IPMAT (I) 2020 QASA

Answer: 34

Explanation :

Let the dividend be N, Quotient be Q and Remainder be R.
Given, Divisor = 5

∴ N = 5 × Q + R

Given, QR = 24 and Q + R = 10
Since Q and R are integers, the only possible value for Q and R will be 6 or 4 in any order.

Here, the divisor is 5 and hence remainder cannot be greater than 5, hence R = 4 and Q = 6.

⇒ N = 5 × 6 + 4 = 34.

Hence, 34.

Answer: 4

Explanation :

The center of the given circle is at (0, 0) and its radius = 1.

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The shortest distance between the point (-4, 3) and the circle = distance between (-4, 3) and the center - radius of the circle.

= OM - ON

= $\sqrt{{(-4-0)}^2+{(3-0)}^2}$ - 1

= 5 - 1

= 4.

Hence, 4.

Answer: 16

Explanation :

Given, $0.{04}^{{\log }_{\sqrt{5}}\left(\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+...\right)}$ 

= $0.{04}^{{\log }_{\sqrt{5}}\left(\frac{1/4}{1-1/2}\right)}$

= $0.{04}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}$

= ${\left(5^{-2}\right)}^{{\log }_{\sqrt{5}}\left(\frac{1}{2}\right)}$

= ${\left(5^{-2}\right)}^{{\log }_{5^{1/2}}\left(\frac{1}{2}\right)}$

= ${\left(5^{-2}\right)}^{2{\log }_5\left(\frac{1}{2}\right)}$

= $5^{-4{\log }_5\left(\frac{1}{2}\right)}$

= $5^{{\log }_5{\left(\frac{1}{2}\right)}^{-4}}$

= $5^{{\log }_{5}16}$

= ${16}^{{\log }_{5}5}$

= 16.

Hence, 16.

Answer: 1

Explanation :

Given, $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& {\mathrm{a}}^3-1\\ \mathrm{b}& {\mathrm{b}}^2& {\mathrm{b}}^3-1\\ \mathrm{c}& {\mathrm{c}}^2& {\mathrm{c}}^3-1\end{array}\right|$ = 0

⇒ $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& {\mathrm{a}}^3\\ \mathrm{b}& {\mathrm{b}}^2& {\mathrm{b}}^3\\ \mathrm{c}& {\mathrm{c}}^2& {\mathrm{c}}^3\end{array}\right|$ + $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& -1\\ \mathrm{b}& {\mathrm{b}}^2& -1\\ \mathrm{c}& {\mathrm{c}}^2& -1\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ - $\left|\begin{array}{ccc}\mathrm{a}& {\mathrm{a}}^2& 1\\ \mathrm{b}& {\mathrm{b}}^2& 1\\ \mathrm{c}& {\mathrm{c}}^2& 1\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ + $\left|\begin{array}{ccc}1& {\mathrm{a}}^2& \mathrm{a}\\ 1& {\mathrm{b}}^2& \mathrm{b}\\ 1& {\mathrm{c}}^2& \mathrm{c}\end{array}\right|$ = 0

⇒ abc × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ - $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ = 0

⇒ (abc - 1) × $\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ = 0

$\left|\begin{array}{ccc}1& \mathrm{a}& {\mathrm{a}}^2\\ 1& \mathrm{b}& {\mathrm{b}}^2\\ 1& \mathrm{c}& {\mathrm{c}}^2\end{array}\right|$ ≠ 0

∴ abc - 1 = 0

⇒ abc = 1

Hence, 1.

Answer: 7

Explanation :

Given, |3 - x| + |2 + x| + |5 - x|

= |x - 3| + |x + 2| + |x - 5|

= |x + 2| + |x - 3| + |x - 5|

Here, the critical points are -2, 3 and 5.

The minimum possible value of |x + 2| + |x - 5| is when x is between -2 and 5.
The minimum possible value of |x - 3| is when x = 3.

∴ x = 3 minimises |x + 2| + |x - 3| + |x - 5|.

⇒ Minimum value of |x + 2| + |x - 3| + |x - 5| = |3 + 2| + |3 - 3| + |3 - 5|

= 5 + 0 + 2 = 7

Hence, 7.

Answer: 100

Explanation :

 Let the price of each pen and pencil be x and y respectively.

1^st Visit: 
Let the number one pens and pencils bougth be 2a and 3a respentively.
⇒ 2a × x + 3a × y = 86   ...(1)

2^nd Visit: 
Number of pens and pencils bougth are 4 and 1 respectively.
⇒ 4 × x + 1 × y = 112   ...(2)

(1) × 2 - (2) × a
⇒ 4ax + 6ay - (4ax + ay) = 172 - 112a
⇒ 5ay = 172 - 112a
⇒ y = (172 - 112a)/5a

Since a and y are positive integers, the only possible value of a = 1.
∴ y = 12

Substituting y = 12 in (2), we get x = 25.

∴ Amount he paid for pens during 2^nd visit = 4x = Rs. 100

Hence, 100.

Answer: 7920

Explanation :

Let the number be 'abcd'. [a, b, c, d and single digit positive integers.]

Given, 

a × d = 0
a cannot be 0 since the given number has to be a 4 digit number, hence d = 0

Also, a - d = 7
⇒ a = 7

Also, b × c = 18   ...(1)

The thousands digit is as much more than the units digit as the hundreds digit is more than the tens digit
⇒ a - d = b - c = 7   ...(2)

From (1) and (2), we get
b = 9 and c = 2.

∴ The number is 7920.

Hence, 7920.

Answer: 20

Explanation :

To maximise the number of students who passed in all the three exams we will assume that no one failed in all the three exams.

∴ M ∪ P ∪ C = M + P + C - M ∩ P - P ∩ C - C ∩ M + M ∩ P ∩ C

⇒ M ∪ P ∪ C - (M + P + C) + M ∩ P + P ∩ C + C ∩ M = M ∩ P ∩ C

To maximise the number of students who passed in all the three exams we need to maximise M ∩ P and P ∩ C and C ∩ M.
Maximum value of M ∩ P =20, that of P ∩ C =20 and that of C ∩ M = 20

∴ 80 - (50 + 30 + 40) + 20 + 20 + 20 + M ∩ P ∩ C

⇒ 20 = M ∩ P ∩ C

Hence, 20.

Answer: 3

Explanation :

Since the ratio of their speeds is 3 : 2, they will meet at only one point on the circular track and that point will be the starting point.

Hence, when they meet for the first time, the faster person travels 3 rounds, the slower person would travel 2 rounds.
i.e., when the faster person travels 3 × 300 = 900 meters, they meet once.

∴ When the faster person travels 2700 (3 × 900) meters, they would have met 3 times.

To meet for the 4th time faster person needs to travel 900 meters more, but he has only 300 meters of the race left.
∴ Now the faster person would complete the race before they meet for the 4t^h time.

Hence, 3.

Answer: 256

Explanation :

Case 1: All 4 objects are distinct.
Number of ways of choosing 4 distinct objects out of 9 = ⁹C₄.
∴ Total number of such ways = 1 way.

Case 2: 3 objects are distinct and 1 is indistinguishable.
Number of ways of choosing 3 distinct objects out of 9 = ⁹C₃.
Number of ways of choosing 1 indistinguishable object out of 4 = 1.
∴ Total number of such ways = ⁹C₃ × 1 = ⁹C₃.

Case 3: 2 objects are distinct and 2 are indistinguishable.
Number of ways of choosing 2 distinct objects out of 9 = ⁹C₂.
Number of ways of choosing 2 indistinguishable object outs of 4 = 1.
∴ Total number of such ways = ⁹C₂ × 1 = ⁹C₂.

Case 4: 1 object is distinct and 3 are indistinguishable.
Number of ways of choosing 1 distinct objects out of 9 = ⁹C₁.
Number of ways of choosing 3 indistinguishable objects out of 4 = 1.
∴ Total number of such ways = ⁹C₁ × 1 = ⁹C₁.

Case 5: All 4 are indistinguishable.
Number of ways of choosing 0 distinct objects out of 9 = ⁹C₀.
Number of ways of choosing 4 indistinguishable objects out of 4 = 1.
∴ Total number of such ways = ⁹C₁ × 1 = ⁹C₁.

⇒ Total number of ways = ⁹C₄ + ⁹C₃ + ⁹C₂ + ⁹C₁ + ⁹C₀ = 256

Hence, 256.