Practice Question On LCM and HCF for Bank , SSC and UPSC

Work And time

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Question no. 1

The H.C.F and L.C.M of two numbers are 21 and 4641 respectively. If one of the numbers lies between 200 and 300, then the two numbers are

looks_one 273 , 357

looks_two 273 , 361

looks_3 273 , 359

looks_4 273 , 363

Option looks_one 273 , 357

Solution :
HCF = 21 , LCM = 4641 = 13 * 3 * 7 * 17
Each number has 21 as factor
First number = 21 * 13 = 273
Second Number = 21 * 17 = 357

Question no. 2

What is the smallest number which when increased by 5 is completely divisible by 8, 11 and 24?

looks_one 264

looks_two 259

looks_3 269

looks_4 None of these

Option looks_two 259

Solution :
LCM ( 8 ,11 , 24) = 264
LCM is the smallest number that is divisible by all their factors
264 is the number completely divisible by 8, 11 and 24
Number is increased by 5 = 264 - 5 = 259

Question no. 3

Find the greatest number that will divide 148,246 and 623 leaving remainders 4, 6 and 11 respectively.

looks_one 11

looks_two 12

looks_3 13

looks_4 14

option looks_two 12

Solution :
148 leaves remainders 4 = 148 - 4 = 144
246 leaves remainders 6 = 246 - 6 = 240
623 leaves remainders 11 = 623 - 11 = 612
HCF (141 , 240 , 612) = 12

Question no. 4

The LCM and HCF of two numbers are 84 and 21, respectively. If the ratio of two numbers be 1:4, then the larger of the two numbers is

looks_one 21

looks_two 48

looks_3 84

looks_4 108

Option looks_3 84

Solution :
LCM = 84 , HCF = 21
First number = x , second number = 4x
product of two number = LCM * HCF
x * 4x = 84 * 21
x² = 21²
x = 21
Largest number = 21 * 4 = 84

Question no. 5

If the L.C.M and H.C.F. oftwo numbers are 2400 and 16, one number is 480; find the second number.

looks_one 40

looks_two 80

looks_3 60

looks_4 None of these

Option looks_two 80

Solution :
LCM = 2400 , HCF = 16
First number = 480 , second number = x
product of two number = LCM * HCF
2400 * 16 = 450 * x
x = 80

Question no. 6

The LCM of two numbers is 280 and their ratio is 7: 8. The two numbers are :

looks_one 70,80

looks_two 54, 68

looks_3 35 , 40

looks_4 28 , 36

Option looks_3 35 , 40

Solution :
LCM = 280
first number = 7x , second number = 8x where x is HCF
product of two number = LCM * HCF
280 * x = 7x * 8x
280 = 56x
x = 5
First number = 7x = 7 * 5 = 35
second number = 8* 5 = 40

Question no. 7

Find the greatest number that will divide 115, 149 and 183 leaving remainders 3, 5, 7 respectively.

looks_one 14

looks_two 16

looks_3 18

looks_4 20

Option looks_two 16

Solution :
115 leaves remainder 3 = 115 - 3 = 112
149 leaves remainder 5 = 149 - 5 = 144
183 leaves reaminder = 183 - 7 = 176
HCF(112 , 144 , 176) = 16
HCF is the greatest number which will divide the three number (112 , 144 , 176)

Question no. 8

The L.C.M. of two number is 630 and their H.C.F. is 9. If the sum of numbers is 153, their difference is

looks_one 17

looks_two 23

looks_3 27

looks_4 33

Option looks_3 27

Solution :
LCM = 630 , HCF = 9
, first number = a , second number = b
Sum of number = a+b = 153
product of two number = ab = 9*630
(a-b)² = (a+b)² - 4 ab = (153)² - 4 (9*630)
(a-b)² = 729
a-b = 27

Question no. 9

Find the greatest number that will divide 55, 127 and 175, so as to leave the same remainder in each case.

looks_one 11

looks_two 16

looks_3 18

looks_4 24

Option looks_4 24

Solution :
127 - 55 = 72
175 - 127 = 48
175 - 55 = 120
HCF (72 , 48 , 120) = 24

Question no. 10

Find the greatest possible rate at which a man should walk to cover a distance of 70km and 245 km in exact number of days?

looks_one 55

looks_two 60

looks_3 35

looks_4 45

Option looks_3 35

Solution :
70 = 2 * 5 * 7
245 = 5 * 7 * 7
HCF(70 , 245) = 35

Question no. 11

The greatest number which will divide 410, 751 and 1030 leaving a remainder 7 in each case is

looks_one 29

looks_two 31

looks_3 17

looks_4 37

Option looks_two 31

Solution :
410 leaves remainder 7 = 410 - 7 = 403
751 leaves reaminder 7 = 751 - 7 = 744
1030 leaves reaminder 7 = 1023
HCF(403 , 744 , 1023) = 31
HCF is the grratest number which will divide 403 , 744 , 1023

Question no. 12

The H.C.F. of two numbers is 23 and the other factors of their L.C.M. are 13 and 14. The larger of the two numbers is:

looks_one 276

looks_two 299

looks_3 322

looks_4 345

Option looks_3 322

Solution :
HCF = 23
LCM = 13 * 14 * 23
first number = 13 * 23 = 299
second number = 14 * 23 = 322

Question no. 13

What will be the least number which when doubled will be exactly divisible by 12,18,21 and 30?

looks_one 196

looks_two 630

looks_3 1260

looks_4 2520

Option looks_two 630

Solution :
12 = 2 * 2 * 3
18 = 2 * 3 * 3
21 = 3 * 7
30 = 2 * 3 * 5
LCM (12 ,18 , 21 , 30) = 2 * 2 * 3 * 3 * 5 * 7 = 1260
2x = 1260
x = 630

Question no. 14

A hall is 13 metres 53 cm long and 8 metres 61cm broad is to be paved with minimum number of square tiles. The number of tiles required is:

looks_one 123

looks_two 77

looks_3 99

looks_4 57

Option looks_two 77

Solution :
length = 1353 cm
breadth = 861 cm
HCF(1353,861) = 123
size of tile = 123 cm
NO. of tiles = total room area / area of tiles = 1353*861/123*123 = 77

Question no. 15

In a school there are 391 boys and 323 girls. These are to be divided into the largest possible equal classes, so that there are equal number of boys and girls in each class. How many classes are possible?

looks_one 32

looks_two 37

looks_3 42

looks_4 49

Option looks_3 42

Solution :
No. of boys = 391
No. of girls = 323
391 = 17 *23
323 = 17 * 19
17 boys with 23 classes , 19 classes with 17 girls
Total number of classes = 19 + 23 = 42