Wipro LCM and HCF Questions and Answers Quiz-1

Let the numbers be 13a and 13b, where a and b are co-primes.

We know that LCM×HCF=Product of two numbers, therefore, we have,

13a\times 13b=13\times 455 \Rightarrow 169ab = 5915 \Rightarrow ab=\frac{1695}{915}​\Rightarrow ab=35

Two co-primes with product 35 are 5 and 7.

Therefore, the required number is (13×5,13×7)=(65,91)

Hence, the number that lies between 75 and 125 is 91.

Option C

Explanation:

L.C.M. of 20, 35, 65 = 1820

Now, 1820= 5 x 4 x 7 x 13

To make it a perfect square, it must be multiplied by 5 x 4 x 7 x 13

So, required number =5 x 4 x 7 x 13 x 5 x 4 x 7 x 13

= 5² x 4² x 7² x 13² = 3312400

Ans: Let the number be n

When divided by 6, n=6K+0 = 6K

Thus, when you divide n by 6, you get K.

 

The resulting number is then divided by 4. Therefore, K divided by 4 will give the remainder 3.

Therefore, it will become K=4A+3

 

The resulting number is again divided by 2. Therefore, A divided by 2 will get the remainder 1.

Therefore, it will become A=2B+1

 

We get the following equations:

 

n=6K

K=4A+3

A=2B+1

 

Replacing n=5K in equation, we get,

 

n=6(4A+3)

=24A+3

 

n= 24(2B+1)+3

=48B+24+3

=48B+27

 

The equation will be satisfied by any number of conditions

i.e, 48B+27 (where B can be 0,1,2,3,...)

 

If B=0, Number is 27.

Dividing 27 by 2, we get Q=13, R=1

Dividing 13 by 4, we get Q=3, R=1

Dividing 3 by 6, we get Q=0, R= 3

The remainder are 1,1,3.

Ans: For the number to be divisible by 7,12 and 32, it has to be divisible by the LCM of 7, 12 and 32.

LCM of 7,12 and 34= 1428

The multiple of 1428 immediately after 3450 is $$1428 \times 3= 4284$$

To find the least number that should be added, we will subtract both.

Therefore, the least number is 4284-3450= 834.

We need to find out the HCF for the given length.
20 meter 15 cm = 2015 cm.
21 meter 55 cm = 2155 cm.
12 meter 65 cm = 1265 cm.
HCF of 2015, 2155 and 1265 is 5.

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.

As given the questions these numbers are co primes, so there is only 1 as their
common factor.
It is also given that two products have the middle number in common.
So, middle number = H.C.F. of 399 and 525 = 21;

So first number is : 399/21 = 19
Third number = 525/21 = 25

So sum of these numbers is = (19 + 21 + 25) = 65

As we know that LCM * HCF = Product of two Numbers

Thus a + b = 11 and ab = 1 x 30 = 30

Thus 1/a + 1/b = (b + a)/(ab) = 11/30

Take the lcm of 3,4,5,6.
3=3×1
4=2×2
5=5×1
6=2×3
LCM=2×2×3×5=60

Therefore 60 is the number that is completely divisible by these numbers.
But we have to find the smallest number that leaves remainder 2.

Answer : 60 + 2=62

LCM of 18, 24 and 32:

18 = 2 x 3 x 3
24 = 2 x 2 x 2 x 3
32 = 2 x 2 x 2 x 2 x 2
LCM = 2 x 2 x 2 x 2 x 2 x 3 x 3 = 288

The smallest four digit number that is divisible by 288:

288 x 3 = 864

288 x 4 = 1152

Since 288 is divisible by 18, 24 and 32, 1152 is also divisible by all
these numbers.

Therefore, 1152 is the smallest four digit number divisible by 18, 24 and 32.