model 3 find hcf of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions

MOST IMPORTANT quantitative aptitude - 5 EXERCISES

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The following question based on LCM & HCF topic of quantitative aptitude

Questions : 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) 6

(b) 9

(c) 12

(d) 10

The correct answers to the above question in:

Answer: (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

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Read more find hcf using formula Based Quantitative Aptitude Questions and Answers

Question : 1

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) 12

b) 6

c) 3

d) 9

Answer: (b)

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

Question : 2

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) 17

b) 19

c) 23

d) 44

Answer: (a)

First of all we find HCF of 391 and 323.

∴ Number of classes = 17

Question : 3

The greatest number, by which 1657 and 2037 are divided to give remainders 6 and 5 respectively, is

a) 305

b) 133

c) 127

d) 235

Answer: (c)

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).)

We have to find HCF of

(1657 – 6 = 1651) and (2037 – 5 = 2032)

1651 = 13 × 127

2032 = 16 × 127

∴ HCF = 127

So, required number will be 127.

Question : 4

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) 8

b) 5

c) 4

d) 6

Answer: (c)

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

Question : 5

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) 21

b) 18

c) 12

d) 6

Answer: (d)

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

Question : 6

H.C.F of $2/3,4/5$and $6/7$ is

a) $24/105$

b) $2/105$

c) $48/105$

d) $1/105$

Answer: (b)

Using Rule 3,

(H.C.F. of fractions =$\text"H.C.F of numerators"/ \text"L.C.M.of denominators"$)

HCF of $2/3,4/5$and $6/7$

= $\text"H.C.F of 2,4 and 6"/ \text"L.C.M.of 3,5 and 7"$

= $2/105$

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