Question 1

If the HCF of 210 and 55 is expressible in the form of 210 x 5 + 55P, then value of P = ?

Accepted Answer

HCF of 210 and 55 is 5 Now, 210x5 + 55P = 5 => 1050 + 55P = 5 => 55P = -1045 => P = -1045/55 => P = -19.

Question 2

The difference of two numbers is 14. Their LCM and HCF are 441 and 7. Find the two numbers ?

Accepted Answer

Since their HCFs are 7, numbers are divisible by 7 and are of the form 7x and 7y Difference = 14 => 7x - 7y = 14 => x - y = 2 product of numbers = product of their hcf and lcm => 7x * 7y = 441 * 7 => x * y = 63 Now, we have x * y = 63 , x - y = 2 => x = 9 , y = 7 The numbers are 7x and 7y => 63 and 49

Question 3

The greatest possible length which can be used to measure exactly the length 7m, 3m 85cm, 12 m 95 cm is

Accepted Answer

Required Length = H.C.F of 700 cm, 385 cm and 1295 c = 35 cm.

Question 4

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:

Accepted Answer

L.C.M. of 5, 6, 7, 8 = 840. Required number is of the form 840k + 3 Least value of k for which (840k + 3) is divisible by 9 is k = 2. Required number = (840 x 2 + 3) = 1683.

Question 5

If the sum of two numbers is 55 and the H.C.F. and L.C.M. of these numbers are 5 and 120 respectively, then the sum of the reciprocals of the numbers is equal to ?

Accepted Answer

Let the numbers be a and b. We know that product of two numbers = Product of their HCF and LCM Then, a + b = 55 and ab = 5 x 120 = 600. => The required sum = (1/a) + (1/b) = (a+b)/ab =55/600 = 11/120

Question 6

A drink vendor has 80 liters of Mazza, 144 liters of Pepsi, and 368 liters of Sprite. He wants to pack them in cans, so that each can contains the same number of liters of a drink, and doesn't want to mix any two drinks in a can. What is the least n…

Accepted Answer

If we want to pack the drinks in the least number of cans possible, then each can should contain the maximum numbers of liters possible.As each can contains the same number liters of a drink, the number of liters in each can is a comman factor for 80,144 and 368; and it is also the highest such factor, as we need to store the maximum number of liters in each can. So, the number of liters in each can = HCF of 80,144 and 368 = 16 liters. Now, number of cans of Maaza = 80/16 = 5 Number of cans of Pepsi = 144/16 = 9 Number of cans of Sprite = 368/16 = 23 Thus, the total number of cans required = 5 + 9 + 23 = 37

Question 7

The H.C.F and L.C.M of two numbers are 84 and 21 respectively. If the ratio of the two numbers is 1 : 4 , then the larger of the two numbers is

Accepted Answer

Let the numbers be x and 4x. Then, x × 4 x = 84 × 21 ⇔ x 2 = 84 × 21 4 ⇔ x = 21 Hence Larger Number = 4x = 84

Question 8

H.C.F of 4 x 27 x 3125, 8 x 9 x 25 x 7 and 16 x 81 x 5 x 11 x 49 is :

Accepted Answer

4 x 27 x 3125 = 2 2 × 3 3 × 5 5 ; 8 x 9 x 25 x 7 = 2 3 × 3 2 × 5 2 × 7 16 x 81 x 5 x 11 x 49 = 2 4 × 3 4 × 5 × 11 × 7 2 H.C.F = 2 2 × 3 2 × 5 = 180.

Question 9

A room is 6 meters 24 centimeters in length and 4 meters 32 centimeters in Width. Find the least number of square tiles of equal size required to cover the entire floor of the room.

Accepted Answer

Let us calculate both the length and width of the room in centimeters. Length = 6 meters and 24 centimeters = 624 cm width = 4 meters and 32 centimeters = 432 cm As we want the least number of square tiles required, it means the length of each square tile should be as large as possible.Further,the length of each square tile should be a factor of both the length and width of the room. Hence, the length of each square tile will be equal to the HCF of the length and width of the room = HCF of 624 and 432 = 48 Thus, the number of square tiles required = (624 x 432 ) / (48 x 48) = 13 x 9 = 117

Question 10

Find the lowest common multiple of 24, 36 and 40?

Accepted Answer

To find the LCM of 24, 36 and 40 24 = 2 x 2 x 2 x 3 36 = 2 x 2 x 3 x 3 40 = 2 x 2 x 2 x 5 Now, LCM of 24, 36 and 40 = 2 x 2 x 2 x 3 x 3 x 5 = 8 x 9 x 5 = 72 x 5 = 360

Question 11

H.C.F of 3240, 3600 and a third number is 36 and their L.C.M is 2 4 x 3 5 x 5 2 x 7 2 . What is the third number?

Accepted Answer

47628

Question 12

Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 10 and 12 seconds respectively. In 30 minutes, how many times do they toll together ?

Accepted Answer

L.C.M. of 2, 4, 6, 8, 10, 12 is 120. So, the bells will toll together after every 120 seconds(2 minutes). In 30 minutes,they will together (30/2)+1=16 times

Question 13

The largest measuring cylinder that can accurately fill 3 tanks of capacity 98, 182 and 266 litres each, is of capacity ?

Accepted Answer

To know the the measuring cylinder that can fill all the given capacities , they must be divisible by the required number. 98,182,266 all are divisible by 14 So 14 litres is the largest cylinder that can fill all the given cylinders. (or) The other method is take HCF of all given capacities i.e 98, 182 and 266.

Question 14

HCF of 1, 2, 3?

Accepted Answer

Given numbers are 1, 2 and 3 Factors of 1 = 1 Factors of 2 = 1, 2 Factors of 3 = 1, 2 and 3 Common factors = 1 Hence, the HCF of 1, 2 and 3 = 1.

Question 15

If N is the greatest number that will divide 1305, 4665 and 6905, leaving the same remainder in each case. What is the sum of the digits of N ?

Accepted Answer

N = H.C.F. of (4665 - 1305), (6905 - 4665) and (6905 - 1305) = H.C.F. of 3360, 2240 and 5600 = 1120. Sum of digits in N = ( 1 + 1 + 2 + 0 ) = 4

Question 16

The G.C.D of 1.08, 0.36 and 0.9 is

Accepted Answer

G.C.D is nothing but H.C.F Given numbers are 1.08 , 0.36 and 0.90 we can take as 108, 36 and 90 Then required H.C.F is 18. Therefore, H.C.F of given numbers is 0.18.

Question 17

Find the greatest number that will divide 964, 1238 and 1400 leaving remainder of 41,31 and 51 respectively ?

Accepted Answer

To find the greatest number which divides the numbers 964, 1238 and 1400 leaving the remainders 41, 31 and 51 is nothing but the HCF of (964 - 41), (1238 - 31), (1400 - 51). Therefore, HCF of 923, 1207 and 1349 is 71.

Question 18

Find the least common multiple of 15 and 12?

Accepted Answer

The least common multiple of 12 and 15 is 12 = 2 x 2 x 3 15 = 3 x 5 LCM of 12 and 15 is 2 x 2 x 3 x 5 = 60.

Question 19

Three numbers are in the ratio of 3:4:5 and their L.C.M is 3600.Their HCF is:

Accepted Answer

Let the numbers be 3x, 4x, 5x. Then, their L.C.M = 60x. So, 60x=3600 or x=60. Therefore, The numbers are (3 x 60), (4 x 60), (5 x 60). Hence,required H.C.F=60

Question 20

In a palace, three different types of coins are there namely gold, silver and bronze. The number of gold, silver and bronze coins is 18000, 9600 and 3600 respectively. Find the minimum number of rooms required if in each room should give the same nu…

Accepted Answer

Gold coins = 18000 , Silver coins = 9600 , Bronze coins = 3600 Find a number which exactly divide all these numbers That is HCF of 18000, 9600& 3600 All the value has 00 at end so the factor will also have 00. HCF for 180, 96 & 36. Factors of 180 = 3 x 3 x 5 x 2 x 2 96 = 2 x 2 x 2 x 2 x 2 x 3 36 = 2 x 2 x 3 x 3 Common factors are 2x2×3=12 Therefore, Actual HCF is 1200 Gold Coins 18000/1200 will be in 15 rooms Silver Coins 9600/1200 will be in 8 rooms Bronze Coins 3600/1200 will be in 3 rooms Total rooms will be (15+8+3) = 26 rooms.

HCF and LCM Questions