A milk vendor has 21 litres of cow milk, 42 litres of toned milk and 63 litres of double toned milk. If he wants to pack them in cans, so that each can contains same litres of milk and does not want to mix any two kinds of milk in a can, then the least number of cans required is = ?
(A) 3
(B) 6
(C) 9
(D) 12
Solution:
For least or minimum number of cans we should have maximum capacity cans for required quantity ⇒ For this we take HCF of given quantities. HCF (21, 42, 63) = 21 ∴ Maximum capacity of a can = 21 litres ∴ Number of cans of cow milk $${\text{ = }}\frac{{21}}{{21}}{\text{ = 1}}$$ ∴ Number of cans of toned milk $${\text{ = }}\frac{{42}}{{21}} = 2$$ ∴ Number of cans of double toned milk $${\text{ = }}\frac{{63}}{{21}} = 3$$ ∴ Total number of cans = 1 + 2 + 3 = 6
3.
The HCF (GCD) of a, b is 12, a, b are positive integers and a > b > 12. The smallest values of (a, b) are respectively ?
(A) 12, 34
(B) 24, 12
(C) 24, 36
(D) 36, 24
Solution:
HCF (GCD) of a, b number is 12 and a > b > 12 (given) ∴ Smallest value of a & b are (36, 24)
4.
Three numbers which are co-prime to each other 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) 75
(B) 81
(C) 85
(D) 89
Solution:
Since the numbers are co-prime, they contain only 1 as the common factor. Also, the given two products have the middle number in common. So, middle number = H.C.F. of 551 and 1073 = 29 $$\eqalign{ & {\text{First}}\,{\text{number}} \cr & = {\frac{{551}}{{29}}} \cr & = 19 \cr & {\text{Third}}\,{\text{number}} \cr & = {\frac{{1073}}{{29}}} \cr & = 37 \cr & \therefore {\text{Required}}\,{\text{sum}} \cr & = {19 + 29 + 37} \cr & = 85 \cr} $$
5.
Three sets of English , Mathematics and Science books containing 336, 240 and 96 books respectively have to be stacked in such a way that all the books are stored subject wise and the height of each stack is the same. Total number of stakes will be ?
(A) 14
(B) 21
(C) 22
(D) 48
Solution:
Number of books in each stack = HCF of 336, 240 and 96 = 48. Hence, total number of stacks $$\eqalign{ & {\text{ = }}\left( {\frac{{336 + 240 + 96}}{{48}}} \right) \cr & = \frac{{672}}{{48}} \cr & = 14 \cr} $$
6.
The LCM of four consecutive numbers is 60. The sum of the first two numbers is equal to the fourth number. What is the sum of four numbers
(A) 17
(B) 14
(C) 21
(D) 24
Solution:
x, x + 1, x + 2, x + 3 1st + 2nd = 4th x + x + 1 = x + 3 x = 2 2, 3, 4, 5 = 2 + 3 + 4 + 5 = 14
7.
Find the greatest number that will divide 964, 1238 and 1400 leaving remainders 41, 31, and 51 respectively = ?
(A) 61
(B) 71
(C) 73
(D) 81
Solution:
Required number = HCF of (964 - 41), (1238 - 31) and (1400 - 51) HCF of 923, 1207 and 1349 = 71
8.
The greatest number of four digits which is divisible by 15, 25, 40 and 75 is:
(A) 9000
(B) 9400
(C) 9600
(D) 9800
Solution:
Greatest number of 4-digits is 9999. L.C.M. of 15, 25, 40 and 75 is 600. On dividing 9999 by 600, the remainder is 399. ∴ Required number (9999 - 399) = 9600.