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) $20/27$
(b) $10/3$
(c) $20/3$
(d) $8/27$
The correct answers to the above question in:
Answer: (c)
Using Rule 2,
$\text"L.C.M of fractions" = \text"L.C.M.of numerators"/\text"H.C.F.of denominators"$
LCM = ${\text"LCM of " 2, 4, 5}/{\text"HCF of " 3, 9, 6}$= $20/3$
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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers
Question : 1
The smallest number, which when divided by 5, 10, 12 and 15, leaves remainder 2 in each case; but when divided by 7 leaves no remainder, is
a) 91
b) 175
c) 182
d) 189
Answer »Answer: (c)
LCM of 5, 10, 12, 15
| 2 | 5, | 10, | 12, | 15 |
| 3 | 5, | 5, | 6, | 15 |
| 5 | 5, | 5, | 2, | 5 |
| 1, | 1, | 2, | 1 |
∴ LCM = 2 × 3 × 5 × 2 = 60
∴ Number = 60k + 2
Now, the required number should be divisible by 7.
Now, 60k + 2 = 7 × 8k + 4k + 2 If we put k = 3, (4k + 2) is equal to 14 which is exactly divisible by 7.
∴ Required number = 60 × 3 + 2 = 182
Question : 2
The least number, which when divided by 18, 27 and 36 separately leaves remainders 5,14, and 23 respectively, is
a) 77
b) 149
c) 113
d) 95
Answer »Answer: (d)
The difference between the divisor and the corresponding remainder is same in each case ie. 18 – 5 = 13, 27 – 14 = 13, 36 – 23 = 13
∴ Required number = (LCM of 18, 27, and 36 ) – 13
= 108 – 13 = 95
Question : 3
Three men step off together from the same spot. Their steps measure 63 cm, 70 cm and 77 cm respectively. The minimum distance each should cover so that all can cover the distance in complete steps is
a) 6950 cm
b) 6930 cm
c) 9360 cm
d) 9630 cm
Answer »Answer: (b)
Required distance = LCM of 63, 70 and 77 cm.
= 6930 cm.
Illustration :
| 7 | 63, | 70, | 77 |
| 9, | 10, | 11 |
∴ LCM = 7 × 9 × 10 × 11
= 6930
Question : 4
Four bells ring at the intervals of 5, 6, 8 and 9 seconds. All the bells ring simultaneously at some time. They will again ring simultaneously after
a) 24 minutes
b) 18 minutes
c) 12 minutes
d) 6 minutes
Answer »Answer: (d)
The LCM of 5, 6, 8 and 9 = 360 seconds = 6 minutes
Question : 5
4 bells ring at intervals of 30 minutes, 1 hour, 1$1/2$ hour and 1 hour 45 minutes respectively. All the bells ring simultaneously at 12 noon. They will again ring simultaneously at :
a) 9 a.m.
b) 6 a.m.
c) 3 a.m.
d) 12 mid night
Answer »Answer: (a)
1$1/2$ hours = 90 minutes
1 hour and 45 minutes = 105 minutes
1 hour = 60 minutes
∴ LCM of 30 minutes, 60 minutes, 90 minutes and 105 minutes
| 3 | 30, | 60, | 90, | 105 |
| 5 | 10, | 20, | 30, | 35 |
| 2 | 2, | 4, | 6, | 7 |
| 1, | 2, | 3, | 7 |
∴ LCM = 3 × 5 × 2 × 2 × 3 × 7 = 1260 minutes
1260 minutes =$ 1260/60$ = 21 hours
∴ The bell will again ring simultaneously after 21 hours.
∴ Time will be
= 12 noon + 21 hours
= 9 a.m
Question : 6
The least multiple of 7, which leaves the remainder 4, when divided by any of 6, 9, 15 and 18, is
a) 364
b) 184
c) 94
d) 76
Answer »Answer: (a)
LCM of 6, 9, 15 and 18
| 2 | 6, | 9, | 15, | 18 |
| 3 | 3, | 9, | 15, | 9 |
| 3 | 1, | 3, | 5, | 3 |
| 1, | 1, | 5, | 1 |
∴ LCM = 2 × 3 × 3 × 5 = 90
∴ Required number = 90k + 4,
which must be a multiple of 7 for some value of k.
For k = 4,
Number = 90 × 4 + 4 = 364,
which is exactly divisible by 7.
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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