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) 9 a.m.
(b) 6 a.m.
(c) 3 a.m.
(d) 12 mid night
The correct answers to the above question in:
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
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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
L.C.M. of $2/3, 4/9, 5/6$ is
a) $20/27$
b) $10/3$
c) $20/3$
d) $8/27$
Answer »Answer: (c)
Using Rule 2,
$\text"L.C.M of fractions" = \text"L.C.M.of numerators"/\text"H.C.F.of denominators"$
LCM = ${\text"LCM of " 2, 4, 5}/{\text"HCF of " 3, 9, 6}$= $20/3$
Question : 3
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 : 4
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.
Question : 5
A number which when divided by 10 leaves a remainder of 9, when divided by 9 leaves a remainder of 8, and when divided by 8 leaves a remainder of 7, is :
a) 1359
b) 359
c) 539
d) 1539
Answer »Answer: (b)
Using Rule 5,
Here, Divisor – remainder = 1
e.g., 10 – 9 = 1, 9 – 8 = 1, 8 – 7 = 1
∴ Required number = (L.C.M. of 10, 9, 8) –1
= 360 – 1 = 359
Question : 6
The number nearest to 10000, which is exactly divisible by each of 3, 4, 5, 6, 7 and 8, is :
a) 10000
b) 9996
c) 10080
d) 9240
Answer »Answer: (c)
LCM of 3, 4, 5, 6, 7, 8 = 840
11$760/840$ = 10000
Since, the remainder 760 is more than half of the divisor 840.
∴ The nearest number
= 10000 + (840 – 760) = 10080
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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