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) 243
(b) 723
(c) 603
(d) 123
The correct answers to the above question in:
Answer: (a)
We find LCM of 5, 6 and 8
5=5
6=3×2
8=$2^3$
= $2^3$ ×3 × 5 = 8 × 15 = 120
Required number = 120K + 3
∴ when K = 2, 120 × 2 + 3 = 243 required no.
It is completely divisible by 9
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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers
Question : 1
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
Question : 2
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 : 3
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 : 4
Find the largest number of four digits such that on dividing by 15, 18, 21 and 24 the remainders are 11, 14, 17 and 20 respectively.
a) 7664
b) 5675
c) 7556
d) 6557
Answer »Answer: (c)
15 = 3 × 5
18 = $3^2$ × 2
21 = 3 × 7
24 = $2^3$ × 3
LCM = 8 × 9 × 5 × 7 = 2520
The largest number of four digits = 9999
3$2439/2520$ = 9999
Required number = 9999 – 2439 – 4 = 7556
(Because 15 - 11 = 4
18 - 14 = 4
21 – 17 = 4
24 – 20 = 4)
Question : 5
Four runners started running simultaneously from a point on a circular track. They took 200 seconds, 300 seconds, 360 seconds and 450 seconds to complete one round. After how much time do they meet at the starting point for the first time ?
a) 4800 seconds
b) 2400 seconds
c) 3600 seconds
d) 1800 seconds
Answer »Answer: (d)
Required time = LCM of 200, 300, 360 and 450 seconds
= 1800 seconds
Question : 6
The least number, which when divided by 12, 15, 20 or 54 leaves a remainder of 4 in each case, is :
a) 544
b) 540
c) 454
d) 450
Answer »Answer: (a)
Using Rule 4,
When a number is divided by a, b or c leaving same remainder ‘r’ in each case then that number must be k + r where k is LCM of a, b and c.
LCM of 15, 12, 20, 54 = 540
Then number = 540 + 4 = 544 [4 being remainder]
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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