the boat can go = 22.5 kmph

Now go from answer options

If 8 boxes are added, the speed after reduction will be (0.9)8×50=<22.5 kmphThus, the maximum number of boxes which can be added will very close to and slightly less than 8. Answer is 7 boxes

### Question 30

In a school of 150 students, 50 students did not play football, 90 played cricket and 20 neither played football nor cricket

How many students played both, football and cricket?

40

50

60

Cannot be determined

#### SOLUTION

Solution :C

50 students did not play football... i.e. 100 played football.

A∪B=A+B−A∩B

130=100+90−x

x = 60

### Question 31

In a school of 150 students, 50 students did not play football, 90 played cricket and 20 neither played football nor cricket

How many students played only football?

40

50

60

Cannot be determined

#### SOLUTION

Solution :A

b + c = 150 - 50 = 100.

a + d = 50 (those who do not play football)

a + b = 90, d = 20

⇒a + 20 = 50 ⇒a = 30

b = 90 - 30 = 60, c = 100 - 60 = 40.

### Question 32

There is a cube Alpha whose side is equal to the diagonal of the cube Beta of side 2. What is the ratio of the volume of the largest sphere Delta, inside the cube Gamma of side equal to the diagonal of cube Alpha and volume of cube Beta?

2π : 9

8π : 9

9π : 8

9π : 2

#### SOLUTION

Solution :D

Option (d)

Side length of Beta cube = 2 cm

Volume of Beta cube = (2)3 = 8 cm3 .........-(1)

Diagonal length of Beta cube = side length of Alpha cube = 2√3 cm

Side length of Gamma cube = diagonal length of Alpha cube = (2 √3)√3 = 6 cm.

Diameter of sphere inside Gamma cube = Length of edge = 6 cm.

Radius of sphere =62 = 3 cm

Volume of sphere S

= (43)×π×(3)3 = 36π.........(2)

Required ratio = S : q = 36π : 8 = 9π : 2

### Question 33

Abhay choses to bat and he scores exactly 50 runs in 12 balls. He makes only zeroes, fours and sixes in these 12 balls. It is also known that he does not score more than 2 zeroes. In how many different ways can he do so?

12C11

11! × 2!

12C10 + 12C11.2C1 + 12C9.3C2

None of these

#### SOLUTION

Solution :D

50 runs can be scored from 12 balls as follows

Option 1:- 11 fours and 1 six = 12C11 .1C1

Option 2:- 1 zero + 8 fours + 3 sixes = 12C8.4C3

Option 3:- 2 zeroes+ 5 fours+ 5 sixes = 12C3.9C2Answer = sum of all 3. option- d

### Question 34

Find the number of ways in which 128 can be written as a sum of two or more consecutive natural numbers.

>2

1

2

None of these

#### SOLUTION

Solution :D

The number of ways of expressing a number as a sum of two or more consecutive numbers = (No. of odd factors of that number -1)

Since 128 is a power of 2, it does not have any odd factor. This means that it cannot be written as a sum of two or more consecutive natural numbers.

Alternate Method:

Sum of the consecutive numbers = 128

⟹n2(2a+n−1)=128 (∵d=1)

⟹n(2a+n−1)=256 (where n>2)

Now, all factors of 256 are even except 1 (n=1 is not possible).

If n is even, then 'a' will never have an integral value. Hence, no such series of consecutive numbers is possible.