Friday, 10 May 2019

CBSE Class 10 - Mathematics - Real Numbers and Polynomials - HCF and LCM related typical problems (#class10Maths)(#cbsenotes)(#eduvictors)

HCF and LCM related Typical Problems

Class 10 - Mathematics - Real Numbers and Polynomials

CBSE Class 10 - Mathematics - Real Numbers and Polynomials - HCF and LCM related typical problems (#class10Maths)(#cbsenotes)(#eduvictors)

   
Q1: Find the HCF of  a²b⁴c³ and a³b²c⁵.

Answer:
H.C.F of  a³ and a² is a²,
H.C.F. of b² and b⁴ is b² and
H.C.F. of c⁵ and c³ is c³

∴ Required H.C.F. is a²b²c³.


Q2:  Find the HCF of 96 and 404 by prime factorisation method. Hence, find their LCM.



Answer: Prime factors of
96  =  2⁵ × 3
404 =  2² × 101

∴ HCF(404,96) = 2² = 4

Now, HCF × LCM = 96 × 404
⇒ LCM = (96 × 404) / 4 = 96 × 101 = 9696         (Answer)


Q3: An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?

Answer:  Clearly, the maximum number of columns in which members can march is the HCF of 616 and 32. So, let us find the HCF of 616 and 32.
Prime Factors 616 =  2³ × 7 × 11
and   32  =  2⁵

∴ HCF(616, 32) = 2³ = 8.

Thus  the required maximum number of columns is 8.


Q4: Find the H.C.F. of  x² + 5x + 6 and x² + x – 6.

Answer: Let us factorise each polynomial
x² + 5x + 6 = x² + 2x + 3x + 6
= x(x + 2) + 3 (x + 2)
= (x + 2)(x + 3)

x² + x – 6 = x² + 3x - 2x – 6
= x(x + 3) -2(x + 3)
= (x + 3)(x - 2)

Only common factor is x + 3
∴ HCF = x + 3


Q5: If H.C.F. of the polynomial  x³ – 2x² + px + 6 and x³ – x² – 5x + q is x – 3, find the value of 3p + 2q².

Answer: As x – 3 is the HCF. of  the given  polynomials.
∴ x – 3 is a factor of x³ – 2x² + px + 6 …(i)
   and x³ – x² – 5x + q …(ii)

∴ When  x – 3 = 0 or x = 3, then the value of polynomials (i) and (ii) is zero. (Factor theorem)


Putting x = 3 in (i),
we get (3)³ – 2(3)² + p(3) + 6 =  0

⇒ 27 – 18 + 3p + 6 = 0

⇒  3p = – 15

⇒  p = – 5

Similarly, putting x = 3 in (ii), we get
(3)³ – (3)² – 5 (3) + q =0

⇒ 27 – 9 – 15 + q =0
⇒ 3 + q = 0
⇒  q = – 3

∴ 3p + 2q² = 3 (– 5) + 2(– 3)² = – 15 + 18 = 3.


Q6: Find the largest possible positive integer that will divide 398, 436, and 542 leaving remainder 7, 11, 15 respectively.  

Answer:
398 - 7  = 391
436 - 11 = 425
542 - 15 = 527

Prime factors of
391 = 17 × 23
425 = 5² × 17
527 = 17 × 31

HCF (391, 425, 527) = 17   (Answer)


Q7: Find the L.C.M. of 12a²b³c² and 18a⁴b²c³

Answer:

12a²b³c² = 2² × 3 × a² × b³ × c²

18a⁴b²c³ = 2 × 3² × a⁴ × b² × c³

∴ LCM = 2² × 3² × a⁴ × b³ × c³ = 36a⁴b³c³


Q8: Find the L.C.M. of the following polynomials : (x - 3)²(x + 4), (x – 3) (x + 4)²

Answer:

Let p(x) = (x - 3)²(x + 4)
    q(x) = (x – 3) (x + 4)²

LCM of  (x - 3)² and (x - 3) is = (x + 3)²

LCM of (x + 4) and (x + 4)² is = (x + 4)²

∴ LCM of p(x) and q(x) = (x + 3)²(x + 4)²                  ...[answer]


Q9: Find the least number that is divisible by all numbers between 1 and 10 (both inclusive). 

Answer:
The required number is the LCM of 1,2,3,4,5,6,7,8,9,10

  ∴ LCM = 2  × 2 × 3 × 2 × 3 × 5 × 7  = 2520


Q10: 144 cartons of Coke Cans and 90 cartons of Pepsi Cans are to be stacked in a Canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have? 

Answer:
Number of cartons of coke cans = 144 Number of cartons of pepsi cans = 90
∴ The greatest number of cartons in one stock = HCF of 144 and 90
By applying Euclid’s division lemma
144 = 90 × 1 + 54 90
= 54 × 1 + 36 54
= 36 × 1 + 18 36
= 18 × 2 + 0
∴ HCF = 18

Hence the greatest number cartons in one stock = 18


Q11:  Sita takes 35 seconds to pack and label a box. For Ram, the same job takes 42  seconds and for Geeta, it takes 28 seconds. If they all start using labeling machines at the same time, after how many seconds will they be using the labeling machines together? 

Answer:
 Required number of seconds is the LCM of 35, 42 and 28
 Prime factors of
  35 = 5 × 7
    42 = 2 × 3 × 7
    28 = 2 × 2 × 7

LCM of 35, 42 and 28 = 2² × 3 × 5 × 7  = 420

Thus Sita, Ram and Geeta will be using the labeling machines together after 420 seconds, i.e 7 minutes.


☛See also:
CH 1: Real Numbers (MCQ)
CH 1: Real  Numbers (Study Points)
CH 1: Real Numbers (NCERT Ex 1.1)
CH 1: Real Numbers (Euclid's Division Lemma - Q & A)
CH 1: Problems on Euclid's Division Algorithm
CH 1: Real Numbers (Problems and Answers)
CH 1: Real Numbers (NCERT Ex 1.2)
CH 1: Real Numbers (NCERT Exemplar Ex 1.1 Q1-3) 
CH 1: Real Numbers (NCERT Exemplar Ex 1.1 Q4-6)
CH 1: Real Numbers (Important Formulae and Key Points)

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