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) 2
(b) 1
(c) –1
(d) –2
The correct answers to the above question in:
Answer: (d)
LCM of x and y = 161
∴ xy = 23 × 7
∴ x = 23; y = 7
∴ 3y – x = 3 × 7 – 23
= 21 – 23 = – 2
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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers
Question : 1
Four bells ring at intervals of 4, 6, 8 and 14 seconds. They start ringing simultaneously at 12.00 O’clock. At what time will they again ring simultaneously ?
a) 12 hrs. 3 min. 44 sec.
b) 12 hrs. 3 min. 20 sec.
c) 12 hrs. 3 min.
d) 12 hrs. 2 min. 48 sec.
Answer »Answer: (d)
LCM of 4, 6, 8, 14
= 168 seconds
= 2 minutes 48 seconds
They ring again at 12 + 2 min. 48 sec.
= 12 hrs. 2 min. 48 sec.
Question : 2
Find the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as remainder in each case.
a) 723
b) 483
c) 243
d) 183
Answer »Answer: (a)
Using Rule 4,
i.e. When a number is divided by x, y or z leaving same remainder ‘R’ in each case then that number must be K + R where k is LCM of x, y and z.
Required number = (LCM of 15, 20, 36 and 48) + 3
2 | 15, | 20, | 36, | 48 |
2 | 15, | 10, | 18, | 24 |
3 | 15, | 5, | 9, | 12 |
5 | 5, | 5, | 3, | 4 |
1, | 1, | 3, | 4 |
∴ LCM = 2 × 2 × 3 × 5 × 3 × 4 = 720
∴ Required number = 720 + 3 = 723
Question : 3
What is the least number which when divided by the numbers 3, 5, 6, 8, 10 and 12 leaves in each case a remainder 2 but when divided by 13 leaves no remainder ?
a) 1586
b) 1562
c) 962
d) 312
Answer »Answer: (c)
LCM of 3, 5, 6, 8, 10 and 12 = 120
∴ Required number = 120x + 2, which is exactly divisible by 13.
120x + 2 = 13 × 9x + 3x + 2
Clearly 3x + 2 should be divisible by 13.
For x=8,3x + 2 is divisible by 13.
∴ Required number = 120x + 2 = 120 × 8 + 2
= 960 + 2 = 962
Question : 4
Five bells begin to toll together and toll respectively at intervals of 6, 7, 8, 9 and 12 seconds. After how many seconds will they toll together again ?
a) 318 Sec.
b) 504 Sec.
c) 612 Sec.
d) 72 Sec.
Answer »Answer: (b)
Required time = LCM of 6, 7, 8, 9 and 12 seconds
= 504 seconds
Question : 5
The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is
a) 5600
b) 5200
c) 4714
d) 1120
Answer »Answer: (c)
2 | 20, | 28, | 32, | 35 |
2 | 10, | 14, | 16, | 35 |
5 | 5, | 7, | 8, | 35 |
7 | 1, | 7, | 8, | 7 |
1, | 1, | 8, | 1 |
∴ LCM = 2 × 2 × 5 × 7 × 8 = 1120
∴ Required number
= 5834 – 1120 = 4714
Question : 6
The largest number of five digits which, when divided by 16, 24, 30, or 36 leaves the same remainder 10 in each case, is :
a) 99350
b) 99269
c) 99370
d) 99279
Answer »Answer: (c)
Using Rule 9,
The n digit largest number which when divided by a, b, c, leaves remainder ‘x’ will be
Required number = [n – digit largest number – R] + x
where, R is the remainder obtained when n – digit largest number is divided by the L.C.M of a, b, c.
We will find the LCM of 16, 24, 30 and 36
2 | 16, | 24, | 30, | 36 |
2 | 8, | 12, | 15, | 18 |
2 | 4, | 6, | 15, | 9 |
3 | 2, | 3, | 15, | 9 |
2, | 1, | 5, | 3 |
∴ LCM = 2 × 2 × 2 × 3 × 2 × 5 × 3 = 720
The largest number of five digits = 99999
On dividing 99999 by 720, the remainder = 639
∴ The largest five-digit number divisible by 720
= 99999 – 639 = 99360
∴ Required number = 99360 + 10
= 99370
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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