Test: Real Number (Easy) - Class 10 MCQ

Step 1: Given integers are 210 and 55 such as 210 > 55 By Euclid division lemma 210 = 55 x 3 + 45 (i)

Step 2 : Since remainder 45 ≠ 0. So for divisor 55 and remainder 45 By Euclid division lemma 55 = 45 x 1 + 10 (ii) Remainder ≠ 0

Step 3 : For division 45 and remainder 10 by Euclid division lemma 45 = 10 x 4 + 5 (iii) Remainder ≠ 0

Step 4 : For new divisor 10 and remainder 5 by Euclid division lemma 10 = 5 x 2 + 0 (iv)

Hence remainder = 0 So, HCF of 210, 55 is 5

In order to arrange the books as required, we have to find the largest number that divides 96, 240 and 336 exactly. Clearly, such a number is their HCF Computation of HCF of 96 and 240: Clearly, HCF of 96 and 240 is 48. Computation of HCF of 48 and 336: Thus, HCF of 96, 240 and 336 is 48. Hence, there must be 48 books in each stack. Now,

Number of stacks of English books = Number of English books / Number of books in each stack = 96 / 48 = 2

Number of stacks of Hindi books = Number of Hindi books / Number of books in each stack = 240 / 48 = 5

and, Number of stacks of Mathematics books = Number of Mathematics books / No. of books in each stack = 336 / 48 = 7

210 x 5 + 55y = 5

55y = 5 - 1050

55y = 1045y

= -19

= 2 × 5 × 637

= 2 × 5 × 7 × 91

= 2 × 5 × 7 × 7 × 13m

= 1n

= 1k

= 2p

= 1m + n + k + p

= 1 + 1 + 2 + 1 = 5

by using Euclid's division lemma,

a = bq + r ; 0 < r < b 53124

= (27727)(1) + 25397 + 27727

= (25397)(1) + 2330 + 25397

= (2330)(10) + 2097 + 2330

= (2097)(1) + 233 + 2097

= (233)(9) + 0

H.C.F is 233

106 = 2 × 53

159 = 3 × 53

265 = 5 × 53

Therefore , HCF (106, 159, 265) = 53

=> 2(1) = 2

=> 2(2) = 4

=> 2(-3) = -6

=> 2(-10) = -20

The above results are all even numbers.

So,For some integer m,every even integer is of the form 2m.

65 = 5 × 13

117 = 3 × 3 × 13

Therefore, HCF of 65 and 117 is 13.

So, 65m − 117 = 13 or, 65m = 130

Therefore, m = 2

225 = 195 × 1 + 30

195 = 30 × 6 + 15

30 = 15 × 2 + 0

Therefore, HCF = 15