Test: CAT Quantitative Aptitude- 2 (October 8) - CAT MCQ

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

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.

First plot the two lines on a graph.
Now, find the point of intersection of y=2x+3 and y=-2x+8

So, the lines intersect at 

If you observe carefully, the figure enclosed in by the lines is a rhombus.
The area of a rhombus is equal to half the product of its diagonals.
The length of the first diagonal is equal to 8 - 3 = 5 units.
The length of the second diagonal is equal to 
Hence, the area of the enclosed figure is equal tosquare units.

Hence, x+y ≥ 18
Hence, option C is the correct answer.

CP = SP + loss
So, CP of 12 apples = SP of 12 apples+ CP of 2 apples
⇒ CP of 10 apples = SP of 12 apples
SP of 12 apples = Rs.600
⇒ CP of 10 apples = Rs.600
CP of 1 apple = Rs. 60
Total CP = Rs. (5 * 12 * 60) = Rs. 3600
Required overall gain = 10%
Required total SP = Rs. 3600 + 10% of Rs. 3600 = Rs. 3960
Remaining apples = 4 dozens
Required SP for 4 dozens = Rs. (3960 - 600) = Rs.3360 
According to the question,
MP - 20% of MP = Rs. 3360
Or, MP = Rs. 4200
Hence, 4200 is the correct answer.

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.

The length of the wire = 4*30 = 120 cm.
Let 'x' be the length of the smallest part of the wire.
Length of the longest part = x + 20
Length of the second longest part*1.25 = Length of the longest part
⇒ Length of the second longest part = 0.8*Length of the longest part
⇒ x + 20 + 0.8*(x+20) + x = 120
⇒ x + 20 + 0.8x + 16 + x = 120
2.8x + 36 = 120
2.8x = 84
⇒ x = 30 cm
The sides of the triangle are 30 cm, 40 cm, and 50 cm.
We can see that the sides of the triangle form a Pythagorean triplet. 
Therefore, the area of the triangle is 0.5*30*40 = 600 cm² 
Therefore, option D is the right answer. 

log m = 300 log(1.003333)

Binomial Expansion of 

=1+1+0.498+0.165+ 0.041....  
(Since and the succeeding terms are very small, they can be ignored because the fraction at the denominator is increasing and at the numerator is decreasing with the subsequent terms of bionomial.
 Further if we observe the fourth term of the binomial expansion:

we can observe that factors and with every subsequent term will get smaller. Thus we can neglect those terms of the expansion. ) 
=1+1+0.498+0.165+ 0.041....  
= {2.704+....}
This can be approximated to 2.7
Hence, C is the correct answer.
In general, when nn is very large, the value of tends to the constant 'e' which is approximately equal to 2.718

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

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. 


From equation (1) and (2) we can say that,

Similarly,

We know that, Area of △PQR =  Area of the △ABC - (Area of the △APR + Area of the △BPQ + Area of the △CRQ)

Hence, option C is the correct answer.