Sign Up

Have an account? Sign In Now

Sign In

Forgot Password?

Don't have account, Sign Up Here

Forgot Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

Have an account? Sign In Now

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

You must login to ask a question.

Forgot Password?

Need An Account, Sign Up Here

Please briefly explain why you feel this question should be reported.

Please briefly explain why you feel this answer should be reported.

Please briefly explain why you feel this user should be reported.

Sign InSign Up

Abstract Classes

Abstract Classes Logo Abstract Classes Logo
Search
Ask A Question

Mobile menu

Close
Ask a Question
  • Home
  • Polls
  • Add group
  • Buy Points
  • Questions
  • Pending questions
  • Notifications
    • Deleted user - voted up your question.September 24, 2024 at 2:47 pm
    • Abstract Classes has answered your question.September 20, 2024 at 2:13 pm
    • The administrator approved your question.September 20, 2024 at 2:11 pm
    • Deleted user - voted up your question.August 20, 2024 at 3:29 pm
    • Deleted user - voted down your question.August 20, 2024 at 3:29 pm
    • Show all notifications.
  • Messages
  • User Questions
  • Asked Questions
  • Answers
  • Best Answers

N.K. Sharma

Ask N.K. Sharma
539 Visits
0 Followers
930 Questions
Home/ N.K. Sharma/Questions
  • About
  • Questions
  • Polls
  • Answers
  • Best Answers
  • Followed
  • Favorites
  • Asked Questions
  • Groups
  • Joined Groups
  • Managed Groups

Abstract Classes Latest Questions

N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Maths

\[ \text { Find the value of } \frac{16}{\sqrt{3}}\left(\cos 50^{\circ} \cos 10^{\circ} \cos 110^{\circ} \cos 60^{\circ}\right) \]

Find the value of `16/sqrt(3) * (cos(50) * cos(10) * cos(110) * cos(60))`.

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 9:38 am

    Given: - We need to find the value of \(\frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right)\). 1. We use the identity \(\cos x \cos(60 - x) \cos(60 + x) = \frac{1}{4} \cos 3x\): \[ \cos x \cos(60 - x) \cos(60 + x) = \frac{1}{4} \cos 3x \] 2. Applying this identitRead more

    Given:
    – We need to find the value of \(\frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right)\).

    1. We use the identity \(\cos x \cos(60 – x) \cos(60 + x) = \frac{1}{4} \cos 3x\):

    \[ \cos x \cos(60 – x) \cos(60 + x) = \frac{1}{4} \cos 3x \]

    2. Applying this identity to \(\cos 50^\circ, \cos 10^\circ, \cos 110^\circ\):

    \[ \cos 50^\circ \cos 10^\circ \cos 110^\circ = \frac{1}{4} \cos 150^\circ \]
    \[ \cos 150^\circ = -\frac{\sqrt{3}}{2} \]
    \[ \cos 50^\circ \cos 10^\circ \cos 110^\circ = -\frac{\sqrt{3}}{8} \]

    3. Also, \(\cos 60^\circ = \frac{1}{2}\).

    4. Substituting these values into the given expression:

    \[ \frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right) \]
    \[ = \frac{16}{\sqrt{3}} \times \left(-\frac{\sqrt{3}}{8}\right) \times \frac{1}{2} \]
    \[ = -1 \]

    Conclusion:
    The value of \(\frac{16}{\sqrt{3}}\left(\cos 50^\circ \cos 10^\circ \cos 110^\circ \cos 60^\circ\right)\) is \(-1\).

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 33
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Maths

There is a piece of land 10,000 metre square which is to be sold at the rate of Rs. 2000 per square metre. If a man has Rs. 2,50,000 with him, find the percentage of land that he can purchase with this amount.

There is a piece of land 10,000 metre square which is to be sold at the rate of Rs. 2000 per square metre. If a man has Rs. 2,50,000 with him, find the percentage of land that he can purchase ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 9:26 am

    Let's calculate the percentage of land that the man can purchase with Rs. 2,50,000. First, we'll find out the total cost of the land: Total cost of the land = Area of the land × Rate per square metre = 10,000 m² × Rs. 2000/m² = Rs. 2,00,00,000 Now, the man has Rs. 2,50,000 with him. The percentage oRead more

    Let’s calculate the percentage of land that the man can purchase with Rs. 2,50,000.

    First, we’ll find out the total cost of the land:

    Total cost of the land = Area of the land × Rate per square metre
    = 10,000 m² × Rs. 2000/m²
    = Rs. 2,00,00,000

    Now, the man has Rs. 2,50,000 with him. The percentage of the land he can purchase with this amount is:

    Percentage of land = (Amount he has / Total cost of the land) × 100
    = (2,50,000 / 2,00,00,000) × 100
    = 0.0125 × 100
    = 1.25%

    Therefore, the man can purchase 1.25% of the land with Rs. 2,50,000.

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 244
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Maths

A student rides on a bicycle at 5 km/hr and reaches his school 3 minute late. The next day he increased his speed to 7 km/hr and reached school 3 min early. Find the distance between his house and the school.

A student rides on a bicycle at 5 km/hr and reaches his school 3 minute late. The next day he increased his speed to 7 km/hr and reached school 3 min early. Find the distance between his house and the ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 8:53 am

    Let's denote the distance between the student's house and the school as \(d\) km. When the student rides at 5 km/hr, he is 3 minutes late. When he rides at 7 km/hr, he is 3 minutes early. Let's denote the time it takes to reach the school on time as \(t\) hours. We can set up two equations based onRead more

    Let’s denote the distance between the student’s house and the school as \(d\) km.

    When the student rides at 5 km/hr, he is 3 minutes late. When he rides at 7 km/hr, he is 3 minutes early. Let’s denote the time it takes to reach the school on time as \(t\) hours.

    We can set up two equations based on the information given:

    1. When riding at 5 km/hr and being 3 minutes late:

    \[ \frac{d}{5} = t + \frac{3}{60} \]

    2. When riding at 7 km/hr and being 3 minutes early:

    \[ \frac{d}{7} = t – \frac{3}{60} \]

    We can solve these two equations simultaneously to find \(d\) and \(t\).

    From equation 1:

    \[ 60d = 300t + 15 \] (1)

    From equation 2:

    \[ 60d = 420t – 21 \] (2)

    Subtracting equation (2) from equation (1):

    \[ 0 = -120t + 36 \]

    \[ 120t = 36 \]

    \[ t = \frac{36}{120} = \frac{3}{10} \text{ hours} \]

    Substituting \(t\) back into either equation (1) or (2) to find \(d\):

    \[ 60d = 300 \times \frac{3}{10} + 15 \]

    \[ 60d = 90 + 15 \]

    \[ 60d = 105 \]

    \[ d = \frac{105}{60} = 1.75 \text{ km} \]

    Therefore, the distance between the student’s house and the school is 1.75 km.

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 39
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Maths

The ratio of quantity of water in fresh fruits to that of dry fruits is 7 : 2. If 400 kg of dry fruits contain 50 kg of water then find the weight of the water in same fruits when they were fresh?

The ratio of quantity of water in fresh fruits to that of dry fruits is 7 : 2. If 400 kg of dry fruits contain 50 kg of water then find the weight of the water in same fruits when ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 8:45 am

    Let's denote the weight of water in the fresh fruits as \(W_f\) kg and the total weight of the fresh fruits as \(F\) kg. According to the given information, the ratio of the quantity of water in fresh fruits to that of dry fruits is 7:2. This means that for every 7 kg of water in fresh fruits, thereRead more

    Let’s denote the weight of water in the fresh fruits as \(W_f\) kg and the total weight of the fresh fruits as \(F\) kg.

    According to the given information, the ratio of the quantity of water in fresh fruits to that of dry fruits is 7:2. This means that for every 7 kg of water in fresh fruits, there are 2 kg of water in dry fruits.

    We know that 400 kg of dry fruits contain 50 kg of water. Using the given ratio, we can find the weight of water in the fresh fruits:

    \[ \frac{W_f}{50 \text{ kg}} = \frac{7}{2} \]

    Solving for \(W_f\):

    \[ W_f = 50 \text{ kg} \times \frac{7}{2} = 175 \text{ kg} \]

    Therefore, the weight of the water in the same fruits when they were fresh is 175 kg.

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 48
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Maths

\[ \text { If } 2^{2 x-1}=\frac{1}{8^{x-3}} \text {, then the value of } x \text { is } \]

If `2^(2x-1) = 1/(8^(x-3))`, then the value of `x` is:

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 8:43 am

    We can rewrite the equation \(2^{2x - 1} = \frac{1}{8^{x - 3}}\) using the properties of exponents: \[2^{2x - 1} = 2^{-3(x - 3)}\] This is because \(8 = 2^3\), so \(8^{x - 3} = (2^3)^{x - 3} = 2^{3(x - 3)}\). Since the bases are the same, we can set the exponents equal to each other: \[2x - 1 = -3(xRead more

    We can rewrite the equation \(2^{2x – 1} = \frac{1}{8^{x – 3}}\) using the properties of exponents:

    \[2^{2x – 1} = 2^{-3(x – 3)}\]

    This is because \(8 = 2^3\), so \(8^{x – 3} = (2^3)^{x – 3} = 2^{3(x – 3)}\).

    Since the bases are the same, we can set the exponents equal to each other:

    \[2x – 1 = -3(x – 3)\]

    Expanding:

    \[2x – 1 = -3x + 9\]

    Adding \(3x\) to both sides and adding 1 to both sides:

    \[5x = 10\]

    Dividing both sides by 5:

    \[x = 2\]

    Therefore, the value of \(x\) is 2.

    See less
    • 1
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 22
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Maths

What would be the measure of the diagonal of a square whose area is equal to 578 sq cm?

What would be the measure of the diagonal of a square whose area is equal to 578 sq cm?

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 8:26 am

    Let's denote the side length of the square as \(s\) cm. The area of the square is given as 578 sq cm, so we can write: \[ \text{Area} = s^2 = 578 \text{ sq cm} \] To find the side length \(s\), we take the square root of the area: \[ s = \sqrt{578} \text{ cm} \] The diagonal of a square is given byRead more

    Let’s denote the side length of the square as \(s\) cm.

    The area of the square is given as 578 sq cm, so we can write:

    \[ \text{Area} = s^2 = 578 \text{ sq cm} \]

    To find the side length \(s\), we take the square root of the area:

    \[ s = \sqrt{578} \text{ cm} \]

    The diagonal of a square is given by \(d = s\sqrt{2}\), so the length of the diagonal is:

    \[ d = \sqrt{578} \times \sqrt{2} = \sqrt{1156} \text{ cm} = 34 \text{ cm} \]

    Therefore, the measure of the diagonal of the square is 34 cm.

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 24
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Exam, SSC Maths

One of the angles of a triangle is two-third of the sum of the adjacent angles of a parallelogram. The remaining angles of the triangle are in the ratio of 5 : 7. What is the value of the second largest angle of the triangle?

One of the angles of a triangle is two-third of the sum of the adjacent angles of a parallelogram. The remaining angles of the triangle are in the ratio of 5 : 7. What is the value of the second ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 6:36 am

    Let's denote the angles of the triangle as \(A\), \(B\), and \(C\), where \(A\) is the angle that is two-third of the sum of the adjacent angles of a parallelogram, and \(B\) and \(C\) are the remaining angles of the triangle in the ratio of 5:7. Since the sum of the angles in a triangle is \(180^\cRead more

    Let’s denote the angles of the triangle as \(A\), \(B\), and \(C\), where \(A\) is the angle that is two-third of the sum of the adjacent angles of a parallelogram, and \(B\) and \(C\) are the remaining angles of the triangle in the ratio of 5:7.

    Since the sum of the angles in a triangle is \(180^\circ\), we can write:

    \[ A + B + C = 180^\circ \]

    Given that \(A\) is two-third of the sum of the adjacent angles of a parallelogram, and we know that the sum of the adjacent angles of a parallelogram is \(180^\circ\), we have:

    \[ A = \frac{2}{3} \times 180^\circ = 120^\circ \]

    Now, given that \(B\) and \(C\) are in the ratio of 5:7, we can write:

    \[ B = 5x \]
    \[ C = 7x \]

    Since the sum of the angles in the triangle is \(180^\circ\), we have:

    \[ 120^\circ + 5x + 7x = 180^\circ \]
    \[ 12x = 60^\circ \]
    \[ x = 5^\circ \]

    Now, we can find the values of \(B\) and \(C\):

    \[ B = 5x = 5 \times 5^\circ = 25^\circ \]
    \[ C = 7x = 7 \times 5^\circ = 35^\circ \]

    So, the second largest angle of the triangle is \(35^\circ\).

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 37
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Exam, SSC Maths

Sindbad rows 24 km against the flow of water and 54 km with the flow of water in 6 hours. He can also row 36 km against the flow and 48 km with the flow in 8 hours. What is his speed in still water?

Sindbad rows 24 km against the flow of water and 54 km with the flow of water in 6 hours. He can also row 36 km against the flow and 48 km with the flow in 8 hours. What ...

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 6:34 am

    Understanding the Problem We are given the rowing speeds of Sindbad in two different scenarios: 1. He rows 24 km upstream and 54 km downstream in 6 hours. 2. He rows 36 km upstream and 48 km downstream in 8 hours. We need to find Sindbad's speed in still water. Let's denote: - Sindbad's speed upstreRead more

    Understanding the Problem

    We are given the rowing speeds of Sindbad in two different scenarios:
    1. He rows 24 km upstream and 54 km downstream in 6 hours.
    2. He rows 36 km upstream and 48 km downstream in 8 hours.

    We need to find Sindbad’s speed in still water.

    Let’s denote:
    – Sindbad’s speed upstream as \(x\) km/h.
    – Sindbad’s speed downstream as \(y\) km/h.

    Solving the Problem

    From the given scenarios, we can write two equations:

    1. For the first scenario:

    \[ \frac{24}{x} + \frac{54}{y} = 6 \]

    Simplifying, we get:

    \[ \frac{4}{x} + \frac{9}{y} = 1 \text{ —- Eqn(1)} \]

    2. For the second scenario:

    \[ \frac{36}{x} + \frac{48}{y} = 8 \]

    Simplifying, we get:

    \[ \frac{9}{x} + \frac{12}{y} = 2 \text{ —- Eqn(2)} \]

    Solving equations (1) and (2), we find:

    \[ x = \frac{11}{2} \text{ km/h} \text{ (Sindbad’s speed upstream)} \]
    \[ y = 33 \text{ km/h} \text{ (Sindbad’s speed downstream)} \]

    Sindbad’s speed in still water is the average of his speeds upstream and downstream:

    \[ \text{Speed in still water} = \frac{x + y}{2} = \frac{\frac{11}{2} + 33}{2} = \frac{77}{4} = 19.25 \text{ km/h} \]

    Conclusion

    Sindbad’s speed in still water is 19.25 km/h.

    See less
    • 2
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 70
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Exam, SSC Maths

The cost price of a table and a chair together is Rs. 690. If the table costs 30% more than the chair, then find the cost price of the table and the chair respectively.

The cost price of a table and a chair together is Rs. 690. If the table costs 30% more than the chair, then find the cost price of the table and the chair respectively.

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 6:19 am

    Understanding the Problem We are given: - The combined cost price of a table and a chair is Rs. 690. - The table costs 30% more than the chair. We need to find the individual cost prices of the table and the chair. Solving the Problem Let's denote the cost price of the chair as \(C\) rupees. Since tRead more

    Understanding the Problem

    We are given:
    – The combined cost price of a table and a chair is Rs. 690.
    – The table costs 30% more than the chair.

    We need to find the individual cost prices of the table and the chair.

    Solving the Problem

    Let’s denote the cost price of the chair as \(C\) rupees.

    Since the table costs 30% more than the chair, the cost price of the table is \(C + 0.30C = 1.30C\) rupees.

    The combined cost price of the table and chair is given as Rs. 690, so we can write the equation:

    \[ C + 1.30C = 690 \]

    Simplifying:

    \[ 2.30C = 690 \]

    Dividing both sides by 2.30:

    \[ C = \frac{690}{2.30} \]

    \[ C = 300 \]

    Now that we know the cost price of the chair is Rs. 300, we can find the cost price of the table:

    \[ \text{Cost price of the table} = 1.30C = 1.30 \times 300 = 390 \]

    Conclusion

    The cost price of the table is Rs. 390, and the cost price of the chair is Rs. 300.

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 29
  • 0
N.K. Sharma
N.K. Sharma
Asked: April 4, 2024In: SSC Exam, SSC Maths

Vasant can do a piece of work in 24 days. He works at it alone for 4 days and his friend Ritu alone finishes the remaining work in 25 days. Both of them together can complete the work in:

Vasant can do a piece of work in 24 days. He works at it alone for 4 days and his friend Ritu alone finishes the remaining work in 25 days. Both of them together can complete the work in:

SSC CGLSSC Maths Practice Questions with Solution
  1. Abstract Classes Power Elite Author
    Added an answer on April 4, 2024 at 6:08 am

    Let's solve this step by step. First, let's find out how much work Vasant can complete in one day. Since Vasant can do the entire piece of work in 24 days, his work rate is: \[ \text{Vasant's work rate} = \frac{1}{24} \text{ (work per day)} \] Vasant works alone for 4 days, so the fraction of the woRead more

    Let’s solve this step by step.

    First, let’s find out how much work Vasant can complete in one day. Since Vasant can do the entire piece of work in 24 days, his work rate is:

    \[
    \text{Vasant’s work rate} = \frac{1}{24} \text{ (work per day)}
    \]

    Vasant works alone for 4 days, so the fraction of the work he completes is:

    \[
    \text{Work done by Vasant} = 4 \times \frac{1}{24} = \frac{1}{6}
    \]

    This means that after Vasant works for 4 days, \(\frac{1}{6}\) of the work is done and \(\frac{5}{6}\) of the work is remaining.

    Ritu completes the remaining \(\frac{5}{6}\) of the work in 25 days. So, Ritu’s work rate is:

    \[
    \text{Ritu’s work rate} = \frac{\frac{5}{6}}{25} = \frac{1}{30} \text{ (work per day)}
    \]

    Now, we want to find out how long it would take for Vasant and Ritu to complete the work together. The combined work rate of Vasant and Ritu is:

    \[
    \text{Combined work rate} = \text{Vasant’s work rate} + \text{Ritu’s work rate} = \frac{1}{24} + \frac{1}{30}
    \]

    To find the time taken for them to complete the work together, we take the reciprocal of their combined work rate:

    \[
    \text{Time taken together} = \frac{1}{\text{Combined work rate}} = \frac{1}{\frac{1}{24} + \frac{1}{30}}
    \]

    Let’s calculate the exact value of the time taken together using this formula.

    The time taken for Vasant and Ritu to complete the work together is:

    \[
    \text{Time taken together} = \frac{1}{\frac{1}{24} + \frac{1}{30}} = \frac{1}{\frac{30 + 24}{24 \times 30}} = \frac{24 \times 30}{54} = \frac{720}{54} = \frac{40}{3} = 13\frac{1}{3} \text{ days}
    \]

    So, Vasant and Ritu together can complete the work in \(13\frac{1}{3}\) days.

    See less
    • 0
    • Share
      Share
      • Share onFacebook
      • Share on Twitter
      • Share on LinkedIn
      • Share on WhatsApp
  • 0
  • 1
  • 51
  • 0

Sidebar

Ask A Question

Stats

  • Questions 21k
  • Answers 21k
  • Popular
  • Tags
  • Abstract Classes

    testing

    • 0 Comments
  • Pushkar Kumar

    Bachelor of Arts (BAM) | IGNOU

    • 0 Comments
  • Pushkar Kumar

    Bachelor of Science (BSCM) | IGNOU

    • 0 Comments
  • Pushkar Kumar

    Bachelor of Arts(Economics) (BAFEC) | IGNOU

    • 0 Comments
  • Pushkar Kumar

    Bachelor of Arts(English) (BAFEG) | IGNOU

    • 0 Comments
Academic Writing Academic Writing Help BEGS-183 BEGS-183 Solved Assignment Critical Reading Critical Reading Techniques Family & Lineage Generational Conflict Historical Fiction Hybridity & Culture IGNOU Solved Assignments IGNOU Study Guides IGNOU Writing and Study Skills Loss & Displacement Magical Realism Narrative Experimentation Nationalism & Memory Partition Trauma Postcolonial Identity Research Methods Research Skills Study Skills Writing Skills

Users

Arindom Roy

Arindom Roy

  • 102 Questions
  • 104 Answers
Manish Kumar

Manish Kumar

  • 49 Questions
  • 48 Answers
Pushkar Kumar

Pushkar Kumar

  • 57 Questions
  • 56 Answers
Gaurav

Gaurav

  • 535 Questions
  • 534 Answers
Bhulu Aich

Bhulu Aich

  • 2 Questions
  • 0 Answers
Exclusive Author
Ramakant Sharma

Ramakant Sharma

  • 8k Questions
  • 7k Answers
Ink Innovator
Himanshu Kulshreshtha

Himanshu Kulshreshtha

  • 10k Questions
  • 11k Answers
Elite Author
N.K. Sharma

N.K. Sharma

  • 930 Questions
  • 2 Answers

Explore

  • Home
  • Polls
  • Add group
  • Buy Points
  • Questions
  • Pending questions
  • Notifications
    • Deleted user - voted up your question.September 24, 2024 at 2:47 pm
    • Abstract Classes has answered your question.September 20, 2024 at 2:13 pm
    • The administrator approved your question.September 20, 2024 at 2:11 pm
    • Deleted user - voted up your question.August 20, 2024 at 3:29 pm
    • Deleted user - voted down your question.August 20, 2024 at 3:29 pm
    • Show all notifications.
  • Messages
  • User Questions
  • Asked Questions
  • Answers
  • Best Answers

Footer

Abstract Classes

Abstract Classes

Abstract Classes is a dynamic educational platform designed to foster a community of inquiry and learning. As a dedicated social questions & answers engine, we aim to establish a thriving network where students can connect with experts and peers to exchange knowledge, solve problems, and enhance their understanding on a wide range of subjects.

About Us

  • Meet Our Team
  • Contact Us
  • About Us

Legal Terms

  • Privacy Policy
  • Community Guidelines
  • Terms of Service
  • FAQ (Frequently Asked Questions)

© Abstract Classes. All rights reserved.