model 3 find hcf of numbers Section-Wise Topic Notes With Detailed Explanation And Example Questions
MOST IMPORTANT quantitative aptitude - 5 EXERCISES
The following question based on LCM & HCF topic of quantitative aptitude
(a) 814
(b) 841
(c) 840
(d) 820
The correct answers to the above question in:
Answer: (a)
Length of the floor
= 15 m 17 cm = 1517 cm
Breadth of the floor
= 9m 2 cm = 902 cm.
Area of the floor
= 1517 × 902 $cm^2$
The number of square tiles will be least, when the size of each tile is maximum.
∴ Size of each tile = HCF of 1517 and 902 = 41
∴ Required number of tiles
=${1517× 902}/{41× 41}$ = 814
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Read more find hcf using formula Based Quantitative Aptitude Questions and Answers
Question : 1
The greatest number by which 2300 and 3500 are divided leaving the remainders of 32 and 56 respectively, is
a) 84
b) 168
c) 136
d) 42
Answer »Answer: (a)
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
Question : 2
Which greatest number will divide 3026 and 5053 leaving remainders 11 and 13 respectively?
a) 60
b) 30
c) 18
d) 45
Answer »Answer: (d)
Using Rule 7,
3026 –11 = 3015 and 5053 –13 = 5040
Required number = HCF of 3015 and 5040
∴ Required number = 45
Question : 3
Two pipes of length 1.5 m and 1.2 m are to be cut into equal pieces without leaving any extra length of pipes. The greatest length of the pipe pieces of same size which can be cut from these two lengths will be
a) 0.41 metre
b) 0.4 metre
c) 0.13 metre
d) 0.3 metre
Answer »Answer: (d)
Maximum length of each piece
= HCF of 1.5 metre and 1.2 metre
= 0.3 metre
Illustration :
∴ HCF of 1.5 and 1.2 metre = 0.3 metre
Question : 4
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 »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 : 5
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 »Answer: (a)
First of all we find HCF of 391 and 323.
∴ Number of classes = 17
Question : 6
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 »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
LCM & HCF Shortcuts and Techniques with Examples
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model 1 Basic formula of LCM & HCF
Defination & Shortcuts … -
model 2 find lcm of numbers
Defination & Shortcuts … -
model 3 find hcf of numbers
Defination & Shortcuts … -
model 4 addition, subtraction, multiplication and division with lcm & hcf
Defination & Shortcuts … -
model 5 lcm & hcf vs ratios
Defination & Shortcuts …
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