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%

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

**Verbal Ability : | Q.01- Q.05 | Q.06- Q.10 | Q.11- Q.15 | Q.16- Q.20 | Q.21- Q.24 | Q.25- Q.29 | Q.30 – Q.34 |**

**Logical Reasoning : | Q.01- Q.04 | Q.05- Q.08 | Q.09- Q.12 | Q.13- Q.16 | Q.17- Q.20 | Q.21- Q.24 | Q.25 – Q.28 | Q.29 – Q.32 |**

**Quantitative Aptitude: | Q.01- Q.05 | Q.06- Q.10 | Q.11- Q.15 | Q.16- Q.20 | Q.21- Q.25 | Q.26- Q.30 | Q.31 – Q.34 |**

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