EQUATIONS AND INEQUATIONS SOLVED QUESTIONS

EQUATIONS AND INEQUATIONS QUESTIONS

Directions: Each question contains three quantities as Quantity I, quantity II and Quantity III. You have to determine the relationship between them and give an answer as, 

Refer the above for the Questions 1 to 5 

1. Quantity I: In a class 45 students, boys and girls are in the ratio of 5: 4. The average age of boys is 36 years and a girl is 22.5 years. Find the average age of the class
Quantity II: The ages of Rahul and Praveen are 40 years and 60 years, respectively. How many years before the ratio of their ages was 3: 5?
Quantity III: The ages of Sam and Sudha are in the ratio of 4: 5. After 12 years the ratio of their ages will be 7: 8. What is the difference in their ages?
Which of the following should be placed in the blank spaces of the expression Quantity I__ Quantity II___Quantity III from left to right with respect to the above statements?   

   A. >, <

   B. <, >

   C. >,>

   D. >, ≥

   E. ≤, ≥

Solution: Option C

Quantity I_>_ Quantity II__>_Quantity III
Quantity I:
----> According to the question,
----> 45 * x = 25 * 36 + 20 * 22.5
----> 45x = 900 + 450
----> x = (1350/45) = 30 yrs

Quantity II:
----> Let x year ago, the ratio of Rahul and Praveen’s ages was (3: 5)
----> (40 – x)/(60 – x) = (3/5)
----> 200 – 5x = 180 – 3x
----> x = 10 years

Quantity III:
----> (4x + 12)/(5x + 12) = 7/8
----> 32x + 96 = 35x + 84
----> 3x = 12
----> x = 4
----> Difference in their ages = (5x – 4x) = x = 4 years

 

2. Quantity I: There are 6 men and 6 women in a class, from these 4 members are to be selected to form a committee. Find the number of ways that at least two women were in the committee.
Quantity II: A and B together can do a piece of work in 6 (6/7) days with the help of C finishes in 5 (1/3) days. If C got Rs.200 as wage, then find the total wages of A, B, and C.
Quantity III: A retailer purchases a sewing machine at a discount of 15% and sells it for Rs.1955. In the bargain, he makes a profit of 15%. How much is the discount which he got from the wholesale?
Which of the following should be placed in the blank spaces of the expression Quantity I__ Quantity II___Quantity III from left to right with respect to the above statements?    

    A. >, <

    B. <, > 

    C. >, =

    D. >, ≥

    E. ≤, ≥

Solution : Option B

Quantity I_<_ Quantity II_>__Quantity III

Quantity I:
-----> Number of ways = (⁶C₂ * ⁶C₂) + (⁶C₃ * ⁶C₁) + (⁶C₄)
-----> = [(6 *5 /1 * 2) * (6 * 5/1(1 * 2)]+ [(6 * 5 * 4/1 * 2 * 3) * 6] + (6 * 5/1 * 2)
-----> = (15 * 15) + (20 * 6) + 15
-----> = 225 + 120 + 15
-----> = 360

Quantity II:
-----> C’s one day work = (3/16) – (7/48)
-----> = ((9 – 7)/48)
-----> = (2/48)
-----> = (1/24)
-----> (A + B)’s one day work: C’s one day work = ((7/48) : (1/24))
-----> = (7: 2)
-----> C’s share = (2/9) of total wage = 200
-----> Total wage = (200/2) * 9 = Rs.900

Quantity III:

-----> Let the marked price be Rs. x
-----> Discount = 15% of Rs. x
-----> C.P. = (x – 15% of x) = Rs. (17x / 20)
-----> Then 15% of (17x / 20) = 1955 – (17x / 20)
-----> (51x/400) + (17x/20) = 1955
-----> x = 2000
-----> Discount received by retailer = (15% of 2000) = Rs.300.

 

3. Quantity I: 6 hrs
Quantity II: A man can cover 48 km distance in 4 hours. If he increases his speed by 20% then find the time taken to cross 72 km.
Quantity III: Pipe A, B, and C together can fill a tank in 8 hrs, while C and A together can fill a tank in 12 hrs. If A is 33(1/3)% efficient of pipe C, then find the time taken by Pipe B and C together to fill the tank?
Which of the following should be placed in the blank spaces of the expression Quantity I__ Quantity II___Quantity III from left to right with respect to the above statements?   

   A. >, <

   B. <, >

   C. >, =

   D. >, ≥

   E. ≤, ≥

Solution: Option A

Quantity I_>_ Quantity II__<_Quantity III
Quantity I: 6 hrs

Quantity II:
-----> Original speed = (48/4) = 12 km/hr
-----> New speed = 12 * (120/100) = 14.4 km/hr
-----> Required time = (72/14.4) = 5 hrs

Quantity III:
-----> A + B + C = 8hrs
-----> A + C = 12hrs
-----> LCM of (8,12) = 24
-----> Number of units of water filled by A, B and C together = (24/8) = 3 units
-----> Number of units of water filled by A and C together = (24/12) = 2 units
-----> (i.e.) Number of units of water filled by B = 3 – 2 = 1 units
-----> Number of units of water filled by C in one hour = x units
-----> Number of units of water filled by A in one hour = 33(1/3)% of x = (x/3)
-----> A + C = 2 units
-----> x + (x/3) = 2
-----> x = 1.5 units
-----> Water filled by B and C together = 1 + 1.5 = 2.5 units
-----> Time = (24/2.5) = 9.6 hrs

4. Quantity I: Marked price of an article is 50% more than the cost price. If the shopkeeper gives a discount of 10% and the cost price of that article is Rs 100 and the sale of his 40 articles, then find the profit earned by him.
Quantity II: Seeta bought an article at a 10% discount on the marked price. Seeta marks 20% above and sells to Geeta at a discount of 20%. If Geeta pays Rs. 864, then find the marked price of an article.
Quantity III: A and B enter into a partnership with a capital ratio of 4: 5. After 3 months, A withdrew 3/4th and receives Rs.3500 from the total profit at the end of the year, then find the total profit.
Which of the following should be placed in the blank spaces of the expression Quantity I__ Quantity II___Quantity III from left to right to the above statements?      

    A. >, <

    B. <, >

    C. >, =

    D. >, ≥

    E. ≤, ≥

Solution : Option A

Quantity I_>_ Quantity II_<__Quantity III
Quantity I:
----> Let CP = 100
----> Then MP = 150
----> Now SP = (90/100) * 150
----> = 135
----> Thus profit percentage = 35%
----> Hence required amount = 100 * 40 * (35/100)
----> = Rs.1400

Quantity II:
----> Let us take marked price of an article be Rs. X
----> According to the question,
----> x * (90/100) * (120/100) * (80/100) = 864
----> x = 864 * (100/90) * (100/120) * (100/80)
----> x = Rs. 1000

Quantity III:
----> Profit ratio of A and B = (4x * 3+ 4x * (1/4) * 9): (5x * 12)
----> = (12x + 9x): 60x
----> = 21x: 60x
----> = (7: 200)
----> Let us take the total profit be x
----> A’s share in the total profit = 3500
----> x * (7/27) = 3500
----> x = 500 * 27 = Rs.13500

5. Quantity I: Nithya can do a job in 15 days alone and Priya can do the same job in 18 days alone. A third person Kalai whose efficiency is (5/11)th of efficiency of both Nithya and Priya together, can do the same job in how many days alone?
Quantity II: A and B together can do a piece of work in 120/11 days. If A and B receive Rs.600 and Rs.500 respectively, then find the number of days taken by B alone.
Quantity III: P and Q together can do a piece of work in 20/3 days. R and S together can do the same work in 10 days. Q and R together can do the same work in 6 days. Find the number of days taken by P alone to complete the work, if S alone can do the same work in 30 days.
Which of the following should be placed in the blank spaces of the expression Quantity I__ Quantity II___Quantity III from left to right to the above statements?    

    A. >, <

    B. <, >

    C. >, =

    D. >, ≥

    E. ≤, ≥

Correct Ans:<, >

Quantity I_<_ Quantity II_>__Quantity III
Quantity I:
----> According to the question,
----> One day work of Nithya and Priya together = (1/15) + (1/18)
----> = (11/900)
----> Thus number of days taken by Kalai to complete the work = (90/11) * (11/5)
----> = 18 days

Quantity II:
----> Efficiency ratio of A and B = (600: 500) = (6: 5)
----> Time ratio of A and B = (5: 6) = (5x : 6x)
----> According to the question,
----> (1/5) x + (1/6) x = (11/120)
----> (11/30) x = (11/120)
----> x = 4
----> Number of days taken by B alone = 6 * 4 = 24 days

Quantity III:
----> (P+ Q)’s one day work = (3/20)
----> (Q+R)’s one day work = (1/6)
----> (R+S)’s one day work = (1/10)
----> R’s one day work = (1/10) – (1/30) = ((3 – 1)/30) = (2/30) = (1/15)
----> Q’s one day work = (1/6) – (1/15) = ((5 – 2)/30) = (3/30) = (1/10)
----> P’s one day work = (3/20) – (1/10) = ((3 – 2)/20) = (1/20)
----> P can complete the work in 20 days

6. Given below are two quantities named A and B. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose between the possible answers.
Quantity A: 4X² + 24X + 36 = 0
Quantity B: 5Y² + 20Y + 20 = 0

    A. Quantity A > Quantity B

    B. Quantity A < Quantity B

    C. Quantity A ≥ Quantity B

    D. Quantity A ≤ Quantity B

    E. Quantity A = Quantity B OR relationship cannot be determined.

Correct Ans:Quantity A < Quantity B

Quantity A: 4X² + 24X + 36 = 0
------> X² + 6X + 9 = 0
------> X² + 3X + 3x + 9 = 0
------> X = -3
Quantity B: 5Y² + 20Y + 20 = 0
------> Y² + 4Y + 4 = 0
------> Y² + 4Y + 4y + 4 = 0
------> Y = -2
------> So, Y > X
Quantity A < Quantity B

7. Given below are two quantities named A and B. Based on the given information, you have to determine the relation between the two quantities. You should use the given data and your knowledge of Mathematics to choose between the possible answers.
Quantity A: X² + 34 X + 289 = 0
Quantity B: Y² + 28 Y + 196 = 0

    A. Quantity A > Quantity B

     B. Quantity A < Quantity B

     C. Quantity A ≥ Quantity B

     D. Quantity A ≤ Quantity B

     E. Quantity A = Quantity B OR relationship cannot be determined.

Correct Ans:Quantity A < Quantity B

----> The roots of 1st equation are →
----> X² + 34 X + 289 = 0
----> So, X =-17
----> And roots for 2nd equation are →
----> Y² + 28 Y + 196 = 0
----> Y = -14
----> So, B is the right choice

8. Given below are two quantities named A and B. Based on the given information, you have to determine the relationship between the two quantities. You should use the given data and your knowledge of Mathematics to choose between the possible answers.
Quantity A: x where x³ – 4x² + 3x = 0
Quantity B: y where 7y³ – 23y² + 6y = 0           

    A. Quantity A > Quantity B

    B. Quantity A < Quantity B

    C. Quantity A ≥ Quantity B

    D. Quantity A ≤ Quantity B

    E. Quantity A = Quantity B OR relationship cannot be determined.

Correct Ans:Quantity A = Quantity B OR relationship cannot be determined.

Quantity A:
----> x where x³ – 4x² + 3x = 0
----> This gives x(x² – 4x + 3) = 0
----> x(x - 3)(x - 1) = 0
----> x = 0, 1 or 3
Quantity B:
----> y where 7y³ – 23y² + 6y = 0
----> y(7y² – 23y + 6) = 0
----> y(7y - 2)(y - 3) = 0
----> y = 0, (2/7) or 3
----> x can be less than y or can be equal or more than y
No relation can be established

9. Direction:Calculate quantity I and quantity II on the basic of the given information then compare them and answer the following questions accordingly.
A can do a piece of work in 10 days, B in 15 days. They work together for 5 days, the rest of the work is finished by C in two more days. They get Rs. 6000 as wages for the whole work.
Quantity I: What is the sum of Rs.100 and the daily wage of B?
Quantity II: What is the daily wage of C?

    A. Quantity I > Quantity II

    B. Quantity I ≥ Quantity II

    C. Quantity II > Quantity I

    D. Quantity II ≥ Quantity I

    E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity I = Quantity II or Relation cannot be established

-----> A’s 5 days work = 50%
-----> B’s 5 days work = 33.33%
-----> C’s 2 days work = 100 - (50+33.33)] = 16.66%
-----> Ratio of the contribution of work of A, B, and C = (3: 2: 1)
-----> A’s total share = Rs. 3000
-----> B’s total share = Rs. 2000
-----> C’s total share = Rs. 1000
-----> A’s one day’s earning = Rs.600
-----> B’s one day’s earning = Rs.400
-----> C’s one day’s earning = Rs.500
-----> Thus, the sum of Rs.100 and the daily wage of B = 500
-----> Therefore, Quantity I = Quantity II or relation cannot be established
-----> So option (5) is the correct answer.

10. Direction: Calculate Quantity I and Quantity II on the basis of the given information then compare them and answer the following questions accordingly.
Quantity I- A takes twice as much time B or thrice as much time to finish a piece of work C. Working together, they can finish the work in 3 days, B can do the work alone in
Quantity II- 12 days

    A. Quantity I > Quantity II

    B. Quantity I ≥ Quantity II

    C. Quantity II > Quantity I

    D. Quantity II ≥ Quantity I 

    E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity II > Quantity I

-----> Quantity I:

-----> Let A, B and C take x, x/2 and x/3 days respectively to finish the work
-----> Then,(1/x+2/x+3/x) = (1/3)
-----> (6/x) = 1/3
-----> X = 18
-----> B = (18/2) = 9 days
-----> Quantity II: 12 days
-----> so, Quantity I < Quantity II

11. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give the answer accordingly.
Quantity I : x² + (343)^(1/3) = 56
Quantity II : (y)^(4/3) * (y)^(5/3) – 295 = 217     

    A. Quantity I > Quantity II

    B. Quantity I ≥ Quantity II

    C. Quantity II > Quantity I

    D. Quantity II ≥ Quantity I

    E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity II > Quantity I

----> Quantity I : x² + (343)^(1/3) = 56
----> x² + 7 = 56
----> x² = 49
----> x = √49 = ±7
----> Quantity II : (y)^(4/3) * (y)^(5/3) – 295 = 217
----> (y)³ = 217 + 295
----> (y)³ = 512 = (8)³
----> or, y = 8
----> Here, Quantity II > Quantity I
----> Hence, option C is correct.

12. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.
Quantity I: 5x² â€“ 34x + 45 = 0
Quantity II: 4y² - 19y + 21 = 0

    A. Quantity I > Quantity II

    B. Quantity I ≥ Quantity II

    C. Quantity II > Quantity I

    D. Quantity II ≥ Quantity I

    E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity I = Quantity II or Relation cannot be established

---> Quantity I
---> 5x² â€“ 34x + 45 = 0
---> 5x² â€“ 25x - 9x + 45 = 0
---> 5x(x-5) – 9(x-5) = 0
---> (x-5) (5x-9) = 0
---> So, x = 5, (5/9)

Quantity II :
---> 4y² - 19y + 21 = 0
---> 4y² - 12y - 7y + 21 = 0
---> 4y(y-3) - 7(y-3) = 0
---> (y-3) (4y-7) = 0
---> So, y = 3, (7/4)
Hence there is no relation between Quantity I and Quantity II

13. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give answer accordingly.
Quantity I: Two men start together to walk a certain distance, one at 4km/hr and another at 3km/hr. The former arrives half an hour before the latter. What is the distance?
Quantity II: A man completes 30 km of a journey at 6 km/hr and the remaining 40 km of the journey in 5 hr. Find the average speed for the whole journey.

    A. Quantity I > Quantity II

    B. Quantity I ≥ Quantity II 

    C. Quantity II > Quantity I

    D. Quantity II ≥ Quantity I 

    E. Quantity I = Quantity II or Relation cannot be established

Correct Ans:Quantity II > Quantity I

---> Quantity I :
---> Let the total distance be x
---> According to question
---> (x/3) – (x/4) = (1/2)
---> x = 6 km
---> Quantity II:
---> Total distance = 70 km
---> Total Time = (30/6) +5 = (60/6) = 10 hr
---> Average speed = Total distance/Total time = (70/10) = 7 km/hr
---> Hence Quantity II > Quantity I

14. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give the answer accordingly.
3^x+y = 81 and 81^x-y = 3
Quantity I: value of x
Quantity II: value of y

     A. Quantity I > Quantity II

     B. Quantity I ≥ Quantity II

     C. Quantity II > Quantity I

     D. Quantity II ≥ Quantity I

     E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity I > Quantity I

----> 3^x+y = 81
----> 3^x+y = 34
----> x +y = 4……. (i)
----> 81^x-y = 3
----> (3⁴) ^x-y = 31
----> x –y = ¼ …….. (ii)
----> On solving, we get
----> x = (17/8)
----> y = (15/8)
----> Hence, Quantity I > Quantity II

15. In the following question, two Quantities numbered are given. You have to solve both the Quantities and find out the relationship between Quantity I and Quantity II. Then give the answer accordingly.
Quantity I : If 47a + 47b = 5452. What is the average of a and b?
Quantity II: The average of four consecutive numbers A, B, C, and D is 49.5. What is the value of B?

      A. Quantity I > Quantity II

      B. Quantity I ≥ Quantity II

      C. Quantity II > Quantity I

      D. Quantity II ≥ Quantity I

      E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity I > Quantity II

---> Quantity I:
---> 47(a+ b) = 5452
---> a + b = (5452/47) = 116
---> Average value = (116/2) = 58
---> Quantity II :
---> Let the numbers A, B, C and D be x, x+1, x+2, x+3
So,
---> x + (x+1) + (x+2) + (x+3)/4 = 49.5
---> 4x + 6 = 198
---> 4x = 192
---> x = 48
---> Value of B = 48+1 = 49
---> Hence Quantity I > Quantity II

16. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives the corresponding answer.
M can do a work in 16 days. N is 60% more efficient than M.
Quantity I: Time taken by M and N together to do the work.
Quantity II: Time taken by M and N to do the work together when M works at doubles his original efficiency and N works at half his original efficiency. 

       A. Quantity I > Quantity II

       B. Quantity I < Quantity II

       C. Quantity I ≥ Quantity II

       D. Quantity I ≤ Quantity II

E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I > Quantity II

Quantity I:
M = 16 days; N = 16 * 100/160 =10 days
M + N together = 1/16 + 1/10
= 26/160 = 13/80
= 80/13 days

Quantity II:
M = 16 days; N = 10 days
M (double efficiency) = 8 day; A (half efficiency) = 20 days
M + N together = 1/8 + 1/20
= 28/160 = 14/80
= 80/14 days
Hence, Quantity I > Quantity II

17. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives the corresponding answer.
Quantity I: The area of a rectangular garden is 2080 Sqm. The length of the rectangular garden is 30 % more than the breadth. Find the perimeter of the garden?
Quantity II: The area of a square park is 2304 Sqm. Find the perimeter of the park? 

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

    C. Quantity I ≥ Quantity II

D. Quantity I ≤ Quantity II

     E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I < Quantity II

Quantity I:
The area of rectangular garden = 2080 Sq m
Length = (130/100)*breadth
l/b = 13/10
l : b = 13 : 10
13x*10x = 2080
130x² = 2080
x² = (2080/130) = 16
x = 4
Length = 52 m, Breadth = 40 m
Perimeter of the garden = 2*(l + b) = 2*(52 + 40) = 2*92 = 184 m

Quantity II:
The area of a square park = 2304 Sq m
Area (a²) = 2304
Side (a) = 48
Perimeter of the park = 4a = 4*48 = 192 m
Hence, Quantity II > Quantity I

18. In the following question, there are two equations. Solve the equations and answer accordingly.
I. x² - 52x + 667 = 0
II. 2y² + 20y + 50 = 0 

     A. x > y

     B. x ≥ y 

     C. x ≤ y

     D. x < y

     E. x = y or relationship cannot be determined.

Correct Ans:x > y

I. x² - 52x + 667 = 0
x² - 23x - 29x + 667 = 0
x(x - 23) - 29(x - 23) = 0
(x - 29)(x - 23) = 0
x = 29, 23

II. 2y² + 20y + 50 = 0
2y² + 10y + 10y + 50 = 0
2y(y + 5) + 10(y + 5) = 0
(2y + 10)(y + 5) = 0
y = -5, -5
Hence, x > y

19. In the following question, there are two equations. Solve the equations and answer accordingly.
I. x² - 28x + 171 = 0
II. y² + 3y - 108 = 0 

     A. x > y

     B. x ≥ y

     C. x ≤ y

     D. x < y

     E. x = y or relationship cannot be determined.

Correct Ans:x ≥ y

I. x² - 28x + 171 = 0
x² - 19x - 9x + 171 = 0
x(x - 19) - 9(x - 19) = 0
(x - 9)(x - 19) = 0
x = 9, 19

II. y² + 3y - 108 = 0
y² + 12y - 9y - 108 = 0
y(y + 12) - 9(y + 12) = 0
(y - 9)(y + 12) = 0
y = 9, -12
So, x ≥ y

20. In the following question, there are two equations. Solve the equations and answer accordingly.
I. x² + 29x + 208 = 0
II. y² + 27y + 176 = 0 

      A. x > y

      B. x ≥ y

      C. x ≤ y

      D. x < y 

      E. x = y or relationship cannot be determined.

Correct Ans:x = y or relationship cannot be determined.

I. x² + 29x + 208 = 0

x² + 13x + 16x + 208 = 0
x(x + 13) + 16(x + 13) = 0
(x + 16) (x + 13) = 0
x = -16, -13

II. y² + 27y + 176 = 0
y² + 11y + 16y + 176 = 0
y(y + 11) + 16(y + 11) = 0
(y + 16) (y + 11) = 0
y = -16, -11
x = y or relationship cannot be determined.

21. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives corresponding answer.
Quantity I: If a shopkeeper to make a 20% profit at a 40% discount, what is the profit he will make if he gives a 40% discount?
Quantity II: Four-fifth of two-fifth of 30%of the number is 60. What is 4% of that number? 

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

  C. Quantity I ≥ Quantity II

     D. Quantity I ≤ Quantity II

     E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I < Quantity II

Quantity I:
MP = x
SP = x * 60/100 = 0.6x
CP = 0.6x * 100/120 = 0.5x
Profit = (0.6 - 0.5)x/0.5x * 100 = 20%

Quantity II:
4/5 * 2/5 * 30/100 * x = 60
x = 625
4/100 * 625 = 25
Hence, Quantity II > Quantity I

22. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives the corresponding answer.
Quantity I: In how many different ways can the letters of the word HOLIDAY can be arranged so that vowels occupy odd positions?
Quantity II: In how many different ways can the letters of the word LEADING can be arranged so that vowels always come together? 

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

      C. Quantity I ≥ Quantity II

     D. Quantity I ≤ Quantity II 

     E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I < Quantity II

Quantity I:
Required no. of ways = 4P₃ * 4P₄
4*3*2*4! = 576 ways

Quantity II:
LEADING = 7 letter
3vowels (EAI) supposed to form one letter
Now the remaining letter arranged in 5!
Vowels arranged in 3!
Therefore 5!*3! = 720 ways
Hence, Quantity I < Quantity II

23. In the following question, there are two equations. Solve the equations and answer accordingly:
I. 3x² - 13x +14 = 0
II. 28y² + 11y + 1 = 0 

     A. x > y

     B. x ≥ y

     C. x ≤ y

     D. x < y 

     E. x = y or relationship cannot be determined.

Correct Ans:x > y

I. 3x² - 13x + 14 = 0
3x² - 6x - 7x + 14 = 0
3x(x - 2) - 7(x - 2) = 0
(3x - 7)(x - 2) = 0
x = (7/3, 2)
II. 28y² + 11y + 1 = 0
28y² + 7y + 4y + 1 = 0
y = (-1/4, -1/7)
Hence, x > y

24. In the following question, two equations I and II are given. You have to solve both the equations and give Answer as,
I. x² - x - 272 = 0
II. y² + y - 156 = 0 

      A. x > y

      B. x ≥ y

      C. x ≤ y

      D. x < y

    E. x = y or relationship cannot be determined.

Correct Ans:x = y or relationship cannot be determined.

I. x² - x - 272 = 0
x² - 17x + 16x - 272 = 0
(x - 17) (x + 16) = 0
x = 17, -16

II. y² + y - 156 = 0
y² + 13y - 12y - 156 = 0
(y + 13) (y - 12) = 0
y = -13, 12
Hence the relationship cannot be determined.

25. In the following question, two equations I and II are given. You have to solve both the equations and give Answer as,
I. 5x = 7y + 21
II. 11x + 4y + 109 = 0 

     A. x > y

     B. x ≥ y

     C. x ≤ y

     D. x < y

     E. x = y or relationship cannot be determined.

Correct Ans:x > y

5x - 7y = 21 -------- (1)
11x +4y = -109 -------- (2)
Multiply equ (1) by 11
55x - 77y = 231 --------- (3)
Multiply equ (2) by 5
55x + 20y = -545 --------- (4)
By solving the equation (3) & (4), we get,
-97y = 776
y = -776/97 = -8
Sub y = -8 in equ (1)
5x - 7(-8) = 21
5x + 56 = 21
5x = 21 - 56
5x = -35
x = -35/5 = -7
x = -7, y = -8
Hence, x > y

26. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives the corresponding answer.
Vessel A contains 60 liters mixture of milk and water and Vessel B contains x liters mixture of milk and water in the ratio of 3:2.
Quantity I: If the quantity of water in Vessel A is equal to the quantity of the milk in vessel B and the difference between the quantity of the milk and water in vessel B is 8 liters, then the quantity of water in vessel A is what percent of the quantity of milk in vessel A?
Quantity II: If the quantity of the water in vessel B is 16 liters, then the total quantity of the mixture in vessel B is what percent of the total quantity of the mixture in vessel A? 

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

     C. Quantity I ≥ Quantity II

     D. Quantity I ≤ Quantity II

     E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I = Quantity II or No relation

Quantity I:
3x - 2x = 8 liters
Quantity of the milk in vessel B = 3 * 8 = 24 liters
Quantity of the water in vessel A = 24 liters
Quantity of the milk in vessel A = 60 - 24 = 36 liters
Required percentage = 24/36 * 100 = 66.67%

Quantity II:
Water quantity in vessel B = 16 liters
Total quantity of vessel B = 5/2 * 16 = 40 liters
Required percentage = 40/60 * 100 = 66.67%
Quantity I = Quantity II

27. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives the corresponding answer.
Quantity I: ‘X’: The cost of 5 bags and 12 notes is Rs. 3100 and the cost of 7 bags and 18 notes is Rs. 4400. What is the cost of 3 notes?
Quantity II: ‘Y’: Y² = 22500 

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

     C. Quantity I ≥ Quantity II

     D. Quantity I ≤ Quantity II

     E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I ≥ Quantity II

Quantity I:
According to the question:
5b + 12n = 3100 —– (i)
7b + 18n = 4400 —– (ii)
After solving equation (i) and (ii),
We get, b= 500 and n= 50
Cost of 3 notes = 50×3= 150

Quantity II:
Y² = 22500
Y = ±150
Quantity I ≥ Quantity II

28. In the following question, there are two equations. Solve the equations and answer accordingly.
I. x² + 23x + 120 = 0
II. y² + 13y + 42 = 0 

     A. x > y

     B. x ≥ y

     C. x ≤ y

     D. x < y

     E. x = y or relationship cannot be determined.

Correct Ans:x < y

I. x² + 23x + 120 = 0
x² + 8x + 15x + 120 = 0
x(x + 8) + 15(x + 8) = 0
(x + 15)(x + 8) = 0
x = -15, -8

II. y² + 13y + 42 = 0
y² + 7y + 6y + 42 = 0
y(y + 7) + 6(y + 7) = 0
(y + 6)(y + 7) = 0
y = -6, -7
Hence x < y

29. In the following question, there are two equations. Solve the equations and answer accordingly.
I. 3x² + 26x + 56 = 0
II. 2y² + 15y + 28 = 0 

     A. x > y

     B. x ≥ y

     C. x ≤ y

     D. x < y

     E. x = y or relationship cannot be determined.

Correct Ans:x ≤ y

I. 3x² + 26x + 56 = 0
3x² + 12x + 14x + 56 = 0
3x(x + 4) + 14(x + 4) = 0
(3x + 14) (x + 4) = 0
x = -14/3, -4 = -4.667, -4

II. 2y² + 15y + 28 = 0
2y² + 8y + 7y + 28 = 0
2y(y + 4) + 7 (y + 4) = 0
(2y + 7) (y + 4) = 0
y = -7/2, -4 = -3.5, -4
Hence, x ≤ y

30. In the following question, there are two equations. Solve the equations and answer accordingly.
I. 20x² - 119x + 176 = 0
II. 45x² + 200x + 155 = 0 

    A. x > y

    B. x ≥ y

     C. x ≤ y

    D. x < y

    E. x = y or relationship cannot be determined.

Correct Ans:x > y

I. 20x² - 119x + 176 = 0
20x² - 64x - 55x + 176 = 0
4x(5x - 16) - 11(5x - 16) = 0
(4x - 11)(5x - 16) = 0
x = 11/4, 16/5

II. 45x² + 200x + 155 = 0
45x² + 45x + 155x + 155 = 0
45x(x + 1) + 155(x + 1) = 0
(45x + 155)(x + 1) = 0
x = -155/45, -1
Hence, x > y

31. Directions: In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer
Quantity 1: Compound interest earned on a certain sum at 10% for 2 years is Rs. 1260. Find the sum.
Quantity 2: Simple interest earned on a certain sum at 12% for 3 years is Rs. 2160. Find the sum.

     A. Quantity 1 > Quantity 2

     B. Quantity 1≥Quantity 2

      C. Quantity 1 < Quantity 2

     D. Quantity 1≤Quantity 2

      E. Quantity 1 = Quantity 2

Correct Ans:Quantity 1 = Quantity 2

Quantity 1:
---> Let, the sum of money = Rs. x
---> Rate of interest = 10%
---> No. of years = 2
---> According to problem,
---> x (1+(10/100))² - x = 1260
---> 1.21x – x = 1260
---> 0.21x = 1260
---> x = (1260/0.21)
---> x = 6000
---> The sum of money = Rs. 6000
---> Quantity 1 = 6000

Quantity 2:
---> Let, the sum of money = Rs. y
---> Rate of interest = 12%
---> Time = 3 years
---> According to problem,
---> y * 12 * (3/100) = 2160
---> 0.36 * y = 2160
---> y = 6000
---> The sum of money = Rs. 6000
---> Quantity 2 = 6000

Hence the answer is Quantity 1 = Quantity 2

32. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
Manoj's monthly salary is 25% more than Mohit's salary. Mayank's monthly salary is Rs. 1750 more than Mohit's monthly salary. Sum of Manoj's, Mayank's and Mohit's yearly salary is Rs. 3,33,000.
Quantity I: Sum of monthly salary of Manoj and Mohit together
Quantity II: Rs. 20,000 

      A. Quantity I > Quantity II

      B. Quantity I ≥ Quantity II

      C. Quantity II > Quantity I

      D. Quantity II ≥ Quantity I

      E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity II > Quantity I

Let the Mohit's monthly salary be Rs. X.
Manoj's monthly salary is 25% more than Mohit's salary.
Manoj's monthly salary = 1.25X
Mayank's monthly salary = X + 1750
Mohit's monthly salary + Manoj's monthly salary + Mayank's monthly salary = Total monthly salary
X + 1.25X + X + 1750 = (333000/12)
3.25X = 27750 - 1750
X = 26000/3.25
X = Rs. 8000
Mohit's monthly salary = Rs. 8000
Manoj's monthly salary = Rs. 10,000
Mayank's monthly salary = Rs. 9750

Quantity I:
Sum of monthly salary of Manoj and Mohit = 10000 + 8000 = Rs. 18,000
Hence, Quantity II > Quantity I.

33. Direction: Given below are two quantities named I and II. Based on the given information, you have to determine the relationship between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.
Quantity I: If a shopkeeper sold 5 oranges at the cost price of 6 oranges, then his profit percentage.
Quantity II: A shopkeeper sold an article for Rs. 276. If the cost price of the article is Rs. 240, then his profit percentage.

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

     C. Quantity 1 ≥ Quantity II

     D. Quantity I ≤ Quantity II

     E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I > Quantity II

Quantity I:
---> SP of 5 oranges = CP of 6 oranges
---> (SP/CP) = (6/5)
---> So, profit percentage gained = (1/5) * 100 = 20 %

Quantity II:
---> Selling price of the article = Rs. 276
---> Cost price of the article = Rs. 240
---> Profit percentage = ((276 - 240) /240) * 100 = 15 %
---> So, Quantity I > Quantity II
Hence the answer is: Quantity I > Quantity II

34. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
Quantity I: Distance traveled by bus to reach point B from point A if a car travels the same distance in 5 hrs and the speed of the bus is 120 km/hr which is 120% of the speed of the car.
Quantity II: Distance traveled by boat to reach point D from point C if the speed of the boat in still water is 15 km/hr and speed of the current is 3 km/hr. It goes from point C to D downstream and return back from point D to C upstream in 25 hrs. 

     A. Quantity I > Quantity II

     B. Quantity I ≥ Quantity II

     C. Quantity II > Quantity I

     D. Quantity II ≥ Quantity I

      E. Quantity I = Quantity II or Relation cannot be established

Correct Ans: Quantity I > Quantity II

Quantity I:
WKT, Distance = Speed x Time
Time taken by car = 5 hrs
Speed of bus = 120 km/hr
Given that speed of the bus is 120% of the speed of the car.
120 = (120/100)*Speed of car
Speed of car = 100 km/hr
Distance = 100 x 5 = 500 km.

Quantity II:
Let the distance traveled be 'd' km.
Speed of the boat in still water = 15 km/hr
Speed of current = 3 km/hr
As per question,
[d/(15 + 3)] + [d/(15 - 3)] = 25
(d/18) + (d/12) = 25
(2 + 3)d/36 = 25
5d/36 = 25
d = (25 x 36)/5
d = 180 km
Hence, Quantity I > Quantity II.

35. Direction: Given below are two quantities named I and II. Based on the given information, you have to determine the relationship between the two quantities. You should use the given data and your knowledge of Mathematics to choose among the possible answers.
Amit can complete a piece of work in 60 days whereas Akash and Sumit working together can complete it in 15 days. When Amit and Sumit alternately work for a day each the work gets completed in 40 days.
Quantity 1:No. of days in which Akash will complete twice the work.
Quantity 2:No. of days in which Sumit will complete twice the work.

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

      C. Quantity 1 ≥ Quantity II

     D. Quantity I ≤ Quantity II

      E. Quantity I = Quantity II or No relation

Correct Ans: Quantity I = Quantity II or No relation

Quantity I:
---> Let total units of work be 60 units.
---> Then units done by Amit in one day = 1 unit
---> For 40 days, Amit and Sumit work alternately.
---> Amit does 20 * 1 = 20 units and Sumit does 40 units in 20 days
---> Sumit does 2 units/day. Akash and Sumit do all the units in 15 days.
---> Which means Sumit does 15 * 2 = 30 units and Akash does the remaining 30 units.
---> Hence, Akash does 2 units/day.
---> Efficiency of Akash & Sumit is the same.
Hence, Quantity 1 = Quantity 2

36. In the following question contains a statement followed by Quantity I and Quantity II. Find both to find the relationship among them. Mark your answer accordingly.
The total surface area of a cylinder is 200% more than that of its sum of the area of base and top of the cylinder. The volume of the cylinder is 2156 cm³.
Quantity I: Volume of the cone, whose base radius and height is the same as that of radius and height of cylinder respectively
Quantity II: Volume of the hemisphere, whose radius is the same as that of the radius of the cylinder. 

      A. Quantity I > Quantity II

      B. Quantity I ≥ Quantity II

       C. Quantity II > Quantity I

      D. Quantity II ≥ Quantity I

       E. Quantity I = Quantity II or Relation cannot be established

Correct Ans:Quantity I = Quantity II or Relation cannot be established

WKT, TSA of cylinder = 2π r(r + h)
Sum of area of base and top of cylinder = 2π r²
TSA of cylinder is 200% more than sum of area of base and top of cylinder.
TSA of cylinder : Sum of area of base and top of cylinder = 300 : 100
TSA of cylinder/Sum of area of base and top of cylinder = 3/1
2π r(r + h)/2π r² = 3/1
r + h = 3r
h = 2r
Volume of cylinder = 2156 cm³
π r²h = 2156
π (2r³) = 2156
r³ = (2156 x 7)/(22 x 2)
r = 7 cm
h = 2(7) = 14 cm

Quantity I:
Volume of cone = (1/3)π r²h
= (1/3)*(22/7)*(7)²*(14)
= 2156/3 ³

Quantity II:
Volume of hemisphere = (2/3)π r³
= (2/3)*(22/7)*(7)³
= 2156/3 ³
Hence, Quantity I = Quantity II.

37. Read the following information carefully & establish a relation between quantity I & quantity II.
There are five munch, four 5- stars, and three KitKat chocolate in a bag. A boy takes out four chocolates randomly from the bag.
Quantity I: Probability of selecting two munch and two five stars
Quantity II: Probability of two munch one five stores and one KitKat.

     A. Quantity I > Quantity II

     B. Quantity I < Quantity II

      C. Quantity I ≥ Quantity II

     D. Quantity I ≤ Quantity II

     E. Quantity I = Quantity II or No relation

Correct Ans:Quantity I < Quantity II

Quantity I:
Total number of chocolates = 12
Required probability = [5C2 x 4C2]/12C4
= [(5 x 4)/2 x (4 x 3)/2]/[(12 x 11)/(2 x 3 x 4)]
= 60/495
= 4/33

Quantity II:
Required probability = [5C2 x 4C1 x 3C1]/12C4
= [(5 x 4)/2 x 4 x 3]/[(12 x 11)/(2 x 3 x 4)]
= 120/495
= 8/33
Hence, Quantity I < Quantity II.

38. Read the following information carefully & establish a relation between quantity I & quantity II.
Quantity I: Profit share of 'A'.
A B and C enter into a partnership. 'A' invests Rs. 4000 for the whole year, 'B' puts in Rs. 6000 at the first and increasing to Rs. 8000 at the end of 4
months, whilst C puts in at first Rs. 8000 but withdraw Rs. 2000 at the end of 9 months. The total annual profit is Rs. 56,500.
Quantity II: Amount which when lending on C.I. at 20% interest being compounded annually for 3 years, gives total interest equal to Rs.9100.

     A. Quantity I > Quantity II

     B. Quantity I ≥ Quantity II

      C. Quantity I < Quantity II

     D. Quantity I = Quantity II or No relation

      E. Quantity I ≤ Quantity II

Correct Ans:Quantity I < Quantity II

Quantity I:
WKT, Ratio of Profit = {A's capital * time} : {B's capital * time} : {C's capital * time}
Ratio of Profit = (4000 x 12) : [(6000 x 4) + (8000 x 8)] ; [(8000 x 9) + (6000 x 3)]
= 48000 : 88000 : 90000
= 24 : 44 : 45
Share of A = (24/113)*56500 = Rs. 12,000

Quantity II:
WKT, Amount = CI + P
Amount = P[1 + (r /100)ⁿ]
CI + P = P[1 + (r /100)ⁿ]
9100 + P = P[1 + (20/100)]³
9100 + P =P[1728/1000]
9100 + P = 1.728P
9100 = 0.728P
P = Rs. 12,500.
Hence, Quantity I < Quantity II.

39. In the following question, there are two equations. Solve the equations and answer accordingly.
I. 3x - √2x = 4
II. (9y - 4)/3 = √2y

     A. x > y

     B. x ≥ y

     C. x < y

     D. x ≤ y

      E. x = y or no relation can be established between x and y

Correct Ans:x ≥ y

I. 3x - √2x = 4
(3x - 4)² = (√2x)²
9x² + 16 - 24x = 2x
9x² - 26x + 16 = 0
9x² - 18x - 8x + 16 = 0
9x(x - 2) - 8(x - 2) = 0
(9x - 8) (x - 2) = 0
x = 8/9; 2

II. (9y - 4)/3 = √2y
(9y - 4)² = (3√2y)²
81y² + 16 - 72y = 18y
81y² - 90y + 16 = 0
81y² - 72y - 18y + 16 = 0
9y(9y - 2) - 8(x - 2) = 0
(9y - 2) (9y - 8) = 0
y = 2/9; 8/9
Hence, x ≥ y.

40. In the following question, there are two equations. Solve the equations and answer accordingly.
I. x² - 5√3x + 18 = 0
II. y² - 3√3y - 30 = 0

     A. x > y

     B. x ≥ y

      C. x < y

     D. x ≤ y

      E. x = y or no relation can be established between x and y

Correct Ans: x = y or no relation can be established between x and y

I. x² - 5√3x + 18 = 0
x² - 2√3x - 3√3x + 18 = 0
(x - 2√3x) (x - 3√3x) = 0
x = 2√3x ; 3√3x

II. y² - 3√3y - 30 = 0
y² - 5√3y + 2√3y - 30 = 0
(y + 2√3y) (y - 5√3y) = 0
y = -2√3y ; 5√3y
(x1, y1) = (2√3x ; -2√3y) = x > y
(x2, y1) = (3√3x ; -2√3y) = x > y
(x1, y2) = (2√3x ; 5√3y) = y > x
(x2, y2) = (3√3x ; 5√3y) = y > x
Hence, no relation can be established.

41. In the following question, there are two equations (I) and (II). Solve the equations and answer accordingly.
I. (10/x²) - (13/x) + 4 = 0
II. (14/y²) + 2 = (11/y)

     A. x ≤ y

     B. x < y

      C. x > y

     D. x ≥ y

      E. x = y or no relation can be established between x and

Correct Ans:x ≤ y

Given, I. (10/x²) - (13/x) + 4 = 0
---> (10 - 13x + 4x²) / x² = 0
---> (10 - 13x + 4x²) = 0
---> 4x² - 13x + 10 = 0
---> 4x² - 8x - 5x + 10 = 0
---> 4x(x - 2) - 5(x - 2) = 0
---> (x - 2) (4x - 5) = 0
---> x = 2, 5/4

II. (14/y²) + 2 = (11/y)
---> (14/y²) + 2 - (11/y) = 0
---> (14 + 2y² - 11y) / y²) = 0
---> (14 + 2y² - 11y) = 0
---> 2y² - 11y + 14 = 0
---> 2y² - 4y - 7y + 14 = 0
---> 2y(y - 2) - 7(y - 2) = 0
---> (y - 2) (2y - 7) = 0
---> y = 2, 7/2

Now, Comparing the values of x and y, we get
For x = 2, y = 2 ---> x = y
For x = 2, y = 7/2 ---> x < y
For x = 5/4, y = 2 ---> x < y
For x = 5/4, y = 7/2 ---> x < y
Hence, x ≤ y

 

42. The solution of the pair of equation (x/2) + y = 0.8 and 7/[x + (y/2)] = 10 is

     A. x = 2/5; y = 3/5

     B. x= 2/3; y = 5

      C. x = 2/5; y = 5/3

     D. x = 3/5; y = 2/5

     E. None of these

Correct Ans:x = 2/5; y = 3/5

Given:
(x/2) + y = 0.8
x + 2y = 1.6
10x + 20y = 16 ....(i)
7/[x + (y/2)] = 10
x + (y/2) = 7/10
20x + 10y = 14 ....(ii)
By solving (i) and (ii),
x = 2/5; y = 3/5.

43. Find the appropriate relation for quantity 1 and quantity 2 in the following question:
Two train going in the opposite direction cross each other in 12 sec.
Quantity 1: Length of train 1 if it crosses the pole in 9 sec
Quantity 2: Length of train 2 if it crosses the pole in 24 sec

     A. quantity 1 > quantity 2

     B. quantity 1 ≥ quantity 2

      C. quantity 1 ≤ quantity 2

     D. quantity 1 < quantity 2

     E. quantity 1 = quantity 2

Correct Ans:quantity 1 > quantity 2

Speed = distance/time
Let the speed and length of 1st train be ‘S₁’ and ‘L₁’ respectively.
Let the speed and length of 2nd train be ‘S₂’ and ‘L₂’ respectively.
Given, trains going in the opposite direction cross each other in 12 sec.
The relative speed between the trains going in the opposite direction = S₁ + S₂
Total distance travelled = L₁ + L₂
∴ S₁ + S₂ = (L₁ + L₂)/12 -------- (1)
Quantity I: Length of train 1 if it crosses the pole in 9 sec
⇒ S₁ = L₁/9
Quantity II: Length of train 2 if it crosses the pole in 24 sec
⇒ S₂ = L₂/24
Substituting the value of S₁ and S₂ in eq1
L₁/9 + L₂/24 = L₁/12 + L₂/12
L₁/9 - L₁/12 = L₂/12 - L₂/24
L₁/36 = L₂/24
L₁/3 = L₂/2
L₁/L₂ = 3/2
L₁ : L₂ = 3 : 2
L₁ > L₂
So, quantity 1 > quantity 2

44. Read the following information carefully & establish a relation between quantity I & quantity II:
Quantity I. p² = 81
Quantity II. q² + 19q + 90 = 0

     A. quantity I ≤ quantity II

     B. quantity I ≥ quantity II

      C. quantity I > quantity II

     D. quantity I < quantity II

     E. quantity I = quantity II

Correct Ans: quantity I ≥ quantity II

Let find the value of Quantity I, Quantity II
Quantity I. p² = 81
p = + 9, - 9;
Quantity II. q² + 19q + 90 = 0
q² + 10q + 9q + 90 = 0
(q+9)(q+10) = 0
q = -10, - 9;
Hence, Quantity I ≥ Quantity II

45. In the following question, two equations numbered I and II are given. You have to solve both equations and give the answer.
I. [1/(x - 3)] + [1/(x + 5)] = 1/3
II. (y + 2)(27 - y) = 210 

    A. x > y

    B. x < y

     C. x ≥ y

    D. x ≤ y

     E. x = y or relationship between x and y cannot be established

Correct Ans:x < y

Given expression: I. [1/(x - 3)] + [1/(x + 5)] = 1/3
---> [(x + 5) + (x - 3)] / [(x - 3) * (x + 5)] = 1/3
---> [2x + 2] / (x² + 5x - 3x -15) = 1/3
---> [2x + 2] / (x² + 2x -15) = 1/3
---> [2x + 2] * 3 = (x² + 2x -15)
---> 6x + 6 = (x² + 2x -15)
---> x² + 2x -15 - 6x - 6 = 0
---> x² - 4x - 21 = 0
---> (x -7) (x + 3) = 0
---> x = 7, -3

Given expression: II. (y + 2)(27 - y) = 210
---> 27y - y² + 54 - 2y - 210 = 0
---> 25y - y² -156 = 0
---> y² - 25y + 156 = 0
---> (y - 13) (y - 12) = 0
---> y = 13, 12

Here, x <y