Q.1. Based on Euclid’s algorithm: a = bq + r ; Using Euclid’s algorithm: Find the HCF of 825 and 175.
Ans:Since 825>175, we division lemma to 825 and 175 to get825 = 175 x 4 + 125.
Since r ≠ 0, we apply division lemma to 175 and 125 to get
175 = 125 x 1 + 50
Again applying division lemma to 125 and 50 we get,
125 = 50 x 2 + 25.
Once again applying division lemma to 50 and 25 we get.
50 = 25 x 2 + 0.
Since remainder has now become 0, this implies that HCF of 825 and 125 is 25.
Q.2. Based on Showing that every positive integer is either of the given forms:
Prove that every odd positive integer is either of the form 4q + 1 or 4q + 3 for some integer q.
Ans: Let a be any odd positive integer (first line of problem) and let b = 4. Using division Lemma we can write a = bq + r, for some integer q, where 0≤r<4. So a can be 4q, 4q + 1, 4q + 2 or 4q + 3. But since a is odd, a cannot be 4q or 4q + 2. Therefore any odd integer is of the form 4q + 1 or 4q + 3.
Q.Find H C F (26,91) if LCM(26,91) is 182
Sol: We know that LCM x HCF = Product of numbers.
or 182 x HCF = 26 x 91
or HCF = 26 x 91 = 13
Hence HCF (26, 91) = 13.
Q. prove that √5 is irrational.
Solution: let us assume on the contrary that √5 is rational. That is we can find co-primes a and b b (≠0) such that √5 = a/b.
Or √5b = a.
Squaring both sides we get
5b2 = a2.
This means 5 divides a2. Hence it follows that 5 divides a.
So we can write a = 5c for some integer c.
Putting this value of a we get
5b2 = (5c)2
Or 5b2 = 25c2
Or b2 = 5b2.
It follows that 5 divides b2. Hence 5 divides b.
Now a and b have at least 5 as a common factor.
But this contradicts the fact that a and b are co-primes.
This contradiction has arisen because of our incorrect assumption that √5 is rational. Hence it follows that √5 is irrational.
Q. prove that product of three consecutive positive integers is divisible by 6?
Ans: Let three consecutive positive integers be, n, n + 1 and n + 2.
Whenever a number is divided by 3, the remainder obtained is either 0 or 1 or 2.
∴ n = 3p or 3p + 1 or 3p + 2, where p is some integer.
If n = 3p, then n is divisible by 3.
If n = 3p + 1, then n + 2 = 3p + 1 + 2 = 3p + 3 = 3(p + 1) is divisible by 3.
If n = 3p + 2, then n + 1 = 3p + 2 + 1 = 3p + 3 = 3(p + 1) is divisible by 3.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 3.
⇒ n (n + 1) (n + 2) is divisible by 3.
Similarly, whenever a number is divided 2, the remainder obtained is 0 or 1.
∴ n = 2q or 2q + 1, where q is some integer.
If n = 2q, then n and n + 2 = 2q + 2 = 2(q + 1) are divisible by 2.
If n = 2q + 1, then n + 1 = 2q + 1 + 1 = 2q + 2 = 2 (q + 1) is divisible by 2.
So, we can say that one of the numbers among n, n + 1 and n + 2 is always divisible by 2.
⇒ n (n + 1) (n + 2) is divisible by 2.
Since, n (n + 1) (n + 2) is divisible by 2 and 3. ∴ n (n + 1) (n + 2) is divisible by 6.
Q. Express HCF of F 65 and 117 in the form of 65m +117n
Ans: Among 65 and 117; 117 > 65
Since 117 > 65, we apply the division lemma to 117 and 65 to obtain
117 = 65 x 1 + 52 … Step 1
Since remainder 52 ≠ 0, we apply the division lemma to 65 and 52 to obtain
65 = 52 x 1 + 13 … Step 2
Since remainder 13 ≠ 0, we apply the division lemma to 52 and 13 to obtain
52 = 4 x 13 + 0 … Step 3
In this step the remainder is zero. Thus, the divisor i.e. 13 in this step is the H.C.F. of the given numbers
The H.C.F. of 65 and 117 is 13
From Step 2:
13 = 65 – 52 x 1 … Step 4
From Step 1:
52 = 117 – 65 x 1
Thus, from Step 4, it is obtained
13 = 65 – (117 – 65 x 1) x 1
⇒13 = 65 x 2 – 117
⇒13 = 65 x 2 + 117 x (–1)
In the above relationship the H.C.F. of 65 and 117 is of the form 65m + 117 n, where m = 2 and n = –1
1oth maths term-1
CLASS X SCIENCE CHEMICAL REACTIONS AND EQUATIONS (solved questions)
1 – Mark questions answer
1. What happens when magnesium ribbon burns in air?
Ans. When magnesium ribbon burns in air, it combines with the oxygen to form magnesium oxide.
2Mg(s) + O2(g) -----> 2MgO(s)
2. Name the gas evolved when zinc reacts with dil. HCl.
Ans. Hydrogen gas is evolved.
3. What is a chemical equation?
Ans. A chemical equation is a symbolic notation that uses formulae instead of words to represent a chemical equation.
4. On what chemical law, balancing of chemical equation is based?
Ans. Balancing of a chemical equation is based on the law of conservation of mass.
5. Represent decomposition of ferrous sulphate with the help of balanced chemical equation.
Ans. 2FeSO4(s) -----> Fe2O3(s) + SO2(g) + SO3(g)
6. When carbon dioxide is passed through lime water, it turns milky, why?
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1. Motion in Living and Non-Living things.
2. Uniform and Non-Uniform Motion.
3. Distance Traversed and Displacement
4. Scalar and Vector Quantities.
5. Speed and velocity
The students will:
1. Understand and distinguish between Rest and Motion
2. Understand Scalar and Vector quantities.
3. Differentiate between Uniform and Non-Uniform motion.
4. Understand and explain the meaning of Average Speed.
Scalars and Vectors, Distance and Displacement, Average Speed, Velocity, Average Velocity
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1. Velocity and Acceleration
Distance – Time
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Speed – Time
Velocity – Time
The students will:
1. Understand the concept of Acceleration.
2. Understand the importance of graphs.
3. Learn how to plot graphs.
4. Compute Speed and Acceleration from graphs.
1. Acceleration 2. Retardation
3. Area under the graph 4. Slope of graph
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2. Numerical based on Equations of Motion
3. Circular Motion (qualitative idea)
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2. Derive the three Equations of Motion.
3. Solve numericals.
4. Explain Circular Motion and identify it as an accelerated motion.
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2. Angular velocity
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Assignment_2012-2013 science 9th chemistry
Assignment_2012-2013 science 9th Physics
The National Council of Educational Research and Training (NCERT) was established by the Government of India in the year 1961 with a view to bringing about qualitative improvement in school education in the country. No sooner the Council was set up than it mounted a number of programmes in this direction. One such programme was to identify and nurture the talented students. This programme took up the shape of a scheme called National Science Talent Search Scheme (NSTSS) in the year 1963 which provided for the identification of talented students and awarding them with scholarships. During the first year of the implementation of the scheme, it was confined to the Union Territory of Delhi wherein only 10 scholarships were awarded to the Class XI students.
Final Result of NTSE 2011 Download
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5. What is the point of intersection of axis called?
6. What is the distance of a point form y -axis called?
7. What is the distance of a point from x-axis called?
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X BOARD SOLVED QUESTIONS Papers 2012
Section – A
1. In fig. AP, AQ and BC are tangents to the circle . If AB = 5 cm , AC = 6 cm and BC = 4cm then length of AP is ( in cm)
(A) 7.5 (B) 15 (C) 10 (D) 9
Ans: (A) 7.5
2. A solid right circular cone is cut into two part at the middle of its height by plane parallel to its base. The ratio of volume of the smaller cone to the whole cone is
(A) 1:2 (B)1:4 (C) 1: 6 (D) 1: 8
Ans: (D) 1: 8
3. A kite is flying at height of 30m from the ground. The length of string from the kite at the ground is (A) 450 (B)300 (C)600 (D)900
4. In fig. P(5,-3),q(3,y) are the points of trisections of the line segments joining A(7,-2) and B(1,-5)Then y equals (A) 2 (B)4 (C)-4 (D)-5/2
5. Card bearing numbers 2,3,4--------------,11 are kept in a bag . A card is drawn at random from the bag . The probability of getting a card with prime number is
(A) 1/ 2 (B) 2/5 (C) 3/10 (D) 5/9
Ans: (A) 1/ 2
6. The distance of point (-3,4) from the x axis is
(A) 3 (B) -3 (C) 4 (D) 5
Ans: (C) 4
7. The circumference of circle is 22cm .The area of its quadrant is
(A) 77/2 (B) 77/4 (C) 77/8 (D) 77/16
Ans: (C) 77/8
8. From a point Q , 13 cm away from the centre of circle ,the length of tangent PQ to the circle is 12 cm . The radius of circle (in cm ) (A) 25 (B) 313 (C) 5 (D) 1
Ans: (C) 5
9. If nth term of AP is (2n+1), then the sum of its first three term
(A) 6n + 3 (B) 15 (C) 12 (D) 21 Ans: (B) 15 10. The root of the quadratic equation 2x2 - x – 6 = 0
(A) -2 , 3/2 (B) 2, -3/2 (C) -2 ,-3/2 (D) 2, 3/2
Ans: (B) 2, -3/2
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