IPMAT 2021 - Quantitative Aptitude

Number of positive Integers dividing a given number = Number of factors of that number
Prime Factorization of 
1890 = 2 × 3³ × 5 × 7
Prime factorization of 
130 = 2 × 5 × 13
Prime factorization of 
170 = 2 × 5 × 17
Prime factorization of 
1890 × 130 × 170 = (2 × 3³ × 5 × 7) × (2 × 5 × 13) × (2 × 5 × 17)
= 2³ × 3³ × 5³ × 7 × 13 × 17
Number of factors of 
1890 × 130 × 170 = (3 + 1) (3 + 1) (3 + 1) (1 + 1) (1 + 1) (1 + 1)
= 4 × 4 × 4 × 2 × 2 × 2
Number of factors multiple of 45    
= 3² × 5 × (2³ × 3 × 5² × 7 × 13 × 17)
= 45 (2³ × 3 × 5² × 7 × 13 × 17)
= (3 + 1) (1 + 1) (2 + 1) (1 + 1) (1 + 1) (1 + 1)
= 4 × 2 × 3 × 2 × 2 × 2
= 192 factors
Number of positive integers that divide 1890 × 130 × 170 are not divisible by 45 is = Total factors − number of factors multiple of 45
= 512 – 192
= 320 factors

As n lines are perpendicular to 5 parallel lines so selecting any 2 lines from n & any 2 lines from 5 will result in a rectangle.
Selecting 2 lines from n lines = ⁿC₂

(n – 9) (n + 8) = 0
n = 9, − 8
Lines can’t be negative in number so discard n= - 8
So value of n = 9

Assumed work = 60 units (take this quantity as LCM for easy calculation)
T1  empties 60 units in = 30 min.
T1 & T2 empties 60 units in = 20 min.
Efficiency of T1 + T2 = 60/20 = 3 units/ min
Efficiency of T1 + Efficiency of T2 = Efficiency of T1 + T2
Efficiency of T2 + 2 = 3
Efficiency of T2 = 1 unit/min.
Time taken by T2 to empty tank = 60/1
= 60 mins

To calculate weight of all possible integers uptill given number than when we can put weights only on left pan is calculated by taking weights of all power of 2 less than n.
Here n = 100
2^x < 100
x = 6
64 < 100
2⁷> 100
128 > 100
So weighs required will be

Hence all integer weights can be calculated by using above taken weights.
∴ Minimum number of weights required = 7

Total students = T

To minimize n(P∩M) we have to maximize students passed in only maths & only physics
100% = 60% + 68% - n(P ∩ M)
n(P ∩ M) = 128 – 100
n(P ∩ M) = 28%

∠B = ∠A + d as A, B, C, are in AP
∠C = ∠A + 2d
∠A + ∠B + ∠C = 180°
∠A + ∠A + d + ∠A + 2d = 180°
3∠A + 3d = 180°
∠A + d = 60°
∠B = ∠A + d = 60°
sin (2A + B) = 1/2 
sinθ = 1/2 
θ = 30° or θ = 150°
Since B = 60° & A can’t be negative so
θ ≠ 30° hence θ = 150°
2A + B = 150°
2A + 60 = 150
2A = 90°
∠A = 45°
∠B = ∠A + d
60 = 45 + d
d = 15°
∠C = 45 + 2 × 15°
∠C = 75°
sin (60 + 2 × 75) = sin( 60 + 150)
sin210° = sin(180 + 30)
= - sin30° = - 1/2

(743)⁸⁵
Unit digit cycle of 3 → 4x + 1 → 3
4x + 2 → 9
4x + 3 → 7
4x → 1
85 = 4x + 1
Unit digit cycle of (743)⁸⁵ = 3
Unit digit cycle of 5
Any power = 5
Unit digit of (525)³⁷ = 5
Unit digit cycle of 7 →
4x + 1 = 7
4x + 2 = 9
4x + 3 = 3
4x = 1
96 = 4x
(987)⁹⁶ = 1
= 3 – 5 + 1
= 4 – 5
But unit digit cannot be negative so we take a carry from previous number so 14 – 5
Unit digit = 9

Ar ?AOB : Ar ?DOC
Property of quadrilateral is that the product of opposite triangles = product of other 2 opposite triangle pair i.e.    
Ar ?AOB × Ar ?DOC = Ar ?AOD × Ar ?BOC 
4K × 9K = xy
xy = 36K²
Total area of quadrilateral = 4K + 9K + x + y 
For x + y to be minimum where product of xy is given for that we use
AM ≥ GM
For AM to be minimum
AM = GM
(x + y)/2 = √(36K²)
(x + y)/2 = 6 K
x + y = 12 K
By putting x+y in area of quadrilateral
= 4K + 9K + 12K
Area = 25 K
For minimum area K = 1
Area = 25 Sq. units

f(x) = ax² + bx + c
g(x) = – 2x
f(x) cuts x axis at (– 2 ,0) hence (– 2, 0) is a point of b(x)
f(– 2) = 4a – 2b + c = 0 ______   (1)
g(x) passes through (a, b)
g(a) = – 2a = b _________ (2)
By putting value of b in (1)
4a – 2 (– 2a) + c     = 0
4a + 4a + c = 0
c = – 8a
f(x) + 9a + 1 = ax² – 2ax – 8a + 9a + 1
= ax² – 2ax + a + 1
= a (x – 1)² + 1
Method 1:
Minimum of (x – 1)² = 0
∴ minimum value of above expression = 1
Alternative Method:
f(x) = ax² – 2ax + a + 1
f’(x) = 2ax – 2a
for minimum value
f’(x) = 0
2ax – 2a = 0
x = 1
f” (x) > 0
f” (x) = 2a > 0 (as given a > 0)
So minimum value of f(x) will be at x = 1
f(1) = a – 2a + a + 1
= 1

Joining any 2 points will give a line so selecting any 2 points from 12 points
= 10C₂
But line formed by joining 1 any two points from 5 co – linear points will result in single line
So we need to subtract 5C₂ from 10C₂ & add 1 for line resulted by joining any of co – linear point.
= 10C₂ – 5C₂ + 1
= (10 × 9)/2 – (5 × 4)/2 + 1
= 45 – 10 + 1
= 36

x + 2y    = 4_____ (1) × 2
2x + 3y    = 6_____ (2)
2x + 4y    = 8_____ (3)
x + 3y    = 6
On subtracting (2) from (3)
2x + 4y    = 8
2x + 3y    = 6
          y   = 2
x + 2 × 2 = 4
x = 0
Point of intersection = (0, 2)
Line passes through point (0, 2)
& perpendicular to 3x – y = 2
Y = 3x – 2
m₁ = 3
For perpendicular lines
m₁m₂ = – 1
3 × m₂ = –1
m₂ = (-1)/3
Eqn of line:-  
y = m₂x + c
y = – x/3 + c
2 = o + c
c = 2
y = – x/3 + 2
3y + x    = 6
x intercept y = 0
x (0) + x = 6
x = 6

If A = 8 points & 0 losses then he have
= 3 wins + 2 draws = 8 points
= 3 x 2 + 2 x 1
If E have less than 5 points & only 1 loss than E have 4 draws
If B have 2 losses than to have 6 points
B have to win 3 matches
If C have 5 points & 2 loss
Then C = 2 wins & 1 draw
If D have 5 points & 1 loss
Then D = 1 win & 3 Draw
Total number of wins = total number of losses
So F must have losses ≥ 3 by table uptill F
wins = 9
losses = 6
So F has 2 cases
Case I : F has 1 wins 4 losses & 0 Draws points = 2
Then total wins of all teams combined = 10
Total losses of all teams combined = 10
But as we can see from table
E have 4 draws & B doesn’t have any draw
So E draw the match with F s F have at least 1 draw
So case I is not valid.
Case II : F has 0 wins 3 losses & 2 draws = 2 points
Total wins combined = Total losses combined

Each Draw is reported twice as in eg:-
If E vs F → Draw then both E & F get 1 point in draw
Number of matches resulted in draw = 12/6 = 6 matches

If A = 8 points & 0 losses then he have
= 3 wins + 2 draws = 8 points
= 3 x 2 + 2 x 1
If E have less than 5 points & only 1 loss than E have 4 draws
If B have 2 losses than to have 6 points
B have to win 3 matches
If C have 5 points & 2 loss
Then C = 2 wins & 1 draw
If D have 5 points & 1 loss
Then D = 1 win & 3 Draw
Total number of wins = total number of losses
So F must have losses ≥ 3 by table uptill F
wins = 9
losses = 6
So F has 2 cases
Case I : F has 1 wins 4 losses & 0 Draws points = 2
Then total wins of all teams combined = 10
Total losses of all teams combined = 10
But as we can see from table
E have 4 draws & B doesn’t have any draw
So E draw the match with F s F have at least 1 draw
So case I is not valid.
Case II : F has 0 wins 3 losses & 2 draws = 2 points
Total wins combined = Total losses combined

From the above table 
Answer is Team E

If A = 8 points & 0 losses then he have
= 3 wins + 2 draws = 8 points
= 3 x 2 + 2 x 1
If E have less than 5 points & only 1 loss than E have 4 draws
If B have 2 losses than to have 6 points
B have to win 3 matches
If C have 5 points & 2 loss
Then C = 2 wins & 1 draw
If D have 5 points & 1 loss
Then D = 1 win & 3 Draw
Total number of wins = total number of losses
So F must have losses ≥ 3 by table uptill F
wins = 9
losses = 6
So F has 2 cases
Case I : F has 1 wins 4 losses & 0 Draws points = 2
Then total wins of all teams combined = 10
Total losses of all teams combined = 10
But as we can see from table
E have 4 draws & B doesn’t have any draw
So E draw the match with F s F have at least 1 draw
So case I is not valid.
Case II : F has 0 wins 3 losses & 2 draws = 2 points
Total wins combined = Total losses combined

Answer is 2

If A = 8 points & 0 losses then he have
= 3 wins + 2 draws = 8 points
= 3 x 2 + 2 x 1
If E have less than 5 points & only 1 loss than E have 4 draws
If B have 2 losses than to have 6 points
B have to win 3 matches
If C have 5 points & 2 loss
Then C = 2 wins & 1 draw
If D have 5 points & 1 loss
Then D = 1 win & 3 Draw
Total number of wins = total number of losses
So F must have losses ≥ 3 by table uptill F
wins = 9
losses = 6
So F has 2 cases
Case I : F has 1 wins 4 losses & 0 Draws points = 2
Then total wins of all teams combined = 10
Total losses of all teams combined = 10
But as we can see from table
E have 4 draws & B doesn’t have any draw
So E draw the match with F s F have at least 1 draw
So case I is not valid.
Case II : F has 0 wins 3 losses & 2 draws = 2 points
Total wins combined = Total losses combined

Team D have only 1 loss & according to question that loss came from team C hence A vs D resulted in a draw.
Answer is Team D

If A = 8 points & 0 losses then he have
= 3 wins + 2 draws = 8 points
= 3 x 2 + 2 x 1
If E have less than 5 points & only 1 loss than E have 4 draws
If B have 2 losses than to have 6 points
B have to win 3 matches
If C have 5 points & 2 loss
Then C = 2 wins & 1 draw
If D have 5 points & 1 loss
Then D = 1 win & 3 Draw
Total number of wins = total number of losses
So F must have losses ≥ 3 by table uptill F
wins = 9
losses = 6
So F has 2 cases
Case I : F has 1 wins 4 losses & 0 Draws points = 2
Then total wins of all teams combined = 10
Total losses of all teams combined = 10
But as we can see from table
E have 4 draws & B doesn’t have any draw
So E draw the match with F s F have at least 1 draw
So case I is not valid.
Case II : F has 0 wins 3 losses & 2 draws = 2 points
Total wins combined = Total losses combined

Team E was defeated by team B as B doesn’t have any draw.
Answer is only team B