Staff Selection Commission Mathematics - LCM and HCF

Q.1

The least number which when divided by 5, 6, 7 and 8 leaves a remainder 3, but when divided by 9 leaves no remainder is

1677

1683

2523

3363

Ans .

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

Q.2

The greatest number of four dig- its which when divided by 12, 16 and 24 leave remainders 2, 6 and 14 respectively is

9974

9970

9807

9998

Ans .

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

Q.3

When a number is divided by 15, 20 or 35, each time the remain- der is 8. Then the smallest num- ber is

428

427

328

338

Ans .

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

Q.4

The smallest number, which, when divided by 12 or 10 or 8, leaves remainder 6 in each case, is

246

186

126

Ans .

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

Q.5

The traffic lights at three differ- ent road crossings change after 24 seconds, 36 seconds and 54 seconds respectively. If they all change simul taneo usly at 10 : 15 : 00 AM, then at what time will they again change simul- taneously?

10:16:54:AM

10:18:36:AM

10:17:02:AM

10:22:12:AM

Ans .

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.

Q.6

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

30 hours

6 hours

10 hours

15 hours

Ans .

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.

Q.7

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 ?

1800 seconds

3600 seconds

2400 seconds

4800 seconds

Ans .

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

Q.8

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 ?

1) 12 hrs. 2 min. 48 sec

(2) 12 hrs. 3 min.

(3) 12 hrs. 3 min. 20 sec.

(4) 12 hrs. 3 min. 44 sec.

Ans .

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

Q.9

4 bells ring at intervals of 30 minutes, 1 hour, 1 1 hour and 1 2 hour 45 minutes respectively. All the bells ring simultaneously at 12 noon. They will again ring si- multaneously at :

(1) 12 mid night

(2) 3 a.m.

(3) 6 a.m.

(4) 9 a.m.

Ans .

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.

Q.10

Four bells ring at the intervals of 5, 6, 8 and 9 seconds. All the bells ring simulataneously at some time. They will again ring simultaneously afte

(1) 6 minutes

(2) 12 minutes

(3) 18 minutes

(4) 24 minutes

Ans .

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

Q.11

Three bells ring simultaneously at 11a.m. They ri ng at regular intervals of 20 mi nutes, 30 minutes, 40 minutes respec- tively. The time when all the three ring together next is

2 p.m.

1 p.m.

1.15 p.m.

1.30 p.m.

Ans .

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

Q.12

The greatest number of four dig- its which when divided by 3, 5, 7, 9 leave remainders 1, 3, 5, 7 respectively is :

9763

9764

9766

9765

Ans .

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

Q.13

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 ?

(1) 72 Sec.

(2) 612 Sec.

(3) 504 Sec.

(4) 318 Sec.

Ans .

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

Q.14

L.C.M.of 2/3, 4/9, 5/6 is

(1) 8/27

(2) 20/3

(3) 10/3

(4) 20/27

Ans .

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

Q.15

The number nearest to 10000, which is exactly divisible by each of 3, 4, 5, 6, 7 and 8, is :

9240

10080

9996

10000

Ans .

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

Q.16

The largest 4-digit number exact- ly divisible by each of 12, 15, 18 and 27 is

9690

9720

9930

9960

Ans .

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.

Q.17

The least number, which is a per- fect square and is divisible by each of the numbers 16, 20 and 24, is

1600

3600

6400

14400

Ans .

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

Q.18

The number nearest to 43582 di- visible by each of 25, 50 and 75 is :

43500

43650

43600

43550

Ans .

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

Q.19

The smallest number, which when increased by 5 is divisible by each of 24,32, 36 and 564, is

869

859

4320

427

Ans .

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

Q.20

The greatest number, which when subtracted from 5834, gives a number exactly divisible by each of 20, 28, 32 and 35, is

1120

4714

5200

5600

Ans .

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

Q.21

The smallest perfect square di- visible by each of 6, 12 and 18 is

196

144

108

Ans .

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

Q.22

The greatest 4-digit number ex- actly divisible by 10, 15, 20 is

9990

9960

9980

9995

Ans .

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

Q.23

Find the least number which when divided separately by 15, 20, 36 and 48 leaves 3 as re- mainder in each case.

183

243

483

723

Ans .

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

Q.24

Three men step off together from the same spot. Their steps mea- sure 63 cm, 70 cm and 77 cm respectively. The minimum dis- tance each should cover so that all can cover the distance in com- plete steps is

9630 cm

9360 cm

6930 cm

6950 cm

Ans .

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

Q.25

Three bells ring at intervals of 36 seconds, 40 seconds and 48 sec- onds respectively. They start ring- ing together at a particular time. They will ring together after ev- ery

6 minutes

12 minutes

18 minutes

24 minutes

Ans .

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

Q.26

The LCM of four consecutive numbers is 60. The sum of the first two numbers is equal to the fourth number. What is the sum of four numbers?

Ans .

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

Q.27

The LCM fo two prime numbers x and y, (x > y) is 161. The value of (3y – x):

-2

-1

Ans .

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

Q.28

. Three electronic devices make a beep after every 48 seconds, 72 seconds and 108 seconds re- spectively. They beeped togeth- er at 10 a.m. The time when they will next make a beep together at the earliest is

10 : 07 : 12 hours

10 : 07 : 24 hours

10 : 07 : 36 hours

Ans .

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

Q.29

The maximum number of stu- dents among whom 1001 pens and 910 pencils can be distributed in such a way that each student gets same number of pens and same number of pencils, is :

910

1001

1911

Ans .

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

Q.30

The greatest number, which when divide 989 and 1327 leave remainders 5 and 7 respectively, is :

Ans .

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

Q.31

H.C.F of 2/3,4/5and6/7 is

48/105

2/105

1/105

24/105

Ans .

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

Q.32

Let N be the greatest number that will divide 1305, 4665 and 6905 leaving the same remain- der in each case. Then, sum of the digits in N is :

Ans .

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

Q.33

What is the greatest number that will divide 307 and 330 leaving remainders 3 and 7 respectively ?

Ans .

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

Q.34

. Which greatest number will di- vide 3026 and 5053 leaving re- mainders 11 and 13 respectively?

Ans .

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

Q.35

The greatest number, by which 1657 and 2037 are divided to give remainders 6 and 5 respectively,is

127

133

235

305

Ans .

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.

Q.36

The largest number, which di- vides 25, 73 and 97 to leave the same remainder in each case, is

Ans .

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

Q.37

What is the greatest number which will divide 110 and 128 leaving a remainder 2 in each case ?

Ans .

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

Q.38

A milkman has 75 litres milk in one can and 45 litres in another. The maximum capacity of con- tainer which can measure milk of either container exact number of times is :

1 litre

5 litres

15 litres

25 litres

Ans .

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

Q.39

What is the least number of square tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad?

840

841

820

814

Ans .

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

Q.40

Three sets of English, Mathemat- ics and Science books contain- ing 336, 240, 96 books respec- tively have to be stacked in such a way that all the books are stored subject-wise and the height of each stack is the same. Total number of stacks will be

Ans .

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

Q.41

A farmer has 945 cows and 2475 sheep. He farms them into flocks, keeping cows and sheep sepa- rate and having the same num- ber of animals in each flock. If these flocks are as large as pos- sible, then the maximum number of animals in each flock and total number of flocks required for the purpose are respectively

15 and 228

9 and 380

45 and 76

46 and 75

Ans .

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

Q.42

A milk vendor has 21 litres of cow milk, 42 litres of toned milk and 63 litres of double toned milk. If he wants to pack them in cans so that each can contains same litres of milk and does not want to mix any two kinds of milk in a can, then the least number of cans required is

Ans .

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

Q.43

The greatest number that divides 411, 684, 821 and leaves 3, 4 and 5 as remainders, respec- tively, is

254

146

136

204

Ans .

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

Q.44

Find the greatest number which will exactly divide 200 and 320

Ans .

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

Q.45

84 Maths books, 90 Physics books and 120 Chemistry books have to be stacked topicwise. How many books will be there in each stack so that each stack will have the same height too ?

Ans .

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

Q.46

The greatest number that will di- vide 729 and 901 leaving remain- ders 9 and 5 respectively, is

Ans .

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

Q.47

Three tankers contain 403 litres, 434 litres, 465 litres of diesel respectively. Then the maximum capacity of a container that can measure the diesel of the three containers exact number of times is

31 litres

62 litres

41 litres

84 litres

Ans .

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

Q.48

There are 24 peaches, 36 apri- cots and 60 bananas and they have to be arranged in several rows in such a way that every row contains the same number of fruits of only one type. What is the minimum number of rows required for this to happen ?

Ans .

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

Q.49

The greatest number by which 2300 and 3500 are divided leav- ing the remainders of 32 and 56 respectively, is

136

168

Ans .

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

Q.50

The product of two 2–digit num- bers is 2160 and their H.C.F. is 12. The numbers are

(12, 60)

(72, 30)

(36, 60)

(60, 72)

Ans .

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

Q.51

Find the greatest number that will divide 390, 495 and 300 without leaving a remainder

Ans .

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

Q.52

In a school, 391 boys and 323 girls have been divided into the largest possible equal classes, so that each class of boys numbers the same as each class of girls. What is the number of classes ?

Ans .

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

Q.53

Two pipes of length 1.5 m and 1.2 m are to be cut into equal pieces without leaving any extra length of pipes. The greatest length of the pipe pieces of same size which can be cut from these two lengths will be

0.13 metre

0.4 metre

0.3 metre

0.41 metre

Ans .

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

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