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Rishabh A.

Science and Math Tutor

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Basic Math

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Question:

Use Euclid's division algorithm to find the HCF of: (i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255

Rishabh A.

Answer:

(i) 135 and 225 Step 1: Since 225 > 135, apply Euclid's division lemma, to a =225 and b=135 to find q and r such that 225 = 135q+r, 0 r<135 On dividing 225 by 135 we get quotient as 1 and remainder as 90 i.e 225 = 135 x 1 + 90 Step 2: Remainder r which is 90 0, we apply Euclid's division lemma to b =135 and r = 90 to find whole numbers q and r such that 135 = 90 x q + r, 0 r<90 On dividing 135 by 90 we get quotient as 1 and remainder as 45 i.e 135 = 90 x 1 + 45 Step 3: Again remainder r = 45 0 so we apply Euclid's division lemma to b =90 and r = 45 to find q and r such that 90 = 90 x q + r, 0 r<45 On dividing 90 by 45 we get quotient as 2 and remainder as 0 i.e 90 = 2 x 45 + 0 Step 4: Since the remainder is zero, the divisor at this stage will be HCF of (135, 225). Since the divisor at this stage is 45, therefore, the HCF of 135 and 225 is 45. (ii) 196 and 38220 Step 1: Since 38220 > 196, apply Euclid's division lemma to a =38220 and b=196 to find whole numbers q and r such that 38220 = 196 q + r, 0 r < 196 On dividing 38220 we get quotient as 195 and remainder r as 0 i.e 38220 = 196 x 195 + 0 Since the remainder is zero, divisor at this stage will be HCF Since divisor at this stage is 196 , therefore, HCF of 196 and 38220 is 196. NOTE: HCF( a,b) = a if a is a factor of b. Here, 196 is a factor of 38220 so HCF is 196. (iii) 867 and 255 Step 1: Since 867 > 255, apply Euclid's division lemma, to a =867 and b=255 to find q and r such that 867 = 255q + r, 0 r<255 On dividing 867 by 255 we get quotient as 3 and remainder as 102 i.e 867 = 255 x 3 + 102 Step 2: Since remainder 102 0, we apply the division lemma to a=255 and b= 102 to find whole numbers q and r such that 255 = 102q + r where 0 r<102 On dividing 255 by 102 we get quotient as 2 and remainder as 51 i.e 255 = 102 x 2 + 51 Step 3: Again remainder 51 is non zero, so we apply the division lemma to a=102 and b= 51 to find whole numbers q and r such that 102 = 51 q + r where 0 r < 51 On dividing 102 by 51 quotient is 2 and remainder is 0 i.e 102 = 51 x 2 + 0 Since the remainder is zero, the divisor at this stage is the HCF Since the divisor at this stage is 51, therefore, HCF of 867 and 255 is 51. Concept Insight: To crack such problem remember to apply Euclid's division Lemma which states that "Given positive integers a and b, there exist unique integers q and r satisfying a = bq + r, where 0 r < b" in the correct order. Here, a > b. Euclid's algorithm works since Dividing 'a' by 'b', replacing 'b' by 'r' and 'a' by 'b' and repeating the process of division till remainder 0 is reached, gives a number which divides a and b exactly. i.e HCF(a,b) =HCF(b,r) Note that do not find the HCF using prime factorization in this question when the method is specified and do not skip steps.

Basic Chemistry

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Question:

Why should a magnesium ribbon be cleaned before burning in air?

Rishabh A.

Answer:

When magnesium ribbon is stored, it reacts with the oxygen of the air to form a layer of magnesium oxide. This layer of magnesium oxide is quite stable and prevents further reaction of magnesium with oxygen. Hence, a magnesium ribbon should be cleaned before burning in air to remove this layer of magnesium oxide. Concept insight: Always remember that magnesium being a reactive metal will react with oxygen if kept in open. So, it has to be cleaned before it is burned in air.

Physics

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Question:

A ray of light traveling in air enters obliquely into the water. Does the light ray bend towards the normal or away from the normal? Why?

Rishabh A.

Answer:

The light ray bends towards the normal. When a ray of light travels from an optically rarer medium to an optically denser medium, it gets bent towards the normal. Since water is optically denser than air, a ray of light traveling from air into the water will bend towards the normal. Concept insight: Air is rarer medium & water is denser medium. The direction of bending of light depends on whether the light is moving from rarer to denser medium or vice versa.

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