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

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

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

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

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

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

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 = $\frac{18}{90}\times 100=20%$

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.

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

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 = $4\times \frac{1}{2}=2$

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:

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

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.

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.

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)

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

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)

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

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

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.

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.

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

We have ,

Taking log both the side we get

Also, we have been given

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

Taking log both the side

From (1) and (2)

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

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

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

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

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 = $\frac{12\pi }{3/4}=16\pi \ km/h$

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 $\Delta APB$

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

Similarly, in Right $\Delta AQB$

Case 1: m is even.

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

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.

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 =$\left( \frac{4}{4+1} \right)\times 100=80%$

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

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

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

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

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,

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.

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.

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

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

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

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

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

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 $\left( \frac{\left( 11/3 \right)t-3t}{3t} \right)\times 100=22.22%$

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 = $\frac{a}{c}=\frac{\sqrt{3}}{3}=\frac{1}{\sqrt{3}}$

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.

Let the amount invested in fixed deposit be x lakhs.

As per the condition:

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

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

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

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

$\Rightarrow x=9$

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 = $\frac{245}{350}\times 100=70%$

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

Therefore,

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

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

$\Rightarrow x=880$

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.

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

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

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