Practice Lcm and hcf - quantitative aptitude Online Quiz (set-1) For All Competitive Exams
Q-1) Let N be the greatest number that will divide 1305, 4665 and 6905 leaving the same remainder in each case. Then, sum of the digits in N is :
(a)
(b)
(c)
(d)
Using Rule 7,
The greatest number N = HCF of (1305 – x ), (4665 – x ) and (6905 – x),
where x is the remainder
= HCF of (4665 – 1305), (6905– 4665) and (6905 – 1305)
= HCF of 3360, 2240 and 5600
∴ N = 1120
Sum of digits = 1 + 1 + 2 + 0 = 4
Q-2) There are 24 peaches, 36 apricots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of only one type. What is the minimum number of rows required for this to happen ?
(a)
(b)
(c)
(d)
Minimum number of rows = Maximum number of fruits in each row
∴ HCF of 24, 36 and 60 = 12
∴ Minimum number of rows
= $24/12 + 36/12 + 60/12$
= 2 + 3 + 5 = 10
Q-3) The largest number, which divides 25, 73 and 97 to leave the same remainder in each case, is
(a)
(b)
(c)
(d)
Using Rule 7,
Let x be the remainder.
Then, (25 – x ), (73 – x ), and (97 – x )
Will be exactly divisible by the required number.
∴ Required number
= HCF of (73 –x ) – (25 –x ), (97 –x ) – (73 –x ) and (97 –x ) – (25 –x )
= HCF of (73 –25), (97 –73), and (97 –25)
= HCF of 48, 24 and 72 = 24
Q-4) The sum of a pair of positive integer is 336 and their H.C.F. is 21. The number of such possible pairs is
(a)
(b)
(c)
(d)
Let the numbers be 21x and 21y
where x and y are prime to each other.
21x + 21y = 336
21 (x + y) = 336
x + y = $336/21$ = 16
∴ Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)
Q-5) The product of the LCM and the HCF of two numbers is 24. If the difference of the numbers is 2, then the greater of the number is
(a)
(b)
(c)
(d)
Using Rule 1,
1st number × 2nd number = L.C.M. × H.C.F
Let the larger number be x.
Smaller number = x – 2
First number × Second number = HCF × LCM
→ x (x – 2) = 24
→ $x^2$ – 2x – 24 = 0
→ $x^2$ – 6x + 4x – 24 = 0
→ x (x – 6) + 4 (x – 6) = 0
→ (x – 6) (x + 4) = 0
→ x = 6 because x ≠ – 4
Q-6) Let x be the least number, which when divided by 5, 6, 7 and 8 leaves a remainder 3 in each case but when divided by 9 leaves no remainder. The sum of digits of x is
(a)
(b)
(c)
(d)
LCM of 5, 6, 7 and 8 = 840
| 2 | 5, | 6, | 7, | 8 |
| 5, | 3, | 7, | 4 |
∴ LCM = 2 × 5 × 3 × 7 × 4 = 840
Required number = 840x + 3
which is divisible by 9 for a certain least value of x.
Now, 840x + 3 = 93x × 9 + 3x + 3
3x + 3, is divisible by 9 for x = 2
Required number = 840 × 2 + 3
= 1680 + 3 = 1683
∴ Sum of digits = 1 + 6 + 8 + 3 = 18
Q-7) Three numbers which are coprime to one another are such that the product of the first two is 551 and that of the last two is 1073. The sum of the three numbers is :
(a)
(b)
(c)
(d)
Let the numbers be x, y and z which are prime to one another.
Now, xy = 551 yz = 1073
∴ y = HCF of 551 and 1073
y = 29
x = $551/29$ = 19
and z = $1073/29$ = 37
∴Sum = 19 + 29 + 37 = 85
Q-8) The HCF of two numbers is 15 and their LCM is 300. If one of the number is 60, the other is :
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM
∴ Second number = ${15 × 300}/60$ = 75
Q-9) The HCF and LCM of two numbers are 13 and 455 respectively. If one of the number lies between 75 and 125, then, that number is :
(a)
(b)
(c)
(d)
HCF = 13
Let the numbers be 13x and 13y.
Where x and y are co-prime.
∴ LCM = 13 xy
∴ 13 xy = 455
∴ xy = $455/13$ = 35 = 5 × 7
∴ Numbers are 13 × 5 = 65 and 13 × 7 = 91
Q-10) The LCM of two numbers is 4 times their HCF. The sum of LCM and HCF is 125. If one of the number is 100, then the other number is
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
Let LCM be L and HCF be H, then L = 4H
∴ H + 4H = 125
⇒ 5H = 125
⇒ H = ${125}/5$ = 25
∴ L = 4 × 25 = 100
∴ Second number = ${L× H}/{First number}$
= ${100 ×25}/100$ = 25
Q-11) The H.C.F. of two numbers, each having three digits , is 17 and their L.C.M. is 714. The sum of the numbers will be :
(a)
(b)
(c)
(d)
Let the numbers be 17x and 17y
where x and y are co-prime.
LCM of 17x and 17y = 17 xy
According to the question,
17xy = 714
⇒ xy = $714/17$ = 42 = 6 × 7
⇒ x = 6 and y = 7
or, x = 7 and y =6.
First number = 17x = 17 × 6 = 102
Second number = 17y =17 × 7 = 119
∴ Sum of the numbers = 102 + 119 = 221
Q-12) The product of two numbers is 4107. If the H.C.F. of the numbers is 37, the greater number is
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
LCM = ${Product of two numbers}/ {HCF}$ = $4107/37$ = 111
Obviously, numbers are 111 and 37 which satisfy the given condition.
Hence, the greater number = 111
Q-13) The sum of two numbers is 45. Their difference is 1 9 of their sum. Their L.C.M. is
(a)
(b)
(c)
(d)
Let the number be x and y.
According to the question,
x + y = 45 ......... (i)
Again, x – y = $1/9$(x + y)
x – y = $1/9$×45
or x – y = 5 ..... (ii)
By (i) + (ii) we have,
x + y = 45
$\text"x – y = 5"/\text"2x"$ = 50
or, x = 25
y = 45 – 25 = 20.
Now, LCM of 25 and 20 = 100.
Q-14) The product of two numbers is 2160 and their HCF is 12. Number of such possible pairs is
(a)
(b)
(c)
(d)
HCF = 12
Numbers = 12x and 12y
where x and y are prime to each other.
∴ 12x × 12y = 2160
⇒ xy = $2160/{12 × 12} $
= 15 = 3 × 5, 1 × 15
Possible pairs = (36, 60) and (12, 180)
Q-15) The L.C.M. of two numbers is 20 times their H.C.F. The sum of H.C.F. and L.C.M. is 2520. If one of the number is 480, the other number is :
(a)
(b)
(c)
(d)
Using Rule 1,
1st number × 2nd number = L.C.M. × H.C.F
Let the H.C.F. be H.
L.C.M. = 20H
Then, H + 20H = 2520
⇒ 21 H = 2520
⇒ H = $2520/21$ = 120
∴ L.C.M. = 20H = 20×120= 2400
As, First number × Second number = L.C.M. × H.C.F.
⇒ 480 × Second number = 2400 × 120
⇒ Second number = ${2400 × 120}/480$ = 600
Q-16) The product of two numbers is 1280 and their H.C.F. is 8. The L.C.M. of the number will be ?
(a)
(b)
(c)
(d)
Product of two numbers = HCF × LCM
1280 = 8 × LCM
160 = LCM
Q-17) The HCF of two numbers is 15 and their LCM is 225. If one of the number is 75, then the other number is ?
(a)
(b)
(c)
(d)
First Number x Second Number = HCF x LCM
Then, 75 x Second Number = 15 x 225
Second Number $latex = ${15 x 225}/75 = 45$
Therefore, other number is 45.
Q-18) The HCF of two numbers is 23 and the other two factors of their LCM are 13 and 14. The larger of the two numbers is :
(a)
(b)
(c)
(d)
HCF is 23. So the other two numbers would be,
(23 * 13) and (23 * 14).
Thus Larger Number = 23 * 14 = 322.
Q-19) The H.C.F. of two numbers is 96 and their L.C.M. is 1296. If one of the number is 864, the other is
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM
⇒ 864 × Second number
= 96 × 1296 ⇒ Second number
${96 × 1296}/864$ = 144
Q-20) The LCM and the HCF of the numbers 28 and 42 are in the ratio :
(a)
(b)
(c)
(d)
L.C.M. of 28 and 42
| 2 | 28, | 42 |
| 2 | 14, | 21 |
| 7 | 7, | 21 |
| 1, | 3 |
= 2 × 2 × 7 × 3 = 84
H.C. F. of 28 and 42
∴ H.C. F = 14
Required ratio = $84/14$ = 6 : 1