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) 5600
(b) 5200
(c) 4714
(d) 1120
The correct answers to the above question in:
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
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Read more find lcm using formula Based Quantitative Aptitude Questions and Answers
Question : 1
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 : 2
The LCM of two prime numbers x and y, (x > y) is 161. The value of (3y – x) :
a) 2
b) 1
c) –1
d) –2
Answer »Answer: (d)
LCM of x and y = 161
∴ xy = 23 × 7
∴ x = 23; y = 7
∴ 3y – x = 3 × 7 – 23
= 21 – 23 = – 2
Question : 3
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 : 4
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 : 5
What is the smallest number which leaves remainder 3 when divided by any of the numbers 5, 6 or 8 but leaves no remainder when it is divided by 9 ?
a) 243
b) 723
c) 603
d) 123
Answer »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
Question : 6
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)
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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