CAT 2022 Question Paper Slot 2

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.

Let the total investment me 15x and the no. of years required be T years

$$\frac{\left(3x\times6\times T\right)}{100}+\frac{\left(5x\times10\times T\right)}{100}+\frac{\left(7x\times1\times T\right)}{100}\ge15x$$

or, $$\frac{75xT}{100}\ge15x$$

or, $$T\ge20$$

So minimum value of T is 20 years

$$a_1+a_2+.....+a_N=300N$$

$$6a_1+a_2+.....+a_N=400N$$

$$5a_1=100N$$

$$a_1=20N$$

As the given sequence of numbers is non-decreasing sequence, N can take values from 2 to 15.

N is not equal to 1, if N = 1, then average of N numbers is 300 wouldn't satisfy.

Therefore, N can take values from 2 to 15, i.e. 14 values.

Case 1: When $$x^2-3x-10=0$$ and $$x^2-10\ne\ 0$$

$$x^2-3x-10=0\ $$or, $$(x-5)(x+2) = 0$$

or, x= 5 or -2

Case 2: $$x^2-10=1$$

$$x^2-11=0$$

No integer solutions

Case 3: $$x^2-10=-1\ and\ x^2-3x-10\ is\ even$$

$$x^2-9=0$$

or, (x+3)(x-3)=0

or, x= -3 and 3

for x= -3 and +3 $$x^2-3x-10$$ is even

In total 4 values of x satisfy the equations

Savings target in a year = 550*12 = Rs 6600

Saving in first 9 months = 9(4000-3500) = Rs 4500

Saving for remaining 3 months should be 6600-4500, i.e. Rs 2100

Savings for each month in last 3 months = $$\frac{2100}{3}$$ = Rs 700

It is given, monthly expenses in last 3 months = Rs 3700

This implies, his monthly earnings from 10th month should be 3700+700, i.e. Rs 4400

The answer is option A.

It is given, Angle BAE = 45 degrees

This implies AE = BE

Let AE = BE = x 

In right-angled triangle ABD, it is given $$\angle ABC=\theta\ $$

$$\sin\theta=\frac{AD}{AB}\ $$

$$\sin\theta=\frac{AD}{x\sqrt{\ 2}}\ $$

$$\sqrt{\ 2}\sin\theta=\frac{AD}{BE}\ $$

The answer is option D.

$$f(x) \geq 0$$for all real numbers $$x$$, so D<=0

Since f(2)=0 therefore x=2 is a root of f(x)

Since the discriminant of f(x) is less than equal to 0 and 2 is a root so we can conclude that D=0

Therefore f(x) = $$a\left(x-2\right)^2$$

f(4)=6

or, 6 = $$a\left(x-2\right)^2$$

a= 3/2

$$f\left(-2\right)=\ -\frac{3}{2}\left(-4\right)^2=24$$