IPMAT 2019 - Quantitative Aptitude

The sum of interior angles of an n sided polygon is given by  

180 n – 360 < 2019
180 n > 2379
n < 13.21
Hence, the maximum possible value for n is 13.

The expression can be written as

f (x) f (y) = f (2xy + 3) + 3f (x + y) - 3f (y) + 6y
Putting x = y = 0,
f (0) f (0) = f (2.0.0 + 3) + 3f (0 + 0) - 3f (0) + 6.0
f (0)²= f (3) + 3f (0) - 3f (0) + 0
f (3) = f (0)²
Putting y = 0,
f (x) f (0) = f (2.x.0 + 3) + 3f (x + 0) - 3f (0) + 6.0
f (x) f (0) = f (3) + 3f (x) - 3f (0)
f (x) f (0) = f (0)² + 3f (x) - 3f (0)
f (x) f (0) - f (0)² = 3f (x) - 3f (0)
f (0) (f (x)- f (0)) = 3 (f (x)- f (0))
Either f(x) = f(0) or f(0) = 3
Since all functions can’t have the same value, f(0)=3
Putting x = 0, y = 3,
f (0) f (3) = f (2.0.3 + 3) + 3f (0 + 3) - 3f (3) + 6.3
f(0) f(3) = f(3) + 3f(3) -3f(3) +18 = f(3) + 18
3 f(3) = f(3) + 18
Therefore, f(3) =9
f(3) = 9.
Putting x = 0, y = 8
f (0) f (8) = f (2.0.8 + 3) + 3f (0+ 8) - 3f (8) + 6.8
3.f(8) = f(3) + f(8) – f(8) + 48
3. f(8) = 9 + 48
f(8) = 19

Let the numbers be 3x, 5x, 7x respectively
∴ 3x + 5x + 7x – 16 = 7x
8x = 16
x = 2
Total apples = 15x = 15 × 2 = 30

A decrease of 25% means a reduction of 1 on 4
An increase of 20% means an increase of 1 on 5

There is a decrease of 2 on 20, i.e., 10%
∴ x = 10

For minimum and maximum value of f(x), f1(x) = 0

x = ± 2
f(2) = 1
f(- 2) = 9
∴ k = 9

It is evident from the graph that the distance AB will be maximum distance = 7

sin x + sin y = sinx cosy + cosx siny
sin x (1 – cosy) + siny = cosx siny
Squaring both sides, 
sin²x (1 – cosy)² + sin² y + 2 sin x sin y (1 – cos y) = cos² x sin² y
Rearranging the terms, 
2 sin x (1 – cos y) (sin x + sin y) = 0
sin x = 0, cos y = 1, sin (x + y) = 0
If sin x = 0, x = 0   ............ (1)
If cos y = 1, y = 0  ............  (2)
If sin(x + y) = 0, x + y = 0, x = - y  ............(3)
Now, |x| + |y| = 1
∴ from (1), (0, 1) and (0, -1)
From (2), (1, 0) and (-1, 0)

According to the question, we can conclude that each row must contain two’s 1 and one -1 and column as well
So the matrix can be

There may be different arrangement possible, but the value of the determinant will not change, so the answer is = 4

It can be seen that Q(k) will always contain composite numbers and never a prime number, for example:
Q(2) will have 4, 6, 8, 10....24
Q(3) will have 6, 9, 12,.....24
It means that P - Q(k) will be the set of all the prime numbers between 2 and 25 (including 2 and 25)
∴ The required set = {2, 3, 5, 7, 11, 13 ,17, 19,  23}
Hence, 9 elements

Let the triangle be of dimensions a, a, a√2, i.e.

Hence, the circular plate well has its radius equal to a√2.
Now, if we take the number of tiles that can be produced to be ‘n,’ then

n = 4π
= 4 × 314 => n = 12.56
Hence, 12 tiles can be produced.

The equation can be rewritten as 

Now, since y is an integer, (x+1)/7 it is also an integer, which means x + 1 is a multiple of 7.
∴ Positive values of x => 6, 13, 20....97 = 14 values
Negative values of x => -1, -8, -15 ...-97 = 15 values
∴ Total = 14 + 15 = 29 values

As we know, that average is nothing but uniform distribution.
Hence, to start with, we can assume the numbers to be 110, 110, 110, and 110.
Now, since the numbers are distinct and we have to maximize a particular number, the final configuration will look like

There are nine single-digit numbers. Hence to write to them all, we’ll need 9 x 1 = 9 digits; similarly, there are 90 two-digit numbers. So, we’ll need 90 × 2 = 180 digits.
For 3 digit numbers, 900 x 3 = 2700 digits
It means that till the point we have written all the 3 digits numbers, we’ve used 9 + 180 + 2700 = 2889 digits.
Now, we have 6389 – 2889 = 3500 digits left.
That means we can write 3500/4 = 875 digits four-digit number, i.e., till 1874 (999 + 875)
Our sequence will look like 1234...........18731874
∴ The last digit will be 4

Let the numbers be x and y
∴ x + y = x × y
(0, 0) clearly satisfies the given equation. Also 

Since y is an integer, (x – 1) is a factor of 1
∴ x – 1 = 1
x = 2 => y = 2
Hence, two solutions (0, 0) and (2, 2)

Let P = set of multiples of 4
Q = set of multiples of 6 
R = set of odd multiples of 9.
∴ P = {4, 8, 12, ...... 996} = 249 elements
Q = {6, 12, 18, ..........996} = 166 elements
R = {9, 27, 45, .............999} = 56 elements
Now, this question can be easily solved through a Venn diagram

P ∩ Q is set of all the multiples of 12 ( 4 & 6) i.e.
{12, 24, 36..........996} = 83 elements
Also P ∩ R, Q ∩ R, P ∩ Q ∩ R will be 0 as P and Q consists of only even numbers while R consists of only odd.
∴ Total numbers = 166 + 83 + 83 + 56 = 388

If 50! is divisible by 21, then it is a multiple of 3 and 7. Since; 7s will be more limited in availability than 3s, we need to count the highest power of 7 in 50!

∴ Required answer = 7 + 1 = 8

Since the number is divisible by ‘9’, the sum of digits must be a multiple of 9.
4530xy = 4 + 5 + 3 + 0 + x + y
=> 12 +  x + y
Now, x + y = 6 or 15
From the options, option (d) can be eliminated as 5 + 4 = 9 and we need the sum to be equal to 6 or 15.
Now, the best way to solve further is by using options and checking the divisibility for 7. 
Option (b) and (c) will be eliminated Hence, option (a)

We need to determine the range of values that ‘x’ can take.
If x => minimum then the area
70 + 50 – x = 100
x = 20%

If x => maximum, then the configuration will be some 
What as shown in figure x = 50%

Hence, option (a)

From the data given in the question, we can make the following arrangement

Let the region denoting only C and H be x then,

∴ 40 + 12 + 3 + 12 – x = 65
x = 2
So the number of students like both cricket and hockey is 5 + 2 = 7

(Multiplying both sides by -1)
Observe discriminant of x² + 2x + 5 is less than 0.
It means that for all x² + 2x + 5 > 0.
Hence we need consider only (x + 1)(x + 2)
(x + 1) (x + 2) ≤ 0
x ∈ [– 2, – 1]
Largest possible difference – 1 – (– 2) = 1

The minimum value that any square can take is 0.

(Taking log on both sides)
(2x² – 3x + 4) log (0.7) < 3 log (0.7)
Cancelling out log (0.7) on both sides,
2x² – 3x + 4 > 3
Please note that the equation has changed as log (0.7) < 0
2x² – 3x + 1 > 0
(2x – 1) (x – 1) > 0

Let the radio of original circle and smaller inscribed circle be 2 and x respectively, then

The side length of square PQRS = √(n² + 1)
Hence, the required ratio

cos x + cos y = 1
Squaring, we get cos²x+ cos²y + 2 cos x cos y = 1  ......(1)
Similarly, let sin x – sin y = p
Squaring, we get sin² x + sin² y – 2sin x sin y = p²    ......(2)
Adding (1) and (2)
cos² x + sin² x + cos² y + sin² y + 2(cos x cos y – sin x sin y) = 1 + p²
2 + 2 cos(x + y) = 1 + p²
p² = 1 + 2 cos(x + y)
Since – 1 ≤ cos (x + y) ≤ 1
1 + 2x(– 1) ≤ p² ≤ 1 + 2 × 1
– 1 ≤ p² ≤ 3
A square can never be negative.
Hence O ≤ P² ≤ 3
Since p² ≤ 3
–√3 ≤ p ≤ √3

As to find the value of x, we will compare the value corresponding to x,

This is  a direct relation that we have already studied in the function module
? f(x) = |x – a| + |x – b| + |x – c|
Where a > b > c
Min f(x) = a – c at x = b
Hence, answer is (d)

α + β = 1, αβ = p
γ + δ = 4, γ + δ = q
Since α, β, γ, δ are in GP, then let us assume them to be equal to α, αr, αr², αr³ respectively.
∴ α + αr = 1 and αr² + αr³= 4
α(r + 1) = 1 and αr² (r + 1) = 4
Diving both, we get r² = 4
r = ± 2
If r = 2

(Not possible as P and Q are integer)
∴r = – 2
α = – 1, β = 2, γ = – 4, δ = 8
P = – 2, Q = – 34
P + Q = – 34

Common difference of 1st AP = 3
Common difference of 2nd AP = 2
∴ Common difference of common terms AP = com (2, 3) = 6
Also 1st term = 5
∴ Required AP = 5, 11, 17....10
No of terms = 17

It can be easily concluded that
f(1/13) > f(2) > f(3)
Hence option (b)

If the sum of three numbers is 15 then the numbers could be;
6 6 3 = 3 permutations
6 5 4 = 6 permutations
5 5 5 = 1 permutations
∴ Sample space = 10
If 4 is placed at 1st place, then there are 2 arrangements possible; 4, 5, 6 or 4, 6, 5
∴ Favourable events = 2
∴ Probability = 2/10 = 1/5

Let the number of families of each kind be ‘n.’
∴ Population with family size ‘2’ = 2n
∴ Population with family size ‘4’ = 4n
Total population = 12n

Now, for 0 < a < 1, log a < 0
logf(x) – log g(x) > 0
∴ f(x) > g(x)         ...(1)
Similarly for a > 1, log a > 0
log f(x) – log g(x) < 0
f(x) > g (x)         ...(2)
From (1) & (2), option (a) is correct

Let the sides be x, y, z then 
6(x² + y² + z²) = 564
x² + y² + z² = 94
Only solutions possible are 81 + 9 + 4 and 49 + 36 + 9
Hence possible sides are 9, 3, 2 and 7, 6, 3
Hence, volume = 9³ + 3³ + 2³ and 7³ + 6³ + 3³
= 764 and 586

Option (d) can be eliminated as 4¹⁰⁰ = 2²⁰⁰, which is lower than option (a). Similarly, option (c) can be eliminated.
Now 2³⁰⁰ = 8¹⁰⁰ and 3²⁰⁰ = 9¹⁰⁰
Hence, option (b) is the greatest number.

If we plot the condition on a graph, it will look like below

Now 0 < x < 31, 0 < y < 31 and 0 < x + y < 31
∴ x + y = 30    29 solutions
x + y  29    28 solutions
x + y = 2     1 solutions
Total solutions = 1 + 2 + 3 + .....29

Let the speeds of faster and slower insects be a and b, respectively.
Hence, relative speeds in case of the same and opposite directions will be a – b and a + b, respectively.

Applying Componendo & Dividendo

Required ratio b : a = (v - u) : (v + u)

Applying the alligation model.

Hence a : b = 1 : 3
Again applying the same model for b: a = 3: 1

∴ Required ratio = 9 : 5

1 digit numbers = 2 (1 or 3)
2 digit numbers = 2 × 2
3 digit numbers = 2 × 2 × 2
6 digit numbers = 2 × 2 × 2 × 2 × 2 × 2
Now for 7 digit numbers = 1222222
We cannot place ‘3’ on the 1st position
∴ Total = 2 + 2¹ + 2³+ .....2⁶ + 2⁶
= 190

Let the value of n = 2. Then a₁ a₂ + a₂ a₁ = 0
No value of a₁, a₂ will satisfy the above equation.
Hence, option (a) is eliminated. A similar pattern option (b) can also be ruled out to create a zero; we need equal occurrences of 1 and -1, and the total number of terms when n = 3 will be 3. Hence obtaining 0 is not possible.
When n = 4, we can easily obtain 0 when the numbers are 1, -1, 1, -1
Similarly, check for n = 8
Hence, the option (c)

Total exports for period of 2006 – 2017
280 + 280 + 230 + 210 + 200 + 220 + 210 + 200 + 200 + 200 + 200 + 220 + 200
Required average = 2650/12 ≈ 220 cr

Percentage decline in exports during 2006 -  2011,

Percentage decline in exports during (2012-17)

= 4.45%
Difference in percentage = 16.5% (approx.)

It can be visually identified that the maximum difference between export & imports is evident in 2014, 120 cr.

So the largest average negative balance of trade is 2010 – 2012.