model 2 find lcm of numbers Practice Questions Answers Test with Solutions & More Shortcuts
LCM & HCF PRACTICE TEST [5 - EXERCISES]
model 1 Basic formula of LCM & HCF
model 2 find lcm of numbers
model 3 find hcf of numbers
model 4 addition, subtraction, multiplication and division with lcm & hcf
model 5 lcm & hcf vs ratios
Question : 41 [SSC Section Officer 2007]
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 : 42 [SSC CGL Prelim 2002]
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 : 43
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 : 44 [SSC CGL 2013]
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 : 45
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
IMPORTANT quantitative aptitude EXERCISES
model 2 find lcm of numbers Shortcuts »
Click to Read...model 2 find lcm of numbers Online Quiz
Click to Start..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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