Test: CAT Quant 2019 (Shift - 2) - CAT MCQ

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 = 90/20=4.5

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

= n (n+3)+4(n+3)/n(n-4)+3(n-4)

= (n+4)(n+3)/(n-4)(n+3)

= (n+4)/(n-4)

=(n-4)+8/(n-4)

= 1 + 8/(n-4) which will be maximum when n-4 =8

n=12

D is the correct answer.

(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.

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

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

a₁- a₂ = 2

a₁ = a₂ + 2

a₁ − a₂ + a₃ = 3

On substituting the value of ai in the above equation, we get

a₃ = 1

a₁ -a₂ + a₃ -a₄ = 4

On substituting the values of a₁, a₃ in the above equation, we get

a₄ = -1

a₁ – a₂ + a₃ – a₄ + a₅ = 5

On substituting the values of a₁, a₃, a₄in the above equation, we get

a₅ = 1

So we can conclude that a₃, a₅,a₇….. a_n+1 = 1 and a₂, a₄, a₆....a_2n = -1

Now we have to find the value of a₅₁ + a₅₂ + ....+ a₁₀₂₃

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

n=973

There will be 486 even and 487 odd terms, so the value of a₅₁ + a₅₂ + .... + a₁₀₂₃ = 486*-1+487*1

Let 'h’ be the height of the triangle ABC, semi perimeter(S) = (4+4+r+4+4+r)/2 = 8 + r,

a=4 +r, b = 4 + r, c = 8

Height (h) = √(8+r)r

Now, h + r = 4

(Considering the height of the triangle)

Alternatively,

AE² + EC² = AC²⟶ 42 + (4 − r)² = (4 + r)²⟶⟶⟶ r = 1

16+4*log²A > 0

log²A > -4

A> 1/16

7a=-b

12a² = c

We have to find the value of b² + c = 49a²+ 12a²=61a²

Now let’s verify the options

61 a² = 3721 ==> a= 7.8 which is not an integer

61 a² = 361 ==> a= 2.42 which is not an integer

61 a² = 427 ==> a= 2.64 which is not an integer

61 a² = 3721 ==> a= 3 which is an integer

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 i.e. OA

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

AB is the height of the equilateral triangle i.e. 10√3

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

OA² + OB² = AB²

h2 + 100= 300

h=10√2

Let p be the length of AP.

It is given that ∠BAC = 90 and ∠APC = 90

Let ∠ABC = ϴ, then ∠BAP= 90 - ϴ and ∠BCA =90 - ϴ

so ∠PAC = ϴ

Triangles BPA and APC are similar

p² = x (20 - x)

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

x=10

4x-x^2/3>=1

4x - x²- 3 >= 0

x² − 4x + 3 =< />

1=< x="" />< />

5x + 3y * 15 = 9686 * 5

5x + 3y * 15 = 48430

16*3y =34992

3y = 2187

y = 7

5x = 13438+2187=15625

x=6

x+y = 13

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 =40,400,180 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 = 70/500 x 100 =14%

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

The number of iron cylinders = 405/27 = 15

The number of aluminium cylinders = 783/27 = 29

The number of copper cylinders = 315/27 = 13

15*πr²h= 405

15*π9*h= 405

πh = 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*(2πrh + 2πr²)

=57(2*3*3 + 2*9*π)

=1026(1 + π)

26x + 23x+2 -21 = 0

= v² + 4v - 21 =0

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

v = 3,-7

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

3x=log₂3

x=log₂3/3

=2⁴ x 3⁵ x 2⁴ * 5⁴

=2⁸ x 3⁵ x 5⁴

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

In 2⁸, the factors which are perfect squares are 2⁰, 2², 2⁴, 2⁶, 2⁸ = 5

Similarly, in 3⁵, the factors which are perfect squares are 3⁰, 3²,3⁴ = 3

In 5⁴, the factors which are perfect squares are 50, 52, 54 = 3

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

=44

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 Mary = 7.5 kmph

Lengths of tracks A and B = 325m

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

On solving we get, 13a=1300

a = 100

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

Mary makes one round of track A =

= 48 sec

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…….. Eq2

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

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

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)*144

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

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

Positive Mark: 3

Negative Mark: 0

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

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 ΔABC = 2* Area of the triangle ABD

Area of ΔABD = Area of ΔAGB + Area of ΔBGD

Since ∠AGB = ∠BGD, 90 (Given)

Area of ΔAGB = ½ x 8 x 6 = 24

Area of ΔBGD = 1/2 x 6 x 4 = 12

Area of ΔABD = 24+12=36

Area of ΔABC = 2 x 36 = 72

B:14,19, 24, 29……, 464

Here the first common term = 19

Common difference = LCM of 5,4=20

19+(n-1)20 ≤ 415

(n-1)20 ≤ 396

(n-1) ≤ 19.8

n=20

Positive Mark: 3

Negative Mark: 0

a²x² + b²y² + 2 abxy = 652

k = ay — bx

k² = a²y² + b²x²- 2abxy

(a² + b²)(x² + y²) = 25 * 169

a²x² + a²y² + b²x² + b²y² = 25 x 169

k² = 652 - (25 x 169)

k = 0

D is the correct answer.

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

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

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

The score of A = 1.0% less than B =90d

90d=72

100d= 72*100/90

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%