InterviewSolution
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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 801. |
Show that (3x – 2y)2 + 24xy = (3x + 2y)2. |
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Answer» LHS (3x – 2y)2 + 24xy = [(3x)2 – 2(3x) (2y) + (2y)2] + 24xy = (9x2 – 12xy + 4y2) + 24xy = 9x2 – 12xy + 24xy + 4y2 = 9x2 + 12xy + 4y2 = (3x)2 + 2 (3x) (2y) + (2y)2 = (3x + 2y)2 (Using identity I) = RHS |
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| 802. |
Find the coefficient of x2 in -3/7 x2y2 . |
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Answer» The coefficient of x2 in -3/7x2y2 = -3/7 y2. |
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| 803. |
Coefficient of x2 in 5x2 will be :(A) 1(B) 4(C) 0(D) 5 |
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Answer» Coefficient of x2 in 5x2 will be 5. |
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| 804. |
Write the coefficient of x2 in the following:(i) −3x2(ii) 5x2yz(iii) 5/7x2z(iv) (-3/2) ax2 + yx |
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Answer» (i) Given −3x2 The numerical coefficient of x2 is -3. (ii) Given 5x2yz The numerical coefficient of x2 is 5yz. (iii) Given 5/7x2z The numerical coefficient of x2 is 57z. (iv) Given (-3/2) ax2 + yx The numerical coefficient of x2 is – (3/2) a. |
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| 805. |
Rakesh’s age is less than Sania’s age by 5 years. The sum of their ages is 27 years. How old are they? |
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Answer» Let the age of Rakesh be x years. ∴ Sania’s age = (x + 5) years. According to the given condition, x + (x + 5) = 27 ∴ 2x + 5 = 27 ∴ 2x = 27 – 5 ∴ 2x = 22 ∴ x = 22/2 = 11 Sania’s age = x + 5 = 11 + 5 = 16 years ∴ The ages of Rakesh and Sania are 11 years and 16 years respectively. |
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| 806. |
Rohan’s mother gave him Rs. 3xy2 and his father gave him Rs. 5(xy2 + 2). Out of this money he spent Rs. (10 – 3xy2) on his birthday party. How much money is left with him? |
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Answer» Given, amount given to Rohan by his mother =Rs. 3xy2 and amount given to Rohan by his father =Rs. 5(xy2 + 2) ∴ Total amount Rohan have = [(3xy2)+(5xy2 + 10)] = Rs. [3xy2 + 5xy2 + 10] = Rs. (8xy2 + 10) Total amount spent by Rohan =Rs. (10 – 3xy2) ∴After spending, Rohan have left money |
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| 807. |
State whether the algebraic expression given below is monomial, binomial, trinomial or multinomial.(i) y2(ii) 4y — 7z(iii) 1 + x + x2(iv) 7mn(v) a2 + b2(vi) 100 xyz(vii) ax + 9(viii) p2 – 3pq + r(ix) 3y2 – x2y2 + 4x(x) 7x2 – 2xy + 9y2 – 11 |
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| 808. |
Add x2 + 2xy + y2 to the sum of x2 – 3y2and 2x2 – y2 + 9. |
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Answer» Given x2 + 2xy + y2, x2 – 3y2and 2x2 – y2 + 9. First we have to find the sum of x2 – 3y2 and 2x2 – y2 + 9 = (x2 – 3y2) + (2x2 – y2 + 9) = x2 + 2x2 – 3y2 – y2+ 9 = 3x2 – 4y2 + 9 Now, required expression = (x2 + 2xy + y2) + (3x2 – 4y2 + 9) = x2 + 3x2 + 2xy + y2 – 4y2 + 9 = 4x2 + 2xy – 3y2+ 9 |
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| 809. |
Write the definition of Algebra? |
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Answer» Algebra is one of the broad parts of mathematics, together with number theory, geometry, and analysis. The more basic parts of algebra are called elementary algebra; the more abstract parts are called abstract algebra or modern algebra. |
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| 810. |
The common factor of 3ab and 2cd is(a) 1 (b) – 1 (c) a (d) c |
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Answer» (a) 1 Considering the two monomials 3ab and 2cd there is no common factor except 1. |
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