Past Papers Solutions: CAT 2021 Quantitative Ability

CAT 2021 Quantitative Ability - Slot 01

2021 QA Slot 01

CAT 2021 Quantitative Ability Slot 01
(Total questions: 22)

Question 01: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Suppose the number is \(abc\) and the number formed by its reversal is \(cba\)

\( \Rightarrow (100c + 10b + a) - (100a + 10b + c) = 198\)

\( \Rightarrow 99(c - a) = 198\)

Hence \(c - a = 2\)

As the number is a three digit number, hence \(a \ge 1\) and \(a \le 7\). The digit at the 10th place, can take any value from 0 to 9, thus there are 10 choices for \(b\).

Total number of such numbers = \(7 \times 10 = 70\)

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Question 02: \(f(x) = \frac{{{x^2} + 2x - 15}}{{{x^2} - 7x - 18}}\) is negative if and only if

A. \(- 5 < x < -2\) or \(3 < x < 9\)
B. \(-2 < x < 3\) or \(x > 9\)
C. \(x < -5\) or \(3 < x < 9\)
D. \(x < -5\) or \(-2 < x < 3\)

A. \(- 5 < x < -2\) or \(3 < x < 9\)

The given function can be factorized as \(\frac{{(x + 5)(x - 3)}}{{(x - 9)(x + 2)}}\)

\( \Rightarrow \frac{{(x + 5)(x - 3)}}{{(x - 9)(x + 2)}} < 0\), the values of \(x\) where Numerator or denominator becomes zero. We have to check the intervals between these values of \(x\), where \(f(x)\) is negative. The sign of the function alternates between the intervals.

When \(x > 9\), the sign of the expression is positive and the sign changes in the alternate interval. Hence \(f(x)\) will be negative in the intervals \(3 < x < 9\) and \( - 5 < x < - 2\)

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Question 03: The number of integers n that satisfy the inequalities |n – 60| < |n – 100| < |n – 20| is:

(1) 21
(2) 18
(3) 20
(4) 19

First we have to find an interval for \(n\), so that the inequalities are satisfied. To solve the inequality, \(|n - 60| < |n - 100|\), let us assume that \(n > 100\), then we have,

\((n - 60) < (n - 100)\), that is not possible, hence \(n < 100\).

Similarly, by taking \(n < 60\) and comparing last two inequalities, we have

\((100 - n) < (n - 20)\) or \(n > 60\).

Thus \(60 < n < 100\), for this interval,

\((n - 60) < (100 - n)\) or \(n < 80\).

So the interval is \(60 < n < 80\), a total of \(19\) values.

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Question 04: Identical chocolate pieces are sold in boxes of two sizes, small and large. The large box is sold for twice the price of the small box. If the selling price per gram of chocolate in the large box is 12% less than that in the small box, then the percentage by which the weight of chocolate in the large box exceeds that in the small box is nearest to:

(1) 127
(2) 135
(3) 124
(4) 144

127

Suppose weights of chocolate in the large and small boxes are \(x\) and \(y\), and their selling prices are 88 and 100, then

\(88x = 2(100y)\)

\( \Rightarrow \frac{x}{y} = \frac{{200}}{{88}} = \frac{{25}}{{11}} = 2.27\)

Percentage difference = \(\frac{{x - y}}{y} \times 100 = 127\)%

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Question 05: \(5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = \log \frac{1}{{\sqrt {1 - {x^2}} }}\), then \(100x\) equals

We can write \(5 = {\log _{10}}{10^5}\), hence the equation can be written as

\(\log \left( {\frac{{{{10}^5} \times {{\left( {\sqrt {1 - x} } \right)}^4}}}{{\sqrt {1 + x} }}} \right) = \log \left( {\frac{1}{{\sqrt {1 - {x^2}} }}} \right)\)

\( \Rightarrow \log \left( {\frac{{{{10}^5} \times {{\left( {\sqrt {1 - x} } \right)}^4}}}{{\sqrt {1 + x} }}} \right) = \log \left( {\frac{1}{{\sqrt {(1 + x)(1 - x)} }}} \right)\)

\( \Rightarrow {10^5}{\left( {\sqrt {1 - x} } \right)^5} = 1\) or \(\sqrt {1 - x} = \frac{1}{{10}}\)

\( \Rightarrow x = \frac{{99}}{{100}}\) or \(100x = 99\)

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Question 06: A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

(1) 12
(2) 13
(3) 11
(4) 10

Let the prices of one apple, one orange and one mango are a, b and c, then

2a + 4b + 6c = a + 4b + 8c = 8b + 7c

Taking the first two expressions,

a = 2c, putting a = 2c, in the last two expressions,

2c + 4b + 8c = 8b + 7c 4b = 3c

Value of one basket in terms of c is, 8b + 7c = 13c.

Hence the number of mangoes in a basket of mangoes that has the same cost as the other baskets is 13.

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Question 07: If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is

(1) 4 √6
(2) 6 √6
(3) 2√6
(4) √6

2√6

We know that area of a hexagon = \(6 \times \frac{{\sqrt 3 }}{4} \times {a^2}\), where \(a\)is the side of the hexagon.

Here \(6 \times \frac{{\sqrt 3 }}{4} \times {a^2} = \frac{{\sqrt 3 }}{4} \times {12^2} \Rightarrow {a^2} = 24\)

Hence \(a = \sqrt {24} = 2\sqrt 6 \)

Question 09: If \(r\) is a constant such that \(|{x^2} - 4x - 13|\, = r\) has exactly three distinct real roots, then the value of \(r\) is

(1) 15
(2) 18
(3) 17
(4) 21

The equation can be written as \({x^2} - 4x - 13 = \pm r\), and we get two equations as

\({x^2} - 4x - 13 - r = 0\)

\({x^2} - 4x - 13 + r = 0\)

Suppose the roots of the first equation are \(\alpha ,\;\beta \) and roots of the second equation are \(\gamma ,\;\delta \), then

\(\alpha + \beta = \gamma + \delta = 4\)

As per the style = "color:#2575be" question there are only 3 roots, hence out of these 4 roots, 2 must be same. There is only one possibility that either \(\alpha = \beta \) or \(\gamma = \delta \). Because if we take \(\alpha = \gamma \), then \(\beta \) will also be equal to \(\delta \) as their sum is same and the number of distinct roots will only be 2.

So we will take \(\alpha = \beta\) and when \(\alpha = \beta = 2\), their product must be 4, so \( - 13 + r = 4\)or \(r = 17\) (as \(r\) is positive)

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Question 10: Amal purchases some pens at ₹ 8 each. To sell these, he hires an employee at a fixed wage. He sells 100 of these pens at ₹ 12 each. If the remaining pens are sold at ₹ 11 each, then he makes a net profit of ₹ 300, while he makes a net loss of ₹ 300 if the remaining pens are sold at ₹ 9 each. The wage of the employee, in INR, is:

1000

Suppose the total number of pens = \(n\) and wage of the employee is \(x\), then

\(100 \times 12 + (n - 100) \times 11 = 8n + x + 300\)

\(100 \times 12 + (n - 100) \times 9 = 8n + x - 300\)

Subtracting the second equation from the first one, we have

\((n - 100) \times 2 = 600 \Rightarrow n = 400\)

Putting the value of \(n\) in any of the equation, we get \(x = 1000\)

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Question 11: Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to

(1) 931.72
(2) 926.84
(3) 929.48
(4) 934.65

931.72

Interests accrued during the second year, \({I_2} = {P_1} \times r\), where \({P_1}\) is the value of the principal amount after 1 year.

Interests accrued during the third year, \({I_3} = {P_2} \times r\), where \({P_2}\) is the value of the principal amount after 2 years.

\( \Rightarrow \frac{{{I_3}}}{{{I_2}}} = \frac{{{P_2}}}{{{P_1}}} = \left( {1 + \frac{r}{{100}}} \right) = \frac{{866.72}}{{866.25}}\)

Similarly, \(\frac{{{I_4}}}{{{I_3}}} = \frac{{{P_3}}}{{{P_2}}} = \left( {1 + \frac{r}{{100}}} \right)\)

Hence \({I_4} = {I_3}\left( {1 + \frac{r}{{100}}} \right) = 866.72 \times \frac{{866.72}}{{806.25}} \approx 931.72\)

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Question 12: The strength of an indigo solution in percentage is equal to the amount of indigo in grams per 100 cc of water. Two 800 cc bottles are filled with indigo solutions of strengths 33% and 17%, respectively. A part of the solution from the first bottle is thrown away and replaced by an equal volume of the solution from the second bottle. If the strength of the indigo solution in the first bottle has now changed to 21% then the volume, in cc, of the solution left in the second bottle is:

200

Suppose the quantity of the solution replaced from the first bottle is \(w\), then

\((800 - w) \times 0.33 + w \times 0.17 = 800 \times 0.21\)

\( \Rightarrow 800 \times 0.33 - 0.16w = 800 \times 0.21\) or \(w = 600\)

Quantity left in the second bottle = 800 – 600 = 200

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Question 13: Amar, Akbar and Anthony are working on a project. Working together Amar and Akbar can complete the project in 1 year, Akbar and Anthony can complete in 16 months, Anthony and Amar can complete in 2 years. If the person who is neither the fastest nor the slowest works alone, the time in months he will take to complete the project is:

Suppose the total work is 48 units, then

work done by Amar and Akbar in one month = \(\frac{{48}}{{12}} = 4\) units.

work done by Akbar and Anthony in one month = \(\frac{{48}}{{16}} = 3\) units.

work done by Anthony and Amar in one month = \(\frac{{48}}{{24}} = 2\) units.

Let us assume works done by Amar, Akbar and Anthony working alone, in one month are \(x,\,y\) and \(z\)

\(x + y = 4,\;\;y + z = 3,\;\;z + x = 2\)

Solving these equations, we have \(x = \frac{3}{2},\;\,y = \frac{5}{2},\;\,z = \frac{1}{2}\)

So Amar is neither the fastest nor the slowest, time taken by him = \(\frac{{48}}{{3/2}} = 32\)

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Question 14: The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

As 3 and 5 are always selected, so we are left with only 6 numbers 1, 2, 4, 6, 7, 8.

Number of groups of three or more distinct numbers (3 and 5 are already there)

= Number of groups of one or more distinct numbers from 1, 2, 4, 6, 7, 8 = \({2^6} - 1 = 63\)

Number of groups where 7 and 8 are also selected along with 3, 5

= number of groups from the remaining 1, 2, 4, 6 = \({2^4} = 16\)

Required number of groups = 63 – 16 = 47.

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Question 15: A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches, then the area of the triangle in square inches is:

120

Suppose the sides AB and AC are \(x,\;y\), then using the formula for in radius,

\(4 = \frac{{x + 10 - y}}{2}\) \( \Rightarrow y = x + 2\)

Using Pythagorean Theorem,

\({(x + 2)^2} = {x^2} + {10^2} \Rightarrow x = 24\)

Area of the triangle = \(\frac{{24 \times 10}}{2} = 120\)

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Question 16: Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. It M is the midpoint of CD, then the length of AM, in cm, is:
(1) √15
(1) √15
(2) √13
(3) √12
(4) √14

√13

In a regular hexagon, the diagonal AC = \(\sqrt 3 \)×side = \(2\sqrt 3 \)

The ▲ACD is a right angled triangle and midpoint of the side CD is M, then

\(A{M^2} = A{C^2} + D{C^2} = {(2\sqrt 3 )^2} + {1^2} = 13\)

\( \Rightarrow AM = \sqrt {13} \)

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Question 17: Suppose hospital A admitted 21 less Covid infected patients than hospital B, and all eventually recovered. The sum of recovery days for patients in hospitals A and B were 200 and 152, respectively. If the average recovery days for patients admitted in hospital A was 3 more than the average in hospital B then the number admitted in hospital A was:

Suppose the number of patients admitted in hospital A and hospital B are \(n\)and \(n + 21\).

Given that, the sum of recovery days for patients in hospitals A and B were 200 and 152, respectively.

The average recovery days for patients admitted in hospital A was 3 more than the average in-hospital B.

\(\frac{{200}}{n} = \frac{{152}}{{n + 21}} + 3\)

\( \Rightarrow \frac{{200}}{n} = \frac{{215 + 3n}}{{n + 21}}\)

\( \Rightarrow 3{n^2} + 15n - 4200 = 0\)

\( \Rightarrow {n^2} + 5n - 1400 = 0\)

\( \Rightarrow (n - 35)(n + 40) = 0\) or \(n = 35\)

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Question 18: The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to

(1) 6119
(2) 7471
(3) 4941
(4) 6090

6119

We know that sum of the first \(n\)odd numbers is \({n^2}\). First group has 1 number, second group has 3 numbers, third group has 5 numbers etc. Hence total numbers used in the first 15 groups = 15² = 225. And total numbers used in the first 14 groups = 14² = 196.

Hence sum of the numbers in 15th group is:

(197 + 198 + 199 + ……. + 225) = \(\frac{{29}}{2}\left( {197 + 225} \right) = 6119\)

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Question 19: Anu, Vinu and Manu can complete a work alone in 15 days, 12 days and 20 days, respectively. Vinu works every day. Anu works only on alternate days starting from the first day while Manu works only on alternate days starting from the second day. Then, the number of days needed to complete the work is
(1)
(1) 8
(2) 6
(3) 5
(4) 7

Let the total work is 60 units, then the work done by Anu, Vinu and Manu in one day are 4, 5 and 3 units respectively.

First day Anu and Vinu are working, so work done is 4 + 5 = 9 units

Second day Vinu and Manu are working, so work done is 5 + 3 = 8 units

Third day Anu and Vinu are working, so work done is 4 + 5 = 9 units

Fourth day Vinu and Manu are working, so work done is 5 + 3 = 8 units

Work done in the first 6 days = 3(9 + 8) = 51 units

Work done on 7th day = 9 units. Hence the work will be completed on 7th day.

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Question 20: Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is:
(1)
(1) 184
(2) 180
(3) 190
(4) 192

Suppose speed of the faster train is \(s\) m/s, then

\(\frac{{160}}{s} = 12 \Rightarrow s = \frac{{40}}{3}\) m/s

Let the length of the other train is \(\ell \) and its speed is \(\frac{{40}}{3} - 6 \times \frac{5}{{18}} = \frac{{35}}{3}\)m/s.

\( \Rightarrow \frac{{160 + \ell }}{{\frac{{40}}{3} + \frac{{35}}{3}}} = 14\) or \(\ell = 190\)

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Question 21: If \({x_0} = 1,\;{x_1} = 2\) and \({x_{n + 2}} = \frac{{1 + {x_{n + 1}}}}{{{x_n}}},\,n = 0,\,\,1,\,\,2,...\), then \({x_{2021}}\) is equal to:

(1) 4
(2) 3
(3) 2
(4) 1

\({x_0} = 1\)

\({x_1} = 2\)

\({x_2} = \frac{{1 + {x_1}}}{{{x_0}}} = \frac{{1 + 2}}{1} = 3\), \({x_3} = \frac{{1 + {x_2}}}{{{x_1}}} = \frac{{1 + 3}}{2} = 2\)

\({x_4} = \frac{{1 + {x_3}}}{{{x_2}}} = \frac{{1 + 2}}{3} = 1\), \({x_5} = \frac{{1 + {x_4}}}{{{x_3}}} = \frac{{1 + 1}}{2} = 1\)

\({x_6} = \frac{{1 + {x_5}}}{{{x_4}}} = \frac{{1 + 1}}{1} = 2\), \({x_7} = \frac{{1 + {x_6}}}{{{x_5}}} = \frac{{1 + 2}}{1} = 3\)

We see that values 1, 2, 3, 2, 1, 1, 2, 3, …. are repeating after an interval of 5. Hence \({x_5} = {x_0},\,\;{x_6} = {x_1}\) etc, in general \({x_{n + 5}} = {x_n}\) or \({x_{2021}} = {x_1} = 2\)

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Question 22: The amount Neeta and Geeta together earn in a day equals what Sita alone earns in 6 days. The amount Sita and Neeta together earn in a day equals what Geeta alone earns in 2 days. The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is:
(1)
(1) 7 : 3
(2) 11 : 3
(3) 11 : 7
(4) 3 : 2

11 : 3

Suppose their earnings are n, g and s, then

n + g = 6s

s + n = 2g

Subtracting second equation from the first, we have

g – s = 6s – 2g or g : s = 7 : 3

Putting these values in the first equation, we get n = 11.

The ratio of the daily earnings of the one who earns the most to that of the one who earns the least is 11 : 3

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CAT 2021 Quantitative Ability - Slot 01

Question 01: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Question 02: \(f(x) = \frac{{{x^2} + 2x - 15}}{{{x^2} - 7x - 18}}\) is negative if and only if

Question 03: The number of integers n that satisfy the inequalities |n – 60| < |n – 100| < |n – 20| is:

Question 05: \(5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = \log \frac{1}{{\sqrt {1 - {x^2}} }}\), then \(100x\) equals

Question 06: A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

Question 07: If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is

Question 09: If \(r\) is a constant such that \(|{x^2} - 4x - 13|\, = r\) has exactly three distinct real roots, then the value of \(r\) is

Question 11: Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to

Question 14: The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

Question 15: A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches, then the area of the triangle in square inches is:

Question 16: Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. It M is the midpoint of CD, then the length of AM, in cm, is:
(1) √15

Question 18: The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to

Question 20: Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is:
(1)

Question 21: If \({x_0} = 1,\;{x_1} = 2\) and \({x_{n + 2}} = \frac{{1 + {x_{n + 1}}}}{{{x_n}}},\,n = 0,\,\,1,\,\,2,...\), then \({x_{2021}}\) is equal to:

CAT 2021 Quantitative Ability - Slot 01

Question 01: How many three-digit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order?

Question 02: \(f(x) = \frac{{{x^2} + 2x - 15}}{{{x^2} - 7x - 18}}\) is negative if and only if

Question 03: The number of integers n that satisfy the inequalities |n – 60| < |n – 100| < |n – 20| is:

Question 05: \(5 - {\log _{10}}\sqrt {1 + x} + 4{\log _{10}}\sqrt {1 - x} = \log \frac{1}{{\sqrt {1 - {x^2}} }}\), then \(100x\) equals

Question 06: A basket of 2 apples, 4 oranges and 6 mangoes costs the same as a basket of 1 apple, 4 oranges and 8 mangoes, or a basket of 8 oranges and 7 mangoes. Then the number of mangoes in a basket of mangoes that has the same cost as the other baskets is

Question 07: If the area of a regular hexagon is equal to the area of an equilateral triangle of side 12 cm, then the length, in cm, of each side of the hexagon is

Question 09: If \(r\) is a constant such that \(|{x^2} - 4x - 13|\, = r\) has exactly three distinct real roots, then the value of \(r\) is

Question 11: Anil invests some money at a fixed rate of interest, compounded annually. If the interests accrued during the second and third year are ₹ 806.25 and ₹ 866.72, respectively, the interest accrued, in INR, during the fourth year is nearest to

Question 14: The number of groups of three or more distinct numbers that can be chosen from 1, 2, 3, 4, 5, 6, 7 and 8 so that the groups always include 3 and 5, while 7 and 8 are never included together is

Question 15: A circle of diameter 8 inches is inscribed in a triangle ABC where ∠ABC = 90°. If BC = 10 inches, then the area of the triangle in square inches is:

Question 16: Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. It M is the midpoint of CD, then the length of AM, in cm, is:(1) √15

Question 18: The natural numbers are divided into groups as (1), (2, 3, 4), (5, 6, 7, 8, 9), ….. and so on. Then, the sum of the numbers in the 15th group is equal to

Question 20: Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is:(1)

Question 21: If \({x_0} = 1,\;{x_1} = 2\) and \({x_{n + 2}} = \frac{{1 + {x_{n + 1}}}}{{{x_n}}},\,n = 0,\,\,1,\,\,2,...\), then \({x_{2021}}\) is equal to:

Question 16: Suppose the length of each side of a regular hexagon ABCDEF is 2 cm. It M is the midpoint of CD, then the length of AM, in cm, is:
(1) √15

Question 20: Two trains cross each other in 14 seconds when running in opposite directions along parallel tracks. The faster train is 160 m long and crosses a lamp post in 12 seconds. If the speed of the other train is 6 km/hr less than the faster one, its length, in m, is:
(1)