IPMAT 2020 - Quantitative Aptitude

? = dQ + R
QR = 24 – 0      -------- 1
Q + R = 10     -------- 2
d = 5
Q = 10 – R
(10 – R) R = 24
10R – R² = 24
R² – 10R + 24 = 0
(R – 6) (R – 4) = 0
R = 6, 4
Q = 4, 6
If division = 5 then Remainder can’t be greater than or equal to 5.
Rem = 4    Q = 6
? = dQ + R
= 5 × 6 + 4
= 30 + 4
? = 34

x² + y² = 1
Centre of circle = (0, 0)
R=1 unit
Shortest distance of any point = ?
Minimum distance of point (-4, 3) from centre = ? +1

Distance can’t be negative
? + 1 = 5
? - 4 = 4 units

Since all numbers are distinct so b-c or a-b ≠ 0 then value of determinant must be 0.

f(x) = |3 – x| + |2 + x| + |5 – x|
x ≤ – 2
= 3 – x + (–(x + 2)) + 5 – x
= 3 – x – x – 2 + 5 – x
= 6 – 3x
Minimum value x = –2
= 6 – 3(–2)
= 6 + 6
= 12
– 2 < x ≤ 3
= 3 – x + 2 + x + 5 – x
= 10 – x
Minimum value at x = 3
= 10 – 3
= 7
3 < x ≤ 5    
= – (3 – x) + (x + 2) + (5 – x)
= – 3 + x + x + 2 + 5 – x
= 4 + x
Minimum value at x = 3
= 4 + 3 = 7
x > 5
– (3 – x) + x + 2 + (– (5 – x)
– 3 + x + x +2 – 5 + x
3 × 5 – 6 = 9
So the minimum value of given expression is 7

First Visit
Ratio of pens = 2: 3
Cost of pens = x
Cost of pencils = y
2kx + 3ky = 86 ________ (i)
Second Visit
Ratio of Pens: Pencils = 4 : 1
4k’x + k’y = 112 ________ (ii)
He purchased 4 pens in 2nd visit     
4k’ = 4
K’ = 1
4x +y = 112  ________ (ii)
2kx + 3ky = 86 ________ (i)
Possible equation of equation 1
2x + 3y = 86
4x + 6y =86 
6x +9y = 86
.
.
.
.
Since 4x + y = 112 & x & y are positive integers then 4x + 6y > 112 hence the only possible equation is 
2x + 3y = 86 ________ (i)
Solving (i) & (ii)
(2x + 3y = 86) × 2
4x + y = 112

4x + 6y =172
4x + y = 112
        5y = 60
         y = 12

4x + 12 = 112
4x = 100
x = 25
The amount paid by Ashok during 2nd visit for pens is 4x i.e. Rs. 100

Let the number be abcd
ad = 0       ________ (i)
|a – d| = 7       ________ (ii)
bc = 18      ________ (iii)
a – d = b – c   ________ (iv)
Since a x d is 0 then at least 1 digit out of a & d must be 0 but a ≠ 0 otherwise the number will not be a 4 digit number 
So, d = 0
a – d = 7
a – 0 = 7
a = 7
bc = 18
b – c = 7
b = c + 7 
(c + 7)(c) = 18
c² + 7c – 18 = 0
c² + qc – 2c – 18 = 0
c(c + q) – 2c – 18 = 0
(c – 2)(c + q) = 0
c = 2, – q
c can’t be negative as it is a digit
c = 2
b = c + 7
b = 2 + 7 
b = 9
Number = abcd = 7920

Total Student = 80
n(m) = 50
n(c) = 40
n(p) = 30
n(m ∩ p) = At most 20
n(p ∩ c) = At most 20
n(m ∩ c) = At most 20

As we need to maximize the students who passed in all 3 exams so the number of students who failed in all 3 = h = 0
a + b + c + 2(d + e + f) + 3g = 50 + 40 + 30 = 120 _________ (i)
a + b + c + d + e + f + g = 80 __________ (ii)
(i) - (ii)
(d + e + f) 2g = 40
Since we need to maximize g we should minimize d + e + f
By taking d + e + f = 0
2g = 40
g = 20
By taking g = 20 no condition for at most 20 students passed in pairs of any of 3 subjects is violated 
So g = 20
Maximum value for students who passed in all 3 subjects is 20

d = 3000m
Length of track = 300m
No. of Rounds = 3000/300 = 10 rounds
Ratio of speed = 3 : 2
When they will meet for the first time then the faster person have covered exactly 1 round extra then the slower person. So if they meet after faster runner have covered n rounds.

Time taken would be same for faster runner completely & slower runner completing (n – 1) rounds.

They will meet after every 3 rounds so number of times the winner passes other is = 10/3
= 3 times

We can choose 4 object in different ways
(i) 0 indistinguishable 4 distinct
= qc₄
(ii) 1 indistinguishable & 3 distinct
1 × qc₃
(iii) 2 indistinguishable & 2 distinct
1 × qc₂ 
(iv) 3 indistinguishable & 1 distinct
1 – qc₁
(v) 4 indistinguishable & 0 distinct
1 × qc₀
Total ways of choosing 4 objects = qc₀ + qc₁ + qc₂ qc₃ + qc₄
= 1 + q + 36 + 84 +126
= 256

10¹⁹ = 2¹⁹ × 5¹⁹
No. of factors of 10¹⁹
= (19 + 1) (19 + 1)
= 20 × 20 = 400
No. of factors of 10¹⁹ that are multiple of 10¹⁵
= 2¹⁹ × 5¹⁹ 
= 2¹⁵ × 5¹⁵ (2⁴ × 5⁴)
= 10¹⁵ (2⁴ × 5⁴)
No. of factors of 10¹⁹ that are multiples of 10¹⁵ are factors of 
= 5 × 5 = 25
No. of factors = (4 + 1 ) (4 + 1)
= 5 × 5 = 25

Perimeter ≤ 100
sides = a – 1 + a + a + 1 ≤ 100
= 3a ≤ 100
a ≤ 100/3
a = 33 
Sides = 32,33,34
Smallest possible triangle = 4, 5, 6 because given triangle is & 3, 4, 5 gives a right triangle
a² + b² > c²
4² + 5² > b²
16² + 25² > 36
41 > 36
Hence 4, 5, 6 is acute triangle
Total No. of Triangles = 32 – 4 + 1
= 29
Set of possible = {(4, 5, 6), (5, 6, 7), (6, 7, 8) ….(32, 33, 34)}

Let us say the line cuts x–axis at (– a, 0) & y–axis at (0, b)
– 5 =  (a + 0)/2
a = 2 × 5 
a = 10
Eq. of line passing through 2 points (– 10, 0) & (– 5, 4) is

Efficiency of 1st worker = 2x Efficiency of 3rd worker
Let efficiency of 3rd worker be x
The efficiency of 2nd worker be Y units 1 hr.
Let the total work be 30 units
Time Taken = (Work Done)/Efficiency     
Efficiency of 1st worker = 2x
Together they take 1 hr. to build a wall i.e. do 30 units of work

Total Student = 140
Only cricket n(c)  = even numbered roll no
Last term = 140
a = first term = 2,  d = 2
140 = 2 + (n – 1)2
138/2 = n – 1
n = 70
n(c) = 70
No. of people who play Basketball but not cricket = {3,9,15,…..135}
Past term = 135,       a = 3, d = 6 
135 = 3 + ( n – 1) 6
132/6 = n – 1
n = 23
No. of people who play football but not cricket  = {5,15,25,35,…..135}
Last term = 135,   a = 5, d = 10
135 = 5 + (n – 1)10
130/10 = n – 1
n = 14
No. of people who play both Basketball & football have been added twice so we need to subtract them once 
No. of people playing both football & not cricket
= {15, 45,75,….135}
a = 15, d = 30    
Last term = 135
135 = 15 + (n – 1)30
120/30 = n – 1
n = 5
Student playing no sport = Total Number of Students – (Student playing only cricket + student playing Basketball not cricket +student playing football but not cricket – student playing Both Basketball & football but not cricket
= 140 – (70 + 23 + 14 – 5)
= 140 – 102
= 38

f(x) = x² + log₃ x 
g(y) = 2y + f(y)
g(3) = 2x³ + f(3)      ______ (1)
f(3) = 3² + log₃ 3 
f(3) = 9 + 1 
f(3) = 10
By putting value of f(3) in (1)
g(3) = 6+10
g(3) = 16

Matrix elements must be distinct
Total Cases    = Selection of any 4 elements from 6 and arranging them
= ⁶C₄ × 4!
= (6×5)/2 × 4 × 3 × 2 × 1 
= 6 × 5 x× 4 × 3
= 360
Favourable Case = Matrix should be singular i.e. value of determinant is 0
= a × c = b × d = 0 & elements are distinct
So we have 2 cases
Case 1: = 1 × 6 – 2 × 3
= 0
So if we take
a = 1 then c = 6
& b & d take any value 2 & 3
So this give 2 possibilities
a = 6 then c = 1
b & d take values again 2 & 3
So this also gives 2 possibilities
a = 2 then c = 3
b & d take value again 1 & 6
interchanging the values so this gives 2 possibilities
a = 3 then c = 2
b& d take value 1, 6 both & result in 2 possibilities
Total possibilities of distinct matrix case from case 1
= 2 + 2 + 2 + 2
= 8
Case 2: = 2 × 6 – 4 × 3
Similarly as Case 1, Case 2 also result in 8 possibilities
Favourable Cases = 8 + 8
= 16
Probability = (Favourable Cases)/(Total Cases)  
= 16/360
= 2/45

Sum of marks of student = 64 × 6
= 384
a + b + c + d + e + f = 384
a + b + c + d + e + = 314
since we need to maximize the difference between second highest & second lowest so we need to assign as minimum values the highest &  second highest value
Minimum value for smallest number = 40
all values should be distinct     
40 + 41 + 42 + d + e = 314
123 + d + e + = 314
d + e  =191
To maximize both assign almost equal values so 
d = 190/2
e = 96
So difference = 95 – 41
= 54

As the shares are purchased on Monday of 3600000 amount then to maximize the profit we need to maximize selling price as amount & no. of shares are constant for ITC 
So max selling price = 253 on Friday 
So to maximize profit I should sell my shares on Friday