model 2 find lcm 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) 723
(b) 483
(c) 243
(d) 183
The correct answers to the above question in:
Answer: (a)
Using Rule 4,
i.e. When a number is divided by x, y or z leaving same remainder ‘R’ in each case then that number must be K + R where k is LCM of x, y and z.
Required number = (LCM of 15, 20, 36 and 48) + 3
2 | 15, | 20, | 36, | 48 |
2 | 15, | 10, | 18, | 24 |
3 | 15, | 5, | 9, | 12 |
5 | 5, | 5, | 3, | 4 |
1, | 1, | 3, | 4 |
∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720
∴ Required number = 720 + 3 = 723
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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers
Question : 1
What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 13 leaves no remainder ?
a) 1586
b) 1562
c) 962
d) 312
Answer »Answer: (c)
LCM of 3, 5, 6, 8, 10 and 12 = 120
∴ Required number = 120x + 2, which is exactly divisible by 13.
120x + 2 = 13 × 9x + 3x + 2
Clearly 3x + 2 should be divisible by 13.
For x=8,3x + 2 is divisible by 13.
∴ Required number = 120x + 2 = 120 × 8 + 2
= 960 + 2 = 962
Question : 2
The smallest perfect square divisible by each of 6, 12 and 18 is
a) 36
b) 108
c) 144
d) 196
Answer »Answer: (a)
The LCM of 6, 12 and 18 = 36 = $6^2$
Question : 3
The least perfect square, which is divisible by each of 21, 36 and 66 is
a) 231444
b) 213444
c) 214434
d) 214344
Answer »Answer: (b)
LCM of 21, 36 and 66
∴ LCM = 3 × 2 × 7 × 6 × 11
= 3 × 3 × 2 × 2 × 7 × 11
∴ Required number
= $3^2 × 2^2 × 7^2 × 11^2$
= 213444
Question : 4
Four bells ring at intervals of 4, 6, 8 and 14 seconds. They start ringing simultaneously at 12.00 O’clock. At what time will they again ring simultaneously ?
a) 12 hrs. 3 min. 44 sec.
b) 12 hrs. 3 min. 20 sec.
c) 12 hrs. 3 min.
d) 12 hrs. 2 min. 48 sec.
Answer »Answer: (d)
LCM of 4, 6, 8, 14
= 168 seconds
= 2 minutes 48 seconds
They ring again at 12 + 2 min. 48 sec.
= 12 hrs. 2 min. 48 sec.
Question : 5
The LCM of two prime numbers x and y, (x > y) is 161. The value of (3y – x) :
a) 2
b) 1
c) –1
d) –2
Answer »Answer: (d)
LCM of x and y = 161
∴ xy = 23 × 7
∴ x = 23; y = 7
∴ 3y – x = 3 × 7 – 23
= 21 – 23 = – 2
Question : 6
Five bells begin to toll together and toll respectively at intervals of 6, 7, 8, 9 and 12 seconds. After how many seconds will they toll together again ?
a) 318 Sec.
b) 504 Sec.
c) 612 Sec.
d) 72 Sec.
Answer »Answer: (b)
Required time = LCM of 6, 7, 8, 9 and 12 seconds
= 504 seconds
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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