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**Answers :-**

*1. (c) LCM × HCF = 1st. Number × 2nd. number or*

*Product of numbers = HCF × LCM*

*→ LCM = 864*

*HCF = 144*

*One number x = 288*

*Thus Let other no. be y*

*Thus,, x y = LCM × HCF*

*→ 288 × y = 864 × 144*

*y =*

*= 432*

*Thus, Other no. will be = 432*

*2. (c) LCM = 225*

*HCF = 5*

*One number = 25*

*Thus, Let other number be y*

*Thus,. 25 × y = 225 × 5*

*y =*

*Thus, Another no. is 45*

*3. (c) LCM = 30*

*HCF = 5 (given)*

*One number = 10*

*Let another number = y*

*Thus, 10y = 30 × 5*

*y = 15*

*4. (b) HCF = 13*

*LCM = 455*

*Thus, Let number be 13x & 13y*

*Thus, LC M = 13 × y*

*Thus, LCM = HCF × Product of other factor*

*13 x y = 455*

*xy =*

*→ x y = 35*

*Possible co-prime Factors of x,y*

*→ (35, 1) (5, 7)*

*Thus, Numbers may be*

*→ 35 × 13, 1 × 13 = (455, 13)*

*or*

*→ 5 × 13, 7 × 13 = (65, 91)*

*→ But it is given that one number lies between*

*(75 & 125) so.*

*→ Numbers are (65, 91) and number between 75*

*& 125 is 91.*

*LCM of (65, 91) and number between 75 & 125*

*is 91.*

*5.*

*(c) LCM of (4, 6 , 8 , 12, 16)*

*→ 16 × 3 = 48*

*Thus, The number when divided by (4, 6 , 8 , 12,*

*16) leaves reminder 2 is = 48 + 2 = 50*

*6. (d) LCM of (12, 15, 20, 54)*

*→ 4 × 3 × 5 × 9 = 540*

*Thus, the required number is*

*540 + 4 = 544*

*→ Because when divided by LCM each is divided*

*completely, By adding 4 in LCM leaves remainder*

*4.*

*7. (a) 1001 pens, 910 pencils (given)*

*HCF of 1001, 910 is = 91*

*Thus, maximum no. of students are = 91*

*8. LCM of 4 , 6 , 8, 14 = 168 seconds*

*2 4 , 6, 8 , 14*

*2 2, 3, 4, 7*

*1, 3. 2 , 7*

*LCM = 3 × 2 × 7 × 2 × 2 = 168 seconds*

*= 2 minute 48 seconds*

*Thus, 1st they start ringing at 12.00 o’clock*

*→ Again they ring all together after 2 minutes 48*

*seconds at 12 hrs. 2 min. 48 seconds.*

*9. (a) LCM × HCF = 24*

*Thus, Product of numbers = 24*

*Let no. be = x, y*

*xy = 24*

*and x – y = 2 (given)*

*Factors of x y = 24 are (4, 6) (12, 2)*

*(8, 3) (24, 1)*

*→ Now difference between numbers be = (x- y)*

*= 2*

*So, Factor is (4, 6)*

*10.*

*(a) LCM = 495*

*HCF = 5 (given)*

*Thus, Let numbers are = 5x & 5y*

*Thus, LCM = 5 x y*

*5 x y = 495*

*x y = 99*

*thus, Possible co-prime factors are*

*1 , 99*

*9 , 11*

*Thus, Possible numbers are*

*5x , 5y = 45, 55*

*5, 495*

*Now, given that sum of numbers = 100*

*so, required numbers are = (45, 55)*

*Thus, Difference of numbers = 55 – 45 = 10*