9
Q:

# 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:

 A) 55/601 B) 601/55 C) 11/120 D) 120/11

Explanation:

Let the numbers be a and b.

Then, a + b = 55 and ab = 5 x 120 = 600.

The required sum =$\inline \fn_jvn \frac{1}{a}+\frac{1}{b}$ = $\inline \fn_jvn \frac{a+b}{ab}$$\inline \fn_jvn \frac{55}{600}$=$\inline \fn_jvn \frac{11}{120}$

Q:

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 number of coins of the same type ?

 A) 26 B) 24 C) 18 D) 12

Explanation:

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

$\inline \fn_jvn \small \therefore$ Actual HCF is 1200

Gold Coins $\inline \fn_jvn \small \left ( \frac{18000}{1200} \right )$ will be in 15 rooms

Silver Coins $\inline \fn_jvn \small \left ( \frac{9600}{1200} \right )$ will be in 8 rooms

Bronze Coins $\inline \fn_jvn \small \left ( \frac{3600}{1200} \right )$ will be in 3 rooms

Total rooms will be (15+8+3)  =  26 rooms.

5 109
Q:

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

 A) 40 B) 60 C) 100 D) 120

Explanation:

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

5 90
Q:

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 number of cans required?

 A) 35 B) 36 C) 37 D) 38

Explanation:

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

12 803
Q:

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.

 A) 107 B) 117 C) 127 D) 137

Explanation:

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

9 719
Q:

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

 A) 1677 B) 1683 C) 2523 D) 3363

Explanation:

L.C.M of 5, 6, 7, 8 = 840

$\inline&space;\fn_jvn&space;\therefore$ Required Number is of the form 840k+3.

Least value of k for which (840k+3) is divisible by 9 is k = 2

$\inline&space;\fn_jvn&space;\therefore$  Required  Number = (840 x 2+3)=1683