**Directions(1-5):** Answer the following questions based on given information.

The graph shows the number of boys and percentage of girls out of the total students who appeared for an exam in 2012 from different schools.

Number of students who appeared for the exam = Number of boys + Number of girls

- Find the ratio of number of students appeared in exam from school D to the number of students appeared in exam from school A.
7:83:29:75:64:5Option D

Ratio = 5:6 - Find the difference between the number of students who appeared in exam from school C and number of girls who appeared in exam from school E.
8050904070Option E

Difference = 160-90 = 70 - The number of students who appeared in the exam from school C in 2015 is 25% more than the number of students who appeared in the from same school in 2012. The number of boys is 50% more than the number of girls appeared in the exam from school C in 2015. Find the difference between number of boys and number of girls who appeared in the exam from school C in 2015.
1535204054Option D

The number of students appeared in the exam from school C in 2015 = 160*1.25 =200

Number of girls appeared in the exam from school C in 2015 be x.

Number of boys appeared in the exam from school C in 2015 =1.5x

Now,

1.5x + x = 200

=>x = 80

Number of girls appeared in the exam from school C in 2015 = 80

Number of boys appeared in the exam from school C in 2015 = 1.5*80 = 120

Difference = 120 – 80 = 40 - Find the average number of girls who appeared for the exam from school A,B and C taken together.
3225223045Option E

Required Average = (45+50+40)/3 = 45 - The number of girls who appeared in exam from school D is what percent of number of students who appeared from school B?
12%20%25%10%18%Option B

Required% = (25/125)*100 = 20% - I.x^2 – 26x + 165= 0

II.y^2 – 26y + 168 = 0

y > xy >= xx >= yx > yNo relationOption E

I.x^2 – 26x + 165= 0

=>x^2 – 15x – 11x + 165 = 0

=>(x-15)(x-11) = 0

=>x = 15,11

II.y^2 – 26y + 168 = 0

=>y^2 – 14y – 12y + 168 = 0

=>(y -14)(y – 12) =0

=>y = 14,12

No relation - I.15x + 8y = 160

II.7x + 7y = 91

y > xNo relationy >= xx >= yx > yOption E

On solving both the equations, we get

x = 8

y = 5

x > y - I. 6x^2 – 29x + 35 = 0

II. 2y^2 – 19y + 35 = 0

x > yy > xy >= xNo relationx >= yOption C

I. 6x^2 – 29x + 35 = 0

=> x = 2.5, 2.33

II. 2y^2 – 19y + 35 = 0

=>y = 7, 2.5

y >= x - I.x^2 – 17x + 72 = 0

II.y^2 – 15y + 56 = 0

y >= xNo relationx > yx >= yy >= xOption D

I.x^2 – 17x + 72 = 0

=>x^2 – 8x – 9x + 72 = 0

=>(x-8)(x-9) = 0

=>x = 8,9

II.y^2 – 15y + 56 = 0

=>y^2 – 8y – 7y + 56 = 0

=>(y-8)(y-7) = 0

=>y = 8,7

x >= y - I.2x^2 – 13x + 21 = 0

II.2y^2 = y + 15

y > xy >= xx > yx >= yNo relationOption D

I.2x^2 – 13x + 21 = 0

=>2x^2 – 7x – 6x + 21 = 0

=>(2x-7)(x-3) = 0

=>x = 3,7/2

II.2y^2 = y + 15

=>2y^2 – 6y + 5y – 15= 0

=>(2y+5)(y-3) = 0

=>y = 3,-5/2

x >= y

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Directions(6-10):** Find the values of x and y and compare their values and choose a correct option.