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

**1. IPMAT (I) 2021 QASA | Algebra - Number Theory**

The number of positive integers that divide (1890) × (130) × (170) and are not divisible by 45 is:

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 × 3^{3} × 5 × 7) × (13 × 2 × 5) × (17 × 2 × 5)

⇒ 1890 × 130 × 170 = 2^{3} × 3^{3} × 5^{3} × 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.

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**2. IPMAT (I) 2021 QASA | Algebra - Progressions**

The sum up to 10 terms of the series 1 × 3 + 5 × 7 + 9 × 11 + . . is

Answer: 5310

**Explanation** :

1 × 3 + 5 × 7 + 9 × 11 +

T_{1} = 1 × 3

T_{2} = 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} = 16n^{2} - 16n + 3

∴ $\underset{n=1}{\overset{10}{\sum {T}_{n}}}$ = 16(1^{2} + 2^{2} + 3^{2} + ... + 10^{2}) + 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.

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**3. IPMAT (I) 2021 QASA | Algebra - Progressions**

It is given that the sequence {x_{n}} satisfies x_{1} = 0, x_{n+1} = x_{n} + 1 + $2\sqrt{1+{x}_{n}}$ for n = 1,2, . . . . . Then x_{31} is

Answer: 960

**Explanation** :

x_{1} = 0 = 1^{2} - 1

x_{2} = x_{1} + 1 + $2\sqrt{1+{x}_{0}}$ = 0 + 1 + $2\sqrt{1+0}$ = 3 = 2^{2} - 1

x_{3} = x_{2} + 1 + $2\sqrt{1+{x}_{1}}$ = 3 + 1 + $2\sqrt{1+3}$ = 8 = 3^{2} - 1

∴ x_{n} = n^{2} - 1

⇒ x_{31} = 31^{2} - 1 = 960

Hence, 960.

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**4. IPMAT (I) 2021 QASA | Modern Math - Permutation & Combination**

There are 5 parallel lines on the plane. On the same plane, there are ‘n’ other lines that are perpendicular to the 5 parallel lines. If the number of distinct rectangles formed by these lines is 360, what is the value of n?

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 = ^{5}C_{2} × ^{n}C_{2}

⇒ 360 = 10 × n(n - 1)/2

⇒ 72 = n(n - 1)

⇒ n = 9

Hence, 9.

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**5. IPMAT (I) 2021 QASA | Time & Work**

There are two taps, T_{1} and T_{2}, at the bottom of a water tank, either or both of which may be opened to empty the water tank, each at a constant rate. If T_{1} is opened keeping T_{2} closed, the water tank (initially full) becomes empty in half an hour. If both T_{1} and T_{2} are kept open, the water tank (initially full) becomes empty in 20 minutes. Then, the time (in minutes) it takes for the water tank (initially full) to become empty if T_{2} is opened while T_{1} is closed is

Answer: 60

**Explanation** :

T_{1} alone can empty the tank in 30 minutes.

T_{1} and T_{2} together can empty the tank in 20 minutes.

Let the time taken by T_{2} alone is t minutes.

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

⇒ t = 60 minutes.

Hence, 60.

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**6. IPMAT (I) 2021 QASA | Average, Mixture & Alligation**

A class consists of 30 students. Each of them has registered for 5 courses. Each course instructor conducts an exam out of 200 marks. The average percentage marks of all 30 students across all courses they have registered for, is 80%. Two of them apply for revaluation in a course. If none of their marks reduce, and the average of all 30 students across all courses becomes 80.02%, the maximum possible increase in marks for either of the 2 students is

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.

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**7. IPMAT (I) 2021 QASA | Algebra - Number Theory**

What is the minimum number of weights which enable us to weigh any integer number of grams of gold from 1 to 100 on a standard balance with two pans? (Weights can be placed only on the left pan)

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.

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**8. IPMAT (I) 2021 QASA | Coordinate Geometry**

If one of the lines given by the equation 2𝑥^{2} + axy + 3y^{2} = 0 coincides with one of those given by 2x^{2}+ b𝑥𝑦 - 3𝑦^{2} = 0 and the other lines represented by them are perpendicular then 𝑎^{2} + 𝑏^{2 }=

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, 2𝑥^{2} + axy + 3y^{2} = 0

⇒ $\frac{2}{3}{x}^{2}$ + $\frac{a}{3}xy$ + y^{2} = (y - mx)(y - nx)

∴ m + n = $\frac{a}{3}$ ...(1) and mn = $\frac{2}{3}$ ...(2)

Also, 2𝑥^{2} + bxy - 3y^{2} = 0

⇒ $-\frac{2}{3}{x}^{2}$ - $\frac{b}{3}xy$ + y^{2} = $\left(y-mx\right)\left(y+\frac{1}{n}x\right)$

∴ - m + $\frac{1}{n}$ = $-\frac{b}{3}$ ...(3) and $-\frac{m}{n}$ = $-\frac{2}{3}$ ...(4)

Multiplying (2) and (4)

⇒ m^{2} = 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: 𝑎^{2} + 𝑏^{2 }= 25 + 1 = 26

Hence, 26.

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**9. IPMAT (I) 2021 QASA | Algebra - Functions & Graphs**

If a function f(a) = max (a, 0) then the smallest integer value of ‘x’ for which the equation f(x - 3) + 2f(x + 1) = 8 holds true is:

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.

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**10. IPMAT (I) 2021 QASA | Miscellaneous**

In a class, 60% and 68% of students passed their Physics and Mathematics examinations respectively. Then atleast ________ percentage of students passed both their Physics and Mathematics examinations.

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.

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