Past Papers Solutions: CAT 2021 Quantitative Ability

CAT 2021 Quantitative Ability - Slot 03

2021 QA Slot 03

CAT 2021 Quantitative Ability Slot 03
(Total questions : 22)

Question 01: If a certain weight of an alloy of silver and copper is mixed with 3 kg of pure silver, the resulting alloy will have 90% silver by weight. If the same weight of the initial alloy is mixed with 2 kg of another alloy which has 90% silver by weight, the resulting alloy will have 84% silver by weight. Then, the weight of the initial alloy, in kg, is

(1) 3
(2) 2.5
(3) 4
(4) 3.5

Suppose the total weight of the alloy is \(w\) and weight of the silver in the alloy is \(x\), then

This Alloy is mixed with 3kgs of Pure Silver to form a new mixture.

Total weight of the mixture = \(w + 3\)

Weight of Silver in the mixture = \(x + 3\)

Since this mixture contains 90% Silver,

\(\frac{{x + 3}}{{w + 3}} = 0.9 \Rightarrow x + 3 = 0.90w + 2.7\)

The initial Alloy is mixed with 2kgs of another alloy having 90% Silver by weight, that means 2×0.90 = 1.8 kg silver is Added.

\(\frac{{x + 1.8}}{{w + 2}} = 0.84 \Rightarrow x + 1.8 = 0.84w + 1.68\)

Subtracting the second equation from the first, we have

\(1.2 = 0.06w + 1.02\)

¬\( \Rightarrow w = \frac{{1.2 - 1.02}}{{0.06}} = 3\)

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Question 02: Anil can paint a house in 12 days while Barun can paint it in 16 days. Anil, Barun, and Chandu undertake to paint the house for ₹ 24000 and the three of them together complete the painting in 6 days. If Chandu is paid in proportion to the work done by him, then the amount in INR received by him is:

3000

Suppose the total work is 48 units, then Anil can paint 4 units per day, Barun can paint 3 units per day and suppose Chandu can paint c units per day.

Total work done in 6 days = 48

Or (4 + 3 + c)×6 = 48

\( \Rightarrow \)c = 1

Work done by Anil, barun and Chandu = 4×6, 3×6 and 1×6 = 24, 18 and 6

Hence the amount received by Chandu = \(\frac{6}{{24 + 18 + 6}} \times 24000 = 3000\)

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Question 03: A four-digit number is formed by using only the digits 1, 2 and 3 such that both 2 and 3 appear at least once. The number of all such four-digit numbers is

Total number of 4 digit numbers = \({3^4} = 81\)

Number of 4 digit numbers when 2 does not appear = \({2^4} = 16\)

Number of 4 digit numbers when 3 does not appear = \({2^4} = 16\)

Number of 4 digit numbers when neither 2 nor 3 appears = 1

Number of 4 digit numbers such that both 2 and 3 appear at least once = 81 –(16 + 16) + 1 = 50

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Question 04: Mira and Amal walk along a circular track, starting from the same point at the same time. If they walk in the same direction, then in 45 minutes, Amal completes exactly 3 more rounds than Mira. If they walk in opposite directions, then they meet for the first time exactly after 3 minutes. The number of rounds Mira walks in one hour is

Suppose the speeds of Amal and Mira are x and y m/min, and the length of the track is r then

\(45(x - y) = 3r\)

\(3(x + y) = r\)

Dividing the first equation by second, we get

\(15\left( {\frac{{x - y}}{{x + y}}} \right) = 3 \Rightarrow x = \frac{3}{2}y\)

Suppose \(x = 3k,\;y = 2k\), then \(r = 15k\)

Time taken by Mira to complete one round =\(\frac{{15k}}{{2k}} = 7.5\)

Number of rounds completed by her in 1 hour =\(\frac{{60}}{{7.5}} = 8\)

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Question 05: In a triangle ABC, ∠BCA = 50°. D and E are points on AB and AC, respectively, such that AD = DE. If F is a point on BC such that BD = DF, then ∠FDE, in degrees, is equal to

(1) 100
(2) 80
(3) 96
(4) 72

Given that AD = DE, so ∠DEA = ∠DAE = y

Similarly, ∠DBF = ∠DFB = x

In the quadrilateral, CEDF, ∠BDF = 360° – (50° + 180° – x + 180° – y) = x + y – 50°

Since x + y = 130°, hence ∠BDF = 130° – 50° = 80°

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Question 06: For a real number\(a\), if \(\cfrac{{{{\log }_{15}}a + {{\log }_{32}}a}}{{{{\log }_{15}}a{{\log }_{32}}a}} = 4\) then \(a\) must lie in the range

(1) \(4 < a < 5\)
(2) \(3 < a < 4\)
(3) \(a > 5\)
(4) \(2 < a < 3\)

\(4 < a < 5\)

The equation can be written as

\(\frac{1}{{{{\log }_{32}}a}} + \frac{1}{{{{\log }_{15}}a}} = 4\)

\( \Rightarrow {\log _a}32 + {\log _a}15 = 4\)

\( \Rightarrow {\log _a}(15 \times 32) = 4\)

\( \Rightarrow {a^4} = 15 \times 32 = 480\)

Thus \(4 < a < 5\)

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Question 07: The arithmetic mean of scores of 25 students in an examination is 50. Five of these students top the examination with the same score. If the scores of the other students are distinct integers with the lowest being 30, then the maximum possible score of the toppers is

Suppose score of the topper is \(x\) and sum of the scores of the remaining 20 is \(y\), then
\(5x + y = 25 \times 50\)

Here note that \(x\) will be maximum when \(y\) is minimum and minimum value of \(y\) is:

30 + 31 + 32 + ………….+ 49 (20 terms)

\( = \frac{{30 + 49}}{2} \times 20 = 790\)

Hence \(5x + 790 = 1250\)

\( \Rightarrow x = 92\)

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Question 08: One part of a hostel's monthly expenses is fixed, and the other part is proportional to the number of its boarders. The hostel collects ₹ 1600 per month from each boarder. When the number of boarders is 50, the profit of the hostel is ₹ 200 per boarder, and when the number of boarders is 75, the profit of the hostel is ₹ 250 per boarder. When the number of boarders is 80, the total profit of the hostel, in INR, will be

(1) 20800
(2) 20200
(3) 20500
(4) 20000

20500

Suppose the fixed part is x and the variable part is y.

When there are 50 boarders, expenses of the hostel per boarder is 1600 – 200 = ₹1400.

50×1400 = x + 50y

When there are 75 boarders, expenses of the hostel per boarder is 1600 – 250 = ₹1350.

75×1350 = x + 75y

Solving these equations, we get, y = 1250 and x = 7500

Total expenses when there are 80 boarders is:

7500 + 80×1250 = 107500

Total collection = 80×1600 = 128000

Total profit of the hostel = 128000 – 107500 = ₹20500

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Question 09: One day, Rahul started a work at 9 AM and Gautam joined him two hours later. They then worked together and completed the work at 5 PM the same day. If both had started at 9 AM and worked together, the work would have been completed 30 minutes earlier. Working alone, the time Rahul would have taken, in hours, to complete the work is

(1) 12
(2) 11.5
(3) 12.5
(4) 10

Suppose one hour’s work of Rahul and Gautam are r and g, then the work is finished when Rahul works for 8 hours and Gautam works for 6 hours. In second case Rahul and Gautam could have finished the work working 7.5 hours together. Thus

Work = 8r + 6g = 7.5r + 7.5g

\( \Rightarrow \)r = 3g

Total work = 8(3g) + 6g = 30g

Time taken by Rahul to complete the work alone = \(\frac{{30{\rm{g}}}}{{3{\rm{g}}}} = 10\)

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Question 10: The total of male and female populations in a city increased by 25% from 1970 to 1980. During the same period, the male population increased by 40% while the female population increased by 20%. From 1980 to 1990, the female population increased by 25%. In 1990, if the female population is twice the male population, then the percentage increase in the total of male and female populations in the city from 1970 to 1990 is

(1) 68.75
(2) 68.50
(3) 68.25
(4) 69.25

68.75

Using allegation method, let us first find the ratio of Male and Female population.

The ratio is (25 – 20) : (40 – 25) = 1 : 3. Suppose males and females populations are 100 and 300 in 1970, then in 1980, these populations increases by 40% and 20%, so the populations in 1980 are 140 and 360.

In 1990 the female population increased by 25% and it is double of male population, hence

In 1990 female population = 360 + 90 = 450 and male population = 225

Total population in 1990 = 450 + 225 = 675.

Percentage increase from 1970 = \(\frac{{675 - 400}}{{400}} \times 100 = \frac{{275}}{4} = 68.75\% \)

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Question 11: Let ABCD be a parallelogram. The lengths of the side AD and the diagonal AC are 10 cm and 20 cm, respectively. If the angle ∠ADC is equal to 30° then the area of the parallelogram, in sq. cm, is

(1) \(\frac{{25\left( {\sqrt 3 + \sqrt {15} } \right)}}{2}\)
(2) \(25(\sqrt 5 + \sqrt {15} )\)
(3) \(\frac{{25(\sqrt 5 + \sqrt {15} )}}{2}\)
(4) \(25(\sqrt 3 + \sqrt {15} )\)

\(25(\sqrt 3 + \sqrt {15} )\)

Using sin 30 and cos 30 in the triangles ADE, we have

\(\sin 30 = \frac{{AE}}{{10}} \Rightarrow AE = 5\), \(\cos 30 = \frac{{DE}}{{10}} \Rightarrow DE = 5\sqrt 3 \)

And \(E{C^2} = {20^2} - {5^2} = 375\) or \(EC = \sqrt {375} \)

Now area of the parallelogram = base×Height

\(\;\;\;\;\;\;\;\; = (DE + EC) \times AE = \left( {5\sqrt 3 + \sqrt {375} } \right) \times 5\)

\(\;\;\;\;\;\;\;\; = 25\left( {\sqrt 3 + \sqrt {15} } \right)\)

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Question 12: A park is shaped like a rhombus and has area 96 sq m. If 40 m of fencing is needed to enclose the park, the cost, in INR, of laying electric wires along its two diagonals, at the rate of ₹125 per m, is

3500

Side of the rhombus = 10

Suppose the diagonals are \({d_1},\;{d_2}\), then

\(\frac{1}{2}{d_1}{d_2} = 96 \Rightarrow {d_1}{d_2} = 192\)

Also \({\left( {\frac{{{d_1}}}{2}} \right)^2} + {\left( {\frac{{{d_2}}}{2}} \right)^2} = 100\) or \(d_1^2 + d_2^2 = 400\)

Now \({\left( {{d_1} + {d_2}} \right)^2} = d_1^2 + d_2^2 + 2{d_1}{d_2}\)

\(\;\;\;\; = 400 + 2 \times 192 = 784\)

Hence sum of the lengths of the diagonals = \(\sqrt {784} = 28\)

The cost of laying electric wires = 125×28 = ₹3500

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Question 13: If \(3x + 2|y| + y = 7\) and \(x + |x| + 3y = 1\), then \(x + 2y\) is

(1) 0
(2) 1
(3) \( - \frac{4}{3}\)
(4) \(\frac{8}{3}\)

First equation can be written in two forms depending upon the sign of \(y\).

When \(y\) is positive, \(3x + 3y = 7\)

When \(y\) is negative, \(3x - 2y + y = 7\) or \(3x - y = 7\)

Second equation can be written in two forms depending upon the sign of \(x\).

When \(x\) is positive, \(2x + 3y = 1\)

When \(x\) is negative, \(x - x + 3y = 1\) or \(3y = 1\)

The intersection point can be found by solving \(2x + 3y = 1\) and \(3x - y = 7\).

At the intersection point \(x = 2,\;y = - 1\), so \(x + 2y = 0\)

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Question 14: In a tournament, a team has played 40 matches so far and won 30% of them. If they win 60% of the remaining matches, their overall win percentage will be 50%. Suppose they win 90% of the remaining matches, then the total number of matches won by the team in the tournament will be

(1) 86
(2) 84
(3) 78
(4) 80

Using allegation method, we can find the ratio of the matches in which 30% and 60% are won. The ratio is 1 : 2. Since it is given that team has played 40 matches so far and won 30% of them, so the number of matches in which team has won 60% is 80.

Now the team has won 90% of the 80 matches = 72. Total wins = 30% of 40 + 72 = 84

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Question 15: A shop owner bought a total of 64 shirts from a wholesale market that came in two sizes, small and large. The price of a small shirt was INR 50 less than that of a large shirt. She paid a total of INR 5000 for the large shirts, and a total of INR 1800 for the small shirts. Then, the price of a large shirt and a small shirt together, in INR, is

(1) 150
(2) 225
(3) 175
(4) 200

200

Suppose the number of large and small shirts are x and (64 – x) and their prices are p and

(p – 50), then

xp = 5000

(64 – x)(p – 50) = 1800

or 64p + 50x – 3200 – xp = 1800

or 64p + 50x – 3200 – 5000 = 1800 (as xp = 5000)

or 64p + 50(5000/p) = 1800 + 5000

or 64p² – 10000p + 250000 = 0

or 4p² – 625p + 25×625 = 0

or (4p – 125)(p – 125) = 0

As p > 50, so p = 125.

Price of the smaller shirt = 125 – 50 = 75.

Sum of the prices = 125 + 75 = 200

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Question 16: Consider a sequence of real numbers \({x_1},{x_2},{x_3},{\rm{ }}{{\rm{x}}_{\rm{n}}}\) such that \({x_{n + 1}} = {x_n} + n - 1\) for all \(n \ge 1\). If \({x_1} = - 1\) then \({x_{100}}\) is equal to

(1) 4949
(2) 4849
(3) 4850
(4) 4950

4850

\({x_2} = {x_1} + 1 - 1\)

\({x_3} = {x_2} + 2 - 1\)

\({x_4} = {x_3} + 3 - 1\)

\({x_{100}} = {x_{99}} + 99 - 1\)

Adding all the equations, we have

\({x_{100}} = {x_1} + (1 + 2 + 3 + ... + 99) - 99\)

\( \Rightarrow {x_{100}} = {x_1} + (1 + 2 + 3 + ... + 98)\)

\( = - 1 + \frac{{98 \times 99}}{2} = 4850\)

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Question 17: A tea shop offers tea in cups of three different sizes. The product of the prices, in INR, of three different sizes is equal to 800. The prices of the smallest size and the medium size are in the ratio 2 : 5. If the shop owner decides to increase the prices of the smallest and the medium ones by INR 6 keeping the price of the largest size unchanged, the product then changes to 3200. The sum of the original prices of three different sizes, in INR, is

Suppose prices of the small and medium size cups are \(2k\) and\(5k\), and price of the largest sized cup is \(x\), then

\((2k)(5k)x = 800\)

Now as per second condition,

\((2k + 6)(5k + 6)x = 3200\)

Dividing first equation by the second equation, we get

\(\frac{{10{k^2}}}{{2(5{k^2} + 21k + 18)}} = \frac{1}{4}\)

\( \Rightarrow 5{k^2} - 7k - 6 = 0\)

\( \Rightarrow (5k + 3)(k - 2) = 0\)

Hence \(k = 2\) and the prices are 4, 10 and 20.

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Question 18: The number of distinct pairs of integers \(\left( {m,{\rm{ }}n} \right)\) satisfying \(|1 + mn|\, < \,\,|m + n|\, < \,5\) is

We know that if \(|x|\, < \,|y|\), then \({x^2} < {y^2}\)

Let us solve the inequality \(|1 + mn|\, < \,|m + n|\)

Squaring both the sides, we get

\(1 + {m^2}{n^2} + 2mn < {m^2} + {n^2} + 2mn\)

\( \Rightarrow {m^2}{n^2} - {m^2} - {n^2} + 1 < 0\)

\( \Rightarrow ({m^2} - 1)({n^2} - 1) < 0\)

Since \(m,\;n\) are integers, so for non-zero values of \(m,\;n\), the values of both \(({m^2} - 1)\) and \(({n^2} - 1)\) will be more than or equal to zero. Therefore for non-zero values of \(m,\;n\) the expression \(({m^2} - 1)({n^2} - 1)\) can never be negative, so either \(m\) or \(n\) has to be 0, both cannot be zero at the same time.

Case 1: \(m = 0\), then \({n^2} - 1 > 0\) and \(|m + n|\, < 5\), so \(n = \pm 2,\,\, \pm 3,\,\, \pm 4\,\)

Case 2: \(n = 0\), then \({m^2} - 1 > 0\) and \(|m + n|\, < 5\), so \(m = \pm 2,\,\, \pm 3,\,\, \pm 4\,\).
A total of 12 solutions.

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Question 19: If \(n\) is a positive integer such that \(\left( {\sqrt[7]{{10}}} \right){\left( {\sqrt[7]{{10}}} \right)^2}{\left( {\sqrt[7]{{10}}} \right)^3}....{\left( {\sqrt[7]{{10}}} \right)^n} > 999\), then the smallest value of \(n\) is

The number is \({10^{\frac{1}{7}}}{10^{\frac{2}{7}}}{10^{\frac{3}{7}}}{.........10^{\frac{n}{7}}} > 999\)

\( \Rightarrow {10^{\frac{{1 + 2 + 3 + ... + n}}{7}}} > 999\)

\( \Rightarrow {10^{\frac{{n(n + 1)}}{{14}}}} > 999\)

Since 1000 > 999, hence \(\frac{{n(n + 1)}}{{14}} \ge 3\)

Or \(n(n + 1) \ge 42 \Rightarrow n \ge 6\)

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Question 20: If \(f\left( x \right) = {x^2} - 7x\) and\(g\left( x \right) = x + 3\), then the minimum value of \(f\left( {g\left( x \right)} \right) - 3x\) is

(1) – 20
(2) – 15
(3) – 12
(4) – 16

– 16

\(f(g(x)) = {\left( {g(x)} \right)^2} - 7g(x)\)

\(\;\;\;\;\;\;\; = {(x + 3)^2} - 7(x + 3)\)

\(\;\;\;\;\;\;\; = {x^2} - x - 12\)

\(\Rightarrow f\left( {g\left( x \right)} \right) - 3x = {x^2} - 4x - 12\)

\(\;\;\;\;\;\;\;\;\; = {(x - 2)^2} - 16\)

Minimum value = – 16

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Question 21: The cost of fencing a rectangular plot is ₹ 200 per ft along one side, and ₹ 100 per ft along the three other sides. If the area of the rectangular plot is 60000 sq. ft, then the lowest possible cost of fencing all four sides, in INR, is

(1) 90000
(2) 160000
(3) 120000
(4) 100000

120000

Suppose length and breadth of the rectangle are \(\ell \) and \(b\), then cost of fencing all four sides will be

\(200\ell + 100(b + b + \ell ) = 300\ell + 200b\)

Also area of the rectangle \(\ell b = 60000\)

Using \(AM \ge GM\), we have

\(\frac{{300\ell + 200b}}{2} \ge \sqrt {300\ell \cdot 200b} \)

\( \Rightarrow 300\ell + 200b \ge 2\sqrt {300 \cdot 200 \cdot \ell b} \)

Putting the value of \(\ell b = 60000\), we get the minimum value of \(300\ell + 200b\) as 120000

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Question 22: Bank A offers 6% interest rate per annum compounded half yearly. Bank B and Bank C offer simple interest but the annual interest rate offered by Bank C is twice that of Bank B. Raju invests a certain amount in Bank B for a certain period and Rupa invests ₹ 10,000 in Bank C for twice that period. The interest that would accrue to Raju during that period is equal to the interest that would have accrued had he invested the same amount in Bank A for one year. The interest accrued, in INR, to Rupa is:

(1) 2436
(2) 3436
(3) 2346
(4) 1436

2436

In Bank C, the rate is twice that of Bank B and Rupa invested in Bank C for twice the period that Raju invested, so in order to get the same interest, Raju must have invested 4 times the money Rupa invested.

Therefore, Raju Invested Rs 40000.

Interest generated in bank A if he has invested in it,
\( = 40000{(1.03)^2} - 40000\)

= 40000×0.0609 = 2436

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