**4 bells toll together ay 9 am. they toll after 7,8 11,12 seconds resp. how many times will they toll together again in the next 3 hrs.?**

**a) 3**

**b)4**

**c)5**

**d)6**

Hello,

Please find the solutions below.

1. The LCM of the individual intervals will give the time at which all the bells will ring together. Which is LCM(7,8,11,12) = 1848sec so the bells would ring together every 1848 secs.

In 3 hours, they would ring (3x60x60)/1848 = 5.8

Thus the bells will toll together 5 times in next 3 hours.'

**Find the greatest no., which will divide 215,167 and 135 so as to leave the same remainder in each case.**

**a) 64**

**b) 32**

**c) 24**

**d)16**

The numbers are 215, 167 and 135.

The difference between the numbers (so that the equal remainder gets cancelled out) is - 48, 80, 32.

Ans = HCF (48, 80, 32) = 16.

**The smallest square no. which is exactly divisible by 2,,3,4,-9,6,18,36&60, is**

**a)900**

**b)1600**

**c)3600**

**d) none of these**

Q3The smallest square no. which is exactly divisible by 2,,3,4,-9,6,18,36&60, is

The required number the smallest multiple of t he LCM of the numbers that forms a perfect square.

LCM of the given numbers – 180 = 2^{2}x3^{2}x5

So the smallest square no. which is exactly divisible by the given numbers is 900.

a,b are factors of 21600.How many pairs of (a,b)are there such that hcf(a,b) = 45 ?

If the sum of two natural numbers and their LCM is 89, then how many such pairs of numbers are possible?

Let the nos. be ax and bx

LCM =abx = 89

ATQ, (a+b)x=abx => a+b=ab

Since, 89 is a prime no, a X b = 1 X 89 != 1+89

Hence, 0 such pairs.

e pair will be (45a, 45b) where a and b will be co-prime to each other. Now 21600 = 2^{5}*3^{3}*5^{2}. To find a and b, we first take the factor of 45 from 21600, which leaves 2^{5}*3*5. Now we need to find the number of co-prime pairs (a, b) that we can make out of 2^{5*}3*5. Let's write down the powers of the prime factors in order to find the co-prime factors:

(2, 2^{2}, 2^{3}, 2^{4}, 2^{5}), 3, 5

Therefore, the number of co-prime pairs is found by various combinations of these prime factors:

(Prime factor, Prime factor) (2, 3), (2^{2}, 3), (2^{3}, 3), (2^{4}, 5), (2^{5}, 5), (3, 5) ----- 11 in number

(Two prime factors, prime factor) (2 * 3, 5), (2^{2} * 3, 5) (2^{4} * 5, 3), (2^{5*}5, 3), (3*5, 2)...(3 *5, 2^{5}) --- 15 in number

(1, prime factor) (1, 2), (1, 2^{2}) .....(1, 2^{5}), (1, 3), (1, 5) --- 7 in number

(1, two prime factors) (1, 2*3), (1, 2^{2*}3)..... (1, 2^{4} * 5), (1, 2^{5} * 5), (1, 3* 5) --- 11 in number

(1, three prime factors)- (1, 2 *3 * 5), (1, 2^{2} * 3 * 5)...... (1, 2^{5} * 3 * 5) --- 5 in number.

Therefore, total number of co-prime pairs (a, b) = 49.

Let the two numbers be x and y

such that HCF(x,y) = h

=> x = ha, y = hb and LCM(x,y) = hab

=> h(a + b + ab) = 89

=> h = 1

and a + b + ab = 89

=> a = (89 - b)/(1 + b) = 90/(1 + b) - 1.

Number of positive divisors of 90 (i.e. 2*5*3^{2}) = 2*2*3 = 12.

So, 1 + b = 1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

=> (a,b) = (89,0), (44,1), (29,2), (17,4), (14,5), (9,8), (8,9), (5,14), (4,17), (2,29), (1,44) or (0,89)

Now a and b are non zero b'coz they are natural numbers.

Also HCF(a,b) = 1.

So final values for (a,b) = **(1,44) , (2,29) , (4,17) , (5,14) , (8,9) i.e. 5 pairs.**

Set A consist of all n positive integers less than 100 such that no two numbers are co prime to each other and no number is multiple of any other number in the group.

I. what is the maximum value of n?

II. What are the possible values?

Hello Apoorva ,

Select all even numbers from 50 - 98

{ 50 , 52 , 54 , ......., 98 }

Hence , maximum value of n = 25

Set A consists of all positive integers les than 100 such that no two numbers are co prime to each other and no. number is a multiple of any other number.

What is the maximum value of n?

Hello Yamin!

Select all even numbers from 50 - 98

{ 50 , 52 , 54 , ......., 98 }

Hence , maximum value of n = 25

A number when divided successively by 5, 8 and 11. It leaves the respectively remainder of 2, 5 and 7. What will be the remainder when such a least possible number is divided by 14?

Hello Surabhi,

Minimum such number = 5 x {8 x (11 x 0+7) + 5} + 2 = 307

307 when divided by 14, remainder = 13

Bhima creates 5 types of sweets, 1452, 1188, 1716, 528 and 792, respectively, in her wedding party. If he packed in cartons to give sweets to all guests in equal number, so that the sweets of all kinds in each carton, then how many maximum cartons to be needed.

Hello Surabhi,

in this case, the maximum number of cartons will be the HCF of the given numbers.

1452 = 2^{2} * 3 * 11^{2}

1188 = 2^{2} * 3^{3} * 11

1716 = 2^{2} * 3 * 11 * 13

528 = 2^{4} * 3 * 11

792 = 2^{3} * 3^{2} * 11

Their HCF = 2^{2} * 3 * 11 = 132, So Maximum number of cartons = 132

**Find the pairs of natural numbers whose LCM is 78 and GCD is 13.**

Hi Nilesh .

Let the numbers be 13a and 13 b.

( Where a and b are coprime )

LCM of 13a and 13b = 13ab

13ab = 78

ab = 6

a = 1 , b = 6 or a = 2 , b = 3

Hence , two possible pairs .