# Free Mixed Bag 7 Practice Test - CAT

### Question 1

In a factory all Supervisors are Engineers. One-third of all Engineers are Supervisors. Half of all Technicians are Engineers. Only one Technician is a Supervisor. Six Technicians are Engineers. If the number of Engineers is 60, how many Engineers are neither Supervisors nor Technicians?

#### SOLUTION

Solution :D

This question can be repressented as a 3 set venn diagram as shown below

As all supervisors are engineers, we see that R = S = 0

One-third of all engineers are supervisors: y + z = (1/3)(x + y + z + p)

Half of all technicians are engineers: z + p = q. Only one technician is a supervisor: z = 1

Six technicians are engineers: p + z =6 → p = 5, q = 6 Number of engineers is 60: x + y + z + p = 60 → Number of supervisors = 1/3 * 60= 20

5 engineers are technicians but not supervisors

Engineers who are neither Supervisors nor Technicians.

Hence the 3 set venn diagram is:

### Question 2

If ST is parallel to QR, and PS = SQ = 3 cm, PT = 4 cm and QR = 10 cm, what is the area of trapezoid STQR?

#### SOLUTION

Solution :D

Option (d).

Triangle PST & PQR are similar. Thus, PSPQ= PTPR ⇒ PR=8Thus, PQR is a right angled triangle with sides-6,8,10

Area of trapezoid STQR= Area of triangle PQR- Area of triangle PST = 12*6*8 - 12*3*4. Answer is 18 cm2

### Question 3

If x,y and z are in AP, then x, y^{-1} and z are in?

#### SOLUTION

Solution :C

Option (c) Take y=2 and = 1 and (y+z)/(1-yz) = 3

When y=2. x= -1/3 and z= 1/7,

= 1/2 = y-1 .. hence they are in HP

### Question 4

Manju, who works at Ganesh Fruit juice center, prepares 51 glasses of milkshake in 4 min, 18 sec in his first shift. After having a sandwich, he prepares 73 glasses in 7 min 13 sec. After another sandwich, he prepares 112 glasses in 12 min, 24 sec. The container in which the milkshake is prepared contains a maximum of 9 glasses of a drink at a time. The approximate average time (to the nearest integer) of preparation of one container of the drink, if he starts with a fresh container each time is (in seconds)

#### SOLUTION

Solution :In the first shift, he uses the container 6 times.

(45 glasses in 5 rounds and the remaining 6 glasses in the 6th round. Similarly, he prepared 73 glasses in 9 rounds, and 112 glasses in 13 rounds.)

Hence, the required average

= total timetotal number of rounds

= (4 min 18 s + 7 min 13 s + 12 min 24 s)/ (6 + 9 +13)

= (23 min+55s)/ 28 = 51 seconds (rounded)

### Question 5

Find the range of p; given that p=xy/z. The range of x,y and z is given as -3<x<3, -5<y<-2, -5<z<0, given that x,y and z are integers

#### SOLUTION

Solution :D

Option (d). To find the range, we need the min value of xy/z and the maximum value of xy/z. xy will ie between -8 and +8, including both.

Z can take the value -1. Hence the range will be -9<xy/z≤8

### Question 6

Aditya and Dhana sit for the selection of a six member team for a project X in company Z. If the team is to be split up into 2 three-person sub-teams, the percent of all the possible subteams that include Aditya also include Dhan is

#### SOLUTION

Solution :Since we have fixed the selection of Aditya, the grouping of the remaining people into 2 groups containing 3 members each can be done in 5!(2!×3!) ways =10 ways

Now, if both Aditya and Dhan are selected, 2 out of the three members in a sub-team are already selected. The third person can be anyone of the remaining 4, which means that we have 4 different possibilities. Thus, the percent of all possible subteams that include Aditya, also include Dhan = 410=40%

### Question 7

Consider an equilateral triangle of side length n, which is divided into unit triangles, as shown. Let f(n) be the number of paths from the triangle in the top row to the middle triangle in the bottom row, such that adjacent triangles in our path share a common edge and the path never travels up (from a lower row to a higher row) revisits a triangle. An example of one such path is illustrated below for n = 5. Determine the value of f(2009)

#### SOLUTION

Solution :D

Define a last triangle of a row as the triangle in the row that the path visits last. From the last triangle in row x, the path must move down and then directly across to the last triangle in row x + 1. Therefore, there is exactly one path through any given set of last triangles. For 1 < m < n - 1, there are m possible last triangles for the mth row. The last triangle of the last row is always in the centre. Therefore, f(n) = (n – 1)! And f(2009) = 2008!. Hence option (e)

Note:- You can also solve this by the Unitary approach

### Question 8

What is the minimum number of positive factors of a 6-digit number of the form abbabb, where a and b represent distinct natural numbers?

#### SOLUTION

Solution :C

A number of the form xyyxyy will be divisible by 7,11 and 13 (difference between triplets at odd and even places is zero).

Also 1001*xyy = xyyxyy (1001= 7*11*13). If we can get a prime number of the form xyy, we can reduce the number of factors.

122 is not a prime number. 233 is a prime number. xyyxyy = 233233 = 7*11*13*233, so the number of factors = 2*2*2*2 = 16. Hence, option (c) is the right answer.

### Question 9

There are four married couples in a club. The number of ways of choosing a committee of three members so that no complete couple appears in the committee is

#### SOLUTION

Solution :D

Option (d)

The answer 8C3=4C1×6 can be reasoned out as 8C3 = all the cases wherein out of 8 people any 3 are chosen. Now choosing any three out of eight will include within it cases where even a married couple is chosen. So, we will subtract all those cases wherein married couples are chosen. The total number of arrangements of married couples will be 4C1×6. This is because out of 4 available, we can choose any one married couple. Now along with this married couple the third person chosen will be any of the 6 remaining people. Thus 4C1×6 number of different possibilities need to be subtracted from 8C3.

### Question 10

If p is the arithmetic mean of the positive numbers a and b, and the reciprocal of the number q is the arithmetic mean of the reciprocals of a and b. then p– q =?

#### SOLUTION

Solution :B

let a=2 and b=4, then p=3

1/2+1/4 = 3/4

Arithmetic mean = 3/8. Hence q= 8/3. P-Q= 3-8/3= 1/3. Look where you are getting 1/3 in the answer options. Answer is option (b)

### Question 11

(22!×P) is a perfect square. Find the least value of P, given that P belongs to a set of natural numbers.

#### SOLUTION

Solution :A

22! can be written as 219×39×54×73×112×13×17×19.

To make it a perfect square, we need to multiply it by 2×3×7×13×17×19.

So, P = 176358.

### Question 12

If p and q are positive and p^{2}q^{2} = 18 – 3pq, then p^{2} =

^{3}

^{2}

^{2}+3q

^{2}

^{2}

#### SOLUTION

Solution :D

Subsitute q=1, the equation reduces to p

^{2}+18 – 3p = 0. Solving this, we get p=-6 & p=3. Hence, p^{2}= 9 (As p is positive)Substitute the value of q=1 in answer options to see where qou get p

^{2}= 9. Answer is option (d)

### Question 13

Six friends have arrived at the theater for the premiere of the film “Love Aaj Kal.” Fred and Jason have are at loggerheads off late. Fred insists on standing behind Jason in line at the ticket stand, though not necessarily right behind him. How many ways can the six arrange themselves in line such that Fred’s requirement is satisfied?

#### SOLUTION

Solution :D

Option (d). Total number of combinations = 6! =720. In exactly half the cases, Jason will be in front of Fred. Answer is 360

### Question 14

The number of students studying C++ but not SQL is

#### SOLUTION

Solution :Students who study C++ but not SQL, c + f = 37. Hence option (e)

### Question 15

The number of students who study only Java is

#### SOLUTION

Solution :Number of students studying only Java = 20.