CAT 2025 Lesson : Bar Graph - Case 6

Read the following information and the accompanying graphs to answer the questions that follow.

www.jay.com spent 5,57,000 USD during the last 12 months for online display advertisements, also called impressions, on five websites (Website A, Website B, Website C, Website D and Website E). In this arrangement, www.jay.com is the Destination site, and the five websites are referred to as the Ad Sites. The allocation of online display advertising expenditure is shown in Graph A. The online display advertisements helped www.jay.com to get visitors on its site. Online visitors, visiting the Ad Sites, are served display advertisements of www.jay.com and on clicking they land on the Destination Site (Graph B). Once on the Destination Site, some of the visitors complete the purchase process (Graph C)




Quality traffic = $\dfrac{\mathrm{No. \ of \ site \ visitors \ who \ start \ purchase \ on \ destination site}}{\mathrm{No. \ of \ visitors \ who \ click \ the \ online \ display \ advertisement}}$

Leakage in online buying = 1 - $\dfrac{\text{Complete buying on the destination website}}{\text{Start buying on the destination website}}$

Efficiency of online display advertising expenditure on an Ad Site = $\dfrac{\mathrm{No. \ of \ Visitors \ from \ the \ Ad \ site \ who \ complete \ the \ purchase \ process} }{\mathrm{Amount \ spent \ on \ the \ Ad \ site}}$

[IIFT 2015]

20) Which of the following Ad Sites provide the facility of least cost per advertisement?

(1) Website A
(2) Website B
(3) Website D
(4) Website E

21) Which Ad Site has provided the maximum quality traffic?

(1) Website A
(2) Website B
(3) Website D
(4) Website E

22) Which Ad Site sent traffic to www.jay.com with maximum leakage?

(1) Website B
(2) Website C
(3) Website D
(4) Website E

23) On which Ad Site is the advertising budget spent most efficiently?

(1) Website A
(2) Website B
(3) Website C
(4) Website E

Solution

20) Cost per advertisement = $\dfrac{\text{Total cost spent on advertisement}}{\text{Number of advertisement}}$

= $\dfrac{\text{Total advertisement expenditure}}{\text{Number of impressions made for website X}}$

For website,

A = $\dfrac{557000 \times \frac{27}{100}}{240}$ = 626.625

B = $\dfrac{557000 \times \frac{22}{100}}{370}$ = 331.19

D = $\dfrac{557000 \times \frac{13}{100}}{300}$ = 219.42

E = $\dfrac{557000 \times \frac{20}{100}}{150}$ = 742.67

Website D has the least cost per advertisement

Answer: (3) Website D

21) Quality Traffic = $\dfrac{\text{No. Of site visitors who start purchase on destination site}}{\text{No. Of visitors who click on online display website}}$

For Website,

A = $\dfrac{2800}{120000}$ = $\dfrac{2.33}{100}$

B = $\dfrac{2500}{60000}$ = $\dfrac{4.167}{100}$

$\dfrac{3000}{80000}$ = $\dfrac{3.7}{100}$

E = $\dfrac{3500}{40000}$ = $\dfrac{8.75}{100}$

The maximum value among these if for Website E

Answer: (4) Website E

22) Leakage in online buying = 1 - $\dfrac{\text{Complete buying on the destination website}}{\text{Start buying on the destination website}}$

For Website

B = 1 – $\dfrac{1200}{2500}$ = 0.52

C = 1 – $\dfrac{900}{2000}$ = 0.55

D = 1 – $\dfrac{1300}{3000}$ = 0.5667

E = 1 – $\dfrac{1600}{3500}$ = 0.5428

The leakage is maximum for Website D

Answer: (3) Website D

23) Efficiency = $\dfrac{\text{No. Of Visitors from Ad site who completed the purchase process}}{\text{Amount spent in the Ad site}}$

For Website

A = $\dfrac{2700}{150399}$ = 0.0179

B = $\dfrac{1200}{122540}$ = 0.009

C = $\dfrac{900}{100260}$ = 0.0089

E = $\dfrac{1600}{111400}$ = 0.014

Website A has the highest efficiency

Answer: (1) Website A

Answer:
20) (3) Website D
21) (4) Website E
22) (3) Website D
23) (1) Website A