CAT 2019 – Slot 1 – Quantitative Ability – : Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks

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Q. 1: Meena scores 40% in an examination and after review, even though her score is increased by 50%, she fails by 35 marks. If her post-review score is increased by 20%, she will have 7 marks more than the passing score. The percentage score needed for passing the examination is
1. 60
2. 75
3. 80
4. 70

Let the total Maximum marks in exam = 100x

Meena’s score = 40% of 100x = 40x

Her marks after review = 40x+ 50% of 40x = 60x

Her marks after post review = 60x + 20% of 60x = 72x

As per question 72x – 7 = 60x+ 35

12x = 42

x = 42/12 = 7/2

so passing marks = 60x + 35 = 60*7/2 + 35 =245

total marks = 100x = 350

Required percentage score needed for passing the examination = 245/350 = 70%

 
 

 

Q. 2: In a race of three horses, the first beat the second by 11 metres and the third by 90 metres. If the second beat the third by 80 metres, what was the length, in metres, of the racecourse? 

Let the length of race = x

When the first beat the second by 11 metres and the third by 90 metres, distance traveled by 2nd and 3rd horse at that time will be x -11 and x -90 respectively.

When the second beat the third by 80 metres traveled by 2nd and 3rd horse at that time will be x and x -80 respectively.

Ratio of speed of 2nd and 3rd horse = ratio of distance traveled = (x-11)/(x-90) = x/(x-80)

(x-11)*(x-80) = x(x-90)

By solving we get x = 880

 

Q. 3: The number of solution to the equation |x| (6x^2 + 1) = 5x^2 is?

case 1) If x >0

X*(6x^2 +1) = 5x^2

6x^3 -5x^2 +x = 0

x(6x^2 -5x +1) = 0

x(6x^2 -3x -2x +1) =0

x(2x -1)*(3x -1) =0

x = 0, ½ or 1/3

number of solutions = 3

case 2) if x < 0

– X*(6x^2 +1) = 5x^2

6x^3 + 5x^2 +x = 0

x(6x^2 +5x +1) = 0

x(6x^2 +3x +2x +1) =0

x(2x +1)*(3x +1) =0

x = 0, -1/2 or -1/3

Thus x can take 5 different values 0,1/2, 1/3, -1/2 and -1/3

Total number of solutions = 5

Q. 4: The product of two positive numbers is 616. If the ratio of the difference of their cubes to the cube of their difference is 157:3, then the sum of the two numbers is
1.   58
2.   50
3.   95
4.   85

Let the number be a and b.

So ab = 616

Given, (a^3 –b^3)/(a-b)^3 = 157/3

(a-b)*(a^2 + b^2 +ab)/(a-b)*(a^2 + b^2 -2ab) = 157/3

Let a^2 + b^2 =k

So (k+ab)/(k-2ab) = 157/3

3k + 3ab = 157k – 314ab

154k = 117ab = 317*616

k = 317*616/154 = 1268

As we know (a+b)^2 = (a^2 + b^2) + 2ab = k + 2ab = 1268 + 2*616

(a+b)^2 =2500 = 50^2

So a+b = 50

 

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