Test: CAT Quant 2017 (Shift - 1) - CAT MCQ

Let Arun's current age be A. Hence, Barun's current age is 2.5A

Let Arun's age be half of Barun's age after X years.

Therefore, 2*(X+A) = 2.5A + X

or, X = 0.5A

Hence, Barun's age increased by 0.5A/2.5A = 20%

Let the rate of work of a person be x units/day. Hence, the total work = 120x.

It is given that one first day, one person works, on the second day two people work and so on.

Hence, the work done on day 1, day 2... will be x, 2x, 3x, ... respectively.

The sum should be equal to 120x.

120x = x * n(n+1)/2

n² + n – 240 = 0

n = 15 is the only positive solution.

Hence, it takes 15 days to complete the work.

It is given that the maximum weight limit is 630. The lightest person's weight is 53 Kg and the heaviest person's weight is 57 Kg.

In order to have maximum people in the lift, all the remaining people should be of the lightest weight possible, which is 53 Kg.

Let there be n people.

53 + n (53) + 57 = 630

n is approximately equal to 9.8. Hence, 9 people are possible.

Therefore, a total of 9 + 2 = 11 people can use the elevator.

We see that the man saves 20 minutes by changing his speed from 12 Km/hr to 15 Km/hr. Let d be the distance

Hence,

d/12 – d/15 = 1/3

d/60 = 1/3

d = 20 Km.

Let his total savings be 100x.

He invests 50x in fixed deposits. 30% of 50x, which is 15x is invested in stocks and 35x goes to savings bank.

It is given 85x = 59500

x = 700

Hence, 100x = 70000

Given,

0.85 M = 1.02 C

M = 1.2 C

If he wants 20% profit he has to sell at 1.2C, which is nothing but the retail price.

Total travel time gets reduced by 75%. So, total travel time becomes 1/4th of the original.
So, s’(New speed) = 4s(Original speed)
Speed of boat : Speed of river = x:1
When the speed of boat is normal
Speed Upstream = x-1
Speed Downstream = x+1
Here, the distance is constant. So, average speed can be found out by the harmonic mean of two speeds = 2ab/ (a+b)
Here, average speed will be 2(x^2-1)/2x = (x^2-1)/x
When the speed of boat is doubled
Speed upstream = 2x-1
Speed downstream = 2x+1
Here, the distance is constant. So, average speed can be found out by the harmonic mean of two speeds = 2ab/(a+b)
Here, average speed will be 2(4x^2-1)/4x = (4x^2-1)/2x
Now, we know that
New speed = 4*Original speed
So,
(4x^2-1)/2x = 4(x^2-1)/x
4x^2 – 1 = 8x^2 – 8
4x^2 = 7
x = √7/2
Ratio = x:1 = √7/2

C1 : C2 : C3 = 9 : 10 : 8 ... i

C2 : C4 : C5 = 18 : 19 : 20 ... ii

Let's multiply i by 9 and ii by 5

C1 : C2 : C3 = 81 : 90 : 72

C2 : C4 : C5 = 90 : 95 : 100

Therefore, C1 : C2 : C3 : C4 : C5 = 81 : 90 : 72 : 95 : 100

Given,

100x - 81x = 19

x = 1

Hence, total profit = 100 + 95 + 72 + 90 + 81 = 438

Girls getting admission = 0.6x

Boys getting admission = 0.45x

Number of students not getting admission = 3x - 0.6x -0.45x = 1.95x

Percentage = (1.95x/3x) * 100 = 65%

Let them be 7x, 17x and 16x

The ratio of L, S, J for chips = 6:15: 14

Let them 6y, 15y and 14y

Given, 40x = 35y, x= 7y/8

Jumbo popcor = 16x = 16 * 7y/8= 14y

Hence, the ratio of jumbo popcorn and jumbo chips = 1: 1

Let the cost of good mangoes be 2x per kg. The cost of medium mangoes be x per kg.

CP of good mangoes = 160x

CP of medium mangoes = 40x

His selling price = 0.9*2x = 1.80x

Therefore, total revenue generated by selling all the mangoes = 120 x 1.8x = 216x

Hence, the profit % =16x/200x * 100= 8%

Given,

0.6 M = 1.2 C

M = 2C

CP of 60 toys = 60C

Now only 50 are remaining.

Hence, M (1 - d) * 50 = 72C

1- d = 0.72

d = .28

Hence 28%

For the first one,

a + 3 = 3b .. i

a + 3 = -3b ... ii

For the second one,

a - 1 = 2(b -1) ... iii

a - 1 = 2 (1 - b) ... iv

we have to solve i and iii, i and iv, ii and iii, ii and iv.

Solving i and iii,

a + 3 = 3b and a = 2b - 1, solving, we get a = 3 and b = 2, which is not what we want.

Solving i and iv

a + 3 = 3b and a = 3 - 2b, solving, we get b = 1.2, which is not possible.

Solving ii and iii

a + 3 = -3b and a = 2b - 1, solving, we get b = 0.4, which is not possible.

Solving ii and iv,

a + 3 = -3b and a = 3 - 2b, solving, we get a = 15 and b = -6 which is what we want.

Thus, a²/b² = 25/4

Therefore the average score of boys in the mid semester exam is G-5.

Hence, the total marks scored by the class is 30*G+20*(G-5) = 50*G – 100

Therefore, the average of the entire class is (50G-100)/50 = G-2

Therefore, the average of girls in the final exam is G-3 and the average of the entire class in the final exam is G.

Hence, the total marks scored by the class is 50G, while the total marks scored by the girls is 30*(G-3)=30G-90

Therefore, the total marks scored by the boys is 50G-(30G-90) = 20G+90

Hence, the average of the boys in the final exam is (20G+90)/20 = G+4.5

Hence, the increase in the average marks of the boys is (G+4.5)-(G-5) = 9.5

The area of the square = (2√2 )² = 8 units.

The perimeter = 100

Semi-perimeter, s = 50

The triangle formed by the centroid and two vertices is removed.

Since the cenroid divides the median in the ratio 2 : 1

The remaining area will be two-thirds the area of the original triangle.

Remaining area = 2 / 3 * 250√3 = 500 / √3.

AB = AC = 6cm. Thus, BC = = 6√2 cm

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

Area of semicircle BQC

Diameter BC = 6√2cm

Radius = 6√2/2 = 3√2 cm

Area = πr²/2 = π*(3√2) /2 = 9π

Area of quadrant BPC

Area = πr²/4 = π * (6)² /4 = 9π

Area of triangle ABC

Area = 1/2 * 6 * 6 = 18

The required area = Area of semi-circle BQC - Area of quadrant BPC + Area of triangle ABC

= 9π - 9 π + 18 = 18

Let the sides be a, a, 2a, 3a and 3a. (x = a³)

Let the side of the original cube be A.

A³ = x+x+8x+27x+27x

A = 4a

Original surface area = 96a²

New surface area = 6(a² + a² + 4a² + 9a² + 9a²) = 144a²

Percentage increase = (144 – 96) / 96 * 100 = 50%

The volume of a cylinder is π *r² *3 = 9π

r = √3 cm.

Radius of the ball is 2 cm. Hence, the ball will lie on top of the cylinder.

Lets draw the figure.

Based on the Pythagoras theorem, the other leg will be 1 cm.

Thus, the height will be 3 + 1 + 2 = 6 cm

The length of the altitude from A to the hypotenuse will be the shortest distance.

This is a right triangle with sides 3:4: 5.

Hence, the hypotenuse = = 25 Km.

Length of the altitude = = 12 Km

The time taken = 12/30 * 60 = 24 minutes

We know that log₃x = a and log₁₂y = a

Hence, x=3^a and y=12^aTherefore, the geometric mean of and equals

This equals = 6^a

Hence, G = 6^aOr, log₆ G = a

We know that x² – x – 1 =0

Therefore x⁴ = (x+1)² = x² + 2x + 1 = x + 1 + 2x + 1 = 3x + 2

Therefore, 2x⁴ = 6x + 4

We know that x>0 therefore, we can calculate the value of x to be 1+√5 /2

Hence, 2x⁴ = 6x + 4 = 3+3√5 +4 = 3√5 + 7

log0.008√5 + log√3 81 – 7

81 = 3⁴ and 0.008

8/1000 = 2³/10³ = 1/5³ = 5^-3

Hence

log0.008√5+ 8 – 7

log5^-35^1/2 + 8 – 7

log5^0.5/log5^-3 + 1

1/-6 + 1

= 5/6

81^x/9 – 81^x/81 = 1944

81^x * [1/9 – 1/81] = 1944

81^x * [8/81] = 243

3^4x = 3⁹

x = 9/4

x - y - z = 25 and x≤40, y≤12, z≤12

If x = 40 then y + z = 15. Now since both y and z are natural numbers less than 12, so y can range from 3 to 12 giving us a total of 10 solutions. Similarly, if x = 39, then y + z = 14. Now y can range from 2 to 12 giving us a total of 11 solutions.

If x = 38, then y + z = 13. Now y can range from 1 to 12 giving us a total of 12 solutions.

If x = 37 then y + z = 12 which will give 11 solutions.

Similarly on proceeding in the same manner the number of solutions will be 10, 9, 8, 7 and so on till 1.

Hence, required number of solutions will be (1 + 2 + 3 + 4 . . . . + 12) + 10 + 11

= 12*13/2 + 21

78 + 21 = 99

(n-5) (n-10)-3(n-2) ≤ 2

=> n² − 15n + 50 − 3n + 6 ≤ 0

=> n² − 18n + 56 ≤ 0

=> (n − 4)(n − 14) ≤ 0

=> Thus, n can take values from 4 to 14. Hence, the required numbers of values are 14 - 4 + 1 = 11.

Positive Mark: 3

Negative Mark: 0

f₁(x) = x² + 11x + n and f₂(x) = x,

f₁(x) = f₂(x)

x² + 11x + n = x

x² + 10x + n = 0

For this equation to have distinct real roots, b² - 4ac>0

10²> 4n

=> n < 100/4

=> n < 25

Thus, largest integral value that n can take is 24.

Positive Mark: 3

Negative Mark: 0

For the value of given expression to be minimum, the values of and should be as close as possible. 30/4 = 7.5. Since each one of these are integers so values must be 8, 8, 7, 7. On putting these values in the given expression, we get

(8 − 8)² + (8 − 7)² + (8 − 7)²

=> 1 + 1 = 2

The total number of given points are 11. (10 on circumference and 1 is the center)

So total possible triangles = 11C3 = 165.

However, AOB, COD, EOF, GOH, JOK lie on a straight line. Hence, these 5 triangles are not possible. Thus, the required number of triangles = 165 - 5 = 160

We can see that the shortest distance of the point (1/2, 1) will be 1 unit.