IPMAT (I) 2020 QA MCQ

Answer: Option D

Explanation :

Total number of factors of 10¹⁹ = 2¹⁹ × 5¹⁹ are (19 + 1) × (19 + 1) = 400

Factors of 2¹⁹ × 5¹⁹ which are also a multiple of 10¹⁵ will have power of both 2 and 5 greater than or equal to 15.
Possible powers of 2 or 5 are 15 or 16 or 17 or 18 or 19 i.e., 5 possibilities.
∴ Number of such factors are 5 × 5 = 25

⇒ Required probability = 25/400 = 1/16

Hence, option (d).

Answer: Option B

Explanation :

Let the sides of the triangle be x, x + 1 and x + 2.

Now, x + x + 1 + x + 2 ≤ 100
⇒ 3x ≤ 97
⇒ x ≤ 32.33   ...(1)

Also, the given triangle is an acute triangle, hence
(x + 2)² < (x + 1)² + x² 
⇒ x² + 4 + 4x < x² + 2x + 1 + x² 
⇒ x² - 2x - 3 > 0
⇒ (x - 3)(x + 1) > 0
⇒ x < -1 or x > 3   ...(2)

From (1) and (2), we get
x can take any value out of 4, 5, 6, ..., 31 and 32 i.e., 29 values.

Hence, option (b).

Answer: Option A

Explanation :

Let the x-intercept and y-intercept of the line are B (a, 0) and A (0, b)

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Since M (-5, 4) is the midpoint of these two intercepts, we have
-5 = (a + 0)/2 and 4 = (0 + b)/2
⇒ a = -10 and b = 8

∴ Equation of the line is: x/-10 + y/8 = 1
⇒ -8x + 10y = 80
⇒ 10y - 8x = 80

Hence, option (a).

Answer: Option C

Explanation :

Given, ${\cos }^2\frac{\mathrm{\pi }}{8}$ + ${\cos }^2\frac{3\mathrm{\pi }}{8}$ + ${\cos }^2\frac{5\mathrm{\pi }}{8}$ + ${\cos }^2\frac{7\mathrm{\pi }}{8}$

= ${\cos }^2\frac{\mathrm{\pi }}{8}$ + ${\cos }^2\frac{3\mathrm{\pi }}{8}$ + ${\cos }^2\left(\frac{\mathrm{\pi }}{2}+\frac{\mathrm{\pi }}{8}\right)$ + ${\cos }^2\left(\frac{\mathrm{\pi }}{2}+\frac{3\mathrm{\pi }}{8}\right)$

= ${\cos }^2\frac{\mathrm{\pi }}{8}$ + ${\cos }^2\frac{3\mathrm{\pi }}{8}$ + ${\sin }^2\frac{\mathrm{\pi }}{8}$ + ${\sin }^2\frac{3\mathrm{\pi }}{8}$

= ${\cos }^2\frac{\mathrm{\pi }}{8}$ + ${\sin }^2\frac{\mathrm{\pi }}{8}$ + ${\cos }^2\frac{3\mathrm{\pi }}{8}$ +  + ${\sin }^2\frac{3\mathrm{\pi }}{8}$

= 1 + 1 = 2

Hence, option (c).

Answer: Option A

Explanation :

If $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... upto ∞ = $\frac{{\mathrm{\pi }}^2}{6}$, then the value of 

Let X = $\frac{1}{1^2}$ + $\frac{1}{3^2}$ + $\frac{1}{5^2}$ + ... upto ∞

Adding and subtracting $\frac{1}{2^2}$ + $\frac{1}{4^2}$ + $\frac{1}{6^2}$ + ... upto ∞, we have

X = $\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... upto ∞ - ($\frac{1}{2^2}$ + $\frac{1}{4^2}$ + $\frac{1}{6^2}$ + ... upto ∞)

⇒ X = $\frac{{\mathrm{\pi }}^2}{6}$ - $\frac{1}{2^2}$($\frac{1}{1^2}$ + $\frac{1}{2^2}$ + $\frac{1}{3^2}$ + ... upto ∞)

⇒ X = $\frac{{\mathrm{\pi }}^2}{6}$ - $\frac{1}{2^2}$($\frac{{\mathrm{\pi }}^2}{6}$)

⇒ X = $\frac{{\mathrm{\pi }}^2}{6}$ - $\frac{1}{4}$($\frac{{\mathrm{\pi }}^2}{6}$)

⇒ X = $\frac{{\mathrm{\pi }}^2}{8}$

Hence, option (a).

Answer: Option E

Explanation :

The answer should be 4/5.

P(A given B) = P(A and B) ÷ P(B)
P(5 given that he says it is a 5) = P(5 and he says it is 5) ÷ P(he says it is 5)

P(5 and he says it is 5) = P(it is 5) × P(he say truth) = 1/6 × 4/5 = 4/30
P(he says it is 5) = P(it is 5) × P(he says the truth) + P(it is not 5) × P(he lies) × P(he lies that it is 5) = 1/6 × 4/5 + 5/6 × 1/5 × 1/5 = 4/30 + 1/30 = 5/30

∴ P(5 given that he says it is a 5) = 4/30 ÷ 5/30 = 4/5.

Hence, 4/5.

Note:
Let's say the die rolls a 4 and the person lies, the person could lie that it is a 1 or 2 or 3 or 5 or 6.
Now, probability that the person lies that is it 5 = 1/5
∴ P(it is not 5) × P(he lies) × P(he lies that it is 5) = 5/6 × 1/5 × 1/5 = 1/30.

In most study material you would find the answer to this question as 4/9.
This is because they assume that when the die does not roll a 5 and he lies, he will always say that it is 5. But this is not true. He could lie that it is a 4 when a 6 rolls. Hence, the difference in the answer.

Answer: Option D

Explanation :

Given, log₅log₈(x² - 1) = 0

⇒ log₈(x² - 1) = 5⁰ = 1

⇒ (x² - 1) = 8¹ 

⇒ x² = 9

⇒ x = +3 or -3.

Hence, option (d).

Answer: Option C

Explanation :

When 0 < x < 1, consider $\frac{1}{1+\mathrm{x}}$ and 1 - x + x²
Let us take x = 0.5
∴ $\frac{1}{1+\mathrm{x}}$ = 0.67 and 1 - x + x² = 1 - 0.5 + 0.25 = 0.75
⇒ $\frac{1}{1+\mathrm{x}}$ < 1 - x + x² 
∴ (i) is correct.

When 0 < x < 1, consider $\frac{1}{1+\mathrm{x}}$ and 1 - x + x²
Let us take x = -0.5
∴ $\frac{1}{1+\mathrm{x}}$ = 2 and 1 - x + x² = 1 + 0.5 + 0.25 = 1.75
⇒ $\frac{1}{1+\mathrm{x}}$ > 1 - x + x² 
∴ (iv) is correct.

Hence, option (d).

Answer: Option B

Explanation :

Fifty litres of a mixture of milk and water contains 30 percent of water.
⇒ Quantity of milk = 70% of 50 = 35 liters
⇒ Quantity of water = 30% of 50 = 15 liters

Eighty litres of another mixture of milk and water that contains 20 percent of water
⇒ Quantity of milk = 80% of 80 = 64 liters
⇒ Quantity of water = 20% of 80 = 16 liters

Let x liters of water is added along with these two solutions to make water 25% of the whole mixture.
⇒ 15 + 16 + x = 25% of (50 + 80 + x)
⇒ 31 + x = 25% of (130 + x)
⇒ 31 + x = 32.5 + 0.25x
⇒ 0.75x = 1.5
⇒ x = 2 liters.

Hence, option (b).

Answer: Option A

Explanation :

Let the time taken by the first worker be x hours.
∴ Time taken by the second worker = x + 1 hours.
∴ Time taken by the third worker = 2x hours.

Together they take 1 hour to complete the task.

⇒ $\frac{1}{\mathrm{x}}+\frac{1}{\mathrm{x}+1}+\frac{1}{2\mathrm{x}}$ = $\frac{1}{1}$

⇒ $\frac{3}{2\mathrm{x}}+\frac{1}{\mathrm{x}+1}$ = $\frac{1}{1}$

⇒ 3(x + 1) + 2x = 2x × (x + 1)

⇒ 5x + 3 = 2x² + 2x

⇒ 2x² - 3x - 3 = 0

⇒ x = $\frac{3\pm \sqrt{9-4\times 2\times -3}}{2\times 2}$ = $\frac{3\pm \sqrt{33}}{4}$

We will accept only +ve value.

∴ x = $\frac{3+\sqrt{33}}{4}$

⇒ Required average = $\frac{\mathrm{x}+\mathrm{x}+1+2\mathrm{x}}{3}$ = $\frac{4\mathrm{x}+1}{3}$ = $\frac{4+\sqrt{33}}{3}$

Hence, option (a).

Answer: Option B

Explanation :

Number of students who opted for cricket = Number of number which are divisible by 2 = Quotient of 140/2 = 70.
Number of students who opted for football = Number of number which are divisible by 5 = Quotient of 140/5 = 28.
Number of students who opted for basketball = Number of number which are divisible by 3 = Quotient of 140/3 = 46.

Number of students who opted for cricket and football = Number of number which are divisible by 2 and 5 = Quotient of 140/10 = 14.
Number of students who opted for basketball and football = Number of number which are divisible by 3 and 5 = Quotient of 140/15 = 9.
Number of students who opted for cricket and basketball = Number of number which are divisible by 2 and 3 = Quotient of 140/6 = 23.

Number of students who opted for cricket and football and basketball = Number of number which are divisible by 2, 5 and 3 = Quotient of 140/30 = 4.

∴ C ∪ F ∪ B = C + F + B - C ∩ F - F ∩ B - B ∩ C + C ∩ F ∩ B

⇒ C ∪ F ∪ B = 70 + 28 + 46 - (14 + 9 + 23) + 4

⇒ C ∪ F ∪ B = 144 - 46 + 4 = 102

∴ Number of students who opted for none of the three sports = 140 - 102 = 38.

Hence, option (b).

Answer: Option A

Explanation :

Given, f(x) = x² + log₃x and g(y) = 2y + f(y), 

∴ g(3) = 2 × 3 + f(3)

⇒ g(3) = 2 × 3 + 3² + log₃3

⇒ g(3) = 6 + 9 + 1 = 16

Hence, option (a).

Answer: Option A

Explanation :

Total number of matrices formed = Number of ways of selecting 4 distinct integer × Arranging these 4 integers in the matrix.
= ⁶C₄ × 4! = 15 × 24 = 360

For a singular matrix $\left|\begin{array}{cc}\mathrm{a}& \mathrm{b}\\ \mathrm{c}& \mathrm{d}\end{array}\right|$, its determinant = 0.
∴ ad - bc = 0
⇒ ad = bc

Now, possible pairs of a and d can be

Case 1: a and b are 2 or 3, 
∴ a × d = b × c = 6
⇒ b and c are 1 or 6.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 2: a and b are 1 or 6,
∴ a × d = b × c = 6
⇒ b and c are 2 or 3.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 3: a and b are 3 or 4, 
∴ a × d = b × c = 12
⇒ b and c are 2 or 6.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

Case 4: a and b are 2 or 6, 
∴ a × d = b × c = 12
⇒ b and c are 3 or 4.
∴ Possible arrangements for a and d = 2! and that for b and c = 2!
∴ Total possible arrangements for a, b, c and d = 2! × 2! = 4.

⇒ Total number of required matrices = 4 + 4 + 4 + 4 = 16

Required probability = 16/360 = 2/45.

Hence, option (a).

Answer: Option C

Explanation :

Let Ashok's investment be Rs. 2x, hence Bharat's investment is Rs. x.
Ashok invested for 12 months whereas Bharat invested for say m months.

⇒ $\frac{\mathrm{Profit} \mathrm{of} \mathrm{Ashok}}{\mathrm{Profit} \mathrm{of} \mathrm{Bharat}}$ = $\frac{\mathrm{Investment} \mathrm{of} \mathrm{Ashok} \times \mathrm{Time} \mathrm{of} \mathrm{Ashok}}{\mathrm{Investment} \mathrm{of} \mathrm{Bharat} \times \mathrm{Time} \mathrm{of} \mathrm{Bharat}}$

⇒ $\frac{3}{1}$ = $\frac{2\mathrm{x}\times 12}{\mathrm{x}\times \mathrm{m}}$

⇒ 3m = 24

⇒ m = 8

∴ If Bharat invested for 8 months it means he invested after 12 - 8 = 4 months of Ashok.

Hence, option (c).

Answer: Option B

Explanation :

Total marks of 6 students = 6 × 64 = 384

Marks of all the students are distinct and greater than 40.

We need to maximise the difference of 2^nd highest and 2^nd lowest marks

For this we need to minimise the 2^nd lowest marks and maximise the 2^nd highest marks.

Now, least possible marks of bottom 2 students = 40 and 41.
Marks of the students is 70.
Let the marks of remaining three students be A, B and C

∴ 40 + 41 + 70 + A + B + C = 384
⇒ A + B + C = 233.

Now, if C is higest and B is second highest, we can maximise B by minimising A.
Least possible marks of A now is 42.
⇒ B + C = 191

For B to be maximum possible, B and C should be as close as possible.
∴ C = B + 1

⇒ B + B + 1 = 191
⇒ B = 95

∴ The 6 numbers are: 40, 41, 42, 70, 95, 96

⇒ Differnce between 2^nd highest and 2^nd lowest marks = 95 - 41 = 54.

Hence, option (b).

Answer: Option A

Explanation :

​​​​​​​To earn maximum profit, the investor should invest in the stock which gives the maximum profit % over the 5 days.

Profit % of

Option (a): Marico = (372/346 - 1) × 100% = 7.5%

Option (b): HUL = (1966/1933 - 1) × 100% = 1.7%

Option (c): ITC = (25/238 - 1) × 100% = 6.3%

Option (d): Brittania = (3140/3046 - 1) × 100% = 3.1%

By investing in Marico, the investor will earn maximum profit.

Hence, option (a).

Answer: Option C

Explanation :

Multiplication for the whole process = 1979/1933 × 3140/3115 = 1.032

∴ Overall % change = (1.032 - 1) × 100% = 3.2%

Hence, option (c).

Answer: Option A

Explanation :

Average spread on

Monday = (2 + 1 + 2 + 1 + 2)/5 = 1.6

Tuesday = (3 + 1 + 3 + 1 + 1)/5 = 1.8

Wednesday = (1 + 3 + 2 + 5 + 5)/5 = 3.2

Thursday = (3 + 3 + 4 + 4 + 2)/5 = 3.2

Friday = (1 + 2 + 3 + 3 + 4)/5 = 2.6

Hence, option (a).

Answer: Option B

Explanation :

Total transation value = 1000 × 3101 + 1000 × 3025 = 1000 × 6126 = 6126000

∴ Commission = 0.1% of 6126000 = 0.1/100 × 6126000 = 6126.

Hence, option (b).

Answer: Option D

Explanation :

To earn maximum possible profit we should sell the stocks when the sellling price is highest.

Of all the 5 days, selling price is highest on Friday i.e., 253.

Hence, option (d).