Home Uncategorized CAT 2019 Quant – Slot 2

CAT 2019 Quant – Slot 2

Q.1) If 5^x-3^y=13438 and 5^(x-1)+3^(y+1)=9686 then find x+y.

a)13

b)14

a)13

Solution:

Here we can go by hit n trial

13438 is 5 digit number so 5^x must be a 5 digit number

5^6=15625

So putting x=6 in the first equation we get 3^y=2187

So y=7

Putting this in eq 2

5^5 +3^8=3125+6561=9686

So x+y=6+7=13

Q.2) 

a)0

b)10

c)1

d)-1

c)1

Solution:

a1=1

a1-a2=2

a2=-1

a1-a2+a3=3

a3=1

a1-a2+a3-a4=4

a4=-1

1,-1,1,-1….so on

so sum of every two consecutive terms will be 0

so only a1023 will remain which is odd numbered term  having value 1

Q.3) Amal invests Rs 12000 at 8% interest, compounded annually, and Rs 10000 at 6% interest, compounded semi-annually, both investments being for one year. Bimal invests his money at 7.5% simple interest for one year. If Amal and Bimal get the same amount of interest, then the amount, in Rupees, invested by Bimal is

a)20920

b)10920

a)20920

Solution:

Interests for both are same so equate their interests

Semi-annually for Amal so rate becomes half  (i.e.3% )and time will become double i.e. 2 years

Interest for Amal= (12000*8*1)/100 +10000[1.03]^2 -10000

Interest for Bimal= (P*7.5*1)/100

960+609=(7.5P)/100

156900=7.5P

P=20920

Q.4) Anil alone can do a job in 20 days while Sunil alone can do it in 40 days. Anil starts the job, and after 3 days, Sunil joins him. Again, after a few more days, Bimal joins them and they together finish the job. If Bimal has done 10% of the job, then in how many days was the job done?

a)14

b)15

c)13

d)12

c)13

Solution:

Work=lcm(20,40,3)=120

A=6 units per day

S=3 units per day

Anil does =6*3=18 units in 3 days

Bimal does 10% of the job=12 units 

So anil and sunil will do 90% of the job=108 units

Anil  has already done 18 units 

So anil and sunil will do 90 units in, Say X days

So

X(A+S)=90

Putting values of A and S

X=10 days

So B will also work for 10 days

So total=3+10=13 days were required to complete the work

Q.5)

a)1

b)2

c)3

d)4

d)4

Solution:

Using options, only option 4 satisfies because (4x-x^2)>0 for those values of x only

Q.6)

a)4

b)5

a)4

Solution:

This can be rewritten as n^2-m^2=105

(n+m)(n-m)=105

We have an odd product in the RHS

So in LHS we must have only odd numbers that give a product of 105

Which are (1,105), (3,35), (5,21), (7,15)

So only 4 pairs will satisfy

iQuanta Shortcut: Number of factors/2= 8/2=4

Q.7) In 2010, a library contained a total of 11500 books in two categories – fiction and non-fiction. In 2015, the library contained a total of 12760 books in these two categories. During this period, there was 10% increase in the fiction category while there was 12% increase in the non-fiction category. How many fiction books were in the library in 2015?

a)6160

b)6600

c)6000

d)5500

b)6600

Solution:

Fiction….F

Non fiction….N

In 2010…..

F be the num of fiction books

11500-F be the number of non-fiction books

In 2015….

1.1F+1.12(11500-F)=12760

Calculating F in the equation above we get

F=6600

[/bg_colllapse]

Q.8) The salaries of Ramesh, Ganesh and Rajesh were in the ratio 6:5:7 in 2010, and in the ratio 3:4:3 in 2015. If Ramesh’s salary increased by 25% during 2010-2015, then the percentage increase in Rajesh’s salary during this period is closest to

a)9

b)8

c)7

d)10

c)7

Solution:

Salaries of Ramesh, Ganesh and Rajesh were  6x:5x:7x in 2010

Ratio of their salaries is 3y:4y:3y in 2015

Salary of Ramesh after 25% increase, salary=6x*1.25=7.5x

So 3y=7.5x

y=2.5x

so Rajesh’s salary in 2015=3y=7.5x

so % increase in Rajesh’s salary=(7.5x-7x)/7x *100=50/7=7.1% so 7% is closest.

Q.9)

a)4851

b)5851

a)4851

Solution:

From 1 to 47 we have 24 terms

48n+12(48)=5280

4n+48=440

4n=392

n=98

so 1+2+3+4+…..+98

98/2(99)=49*99=4851

Q.10) Two circles, each of radius 4 cm, touch externally. Each of these two circles is touched externally by a third circle. If these three circles have a common tangent, then the radius of the third circle, in cm, is

a)1

b)2

c)4

d)5

a)1

Solution:

r is the radius of the smaller circle.

In right triangle ABC

AB^2+BC^2=AC^2

(4-r)^2+4^2=(4+r)^2

r=1

Q.11) Let A and B be two regular polygons having a and b sides, respectively. If b = 2a and each the interior angle of B is 3/2 times each interior angle of A, then each interior angle, in degrees, of a regular polygon with a + b sides is

a)140

b)150

b)150

Solution:

the angle of B=1.5 * angle of A

so the angle of B is greater

(b-2)*180/b=1.5*(a-2)*180/a

(b-2)/b=1.5(a-2)/a

ab-2a=1.5ab-3b

0.5ab=3b-2a

½=3/a-2/b

b=2a

so a=4….b=8

a+b=12

so each interior angle of polygon having 12 sides is (12-2)*180/12=150

Q.12) John jogs on track A at 6 kmph and Mary jogs on track B at 7.5 kmph. The total length of tracks A and B is 325 metres. While John makes 9 rounds of track A, Mary makes 5 rounds of track B. In how many seconds will Mary make one round of track A?

a)34

b)48

c)45

d)50

b)48

Solution:

A+B=325…eq1

9A/6=5B/7.5

A=4B/9

So putting in eq1

4B/9+B=325

B=225m

A=100 m

Speed of Mary=7.5 kmph=37.5/18  m/sec

So Mary completes one round of A i.e 100 m in =100/(37.5/18)=18000/375=48 sec

Q.13) In a triangle ABC, medians AD and BE are perpendicular to each other, and have lengths 12 cm and 9 cm, respectively. Then, the area of triangle ABC, in sq cm, is

a)50

b)60

c)72

d)80

c)72

Solution:

Area(ABC)=2*area(ADB)

Since median intersect at centroid and centroid divides median into 2:1

So 

AG=8….GD=4

BG=6…GE=3

Area of AGB=1/2*AD*BG=1/2*12*6=36

So Area of ABC=2*36=72

Q.14)

a)k>5/7

b)k<=5/7

c)k=0

d)None of these

c)k=30

Solution:

a,b=+-3, +-4

x,y=+-12, +-5

But no combination of these values gives us 65

But a=5, b=0…x=13….y=0

This satisfies ax+by=65

So k=0

Q.15)The base of a regular pyramid is a square and each of the other four sides is an equilateral triangle, length of each side being 20 cm. The vertical height of the pyramid, in cm, is

a)10√2

b)10

c)20

d)20√2

a)10√2

Solution:

Since each side is an equilateral triangle of side 20 cm , 

Hence the side of base square will also be 20 each.

So in right triangle ABC 

BC^2=AB^2+AC^2

AC= ½ * diagonal of square

Diagonal of square= 20√2

So AC=10√2

So AB^2=20^2-(10√2)^2=200

AB=10√2

Q.16) The quadratic equation x^2+ bx + c = 0 has two roots 4a and 3a, where a is an integer. Which of the following is a possible value of b^2+ c?

 a)427

b)549

c)366

d)305

b)549

Solution:

So roots will be 4a,3a

So equation will be x^2-7ax+12a^2=0

b^2+c=(7a)^2+12a=49a^2+12a^2=61a^2

so answer must be a multiple of 61 multiplied by a perfect square which is only among the 549

Q.17) Two ants A and B start from a point P on a circle at the same time, with A moving clock-wise and B moving anti-clockwise. They meet for the first time at 10:00 am when A has covered 60% of the track. If A returns to P at 10:12 am, then B returns to P at

a)10:14

b)10:27

c)10:15

d)10:29

b)10:27

Solution: 

Q.18) Let f be a function such that f (mn) = f (m) f (n) for every positive integers m and n. If f(1), f (2) and f (3) are positive integers, f (1) < f (2), and f (24) = 54, then f (18) equals

a)18

b)52

c)54

d)67

a)18

Solution:

f(24)=54

f(24)=f(2*2*2*3)=f(2)*f(2)*f(2)*f(3)=54=3*3*3*2

so comparing LHS and RHS

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

so

f(18)=f(2*3*3)=f(2)*f(3)*f(3)=3*3*2=18

54/f(18)=f(2)(f(2)/f(3)

Q.19) John gets Rs 57 per hour of regular work and Rs 114 per hour of overtime work. He works altogether 172 hours and his income from overtime hours is 15% of his income from regular hours. Then, for how many hours did he work overtime?

a)10

b)11

c)14

d)12

d)12

Solution:

X hrs normally

172-x hrs overtime

Income from normal working =57x

Income from overtime=114(172-x)

Given income from overtime hours is 15% of his income from regular hours

So 114(172-x)=(15*57x)/100

200(172-x)=15x

40(172-x)=3x

43x=40*172

X=160

So he worked overtime for 172-x i.e. 172-160=12 hrs

Q.20) In an examination, the score of A was 10% less than that of B, the score of B was 25% more than that of C, and the score of C was 20% less than that of D. If A scored 72, then the score of D was

a)50

b)60

c)70

d)80

d)80

Solution:

Given

A=0.9B

B=1.25C

C=0.8D

A=72…putting this in first eq above

So B=80

C=64

D=80

Q.21) The number of common terms in the two sequences: 

15, 19, 23, 27, . . . . , 415 and 14, 19,24, 29, . . . , 464 is

a)20

b)21

c)18

d)23

a)20

Solution:

Common difference of 1st series is 4

Common difference of 2nd series is 5

So series of common terms will have common difference =LCM(4,5)=20

Now find the first common terms in both the series

And that is 19

So series of common terms will be like 19, 39,, 59…. till 415 because first series has terms only till 415

So 415=19+(n-1)20

20(n-1)=396

n=20.xx

so 20 terms will be common  

Q.22)