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
