KINGS: CAT 2020 Quantitative Ability (slot 1)

CAT 2020 Quantitative Ability (slot 1)

Question 01: How many 3-digit numbers are there, for which the product of their digits is more than 2 but less than 7?

Suppose the number is abc, then the product of the digits can be 3, 4, 5, 6.

Product = 3: Possible digits are (1, 1, 3), and their permutations are 3!/2! = 3

Product = 4: Possible digits are (1, 1, 4) or (2, 2, 1) and their permutations are (3!/2!) + (3!/2!) = 6

Product = 5: Possible digits are (1, 1, 5) and their permutations are (3!/2!) = 3

Product = 6: Possible digits are (1, 1, 6) or (1, 2, 3) and their permutations are (3!/2!) + 3! = 9

Total number of such numbers = 3 + 6 + 3 + 9 = 21

Question 02: If \(f(5 + x) = f(5 - x)\) for every real \(x\) and \(f(x) = 0\) has four distinct real roots, then the sum of the roots is

(1) 0
(2) 40
(3) 10
(4) 20

It is given that \(f(5 + x) = f(5 – x)\), that means \(f(6) = f(4), f(7) = f(3)\) etc.

Let us assume that \(f(5 + \alpha) = 0\), then \(f(5 – \alpha)\) is also 0. In other words, if one of the root is \(5 + \alpha\), then the other root is \(5 – \alpha\). Similarly if one of the root is \(5 + \beta\), then the other root is \(5 – \beta\). Given that there are 4 distinct roots, so their sum is \((5 + \alpha) + (5 - \alpha) + (5 + \beta) + (5 - \beta) = 20\).

Question 03: Veeru invested Rs 10000 at 5% simple annual interest, and exactly after two years, Joy invested Rs 8000 at 10% simple annual interest. How many years after Veeru’s investment, will their balances, i.e., principal plus accumulated interest, be equal?

Suppose t years after Veeru's investment, their balances will be same, then Veeru invested for t years and Joy invested for (t – 2) years.

10000 + 10000×5×t/100 = 8000 + 8000×10×(t – 2)/100

=> 10000 + 500t = 8000 + 800(t – 2)

or t = 12 years

Question 04: A train travelled at one-thirds of its usual speed, and hence reached the destination 30 minutes after the scheduled time. On its return journey, the train initially travelled at its usual speed for 5 minutes but then stopped for 4 minutes for an emergency. The percentage by which the train must now increase its usual speed so as to reach the destination at the scheduled time, is nearest to:

(1) 58

(2) 67

(3) 50

(4) 61

When train travelled at one third speed, then the time taken would have been three times of the usual time. Suppose the usual time is T, then

3T – T = 30

or T = 15 minutes.

Let us assume total distance is \(15x\) and each minute train covers \(x\) units at usual speed. In the return journey, in the first 9 minutes, the distance covered is \(5x\). So in the remaining 6 minutes, the train has to cover the distance of \(10x\). So the speed required = \(\frac{10x}{6} = \frac{5x}{3}\) and percentage increase in the speed \(=\frac{(5x/3)-x}{x}\times 100 =66.67\).

Nearest integer = 67%

Question 05: If \({\log _4}5 = \left( {{{\log }_4}y} \right)\left( {{{\log }_6}\sqrt 5 } \right)\) then \(y\) equals

The equation can be written as,

\(\frac{{\log 5}}{{\log 4}} = \left( {\frac{{\log y}}{{\log 4}}} \right)\left( {\frac{{\log \sqrt 5 }}{{\log 6}}} \right)\)

\( \Rightarrow \frac{{\log 5}}{{\log 4}} = \left( {\frac{{\log y}}{{\log 4}}} \right)\left( {\frac{{\frac{1}{2}\log 5}}{{\log 6}}} \right)\)

\( \Rightarrow \frac{1}{2}\log y = \log 6\)

Hence \(\log \sqrt y = \log 6 \Rightarrow y = 36\)

Question 06: The number of real-valued solutions of the equation 2^x + 2^-x = 2 – (x – 2)² is

(1) 0

(2) 1

(3) 2

(4) infinite

We know that for a positive number a,
a + (1/a) ≥ 2

And the value is 2 at a = 1.

Here (2^x + 1/2^x) ≥ 2 and the value is 2 at x = 0.
But the value of 2 – (x – 2)² is always less than or equal to 2.
Therefore the values of 2^x + 2^-x and 2 – (x – 2)² can never be same and so there is no solution to the equation.

Question 07: A straight road connects points A and B. Car 1 travels from A to B and Car 2 travels from B to A, both leaving at the same time. After meeting each other, they take 45 minutes and 20 minutes, respectively, to complete their journeys. If Car 1 travels at the speed of 60 km/hr, then the speed of Car 2, in km/hr, is:

(1) 90
(2) 80
(3) 70
(4) 100

Suppose the cars meet at the point C, where AC = x and CB = y. Let the speeds of the car1 and car 2 are v₁ and v₂, and the cars meet at the point C after T minutes, then

x = v₁×T and y = v₂×T
=> (x/y) = v₁/v₂

After meeting at C, the cars take 45 and 20 minutes to reach the points B and A. y = v₁×45 and x = v₂×20
= x/y = (20v₂)/(45v₁)

equating x/y, we get
v₁/v₂ = (20v₂)/(45v₁)
=> (v₁/v₂)² = 20/45 or v₁/v₂ = 2/3

Given that v₁ = 60, hence v₂= 90

Question 08: Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is

(1) 6
(2) 4
(3) 7
(4) 5

From the first condition,
A + (B + C)/2 = 5 or 2A + B + C = 10

From the second condition,
B + (A + C)/2 = 7 or 2B + A + C = 14

Subtracting the first equation from the second, we have B – A = 4

The possible values of (B, A) are (5, 1), (6, 2), (7, 3) etc. Note that if we take B = 6 and A = 2, then the value of C becomes zero or negative, hence (B, A) can only be (5, 1).
Thus A + B = 6.

Question 09: If \(x = {(4096)^{7 + 4\sqrt 3 }}\), then which of the following equals 64?

(1) \(\frac{{{x^{7/2}}}}{{{x^{4/\sqrt 3 }}}}\)

(2) \(\frac{{{x^7}}}{{{x^{4\sqrt 3 }}}}\)

(3) \(\frac{{{x^{7/2}}}}{{{x^{2\sqrt 3 }}}}\)

(4) \(\frac{{{x^7}}}{{{x^{2\sqrt 3 }}}}\)

We know that \(4096 = {64^2}\)

\(x = {(4096)^{7 + 4\sqrt 3 }} = {(64)^{2(7 + 4\sqrt 3 )}}\)

\( \Rightarrow 64 = {x^{\frac{1}{{2(7 + 4\sqrt 3 )}}}} = {x^{\frac{{(7 - 4\sqrt 3 )}}{{2(7 + 4\sqrt 3 )(7 - 4\sqrt 3 )}}}}\)

\( \Rightarrow 64 = {x^{\frac{7}{2} - 2\sqrt 3 }} = \frac{{{x^{7/2}}}}{{{x^{2\sqrt 3 }}}}\)

Question 10: The mean of all 4 digit even natural numbers of the form aabb, where a > 0, is

(1) 5544
(2) 4466
(3) 4864
(4) 5050

In the numbers of the type aabb, the digit b can be 0, 2, 4, 6, 8. The corresponding value of a can be 1, 2, 3, .....9.

Total number of such numbers = 9×5 = 45

Sum of the unit digits = (0 + 2 + 4 + 6 + 8)×9 = 180

Sum of the digits at 10th place = (0 + 2 + 4 + 6 + 8)×9 = 180

Sum of the digits at 100th place = (1 + 2 + 3 +....... + 9)×5 = 225

Sum of the digits at 1000th place = (1 + 2 + 3 +....... + 9)×5 = 225

Sum of all the numbers = 180 + 180×10 + 225×100 + 225×1000

Average of these numbers \( = \frac{{180 + 180 \times 10 + 225 \times 100 + 225 \times 1000}}{{45}} = 5544\)

Question 11: The number of distinct real roots of the equation \({\left( {x + \cfrac{1}{x}} \right)^2} - 3\left( {x + \cfrac{1}{x}} \right) + 2 = 0\) equals

Given that \({\left( {x + \frac{1}{x}} \right)^2} - 3\left( {x + \frac{1}{x}} \right) + 2 = 0\)

Suppose \(x + \frac{1}{x} = t\), then \({\left( {x + \frac{1}{x}} \right)^2} = {t^2}\)

So the equation becomes, \({t^2} - 3t + 2 = 0\) or \(t = 1,2\)

Since \(x + \frac{1}{x} \le - 2\) or \(x + \frac{1}{x} \ge 2\)

Hence \(t\)can take only one value i.e. 2, or \(x\) = 1.

Question 12: A person spent Rs 50000 to purchase a desktop computer and a laptop computer. He sold the desktop at 20% profit and the laptop at 10% loss. If overall he made a 2% profit then the purchase price, in rupees, of the desktop is:

Suppose the purchase price of the desktop and laptop are \(x\) and \(y\), then

\(\frac{{0.20x - 0.10y}}{{x + y}} \times 100 = 2\)

\( \Rightarrow 20x - 10y = 2x + 2y\)

Thus \(x : y\) = 2 : 3. So the purchase price of the desktop = Rs. 20000.

Question 13: Among 100 students, x₁ have birthdays in January, x₂ have birthdays in February, and so on. If x₀ = max(x₁, x₂, ..., x₁₂), then the smallest possible value of x₀ is:

(1) 8

(2) 10

(3) 12

(4) 9

Since x₀ = max(x₁, x₂, ..., x₁₂) and we want smallest possible value of x0, so we have to keep all x₁, x₂, ..., x₁₂ as low as possible.

Also x₁ + x₂ + ...+ x₁₂ = 100

If we keep all x₁, x₂, ..., x₁₂ ≤ 8, then their sum will not reach to 100 (as 12×8 = 96) and hence it is not possible. In other words some of x₁, x₂, ..., x₁₂ will be more than 8.

We can manage the sum as 100 by taking some of x₁, x₂, ..., x₁₂ as 9.

For example 8,8,8,8,8,8,8,8,9,9,9,9.

So minimum value of x₀ = 9

Question 14: Two persons are walking beside a railway track at respective speeds of 2 and 4 km per hour in the same direction. A train came from behind them and crossed them in 90 and 100 seconds, respectively. The time, in seconds, taken by the train to cross an electric post is nearest to:

(1) 87
(2) 82
(3) 78
(4) 75

Suppose the length of the train is \(x\) and the speed is \(v\), then the time taken to cross the men will be \(\frac{x}{{v - 2}}\) and \(\frac{x}{{v - 4}}\)

Ratio of the time taken to cross the men is

\(\frac{{\,\frac{x}{{v - 2}}\,}}{{\frac{x}{{v - 4}}}} = \frac{{90}}{{100}} \Rightarrow \frac{{v - 4}}{{v - 2}} = \frac{9}{{10}}\)

Or \(v = 22\;{\rm{kmph}}\)

Now time taken to cross a post \(=\frac{x}{v} = \frac{x}{{22}}\)

Now time taken to cross the first man \(=\frac{x}{{v - 2}} = \frac{x}{{20}}\)

Now time taken to cross a post \(=90 \times \frac{{20}}{{22}} = 82\;{\rm{sec}}\)

Question 15: How many distinct positive integer-valued solutions exist to the equation \({({x^2} - 7x + 11)^{{x^2} - 13x + 42}} = 1\)

(1) 6
(2) 2
(3) 4
(4) 8

Given that \({({x^2} - 7x + 11)^{{x^2} - 13x + 42}} = 1\)

We know that \(a^0 = 1\), when \(a\) is a non zero number.

case 1: \(x^2 - 13x + 42 = 0\)

or \((x – 6)(x – 7) = 0\) or \(x = 6, 7\)

case 2: We know that 1^{any number} = 1, hence

\(x^2 - 7x + 11 = 1\) or \(x^2 - 7x + 10 = 0\)

or \(x = 2, 5\)

case 3: We know that (– 1)^{even number} = 1, hence

\(x^2- 7x + 11 = - 1\) or \(x^2-7x + 12 = 0\)

or \(x = 3, 4\)

Note that at \(x =3, 4\) the value of \(x^2-13x + 42\) is even. Hence there are 6 solutions.

Question 16: The area of the region satisfying the inequalities |x| - y ≤ 1, y ≥ 0, and y ≤ 1 is

There are 3 graphs y ≥ |x| – 1, y ≥ 0 and y ≤ 1

When y = 1, |x| – 1 = 1 or x = 2, – 2

Area of the trapezium ABDE = Area of the ▲ ABC – Area of the ▲ CDE = (4×2)/2 – (2×1)/2 = 3

Question 17: A solid right circular cone of height 27 cm is cut into 2 pieces along a plane parallel to its base at a height of 18 cm from the base. If the difference in the volume of the two pieces is 225 cc, the volume, in cc, of the original cone is:

(1) 264
(2) 232
(3) 243
(4) 256

The height of the upper part is 27 – 18 = 9. Suppose volume of the upper part and lower parts of the cone = \({v_1}\)and \({v_2}\).

We know that ratio of the volumes of two similar cones = cube of ratio of their heights

\(\frac{{{v_1}}}{{{v_1} + {v_2}}} = {\left( {\frac{9}{{27}}} \right)^3} = \frac{1}{{27}}\)

\( \Rightarrow {v_2} = 26{v_1}\)

Given that difference in their volumes = 225

\(26{v_1} - {v_1} = 225 \Rightarrow {v_1} = 9\)

Total volume of the cone \( = {v_1} + 26{v_1} = 27{v_1} = 243\)

Question 18: A circle is inscribed in a rhombus with diagonals 12 cm and 16 cm. The ratio of the area of the circle to the area of the rhombus is:

(1) 2π/15
(2) 6π/25
(3) 3π/25
(4) 5π/18

We know that diagonals of a rhombus intersect at right angles.

Therefore EA = 8 and ED = 6, and AD = √(8² + 6²) = 10 EF is the radius and EF is perpendicular to AD.

Area of ▲ EAD = (8×6)/2 = 24 => (AD×r)/2 = 24 or r = 24/5 = 4.8

Area of the rhombus = (1/2)×16×12 = 96

Area of the circle = πr² = π(24/5)² = 576π/25

The ratio of areas = 6π : 25

Question 19: If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is

(1) 56
(2) 49
(3) 46
(4) 59

Given that ab = 432 and bc = 96, thus a : c = 432 : 96 = 9 : 2

Let us assume a = 9k, c = 2k and b = 48/k

=> a + b + c = 11k + 48/k

The value of 11k + 48/k will be minimum when 11k ≈ 48/k or k = 2

=> a + b + c = 11×2 + 48/2 = 46

Alternate Solution: Given that c < 9, hence k ≤ 4.

k = 1, a + b + c = 11 + 48 = 59

k = 2, a + b + c = 11×2 + 48/2 = 46

k = 3, a + b + c = 11×3 + 48/3 = 49

k = 4, a + b + c = 11×4 + 48/4 = 56

Smallest value = 46

Question 20: Leaving home at the same time, Amal reaches office at 10:15 am if he travels at 8 kmph, and at 9:40 am if he travels at 15 kmph. Leaving home at 9:10 am, at what speed, in kmph, must he travel so as to reach office exactly at 10:00 am?

(1) 12
(2) 11
(3) 13
(4) 14

When his speed becomes 8 to 15, the time taken in the journey will become 8/15. If the time taken when the speed is 8 = T, then time taken now will be 8T/15

Saving in time = T – 8T/15 = 7T/15

Here the time decreases by 35 minutes, thus

7T/15 = 35 or T = 75 minutes.

Now we need that the time taken in the journey should be 10: 00 – 9:10 = 50 minutes, therefore the speed required = (75/50)×8 = 12 kmph