IPMAT 2024 - Quantitative Aptitude

1800 = 2³ × 3² × 5²
= 2 × 3x (2² × 3¹ × 5²)
= 6x (2² × 3¹ × 5²)
Number of factors of 1800 which are also multiple of 6 = Number of factors of the highlighted part of the number = (2 + 1)(1 + 1)(2 + 1)
= 3 × 2 × 3 = 18

For real solution there are three cases possible:
Case I: (any integer)⁰ = 1
∴ x² − 5x + 6 = 0
(x − 2)(x − 3) = 0
x = 2 or 3
Case II: (1)^{any integer} = 1
∴ x² − 15x + 55 = 1
x² − 15x + 54 = 0
(x − 9)(x − 6) = 0
x = 6 or 9
Case III: (− 1)^{any even integer} = 1
∴ x² − 15x + 55 = − 1
x² − 15x + 56 = 0
(x − 7)(x − 8) = 0
x =7 or 8
But we need to check whether the exponent x² − 5x + 6 is even or not for x = 7 & 8.
Putting x = 7 in x² − 5x + 6 we get 20,which is even.
Putting x = 8 in x² − 5x + 6 we get 30, which is again even.
Thus both 7 and 8 are valid solutions,as well.
∴ Total the total number of real solutions for the given equation is 6 i.e (2, 3, 6, 9, 7, 8)

We know, n(T ∪ J ∪ C) = x + y + z …(i)
And also, n(T) + n(J) + n(C) = x + 2y + 3z …(ii)
Where, x, y and z represent number of students who drink exactly, one, exactly two and all three drinks.
We have to find out the maximum number of students that like more than one drink.
It means those students who like exactly two or exactly three drinks i.e. the maximum possible value of y + z.
Given n(T ∪ J ∪ C) = 150; n(T) = 52; n(J) = 48; n(C) = 62
Subtracting (i) from (ii), we get
n(T) + n(J) + n(C) – n(T ∪ J ∪ C) = y + 2z
162 – 150 = (y + z) + z
12 = (y + z) + z
The maximum value of (y + z) can be 12, when is z is 0 & y is 12.

Let BD = x, BF = y
We have to find maximum value of area of rectangle BDEF i.e. maximum value of x.y.
Let the area of rectangle BDEF, A = x ⋅ y  …... (i)
?AFE ~?ABC (AA criteria)
∴ AF/AB = FE/BC
(18 − y)/18 = x/18
y = x − 18 .….. (ii)
Putting y = x − 18 in eqn. (i), we per–
A = x(x − 18)
A = x² − 18x
Differentiating A w.r.t x, we get dA/dx = 2x − 18
Equating dA/dx to 0, in order to get the value x for which area,
A will be maximum, we get
dA/dx = 2x − 18 = 0
x = 9
∴ x = y = 9
∴ Maximum area of rectangle BDEF = x.y
= 9 × 9 = 81

Let the number of oranges, apples and bananas be 3x, 6x and 7x.
As the number of oranges is multiple of both 5 & 6 i.e. 30 (LCM of 5 & 6 is 30).
Their minimum number could be 30 only.
∴ 3x = 30
Hence x = 10 (minimum)
For minimum number of total fruits = 3x + 6x + 7x = 16x
16 × 10 = 160

Taking x – 5 = X and y – 5 = Y, we get |X|+|Y| ≤ 6
Here, if x and y are integers, then X and Y will also be integers.
Case I: X + Y ≤ 6
Case II: X – Y ≤ 6
Case III: – X + Y ≤ 6
Case IV: – X – Y ≤ 6
Plotting the graphs of above inequalities:

Counting the number of integral coordinates (X, Y) in the 1st quadrant, we get 5 + 4 + 3 + 2 + 1 = 15 points. (as shown in the diagram)
So, in all 4 quadrants, total number of points with integral coordinates will be 15 × 4 = 60 points.
Now, counting the number of points with integral coordinate (X, Y) on + x-axis, we get, 6 points.
So, total points on all 4-axes, 6 × 4 = 24 points. Also, the origin (0, 0).
Total number of points (X, Y) with integral coordinates satisfying all the 4 inequalities = 60 + 24 + 1 = 85
Thus, the corresponding pairs of (x, y) will also be 85

Let the original price of chocolates be P.
Final price = Original Price x MF₁ x MF₂
Here, MF is multiplying factor and it is equals to (1 + % change) in case of increase & equals (1 − % change) in case of decrease.
∴ Final price = Original price x (1 + x%) x (1 − x%)
96.76%P = P [1−(x%)²]
96.76% P = P – P (x%)²
96.76% P = 100% P – (x%)²P
96.76 = 100 – x²%
x²% = 3.24
x² = 324
x = 18

f(x) = |x +|x|| and g(x) = 1/x for x ≠ 0
Given f(a) + g(f(a)) = 13/6
f(a) + 1/f(a) = 13/6
[f(a)]² + 1 = f(a) × 13/6
6[f(a)]² + 6 = 13 × f(a)
Let f(a) = t
∴ 6t² − 13t + 6 = 0
Solving above quadratic equation, we get, t = f(a) = 3/2 or 2/3
Case I:
If f(a) = 3/2
∴|x +|x||= 3/2
x +|x|= 32 or − 32 ........ (i)
We know |x| = +x for x > 0 & |x|= − x for x < 0,
Taking x < 0 will not lead to any solution of x.
So, taking x > 0, and substituting |x| with +x in eqn.(i), we get
x + x = 2x = 3/2 or − 3/2
We will get, x = 3/4 (valid) & x = −3/4 (invalid)
Case II:
If f(a) = 2/3
∴|x +|x||= 2/3
x +|x|= 2/3 or − 2/3
When x > 0, |x| = +x,
Substituting in eqn.(i) we get, x + x = 2x = 2/3 or = −2/3
x = 2/6 = 1/3 (valid) or −2/6 = −1/3 (invalid)
Now, f(g(a)) =|g(a) +|g(a)|| = |1/x +|1/x||
For x = 3/4
f(g(a)) = |4/3 +|4/3|| = 8/3 and for x = 1/3
f(g(a)) = |3 + |3||= 6
So, the maximum value of f(g(a)) = 6

Let the first portion borrowed at 3% p.a. SI be Rs x
And the other portion borrowed at 4% p.a. be Rs (4000 − x)
∴ Total interest = SI₁ + SI₂

13000 = 3x + 16000 − 4x
x = 3000
∴ Amount borrowed at 3% p.a.SI was Rs 3000

One equilateral triangle with sides (5, 5, 5)
Three isosceles triangles (7, 7, 1), (6, 6, 3) and (4, 4, 7)
Reason: The measure of two equal sides cannot be less than 4 or more than 7. Taking two equal sides as 8 will make sum of these two sides only as 16, more than the perimeter. Taking two equal sides as 3 each, the third side will be 9, eventually not following the property of triangle.
Three scalene triangles (7, 6, 2), (7, 5, 3), (6, 5, 4)
Reason: The maximum measure of two equal sides of the triangle can be 7 or 6 only. Taking 8 as the largest side of the triangle will leave us with the sum of other two sides as 7. Taking 5 as the largest side of triangle we end up getting an equilateral triangle which we have already considered before. 4 cannot be the largest side.
Total number of possible triangles is 7.

Let the sum of all three elements along any row, column or diagonal be S.
Then, f = S – (8 + 3) = S – 11 …...(i) (from the bottom row)
d = S – (7 + f) …....(ii) (from one of the diagonals)
From equations (i) and (ii), we get
d = S – (7 + S – 11)
∴ d = 4
Also, b = S – (8 + d) = S – 12 (from the middle column) …....(iii)
a = S – (b + 7) (from the first row)…....(iv)
From equations (iii) and (iv), we get a = 5 & b = 0
∴ S = 12
∴ The matrix A comes out to be
Determinant |A| = 5(4 × 3 − 8 × 2) + 0(2 × 1 − 6 × 3) + 7(6 × 8 − 1 × 4)
= 5(12 − 16) + 0 + 7(48 − 4)
= − 20 + 0 + 308 = 288

Median is defined as the positional mean of the given values.

Highest possible age of an employee of company A can be 29.
This is possible when the least possible age of any employee is 30 years, which is feasible in the given scenario.
(when the age of 15 employees each is equal to 30).

Total number of employees = 32 + 28 + 43 + 39 + 35 + 29 + 23 + 16
= 245
Median age of all these employees = age of the employee at the middle most position when all the employees are arranged in ascending/ descending order of their age.
Middle position will be (245 + 1)/2 = 123
∴ The age of employee whose position is 123rd.
Adding, number of employees of companies A, B and C = 32 + 28 + 43 = 103.
To reach 123rd position, we need to add 20 employees of company D.
We know, the age of 20th employee of company D = median age of the employees of the company D = 45 years. (Here, 20th position will be the middle most position when all 39 employees are arranged in ascending / descending order of their age)
Therefore, median age of an employee across all eight companies = 45 years

Lowest possible sum of ages of all the employees of company F, is only possible when the maximum possible age of any employee of company E is 49 and not more.
In that case, the minimum possible age of employees of company F, can be 50.
Now, company F has 29 employees with median 54.
That means there can be 14 employees with age 50 each and 15 employees with age 54 each.
∴ Sum total of age of 29 employees company F = 14 × 50 + 15 × 54 = 700 + 810 = 1510

Given that |x + 1| + (y + 2)² = 0
Sum of an absolute value & a perfect square value can be equal to 0 only when each of these terms is equal to 0.
∴ |x + 1| = 0
x = − 1
And (y + 2)² = 0
y + 2 = 0
y = −2
Putting the values of x and y in the equation, ax − 3ay
= 1, we get
a (− 1) − 3a (− 2) = 1
− a + 6a = 1
a = 1/5

In ?ABD, ∠ABD = ∠ACD = x (let′s say) (? AB = AC)
Now, take ∠CDE = y
∠AED = x + y (Using exterior angle property)
∴ ∠ADE = x + y (ADE is an isosceles ?, AD = AE)
∴ ∠DAE = 180° − (2x + 2y)
Using angle sum property, in ?ABC
A + ∠B + ∠C =
x + (30° + 180 − 2x − 2y) + x = 180°
2y = 30°
y = ∠CDE = 15°

“At Least one boy sitting between any two girls” can be inferred as “No two girls should sit together”.

First, arrange 5 boys around a circular table. This can be done in (5 – 1)! = 4!
Now, there are 5 gaps available where we can make 3 girls sit in ⁵P₃ ways.
In this arrangement, no two girls will be sitting together.
Total ways = 4! × ⁵P₃
Probability 

Given: BA = ‘c’
BD = ‘k’
∠BDA = θ (acute angle)
Now in ?BDP, BP = k sinθ
Now in ?ABD,

Given,

Given that

Also ac > 0. It means,
(i) either both a and c are positive values, or
(ii) both a ad c are negative values.
Let’s take case-(i), a and c are +ve values.
From equation (3), we can say c is + 4x then b = − 3x
Consequently, from equation (1), a = + x (a & c should be of the same sign) 

Let’s take case-(ii), a and c are -ve values
From equation (3), we can say c is − 4x then b = +3
Consequently, from equation (1), a = − x (a & c should be of the same sign)

In either of the cases considered above, the value of the expression is equal to 1.

(?² − 5)⁴ + (?² − 7)⁴ = 16
(?² − 6 + 1)⁴ + (?² − 6 − 1)⁴ = 16
Take x² − 6 = t….(i)
(t + 1)⁴ + (t − 1)⁴ = 16
(t⁴ + 4 t³ + 6 t² + 4t + 1) + (t⁴ − 4t³ + 6t² − 4t + 1) = 16
2(t⁴ + 6t² + 1) = 16
t⁴ + 6t² − 7 = 0
Let t⁴ = k²
k² + 6k − 7 = 0
(k + 7)(k − 1) = 0
k = − 7, k = 1
k = t² = − 7 (not possible, because a perfect square can't be negative)
∴ t² = 1
t = ± 1
Putting t = 1 in the equation (i), we get
x² − 6 = 1
x² = 7
x = ± √7
Putting t = − 1 in the equation (i), we get
x² − 6 = − 1
x² = 5
x = ± √5
Difference between maximum root (+ √7) and minimumroot − √7) is
√7 − (− √7) = 2√7

For number of digits of 2²⁰²⁴
log 2²⁰²⁴ = 2024 log 2 = 2024 × 0.301 = 609.28
Number of digits in 2²⁰²⁴ = 609 + 1 = 610
For number of digits of 5²⁰²⁴
log 5²⁰²⁴ = 2024 log 5 = 2024 × 0.699 = 1414.71
Number of digits in 52024 = 1414 + 1 = 1415
So, the total digits when 2²⁰²⁴ + 5²⁰²⁴ are written out consecutively is 610 + 1415 = 2025

Upstream Speed = (Distance travelled upstream)/(Time taken) = 96/8 = 12 km/h
Downstream speed = (Distance travelled downstream/(Time taken) = 96/6 = 16 km/h
In one trip upstream and downstream, travelling for an hour each, the boat will be effectively 4 km (12km up followed by 16 down) downstream of A.
So, in total 5 such trip the boat will be 4 × 5 = 20 km downstream of A

Case I: Point C lies anywhere outside the circle.
AC = 9 cm
BC = 4 cm
AC = 2r + BC = 9 cm
2r + 4 = 9 cm
r = 2.5 cm
Case II:

Point C lies anywhere inside the circle.
BC = 4 cm
AC =  9cm
2r = AC + BC = 9 + 4 = 13 cm
2r = 13 cm
r = 6.5 cm
So, the radius of the circle can be 2.5 cm or 6.5 cm.

Let, n(FW) – Number of people who owns 4-wheelers, n(TW) – Number of people who owns 2-wheelers, n(FW ∪ TW) – Number of people who owns at least one of the vehicles, and n(FW ∩ TW) – Number of people who owns both type of vehicles
Now, the number of people who owns at least one of the vehicles = Total people surveyed – Number of people who owns none
(FW ∪ TW) = 500 – 100 = 400
Using the standard formula in set theory,
n(FW ∪ TW) = n(FW) + n(TW) – n(FW ∩ TW)
400 = 250 + 100 – n(FW ∩ TW)
So, the number of people who owned both are (FW ∩ TW) = 50

Let the series be a, ar, ar² …..
Given: Sum_∞ = a/(1 − r) = 80 …(i)
Also, a + ar = a(1 + r) = 35 …(ii)
80(1 – r) (1 + r) = 35
1 – r² = 35/80 = 7/16
r² = 9/16
r = ± 3/4
Substituting r = 3/4 in eq.(i) we get a = 20
These values of a and r, won’t lead the sum of series to 100 as sum of infinite terms is 80 only.
Substituting r = − 3/4 in eq.(i) we get a = 140.
a₁ = 140;
a₂ = 140 × − 3/4 = −105;
a₃ = − 105 × − 3/4 = 78.75;
a₄ = − 78.75 × − 3/4 = − 59.0625;
a₅ = − 59.0625 × − 3/4 = 44.296875 & so on.
Adding first 5 terms of the series, we get
a₁ + a₂ + a₃ + a₄ + a₅ = 98.984375 ≈ 100
So, the sum of first 5 terms of the series when r = − 3/4 and a = 140 gives the sum close to 100.

Let us give one balloon to each one of them.
So, we are now left with 12 balloons to be distributed.
CASE I:
G1 G2 G3 G4 G5 B1 B1 B1
Giving no balloons to any boy and distributing remaining 12 balloons to the 5 girls.
Total ways = ¹²⁺⁵⁻¹C₅₋₁ = ¹⁶C₄ = 1820 ways.
[By using the formula ^r−1C_n+r−1]
CASE II:
Giving 1 balloon to one of the boys, which can be done in 3 ways
And distributing remaining 11 balloons to the 5 girls, such that each girl should get at least one balloon.
This can be done in ¹⁰C⁴ = ways.
∴ Total ways 3 × ¹⁰C₄ = 3 × 210 = 630 ways.
CASE III:
Giving 1 balloon each to 2 boys, which again can be done in 3 ways.
And distributing remaining 10 balloons to the 5 girls, such that each girl should get at least one balloon.
This can be done in ⁹C₄ ways.
∴ Total ways 3 × ⁹C₄ = 3 × 126 = 378 ways.
CASE IV:
Given 1 balloon to each of the 3 boys and distributing remaining 9 balloons to the 5 girls such that each girl should get at least one balloon.
∴ Total way = 1 × ⁸C₄ = 70 way
CASE V:
Giving 2 balloon to exactly one boy which can be done in 3 ways and distributing remains 10 balloons to the 5 girls such that each get at least 2 balloons which can be done in only 1 way.
∴ Total ways = 3 × 1 = 3 ways
Adding all the possible ways obtained in all the 5 cases, we get
1820 + 630 + 378 + 70 + 3 = 2901 ways.

Option (a) 3²⁰⁰ = (3²)¹⁰⁰ = 9¹⁰⁰
Option (c) 4¹⁰⁰
Option (d) 2³⁰⁰ = (2³)¹⁰⁰ = 8¹⁰⁰
Clearly, among the option (a), (c) and (d) the exponent is same.
So, the option with largest base will be largest.
So, among these three options, option (a) is largest.
Now option (b) is 2¹⁰⁰ + 3¹⁰⁰ < 3¹⁰⁰ + 3¹⁰⁰
Or, 2¹⁰⁰ + 3¹⁰⁰ < 2x 3¹⁰⁰
And, 2x3¹⁰⁰ is certainly less than 3²⁰⁰ i.e. option (a)
That means, option (b) is less than option (a).
∴ Option (a) is the highest.

Th given expression can also be written as ^17−xC_3x+1.
At x = 0, we get ¹⁷C₁ (an integer)
At x = 1, we get ¹⁶C₄ (an integer)
At x = 2, we get ¹⁵C₇ (an integer)
At x = 3, we get ¹⁴C₁₀ (an integer)
At x = 4, we get ¹³C₁₃ (an integer)
At x = 5, we get ¹²C₁₆ (the expression is not defined)
After putting x = 5 or more, the expression will not be defined.
So, x can take only 5 values i. e. 0, 1, 2, 3, 4.

Given x₁ + x₂ + x₃ + x₄ = 50 …(i)
And x₁ ≥ 1 or x₁ − 1 ≥ 0 |x₂ ≥ 2 or x₂ − 2 ≥ 0 |x₃ ≥ 0|x₄ ≥ 0
Let x₁ − 1 = a |x₂ − 2 = b |x₃ = c |x₄ = d
So, now we have a, b, c, d ≥ 0
Equation (i) can be written as (a + 1) + (b + 2) + c + d = 50
Or, a + b + c + d = 47
We have to find non-negative integer solution of above equation.
It can be find be find out using the formula ^n+r−1C_r−1 where, n = 47 & r = 4
∴ Total number of solutions = ⁴⁷⁺⁴⁻¹C₄₋₁ = ⁵⁰C₃

= 19600

Let the amount invested in scheme B @12% S.I. per annum be x and amount invested in scheme A @10% C.I. per annum, compounded half-yearly be (10000 − x).

Total simple interest e arned from scheme B = 6000 × 12/100 = 720 rupees
Total compound interest earned from scheme A = 4000 × (21/20)² − 4000
= 

∴ Total interest earned = 720 + 410 = 1130 rupees. 

Let the number of apples and oranges be x and y.
Then the number of bananas = x + 10
Total number of fruits = 2x + y + 10
As per the question, x + (x +10) = 11k (where k is a multiple of 11)
2x + 10 = 11k
k = (2x + 10)/11   ...….(i)
Also, total fruits sold by the seller can be expressed in two ways.
Equating both the expressions, we get

14x + 12x + 120 + 10y = 22x + 11y + 110
4x + 10 = y  ...…(ii)
Now, it is given that the fruit seller sold 70% of apples i.e. 7x/10
In order to keep this number an integer, x must be a multiple of 10.
The smallest value of x in eqn.(i), for which k = (2x+10)/11 will be an integer is 50, a multiple of 10.
Substituting, x = 50 in eqn.(ii), we get
y = 4(50) + 10 = 210 oranges.

We know nth term of a G.P. is given by the formula,

x² − 10|x| − 56 = 0
We know, if x ≥ 0, then |x| = + x
x² − 10x − 56 = 0
(x − 14)(x + 4) = 0
x = 14 or − 4 (only one root, + 14 is valid)
Also, if x < 0, then |x| = − x
x² − 10(− x) − 56 = 0
x² + 10x − 56 = 0
(x + 14)(x − 4) = 0
x = − 14 or + 4 (only one root, − 14 is valid)
So, only 2 real solutions.

Let the number of girls be G and total number of students be N.
Given that, 40% N < G < 50%
12N/25 < G < 1N/2
Multiplying throughout by 50, we get
24N < 50G < 25N
Here 50G is a number which is multiple of 50.
Now checking with the options:
Option (a), 25 (starting with the smallest of all the given options)
24 × N = 24 × 25 = 600
And, 25 × N = 25 × 25 = 625
So, there is no multiple of 50 in this range.
Option (c), 27 (the second smallest value in the given options)
24 × N = 24 × 27 = 648
And, 25 × N = 25 × 25 = 675
So, there lies a multiple of 50 in this range i.e. 650.
So, option (c) satisfies. Ans.
Verification:
50G = 650
G = 13, N = 27
13/27 × 100 = 48.14% which lies between 48% and 50%.

From point (2), we can conclude that C must have 8 votes each in S3 and S4, that so as up to 16.
It also means C did not get any votes in S1, S2 and S5.
It also implied that B must have win S4 and S5 constituencies as ‘A’ has won only in S1.
Now in S3 and S4, possible combination of votes A and B can be (5, 7), (7, 5) or (6, 6) in that order respectively.
By doing hit and trial, we can come to a conclusion that B must have win 11 and 12 votes in S4 and S5 respectively.
And also, it must have received 1, 5 and 6 votes in S1, S2 and S2 constituency from point (3).
From the completed table we can see, B got lower number of votes compared to A and C only in S2.

From point (2), we can conclude that C must have 8 votes each in S3 and S4, that so as up to 16.
It also means C did not get any votes in S1, S2 and S5.
It also implied that B must have win S4 and S5 constituencies as ‘A’ has won only in S1.
Now in S3 and S4, possible combination of votes A and B can be (5, 7), (7, 5) or (6, 6) in that order respectively.
By doing hit and trial, we can come to a conclusion that B must have win 11 and 12 votes in S4 and S5 respectively.
And also, it must have received 1, 5 and 6 votes in S1, S2 and S2 constituency from point (3).
From the completed table we can see, B got lower number of votes compared to A and C only in S2.

From point (2), we can conclude that C must have 8 votes each in S3 and S4, that so as up to 16.
It also means C did not get any votes in S1, S2 and S5.
It also implied that B must have win S4 and S5 constituencies as ‘A’ has won only in S1.
Now in S3 and S4, possible combination of votes A and B can be (5, 7), (7, 5) or (6, 6) in that order respectively.
By doing hit and trial, we can come to a conclusion that B must have win 11 and 12 votes in S4 and S5 respectively.
And also, it must have received 1, 5 and 6 votes in S1, S2 and S2 constituency from point (3).
From the completed table we can see, B got lower number of votes compared to A and C only in S2.

From point (2), we can conclude that C must have 8 votes each in S3 and S4, that so as up to 16.
It also means C did not get any votes in S1, S2 and S5.
It also implied that B must have win S4 and S5 constituencies as ‘A’ has won only in S1.
Now in S3 and S4, possible combination of votes A and B can be (5, 7), (7, 5) or (6, 6) in that order respectively.
By doing hit and trial, we can come to a conclusion that B must have win 11 and 12 votes in S4 and S5 respectively.
And also, it must have received 1, 5 and 6 votes in S1, S2 and S2 constituency from point (3).
From the completed table we can see, B got lower number of votes compared to A and C only in S2.

From point (2), we can conclude that C must have 8 votes each in S3 and S4, that so as up to 16.
It also means C did not get any votes in S1, S2 and S5.
It also implied that B must have win S4 and S5 constituencies as ‘A’ has won only in S1.
Now in S3 and S4, possible combination of votes A and B can be (5, 7), (7, 5) or (6, 6) in that order respectively.
By doing hit and trial, we can come to a conclusion that B must have win 11 and 12 votes in S4 and S5 respectively.
And also, it must have received 1, 5 and 6 votes in S1, S2 and S2 constituency from point (3).
From the completed table we can see, B got lower number of votes compared to A and C only in S2.