Test Preparation on Quantitative Aptitude ( HCF , LCM , Permutation and Combinations)
The sum of the two numbers is 468 and their HCF is 39. How many pairs of such numbers can be formed?
468 + 39 = 12
Therefore, 11 numbers are available and the number of pair of numbers having HCF = 39 is
= 66 numbers or 33 pairs of numbers
Find the sum of the numbers between 500 and 700 such that when they are divided by 6, 8 and 12
(i) it leaves no remainder
The least number which is divisible by 6, 8 and 12 is the LCM of 6, 8 and 12 = 24.
The numbers greater than 500 and less than 700 which are divisible by 24 are
504, 528, 552, .., 696.
The sum of the no.= 9 x 600 = 5400
Is it possible to divide 1,000 into two parts such that their HCF is 15?
Since 1000 is not divisible by 15, it is not possible to divide 1000 into two parts such that their HCF may be 15. Hence the correct answer is (d).
In a school, 442 boys and 374 girls have been divided into the largest possible equal classes, so that each class of boys numbers the same as each class of girls. What is the number of classes?
Number of boys = 442; Number of girls = 374
As the number of boys in the class is the same as that of girls. Strength of one class = HCF of 442 and 374 = 34
Number of classes of boys = 442 + 34 = 13
Number of classes of girls = 374 + 34 = 11
Required number of classes = 13 + 11 = 24
Find the greatest number of 4 digits and the least number of 5 digits that have 196 as their HCF.
The greatest number of 4 digits = 9999
On dividing 9999 by 196, 3 will be left as remainder.
The required greatest number = 9996. Least number of 5 digits = 10000
On dividing 10000 by 196, 4 will be left as remainder.
The product of two numbers is 4212 and their HCF is 9. Find the possible pair of numbers.
Let x and y be the two numbers and their HCF = 9; product = 4212.
x x y = 92 (product in lower terms) i.e. 4212 = 81 (product in lower terms) Product in lower terms = 52
The HCF and LCM of two numbers are 36 and 756 respectively. Find the possible pair of numbers.
Given HCF is 36 and LCM is 756.
Product of two numbers = HCF x LCM
= 36 x 756
(Product in lower terms) x (HCF)2 = HCF x LCM
The numbers are multiples of 3 and 7.
The required numbers are 3 x 36 and 7 x 36, i.e. 108, 252.
Find the maximum possible length which can be used to measure exactly the lengths 6 m, 4 m 75 cm, and 10 m 25 cm.
Required maximum length = HCF of 6 m, 4 m 75 cm and 10 m 25 cm i.e. the HCF of 600 cm, 475 cm and 1025 cm = 25 cm
The LCM of two numbers is 1008 and their HCF is 36. If one of the two numbers is 144, find the other number.
. Let the second number be x.
x x the first number = HCF x LCM
x x 144 = 36 x 1008
x=x(1008)/144=252
Required number = 252
Required number = 252
Three bells ring together and they will ring at intervals of 6 seconds, 7 seconds and 8 seconds respectively. After what interval will they again ring together?
Since the bells ring at the intervals of 6 seconds, 7 seconds and 8 seconds, respectively, next time they will ring together after an interval equal to the LCM of 6, 7 and 8 seconds. Therefore, the required time = LCM of 6, 7 and 8 seconds
= 168 seconds
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