X Real Number MCQ Assignments in Mathematics Class X (Term I)
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![]() 1. Euclid’s division algorithm can be applied to : |
(a) only positive integers (b) only negative integers
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(c) all integers (d) all integers except 0.
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2. For some integer m, every even integer is of the form :
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(a) m (b) m + 1 (c) 2m (d) 2m + 1
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3. If the HCF of 65 and 117 is expressible in the form 65m – 117, then the value of m is :
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(a) 1 (b) 2 (c) 3 (d) 4
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4. If two positive integers p and q can be expressed as p = ab2 and q = a3b, a; b being prime numbers, then LCM (p, q) is :
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(a) ab (b) a2b2 (c) a3b2 (b) a3b3
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5. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) is :
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(a) 10 (b) 100 (c) 504 (d) 2520
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6. 7 × 11 × 13 × 15 + 15 is :
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(a) composite number (b) prime number
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(c) neither composite nor prime (d) none of these
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7. 1.23
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(a) an integer (b) an irrational number (c) a rational number (d) none of these
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8. If two positive integers p and q can be expressed as p = ab2 and q = a2b; a, b being prime numbers, then LCM (p, q) is :
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(a) a2b2 (b) ab (c) ac3b3 (d) a3b2
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9. Euclid’s division lemma states that for two positive integers a and b, there exist unique integers q and r such that a = bq + r, where :
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(a) 0 < r ≤ b (b) 1 < r < b (c) 0 < r < b (d) 0 ≤ r < b
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10. 3.24636363... is :
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(a) a terminating decimal number (b) a non-terminating repeating decimal number
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(c) a rational number (d) both (b) and (c)
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11.(n + 1)2 – 1 is divisible by 8, if n is :
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(a) an odd integer (b) an even integer (c) a natural number (d) an integer
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12. The largest number which divides 71 and 126, leaving remainders 6 and 9 respectively is :
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(a) 1750 (b) 13 (c) 65 (d) 875
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13. For some integer q, every odd integer is of the form :
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(a) 2q (b) 2q + 1 (c) q (d) q + 1
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14. If the HCF of 85 and 153 is expressible in the form 85m – 153, then the value of m is :
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(a) 1 (b) 4 (c) 3 (d) 2
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15. According to Euclid’s division algorithm, HCF of any two positive integers a and b with a > b is obtained by applying Euclid’s division lemma to a and b to find q and r such that a = bq + r, where r must satisfy :
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(a) 1 < r < b (b) 0 < r < b (c) 0 ≤ r < b (d) 0 < r ≤ b
10th Real Numbers CBSE Test Papers Download File |
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....

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