Quantitative Aptitude (QA) Test - 19

Let 'x' and 'y' be the number of burgers and rolls bought by Ramesh in the first case.

It is given that: 80x + 60y < 3000

⇒ 4x+3y < 150 ... (1)

We are given that in the 2nd case he spends Rs. 80x on purchasing rolls and Rs. 60y on burgers.

It is known that in the 2nd case Ramesh was able to buy 3 more items in the same amount of money. Hence,

At y = 4, x = 12

At y = 8, x = 15

At y = 12, x = 18

At y = 16, x = 21

At y = 20, x can't be 24 because in that case 4x+3y > 150. Hence, we can say that x_max = 21.

Let 'V', 'R' and 'S' be the number of runs scored by Virat, Rohit and Shikar respectively.

It is given that, 

⇒ 4R = R + 150 + R + 300

⇒ R = 225 Runs.

Therefore, V = R + 150 or 375 runs,   S = R + 300 = 525 runs.

Hence, the total number of runs scored by all 3 players combined = 375 + 225 + 525 = 1125. 

Given, Set P has 4 elements and set Q has 5 elements

⇒ The numbers of injections are defined from P to Q = ⁵P₄ = 5!/(5−4)!

⇒ The numbers of injections are defined from P to Q = 5! = 120

f(a)f(b) = f(a) + f(b) + f(ab) - 2

Put a = b = 1.

[f(1)]² =3f(1)−2 ⇒ f(1) = 1 (or) 2.

Let's assume f(1) = 1

Now, put b = 1.

f(a) = 2f(a) - 1

⇒ f(a) = 1 ⇒ For all values of a, f(a) = 1.

This is false because f(4) = 17.

⇒ f(1) = 2 is the correct value.

Now put b = 1/a

f(a)f(1/a) = f(a) + f(1/a) + 2 - 2

⇒ f(a)f(1/a) = f(a) + f(1/a)

So taking RHS terms to LHS and adding 1 to both sides we get

f(a)f(1/a) - f(a) - f(1/a) +1 = 1

(f(a) - 1) (f(1/a)-1) = 1

Let g(x) = f(x)-1

So g(x)*g(1/x) = 1

So g(x) is of the form ±xⁿ

So f(x) is of the form ±xⁿ +1.

f(a) = aⁿ +1 satisfies the above condition.

−4ⁿ +1 = 17 ⇒ -4ⁿ−4 n  = 16, which is not possible.

4^a +1 = 17 ⇒ n = 2

⇒ f(a) = a² + 1a 2 +1

⇒ f(7) = 7² + 1 = 50

Using the identity a³ + b³ + c³ - 3abc = (a + b + c) × (a² + b² + c² - ab - bc - ac).

We find that if a + b + c = 0.

Then, a³ + b³ + c³ = 3abc.

Therefore, a³ + b³ + c³ is divisible by a, b, c and 3.

Now, here a = 82, b = - 62 and c = - 20.

i.e. a + b + c = 0

Therefore, N is divisible by 3, 82, 62 and 20.

Thus, N can be expressed as a product of 3 and 82 × (- 62) × (- 20)

Thus, the answer is 82 × (- 62) × (- 20) = 1,01,680

In order to minimize the total number of contracts, the number of contracts received by the companies must be consecutive integers because each of the companies got a distinct number of contracts.

So, let the number of contracts of different companies be a-4,a-3,a-2,a-1,a,a+1,a+2,a+3 and a+4.

To maintain the sum of the number of contracts given to any five companies is greater than the sum of the number of contracts given to the remaining four companies, the sum of the lowest five companies must exceed the amount of top four companies. If such a condition is maintained, then any set of 5 companies will exceed any set of 4 companies.
It is the limiting condition for five companies to exceed four companies.

5a - 10 > 4a + 10

a > 20 ⇒ Minimum value of a is 21 ⇒ Minimum value of the number of contracts got by the company which was given the least number of contracts among the 9 companies = a - 4 = 21 - 4 = 17.

To get the value of  [a/3]-[a/5]=2, [a/3] should be ≥2 because [a/3] ≥ [a/5]

At a=6,  [a/3]=2 but [a/5]= 1. So [a/3]-[a/5] = 1

We will take the next multiple of 3.

At a=9, [a/3]=3 but [a/5]= 1. So [a/3]-[a/5] = 2.............(1st case)

In this question, we will check the multiples of 3 and 5 as these points will change the values of [a/3] and [a/5].

Given ‘a’ is a positive integer.

We can create a following table:

For any value greater than 20 the equation [a/3]-[a/5]=2 will be greater than 3 

Hence total of 8 values.

Answer A

Let us draw the diagram according to the information given in the question. 

Let '8x', '8y' and '8z' be the length of side AB, BC and CA respectively. 

Hence, option C is the correct answer.