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@shattereddream said:@Raghu5janA 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other parameter if the parameters are inversely proportional.- This is also one f the short cut method for these questions
THanks buddy
The HCF of 2472, 1284 & a third number "N" is 12. If their LCM is 2^3 x 3^2 x 5 x 103 x 107, then the number "N" is:
a) 2^2 x 3^2 x 7 b) 2^2 x 3^3 x 103 c) 2^2 x 3^2 x 5 d) None of these
Do share the approach
a) 2^2 x 3^2 x 7 b) 2^2 x 3^3 x 103 c) 2^2 x 3^2 x 5 d) None of these
Do share the approach
@anilapex the answer will be (C) because if you will factor 2472-->103*2^3*3
1284=107*2^2*3 and since LCM is containing one 5 so. the N will contains 5 so by seeing option you can easuily say the answer will be (C) because it also statisfy the condition of HCF=12
Find the greatest number, which will divide 215, 167 & 135 so as to leave the same remainder in each case?
a) 64
b) 32
c) 24
d) 16
Do share the approach.
a) 64
b) 32
c) 24
d) 16
Do share the approach.
What is the number of nines used in numbeing a 453 page book?
how many four-digit numbers are divisible by 5 but not by 25
@bpolagani 14400 isit right?
@shattereddream said: A 1/x increase in one of the parameters will result in a 1/(x+1) decrease in the other parameter if the parameters are inversely proportional.- This is also one f the short cut method for these questions
That was really great. But I have not grasped the concept upto the mark. Can you please give more example(s) that can be solved with that approach?
I would be thankful to you bro if you can help me.
Thanks in advance.
@amitranjan2311 said:@anilapex16
Can you please share your approach,bro?
It will be more helpful if you elaborate it.
Thanks.
@amitranjan2311
Find the greatest number, which will divide 215, 167 & 135 so as to leave the same remainder in each case?
a) 64
b) 32
c) 24
d) 16
You answered 16.
Can you please elaborate the approach?
Thank you in advance.
@cadmium said:@amitranjan2311Find the greatest number, which will divide 215, 167 & 135 so as to leave the same remainder in each case?a) 64b) 32c) 24d) 16You answered 16.Can you please elaborate the approach?Thank you in advance.
ans iss 24 yaar..
Ok. But how 24?
Elaborated approach expected.
Thanks.
rkshtsurana
@rkshtsurana
Ok. But how 24?Elaborated approach expected.Thanks.
@cadmium said:@amitranjan2311Find the greatest number, which will divide 215, 167 & 135 so as to leave the same remainder in each case?a) 64b) 32c) 24d) 16You answered 16.Can you please elaborate the approach?Thank you in advance.
largest number which will divide p,q,r and leave same remainder in each case is given by
hcf of (p ~ q ) and (q ~ r ) = hcf of (p ~ q ) and (p ~ r ) = hcf of (p ~ r ) and (p ~ r )
hcf of (p ~ q ) and (q ~ r ) = hcf of (p ~ q ) and (p ~ r ) = hcf of (p ~ r ) and (p ~ r )
@hedonistajay
Thanks bro.!!!
@anilapex said: For the number 7200 find, the sum & no.of factors not divisible by 75?
ďťż
7200 = 2^5 * 3^2 * 5^2
75 = 3 * 5^2
For not divisible by 75, we need to remove terms containing >= 3^1 i.e. 3^1 & 3^2
We also need to remove term containing >= 5^2 i.e 5^2
Thus,
(2^0 * 2^1 * 2^2 * 2^3 * 2^4 * 2^5)(3^0 * 3^1 * 3^2)(5^0 * 5^1 * 5^2)
will become
Sum of factors not divisible by 75 =
(2^0 * 2^1 * 2^2 * 2^3 * 2^4 * 2^5)(3^0)(5^0 * 5^1) = 378
No.of factors not divisible by 75 = 6 * 1 * 2 = 12
ďťż
@ashishgupta144 said:@anilapex number of factors is 18 ?
@sonic_boom said:no. of factors 12sum of factor 252whats the ans?
7200 = 2^5 * 3^2 * 5^2
75 = 3 * 5^2
For not divisible by 75, we need to remove terms containing >= 3^1 i.e. 3^1 & 3^2
We also need to remove term containing >= 5^2 i.e 5^2
Thus,
(2^0 * 2^1 * 2^2 * 2^3 * 2^4 * 2^5)(3^0 * 3^1 * 3^2)(5^0 * 5^1 * 5^2)
will become
Sum of factors not divisible by 75 =
(2^0 * 2^1 * 2^2 * 2^3 * 2^4 * 2^5)(3^0)(5^0 * 5^1) = 378
No.of factors not divisible by 75 = 6 * 1 * 2 = 12