model 4 addition, subtraction, multiplication and division with lcm & hcf 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) 3
(b) 6
(c) 4
(d) 8
The correct answers to the above question in:
Answer: (b)
Using Rule 1,
1st number × 2nd number = L.C.M. × H.C.F
Let the larger number be x.
Smaller number = x – 2
First number × Second number = HCF × LCM
→ x (x – 2) = 24
→ $x^2$ – 2x – 24 = 0
→ $x^2$ – 6x + 4x – 24 = 0
→ x (x – 6) + 4 (x – 6) = 0
→ (x – 6) (x + 4) = 0
→ x = 6 because x ≠ – 4
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Question : 1
The sum of two numbers is 84 and their HCF is 12. Total number of such pairs of number is
a) 2
b) 4
c) 3
d) 5
Answer »Answer: (c)
HCF = 12
∴ Numbers = 12x and 12y where x and y are prime to each other.
∴ 12x + 12y = 84
⇒ 12 (x + y) = 84
⇒ x + y = $84/12$ = 7
∴ Possible pairs of numbers satisfying this condition
= (1,6), (2,5) and (3,4). Hence three pairs are of required numbers.
Question : 2
L.C.M. of two numbers is 120 and their H.C.F. is 10. Which of the following can be the sum of those two numbers ?
a) 140
b) 60
c) 80
d) 70
Answer »Answer: (d)
Let the numbers be 10x and 10y
where x and y are prime to each other.
∴ LCM = 10 xy
→ 10xy = 120
→ xy = 12
Possible pairs = (3, 4) or (1, 12)
∴ Sum of the numbers = 30 + 40 = 70
Question : 3
Sum of two numbers is 384. H.C.F. of the numbers is 48. The difference of the numbers is
a) 100
b) 288
c) 192
d) 336
Answer »Answer: (b)
Let the numbers be 48x and 48y
where x and y are co-primes.
48x + 48y = 384
→ 48 ( x + y) = 384
→ x + y = $384/48$ = 8 ........... (i)
Possible and acceptable pairs of x and y satisfying this condition are : (1, 7) and (3, 5).
∴ Numbers are : 48 × 1 = 48 and 48 × 7 = 336
and 48 × 3 = 144 and 48 × 5 = 240
∴ Required difference = 336 – 48 = 288
Question : 4
HCF and LCM of two numbers are 7 and 140 respectively. If the numbers are between 20 and 45, the sum of the numbers is :
a) 70
b) 63
c) 77
d) 56
Answer »Answer: (b)
Let the numbers be 7x and 7y
where x and y are co-prime.
Now, LCM of 7x and 7y = 7xy
∴ 7xy = 140 ⇒ xy = $140/7$ = 20
Now, required values of x and y whose product is 50 and are coprime, will be 4 and 5.
∴ Numbers are 28 and 35 which lie between 20 and 45.
∴ Required sum = 28 + 35 = 63.
Question : 5
The sum of a pair of positive integer is 336 and their H.C.F. is 21. The number of such possible pairs is
a) 2
b) 4
c) 3
d) 5
Answer »Answer: (b)
Let the numbers be 21x and 21y
where x and y are prime to each other.
21x + 21y = 336
21 (x + y) = 336
x + y = $336/21$ = 16
∴ Possible pairs = (1, 15), (5, 11), (7, 9), (3, 13)
Question : 6
A number between 1000 and 2000 which when divided by 30, 36 and 80 gives a remainder 11 in each case is
a) 1451
b) 1712
c) 1641
d) 1523
Answer »Answer: (a)
We find LCM of 30, 36 and 80.
2 | 30, | 36, | 80 |
2 | 15, | 18, | 40 |
3 | 15, | 9, | 20 |
5 | 5, | 3, | 20 |
1, | 3, | 4 |
LCM = 2 × 2 × 3 × 3 × 4 × 5 = 720
∴ Required number = 2 × 720 + 11
= 1440 + 11 = 1451
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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