So much spamming on Quant Thread ? Move to SB. Question:A not-so-good clockmaker has four clocks on display in the window. Clock #1 loses 15 minutes every hour. Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15).Clock #3 loses 20 minutes every hour relative to Clock #2. Finally, Clock #4 gains 20 minutes every hour relative to Clock #3. If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day) ?P.S. Spam Posts delete kar dena!
What a slow start to the sunday...post karo yaar questions.. Q: Amit has 2n marbles numbered 1 through 2n. He removes n marbles that are numbered consecutively. The sum of the numbers on the remaining marbles is 1615. Find all possible values of n.
(n)(2n+1)-n/2(m+1+m+n)=1615 n(2n+1-m-n/2-1/2)=1615 n(3n/2-m-1/2)=1615 n(3n-2m-1)=3230 3230=17*19*5*2 1615*2=x(x+1) x(x+1)=3230 x is close to 59 atleast n has to be 59 values 85 95 170 190 323 646 1615 3230 ?
amit threw five standard dice simultaneously. He found that the product of the numbers on the top faces was 216. which of the following couldn't be the sum of numbers on top five faces:
a: 17
b: 19
c: 18
d: 20
Q2:
The number 2006! is written in base 22. How many zeroes are there at the end?
a: 500
b: 450
c: 200
d: 199
Q3:
X & Y are two natural no's such that X+Y=949. LCM of X & Y is 2628. What is the HCF of X & Y??
Q1:amit threw five standard dice simultaneously. He found that the product of the numbers on the top faces was 216. which of the following couldn't be the sum of numbers on top five faces:a: 17b: 19c: 18d: 20Q2:The number 2006! is written in base 22. How many zeroes are there at the end?a: 500b: 450c: 200d: 199Q3:X & Y are two natural no's such that X+Y=949. LCM of X & Y is 2628. What is the HCF of X & Y??a: 23b: 73c: 69d: noneAnswer along with approach is must
What a slow start to the sunday...post karo yaar questions.. Q: Amit has 2n marbles numbered 1 through 2n. He removes n marbles that are numbered consecutively. The sum of the numbers on the remaining marbles is 1615. Find all possible values of n.
2n*(2n + 1)/2 - {[n*(n + 1)/2] + nx} = 1615
=> 3n^2 + n - 2nx = 3230
=> when n = 34 , x = 4
=> when n = 38 , x = 15
it means , (sum of 1 to 68) - (sum of 5 to 38) = 1615
(n)(2n+1)-n/2(m+1+m+n)=1615n(2n+1-m-n/2-1/2)=1615n(3n/2-m-1/2)=1615n(3n-2m-1)=32303230=17*19*5*21615*2=x(x+1)x(x+1)=3230x is close to 59atleast n has to be 59values 85 95 170 190 323 646 1615 3230 ?
2n*(2n + 1)/2 - {[n*(n + 1)/2] + nx} = 1615 => 3n^2 + n - 2nx = 3230 => when n = 34 , x = 4 => when n = 38 , x = 15 it means , (sum of 1 to 68) - (sum of 5 to 38) = 1615 (sum of 1 to 76) - (sum of 16 to 53) = 1615 sum of all the possible values of n = 34 + 38 = 72
Q1:amit threw five standard dice simultaneously. He found that the product of the numbers on the top faces was 216. which of the following couldn't be the sum of numbers on top five faces:a: 17b: 19c: 18d: 20
1)216=2^3*3^320=9+4+3+2+1+117=3+3+3+2+2+219=6+6+2+3+1+118 not possible2)22=11*2no of 11's in 2006=182+16+1=1993)X+Y=949h(a+b)=949h(a*b)=2628h(a*b-a-b)=1679=23*73if h=23a+b is not an integerhence h=73
Q1:amit threw five standard dice simultaneously. He found that the product of the numbers on the top faces was 216. which of the following couldn't be the sum of numbers on top five faces:a: 17b: 19c: 18d: 20Q2:The number 2006! is written in base 22. How many zeroes are there at the end?a: 500b: 450c: 200d: 199Q3:X & Y are two natural no's such that X+Y=949. LCM of X & Y is 2628. What is the HCF of X & Y??a: 23b: 73c: 69d: noneAnswer along with approach is must
1) 216 = 2^3*3^3
18 -> 6+6+3+2+1 or 8+3+3+3+1 (possible)
20 -> 6+6+6+1+1 (possible)
17 -> 6+4+3+3+1 (possible)
hence 19 not possible @Subhashdec2@joyjitpal i guess u guys took 6 dices plus the product of ur numbers is also not coming out to be 216 (at places)...pls check...
amit threw five standard dice simultaneously. He found that the product of the numbers on the top faces was 216. which of the following couldn't be the sum of numbers on top five faces:
a: 17
b: 19
c: 18
d: 20
Q2:
The number 2006! is written in base 22. How many zeroes are there at the end?
a: 500
b: 450
c: 200
d: 199
Q3:
X & Y are two natural no's such that X+Y=949. LCM of X & Y is 2628. What is the HCF of X & Y??
a: 23
b: 73
c: 69
d: none
Answer along with approach is must
for first
216=6*6*6*1 , 6*6*2*3*1 , 4*3*3*6*1
so it eliminates 20,18 and 17
so only 19 left which is not possible
for 2nd
for base 10 number of 0s depend on power of 5 so in base 22 it will depend on power of 11
so power of 11 in 2006! = 199 [2006/11]+[2006/121] +[2006/1331] where [] represents greatest integer
A not-so-good clockmaker has four clocks on display in the window.
Clock #1 loses 15 minutes every hour.
Clock #2 gains 15 minutes every hour relative to Clock #1 (i.e., as Clock #1 moves from 12:00 to 1:00, Clock #2 moves from 12:00 to 1:15).
Clock #3 loses 20 minutes every hour relative to Clock #2.
Finally, Clock #4 gains 20 minutes every hour relative to Clock #3.
If the clockmaker resets all four clocks to the correct time at 12 noon, what time will Clock #4 display after 6 actual hours (when it is actually 6:00 pm that same day) ?
P.S. Spam Posts delete kar dena!
in 1 actual hour
clock 1 goes 45
clock 2 goes 75 when clock 1 goes 60 so when clock 1 goes 45 clock 2 goes 75*45/60=225/4
clock 3 goes 40 when clock 2 goes 60 so when clock 3 goes 225/4 then 40*225/4*60=75/2
clock 4 goes 80 when clock 3 goes 60 so when clock 3 goes 75/2 clock 4 goes 80*75/2*60=50
Find the area of the traingle inscribed in a circle circumscribed by a square made by joining the mid points of the adjacent sides of a square of side 'A' ?
Find the area of the traingle inscribed in a circle circumscribed by a square made by joining the mid points of the adjacent sides of a square of side 'A' ?
