# CAT 2019 Quant Question Paper with Solution (Slot -1)

#### [SLOT 1]

Question 1:
If m and n are integers such that $(\sqrt{2})^{19} 3^{4} 4^{2} 9^{m} 8^{n}=3^{n} 16^{m}(\sqrt[4]{64})$ then m is
1. $-20$
2. $-12$
3. $-24$
4. $-16$

$(\sqrt{2})^{19} 3^{4} 4^{2} 9^{m} 8^{n}=3^{n} 16^{m}(\sqrt[4]{64})$

$\Rightarrow 2^{19 / 2} \times 3^{4} \times 2^{4} \times 3^{2 m} \times 2^{3 n}=3^{n} \times 2^{4 m} \times 2^{3 / 2}$

$\Rightarrow {{2}^{(19/2+4+3n)}}\times {{3}^{(4+2m)}}={{2}^{(4m+3)}}\times {{3}^{n}}$

Comparing the powers of same bases we get

$\frac{19}{2}+4+3 n=4 m+\frac{3}{2} \cdots(1)$

$4+2 m=n \cdots(2)$

Substitute the value of n from (2) in (1) and solving for m, we get m = -12

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Question 2:
The income of Amala is 20% more than that of Bimala and 20% less than that of Kamala. If Kamala's income goes down by 4% and Bimala's goes up by 10%, then the percentage by which Kamala's income would exceed Bimala's is nearest to
1. 31
2. 28
3. 32
4. 29

Let the income of Bimla be Rs. 100

Amla’s income = 1.2*100 = Rs. 120

Also, Amla’s income = $\frac{4}{5}\times$ Kamla’s income

Therefore, Kamla’s income = $\frac{5}{4}\times 120$=Rs. 150

Kamala's new income with 4% decrease = 0.96×150 =Rs. 144

Bimla’s new income with 10% increase = 1.1×100= Rs. 110

Required percentage increase = $\left( \frac{144-110}{110} \right)\times 100\approx 31%$

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Question 3:
In a class, 60% of the students are girls and the rest are boys. There are 30 more girls than boys. If 68% of the students, including 30 boys, pass an examination, the percentage of the girls who do not pass is

Let number of girls and boys be 3x and 2x respectively.

Given,

3x-2x=30

Or x = 30. So the number of students = 3x+2x=5x =150.

Also, number of girls (3x) and boys (2x) = 90 and 60 respectively.

Number of students pass the exam = 68% of 150=102.
Number of girls pass the exam = 102-30 = 72 .

Therefore, number of girls who fail in the exam = 90-72=18

Hence, percentage of girls fail the exam = $\frac{18}{90}\times 100=20%$

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Question 4:
On selling a pen at 5% loss and a book at 15% gain, Karim gains Rs. 7. If he sells the pen at 5% gain and the book at 10% gain, he gains Rs. 13. What is the cost price of the book in Rupees?
1. 80
2. 85
3. 95
4. 100

Let the cost prices of a pen and a books be x and y respectively.

From 1st condition,

0.95x + 1.15y = x+y+7 ... (1)

From 2nd condition,

1.05x + 1.1y = x + y +13 ... (2)

Solving both (1) and (2) for y, we get y=80.

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Question 5:
Corners are cut off from an equilateral triangle T to produce a regular hexagon H. Then, the ratio of the area of H to the area of T is
1. 5: 6
2. 4: 5
3. 3: 4
4. 2: 3

Refer to the below figure:

As we can see, the triangle can be divided into 9 concurrent smaller equilateral triangles.
Also, hexagon contains 6 of them.

Hence the required ration = 6/9 =2/3

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Question 6:
Let S be the set of all points (x, y) in the x-y plane such that | x | + | y | ≤ 2 and | x | ≥ 1. Then, the area, in square units, of the region represented by S equals

The below figure shows the graph of both the expressions.

The required area = $4 \times$ the area of red triangle.

Area of red triangle =$\frac{1}{2}\times 1\times 1=\frac{1}{2}$ sq. units

Therefore, the required area = $4\times \frac{1}{2}=2$

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Question 7:
Ramesh and Gautam are among 22 students who write an examination. Ramesh scores 82.5. The average score of the 21 students other than Gautam is 62. The average score of all the 22 students is one more than the average score of the 21 students other than Ramesh. The score of Gautam is
1. 49
2. 48
3. 51
4. 53

Let Gautam’s score be g. Also, let x be the average of all 22 students so that the total scores of all 22 students =22x.

From the first condition:

22x-g = 21×62 … (1)

From the second condition:

\begin{align} & 22x\text{ }-82.5\text{ }=\text{ }21\left( x-1 \right) \\ & \Rightarrow 22x-21x=82.5-21 \\ & \Rightarrow x=61.5 \\ \end{align}

Putting x=61.5 in (1), we get g = 51.

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Question 8:
At their usual efficiency levels, A and B together finish a task in 12 days. If A had worked half as efficiently as she usually does, and B had worked thrice as efficiently as he usually does, the task would have been completed in 9 days. How many days would A take to finish the task if she works alone at her usual efficiency?
1. 24
2. 18
3. 12
4. 36

Let the work be LCM of (9 and 12) = 36 units.

Let the amount of work done in one day with their normal efficiencies by A and B be x and y units respectively.

Therefore, (x+y)×12=36

Or x+y =3 … (1)

Similarly,

(x/2+3y)×9=36

Or x/2 +3y = 4 … (2)

Solving (1) and (2) for x, we get x =2 units

Hence, A alone would take 36/x = 36/2 = 18 days to complete the work with her normal efficiency.

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Question 9:
If $a_{1}, a_{2}, \ldots$ are in A.P., then, $\frac{1}{\sqrt{a_{1}}+\sqrt{a_{2}}}+\frac{1}{\sqrt{a_{2}}+\sqrt{a_{3}}}+\cdots+\frac{1}{\sqrt{a_{n}}+\sqrt{a_{n+1}}}$ is equal to
1. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n-1}}}$
2. $\frac{n}{\sqrt{a_{1}}+\sqrt{a_{n+1}}}$
3. $\frac{n-1}{\sqrt{a_{1}}+\sqrt{a_{n}}}$
4. $\frac{n}{\sqrt{a_{1}}-\sqrt{a_{n+1}}}$

Shortcut:

For such questions, we can take value of n =1. The right option must give the first term i.e. $\frac{1}{\sqrt{{{a}_{1}}}+\sqrt{{{a}_{2}}}}$

Only option (2) satisfies.

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Question 10:
In a circle of radius 11 cm, CD is a diameter and AB is a chord of length 20.5 cm. If AB and CD intersect at a point E inside the circle and CE has length 7 cm, then the difference of the lengths of BE and AE, in cm, is
1. 2.5
2. 3.5
3. 0.5
4. 1.5

Refer to the diagram below:

Applying chord chord power theorem

$AE\times BE=CE\times DE$

$\Rightarrow AE\times BE=7\times 15=105$ …(1)

Also, it is given that AE+BE=20.5 ...(2)

${{\left( AE-BE \right)}^{2}}={{\left( AE+BE \right)}^{2}}-4\times AE\times BE$

\begin{align} & \Rightarrow {{\left( AE-BE \right)}^{2}}={{\left( 20.5 \right)}^{2}}-4\times 105 \\ & \Rightarrow {{\left( AE-BE \right)}^{2}}=420.25-420=0.25 \\ & \Rightarrow \left( AE-BE \right)=0.5 \\ \end{align}Online CAT 2022 Quant Course

Question 11:
With rectangular axes of coordinates, the number of paths from (1,1) to (8,10) via (4,6), where each step from any point (x, y) is either to (x, y+1) or to (x+1, y), is

Let A, B and C represent the coordinates (1, 1), (8,10) and (4,6) respectively.

The number of ways to go from A to B via C = (The number of ways to go from A to C)×(The number of ways to go from C to B)

$={}^{8}{{C}_{5}}\times {}^{8}{{C}_{4}}=3920$

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Question 12:
Amala, Bina, and Gouri invest money in the ratio 3 : 4 : 5 in fixed deposits having respective annual interest rates in the ratio 6 : 5 : 4. What is their total interest income (in Rs) after a year, if Bina's interest income exceeds Amala's by Rs 250?
1. 6350
2. 7250
3. 7000
4. 6000

Ratio of their incomes = 3:4:5

Ratio of their interests = 6:5:4

Therefore, the ratio of their interest income = $\left( 3\times 6 \right):\left( 4\times 5 \right):\left( 5\times 4 \right)=18:20:20$

Let the interest incomes of Amala, Bina, and Gouri be 18x, 20x, and 20x respectively.

Given,

Bina's interest income exceeds Amala's by Rs 250

Therefore, 20x-18x=2x=250

Or x = 125.

Total interest incomes = 18x+20x+20x=58x = 58×125 =7250

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Question 13:
A club has 256 members of whom 144 can play football, 123 can play tennis, and 132 can play cricket. Moreover, 58 members can play both football and tennis, 25 can play both cricket and tennis, while 63 can play both football and cricket. If every member can play at least one game, then the number of members who can play only tennis is
1. 45
2. 38
3. 32
4. 43

Let the number of members playing all three games be x.

Given, that all the members play at least one of these three games, hence the union of these three sets = 256.

Therefore,

256 =144+123+132)-(58+25+63)+x

Or x = 3.

Fitting the numbers in the venn diagram, we get

Clearly, the number of members playing only tennis = 43.

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Question 14:
Let T be the triangle formed by the straight line 3x + 5y - 45 = 0 and the coordinate axes. Let the circumcircle of T have radius of length L, measured in the same unit as the coordinate axes. Then, the integer closest to L is

Clearly, the triangle will be right angled triangle and the hypotenuse would be the diameter of the circumcircle.

Getting the coordinates by substituting x and y with 0 alternatively in the equation 3x+5y-45=0,  we have

Hypotenuse = $\sqrt{{{9}^{2}}+{{15}^{2}}}=\sqrt{306}\approx 17.5$

Therefore, the radius =$\frac{17.5}{2}\approx 8.7$

Hence, the closest integer = 9.

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Question 15:
Three men and eight machines can finish a job in half the time taken by three machines and eight men to finish the same job. If two machines can finish the job in 13 days, then how many men can finish the job in 13 days?

Let one machine completes 1 unit of work per day.

Given, two machines can finish the job in 13 days

Therefore, the work of 2×1×13 = 26 units.

Also, let on man completes m units of work per day.

From the given condition:

3m+8×1 = 2(8m+3×1)

Or m = $\frac{2}{13}units$

Let it require ‘x’ number of men to complete the work in 13 days.

Therefore, xm×13 =26 units

Or x = 13 men

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Question 16:
Two cars travel the same distance starting at 10:00 am and 11:00 am, respectively, on the same day. They reach their common destination at the same point of time. If the first car travelled for at least 6 hours, then the highest possible value of the percentage by which the speed of the second car could exceed that of the first car is
1. 30
2. 25
3. 10
4. 20

Question 17:
If $(5.55)^{x}=(0.555)^{y}=1000,$ then the value of $\frac{1}{x}-\frac{1}{y}$ is
1. 3
2. 1
3. $\frac{1}{3}$
4. $\frac{2}{3}$

We have ,

\begin{align} & {{(5.55)}^{x}}=1000 \\ & \Rightarrow {{(5.55)}^{x}}={{10}^{3}} \\ \end{align}

Taking log both the side we get

\begin{align} & x{{\log }_{10}}(5.55)=3 \\ & \Rightarrow {{\log }_{10}}(5.55)=\frac{3}{x} \\ & \Rightarrow {{\log }_{10}}(10\times 0.555)=\frac{3}{x} \\ & \Rightarrow {{\log }_{10}}(0.555)+1=\frac{3}{x}...(1) \\ \end{align}

Also, we have been given

${{(0.555)}^{y}}=1000$

Taking log both the side

\begin{align} & y{{\log }_{10}}(0.555)=3 \\ & \Rightarrow {{\log }_{10}}(5.55)=\frac{3}{y}...(2) \\ \end{align}

From (1) and (2)

\begin{align} & \frac{3}{y}+1=\frac{3}{x} \\ & \Rightarrow \frac{1}{x}-\frac{1}{y}=\frac{1}{3} \\ \end{align}Online CAT 2022 Quant Course

Question 18:
The product of the distinct roots of $\left|x^{2}-x-6\right|=x+2$ is
1. $-8$
2. $-24$
3. $-4$
4. $-16$

${{x}^{2}}-x-6=(x+2)(x-3)$

Case 1:  ${{x}^{2}}-x-6<0$

i.e. (x+2)(x-3)<0

$\Rightarrow -2<x<3$ and $\left| {{x}^{2}}-x-6 \right|=-\left( {{x}^{2}}-x-6 \right)$

Therefore, $\left|x^{2}-x-6\right|=x+2$

\begin{align} & =-(x+2)(x-3)=x+2 \\ & \Rightarrow (x-3)=-1\Rightarrow x=2 \\ \end{align}

Case 2:${{x}^{2}}-x-6\ge 0$

i.e. $\left( x+2 \right)\left( x-3 \right)\ge 0$

$\Rightarrow x\le -2\ or\ x\ge 3$

Checking for boundary conditions:

For x=-2, $\left|x^{2}-x-6\right|=x+2$, therefore, x=-2 is also the root. But for x=3, $\left| {{x}^{2}}-x-6 \right|\ne x+2$.

Hence x=3 is NOT the root.

And for the interval$x<-2\ or\ x>3$ the expression$\left| {{x}^{2}}-x-6 \right|={{x}^{2}}-x-6$

Therefore, $\left|x^{2}-x-6\right|=x+2$

\begin{align} & =(x+2)(x-3)=x+2 \\ & \Rightarrow (x-3)=1\Rightarrow x=4 \\ \end{align}

Therefore, the root are -2, 2, and 4. So the required product = (2)(-2)(4)=-16

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Question 19:
The wheels of bicycles A and B have radii 30 cm and 40 cm, respectively. While traveling a certain distance, each wheel of A required 5000 more revolutions than each wheel of B. If bicycle B traveled this distance in 45 minutes, then its speed, in km per hour, was
1. $18 \pi$
2. $12 \pi$
3. $16 \pi$
4. $14 \pi$

The distance travelled by bicycle A in one revolution = $2\pi {{r}_{a}}=2\pi \times 30=60\pi \ cm$

The distance travelled by bicycle B in one revolution = $2\pi {{r}_{b}}=2\pi \times 40=80\pi \ cm$

Let B makes ‘n’ revolutions to cover the distance. Then, A would make (n+5000) to cover the same distance.

$\therefore n\times 80\pi =(n+5000)\times 60\pi$ $\Rightarrow n=15000$

Distance travelled by B = $n\times 80\pi \ cm=\frac{15000\times 80\pi }{{{10}^{5}}}\ km=12\ km$

Time taken by B= 45 min = $\frac{45}{60}=\frac{3}{4}hrs$

Hence the speed of B = $\frac{12\pi }{3/4}=16\pi \ km/h$

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Question 20:
AB is a diameter of a circle of radius 5 cm. Let P and Q be two points on the circle so that the length of PB is 6 cm, and the length of AP is twice that of AQ. Then the length, in cm, of QB is nearest to
1. 7.8
2. 8.5
3. 9.1
4. 9.3

Refer to the figure below:

$\angle APB=\angle AQB={{90}^{0}}$ {angle in a semicircle is a right angle}

Also, let AQ=x, so AP=2x

In Right $\Delta APB$

$A{{P}^{2}}=A{{B}^{2}}-B{{P}^{2}}$

$A{{P}^{2}}=A{{B}^{2}}-B{{P}^{2}}$ \begin{align} & \Rightarrow A{{P}^{2}}={{10}^{2}}-{{6}^{2}}={{8}^{2}} \\ & \Rightarrow AP=8\Rightarrow 2x=8 \\ & \Rightarrow x=4 \\ \end{align}

Similarly, in Right $\Delta AQB$

\begin{align} & B{{Q}^{2}}=A{{B}^{2}}-A{{Q}^{2}} \\ & \Rightarrow B{{Q}^{2}}={{10}^{2}}-{{4}^{2}}=84 \\ & \Rightarrow BQ=\sqrt{84}\approx 9.1 \\ \end{align}Online CAT 2022 Quant Course

Question 21:
For any positive integer n, let f(n) = n(n + 1) if n is even, and f(n) = n + 3 if n is odd. If m is a positive integer such that 8f(m + 1) - f(m) = 2, then m equals

Case 1: m is even.

Given, 8f(m + 1) - f(m) = 2

\begin{align} & \Rightarrow 8\left( m+1+3 \right)-m\left( m+1 \right)=2 \\ & \Rightarrow 8m+32-{{m}^{2}}-m=2 \\ & \Rightarrow {{m}^{2}}-7m+30=0 \\ & \Rightarrow (m-10)(m+3)=0 \\ & \Rightarrow m=10\ or\ -3 \\ \end{align}

As m is positive integer, the only possible value of m =10.

Case 2:

If m is odd, then we would not be getting positive solution.

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Question 22:
A chemist mixes two liquids 1 and 2. One litre of liquid 1 weighs 1 kg and one litre of liquid 2 weighs 800 gm. If half litre of the mixture weighs 480 gm, then the percentage of liquid 1 in the mixture, in terms of volume, is
1. 85
2. 70
3. 75
4. 80

Weight of liquid 1 per litre = 1000 gm

Weight of liquid 2 per litre = 800 gm

Weight of mixture per litre = 2×480=960 gm

Applying alligation rule

$\frac{Quantity\ of\ liquid\ 1}{Quantity\ of\ liquid\ 2}=\frac{960-800}{1000-960}=\frac{4}{1}$

Therefore, the liquids are mixed in 4:1.

Hence, the percentage of liquid 1 =$\left( \frac{4}{4+1} \right)\times 100=80%$

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Question 23:
If $a_{1}+a_{2}+a_{3}+\ldots+a_{n}=3\left(2^{n+1}-2\right),$ for every $n \geq 1,$ then $a_{11}$ equals

If n =1, ${{a}_{1}}=3\left( {{2}^{1+1}}-2 \right)=6=3\times {{2}^{1}}$

If n=2, ${{a}_{1}}+{{a}_{2}}=3\left( {{2}^{2+1}}-2 \right)=18$$\Rightarrow {{a}_{2}}=18-{{a}_{1}}=12=3\times {{2}^{2}} If n=3, {{a}_{1}}+{{a}_{2}}++{{a}_{3}}=3\left( {{2}^{3+1}}-2 \right)=42$$\Rightarrow {{a}_{3}}=42-\left( {{a}_{1}}+{{a}_{2}} \right)=24=3\times {{2}^{3}}$

Following the pattern, ${{a}_{n}}=3\times {{2}^{n}}$

Therefore, ${{a}_{11}}=3\times {{2}^{11}}=6144$

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Question 24:
Consider a function $f$ satisfying $f(x+y)=f(x) f(y)$ where x, y are positive integers, and $f(1)=2$. If $f(a+1)+f(a+2)+\ldots+f(a+n)=16\left(2^{n}-1\right)$ then a is equal to

$f(a+1)+f(a+2)+\ldots+f(a+n)=16\left(2^{n}-1\right)$

$\Rightarrow f(a)f(1)+f(a)f(2)+...+f(a)f(n)=16\left( {{2}^{n}}-1 \right)$

$\Rightarrow f(a)\left( f(1)+f(2)+...+f(n) \right)=16\left( {{2}^{n}}-1 \right)$

Take n=1,

$\Rightarrow f(a)f(1)=16\left( {{2}^{1}}-1 \right)=16$

$\Rightarrow f(a)\times 2=16\Rightarrow f(a)=8$

Therefore,

\begin{align} & f(a)\left( f(1)+f(2)+...+f(n) \right)=16\left( {{2}^{n}}-1 \right) \\ & \Rightarrow f(1)+f(2)+...+f(n)=2\left( {{2}^{n}}-1 \right) \\ \end{align}

If n=2, then $f(1)+f(2)=2\left( {{2}^{2}}-1 \right)=6$

$\Rightarrow f(2)=6-f(1)=6-2=4$

If n=3, then $f(1)+f(2)+f(3)=2\left( {{2}^{3}}-1 \right)=14$

$\Rightarrow f(3)=6-f(1)-f(2)=14-2-4=8=f(a)$

Hence a =3.

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Question 25:
The number of the real roots of the equation $2 \cos (x(x+1))=2^{x}+2^{-x}$ is
1. 2
2. 1
3. infinite
4. 0

For any real value of x, the expression $2 \cos (x(x+1))=2^{x}+2^{-x}$ would always be positive.

Lets find the maximum value of $2 \cos (x(x+1))=2^{x}+2^{-x}$.

Applying AM-GM inequality we have

$\frac{{{2}^{x}}+{{2}^{-x}}}{2}\ge \sqrt{{{2}^{x}}\times {{2}^{-x}}}$

$\Rightarrow {{2}^{x}}+{{2}^{-x}}\ge 2\sqrt{{{2}^{0}}}$

$\Rightarrow {{2}^{x}}+{{2}^{-x}}\ge 2$

Therefore, $2\cos \left( x\left( x+1 \right) \right)\ge 2$

It is known that $-1\le \cos \theta \le 1$

$\Rightarrow 2\cos \left( x\left( x+1 \right) \right)=2$

Hence, the expression is valid only if ${{2}^{x}}+{{2}^{-x}}=2$, which is true for only one value of x i.e. 0.

Therefore, the expression has only one real solution.

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Question 26:
Let $x$ and $y$ be positive real numbers such that $\log _{5}(x+y)+\log _{5}(x-y)=3,$ and $\log _{2} y-\log _{2} x=1-\log _{2} 3 .$ Then xy equals
1. 250
2. 25
3. 100
4. 150

${{\log }_{5}}(x+y)+{{\log }_{5}}(x-y)=3$

$\Rightarrow {{\log }_{5}}\left[ \left( x+y \right)\left( x-y \right) \right]=3$

$\Rightarrow \left( x+y \right)\left( x-y \right)={{5}^{3}}=125$

$\Rightarrow {{x}^{2}}-{{y}^{2}}=125...(1)$

And $\log _{2} y-\log _{2} x=1-\log _{2} 3 .$

$\Rightarrow \log _{2}\left(\frac{y}{x}\right)=\log _{2} 2-\log _{2} 3$

$\Rightarrow \log _{2}\left(\frac{y}{x}\right)=\log _{2}\left(\frac{2}{3}\right)$

$\Rightarrow \frac{y}{x}=\frac{2}{3}$

Let $x = 3k$ and $y=2k$. Putting the values in (1)

$(3k)^{2}-(2k)^{2}=125$

$\Rightarrow 5k^{2}=125$

$\Rightarrow k=5$

Hence $x \times y = 3k \times 2k =6 \times 25 = 150$

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Question 27:
One can use three different transports which move at 10, 20, and 30 kmph, respectively. To reach from A to B, Amal took each mode of transport 1/3 of his total journey time, while Bimal took each mode of transport 1/3 of the total distance. The percentage by which Bimal’s travel time exceeds Amal’s travel time is nearest to
1. 21
2. 22
3. 20
4. 19

Let the total journey time taken by Amal be 3t hours.

Therefore, the total distance = 10t + 20t + 30t = 60t kms

Bimal took each mode of transport 1/3 of the total distance.

Therefore, total time taken by him = $\frac{20t}{10}+\frac{20t}{20}+\frac{20t}{30}=\frac{11t}{3}\ hours$

The percentage by which Bimal’s travel time exceeds Amal’s travel time $\left( \frac{\left( 11/3 \right)t-3t}{3t} \right)\times 100=22.22%$

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Question 28:
If the rectangular faces of a brick have their diagonals in the ratio $3: 2 \sqrt{3}: \sqrt{15},$ then the ratio of the length of the shortest edge of the brick to that of its longest edge is
1. $\sqrt{3}: 2$
2. $2: \sqrt{5}$
3. $1: \sqrt{3}$
4. $\sqrt{2}: \sqrt{3}$

Let the edges of the brick be a, b, and c such that $a<b<c$

${{a}^{2}}+{{b}^{2}}={{3}^{2}}=9....(1)$

${{a}^{2}}+{{c}^{2}}={{\left( 2\sqrt{3} \right)}^{2}}=12...(2)$

${{b}^{2}}+{{c}^{2}}={{\left( \sqrt{15} \right)}^{2}}=15...(3)$

Adding all three equations. We get

$2\left( {{a}^{2}}+{{b}^{2}}+{{c}^{2}} \right)=9+12+15=36$

${{a}^{2}}+{{b}^{2}}+{{c}^{2}}=18...(4)$

From (1) and (4), $c=3$

From (3) and (4), $a=\sqrt{3}$

Therefore, required ratio = $\frac{a}{c}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$

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Question 29:
If the population of a town is p in the beginning of any year then it becomes 3+2p in the beginning of the next year. If the population in the beginning of 2019 is 1000, then the population in the beginning of 2034 will be
1. $(997) 2^{14}+3$
2. $(1003)^{15}+6$
3. $(1003) 2^{15}-3$
4. $(997)^{15}-3$

We can transform each of the options for ‘n’ years.

$(997){{2}^{14}}+3\equiv \left( p-3 \right){{2}^{n-1}}+3$

$(1003){{2}^{15}}+6\equiv \left( p+3 \right){{2}^{n}}+6$

$(1003){{2}^{15}}-3\equiv \left( p+3 \right){{2}^{n}}-3$

${{(997)}^{15}}-3\equiv {{\left( p-3 \right)}^{n}}-3$

As per the condition, in one year, the population ‘p’ becomes ‘3+2p’

Putting the value of n =1 in each option, and checking to get 3+2p, we have

$\left( p-3 \right){{2}^{n-1}}+3\equiv 3\ne 3+2p$

$\left( p+3 \right){{2}^{n}}+6\equiv \left( p+3 \right)2+6\ne 3+2p$

$(1003){{2}^{15}}-3\equiv \left( p+3 \right)2-3=3+2p$

${{\left( p-3 \right)}^{n}}-3\equiv \left( p-3 \right)-3\ne p-6$

Hence, the right answer is option 3.

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Question 30:
A person invested a total amount of Rs 15 lakh. A part of it was invested in a fixed deposit earning 6% annual interest, and the remaining amount was invested in two other deposits in the ratio 2 : 1, earning annual interest at the rates of 4% and 3%, respectively. If the total annual interest income is Rs 76000 then the amount (in Rs lakh) invested in the fixed deposit was

Let the amount invested in fixed deposit be x lakhs.

As per the condition:

$x\times \frac{6}{100}+\frac{2}{3}\times \left( 15-x \right)\times \frac{4}{100}+\frac{1}{3}\times \left( 15-x \right)\times \frac{3}{100}=\frac{76000}{{{10}^{5}}}$

$\Rightarrow 6x-\frac{8}{3}x-x=76-\frac{8}{3}\times 15-15$

$\Rightarrow \frac{7}{3}x=\frac{76\times 3-11\times 15}{3}$

$\Rightarrow \frac{7}{3}x=\frac{63}{3}$

$\Rightarrow x=9$

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Question 31:
Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is
1. 70
2. 60
3. 75
4. 80

Let the total score of the exam be 100x.

Meena’s before review = 40% of 100x = 40x.

Her score after review = 40x + 50%of 40x = 60x.

Passing marks = 60x+35. .. (1)

Her score after increasing it by 20% of post revie score = 60x+20% of 60x = 72x

Passing marks = 72x-7. … (2)

Equating (1) and (2)

72x-7 = 60x + 35

$\Rightarrow x=3.5$

Hence, Total Marks = 100x= 350 and passing marks = 60x+35 = 245

Therefore, passing percentage = $\frac{245}{350}\times 100=70%$

Online CAT 2022 Quant Course

Question 32:
In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse?

Let the first, second, and third horses be A, B and C respectively. Also, let the length of the racecourse be x.

Therefore,

$\frac{\left( x-11 \right)}{\left( x \right)}=\frac{\left( x-90 \right)}{\left( x-80 \right)}$

$\Rightarrow {{x}^{2}}-91x+880={{x}^{2}}+90x$

$\Rightarrow x=880$

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Question 33:
The number of solutions to the equation $|x|\left(6 x^{2}+1\right)=5 x^{2}$ is

case I: x=0.

Clearly, $x=0$ satisfy the equation.

case II: x>0

$|x|\left(6 x^{2}+1\right)=5 x^{2}$

$\Rightarrow x\left(6 x^{2}+1\right)=5 x^{2}$

$\Rightarrow 6 x^{2}+1-5 x=0$

On solving the quadratic equation, we get $x=\frac{1}{2},\frac{1}{3}$ (both valid)

Case III: x<0

$|x|\left(6 x^{2}+1\right)=5 x^{2}$

$\Rightarrow \quad-x\left(6 x^{2}+1\right)=5 x^{2}$

$\Rightarrow \quad 6 x^{2}+5 x+1=0$

On solving the quadratic equation, we get $x=\frac{-1}{2}\text{ ,}\frac{-1}{3}$ (both valid)

Hence there are 5 solutions.

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Question 34:
The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is
1. 58
2. 50
3. 95
4. 85

Let the two numbers be x and y

Given,

$x \times y=616$

Also, $\frac{x^{3}-y^{3}}{(x-y)^{3}}=\frac{157}{3}$

Let $x^{3}-y^{3}=157 k$ and $(x-y)^{3}=3 k$

we know that

$(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)$

$\Rightarrow (3 k)^{3}=157 k-3 \times 616(3 k)^{1 / 3}$

$\Rightarrow 154 k=3 \times 616 \times(3 k)^{1 / 3}$

$\Rightarrow k=\frac{3 \times 616}{154} \times(3 k)^{1/3}$

$\Rightarrow k= 12 \times(3 k)^{1 / 3}$

$\Rightarrow k^{3}= 12^{3} \times 3 \times k$

$\Rightarrow k^{2}= 3 \times 12^{3}$

$\Rightarrow k= 72$

Therefore, $x-y={{(3k)}^{1/3}}={{(3\times 72)}^{1/3}}=6$

Also, ${{\left( x+y \right)}^{2}}={{\left( x-y \right)}^{2}}+4xy$

$\Rightarrow {{\left( x+y \right)}^{2}}={{6}^{2}}+3\times 616=2500$

$\Rightarrow \left( x+y \right)=50$

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