IPMAT (I) 2023 QASA | Previous Year IPMAT - Indore Paper

Answer: 8

Explanation :

Let there be n players in each of these 4 groups.
∴ There are total 4n players.

Let the groups be called G1, G2, G3 & G4.

Now, a player from G1, say A, plays with other (n - 1) players from G1 exactly once.
Hence, he plays (n -1) games.

A will play with each of n players of G2 twice, hence total 2n games.

A will play with each of n players of G3 thrice, hence total 3n games.

A will play with each of n players of G4 four times, hence total 4n games.

∴ A will play a total of (n - 1) + 2n + 3n + 4n = 10n - 1 games.

Now, each of the 4n players will play (10n - 1) games.
∴ There will be total 4n × (10n - 1)/2 games [/2 because each game is counted twice]
⇒ There are total 2n(10n - 1) games

⇒ 2n(10n - 1) > 1000
⇒ n(10n - 1) > 500

The least possible value of n satisfying the above inequality is 8.

Hence, 8.

Answer: 3

Explanation :

Here, factorials of number 5 and above are always divisible by 15, hence we can ignore them.

So the question simplifies to finding remainder of 1! + 2! + 3! + 4! when divided by 15.

= Remainder of (1 + 2 + 6 + 24) / 15

= Remainder of (33) / 15

= 3

Hence, 3.

Answer: 24

Explanation :

Let the efficiency of Amish be 2 units/day.

If she can do the entire work in 20 days.
⇒ Total work to be done = 2 × 20 = 40 units.

Now, she works with full efficiency for first 4 days and with half efficiency for another 12 days.
∴ Work done by Amisha = 2 × 4 + 1 × 12 = 20 units.

⇒ Amisha's co-worker completed the remaining (40 - 20 =) 20 units of work in the last 12 days.

Hence, to complete the entire 40 units of work the co-worker will take 24 days.

Hence, 24.

Answer: 140

Explanation :

​​​​​​​

y = 4970 - |x| & y = x²

To find intersection point we equate both equations.
⇒ 4970 - |x| = x² 
⇒ x² + |x| - 4970 = 0
⇒ |x|² + |x| - 4970 = 0 [x² = |x|²]
⇒ |x|² + 71|x| - 70|x| - 4970 = 0
⇒ (|x| + 71)(|x| - 70) = 0
⇒ |x| = - 71 or 70 [- 71 rejeceted as |x| cannot be negative]

⇒ x = + 70 or - 70

∴ Line joining the intersection point of these graphs will be a horizontal line. The x coordinates of its end points are +70 & -70.
∴ Distance between the two points = 70 + 70 = 140.

Hence, 140.

Answer: 3034

Explanation :

Given, f(1) = 1, f(n) = 3n - f(n - 1)

Substituting,
n = 2, f(2) = 3 × 2 - f(1) = 5
n = 3, f(3) = 3 × 3 - f(2) = 4
n = 4, f(4) = 3 × 4 - f(3) = 8
n = 5, f(5) = 3 × 5 - f(4) = 7
n = 6, f(6) = 3 × 6 - f(5) = 11
n = 7, f(7) = 3 × 7 - f(6) = 10

Here, f(1), f(3), f(5), and so on, i.e., f(odd number)  form an arithmetic progression with common difference of 3.

We need to find f(2023). Now, how many odd numbers are there from 1 till 2023 = (2023 + 1)/2 = 1012

∴ f(2023) = f(1) + (1012- 1) × 3
⇒ f(2023) = 1 + 1011 × 3 = 3034

Hence, 3034.

Answer: 32

Explanation :

Here, let log₂x = a

The given equation can be written as:
a² × log₂2 - 5a + 6 = 0
⇒ a² - 5a + 6 = 0 [log₂2 = 1]
⇒ a² - 3a - 2a + 6 = 0
⇒ a(a - 3) - 2(a - 3) = 0
⇒ (a - 3)(a - 2) = 0
⇒ a = 2 or 3

⇒ log₂x = 2 or log₂x = 3
⇒ x = 2² or x = 2³ 
⇒ x = 4 or x = 8 

∴ Required product of the roots = 4 × 8 = 32

Hence, 32.

Answer: 2500

Explanation :

Ankita earns Rs. 10,000 at the end of year 1 which is invested at 0.5% for 4 years.
∴ Interest earned = 10,000 × 0.5% × 4 = 200

Ankita earns Rs. 20,000 at the end of year 2 which is invested at 1% for 3 years.
∴ Interest earned = 20,000 × 1% × 3 = 600

Ankita earns Rs. 30,000 at the end of year 1 which is invested at 1.5% for 2 years.
∴ Interest earned = 30,000 × 1.5% × 2 = 900

Ankita earns Rs. 40,000 at the end of year 1 which is invested at 2% for 1 years.
∴ Interest earned = 40,000 × 2% × 1 = 800

∴ Total interest earned = 200 + 600 + 900 + 800 = 2500

Hence, 2500.

Answer: 5

Explanation :

Concept: Remainder of f(x) when divided by x - a is f(a).

Here f(x) = 4x¹⁰ - x⁹ + 3x⁸ – 5x⁷ + cx⁶ + 2x⁵ – x⁴ + x³ – 4x² + 6x – 2

It has to be divided by x - 1.

∴ Remainder will be f(1)

⇒ f(1) = 4 × 1¹⁰ - 1⁹ + 3 × 1⁸ – 5 × 1⁷ + c × 1⁶ + 2 × 1⁵ – 1⁴ + 1³ – 4 × 1² + 6 × 1 – 2 = 2
⇒ 4 - 1 + 3 – 5 + c + 2 – 1 + 1 – 4 + 6 – 2 = 2
⇒ c + 3 = 2
⇒ c = - 1

∴ c + 6 = -1 + 6 = 5

Hence, 5.

Answer: 5000

Explanation :

Let the total number of voters in the list be 100x.

Total voters who voted is 85% = 85x
Invalid votes = 50
∴ Valid votes = 85x - 50

Winner gets 44% of total votes = 44x
∴ Second candidates gets = 85x - 50 - 44x = 41x - 50 votes

Given in the question
44x - (41x - 50) = 200
⇒ 3x + 50 = 250
⇒ 3x = 150
⇒ x = 50

∴ Number of voters in the voting list = 100x = 5000.

Hence, 5000.

Answer: 199

Explanation :

Given, f(n) = 1 + 2 + 3 + ... + (n+1) = $\frac{(\mathrm{n}+1)(\mathrm{n}+2)}{2}$

Now,

$\underset{k=1}{\overset{k=n}{\sum \frac{1}{f\left(k\right)}}}$ = $\frac{1}{\mathrm{f}\left(1\right)}$ + $\frac{1}{\mathrm{f}\left(2\right)}$ + $\frac{1}{\mathrm{f}\left(3\right)}$ + ... + $\frac{1}{\mathrm{f}\left(\mathrm{n}\right)}$

= $\frac{2}{2\times 3}$ + $\frac{2}{3\times 4}$ + $\frac{2}{4\times 5}$ + ... + $\frac{2}{\left(\mathrm{n}\right)\times (\mathrm{n}+1)}$ + $\frac{2}{(\mathrm{n}+1)\times (\mathrm{n}+2)}$

= $2\left[\frac{1}{2\times 3}+\frac{1}{3\times 4}+\frac{1}{4\times 5}+...+\frac{1}{\mathrm{n}\times (\mathrm{n}+1)}+\frac{1}{(\mathrm{n}+1)\times (\mathrm{n}+2)}\right]$

= $2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{\mathrm{n}}-\frac{1}{\mathrm{n}+1}+\frac{1}{(\mathrm{n}+1)}-\frac{1}{\mathrm{n}+2}\right]$

= $2\left[\frac{1}{2}-\frac{1}{\mathrm{n}+2}\right]$

= $2\left[\frac{\mathrm{n}}{2(\mathrm{n}+2)}\right]$

= $\frac{\mathrm{n}}{(\mathrm{n}+2)}$

Now, this should be greater than 99/100

⇒ $\frac{\mathrm{n}}{(\mathrm{n}+2)}$ > $\frac{99}{100}$

⇒ 100n > 99n +198

⇒ n > 198

∴ Least possible integer value of n = 199.

Hence, 199.

Answer: 312

Explanation :

The speed of the car in the fourth gear is five times its speed in the first gear
⇒ Let the speed in 1^st gear be x kmph, hence speed in 4^th gear will be 5x kmph.

The car takes twice the time to travel a certain distance in the second gear as compared to the third gear
⇒ Speed in 3^rd gear is twice the speed in 2^nd gear.
⇒ Let speeds in 2^nd and 3^rd gears be y & 2y kmph.

In a 100 km journey, if Vinita travels equal distances in each of the gears, she takes 585 minutes
⇒ 25/x + 25/5x + 25/y + 25/2y = 585/60
⇒ 25/x + 5/x + 50/2y + 25/2y = 39/4
⇒ 30/x + 75/2y = 39/4
⇒ 60/x + 75/y = 39/2
⇒ 20/x + 25/y = 13/2
⇒ 5(4/x + 5/y) = 13/2
⇒ 4/x + 5/y = 13/10  ...(1)

Now, we need to find the time taken if Vinita travels 4, 4, 32, 60 kms at 1^st, 2^nd, 3^rd & 4^th gear respectively.
⇒ 4/x + 4/y + 32/2y + 60/5x
= 4/x + 4/y + 16/y + 12/x
= 16/x + 20/y
= 4(4/x + 5/y)
= 4 × 13/10
= 5.2 hours
= 5.2 × 60 minutes
= 312 minutes

Hence, 312.

Answer: 1160

Explanation :

We know number of ways of distributing n identical objects among r different groups such that each groups gets at least one object is ^n-1C_r-1.

Now,

Case 1: a + b + c = 21.
∴ We need to distribute 21 among 3 different variables a, b & c.
Number of ways of doing this = ^21-1C_3-1 = ²⁰C₂ = 190

Case 2: a + b + c = 22.
Number of ways of doing this = ^22-1C_3-1 = ²¹C₂ = 210

Case 3: a + b + c = 23.
Number of ways of doing this = ^23-1C_3-1 = ²²C₂ = 231

Case 4: a + b + c = 24.
Number of ways of doing this = ^24-1C_3-1 = ²³C₂ = 253

Case 5: a + b + c = 25.
Number of ways of doing this = ^25-1C_3-1 = ²⁴C₂ = 276

∴ Total number of ways = 190 + 210 + 231 + 253 + 276 = 1160

Hence, 1160.

Answer: 1442

Explanation :

Given, a : b = 2 : 5 and c: d = 5 : 2

⇒ a = 2x, b = 5x, c = 5y and d = 2y

Also, a + b + c + d = 2023
⇒ 2x + 5x + 5y + 2y = 2023
⇒ 7x + 7y = 2023
⇒ x + y = 289

Now, we need to maximise a + c = 2x + 5y = 2(x + y) + 3y = 2 × 289 + 3y.

This can be maximised by maximising y (higher coefficient).
​​​​​​​Since x & y have to be natural numbers, least value of x has to be 1, hence maximum value of y will be 288.

⇒ Maximum value of a + c = 2 × 1 + 5 × 288 = 2 + 1440 = 1442

Hence, 1442.

Answer: 14

Explanation :

​​​​​​​

AB = 4

Based on the figure, area of triangel ABC = 1/2 × AB × |y| = 2|y|

∴ 2|y| must be an integer.

Hence, y can be

Case 1: y = ± 1/2
⇒ x² + (1/2)² = 4 
⇒ x² = 15/4
⇒ x = ± √15/2
∴ 4 points i.e, (√15/2 ,1/2), (-√15/2 ,1/2), (√15/2 ,-1/2) & (-√15/2 ,-1/2)

Case 2: y = ± 1
⇒ x² + (1)² = 4 
⇒ x² = 3
⇒ x = ± √3
∴ 4 points i.e, (√3 ,1), (-√3 ,1), (√3 ,-1) & (-√3 ,-1)

Case 1: y = ± 3/2
⇒ x² + (3/2)² = 4 
⇒ x² = 7/4
⇒ x = ± √7/2
∴ 4 points i.e, (√7/2 ,3/2), (-√7/2 ,3/2), (√7/2 ,-3/2) & (-√7/2 ,-3/2)

Case 1: y = ± 2
⇒ x² + (2)² = 4 
⇒ x² = 0
⇒ x = 0
∴ 2 points i.e, (0 ,2) & (0 ,-2)

∴ Total 4 + 4 + 4 + 2 = 14 points

Hence, 14.

Answer: 39

Explanation :

We know (x + y)ⁿ = ⁿC₀ × xⁿ + ⁿC₁ × x^n-1 × y¹ + ⁿC₂ × x^n-2 × y² + ... + ⁿC_n × yⁿ

Let the three consecutive coefficients be ⁿC_r-1, ⁿC_r & ⁿC_r+1

Now, ⁿC_r-1 : ⁿC_r = 1 : 9
⇒ $\frac{\mathrm{n}!}{(\mathrm{r}-1)!\times (\mathrm{n}-\mathrm{r}+1)!}$ : $\frac{\mathrm{n}!}{\left(\mathrm{r}\right)!\times (\mathrm{n}-\mathrm{r})!}$ = 1 : 9

⇒ $\frac{\mathrm{r}}{\mathrm{n}-\mathrm{r}+1}$ = $\frac{1}{9}$

⇒ 9r = n - r + 1

⇒ 10r = n + 1   ...(1)

Also, ⁿC_r : ⁿC_r+1 = 9 : 63 = 1 : 7
⇒ $\frac{\mathrm{n}!}{\left(\mathrm{r}\right)!\times (\mathrm{n}-\mathrm{r})!}$ : $\frac{\mathrm{n}!}{(\mathrm{r}+1)!\times (\mathrm{n}-\mathrm{r}-1)!}$ = 1 : 7

⇒ $\frac{\mathrm{r}+1}{\mathrm{n}-\mathrm{r}}$ = $\frac{1}{7}$

⇒ 7r + 7 = n - r

⇒ 8r = n - 7   ...(2)

Solving (1) & (2), we get

n = 39 & r = 4

Hence, 39.