Practice Find hcf using formula - 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)

Explanation:

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)

Explanation:

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)

Explanation:

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)   84 Maths books, 90 Physics books and 120 Chemistry books have to be stacked topicwise. How many books will be there in each stack so that each stack will have the same height too ?

(a)

(b)

(c)

(d)

Explanation:

As the height of each stack is same,

the required number of books in each stack

= HCF of 84, 90 and 120

84 = 2 × 2 × 3 × 7

90 = 2 × 3 × 3 × 5

120 = 2 × 2 × 2 × 3 × 5

∴ HCF = 2 × 3 = 6


Q-5)   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)

(b)

(c)

(d)

Explanation:

Maximum quantity in each can

= HCF of 21, 42 and 63 litres

= 21 litres

Required least number of cans

= $21/21 + 42/ 21 + 63/21$

= 1 + 2 + 3 = 6


Q-6)   In a school, 391 boys and 323 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 ?

(a)

(b)

(c)

(d)

Explanation:

First of all we find HCF of 391 and 323.

∴ Number of classes = 17


Q-7)   The greatest number by which 2300 and 3500 are divided leaving the remainders of 32 and 56 respectively, is

(a)

(b)

(c)

(d)

Explanation:

Using Rule 7,

(The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r). )

Required number

= HCF of 2300 – 32

= 2268 and 3500 – 56 = 3444

∴ HCF = 84


Q-8)   Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?

(a)

(b)

(c)

(d)

Explanation:

Using Rule 7,

3026 –11 = 3015 and 5053 –13 = 5040

Required number = HCF of 3015 and 5040

∴ Required number = 45


Q-9)   Find the greatest number that will divide 390, 495 and 300 without leaving a remainder.

(a)

(b)

(c)

(d)

Explanation:

Required number = HCF of 390, 495 and 300 = 15

Illustration :

HCF of 15 and 300 = 15


Q-10)   The greatest number, which when divide 989 and 1327 leave remainders 5 and 7 respectively, is :

(a)

(b)

(c)

(d)

Explanation:

Using Rule 7,

(The largest number which when divide the numbers a, b and c give remainders as p, q, r respectively is given by H.C.F. of (a – p), (b – q) and (c – r). )

Required number

= HCF of (989 – 5) and (1327 – 7)

= HCF of 984 and 1320 = 24

∴ HCF = 24