IPMAT (I) 2021 QASA | Previous Year IPMAT - Indore Paper

Answer: 320

Explanation :

We first need to find all factors of (1890) × (130) × (170) and then remove those factors which are divisible by 45.

(1890) × (130) × (170) = (2 × 3³ × 5 × 7) × (13 × 2 × 5) × (17 × 2 × 5)

⇒ 1890 × 130 × 170 = 2³ × 3³ × 5³ × 7 × 13 × 17

∴ Total factors = (3 + 1)(3 + 1)(3 + 1)(1 + 1)(1 + 1)(1 + 1) = 512

Now for factors that are divisible by 45, least power of 3 should be 2 and that of 5 should be 1. Powers of 2, 7, 13 and 17 can be anything.

∴ Total factors which are divisible by 45 = 4 × 2 × 3 × 2 × 2 × 2 = 192

⇒ Number of factors which are not divisible by 45 = 512 - 192 = 320.

Hence, 320.

Answer: 5310

Explanation :

1 × 3 + 5 × 7 + 9 × 11 + 

T₁ = 1 × 3
T₂ = 5 × 7 and so on

In each of these terms, the first number forms an AP whose first term is 1 and common difference is 4.
∴ first number of n^th term = 1 + (n - 1) × 4 = 4n - 3

In each of these terms, the second number forms an AP whose first term is 3 and common difference is 4.
∴ second number of n^th term = 3 + (n - 1) × 4 = 4n - 1

⇒ T_n = (4n - 3) × (4n - 1)
⇒ T_n = 16n² - 16n + 3

∴ $\underset{n=1}{\overset{10}{\sum T_n}}$ = 16(1² + 2² + 3² + ... + 10²) + 16(1 + 2 + 3 + 3 + ... + 10) + (3 + 3 + 3 + ... + 3)

= $16\left(\frac{10\times 11\times 21}{6}\right)$ - $16\left(\frac{10\times 11}{2}\right)$ + 30

= 6160 - 880 + 30 = 5310

Hence, 5310.

Answer: 960

Explanation :

x₁ = 0 = 1² - 1

x₂ = x₁ + 1 + $2\sqrt{1+x_0}$ = 0 + 1 + $2\sqrt{1+0}$ = 3 = 2² - 1

x₃ = x₂ + 1 + $2\sqrt{1+x_1}$ = 3 + 1 + $2\sqrt{1+3}$ = 8 = 3² - 1

∴ x_n = n² - 1

⇒ x₃₁ = 31² - 1 = 960

Hence, 960.

Answer: 9

Explanation :

There are 5 parallel lines and there are another 'n' lines which are parallel but perpendicular to the first 5 lines.

∴ Each of the first 5 lines (say they are all horizontal lines) are perpendicular to each of the other n lines (say they are vertical lines).

To form a rectangle we need two horizontal and two vertical lines.

∴ Number of rectanges = (number of ways of selecting 2 horizontal lines) × (number of ways of selecting 2 vertical lines) 

⇒ 360 = ⁵C₂ × ⁿC₂ 

⇒ 360 = 10 × n(n - 1)/2 

⇒ 72 = n(n - 1)

⇒ n = 9

Hence, 9.

Answer: 60

Explanation :

T₁ alone can empty the tank in 30 minutes.
T₁ and T₂ together can empty the tank in 20 minutes.
Let the time taken by T₂ alone is t minutes.

⇒ $\frac{1}{30}$ + $\frac{1}{t}$ = $\frac{1}{20}$

⇒ t = 60 minutes.

Hence, 60.

Answer: 6

Explanation :

Initial sum of the marks of all students across all subjects = 30 × 80% × 200 × 5 = 24,000

Sum of the marks of all students across all subjects after revaluation = 30 × 80.02% × 200 × 5 = 24,006

∴ Marks of these students increased by 6.

Least increase in marks of one of the students can be 0, hence maximum increase in marks of the other student must be 6.

Hence, 6.

Answer: 7

Explanation :

If weight can be placed only on one pan, we need to take weights which are powers of 2.

Weights required are 1 kg, 2 kg, 4 kg, 8 kg, 16 kg, 32 kg and 64 kg i.e., 7 weights.

Hence, 7.

Answer: 26

Explanation :

The two given equations contain 3 lines.

Let the slope of common line between 2 equations be 'm'
Let the slope of remaining two perpendicular lines be b and -1/n.

Given, 2𝑥² + axy + 3y² = 0
⇒ $\frac{2}{3}{x}^2$ + $\frac{a}{3}xy$ + y² = (y - mx)(y - nx)
∴ m + n = $\frac{a}{3}$ ...(1) and mn = $\frac{2}{3}$   ...(2)

Also, 2𝑥² + bxy - 3y² = 0
⇒ $-\frac{2}{3}{x}^2$ - $\frac{b}{3}xy$ + y² = $\left(y-mx\right)\left(y+\frac{1}{n}x\right)$
∴ - m + $\frac{1}{n}$ = $-\frac{b}{3}$   ...(3)  and $-\frac{m}{n}$ = $-\frac{2}{3}$   ...(4)

Multiplying (2) and (4)
⇒ m² = 4/9

Case 1: m = + 2/3 ⇒ n = 1
⇒ a/3 = m + n ⇒ a = 5
⇒ - b/3 = - m + 1/n ⇒ b = 1

Case 2: m = - 2/3 ⇒ n = - 1
⇒ a/3 = m + n ⇒ a = - 5
⇒ -b/3 = - m + 1/n ⇒ b = - 1

In both cases: 𝑎² + 𝑏² = 25 + 1 = 26

Hence, 26.

Answer: 3

Explanation :

Case 1: x > 3
⇒ f(x - 3) = max(x - 3, 0) = x - 3
⇒ f(x + 1) = max(x + 1, 0) = x + 1
Given, f(x - 3) + 2f(x + 1) = 8
⇒ x - 3 + 2(x + 1) = 8
⇒ 3x = 9
⇒ x = 3

Case 2: -1 < x ≤ 3
⇒ f(x - 3) = max(x - 3, 0) = 0
⇒ f(x + 1) = max(x + 1, 0) = x + 1
Given, f(x - 3) + 2f(x + 1) = 8
⇒ 0 + 2(x + 1) = 8
⇒ 2x = 6
⇒ x = 3

Case 3: x < -1
⇒ f(x - 3) = max(x - 3, 0) = 0
⇒ f(x + 1) = max(x + 1, 0) = 0
Given, f(x - 3) + 2f(x + 1) = 8
⇒ 0 + 0 = 8
Not possible

∴ x = 3 is the only value that satisfies the given function.

Hence, 3.

Answer: 28

Explanation :

Let the percentage of students who passed in both the subjects is 'x' and those who failed in both the subjects is 'n'.

Now, P ∪ M = P + M - P ∩ M

⇒ 100 - n = 60 + 68 - x

⇒ x = n + 28

x will be least when n is least. Least value of n can be 0, hence least value of x = 28

Hence, 28.