IF AB+XY=1XP,WHERE A IS NOT EQUAL TO 0 AND ALL LETTERS SIGNIFY DIFFERENT DIGITS FROM 0 TO 9,THEN THE VALUE OF A IS
a.6
b.7
c.9
d.8
e.any value above 6
1) 8 orators A,B,C,D,E,F,G and H, In how many ways can the arrangements be made so that A always comes before B and B always comes before C.
A) 8!/3! B) 8!/6! C) 5!.3! D) 8!/(5!.3!)
2) A train is running between Patna and Howrah. Seven people enter the train somewhere between patna and howrah. It is given that nine stops are there in between
Patna and Howrah. In how many ways can the tickets be purchased if no restriction is there with respect to the number of tickets at any station.
A) 45C7 B) 63C7 C) 56C7 D) 52C7
Please post the approach also...
1.Number of way to arrange 8 orators = 8!
A always comes before B and B always comes before C i.e ....A.... B ......C.. thats means it cant be BCA format or CAB..i.e 3!
Number of way = 8!/3!
2. PNBE --1-2-|3-|4-|5-|6-|7-8-9--HWH (10)
From Station 1 we have 9 types of tickets for st 2,3,4,....st 10
From Station 2 we have 8 types of tickts f0r 3,4,5,.....st 10
similiarliy for station 9 wh have only 1 types(for HWH only)
Hence total types of tckts = 9+8+..+1 = 45
7 people enter the train somewhere between patna and howrah, so we have 45C7 combination
IF AB+XY=1XP,WHERE A IS NOT EQUAL TO 0 AND ALL LETTERS SIGNIFY DIFFERENT DIGITS FROM 0 TO 9,THEN THE VALUE OF A IS
a.6
b.7
c.9
d.8
e.any value above 6
Ans C i.e 9
AB + XY =1XP
=> 10A + B + 10X + Y = 100 + 10X +P
=> 10A + B +Y = 100 + p
if A=8
then 80 + B + Y = 100 +P
=> B + Y = 20 +P
B , Y = 9, 7 MAX but not valid
if A= 9
then this equ only valid B+Y = 10 + P
hence A=9
12^55/3^11+8^48/16^18 will give the digit at unit place as
a.4
b.6
c.8
d.0
e.5
12^55/3^11+8^48/16^18 will give the digit at unit place as
a.4
b.6
c.8
d.0
e.5
Is the answer 0?? I'm not quite sure about it. If it's correct, I'll post my approach.
12^55/3^11+8^48/16^18 will give the digit at unit place as
a.4
b.6
c.8
d.0
e.5
I m getting the answer as d i.e 0.Is it correct?
Individualist SaysI m getting the answer as d i.e 0.Is it correct?
thnx 4 answering but the answer is not correct the answer is 4.
thnx 4 answering but the answer is 4 as per the book.
avinav2712 SaysIs the answer 0?? I'm not quite sure about it. If it's correct, I'll post my approach.
I get the answer to be e. 5
12^55/3^11+8^48/16^18
will give
8/7 + 6/6
Now, 8 comes only when 7 is multiplied by a 4 in the units digits which gives 4 + 1 = 5
I am sure this is not the way to do this.
Roshan Jha
IIMA ASPIRANT
12^55/3^11+8^48/16^18 will give the digit at unit place as
a.4
b.6
c.8
d.0
e.5
12^55/3^11+8^48/16^18 = (3^55 * 4^55 / 3^11) + (2^3)^48 / (2^4)^18
rem(3^55 * 4^55 / 3^11) = 0
rem(2^144 / 2^72) =0
so I think ans sud be 0
1.the expression 333^555+555^333 is divisible by
a.2
b.3
c.37
d.11
2. denotes the gratest integer value just below x and {x} its fractional value. The sum ^2 and {x}^1 is 25.16 .find x
a.5.16
b.-4.84
c.both a and b
d.4.84
e.cannot determined
e.all of these
1.the expression 333^555+555^333 is divisible by
a.2
b.3
c.37
d.11
2. denotes the gratest integer value just below x and {x} its fractional value. The sum ^2 and {x}^1 is 25.16 .find x
a.5.16
b.-4.84
c.both a and b
d.4.84
e.cannot determined
e.all of these
Q1. 333^555+555^333 is divisible by 2, 3, and 37. So, I think the question should be 333^555+555^333 is not divisible by:
Q2. + {x} = 25.16
As, is an integer, {x} = 0.16
=> = 25
=> = 5 or -5
We know that x = + {x}
=> x = 5 + 0.16 = 5.16 or -5 + 0.16 = -4.84
So option c) both a and b is true
1.the expression 333^555+555^333 is divisible byCan someone post the approach for this question...:-(:-(:-(
a.2
b.3
c.37
d.11
I think question is right. and only 37 & 3 is the answer. because 333 & 555 are divisible by 3 &37 as well. hence the term 333^555 or 555^333 are also divisible by 37 & 3 . And their additions also become divisible. Lets have a simple ex (9/3)+(27/3)=36/3=12 Above 9 & 27 are divisible by same number that is 3 and hence their addition is also divisible giving the same answer.
pls tell me is it right or not?
Ganu02 SaysCan someone post the approach for this question...:-(:-(:-(
333^555 + 555^333
Now we can see that both the terms are odd, so odd + odd = even and hence it will be divisible by 2.
333 = 3*3*37
555 = 3*5*37
It mean the expression will be divisible by 3 and 37 also.
Eulers number for 11 is 10
=> 333^555 = 333^5(mod 11) = 3^5 (mod 11) = 1 (mod 11)
Similarly, 555^333 = 555^3(mod 11) = 5^3 (mod 11) = 4(mod 11)
=> 333^555 + 555^333 = 5(mod 11)
hence not divisible by 11.
A precious stone weighing 35 grams worth Rs. 12250 is accidentally dropped and gets broken into two pieces having weights in the ratio of 2 : 5. If price varies as square of the weight find the loss incured.
a)5750
b)6000
c)5500
d)5000
A precious stone weighing 35 grams worth Rs. 12250 is accidentally dropped and gets broken into two pieces having weights in the ratio of 2 : 5. If price varies as square of the weight find the loss incured.
a)5750
b)6000
c)5500
d)5000
Is the answer 5000 d??
yes its b, how did u solve it ?
EagleMenace Saysyes its b, how did u solve it ?
Intially it was 35 gms and price was 12250 (35^2=1225)prices varies with square of the weight
Later 2 : 5=10gms and 25 gms so their price will be 1000+6250=7250
so loss=12250-7250