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CBSE NCERT 10th MATH REAL NUMBER GUESS PAPERS

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Gist of lesson REAL NUMBER CLASS 10th
• Euclid’s Division Lemma : Given two positive integers a and b, there exist unique integers q and r satisfying a = bq + r, 0 ≤ r < b.
• Euclid’s Division Algorithm to obtain the HCF of two positive integers, say c and d, c > d.
Step 1 : Apply Euclid’s division lemma to c and d, to find whole numbers q and r,
such that c = dq + r, 0 ≤ r < d.
Step 2 : If r = 0, d is the HCF of c and d. If r ¹ 0, apply the division lemma to
d and r.
Step 3 : Continue the process till the remainder is zero. The divisor at this stage
will be the required HCF.
• Fundamental Theorem of Arithmetic : Every composite number can be expressed
as a product of primes, and this expression (factorisation) is unique, apart from the
order in which the prime factors occur.
• Let p be a prime number. If p divides a2, then p divides a, where a is a positive
integer.
• √2 , √ 3 , √ 5 are irrational numbers.
• The sum or difference of a rational and an irrational number is irrational.
• The product are quotient of a non-zero rational number and an irrational number is irrational.
• For any two positive integers a and b, HCF (a, b) × LCM (a, b) = a × b.
Let x = p/q , p and q are co-prime, be a rational number whose decimal expansion
terminates. Then, the prime factorisation of q is of the form 2m.5n; m, n are
non-negative integers.
• Let x = p/q be a rational number such that the prime factorisation of q is not of the form 2m.5n; m, n being non-negative integers. Then, x has a non-terminating repeating decimal expansion

Downloadable study files

 Real Numbers CBSE test paper-1

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 Real Numbers CBSE test paper-2

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 Real Numbers CBSE test paper-3 (solved)-3

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 Real Numbers CBSE test paper-4 objectives questions


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10th Real Numbers CBSE Test Papers  
   
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  1. J.Sunil SirApril 02, 2013

    find all positive integral values of n for which n2+96 is a perfect square.
    Let n2 + 96 = x2
    ⇒ x2 – n2 = 96
    ⇒ (x – n) (x + n) = 96
    ⇒ both x and n must be odd or both even
    on these condition the cases are
    x – n = 2, x + n = 48
    x – n = 4, x + n = 24
    x – n = 6, x + n = 16
    x – n = 8, x + n = 12
    and the solution of these equations can be given as
    x = 25, n = 23
    x = 14, n = 10
    x = 11, n = 5
    x = 10, n = 2
    So, the required values of n are 23, 10, 5, and 2.

    For more solved questions visit:
    http://cbseadda.blogspot.in/2011/04/cbse-test-paper-mathematics-class-10.html?

    ReplyDelete

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CBSE ADDA :By Jsunil Sir : Your Ultimate Destination for CBSE Exam Preparation and Academic Insights

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