Test: CAT Quantitative Aptitude- 2 - CAT MCQ

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, 
⇒  V - R = 150 or V = R + 150 ... (1)

⇒ 2R + 300 = R + S
⇒ S = R + 300 ... (2)
Also, 
⇒ 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. 


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. 

We are given that ∠PST = ∠STR ,therefore, we can say that PS ║ TR 
Similarly, ∠STR = ∠TRQ, TS ║QR

In triangles PST and TRQ,
 TS ║QR PS ║TR and ∠PST = ∠TRQ
Hence, we can say that △PST ∼ △TRQ.
Therefore, we can say that 
⇒ ST*TR = PS*RQ  = 16*24 = 384 sq. cm
We know that area of triangle TRS = cm
Thus, option D is the correct answer. 

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. 

The sum of the exterior angles of any polygon is 360 degrees. So, there will be no change in the sum of the exterior angles. The sum of the interior angles of an 'n' sided polygon is (n-2)*180. Hence, if the number of sides is increased by 1, the sum of the interior angles increases by 180.
Therefore, A = 0 and B = 180.

4x + 9y = 36
Both x and y are positive in the first quadrant ⇒ AM ≥ GM inequality holds true.

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.

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

We are given that abc = 10⁹⁹

We know that the number is a multiple of 72. Therefore, the number must be a multiple of both 8 and 9. 
If the number contains all the digits from 0 to 9, the sum of the digits will be 9*10/2 = 45.
Therefore, a number containing all the ten digits will be divisible by 9.
For a number to be divisible by 8, the last 3 digits should be divisible by 8.
Now, we have to find the largest possible number. Therefore, the left-most digit must be 9, the second digit from the left should be 8 and so on. 
We'll get 9876543210 as the number. However, the last 3 digits are not divisible by 8. We must try to make the last 3 digits divisible by 8 without altering the position of other numbers. '120' is divisible by 8.
Therefore, the largest number with different digits divisible by 72 is 9876543120.

 Let 2a+1, 2b+1, 2c+1 be the number of pens received by each of the three boys.
2a+1+2b+1+2c+1=33  [0 ≤ a, b, c ≤ 15]
2a+2b+2c=30
a+b+c=15
Required number of ways  = ^{15 + 3 - 1}C_{3 - 1}
= ¹⁷C₂
= 136
A 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.

To make the calculation easier, let us take the amount the investor wants to invest as x.
Scheme A :
Amount of the investment after 2 years, a₂ = x * (1.1)² = 1.21x
Amount of investment after 5 years, a5 = 1.21x * (1 + (0.2 x 3)) = 1.936x
Since the increase is more than 60%, the additional 10% won’t be added to it.
Scheme B :
Amount of investment after 2 years, b2 = x * (1 + (0.1 x 2)) = 1.2x
Amount of investment after 5 years, b₅ = 1.2x (1.2)³ = 2.0736x
Since the increase is more than 50%, the additional 15% won’t be added to it.
After looking at the interest earned, we can say that the investor should choose scheme B.
Amount earned by the investor = 10⁶
 ×(2.0736−1)=1073600

Let us consider the first term of GP=a
Common ratio=r

0 < a < 20 but the value of a cannot be 10
Sum of possible integral values of a =Sum of first 19 Natural Numbers - 10

=180