# CAT DI with Data Connected Sets Questions PDF

**DI With Data Connected Sets **is one of the important topics in the **CAT** **LRDI **section. These questions can be a bit challenging, so ensure that you are aware of the **Important ways to represent and solve DI With Data Connected Sets in LRDI**. You can check out these CAT DI With Data Connected Sets LRDI questions from the **CAT Previous year papers**. This post will look into some important LRDI DI With Data Connected Sets for CAT. These are a good source of practice for CAT preparation; If you want to practice these questions, you can download these Important **DI With Data Connected Sets LRDI Questions for CAT** (with detailed answers) **PDF** below, which is completely Free.

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**Instructions**

**Question 1:Â **Which retail format gives the highest profit for BnC?

a)Â BookStore

b)Â Supermarket

c)Â Online

d)Â All are equally profitable.

e)Â Cannot be determined from given information.

**1)Â AnswerÂ (A)**

**Solution:**

~~Sales from online = 50% of total sales = 50% of 60000 = 30000~~

The ratio of supermarket sales and book store sales is 1:2

Supermarket sales = (30000*1)/3 = 10000

Bookstore sales =Â (30000*2)/3 = 20000

Initial costs for online = 50% of total initial costsÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Initial cost is allocated in ratio of sales)

= 50% of 39000 = 19500

Initial costs for supermarket =Â (19500*1)/3 = 6500Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Initial cost is allocated in ratio of sales)

Initial costs for bookstore =Â (19500*2)/3 = 13000

For online:

Additional costs = Red + Yellow + Green + VioletÂ Â Â Â Â Â Â Â for online)Â Â Â ….(1)

Total red costs = 5500

Red costs for online = 5500*(200/275) = 4000

Similarly,

Yellow costs for online = 3100*(52/100) =Â 1612

Green costs for online = 4800*(50/80)Â = 3000

Violets costs for online = 2400*(50/80)Â = 1500

Addition costs for online = 4000+1612+3000+1500 = 10112Â Â Â Â Â Â Â Â Â Â From (1)

Profit = Sales – Initial costs – Additional costs = 30000-19500- 10112 = 388

For Supermarket:

Additional costs = Red + Yellow + Green + Violet

=Â 5500*(65/275) + 3100*(20/100) +4800*(21/80)+2400*(21/80) = 3810

Profit = 10000-6500- 3810

= -310

For Bookstore:

Additional costs = Red + Yellow + Green + Violet

=Â 5500*(10/275) + 3100*(30/100) +4800*(9/80)+2400*(9/80)

Profit = 20000-13000-1940 = 5060

~~Bookstore gives highest profit for BnC.~~

**Question 2:Â **Which retail format is least profit making for BnC?

a)Â Online

b)Â Supermarket

c)Â Book Store

d)Â All formats are loss making. E. All formats are profit making.

**2)Â AnswerÂ (B)**

**Solution:**

Sales from online = 50% of total sales = 50% of 60000 = 30000

The ratio of supermarket sales and book store sales is 1:2

Supermarket sales = (30000*1)/3 = 10000

Bookstore sales =Â (30000*2)/3 = 20000

Initial costs for online = 50% of total initial costsÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Initial cost is allocated in ratio of sales)

= 50% of 39000 = 19500

Initial costs for supermarket =Â (19500*1)/3 = 6500Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Initial cost is allocated in ratio of sales)

Initial costs for bookstore =Â (19500*2)/3 = 13000

For online:

Additional costs = Red + Yellow + Green + VioletÂ Â Â Â Â Â Â Â for online)Â Â Â ….(1)

Total red costs = 5500

Red costs for online = 5500*(200/275) = 4000

Similarly,

Yellow costs for online = 3100*(52/100) =Â 1612

Green costs for online = 4800*(50/80)Â = 3000

Violets costs for online = 2400*(50/80)Â = 1500

Addition costs for online = 4000+1612+3000+1500 = 10112Â Â Â Â Â Â Â Â Â Â From (1)

Profit = Sales – Initial costs – Additional costs = 30000-19500- 10112 = 388

For Supermarket:

Additional costs = Red + Yellow + Green + Violet

=Â 5500*(65/275) + 3100*(20/100) +4800*(21/80)+2400*(21/80) = 3810

Profit = 10000-6500- 3810

= -310

For Bookstore:

Additional costs = Red + Yellow + Green + Violet

=Â 5500*(10/275) + 3100*(30/100) +4800*(9/80)+2400*(9/80)

Profit = 20000-13000-1940 = 5060

Supermarket is making least profit for BnC.

**Question 3:Â **What is the profit/loss from “online” sales? (Use data from previous question if needed)

a)Â 0

b)Â -310

c)Â +20

d)Â +388

e)Â Cannot be determined from given information

**3)Â AnswerÂ (D)**

**Solution:**

Sales from online = 50% of total sales = 50% of 60000 = 30000

Initial costs for online = 50% of total initial costsÂ Â Â Â Â Â Â Â Â Â Â Â Â Â Â (Initial cost is allocated in ratio of sales)

= 50% of 39000 = 19500

Additional costs = Red + Yellow + Green + VioletÂ Â Â Â Â Â Â Â for online)Â Â Â ….(1)

Total red costs = 5500

Red costs for online = 5500*(200/275) = 4000

Similarly,

Yellow costs for online = 3100*(52/100) =Â 1612

Green costs for online = 4800*(50/80)Â = 3000

Violets costs for online = 2400*(50/80)Â = 1500

Addition costs for online = 4000+1612+3000+1500 = 10112Â Â Â Â Â Â Â Â Â Â From (1)

Profit = Sales – Initial costs – Additional costs = 30000-19500- 10112 = 388

**Question 4:Â **Read the following statements:

Statement I. Online store accounted for 50% of the sales at BnC and the ratio of supermarket sales and book store sales is 1:2.

Statement II. Initial Cost is allocated in the ratio of sales. If you want to calculate the profit/loss from the different retail formats, then

a)Â Statement I alone is sufficient to calculate the profit/loss.

b)Â Statement II alone is sufficient to calculate the profit/loss.

c)Â Both statements I and Il are required to calculate the profit/loss.

d)Â Either of the two statements is sufficient to calculate the profit/loss.

e)Â Neither Statement I nor Statement Il is sufficient to calculate the profit/loss.

**4)Â AnswerÂ (C)**

**Solution:**

Sales for online = 50% of 60000 = 30000

Remaining sales = 60000-30000 = 30000

The ratio of supermarket sales and book store sales is 1:2

Supermarket sales = (30000*1)/3 = 10000

Bookstore sales =Â (30000*2)/3 = 20000

Profit = Sales – Initial costs – Additional costs

Sales and Additional costs can be calculated from the table. But no information about distribution of initial costs among different retails formats. From statement 2 the distribution of initial costs is clear.

Hence both statements will be used to calculate profit for different retail formats.

**Instructions**

Answer the questions based on the following information. Mulayam Software Co., before selling a package to its clients, follows the given schedule.

The number of people employed in each month is:

**Question 5:Â **Which five consecutive months have the lowest average cost per man-month under the new technique?

a)Â 1-5

b)Â 9-13

c)Â 11-15

d)Â None of these

**5)Â AnswerÂ (C)**

**Solution:**

Lowest cost per person is for coding and maintenance i.e. 10 rs.

But for consecutive five months avg. for coding and any other will be more than consecutive 5 months for maintenance.

So avg. cost for five months of maintenance is = $\frac{(3+3+1+1+1) \times 10}{9} = 10$

Which is lowest among all others

Checkout: **CAT Free Practice Questions and Videos**

**Question 6:Â **Under the new technique, which stage of software development is most expensive for Mulayam Software Co.?

a)Â Testing

b)Â Specification

c)Â Coding

d)Â Design

**6)Â AnswerÂ (D)**

**Solution:**

Total incurred cost for testing = $(4+1)\times 15 = 75$

Incurred cost for specification = $(2+3) \times 40 = 200$

Incurred cost for coding = $(4+5+5) \times 10 = 140$

Incurred cost for design = $(3+4+5) \times 20 = 240$

Hence design is most expensive for the company.

**Question 7:Â **What is the difference in cost between the old and the new techniques?

a)Â Rs. 30,000

b)Â Rs. 60,000

c)Â Rs. 70,000

d)Â Rs. 40,000

**7)Â AnswerÂ (B)**

**Solution:**

In new technique, increment in 5th month cost (‘000) Â = $(5 \times 20 – 4 \times 10)$ = 60

**Question 8:Â **With reference to the above question, what is the cost incurred in the new â€˜codingâ€™ stage? (Under the new technique, 4 people work in the sixth month and 5 in the eighth.)

a)Â Rs. 1,40,000

b)Â Rs. 1,50,000

c)Â Rs. 1,60,000

d)Â Rs. 1,70,000

**8)Â AnswerÂ (A)**

**Solution:**

As coding is done in 6-8 months,

So total person employed = 4+5+5 = 14

cost per person for 6-8 month = 10000

Total cost will be = $(14 \times 10000)$ = 140,000

**Question 9:Â **Due to overrun in â€˜designâ€™, the design stage took 3 months, i.e. months 3, 4 and 5. The number of people working on design in the fifth month was 5. Calculate the percentage change in the cost incurred in the fifth month. (Due to improvement in â€˜codingâ€™ technique, this stage was completed in months 6-8 only.)

a)Â 225%

b)Â 150%

c)Â 275%

d)Â 240%

**9)Â AnswerÂ (B)**

**Solution:**

In given table, value of cost for 5th month = $(N) \times (C)$ (No.of employer for 5th month = N andÂ cost per person for 5th month = C)

So value of cost for 5th month before overrun = $4 \times 10 = 40$

value of cost for 5th month after overrun = $ 5 \times 20 = 100 $

Change = 60

%change = $\frac{60}{40} \times 100$ = 150

**Question 10:Â **If the sum of the roots of the equation $\frac{(x+p)}{(x+p+r)}$ + $\frac{(x-q)}{(x-q+r)}$ = 1 is zero, which of the following options is correct?

a)Â p + q = 0

b)Â p + q + r = 0

c)Â p + q = 2r

d)Â p = q

**10)Â AnswerÂ (D)**

**Solution:**

Expanding the equation on both sides,

$\frac{(x+p)}{(x+p+r)}$ + $\frac{(x-q)}{(x-q+r)}$ = 1

1 – $\frac{(r)}{(x+p+r)}$ + 1 –Â $\frac{(r)}{(x-q+r)}$ = 1

$\frac{(r)}{(x+p+r)}$ +Â $\frac{(r)}{(x-q+r)}$ = 2-1 = 1

$\frac{(r)(2x+2r+p-q)}{(x+p+r)(x-q+r)}$ = 1

2rx + $2r^2$ + pr – qr = $x^2$ – qx + rx + px – pq + pr + rx – qr +Â $r^2$

$x^2$+ (p-q)x – pq- $r^2$ = 0

Sum of the roots = 0

-(p-q)/1 = 0

q = p

**Question 11:Â **If y is a real number, what is the difference in the maximum and minimum values obtained by $\frac{y+5}{y^2+5y+25}$ ?

a)Â 2/15

b)Â 4/15

c)Â 1/5

d)Â 1/15

**11)Â AnswerÂ (B)**

**Solution:**

Let $\frac{y+5}{y^2+5y+25} = k$

So, $ky^2+5ky+25k-y-5=0$

For the above quadriatic equation to have real roots, its discriminant should be greater than or equal to zero.

$(5k-1)^2 \geq 4*k*5*(5k-1)$

$25k^2-10k+1\ge20k\left(5k-1\right)$

$25k^2-10k+1\ge100k^2-20k$

$75k^2-10k-1\le0$

solving we get

$(15k+1)(5k-1)\leq 0 $

So $-1/15 \leq k \leq 1/5$

So, the difference in the maximum and minimum value is 4/15

**Question 12:Â **How many integral solutions exist for the inequality |x-5| + |y-4| < 4?

a)Â 10

b)Â 15

c)Â 25

d)Â 33

**12)Â AnswerÂ (C)**

**Solution:**

Let x-5 = a and y-4 = b.

So, the number of solutions to the inequality equals that of the inequality |a|+|b|<4

If |a|+|b|=0, number of integral solutions is 1 i.e. {0,0}

If |a|+|b|=1, number of integral solutions is 4 i.e. {1,0},{0,1},{-1,0},{0,-1}

If |a|+|b|=2, number of integral solutions is 8Â i.e. {2,0},{1,1},{0,2},{1,-1},{-1,1},{-1,-1},{-2,0},{0,-2}

If |a|+|b|=3, number of integral solutions is 12Â i.e. {3,0}x2, {0,3}x2, {2,1}x4, {1,2}x4

Total is 25

**Question 13:Â **If a,b and c are three whole numbers such that |a-1|+|b-2|+|c-3|=3, what is the maximum value of (a+1)(b+2)(c+3)?

a)Â 120

b)Â 144

c)Â 112

d)Â 105

**13)Â AnswerÂ (A)**

**Solution:**

There are fifteen solutions to the equation |a-1|+|b-2|+|c-3|=3.

Let |a-1| be x, |b-2| be y and |c-3| be z.

x, y and z are whole numbers.

x +Â y +Â z = 3.

Therefore, the number of solutions to this equation will be n+r-1 C r-1 = 3+3-1C2 = 5C2 = 10.

There are 10 solutions to the equation x+y+z = 3.

After enumerating them, we find that (a+1)(b+2)(c+3) is maximum when a=b=c=3 and its maximum value is 120.

**Instructions**

The first table gives the number of saris (of all the eight colours) stocked in six regional showrooms. The second gives the number of saris (of all the eight colours) sold in these six regional showrooms. The third table gives the percentage of saris sold to saris stocked for each colour in each region. The fourth table gives the percentage of saris of a specific colour sold within that region. The fifth table gives the percentage of saris of a specific colour sold across all the regions. Study the tables and for each of the following questions, choose the best alternative.

Table 1

Table 2

Table 3

Table 4

Table 5

**Question 14:Â **In which region is the maximum percentage of blue saris sold?

a)Â 2

b)Â 3

c)Â 1

d)Â 4

**14)Â AnswerÂ (A)**

**Solution:**

From Table 5, we can see that out of all the regions, the maximum number of blue saries were sold in region 2

**Question 15:Â **Out of its total sales, which region sold the minimum percentage of green saris?

a)Â 1

b)Â 4

c)Â 6

d)Â 2

**15)Â AnswerÂ (B)**

**Solution:**

Given we have 6 regions where sales are defined in the table number two . Now given that we need to find the region where green saris as a percentage of the total sales is lowest . In order to do that we calculate Green saris sold in a region/ Total sales of a particular region by considering the values from table 2 .

Now for option 1 which is region 1 the value is (164/788)*100Â = 20.81 %

For option 2 which is region 4 the value is (85/511)*100 = 16.63 %

For option 1 which is region 6 the value is (3/18)*100 = 16.66 %

For option 1 which is region 2Â the value is (200/676)*100 = 29.58Â % .

Hence among the given options the percentage of region 4 is lowest which is option 2 .

**Question 16:Â **Which region sold the maximum percentage of magenta saris out of the total sales of magenta saris?

a)Â 3

b)Â 4

c)Â 2

d)Â 1

**16)Â AnswerÂ (D)**

**Solution:**

Since table 2 gives us the sales values of saris

Hence, for the magentaÂ sari, Region1 has the maximum number of sales out of total magenta sales.

I.e. 71 out of 161.

**Question 17:Â **Which colour is the most popular in region1?

a)Â Blue

b)Â Brown

c)Â Green

d)Â Violet

**17)Â AnswerÂ (B)**

**Solution:**

As the most popular colours will have more sale than any other colour

Since table 2 gives us value of sales, Brown colour in region 1 has maximum no. of sales

Hence, our answer will be Brown.

**Question 18:Â **Which region-colour combination accounts for the highest percentage of sales to stock?

a)Â (1, Brown)

b)Â (2, Yellow)

c)Â (4, Brown)

d)Â (5, Red

**18)Â AnswerÂ (C)**

**Solution:**

Sales to stock percentage values are given in table 3

And at region 4, Brown colour, percentage is maximum compared to all.

Hence answer will be C