Problems On H.C.F And L.C.M - Aptitude MCQ Questions and Solutions with Explanations

$$\eqalign{ & {\text{Let HCF be }}h{\text{ and LCM be }}l \cr & {\text{Then }}l = {\text{ }}12h{\text{ and}} \cr & {\text{ }}l + h{\text{ }} = {\text{ }}403 \cr & \therefore 12h + h = 403{\text{ or }}h = {\text{ }}31 \cr & {\text{So }}l{\text{ = }}\left( {403 - 31} \right) = 372 \cr & {\text{Hence other number}} \cr & {\text{ = }}\left( {\frac{{31 \times 372}}{{93}}} \right) = 124 \cr} $$

$$\eqalign{ & {\text{LCM of }}\left( {9,10,15} \right) \cr & = 3 \times 3 \times 10 \cr & = 90 \cr & \frac{{1936}}{{90}},{\text{remainder = 46}} \cr} $$
Least number when is subtracted from 1936 which gives remainder 7 when divided by (9, 10, 15) is = (46 - 7) = 39

LCM of (24, 32, 36, 64)
⇒ 8 × 3 × 4 × 3 × 2 = 576
Required number is 576 - 5 = 571

$$\eqalign{ & {\text{Let HCF = }}x \cr & {\text{LCM = 44}}x \cr & {\text{given HCF + LCM}} \cr & = 44x + x = 45x \cr & 45x = {\text{ }}112{\text{5}} \cr & x = \frac{{1125}}{{45}} = 25 \cr & \therefore {\text{HCF}} = {\text{25}} \cr & {\text{LCM = 25}} \times {\text{44}} \cr & {\text{also given that one number}} \cr & {\text{ = 25}} \cr & {\text{Let another number = }}y \cr & \therefore 25y = 25 \times 25 \times 44 \cr & y = \frac{{25 \times 25 \times 44}}{{25}} \cr & y = 1100 \cr} $$

Since HCF is always a factor of LCM, we cannot have three numbers with HCF 35 and LCM 120.

12 - 2 = 10
16 - 6 = 10
24 - 14 = 10
LCM of (12, 16, 24) = 6 × 2 × 4 × 1 = 48
Greatest number of four digits = 9999
∴ When it is divided by 48 we get remainder = 15
⇒ The greatest number of 4 digits which completely divides the given number is = 9999 - 15 = 9984
∴ Number is = 9984 - 10 = 9974

Let the numbers are = 3x, 4x respectively
∴ HCF = x
LCM = 3 × 4 × x = 12x
given that = HCF × LCM
=> x × 12x = 2028
=> 12x² = 2028
=> x² = 169
=> x = 13
∴ sum of numbers
= 3x + 4x
= 7x
= 7 × 13
= 91

HCF = 48
Let number are 48x and 48y respectively
$$\eqalign{ & \Rightarrow 48x + 48y = 384 \cr & \left( {x + y} \right) = \frac{{384}}{{48}} = 8 \cr} $$
So, possible pairs of co-prime number are (1, 7) (3, 5)
numbers are (48, 336) or (144, 240)
∴ difference between numbers is = 336 - 48
= 288 and 240 - 144 = 96

Product of number = 21 × 4641 = 97461
Let the numbers be 21a and 21b
Then 21a × 21b = 97461
ab = 221
Now, co-primes with product 221 are (1, 221) and (13, 17)
So, the numbers are (21 × 1, 21 × 221) and (21 × 13, 21 × 17)
Since one number lies between 200 and 300, the suitable pair is (273, 357)

Let number be 12x and 12y respectively
LCM = 12xy
12xy = 1056 (given)
xy = 88
∴ Possible pairs are (1, 88), (8, 11)
Possible numbers are = (12, 1056)(96, 132)
given that one number is 132
so other is 96