CBSE ADDA

Class 10 Statics mean median mode and Ogive MULTIPLE CHOICE QUESTIONS 1. If 35 is the upper limit of the class-interval of class-size 10, then the lower limit of the class-interval is : (a) 20 (b) 25 (c) 30 (d) none of these 2. In the assumed mean method, if A is the assumed mean, than deviation di is : (a) xi + A (b) xi – A  (c) A – xi (d) none of these 3. Mode is: (a) Middle most value (b) least frequent value (c) most frequent value (d) none of these 4. While computing mean of grouped data, we assume that the frequencies are : (a) evenly distributed over all the classes (b) centred at the class-marks of the classes (c) centred at the upper limits of the classes (d) centred at the lower limits of the classes 5. The curve drawn by taking upper...

X Class 10 Triangle Case based MCQs 1 mark each Q1. If in two triangles ABC and PQR, AB/ QR =BC/ PR= CA/ PQ , then (A) D PQR ~ D CAB (B) D PQR ~ D ABC  (C) D CBA ~ D PQR (D) D BCA ~ D PQR Q2. In ΔABC, DE II BC intersecting AB at D and AC at E, AD = 1cm, DB = 3cm, AE  = 1.5cm, AC =? (A) 6 cm (B) 10 cm  (C) 8 cm (D) None of these Q3. In ΔABC, D is a point on AB and E is a point on AC, DE is joined. AD = 2, DB  = 3, AE = 3 cm, EC = 4.5. Is DE II BC? Q4. The lengths of the diagonals of a rhombus are 16 cm and 12 cm. Then, the  length of the side of the rhombus is (A) 9 cm (B) 10 cm  (C) 8 cm (D) 20 cm Q5. In triangles ABC and DEF, Ð B = Ð E, Ð F = Ð C and AB = 3 DE. Then, the two  triangles are (A) congruent but not similar (B) similar but not congruent (C) neither congruent nor similar (D) congruent as well as similar Q.6 The perimeters of two similar triangles ABC and PQR are respectively 36cm  and 48cm. If  PQ = 12cm, then AB = (a) 16cm (b...

X Real Number MCQ  Assignments in Mathematics Class X (Term I)   1. Euclid’s division algorithm can be applied to : (a) only positive integers            (b) only negative integers (c) all integers                            (d) all integers except 0.  2. For some integer m , every even integer is of the form : (a) m (b) m + 1 (c) 2 m (d) 2 m + 1 3. If the HCF of 65 and 117 is expressible in the form 65 m – 117, then the value of m is : (a) 1 (b) 2 (c) 3 (d) 4 4. If two positive integers p and q can be expressed as p = ab 2 and q = a 3 b , a ; b being prime numbers, then LCM ( p, q ) is : (a) ab (b) a 2 b 2 (c) a 3 b 2 (b) a 3 b 3 5. The least number that is divisible by all...

CBSE_NCERT class 10  Chapter Real Numbers Practice questions 1. The product of three integer p,q,r is 72 where p, q, rare positive integers. HCF of p and q is 2. HCF of p and r is 1. Find the LCM of p,q,r 2. Find the LCM and HCF of 12, 15 and 21 by the prime factorization method. 3. Find the LCM and HCF of 6 and 20 by the prime factorization method. 4. State whether 13/3125 will have a terminating decimal expansion or a non-terminating repeating decimal. 5. State whether 17/8 will have a terminating decimal expansion or a non-terminating repeating decimal. 6. Find the LCM and HCF of 26 and 91 and verify that LCM × HCF = product of the two numbers. 7. Use Euclid’s division algorithm to find the HCF of 135 and 225 8. Use Euclid’s division lemma to show that the square of any positive integer is either of the form 3m or 3m + 1 for some integer m 9. Prove that √3 is irrational. 10. Show that 5 – √3 is irrational 11. Show th...

Self Evaluation Tests Paper For Maths-1 1. The values of the remainder r, when a positive integer a is divided by 3 are 0 and 1 only. Justify your answer No. According to Euclid’s division lemma, a = 3q + r, where 0 ≤ r < 3 and r is an integer. Therefore, the values of r can be 0, 1 or 2. 2. Can the number 6n, n being a natural number, end with the digit 5? Give reasons. No, because 6 n = (2 × 3) n = 2 n × 3 n , so the only primes in the factorization of 6 n are 2 and 3, and not 5. Hence, it cannot end with the digit 5. 3. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer. No, because an integer can be written in the form 4q, 4q+1, 4q+2, 4q+3. 4. “The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons. True, because n (n+1) will always be even, as one out of n or (n+1) must be even 5. “The product of three consecutive...

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 i...

1. Write whether every positive integer can be of the form 4q + 2, where q is an integer. Justify your answer. 2. “The product of two consecutive positive integers is divisible by 2”. Is this statement true or false? Give reasons. 3. “The product of three consecutive positive integers is divisible by 6”. Is this statement true or false”? Justify your answer. 4. Write whether the square of any positive integer can be of the form 3m + 2, where m is a natural number. Justify your answer. 5. A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer. 6. The numbers 525 and 3000 are both divisible only by 3, 5, 15, 25 and 75. What is HCF (525, 3000)? Justify your answer. 7. Explain why 3 × 5 × 7 + 7 is a composite number. 8. Can two numbers have 18 as their HCF and 380 as their LCM? Give reasons. 9. Without actually performing the long divisio...

X Real Numbers : Topics: 1. Euclid's Division Lemma/Algorithm 2. Fundamental Theorem of Arithmetic 3. Irrational Numbers 4. Decimal expression of Rational Number Q.1. Based on Euclid’s algorithm: a = bq + r where  0 ≤ r Ð b Solved example:Using Euclid’s algorithm: Find the HCF of 825 and 175. Explanation: Step 1. Since 825>175. Divide 825 by 175. We get, quotient = 4 and remainder = 125. This can be written as 825 = 175 x 4 + 125 Step II. Now divide 175 by the remainder 125. We get quotient = 1 and remainder = 50. So we write 175 = 125 x 1 + 50. Step III. Repeating the above step we now divide 125 by 50 and get quotient = 2 and remainder = 25. so 125 = 50 x 2 + 25 Step IV. Now divide 50 by 25 to get quotient = 2 and remainder 0. Since remainder has become zero we stop here. Since divisor at this stage is 25, so the HCF of 825 and 175 is 25. Solution: This is how a student should write answer in his answer sheet: Since ...