Skip to main content

CBSE NCERT 10th MATH REAL NUMBER GUESS PAPERS

Image
By
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

 Download File

 Real Numbers CBSE test paper-2

 Download File

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

 Download File

 Real Numbers CBSE test paper-4 objectives questions


Download File
10th Real Numbers CBSE Test Papers  
   
Download File
Labels:

Comments

  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

Post a Comment

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

Living science ratna sagar class 6 solutions

By
Ratna sagar living science 6 answers by jsunil. Class6 Living science solution Term-1 Living Science Solution chapter-1 Source of food Download File Living Science Solution chapter-2 Component of food Download File Living Science Solution chapter-3 Fibre to fabric Download File Living Science Sol ch-4 Sorting of material into group Download File Living Science Soln ch-5 Separation of substance Download File Living Science Solution chapter-6 Change around Us Download File Living Science Solution ch-7 Living and Non Living Download File Living Science Solution ch-8 Getting to Know Plants Download File Living Science Sol ch-9 The Body and Its movements Download File Visit given link for full answer Class6 Living science solution Term-II
Post a Comment
Read more

Class 10 Chapter 02 Acid Bases and Salts NCERT Activity Explanation

By
NCERT Activity Chapter 02 Acid Bases and Salt Class 10 Chemistry Activity 2.1 Indicator Acid Base Red litmus No Change Blue Blue Litmus Red No change Phenolphthalein Colourless Pink Methyl Orange Pink   Yellow Indictors are substance which change colour in acidic or basic media. Activity 2.2 There are some substances whose odour changes in in acidic or basic media. These are called olfactory indicators. Like onion vanilla, onion and clove. These changes smell in basic solution. Activity 2.3 Take about 5 mL of dilute sulphuric acid in a test tube and add few pieces of zinc granules to it. => You will observe bubbles of hydrogen gas on the surface of zinc granules. Zn + H2SO4 --> ZnSO4 + H2 => Pass the Hydrogen gas through the soap solution. Bubbles formed in the soap solution as Hydrogen gas it does not get d...
Post a Comment
Read more

Class 10 Metal and Non MetalsChapter 03 NCERT Activity Solutions

By
X Class 10 NCERT Activity Explanation Class 10 Metals and Non Metals Activity 3.1 Page No. 37 Take samples of iron, copper, aluminium and magnesium. Note the appearance of each sample. They have a shining surface. Clean the surface of each sample by rubbing them with sand paper and note their appearance again. They become more shiny. => Freshly cut Metal have shiny surface Activity 3.2 Page No. 37 Take small pieces of iron, copper, aluminium, and magnesium. Try to cut these metals with a sharp knife and note your observations. They are very hard to cut. Hold a piece of sodium metal with a pair of tongs and try to cut it with a knife. Sodium can be cut easily with knife. Hence K and Na are soft metal cut with knife Activity 3.3 Page No. 38 Take pieces of iron, zinc, lead and copper try to strike it four or five times with a hammer. These metals are beaten into thin sheet on hammering. This property of metal is called malleability and metals are called malleable. Activity 3.4 Page ...
Post a Comment
Read more

Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination(Solved)

By
Class 09 Atoms and Molecules Numerical Problem based on Law of chemical Combination Law of conservation of mass Law of constant proportion Empirical formula 1. If 10 grams of CaCO 3 on heating gave 4.4g of CO 2 and 5.6g of CaO, show that these observations are in agreement with the law of conservation of mass.(Based on Law of conservation of mass) Solution:  Mass of the reactants = 10g ;  Mass of the products = 4.4 + 6.6g = 10g Since the mass of the reactants is equal to the mass of the products, the observations are in agreement with the law of conservation of mass. 2. 1.375 g of cupric oxide was reduced by heating and the weight of copper that remained was 1.098g.  In another experiment 1.179 g of copper was dissolved in nitric acid and the resulting copper nitrate was converted into cupric oxide by ignition . The weight of cupric oxide formed was 1.476 g. which law of chemical combinations does this data state? Solution: in first experiment: Copper oxide = 1....
Post a Comment
Read more

CBSE I NCERT 10th Numerical Problem solved Reflection and reflection of light

By
Q. 1. A concave mirror of focal length 20cm is placed 50 cm from a wall. How far from the wall an object be placed to form its real image on the wall?  Solution: V= -50 cm F= -20cm From mirror formula 1/u = 1/f – 1/v = -1/20+ 1/50 = - 3/100  U = - 33.3 cm Therefore, the distance of the object from the wall x =  50 – u X = 50 – 33.3 = 16.7 cm. Q.2. An object is placed at a distance of 40cm from a concave mirror of focal length 15cm. If the object is displaced through a distance of 20 cm towards the mirror, By how much distance is the image displaced? Answer: Here f = - 15 cm, u = - 40 cm Now 1/f = 1/u + 1/v Then 1/v = 1/f – 1/u Or V= uf/u-f =( - 40 x -15)/25 = -24 cm Then object is displaced towards the mirror let u1 be the distance object from the Mirror in its new position. Then u1 = -(40-20) = -20cm If the image is formed at a distance u1 from the mirror then v1 = u1f/u1-f = -20X-15/-20+15 = -60 cm. = - 20 x-15/-20+15 = -60 ...
2 comments
Read more