Practice Lcm and hcf - quantitative aptitude Online Quiz (set-2) For All Competitive Exams
Q-1) The sum of two numbers is 84 and their HCF is 12. Total number of such pairs of number is
(a)
(b)
(c)
(d)
HCF = 12
∴ Numbers = 12x and 12y where x and y are prime to each other.
∴ 12x + 12y = 84
⇒ 12 (x + y) = 84
⇒ x + y = $84/12$ = 7
∴ Possible pairs of numbers satisfying this condition
= (1,6), (2,5) and (3,4). Hence three pairs are of required numbers.
Q-2) If the product of three consecutive numbers is 210 then sum of the smaller number is :
(a)
(b)
(c)
(d)
| 2 | 210 |
| 3 | 105 |
| 5 | 35 |
| 7 |
210 = 2 × 3 × 5 × 7 = 5 × 6 × 7
∴ Required answer = 5 + 6 = 11
Q-3) The sum of two numbers is 216 and their HCF is 27. How many pairs of such numbers are there?
(a)
(b)
(c)
(d)
HCF of two numbers = 27
Let the numbers be 27x and 27y
where x and y are prime to each other.
According to the question,
27x + 27y = 216
27 (x + y) = 216
x + y = $216/27$ = 8
Possible pairs of x and y = (1, 7) and (3, 5)
Numbers =(27, 189) and (81, 135)
Q-4) The sum of two numbers is 36 and their H.C.F. is 4. How many pairs of such numbers are possible ?
(a)
(b)
(c)
(d)
HCF of two numbers = 4.
Hence, the numbers can be given by 4x and 4y
where x and y are co-prime.
Then, 4x + 4y = 36
4 (x + y) = 36
x + y = 9
Possible pairs satisfying this condition are : (1, 8), (4, 5), (2, 7)
Q-5) Let x be the smallest number, which when added to 2000 makes the resulting number divisible by 12, 16, 18 and 21. The sum of the digits of x is
(a)
(b)
(c)
(d)
| 2 | 12, | 16, | 18, | 21 |
| 2 | 6, | 8, | 9, | 21 |
| 3 | 3, | 4, | 9, | 21 |
| 1, | 4, | 3, | 7 |
LCM = 2 × 2 × 3 × 4 × 3 × 7 = 1008
Multiple of 1008 = 2016
∴ Required number = 2016 – 2000 = 16 = x
∴ Sum of digits of x = 1 + 6 = 7
Q-6) The LCM of two numbers is 520 and their HCF is 4. If one of the number is 52, then the other number is
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × second number = HCF × LCM
⇒ 52 × second number = 4 × 520
⇒ Second number =${4 × 520}/52$ = 40
Q-7) The LCM of two positive integers is twice the larger number. The difference of the smaller number and the GCD of the two numbers is 4. The smaller number is :
(a)
(b)
(c)
(d)
Let the numbers be x H and yH
where H is the HCF and yH > x H.
LCM = xy H
xyH = 2yH
→ x = 2
Again, x H – H = 4
→ 2H – H = 4 ⇒ H = 4
∴ Smaller number = x H = 8
Q-8) The H.C.F and L.C.M of two numbers are 12 and 336 respectively. If one of the number is 84, 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
⇒ 84 × second number = 12 × 336
∴ Second number = ${12 × 336}/84$ = 48
Q-9) If the HCF and LCM of two consecutive (positive) even numbers be 2 and 84 respectively, then the sum of the numbers is
(a)
(b)
(c)
(d)
Let the numbers be 2x and 2y
where x and y are prime to each other.
∴LCM = 2xy
⇒ 2xy = 84
⇒ xy = 42 = 6 × 7
∴ Numbers are 12 and 14.
Hence Sum = 12 + 14 = 26
Q-10) 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)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number × Second number = HCF × LCM
⇒ 75 × Second number = 15 × 225
∴ Second number = ${15 × 225}/75$ = 45
Q-11) The HCF and LCM of two numbers are 18 and 378 respectively. If one of the number is 54, then the other number is
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
Second number =${HCF × LCM}/{First number} $
= ${18 ×378}/54$ = 126
Q-12) 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)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
HCF × LCM = Product of two numbers
⇒ 8 × LCM = 1280
⇒ LCM = $1280/8$ =160
Q-13) The HCF and LCM of two numbers are 12 and 924 respectively. Then the number of such pairs is
(a)
(b)
(c)
(d)
Let the numbers be 12x and 12y where x and y are prime to each other.
∴ LCM = 12xy
∴ 12xy = 924
⇒ xy = 77
∴ Possible pairs = (1,77) and (7,11)
Q-14) The H.C.F. and L.C.M. of two numbers are 44 and 264 respectively. If the first number is divided by 2, the quotient is 44. The other number is
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
First number = 2 × 44 = 88
∴ First number × Second number
= H.C.F. × L.C.M.
⇒ 88 × Second numebr
= 44 × 264
⇒ Second number = ${44 × 264}/88$ = 132
Q-15) The smallest number, which when increased by 5 is divisible by each of 24,32, 36 and 564, is
(a)
(b)
(c)
(d)
Required number = (LCM of 24, 32, 36 and 54) – 5
Now,
| 2 | 24, | 32, | 36, | 54 |
| 2 | 12, | 16, | 18, | 27 |
| 2 | 6, | 8, | 9, | 27 |
| 3 | 3, | 4, | 9, | 27 |
| 3 | 1, | 4, | 3, | 9 |
| 1, | 4, | 1, | 3 |
LCM = 2 × 2 × 2 × 3 × 3 × 3 × 4 = 864
∴ Required number = 864 – 5
= 859
Q-16) The smallest five digit number which is divisible by 12,18 and 21 is :
(a)
(b)
(c)
(d)
| 2 | 12, | 18, | 21 |
| 3 | 6, | 9, | 21 |
| 2, | 3, | 7 |
LCM of 12, 18 and 21
= 2 × 3 × 2 × 3 × 7 = 252
Of the options, 10080 ÷ 252 = 40
Q-17) The LCM of two numbers is 1920 and their HCF is 16. If one of the number is 128, find the other number.
(a)
(b)
(c)
(d)
Using Rule 1 :
1st number × 2nd number = L.C. M. × H.C.F,
We have,
First number × second number = LCM × HCF
∴ Second number = ${1920 × 16}/128$ = 240
Q-18) 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-19) The least number, which is a perfect square and is divisible by each of the numbers 16, 20 and 24, is
(a)
(b)
(c)
(d)
The smallest number divisible by 16, 20 and 24
= LCM of 16, 20 and 24
| 2 | 16, | 20 | 24 |
| 2 | 8, | 10, | 12, |
| 2 | 4, | 5, | 6, |
| 2, | 5, | 3 |
∴LCM = 2×2×2×2×5×3
= $2^2$ × $2^2$ × 5 × 3
∴Required complete square number =$2^2$ × $2^2$ × $5^2$ × $3^2$ = 3600
Q-20) Find the greatest number of five digits which when divided by 3, 5, 8, 12 have 2 as remainder :
(a)
(b)
(c)
(d)
Using Rule 4,
i.e. When a number is divided by x, y or z leaving same remainder ‘R’ in each case then that number must be K + R where k is LCM of x, y and z.
The greatest number of five digits is 99999.
LCM of 3, 5, 8 and 12
| 2 | 3, | 5, | 8, | 12 |
| 2 | 3, | 5, | 4, | 6 |
| 3 | 3, | 5, | 2, | 3 |
| 1, | 5, | 2, | 1 |
∴ LCM = 2 × 2 × 3 × 5 × 2 = 120 After dividing 99999 by 120, we get 39 as remainder 99999 – 39 = 99960 = (833 × 120)
99960 is the greatest five digit number divisible by the given divisors.
In order to get 2 as remainder in each case we will simply add 2 to 99960.
∴ Greatest number = 99960 + 2 = 99962