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) 6950 cm
(b) 6930 cm
(c) 9360 cm
(d) 9630 cm
The correct answers to the above question in:
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
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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
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 : 3
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
Question : 4
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 : 5
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 : 6
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 = ${\text"LCM of " 2, 4, 5}/{\text"HCF of " 3, 9, 6}$= $20/3$
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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