Directions(1-5): Solve each equations individually and compare the values of x and y and choose accordingly.
- I.3x + 7y = 25
II.7x + 6y = 48y>=xy>xx>=yx>yRelation cannot be established.Option D
On solving both the equations, we get
x = 6
y = 1
x>y - I.16x^2 – 88x + 117 = 0
II.25y^2 – 125y + 156 = 0y>xx>yRelation cannot be established.y>=xx>=yOption C
I.16x^2 – 88x + 117 = 0
=>16x^2 – 36x – 52x + 117 = 0
=>(4x – 9)((4x – 13) = 0 =>x = 13/4,9/4
II.25y^2 – 125y + 156 = 0
=>25y^2 – 65y – 60y + 156 = 0
=>(5y – 12)(5y – 13) =0
=>y = 12/5,13/5
Relation cannot be established. - I.3x + 4y = 24
II.2x + 3y = 17Relation cannot be established.y>=xx>=yy>xx>yOption E
On solving both the equations, we get
x =4
y = 3
x>y - I.2x^2 + 11x – 195 = 0
II.3y^2 + 10y – 125 =0x>yx>=yRelation cannot be established.y>=xy>xOption C
I.2x^2 + 11x – 195 = 0
=>2x^2 + 26x – 15x – 195 = 0
=>(x+13)(2x – 15) =0
=>x = -13,15/2
II.3y^2 + 10y – 125 =0
=>3y^2 + 25y – 15y – 125 = 0
=>(3y + 25)(y – 5) = 0
=>y = -25/3,5
Relation cannot be established. - I.x^2 + 17x + 52 = 0
II.y^2 +27y +182 = 0y>=xRelation cannot be established.x>yy>xx>=yOption E
I.x^2 + 17x + 52 = 0
=>x^2 + 13x + 4x + 52 = 0
=>(x+4)(x+13) = 0
=> x = -4, – 13
II. y^2 +27y +182 = 0
=>y^2 + 14y + 13y + 182 = 0
=>(y+14)(y+13) = 0
=>y = -14, – 13
x>=y - What is the ratio of 2/5 of students enrolled college A and C together to 5/3 of students enrolled in college D and E together.
41:22449:22051:22247:22543:221Option D
Students enrolled in college A and C together = 350+ 400+300+ 450+400+450 = 2350
Students enrolled in college D and E together = 500+400+350+400+600+450 = 2700
Required ratio = 47:225 - If 76/5% of students enrolled in college D leaved the college and number of students from all three disciplines (BA,BSC and BCOM) who leave college D are in the ratio 8:6:5, then the remaining student from BCOM in college D are what percent of students from BSC discipline from college E.
80%70%40%60%50%Option E
Student leaves college D = 76/5% of (500+400+350)
= 190
Remaining students in BCOM from D = 350 – 5/19*190 = 300
Required% = 300/600*100 = 50% - Total students from BA discipline from all colleges together what percent more or less than the total students from BSC discipline from all colleges together?
10% more9% less7% less5% less7% moreOption C
Total students in BA from all colleges together = 350+300+450+500+400 = 2000
Total students enrolled in BSC from all colleges = 400+350+400+400+600 = 2150
Required% = (2150-2000)/2150*100 = 7% less - If 140/9% of students enrolled for BA disciplines in college C are females then number of males in BA disciplines in college are what percent of females in college D for BA discipline if females in BA discipline of college D are 2000/7% more than females in BA discipline of college C.
150%140%120%100%90%Option B
Number of female students in BA discipline from college C = 140*150/9*100 = 70
Number of male in BA from college C = 450 – 70 = 380
Female in BA from college D = (1+20/7)*70 = 270
Required% = 380/270*100 = 140% - Number of students enrolled in BSC from C and E together are what percent of the total number of students BCOM from all colleges together?
44%37%40%49%38%Option D
Number of students enrolled in BSC from college C and E together = 400+600 = 1000
Total number of students enrolled in BCOM from all college together = 2050
Required% = 1000/2050*100 = 49%
Directions(6-10): Study the following graph and answer the questions.
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