CAT 2022 Question Paper DILR Slot 2

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be 'x'.
Total number of households each of them met in two days will be '2x'.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.
Let the number of households Tohri met on day 1 be 'a'
It is given,
a + a + 30 = 2x
a + 15 = x
a = x - 15

In statement 5, it is given that

$$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$$ ...... (1)

$$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$$ ...... (2)

$$\frac{100}{x}=\ \frac{\ y}{x-15}$$

$$\frac{100}{y}=\ \frac{\ x}{x-15}$$

$$\frac{y}{100}=\ \ 1-\frac{15}{x}$$

$$x=\frac{1500}{100-y}$$

Substituting x in (2), we get

y = 40 and x = 25

Final Table:

The total number of households met by Tohri, Hokli and Lahur on the first day is 10 + 49 + 25, i.e. 84.

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be 'x'.
Total number of households each of them met in two days will be '2x'.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.
Let the number of households Tohri met on day 1 be 'a'
It is given,
a + a + 30 = 2x
a + 15 = x
a = x - 15

In statement 5, it is given that

$$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$$ ...... (1)

$$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$$ ...... (2)

$$\frac{100}{x}=\ \frac{\ y}{x-15}$$

$$\frac{100}{y}=\ \frac{\ x}{x-15}$$

$$\frac{y}{100}=\ \ 1-\frac{15}{x}$$

$$x=\frac{1500}{100-y}$$

Substituting x in (2), we get

y = 40 and x = 25

Final Table:

The number of items sold by Tohri on the first day is 40.

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be 'x'.
Total number of households each of them met in two days will be '2x'.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.
Let the number of households Tohri met on day 1 be 'a'
It is given,
a + a + 30 = 2x
a + 15 = x
a = x - 15

In statement 5, it is given that

$$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$$ ...... (1)

$$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$$ ...... (2)

$$\frac{100}{x}=\ \frac{\ y}{x-15}$$

$$\frac{100}{y}=\ \frac{\ x}{x-15}$$

$$\frac{y}{100}=\ \ 1-\frac{15}{x}$$

$$x=\frac{1500}{100-y}$$

Substituting x in (2), we get

y = 40 and x = 25

Final Table:

Lahur met 25 households on day 2. The answer is option A.

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be 'x'.
Total number of households each of them met in two days will be '2x'.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.
Let the number of households Tohri met on day 1 be 'a'
It is given,
a + a + 30 = 2x
a + 15 = x
a = x - 15

In statement 5, it is given that

$$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$$ ...... (1)

$$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$$ ...... (2)

$$\frac{100}{x}=\ \frac{\ y}{x-15}$$

$$\frac{100}{y}=\ \frac{\ x}{x-15}$$

$$\frac{y}{100}=\ \ 1-\frac{15}{x}$$

$$x=\frac{1500}{100-y}$$

Substituting x in (2), we get

y = 40 and x = 25

Final Table:

Tohri met 10 households on day 1. The answer is option D.

In statement 1, it is given that all three of them met the same total number of households, and each of them sold a total of 100 items in two days. In statement 2, it is given that on both days, Lahur met the same number of households and sold the same number of items. This implies he sold 50 items per day. Let the number households Lahur met in a day be 'x'.
Total number of households each of them met in two days will be '2x'.

In statement 3, it is given that Hokli could not sell any item on the second day because the first household he met on that day complained against him. This implies he met only 1 household on day 2.

In statement 4, it is given that Tohri met 30 more households on the second day than on the first day.
Let the number of households Tohri met on day 1 be 'a'
It is given,
a + a + 30 = 2x
a + 15 = x
a = x - 15

In statement 5, it is given that

$$2\left(\frac{50}{x}\right)=\frac{y}{x-15}$$ ...... (1)

$$\frac{3}{4}\left(\frac{50}{x}\right)=\ \frac{\ 100-y}{x+15}$$ ...... (2)

$$\frac{100}{x}=\ \frac{\ y}{x-15}$$

$$\frac{100}{y}=\ \frac{\ x}{x-15}$$

$$\frac{y}{100}=\ \ 1-\frac{15}{x}$$

$$x=\frac{1500}{100-y}$$

Substituting x in (2), we get

y = 40 and x = 25

Final Table:

Among the three, Tohri had the highest success rate on the second day - this statement is incorrect. On day 2, Lahur had the highest success rate, i.e. 2 whereas Tohri's success rate is 1.5.
Tohri had a higher success rate on the first day compared to the second day - this statement is correct.
Tohri's day 1 success rate is 4 and day 2 success rate is 1.5.
Among the three, Tohri had the highest success rate on the first day - this statement is correct.
Tohri's success rate on day 1 is 4.
Hokli's success rate on day 1 is 2.04.
Lahur's success rate on day 1 is 2.
Among the three, Lahur had the lowest success rate on the first day - this statement is correct.
Tohri's success rate on day 1 is 4.
Hokli's success rate on day 1 is 2.04.
Lahur's success rate on day 1 is 2.
The answer is option A.

Points to be noted:
1. Starts from the warehouse and ends at a location after visiting all the locations exactly once.
2. While making the route plan, the supplier goes to the locations in decreasing order of demand. If equal demand, goes to the nearest ones first.
3. While creating the route, the supplier can either follow the direct path (if available) from one location to another or can take the path via the warehouse(Prefers minimum distance).
In the question, it is given that last location is A. The demand in the remaining places should be greater than A. This implies A demand cannot be 70 units. Therefore, it is 50 units.
The demand of the location D is 30 or 50 units. This implies this should be placed before D.
The demand of the location placed before D should be greater than or equal to 50 units. Location supplier visited before D is B(60 units of demand). It cannot be C because values of C is greater than the values of B.
Therefore, order is C - B - D - A.
From warehouse to C - 12 km
C to B - 4 km
B to D - 12 km
D to A - 7 km (through warehouse)
Total distance covered = 12 + 4 + 12 + 7 = 35 km

This question is removed from the paper because if the order is CBDA, the supplier will go to A from B (and hence the last city visited will be D).

Points to be noted:
1. Starts from the warehouse and ends at a location after visiting all the locations exactly once.
2. While making the route plan, the supplier goes to the locations in decreasing order of demand. If equal demand, goes to the nearest ones first.
3. While creating the route, the supplier can either follow the direct path (if available) from one location to another or can take the path via the warehouse(supplier prefers minimum distance).

In the question, it is given that total number of units delivered is 250 units.

Maximum number of widgets that can be delivered is 70 units(A) + 50 units(D) + 60 units(B) + 100 units(C) = 280 units

From this 30 units should be decreased. 30 units can be decreased only when C's demand decreases to 70 units(because the difference for remaining locations is 20 units)

Therefore, the only possibility is 70 units(A) + 50 units(D) + 60 units(B) + 70 units(C)

From statement 1, the order should be A - C - B - D(as A is nearer to warehouse than C)

Distance from Warehouse to A is 5 km

Distance from A to C is 17 km

Distance from C to B is 4 km

Distance from B to D is 12 km

Total distance covered = 5 + 17 + 4 + 12 = 38 km

Maximum number of widgets delivered in a day is 100 units(C) + 70 units(A) + 60 units(B) + 50 units(D), i.e. 280 units

Given, total number of widgets delivered is 260 units. This implies 20 units must be decreased from any one of the locations.

A - (70,50), B - (60,40), C - (100,70) and D - (50,30)

20 units can be decreased from A, B or D. 

Demand at location C will be 100 units and supplier first visits C.

In the question, it is also given that the route ends at B.

C(100 units), _, _, B 

If B's demand is 60 units, D's demand should be more than 60 units which is not possible. Therefore, B's demand should be 40 units.

20 units is decreased at location B. This implies demand at location A is 70 units and at location D is 50 units.

Order will be C(100 units) - A(70 units) - D(50 units) - B(40 units).

It is given,

C - 100 units - 70%

A - 70 units - 60%

D - 50 units - 60%

B - 40 units - 30%

Required value = 0.7*0.6*0.6*0.3 = 0.0756 = 7.56%

The answer is option D.

It is given that the first location visited from the warehouse is A. 

If A's demand is 50 units, C's demand should be less than 50 units which is not possible. This implies demand of locations A and C is 70 units.

A(70 units) -> C(70 units) 

Warehouse to A - 5km

A to C - 17 km

Distance covered = 5 + 17 = 22 km

Remaining distance = 40 - 22 = 18 km

C to B - 4 km

B to D - 12 km

Distance covered = 4 + 12 = 16 km $$\ne\ $$ 18 km

C to D - 6 km

D to B - 12 km

Distance covered = 6 + 12 = 18 km

Therefore, supplier can cover distance 18 km if he visits D before B, i.e. demand of D should be more than demand of B. This is only possible when D's demand is 50 units and B's demand is 40 units.

It is given,

D - 50 units - 60% probability

B - 40 units - 30% probability

Required value = 0.6*0.3 = 0.18 = 18%

The answer is option A.

Points to be noted:
1. Starts from the warehouse and ends at a location after visiting all the locations exactly once.
2. While making the route plan, the supplier goes to the locations in decreasing order of demand. If equal demand, goes to the nearest ones first.
3. While creating the route, the supplier can either follow the direct path (if available) from one location to another or can take the path via the warehouse(Prefers minimum distance).
Demand in all other locations should be less than or equal to the demand in first location. In the question, it is given that A is not first location. B and D cannot be first location. This implies C should be the first location.
It is given, total route distance is 29 km.
Warehouse to C is  12 km.
This implies remaining distance should be 29-12, i.e. 17km
This is only possible when C visits B, A and D later(4 + 6 + 7).
The order should be C - B - A - D
C's demand can be 70/100, B's demand should be 60, A's demand should be 50 and D's demand can be 30/50.
The possible number of widgets delivered can be
70 + 60 + 50 + 30 = 210
70 + 60 + 50 + 50 = 230
100 + 60 + 50 + 30 = 240
100 + 60 + 50 + 50 = 260
The answer is option A.

It is given,

Company A:

Revenue = 240 and cost incurred = 180

Profit = 240 - 180 = 60

Company B:

Revenue = 220 and cost incurred = 145

Profit = 220 - 145 = 75

Company C:

Revenue = 195 and cost incurred = 110

Profit = 195 - 110 = 85

Company D:

Revenue = 140 and cost incurred = 160

No profit.

Company C had the highest annual profit.

The answer is option D.

For all the companies in all three years, cost incurred is less than Revenue except for D in 2020.

Revenue is 20 and cost incurred is 50

Company D experienced the highest annual loss in 2020.

The answer is option D.

It is given,

Measure of A's performance in 2019 = $$\frac{90-85}{85}=\frac{5}{85}=0.06$$

Measure of B's performance in 2019 = $$\frac{100-75}{75}=\frac{25}{75}=0.33$$

Measure of C's performance in 2019 = $$\frac{25-20}{25}=\frac{5}{25}=0.2$$

Measure of D's performance in 2019 = $$\frac{50-40}{40}=\frac{10}{40}=0.25$$

Company A had the lowest value.

The answer is option A.

It is given,

Company A:

The number of employees in the beginning of 2019 = 150

The number of employees hired in 2019 = 20

The number of employees should be at the beginning of 2020 is 150+20, i.e. 170 but there are 140 only. This implies 30 left company A in 2019.

The number of employees in the beginning of 2020 = 140

The number of employees hired in 2020 = 35

The number of employees should be at the beginning of 2021 is 140+35, i.e. 175 but there are 150 only. This implies 25 left company A in 2020.

The number of employees left company A in 2019 and 2020 = 30 + 25 = 55

Company B:

Similarly, the number of employees left company B in 2019 = 210 + 35 - 240 = 5

The number of employees left company B in 2020 = 240 + 45 - 250 = 35

The number of employees left company B in 2019 and 2020 = 5 + 35 = 40

Company C:

Similarly, the number of employees left company C in 2019 = 320 + 45 - 320 = 45

The number of employees left company C in 2020 = 320 + 40 - 320 = 40

The number of employees left company C in 2019 and 2020 = 45 + 40 = 85

Company D:

Similarly, the number of employees left company D in 2019 = 400 + 30 - 410 = 20

The number of employees left company D in 2020 = 410 + 35 - 400 = 45

The number of employees left company D in 2019 and 2020 = 20 + 45 = 65

The total number of employees lost in 2019 and 2020 is least for company B.

The answer is option A.

It is given,

Company A = $$\frac{90-65}{145}=\frac{25}{145}=\frac{50}{290}$$

Company B = $$\frac{90-40}{245}=\frac{50}{245}$$

Company C = $$\frac{70-60}{320}=\frac{10}{320}$$

Company B had the highest profit per employee.

The answer is option C.

It is given that the total number of products supermarket sells is 320.

cosmetic + nutrition = foreign + domestic = FDA + EU = 320 products

In statement 1, it is given that the number of foreign products is equal to the number of domestic products.

Foreign products = Domestic products = 320/2 = 160

In statement 2, it is given that half of the domestic products were FDA approved cosmetic products, i.e. domestic, cosmetic and FDA = 80

In statement 4, it is given that there were 140 nutrition products, half of them were foreign products. This implies remaining half are domestic.

In statement 5, it is given that there are 200 FDA approved products out of which 70 are foreign products and 120 are cosmetic products.
If 70 are foreign products, remaining 130 should be domestic products. In domestic products, FDA approved cosmetic products are 80. This implies FDA approved nutrition products are 130-80, i.e. 50.

There are 120 FDA approved cosmetic products.

Domestic, cosmetic and FDA approved = 80

This implies, Foreign, cosmetic and FDA approved is 120-80, i.e. 40.

There are 70 FDA approved foreign products.

This implies Foreign, nutrition and FDA approved is 70-40, i.e. 30.

Domestic and Cosmetic = 90

Domestic, comestic and FDA approved = 80

This implies, Domestic, cosmetic and FDA not approved is 90-80, i.e. 10.

Therefore, (domestic, cosmetic and only EU) = 10

Similarly, we get (domestic, nutrition and only EU) = 70-50 = 20

The number of foreign, cosmetic and FDA approved products is 40.

The answer is 40.

It is given that the total number of products supermarket sells is 320.

cosmetic + nutrition = foreign + domestic = FDA + EU = 320 products

In statement 1, it is given that the number of foreign products is equal to the number of domestic products.

Foreign products = Domestic products = 320/2 = 160

In statement 2, it is given that half of the domestic products were FDA approved cosmetic products, i.e. domestic, cosmetic and FDA = 80

In statement 4, it is given that there were 140 nutrition products, half of them were foreign products. This implies remaining half are domestic.

In statement 5, it is given that there are 200 FDA approved products out of which 70 are foreign products and 120 are cosmetic products.
If 70 are foreign products, remaining 130 should be domestic products. In domestic products, FDA approved cosmetic products are 80. This implies FDA approved nutrition products are 130-80, i.e. 50.

There are 120 FDA approved cosmetic products.

Domestic, cosmetic and FDA approved = 80

This implies, Foreign, cosmetic and FDA approved is 120-80, i.e. 40.

There are 70 FDA approved foreign products.

This implies Foreign, nutrition and FDA approved is 70-40, i.e. 30.

Domestic and Cosmetic = 90

Domestic, comestic and FDA approved = 80

This implies, Domestic, cosmetic and FDA not approved is 90-80, i.e. 10.

Therefore, (domestic, cosmetic and only EU) = 10

Similarly, we get (domestic, nutrition and only EU) = 70-50 = 20

The number of cosmetic products which do not have FDA approval = domestic only EU + foreign EU = 10 + 50 = 60

The answer is option D.

It is given that the total number of products supermarket sells is 320.

cosmetic + nutrition = foreign + domestic = FDA + EU = 320 products

In statement 1, it is given that the number of foreign products is equal to the number of domestic products.

Foreign products = Domestic products = 320/2 = 160

In statement 2, it is given that half of the domestic products were FDA approved cosmetic products, i.e. domestic, cosmetic and FDA = 80

In statement 4, it is given that there were 140 nutrition products, half of them were foreign products. This implies remaining half are domestic.

In statement 5, it is given that there are 200 FDA approved products out of which 70 are foreign products and 120 are cosmetic products.
If 70 are foreign products, remaining 130 should be domestic products. In domestic products, FDA approved cosmetic products are 80. This implies FDA approved nutrition products are 130-80, i.e. 50.

There are 120 FDA approved cosmetic products.

Domestic, cosmetic and FDA approved = 80

This implies, Foreign, cosmetic and FDA approved is 120-80, i.e. 40.

There are 70 FDA approved foreign products.

This implies Foreign, nutrition and FDA approved is 70-40, i.e. 30.

Domestic and Cosmetic = 90

Domestic, comestic and FDA approved = 80

This implies, Domestic, cosmetic and FDA not approved is 90-80, i.e. 10.

Therefore, (domestic, cosmetic and only EU) = 10

Similarly, we get (domestic, nutrition and only EU) = 70-50 = 20

In statement 3, it is given that the number of domestic products which have both the approvals = 60

In the question, it is given that a + c = 60

To find the minimum value of a, we need to maximise c.

Maximum value c can take is 50

Therefore, minimum value of a is 60-50, i.e. 10.

To find the maximum value of a, we need to minimise c.

Maximum value c can take is 0.

Therefore, maximum value of a is 60-0, i.e. 60.

a is minimum:

a is maximum:

Therefore, the number of domestic cosmetic products that had both the approvals is at least 10 and at most 60.

The answer is option A.

It is given that the total number of products supermarket sells is 320.

cosmetic + nutrition = foreign + domestic = FDA + EU = 320 products

In statement 1, it is given that the number of foreign products is equal to the number of domestic products.

Foreign products = Domestic products = 320/2 = 160

In statement 2, it is given that half of the domestic products were FDA approved cosmetic products, i.e. domestic, cosmetic and FDA = 80

In statement 4, it is given that there were 140 nutrition products, half of them were foreign products. This implies remaining half are domestic.

In statement 5, it is given that there are 200 FDA approved products out of which 70 are foreign products and 120 are cosmetic products.
If 70 are foreign products, remaining 130 should be domestic products. In domestic products, FDA approved cosmetic products are 80. This implies FDA approved nutrition products are 130-80, i.e. 50.

There are 120 FDA approved cosmetic products.

Domestic, cosmetic and FDA approved = 80

This implies, Foreign, cosmetic and FDA approved is 120-80, i.e. 40.

There are 70 FDA approved foreign products.

This implies Foreign, nutrition and FDA approved is 70-40, i.e. 30.

Domestic and Cosmetic = 90

Domestic, comestic and FDA approved = 80

This implies, Domestic, cosmetic and FDA not approved is 90-80, i.e. 10.

Therefore, (domestic, cosmetic and only EU) = 10

Similarly, we get (domestic, nutrition and only EU) = 70-50 = 20

In the question, it is given that 70 cosmetic products did not have EU approval.

In foreign, 40 cosmetic products did not have EU approval. This implies 30 cosmetic products should have only FDA approval in domestic products.

According to the above statement, b = 30

a = 80 - 30 = 50

Given, a + c = 60

c = 60 - 50 = 10

Therefore, the number of nutrition products which had both the approvals is 10.

The answer is option C.

It is given that the total number of products supermarket sells is 320.

cosmetic + nutrition = foreign + domestic = FDA + EU = 320 products

In statement 1, it is given that the number of foreign products is equal to the number of domestic products.

Foreign products = Domestic products = 320/2 = 160

In statement 2, it is given that half of the domestic products were FDA approved cosmetic products, i.e. domestic, cosmetic and FDA = 80

In statement 4, it is given that there were 140 nutrition products, half of them were foreign products. This implies remaining half are domestic.

In statement 5, it is given that there are 200 FDA approved products out of which 70 are foreign products and 120 are cosmetic products.
If 70 are foreign products, remaining 130 should be domestic products. In domestic products, FDA approved cosmetic products are 80. This implies FDA approved nutrition products are 130-80, i.e. 50.

There are 120 FDA approved cosmetic products.

Domestic, cosmetic and FDA approved = 80

This implies, Foreign, cosmetic and FDA approved is 120-80, i.e. 40.

There are 70 FDA approved foreign products.

This implies Foreign, nutrition and FDA approved is 70-40, i.e. 30.

Domestic and Cosmetic = 90

Domestic, comestic and FDA approved = 80

This implies, Domestic, cosmetic and FDA not approved is 90-80, i.e. 10.

Therefore, (domestic, cosmetic and only EU) = 10

Similarly, we get (domestic, nutrition and only EU) = 70-50 = 20

In the question, it is given that 50 nutrition products did not have EU approval. 

In Foreign products, there are 30 nutrition products which do not have EU approval. This implies 20 nutrition products do not have EU(have only FDA) approval in domestic products.

It is given, d = 20

c = 50 - 20 = 30

It is given, a + c = 60

a = 60 - 30 = 30

b = 80 - 30 = 50

Therefore, the number of domestic cosmetic products did not have EU(only FDA) approval is 50.