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 : 46 [SSC CPO 2008]
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 : 47 [SSC CGL Tier-II 2014]
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 : 48 [SSC CGL Prelim 2003]
From a point on a circular track 5 km long A, B and C started running in the same direction at the same time with speed of 2$1/2$ km per hour, 3 km per hour and 2 km per hour respectively. Then on the starting point all three will meet again after
a) 15 hours
b) 10 hours
c) 6 hours
d) 30 hours
Answer »Answer: (b)
A makes one complete round of the circular track in $5/{5/2}$ = 2 hours,
B in $5/3$ hours and C in $5/2$ hours.
That is after 2 hours A is at the starting point, B after $5/3$ hours and C after $5/2$ hours.
Hence the required time = LCM of 2, $5/3$ and $5/2$ hours
= ${\text"LCM of " 2, 5, 5}/ {\text"HCF of " 3, 2}$
= $10/1$ = 10 hours.
Question : 49
The smallest number, which when divided by 12 and 16 leaves remainder 5 and 9 respectively, is :
a) 29
b) 39
c) 41
d) 55
Answer »Answer: (c)
Using Rule 5,
When a number is divided by a, b or c, leaving remainders p, q or r respectively such that the difference between divisor and remainder in each case is the same
i.e., (a – P) = (b – q) = (c – r) = t (say) then that (least) number must be in the form of (k – t),
Where k is LCM of a, b and c
Here, 12 – 5 = 7, 16 – 9 = 7
∴ Required number = (L.C.M. of 12 and 16) – 7
= 48 – 7 = 41
Question : 50 [SSC CGL Prelim 2008]
What least number must be subtracted from 1936 so that the resulting number when divided by 9, 10 and 15 will leave in each case the same remainder 7 ?
a) 30
b) 39
c) 36
d) 37
Answer »Answer: (b)
LCM of 9, 10 and 15 = 90
⇒ The multiple of 90 are also divisible by 9, 10 or 15.
∴ 21 × 90 = 1890 will be divisible by them.
∴ Now, 1897 will be the number that will give remainder 7.
1936 – 1897
Required number = 1936 – 1897 = 39
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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