Along a road lie an odd number of stones placed at intervals of 10 m.These stones have to be assembled around the middle stone .A person can carry only one stone at a time. A man carried out the job starting with stone in middle carrying stones in succession thereby covering a distance =4.8 km.no of stones is=
a.35 b.15 c.29 d.31
ans=d
puys ! help me solve this
Along a road lie an odd number of stones placed at intervals of 10 m.These stones have to be assembled around the middle stone .A person can carry only one stone at a time. A man carried out the job starting with stone in middle carrying stones in succession thereby covering a distance =4.8 km.no of stones is=
a.35 b.15 c.29 d.31
ans=d
puys ! help me solve this
let the number of stones be 2x+1
1 is the middle stone and x stones are organised on both sides of the middle stone.
so, first stone is located at distance of 10 meters and second at 20 and so on. Total of 4800 meters. To travel from the center stone and come back to it, he must travel 2* the distance. This again for both the sides. So, we multiply it by 4.
So, 2*2*(10+20+...x)=4800
20*x*(x+1)=4800
x(x+1)=240
x=15.
So, total number of stones= 2x+1=31
Option (D)
Along a road lie an odd number of stones placed at intervals of 10 m.These stones have to be assembled around the middle stone .A person can carry only one stone at a time. A man carried out the job starting with stone in middle carrying stones in succession thereby covering a distance =4.8 km.no of stones is=
a.35 b.15 c.29 d.31
ans=d
puys ! help me solve this
ans :31
2400 m on either side he covers
15*150=2250
2250 +150 metres=2400
hi all,
6 P + 4 S = 4 N
where
N is any no. b/w 10 - 1000
P is the product of digits of the no.
S is the sum of the digits of the no.
How many values can N take ?
I know the answer & the soln. as well .
Got a good quest. so posted it..
Along a road lie an odd number of stones placed at intervals of 10 m.These stones have to be assembled around the middle stone .A person can carry only one stone at a time. A man carried out the job starting with stone in middle carrying stones in succession thereby covering a distance =4.8 km.no of stones is=
a.35 b.15 c.29 d.31
ans=d
puys ! help me solve this
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10 m intervals...???
Could anyone help me solve the following
1.E(N) is defined as
Could anyone help me solve the following
1.E(N) is defined as Square root of(1+(1/N^2)+(1/(N-1)^2)) where N is a natural number which is atleast 2. Find Sigma(i=2 to 1005) E(i).
2.What is the value of the expression
[ (15/(62-1)) + (1/(82-1)) + (1/(102-1)) +.. +(1/(162-1)) ]
3.What is the value of sqrt(2+(sqrt2+(sqrt2+(sqrt2+)))).where sqrt means square root
4.Is there any simple way to calculate 2+(1/(2+(1/2+(1/(2+1/11)))))
1.In a four digit number ,the sum of the digits in the units place and the tens place is equal to the sum of the digits in the hundreds and thousands places.the sum of the digits in the tens and hundreds places is twice the sum of the other two digits.if the sum of the digits of the number is more than 20,then the digit in the units place can be
1)6 5)2
2)7
3)8
4)5
2.The value of a diamond varies directly with the square if its weight.The diamond broke into 3 pieces whose weights were in the ratio 32:24:9.The loss caused due to the breakage was Rs.25.44 lakhs.Find the initial value of the diamond(in lakhs of rupees)
1)33.62 4)8.405
2)16.81 5)84.50
3)42.25
3.if (p+q)/r =(q+r)/p=(p+r)/q=k,then find k.
thank u
how can we solve the questions like "how many 3 digit nos. when divided by 4 and 7 leaves remainder 3 and 5 respectively." Plz help puys.
on the edge Sayshow can we solve the questions like "how many 3 digit nos. when divided by 4 and 7 leaves remainder 3 and 5 respectively." Plz help puys.
4x + 3 = 7y + 5
4x + 3 numbers : 7,11,15,19,23,27
7y + 5 numbers : 12,19
so numbers will follow a series LCM (7.4) +19
now 28k + 19
now question is like how many numbers of form 28 k + 17 is three digit number
900/28 is approx 32 .. so answer should 33 or 32 .. jus calculate
I have to divide 2860 by 3186. The question gives only 2 minutes and that division is only half part of question. Now I can't possibly make that division in or less than 2 minutes by applying traditional methods, which I can't apply on that division anyways.
So anyone can perform below division using faster technique?
2860/3186
Thanks for reading, hoping to get some answers. 😃
I have to divide 2860 by 3186. The question gives only 2 minutes and that division is only half part of question. Now I can't possibly make that division in or less than 2 minutes by applying traditional methods, which I can't apply on that division anyways.
So anyone can perform below division using faster technique?
2860/3186
Thanks for reading, hoping to get some answers. :)
Here is a method 4 u...
v convert denominator into number by which we can convenintly divide.
I will make it here 3000.
to make it 3000, i need to subtract 186.
Immediatly, i could see 3186 * 5% = 159.** (few seconds-depends on ur practice : i took 5 to 10 seconds)
stil got 26.** places to go...! here i could see 1% =31.86. so, 5 places less.
approx .8% or just using the mind ..u can assume if M taking 5 less here, then above I can take 4 less depending on the number.here i assume 4. (not more than 10 seconds)
Now come to numerator : 2860 - 2860*5% = 143 and 1% = 28.60 - 4 =24.60.
2860 - 167 = 2693.(10 to 20 seconds)
2693/3000 = .897
Note : Apply this once u do enough practice. U can use this method in 95% calculations saving a lot of time.
hope it helps 😃
1.In a four digit number ,the sum of the digits in the units place and the tens place is equal to the sum of the digits in the hundreds and thousands places.the sum of the digits in the tens and hundreds places is twice the sum of the other two digits.if the sum of the digits of the number is more than 20,then the digit in the units place can be
1)6 5)2
2)7
3)8
4)5
2.The value of a diamond varies directly with the square if its weight.The diamond broke into 3 pieces whose weights were in the ratio 32:24:9.The loss caused due to the breakage was Rs.25.44 lakhs.Find the initial value of the diamond(in lakhs of rupees)
1)33.62 4)8.405
2)16.81 5)84.50
3)42.25
3.if (p+q)/r =(q+r)/p=(p+r)/q=k,then find k.
thank u
1. Let the number be
XYZW
NOW, Z+W=X+Y
AND Y+Z =2X+2W
SO, Y+Z+W=2X+3W
=2X+Y=2X+3W
THEREFORE, Y=3W
SO, unit digit can't be only 1,2,3
as per options answer should be option 5 i.e 2.
2.let price of diamond be P and weight be W
so, P=k*w^2
initial price=k*65^2
=4225k
now when the diamond is broken
seperate prices will be
1024k,576k and 81k
adding
1681k=4225 k-25.44
=k=1/100
so initial price=42.25
3. P+q/r =q+r/p =p+r/q =k
p+q=kr
q+r=kp
p+r=kq
adding all
2(p+q+r)=k(p+q+r)
so k=2.
Here is a method 4 u...
v convert denominator into number by which we can convenintly divide.
I will make it here 3000.
to make it 3000, i need to subtract 186.
Immediatly, i could see 3186 * 5% = 159.** (few seconds-depends on ur practice : i took 5 to 10 seconds)
stil got 26.** places to go...! here i could see 1% =31.86. so, 5 places less.
approx .8% or just using the mind ..u can assume if M taking 5 less here, then above I can take 4 less depending on the number.here i assume 4. (not more than 10 seconds)
Now come to numerator : 2860 - 2860*5% = 143 and 1% = 28.60 - 4 =24.60.
2860 - 167 = 2693.(10 to 20 seconds)
2693/3000 = .897
Note : Apply this once u do enough practice. U can use this method in 95% calculations saving a lot of time.
hope it helps :)
Thanks man. 😃
Please help me with the solutions. Thank you. :)
1) Number of consecutive zeroes at the end of the folln nos:
a) 77! * 42!
b) 1^1 * 2^2 * .... *10^10
Ans- 27, 15
2)Find the remainder when:
a) 75^80 is divided by 7
b) 83^261 is divided by 17
c) 9+ 9^2 + 9^3 +....+ 9^(2n+1) is divided by 6
d) 43^101 + 23^101 is divided by 66
e) 2222^5555 + 5555^2222 is divided by 7
f) 33^34^35 is divided by 7
g) 2^2+22^2+....+(222...49 twos)^2 is divided by 9
Ans-4, 2, 3, 0, 0, 2, 6
3) A hundred and twenty digit no. is formed by writing the first x natural no. infront of each other as 12345678910111213. Find the remainder when this no. is divided by 8.
Ans- 6
4)Find the least number which must be subtracted from 7147 to make it a perfect square.
Ans-91 (Is there any trick involved pertaining to finding out perfect squares quickly?)
5) What is the total no. of divisors of the number
12^33 * 34^23 * 2^47 ?
6) How many integer values of x and y are such that 4x+7y=3, while x
Ans-143
7) Find the sum of all 3 digit nos that give a remainder of 4 when they are divided by 5.
Ans- 99270
Find the last digit of no- 1^2 + 2^2 +...+ 99^2.
Ans-0
9) n is a number such that 2n has 28 factors and 3n has 30 factors.6n has?
Ans-35
Please help me with the solutions. Thank you. :)
1) Number of consecutive zeroes at the end of the folln nos:
b) 1^1 * 2^2 * .... *10^10
We will have zeros by having a 5 and 2 that gives us one zero so to know number of zeros we only need to know how many 5s and 2s are given. the fast way is to check only number of 5s cause it's obvious, there are more number of 2s and less number of 5s and we won't get zero only by 2s, we need 5s also.
for eg- four 2s and one 5 will only give me one zero.
So in this ques check number of 5s. we get 5 from 5^5 and 10^10.
from 5^5 we have five 5s and from 10^10- break 10 as 2*5 and we get (2^10)*(5^10) that means we have total as= 5+10= fifteen 5s.
Hense Ans-15
Remainder Questions:
for all questions i'll use Euler theorem.
2) a)75^80 divided by 7:
euler no is 6 so 75^78 mod 7 = 1
now left part is 75^2 mod 7
so after dividing we have the remainder 5*5 mod 7 and 25 mod 7= 4 so the remainder is 4
b) 83^261 divided by 17
euler no is 16 so 83^256 mod 17 = 1
now left part is 83^5 mod 7
after dividing we have the negative remainder as (-2)^5 mod 17 and 32 divided by 17 the remainder is 15. since it was the negative remainder and power is odd so we have the remainder as -15 and to change the remainder to positive we add 17. so we have -15+17= 2
so the remainder is 2
d) 43^101 + 23^101 divided by 66
euler no of 66 is 20
separating the sum we take 43^101 mod 66
we have 43^100 mod 66 = 1
and now we left with 43 mod 66
similary we'll have 23 mod 66
adding the two remainders we have 43+23 mod 66 is 66 mod 66 so the remainder is 0.
e) 2222^5555 + 5555^2222 mod 7
euler no of 7 is 6
again separating the sum we take 2222^5555 mod 7
so 2222^5550 mod 7 = 1
and 2222^5 mod 7 = 3^5 mod 7 = 5
and 5555^2219 mod 7 = 1
and 5555^3 mod 7 = 4^3 mod 7 = 1
so adding the two we get 5+1=6 as the remainder
f) 33^34^35 mod 7
euler no of 7 is 6
33^30 mod 7 = 1
and now 33^4 mod 7 = 5^4 mod 7 = 9 mod 7 = 2
now we have 2^35 mod 7
2^30 mod 7= 1
and 2^5 mod 7 = 4
so the remainder is 4 (which i think is wrong :|)
2)Find the remainder when:
a) 75^80 is divided by 7
5^80/7
now u can use fermat little theorem
5^6/7=1, so 5^78/7=1
5^2/7=25/7=4
b) 83^261 is divided by 17
again fermat little theorem
83^16/17=1
83^5/17=15x15x15x15x15/17
=3
c) 9+ 9^2 + 9^3 +....+ 9^(2n+1) is divided by 6
0 bcoz divisibilty rule of 6 is sum of all numbers shd divible by 2 and3.
d) 43^101 + 23^101 is divided by 66
again 0
a^n+b^n is divible by a+b if when n is odd
e) 2222^5555 + 5555^2222 is divided by 7
3^2222+4^5555/7
so 2
f) 33^34^35 is divided by 7
5^34^35
from fermat little theorem
5^6/7=1
divide 34/35 by 6
4^35/6=4, so this can be put in the form 6k+4
5^6k+4/7
5^4/7
=625/7=2
3) A hundred and twenty digit no. is formed by writing the first x natural no. infront of each other as 12345678910111213. Find the remainder when this no. is divided by 8.
Ans- 6
Divisiblity rule for 8 is last 3 digit must be divisible by 8
last three digit of number would be 646...646/8=6 is the remainder
4)Find the least number which must be subtracted from 7147 to make it a perfect square.
Ans-91 (Is there any trick involved pertaining to finding out perfect squares quickly?)
90x90=8100 so its somewhere between 80-80.. try 85x85=7225...so it shd be 84x84=7056.. I guess there are no such shortcuts, for number to be perfect square all prime factors should be even..
What is the total no. of divisors of the number
12^33 * 34^23 * 2^47 ?
(2x2x3)^33 x (17x2)^23 x 2^47
2^126 x 3^33 x 17^46 use the formula and find now
Post got too long :D
7) Find the sum of all 3 digit nos that give a remainder of 4 when they are divided by 5.
Ans- 99270
since the remainder is 4 last digit of the number end with 4 or 9..
Numbers are 104+109+114+119+....................+999
its in Ap, L = a + (n - 1) d
999 = 104 +( n - 1)5
n = 180
Or u can also find using pnc
first place can be filled in 9ways(0excluded) x second place can be filled in 10 ways x last digit in 2 ways(4or9)
so 9x10x2=180 digits
S=(104+999)180/2
s=99270
Find the least number which when divided by 12, 16, 18, 20 leaves remainder 4 in each case, but it is completely divisible by 7. There are no options, its from the numbers exercise in base builder of career forum. Can anyone show me a method to solve the problem. I found the LCM of 12, 16, 18 and 20 as 720, but I'm not quite sure about the method which finds the number which will be divisible by 7.
Thanks in advance. 😃
All numbers give same remainder, also its completely divisible by 7..
so lcm(12,16,18,20)k+4, where k=1,2,3,4....... we get the required answer when k=4, so 2880+4=2884 is the answer
All numbers give same remainder, also its completely divisible by 7..
so lcm(12,16,18,20)k+4, where k=1,2,3,4....... we get the required answer when k=4, so 2880+4=2884 is the answer
Thanks for the reply man. Yes, I'm aware of the formula and usage of 'K'. But isn't that method trial and error? Ain't there a definite method to find out the answer, especially when the number is large and we need to specifically check if its divisible by some number with hardest possible divisibility rule. 😛