Find the LCM and HCF of the following pairs of integers and verify that LCM × HCF = product of the two numbers.
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
(i) 26 and 91
First we have to find the L.C.M and H.C.F of 26, 91 using fundamental theorem of arithmetic,
26 = 13 × 2
91 = 13 × 7
For L.C.M - list all the prime factors (only once) of 26, 91 with their greatest power.
L.C.M (26, 91) = 13 × 2 × 7 = 182
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (26, 91) = 13
Verification : L.C.M (26, 91) × H.C.F (26, 91) = 26 × 91
182 × 13 = 2366, => 2366 = 2366
Hence, proved.
(ii) 510 and 92
First we have to find the L.C.M and H.C.F of 510, 92 using fundamental theorem of arithmetic,
510 = 2 × 3 × 5 × 17
92 = 2 × 2 × 23
For L.C.M - list all the prime factors (only once) of 510, 92 with their greatest power.
L.C.M (510, 92) = 2 × 2 × 3 × 5 × 17 × 23 = 23460
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (510, 92) = 2
Verification : L.C.M (510, 92) × H.C.F (510, 92) = 510 × 92
23460 × 2 = 46920, => 46920 = 46920
Hence, proved.
(iii) 336 and 54
First we have to find the L.C.M and H.C.F of 336, 54 using fundamental theorem of arithmetic,
336 = 2 × 2 × 2 × 2 × 3 × 7 = 24 × 3 × 7
54 = 2 × 3 × 3 × 3 = 2 × 33
For L.C.M - list all the prime factors (only once) of 336, 54 with their greatest power.
L.C.M (336, 54) = 3024
For H.C.F – write all the common factors (only once) with their smallest exponent.
H.C.F (336, 54) = 2 × 3 = 6
Verification : L.C.M (336, 54) × H.C.F (336, 54) = 336 × 54
3024 × 6 = 18144 => 18144 = 18144
Hence, proved.
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(ii) The probability of an event that cannot happen is . Such an event is called .
(iii) The probability of an event that is certain to happen is . Such an event is called .
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(v) The probability of an event is greater than or equal to and less than or equal to .
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(ii) The product of two consecutive positive integers is 306. We need to find the integers.
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Represent the following situations in the form of quadratic equations :
(i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
(iii) Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
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(i) (x + 1)2 = 2(x – 3) (ii) x2 – 2x = (–2) (3 – x) (iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x2+ 3x + 1 = (x – 2)2 (vii) (x + 2)3 = 2x (x2 – 1) (viii) x3 – 4x2 – x + 1 = (x – 2)3
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completing the square:
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