CAT 2019 Slot 2 Quant Question Paper

It is given that the average of the 30 integers = 5

Sum of the 30 integers = 30*5=150

There are exactly 20 integers whose value is less than 5.

To maximise the average of the 20 integers, we have to assign minimum value to each of the remaining 10 integers

So the sum of 10 integers = 10*6=60

The sum of the 20 integers = 150-60= 90

Average of 20 integers = $$\ \frac{\ 90}{20}$$ = 4.5

The amount with Amal at the end of 1 year = 12000*1.08+10000*1.03*1.03=23569

Interest received by Amal = 23569-22000=1569

Let the amount invested by Bimal = 100b

Interest received by Bimal = 100b*7.5*1/100=7.5b

It is given that the amount of interest received by both of them is the same 

7.5b=1569

b=209.2

Amount invested by Bimal = 100b = 20920

$$\ \frac{\ n^2+3n+4n+12}{n^2-4n+3n-12}$$

=$$\ \frac{\ n^{ }\left(n+3\right)+4\left(n+3\right)}{n^{ }\left(n-4\right)+3\left(n-4\right)}$$

=$$\ \frac{\left(\ n+4\right)\left(n+3\right)}{\left(n-4\right)\left(n+3\right)}$$

=$$\ \frac{\left(\ n+4\right)}{\left(n-4\right)}$$

=$$\ \frac{\left(\ n-4\right)+8}{\left(n-4\right)}$$

=$$\ 1+\frac{8}{\left(n-4\right)}$$ which will be maximum when n-4 =8

n=12

D is the correct answer.

$$n^2-m^2=105$$

(n-m)(n+m) = 1*105, 3*35, 5*21, 7*15, 15*7, 21*5, 35*3, 105*1.

n-m=1, n+m=105  ==> n=53, m=52

n-m=3, n+m=35 ==> n=19, m=16

n-m=5, n+m=21  ==> n=13, m=8

n-m=7, n+m=15 ==> n=11, m=4

n-m=15, n+m=7 ==> n=11, m=-4

n-m=21, n+m=5 ==> n=13, m=-8

n-m=35, n+m=3 ==> n=19, m=-16

n-m=105, n+m=1 ==> n=53, m=-52

Since only positive integer values of m and n are required. There are 4 possible solutions.

When A and B met for the first time at 10:00 AM, A covered 60% of the track.

So B must have covered 40% of the track.

It is given that A returns to P at 10:12 AM i.e A covers 40% of the track in 12 minutes

60% of the track in 18 minutes

B covers 40% of track when A covers 60% of the track.

B covers 40% of the track in 18 minutes.

B will cover the rest 60% in 27 minutes, hence it will return to B at 10:27 AM

$$a_1 - a_2 + a_3 - a_4 + .... + (-1)^{n - 1} a_n = n$$

It is clear from the above equation that when n is odd, the co-efficient of a is positive otherwise negative.

$$a_1 - a_2 = 2$$

$$a_1 = a_2 + 2$$   

$$a_1 - a_2 + a_3 = 3$$

On substituting the value of $$a_1$$ in the above equation, we get

$$a_3$$ = 1

$$a_1 - a_2 + a_3 - a_4 = 4$$

On substituting the values of $$a_1, a_3$$ in the above equation, we get

$$a_4$$ = -1

$$a_1 - a_2 + a_3 - a_4 +a_5 = 5$$

On substituting the values of $$a_1, a_3, a_4$$ in the above equation, we get

$$a_5$$ = 1

So we can conclude that $$a_3, a_5, a_7....a_{n+1}$$ = 1 and $$a_2, a_4, a_6....a_{2n}$$ = -1

Now we have to find the value of $$a_{51} + a_{52} + .... + a_{1023}$$

Number of terms = 1023=51+(n-1)1

n=973

There will be 486 even and 487 odd terms, so the value of $$a_{51} + a_{52} + .... + a_{1023}$$ = 486*-1+487*1=1

Let 'h' be the height of the triangle ABC, semiperimeter(S) $$= \frac{4+4+r+4+4+r}{2} = 8+r$$, 
$$a=4+r, b=4+r, c=8$$

Area of triangle ABC $$=\ \ \sqrt{\ s\cdot\left(s-a\right)\left(s-b\right)\left(s-c\right)}=$$
$$= \sqrt{\left(\ 8+r\right)\times\ 4\times\ 4\times\ r}$$ = $$\ \frac{\ 1}{2}\times\ \left(4+4\right)\times\ height$$

Height (h) = $$\sqrt{\ \left(8+r\right)r}$$

Now, $$ h + r = 4 \longrightarrow  \sqrt{\ \left(8+r\right)r} + r = 4$$ (Considering the height of the triangle)

$$\sqrt{\ \left(8+r\right)r}$$=4-r

16r=16

r=1


Alternatively,

$$\text{AE}^@+\text{EC}^2=\text{AC}^2 \longrightarrow 4^2+\left(4-r\right)^2 = \left(4+r\right)^2 \longrightarrow\ \longrightarrow\ \longrightarrow\ r=1$$

The roots of $$x^2 - 4x - log_{2}{A} = 0$$ will be real and distinct if and only if the discriminant is greater than zero

16+4*$$log_{2}{A}$$ > 0

$$log_{2}{A}$$ > -4

A> 1/16

Given,

The quadratic equation $$x^2 + bx + c = 0$$ has two roots 4a and 3a

7a=-b

12$$a^2$$ = c

We have to find the value of  $$b^2 + c = 49a^2+ 12a^2=61a^2$$

Now lets verify the options 

61$$a^2$$ = 3721 ==> a= 7.8 which is not an integer

61$$a^2$$ = 361 ==> a= 2.42 which is not an integer

61$$a^2$$ = 427 ==> a= 2.64 which is not an integer

61$$a^2$$ = 549 ==> a= 3 which is an integer

It is given that the base of the pyramid is square and each of the four sides are equilateral triangles.

Length of each side of the equilateral triangle = 20cm

Since the side of the triangle will be common to the square as well, the side of the square = 20cm

Let h be the vertical height of the pyramid ie OA

OB = 10 since it is half the side of the square 

AB is the height of the equilateral triangle i.e 10$$\sqrt{\ 3}$$

AOB is a right angle, so applying the Pythagorean formula, we get

$$OA^2+ OB^2= AB^2$$

$$h^2$$ + 100= 300

h=10$$\sqrt{\ 2}$$

Let p be the length of AP.

It is given that $$\angle\ BAC\ =90\ $$ and $$\angle\ APC\ =90\ $$

Let $$\angle\ ABC\ =\theta\ $$, then $$\angle\ BAP\ =90-\theta\ $$ and $$\angle\ BCA\ =90-\theta\ $$

So $$\angle\ PAC\ =\theta\ $$

Triangles BPA and APC are similar

$$p^2=x\left(20-x\right)$$

We have to maximize the value of p, which will be maximum when x=20-x

x=10

$$\sqrt{\log_{e}{\dfrac{4x - x^2}{3}}}$$ will be real if $$\log_e\ \dfrac{\ 4x-x^2}{3}\ \ge\ 0$$

$$\dfrac{\ 4x-x^2}{3}\ \ge\ 1$$

$$\ 4x-x^2-3\ \ge\ 0$$

$$\ x^2-4x+3\ \le\ 0$$

$$1\le\ x\le\ 3$$

$$5^x - 3^y = 13438$$ and $$5^{x - 1} + 3^{y + 1} = 9686$$

$$5^{x} + 3^{y}*15 = 9686*5$$

$$5^{x} + 3^{y}*15 = 48430$$

16*$$3^y$$=34992

$$3^y$$ = 2187

y = 7

$$5^x$$=13438+2187=15625

x=6

x+y = 13

Each of three vessels A, B, C contains 500 ml of salt solution of strengths 10%, 22%, and 32%, respectively.

The amount of salt in vessels A, B, C = 50 ml, 110 ml, 160 ml respectively.

The amount of water in vessels A, B, C = 450 ml, 390 ml, 340 ml respectively.

In 100 ml solution in vessel A, there will be 10ml of salt and 90 ml of water

Now, 100 ml of the solution in vessel A is transferred to vessel B. Then, 100 ml of the solution in vessel B is transferred to vessel C. Finally, 100 ml of the solution in vessel C is transferred to vessel A

i.e after the first transfer, the amount of salt in vessels A, B, C = 40, 120, 160 ml respectively.

after the second transfer, the amount of salt in vessels A, B, C =40, 100, 180 ml respectively.

After the third transfer, the amount of salt in vessels A, B, C = 70, 100, 150 respectively.

Each transfer can be captured through the following table.

Percentage of salt in vessel A =$$\ \frac{\ 70}{500}\times\ 100$$

=14%

It is given that starting from 10:01 am, every minute a motorcycle leaves A and moves towards B.

Forty-five such motorcycles reach B by 11 am.

It means that the forty-fifth motorcycle starts at 10:45 AM at A and reaches B by 11:00 AM i.e 15 minutes.

Since the speed of all the motorcycles is the same, all the motorcycles will take the same duration i.e 15 minutes.

If the cyclist doubles the speed, then he will reach B by 10:30 AM. (Since if the speed is doubled, time is reduced by half)

Since each motorcycle takes 15 minutes to reach B, 15 motorcycles would have reached B by the time the cyclist reaches B

It is given that the volume of all the cylinders is the same, so the volume of each cylinder = HCF of (405, 783, 351)

=27

The number of iron cylinders = $$\ \frac{\ 405}{27}$$ = 15

The number of aluminium cylinders = $$\ \frac{\ 783}{27}$$ = 29

The number of copper cylinders = $$\ \frac{351\ }{27}$$ = 13

15*$$\pi\ r^2h=$$ 405  

15*$$\pi\ 9*h=$$ 405

$$\pi\ h=3$$

Now we have to calculate the total surface area of all the cylinders 

Total number of cylinders = 15+29+13 = 57

Total surface area of the cylinder = 57*($$2\pi\ rh+2\pi\ r^2$$)

=57(2*3*3 + 2*9*$$\pi\ $$)

=$$1026(1 + \pi)$$

Let $$2^{3x}$$ = v

$$2^{6x} + 2^{3x + 2} - 21 = 0$$

= $$v^2+4v-21=0$$

=(v+7)(v-3)=0

v=3, -7

$$2^{3x}$$ = 3 or $$2^{3x}$$ = -7(This can be negated)

3x=$$\log_23$$

x=$$\log_23$$/3

$$2^4 \times 3^5 \times 10^4$$

=$$2^4 \times 3^5 \times 2^4*5^4$$

=$$2^8 \times 3^5 \times 5^4$$

For the factor to be a perfect square, the factor should be even power of the number.

In $$2^8$$, the factors which are perfect squares are $$2^0, 2^2, 2^4, 2^6, 2^8$$ = 5

Similarly, in $$3^5$$, the factors which are perfect squares are $$3^0, 3^2, 3^4$$ = 3

In $$5^4$$, the factors which are perfect squares are $$5^0, 5^2, 5^4$$ = 3

Number of perfect squares greater than 1 = 5*3*3-1

=44

Let the six-digit number be ABCDEF

F = A+B+C, E= A+B, C=A, B= 2A, D= E+F.

Therefore D = 2A+2B+C = 2A + 4A + A= 7A.

A cannot be 0 as the number is a 6 digit number.

A cannot be 2 as D would become 2 digit number.

Therefore A is 1 and D is 7.

Speed of John = 6kmph 

Speed of Mary = 7.5 kmph 

Lengths of tracks A and B = 325 m

Let the length of track A be a, then the length of track B = 325-a

9 rounds of John on track A = 5 rounds of Mary on track B

$$\ \frac{\ 9\times\ a}{6\ \times\ \ \frac{\ 5}{18}}\ =\ \ \frac{\ 5\cdot\left(325-a\right)}{7.5\times\ \ \frac{\ 5}{18}}$$

On solving we get , 13a=1300

a=100

The length of track A = 100m, track B = 225m

Mary makes one round of track A = $$\ \frac{\ 100}{7.5\times\ \ \frac{\ 5}{18}}$$

= 48 sec

Let the number of fiction and non-fiction books in 2010 = 100a, 100b respectively

It is given that the total number of books in 2010 = 11500

100a+100b = 11500            -------Eq 1

The number of fiction and non-fiction books in 2015 = 110a, 112b respectively

110a+112b = 12760           -------Eq 2

On solving both the equations we get, b=55, a= 60

The number of fiction books in 2015 = 110*60=6600

It is given that John works altogether 172 hours i.e including regular and overtime hours.

Let a be the regular hours, 172-a will be the overtime hours

John's income from regular hours = 57*a 

John's income for working overtime hours = (172-a)*114

It is given that his income from overtime hours is 15% of his income from regular hours

a*57*0.15 = (172-a)*114

a=160

The number of hours for which he worked overtime = 172-160=12 hrs

Let us first find the number of terms 

47=1+(n-1)2

n=24

24*2n+1+3+5+....47=5280

48n+576=5280

48n=4704

n=98

Sum of first 98 terms = 98*99/2

=4851

CP of the table at which the shopkeeper procured each table = p

It is given that shopkeeper sold the tables to Amal and Asim at a profit of 20% and at a loss of 20%, respectively

The selling price of the tables = 1.2p and 0.8p to Amal and Asim respectively.

Amal sells his table to Bimal at a profit of 30%

So, CP of the table by Bimal (x)= 1.2p*1.3 = 1.56p

Asim sells his table to Barun at a loss of 30%

So, CP of the table by Barun (y)= 0.7*0.8p = 0.56p

(x-y)/p = (1.56p-0.56p)/p = p/p=1

It is given that AD and BE are medians which are perpendicular to each other.

The lengths of AD and BE are 12cm and 9cm respectively.

It is known that the centroid G divides the median in the ratio of 2:1

Area of $$\triangle$$ ABC = 2* Area of the triangle ABD

Area of $$\triangle\ $$ABD = Area of $$\triangle\ $$ AGB + Area of $$\triangle\ $$ BGD

Since $$\angle\ AGB\ =\ \angle\ BGD\ =\ 90$$ (Given)

Area of $$\triangle\ $$ AGB = $$\ \frac{\ 1}{2}\times\ 8\times\ 6$$ = 24

Area of $$\triangle\ $$ BGD = $$\ \frac{\ 1}{2}\times\ 6\times\ 4$$ = 12

 Area of $$\triangle\ $$ABD = 24+12=36

Area of $$\triangle\ ABC\ =\ 2\times\ 36=72$$

A: 15, 19, 23, 27, . . . . , 415

B: 14, 19, 24, 29, . . . , 464

Here the first common term = 19

Common difference = LCM of 5, 4=20

19+(n-1)20 $$\le\ $$ 415

(n-1)20 $$\le\ $$ 396

(n-1) $$\le\ $$ 19.8

n=20

$$\left(ax+by\right)^2=65^2$$

$$a^2x^2\ +\ b^2y^2+\ 2abxy\ =\ 65^2$$
$$k = ay - bx$$

$$k^2\ =\ a^2y^2+b^2x^2-2abxy$$

$$(a^2 + b^2)(x^2 + y^2 )= 25* 169$$

$$a^2x^2+a^2y^2+b^2x^2+b^2y^2=\ 25\times\ 169$$

$$k^2=\ 65^2\ -\ \left(25\times\ 169\right)$$

k = 0

D is the correct answer.

Let the cost of each bicycle= 100b

CP of 10 bicycles = 1000b

It is given that he sold six of these at a profit of 25% and the remaining four at a loss of 25%

SP of 10 bicycles = 125b*6+75b*4 

=1050b

Profit = 1050b-1000b =50b

50b=2000

CP = 100b = 4000

Let the score of D = 100d

The score of C = 20% less than that of D = 80d

The score of B = 25% more than C = 100d

The score of A = 10% less than B =90d

90d=72

100d= 72*100/90

= 80

Let the salaries of Ramesh, Ganesh and Rajesh in 2010 be 6x, 5x, 7x respectively

Let the salaries of Ramesh, Ganesh and Rajesh in 2015 be 3y, 4y, 3y respectively

It is given that Ramesh’s salary increased by 25% during 2010-2015,3y = 1.25*6x

y=2.5x

Percentage increase in Rajesh's salary = 7.5-7/7=0.07

=7%

Each interior angle in an n-sided polygon = $$\frac{\left(n-2\right)180\ }{n}$$

It is given that each interior angle of B is $$\frac{3}{2}$$ times each interior angle of A and b = 2a

$$\frac{\left(b-2\right)180\ }{b}$$ = $$\ \frac{\ 3}{2}\times\ $$ $$\frac{\left(a-2\right)180\ }{a}$$

$$2\times\ \left(b-2\right)\times\ a\ =\ 3\times\ \left(a-2\right)\times\ b$$

2(ab-2a) = 3(ab-2b)

ab-6b+4a=0

a*2a-12a+4a=0

$$2a^2-8a=0$$

a(2a-8) = 0

a cannot be zero so 2a=8

a=4, b = 4*2=8

a+b = 12

Each interior angle of a regular polygon with 12 sides = $$\ \frac{\ \left(12-2\right)\times\ 180}{12}$$

=150

Given, f(mn) = f(m)f(n)
when m= n= 1, f(1) = f(1)*f(1) ==> f(1) = 1

when m=1,  n= 2, f(2) = f(1)*f(2) ==> f(1) = 1

when m=n= 2, f(4) = f(2)*f(2) ==> f(4) = $$[f(2)]^2$$

Similarly f(8) = f(4)*f(2) =$$[f(2)]^3$$

f(24) = 54

$$[f(2)]^3$$ * $$[f(3)]$$ = $$3^3*2$$

On comparing LHS and RHS, we get 

f(2) = 3 and f(3) = 2

Now we have to find the value of f(18)

f(18) = $$[f(2)]$$ * $$[f(3)]^2$$

= 3*4=12

Let the total work be LCM of 20, 40 = 40 units 

Efficiency of Anil and Sunil is 2 units and 1 unit per day respectively.

Anil works alone for 3 days, so Anil must have completed 6 units.

Bimal completes 10% of the work while working along with Anil and Sunil.

Bimal must have completed 4 units.

The remaining 30 units of work is done by Anil and Sunil

Number of days taken by them 30/3=10

The total work is completed in 3+10=13 days

Let the scores of Rama, Anjali and Mohan be r, a, m.

It is given that Rama's score was one-twelfth of the sum of the scores of Mohan and Anjali

r=$$\ \frac{\ m+a}{12}$$ ———-(1)

The scores of Rama, Anjali and Mohan after review = r+6, a+6, m+6

a+6:m+6:r+6 = 11:10:3

Let a+6 = 11x => a= 11x-6

      m+6=10x => m=10x-6

      r+ 6 =3x => r = 3x-6 
   Substituting these values in equation (1), we get   

3x-6=$$\ \frac{\ 21x-12}{12}$$

12(3x-6) = 21x-12

x=4

Anjali's score exceeds Rama's score by (a-r)=8x=32