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Equations and Inequations Questions: Solved 316 Equations and Inequations Questions and answers section with explanation for various online exam preparation, various interviews, Aptitude Category online test. Category Questions section with detailed description, explanation will help you to master the topic.

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 answer as, 

Refer the above for the Questions 1 to 0
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?   




SHOW ANSWER
Correct Ans:>,>
Explanation:
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
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Directions : Each question contains three quantities as Quantity I, quantity II and Quantity III. You have to determine the relationship between them and give answer as, 

Refer the above for the Questions 2 to 1
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 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?    




SHOW ANSWER
Correct Ans:<, >
Explanation:
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.
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Directions : Each question contains three quantities as Quantity I, quantity II and Quantity III. You have to determine the relationship between them and give answer as, 

Refer the above for the Questions 3 to 2
3. Quantity I: 6 hrs

Quantity II: A man can covers 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 the 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?   




SHOW ANSWER
Correct Ans:>, <
Explanation:
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
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Directions : Each question contains three quantities as Quantity I, quantity II and Quantity III. You have to determine the relationship between them and give answer as, 

Refer the above for the Questions 4 to 3
4. Quantity I: Marked price of an article is 50% more than the cost price. If the shopkeeper gives discount of 10% and cost price of that article is Rs 100 and he sale his 40 articles, then find the profit earned by him.

Quantity II: Seeta bought an article at 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 with respect to the above statements?      




SHOW ANSWER
Correct Ans:>, <
Explanation:
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
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Directions : Each question contains three quantities as Quantity I, quantity II and Quantity III. You have to determine the relationship between them and give answer as, 

Refer the above for the Questions 5 to 4
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 receives 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 with respect to the above statements?    




SHOW ANSWER
Correct Ans:<, >
Explanation:
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
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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




SHOW ANSWER
Correct Ans:Quantity A < Quantity B
Explanation:
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
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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




SHOW ANSWER
Correct Ans:Quantity A < Quantity B
Explanation:
----> 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
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8. 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: x where x3 – 4x2 + 3x = 0

Quantity B: y where 7y3 – 23y2 + 6y = 0           




SHOW ANSWER
Correct Ans:Quantity A = Quantity B OR relationship cannot be determined.
Explanation:
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
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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?




SHOW ANSWER
Correct Ans:Quantity I = Quantity II or Relation cannot be established
Explanation:
-----> 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 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.
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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




SHOW ANSWER
Correct Ans:Quantity II > Quantity I
Explanation:
-----> 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
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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 answer accordingly.

Quantity I : x2 + (343)(1/3) = 56

Quantity II : (y)(4/3) * (y)(5/3) – 295 = 217     




SHOW ANSWER
Correct Ans:Quantity II > Quantity I
Explanation:
----> 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.
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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 "“ 34x + 45 = 0
Quantity II: 4y2 - 19y + 21 = 0




SHOW ANSWER
Correct Ans:Quantity I = Quantity II or Relation cannot be established
Explanation:
---> Quantity I

---> 5x2 – 34x + 45 = 0

---> 5x2 – 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
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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.




SHOW ANSWER
Correct Ans:Quantity II > Quantity I
Explanation:
---> 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

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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 answer accordingly.

3x+y = 81 and 81x-y = 3

Quantity I: value of x
Quantity II: value of y




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
----> 3x+y = 81

----> 3x+y = 34

----> x +y = 4……. (i)

----> 81x-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
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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 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?




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
---> 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
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16. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives 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. 




SHOW ANSWER
Correct Ans:Quantity I > Quantity II
Explanation:
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
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17. In the following question, two quantities are given as Quantity I and Quantity II. By finding these quantities gives corresponding answer.

Quantity I: The area of rectangular garden is 2080 Sq m. 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 Sq m. Find the perimeter of the park? 




SHOW ANSWER
Correct Ans:Quantity I < Quantity II
Explanation:
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
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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 




SHOW ANSWER
Correct Ans:x > y
Explanation:
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
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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 




SHOW ANSWER
Correct Ans:x ≥ y
Explanation:
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
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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 




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