- Staff Selection Commission Mathematics 1999 to 2017 - LCM and HCF Part 3

# Staff Selection Commission Mathematics - LCM and HCF

Ans .

2

1. Explanation :

The LCM of 5, 6, 7 and 8 = 840
Required number = 840 k + 3 which is exactly divisible by 9 for some value of k. Now, 840 k + 3 = 93 × 9 k + (3k + 3) When k = 2, 3k + 3 = 9, which is divisible by 9.
Required number = 840 × 2 + 3 = 1683

Ans .

1

1. Explanation :

Using Rule 5, Here,12 – 2 = 10; 16 – 6 = 10; 24 – 14 = 10
Now, LCM of 12, 16 and 24 = 48
The greatest 4–digit number exactly divisible by 48 = 9984
Required number = 9984 – 10 = 9974

Ans .

1

1. Explanation :

Using Rule 5, LCM of 15, 20 and 35 = 420
Required least number = 420 + 8 = 428

Ans .

3

1. Explanation :

Using Rule 5, The smallest number divisible by 12 or 10 or 8 = LCM of 12, 10 and 8 = 120
Required number =120 + 6 = 126

Ans .

2

1. Explanation :

LCM of 24, 36 and 54 seconds = 216 seconds = 3 minutes 36 seconds
Required time = 10 : 15 : 00 + 3 minutes 36 seconds = 10 : 18 : 36 a.m.

Ans .

3

1. Explanation :

A makes one complete round of the circular track in 5= 2 hours,
5
2
B in 5
3
hours and C in 5 hours.
2
That is after 2 hours A is at the starting point, B after 5 hours
3
and C after 5 hours.
2
Hence the required time = LCM of 2, 5
3 and 5
2 hours = LCM of 2,5,5
HCFof3,2
=10
1 = 10 hours.

Ans .

1

1. Explanation :

Required time = LCM of 200, 300, 360 and 450 seconds = 1800 seconds

Ans .

1

1. Explanation :

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

Ans .

4

1. Explanation :

1 1 hours = 90 minutes
2
1 hour and 45 minutes = 105 minutes
1 hour = 60 minutes
LCM of 30 minutes, 60 minutes, 90 minutes and 105 minutes.
3   30, 60, 90, 105
5   10, 20, 30, 35
2   2, 4, 6, 7
1, 2, 3, 7
LCM = 3 × 5 × 2 × 2 × 3 × 7 = 1260 minutes 1260 minutes= 1260 = 21 hours
60
The bell will again ring simultaneously after 21 hours.
Time will be = 12 noon + 21 hours = 9 a.m.

Ans .

1

1. Explanation :

The LCM of 5, 6, 8 and 9 = 360 seconds = 6 minutes

Ans .

2

1. Explanation :

LCM of 20, 30 and 40 minutes = 120 minutes Hence, the bells will toll together again after 2 hours i.e. at 1 p.m

Ans .

1

1. Explanation :

The difference between divisor and the corresponding remainder is equal. LCM of 3, 5, 7 and 9 = 315 Largest 4-digit number = 9999
315)9999(31
945
549
315
234
Number divisible by 315 = 9999 – 234 = 9765 Required number = 9765 – 2 = 9763

Ans .

3

1. Explanation :

Required time = LCM of 6, 7, 8, 9 and 12 seconds = 504 seconds

Ans .

2

1. Explanation :

Using Rule 2, LCM = LCM of 2, 4, 5
HCF of 3, 9, 6
= 20
3

Ans .

2

1. Explanation :

LCM of 3, 4, 5, 6, 7, 8 = 840
840 ) 10000 ( 11
840
1600
840
760
Since, the remainder 760 is more than half of the divisor 840. The nearest number = 10000 + (840 – 760) = 10080

Ans .

2

1. Explanation :

Using Rule 8, The largest number of 4-digits is 9999. L.C.M. of divisors
2   12, 15, 18, 27
3   6, 15, 9, 27
3   2, 5, 3, 9
2, 5, 1, 3
LCM = 2 × 2 × 3 × 3 × 3 × 5 = 540
Divide 9999 by 540, now we get 279 as remainder. 9999 – 279 = 9720 Hence, 9720 is the largest 4-digit number exactly divisible by each of 12, 15, 18 and 27.

Ans .

2

1. Explanation :

The smallest number divisible by 16, 20 and 24 = LCM of 16, 20 and 24
2   16, 20, 24
2   8, 10, 12
2   4, 5, 6
2, 5, 3
LCM = 2×2×2×2×5×3 = 22 × 22 × 5 × 3
Required complete square number = 22 × 22 ×52 × 32 = 3600

Ans .

2

1. Explanation :

LCM of 25, 50 and 75 = 150
On dividing 43582 by 150, remainder = 82
150 ) 43582 ( 290
300
1358
1350
82
Required number = 43582 + (150 – 82) = 43650

Ans .

2

1. Explanation :

Required number = (LCM of 24, 32, 36 and 54) – 5
Now,
2   24, 32, 36, 54
2   12, 16, 18, 27
2   6, 8, 9, 27
3   3, 4, 9, 27
3   1, 4, 3, 9
1, 4, 1, 3
LCM = 2 × 2 × 2 × 3 × 3 × 3 × 4 = 864
Required number = 864 – 5 = 859

Ans .

2

1. Explanation :

2   20, 28, 32, 35
2   10, 14, 16, 35
5   5, 7, 8, 35
7   1, 7, 82, 7
1, 1, 8, 1

LCM = 2 × 2 × 5 × 7 × 8 = 1120
Required number = 5834 – 1120 = 4714

Ans .

4

1. Explanation :

The LCM of 6, 12 and 18 = 36 = 62

Ans .

2

1. Explanation :

Using Rule 8, LCM of 10, 15 and 20 = 60
Largest 4-digit number = 9999
60 ) 9999 ( 166
60
399
360
399
360
39
Required number = 9999 – 39 = 996

Ans .

4

1. Explanation :

Using Rule 4,
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

Ans .

3

1. Explanation :

Required distance = LCM of 63, 70 and 77 cm. = 6930 cm.
7   63, 70, 77
9, 10, 11
LCM = 7 × 9 × 10 × 11 = 6930

Ans .

2

1. Explanation :

Required answer = LCM of 36, 40 and 48 seconds = 720 seconds
= 720 minutes = 12 minutes
60
2   36, 40, 48
2   18, 20, 24
2   9, 10, 12
3   9, 5, 6
3, 5, 2
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 5 = 720

Ans .

2

1. Explanation :

2   60
2   30
3   15
5
60 = 2 × 2 × 3 × 5 i.e.,
Numbers = 2, 3, 4 and 5
Required sum = 2 + 3 + 4 + 5 = 14

Ans .

1

1. Explanation :

LCM of x and y = 161
xy = 23 × 7
x = 23; y = 7
3y – x = 3 × 7 – 23 = 21 – 23 = – 2

Ans .

1

1. Explanation :

Required time = LCM of 48, 72 and 108 seconds
2   48, 72, 108
2   24, 36, 54
2   12, 18, 54
3   6, 9, 27
3   2, 3, 9
2, 1, 3
LCM = 2 × 2 × 2 × 2 × 3 × 3 × 3 = 432 seconds = 7 minutes 12 second
Required time = 10 : 07 : 12 hours

Ans .

1

1. Explanation :

Maximum number of students = The greatest common divisor = HCF of 1001 and 910 = 91

Ans .

3

1. Explanation :

Using Rule 7, Required number = HCF of (989 – 5) and (1327 – 7) = HCF of 984 and 1320 = 24 \ HCF = 24

Ans .

2

1. Explanation :

Using Rule 3,
HCF of 2 4 , and 6
3   5      7
= HCF of 2, 4 and 6
LCM of 3, 5 and 7\
= 2
105

Ans .

1

1. Explanation :

Using Rule 7, The greatest number N = HCF of (1305 – x ), (4665 – x) and (6905 – x), where x is the remainder = HCF of (4665 – 1305), (6905– 4665) and (6905 – 1305) = HCF of 3360, 2240 and 5600
2240 ) 3360 ( 1
2240
1120 ) 2240 ( 2
2240
x
N = 1120 Sum of digits = 1 + 1 + 2 + 0 = 4

Ans .

1

1. Explanation :

Using Rule 7,
The number will be HCF of 307 – 3 = 304 and 330 – 7 = 323.
304 ) 323 ( 1
304
19 )304( 16
19
114
114
x
Required number = 19

Ans .

3

1. Explanation :

Using Rule 7, 3026 –11 = 3015 and 5053 –13 = 5040 Required number = HCF of 3015 and 5040
3015 ) 5040 ( 1
3015
2025 ) 3015 ( 1
2025
990 )2025 ( 2
1980
45 )990( 22
90
90
90
x
Required Number = 45

Ans .

1

1. Explanation :

Using Rule 7, We have to find HCF of (1657 – 6 = 1651) and (2037 – 5 = 2032) 1651 = 13 × 127 2032 = 16 × 127
HCF = 127 So, required number will be 127.

Ans .

1

1. Explanation :

Using Rule 7, Let x be the remainder. Then, (25 – x), (73 – x ), and (97 – x ) Will be exactly divisible by the required number.
Required number = HCF of (73 –x ) – (25 –x ), (97 –x ) – (73 –x ) and (97 –x ) – (25 –x ) = HCF of (73 –25), (97 –73), and (97 –25) = HCF of 48, 24 and 72 = 24

Ans .

2

1. Explanation :

Using Rule 7, Required number = HCF of (110 – 2) and (128 – 2) = HCF of 108 and 126 = 18

Ans .

3

1. Explanation :

Required maximum capacity of container = HCF of 75 l and 45 l Now, 75 = 5 × 5 × 3 45 = 5 × 3 × 3
HCF = 15 litres

Ans .

4

1. Explanation :

Length of the floor = 15 m 17 cm = 1517 cm Breadth of the floor = 9m 2 cm = 902 cm. Area of the floor = 1517 × 902 cm2 The number of square tiles will be least, when the size of each tile is maximum.
Size of each tile = HCF of 1517 and 902 = 41
Required number of tiles = 1517 x 902 = 814
41 x 41

Ans .

1

1. Explanation :

Number of books in each stack = HCF of 336, 240, 96 = 48
240 ) 336 ( 1
240
96 ) 240 ( 2
192
48 ) 96 ( 2
96
x
Total number of stacks =
360 - 240 + 96
48  48  48 =7 + 5 + 2 = 14

Ans .

3

1. Explanation :

First of all we find the HCF of 945 and 2475. HCF = 45 Illustration :
945 )2475( 2
1890
585 ) 945 ( 1
585
360 ) 585 ( 1
360
225 ) 360 ( 1
225
135) 225 ( 1
135
90 ) 135 ( 1
90
45 ) 90 ( 2
90
x
Maximum number of animals in each flock = 45
Required total number of flocks
945 + 2475 = 21 + 55 = 76
45 45

Ans .

2

1. Explanation :

Maximum quantity in each can = HCF of 21, 42 and 63 litres = 21 litres Required least number of cans
21 + 42 + 63
21 21 21
= 1 + 2 + 3 = 6

Ans .

3

1. Explanation :

Using Rule 7, Required number = HCF of 411 – 3 = 408; 684 – 4 = 680 and 821 – 5 = 816 HCF of 408 and 816 = 408 HCF of 408 and 680
408 ) 680 ( 1
408
272 ) 408 ( 1
272
136 ) 272 ( 2
272
x
Required number = 136

Ans .

4

1. Explanation :

Required number = HCF of 200 and 320 = 40
200 ) 320 ( 1
200
120 ) 200 ( 1
120
80 ) 120 ( 1
80
40 ) 80 ( 2
80
x

Ans .

3

1. Explanation :

As the height of each stack is same, the required number of
books in each stack
= HCF of 84, 90 and 120
84 = 2 × 2 × 3 × 7
90 = 2 × 3 × 3 × 5
120 = 2 × 2 × 2 × 3 × 5
HCF = 2 × 3 = 6

Ans .

2

1. Explanation :

Using Rule 7,
Required number
= HCF of (729 – 9)
= 720 and (901 – 5)
= 896
720 ) 896 ( 1
720
176 ) 720 ( 4
704
16 ) 176 ( 11
16
16
16
x

Ans .

1

1. Explanation :

Greatest capacity of measuring vessel
= HCF of 403 litres, 434 litres
and 465 litres
= 31 litres
HCF of 403 and 434
403 ) 434 ( 1
403
31 ) 403 ( 13
31
93
93
x
HCF of 31 and 465
31) 465 ( 15
31
155
155
x

Ans .

4

1. Explanation :

Minimum number of rows = Maximum number of fruits in each row

HCF of 24, 36 and 60 = 12
Minimum number of rows 24 + 36 + 60
12  12  12
2 + 3 + 5 = 10

Ans .

4

1. Explanation :

Using Rule 7, Required number = HCF of 2300 – 32 = 2268 and 3500 – 56 = 3444
2268 ) 3444 ( 1
2268
1176 ) 2268 ( 1
1176
1092 ) 1176 ( 1
1092
84 ) 1092 ( 13
84
252
252
x
HCF = 84

Ans .

3

1. Explanation :

HCF of numbers = 12 Let the numbers be 12x and 12y
where x and y are co–prime.
According to the question,
12x × 12y = 2160
xy =
Required numbers = 12 × 3 = 36 and 12 × 5 = 60

Ans .

2

1. Explanation :

Required number = HCF of 390, 495 and 300 = 15
390 ) 495 ( 1
390
105 ) 390 ( 3
315
75 ) 105 ( 1
75
30 ) 75 ( 1
60
15 ) 30 ( 2
30
x
HCF of 15 and 300 = 15

Ans .

4

1. Explanation :

First of all we find HCF of 391 and 323.
323 ) 391 ( 1
323
68 ) 323 ( 4
272
51 ) 68 ( 1
51
17 ) 51 ( 3
51
x
Number of classes = 17

Ans .

3

1. Explanation :

Maximum length of each piece = HCF of 1.5 metre and 1.2 metre = 0.3 metre
12 ) 15 ( 1
12
3 ) 12 ( 4
12
x
HCF of 1.5 and 1.2 metre = 0.3 metre