@rubikmath said:shudnt it be 2*51C3 , since b or c > 50 ??
if a>50,b>50 no way c can be >50
@krum said:if a>50,b>50 no way can be >50
yup but a>50andb>50 or a>50andc>50 ryt ??
while a>50 in 50C3 times b>50
another 50C3 times c>50 (now b??
while a>50 in 50C3 times b>50
another 50C3 times c>50 (now b??
@krum said:seems like u r right, i'm confused now, but have to finish my movie
ur soln. was right .
its 3(102C3*-51C3) only .
when a>50:
b>50--> 51c3
c>50--> 51c3
when b>50:
a>50-->51c3
c>50-->51c3
when c>50:
a>50-->51c3
b>50-->51c3
so 3*2*51C3 - 3*51C3 to get all cases includin a>50 ;b>50
so on.. i'm confused too
but think this is ryt .
its 3(102C3*-51C3) only .
when a>50:
b>50--> 51c3
c>50--> 51c3
when b>50:
a>50-->51c3
c>50-->51c3
when c>50:
a>50-->51c3
b>50-->51c3
so 3*2*51C3 - 3*51C3 to get all cases includin a>50 ;b>50
so on.. i'm confused too
but think this is ryt . @rubikmath said:yup but a>50andb>50 or a>50andc>50 ryt ??while a>50 in 50C3 times b>50 another 50C3 times c>50 (now b??
a>50, b>50
a>50, c>50
b>50, c>50
there should be only 3 such cases, i should put my answer in a different way i think
a>50, c>50
b>50, c>50
there should be only 3 such cases, i should put my answer in a different way i think
@krum said:a>50, b>50a>50, c>50b>50, c>50there should be only 3 such cases, i should put my answer in a different way i think
exactly ,only 3 cases 
@insane.vodka said:Find the smallest positive integer such that when its leftmost digit is removed, the resulting number becomes 1/57th of original number.
uvxy
100v+10x+y=1/57(y+10x+100v+1000u)
56y+560x+5600v=y+10x+100v+1000u
u=56(y+10x+100v)/1000
u=7*(y+10x+100v)/125
u=7
v=1
x=2
y=5
7125
@pavimai said:the digits of a 3-digit number in Base 4 get reversed when it is converted into Base 3.how many such digits exist??
16a+4b+c=9c+3b+a
15a+b=8c
1 1 2-----1 possibility
2 2 4---2 possibility
3 3 6
4 4 8
4 possibilities
15a+b=8c
1 1 2-----1 possibility
2 2 4---2 possibility
3 3 6
4 4 8
4 possibilities
@techsurge said:As discussed above the number cannot be 3 digit1000a+100b+10c+d /57 = 100a+10b+c1000a =5600b+560c+56d250a=1400b+140c+14db=1 => 140c +14d should be multiple of 350,600,850 [LHS is multiple of 250]this can be achived by putting c=2 and d=5 lowest => a=7so 7125
sahi bhai
A grandfather bought a certain no. of oranges from a market to give them to his three grandsons. Their Grandmother placed the oranges in a big basket. After some time, the first grandson came and divided the oranges into three equal parts and found one extra. He hid one of those parts and the extra orange and left, Then the second grandson came and did the same, After dividing the oranges into three equal parts he found one extra. He hid one of those parts and the extra orange in a different place and left,. then , the third grandson came and did the same (after diving the oranges into three equal parts. He hid one of those parts and the extra orange and left). Next morning, their grandmother took the remaining oranges and divided them into three equal parts and found out that each part had Z oranges and as in previous cases there was an extra orange. She kept that orange for herself and gave one part (i.e. Z oranges) each to her three grandsons. Find the minimum possible value of Z?
@gupanki2 said:A grandfather bought a certain no. of oranges from a market to give them to his three grandsons. Their Grandmother placed the oranges in a big basket. After some time, the first grandson came and divided the oranges into three equal parts and found one extra. He hid one of those parts and the extra orange and left, Then the second grandson came and did the same, After dividing the oranges into three equal parts he found one extra. He hid one of those parts and the extra orange in a different place and left,. then , the third grandson came and did the same (after diving the oranges into three equal parts. He hid one of those parts and the extra orange and left). Next morning, their grandmother took the remaining oranges and divided them into three equal parts and found out that each part had Z oranges and as in previous cases there was an extra orange. She kept that orange for herself and gave one part (i.e. Z oranges) each to her three grandsons. Find the minimum possible value of Z?
is it 7??
approach plssss
@gupanki2 said:A grandfather bought a certain no. of oranges from a market to give them to his three grandsons. Their Grandmother placed the oranges in a big basket. After some time, the first grandson came and divided the oranges into three equal parts and found one extra. He hid one of those parts and the extra orange and left, Then the second grandson came and did the same, After dividing the oranges into three equal parts he found one extra. He hid one of those parts and the extra orange in a different place and left,. then , the third grandson came and did the same (after diving the oranges into three equal parts. He hid one of those parts and the extra orange and left). Next morning, their grandmother took the remaining oranges and divided them into three equal parts and found out that each part had Z oranges and as in previous cases there was an extra orange. She kept that orange for herself and gave one part (i.e. Z oranges) each to her three grandsons. Find the minimum possible value of Z?
total=n
after 1st - (n-1)*2/3
after 2nd - [(n-1)*2/3-1]*2/3=[2/3n-5/3]*2/3=4/9n-10/9
after 3rd - [4/9n-10/9-1]*2/3=8/27n-38/27
Z=[8/27n-38/27-1]*1/3=8/81n-65/81=(8n-65)/81
n=79
=>z=7
OA?
after 1st - (n-1)*2/3
after 2nd - [(n-1)*2/3-1]*2/3=[2/3n-5/3]*2/3=4/9n-10/9
after 3rd - [4/9n-10/9-1]*2/3=8/27n-38/27
Z=[8/27n-38/27-1]*1/3=8/81n-65/81=(8n-65)/81
n=79
=>z=7
OA?
bhai ulta sove kr rha hun..
grandmothr has given Z to each of three , she has kept 1 with hersef
distribution: z...........z..........z......(1 balance)
hence after 3rd boy there were 3z+1 oranges
but these oranges were kept in the group of two:i.e.,(3z+1/2& 3z+1/2),in front of grand mother
distribution:3z+1/2..........3z+1/2.........3z+1/2........(1 balance)
hence after 2nd boy there were 3(3z+1/2) +1 oranges=9z+5/2
proceeding as above after 1st boy there were 3(9z+5/4)+1=27z+19/4
similarly ,initially there were 3(27z+19/8)+1=81z+65/8
for 81z+65/8 to be integer z(min)=7
grandmothr has given Z to each of three , she has kept 1 with hersef
distribution: z...........z..........z......(1 balance)
hence after 3rd boy there were 3z+1 oranges
but these oranges were kept in the group of two:i.e.,(3z+1/2& 3z+1/2),in front of grand mother
distribution:3z+1/2..........3z+1/2.........3z+1/2........(1 balance)
hence after 2nd boy there were 3(3z+1/2) +1 oranges=9z+5/2
proceeding as above after 1st boy there were 3(9z+5/4)+1=27z+19/4
similarly ,initially there were 3(27z+19/8)+1=81z+65/8
for 81z+65/8 to be integer z(min)=7
@gupanki2 said:@19rsb Itneyy badeyyyy question ka itna chota answer...BUT YES !! Perfect approach plssss
There were 4 parcels all of whose weights were integers (in kg). The weights of all the possible pairs of parcels were noted down and amongst these the distinct values observed were 94 kg, 97 kg, 101 kg and 104 kg. Which of the following can be the weight of one of the parcels?
(a) 40 kg (b) 45 kg (c) 48 kg (d) 53 kg
(a) 40 kg (b) 45 kg (c) 48 kg (d) 53 kg
@rkshtsurana said:sum of angles = 1080avg = 135so angles be135 - 7x , 135 - 5x, 135 - 3x , 135 - x, 135 + x , 135 + 3x, 135 + 5x , 135 + 7x135 + 7x 7x xx = 6two grtest angles are - 165 and 177so 342
surana bhai,
bhai ek doubt tha..
i approached the qns like this
4(2a+7d)=1080
that gave me d=2
and subsequently, a=128
therefore the greatest angle will be a+7d=142 (which is
so, we require (a+6d)+(a+7d)= 2a+13d=282..
where am i going wrong?? 
@rkshtsurana said:hga ..iske total 6 solution ayenge..when x = 1 , 2, 3 ,4 5, 6but we want max..or acc to options...i calc max sum..@maddy2807 ques me it shud be max sum..otherwise 6 diff APs are possible..
@grkkrg said:There were 4 parcels all of whose weights were integers (in kg). The weights of all the possible pairsof parcels were noted down and amongst these the distinct values observed were 94 kg, 97 kg,101 kg and 104 kg. Which of the following can be the weight of one of the parcels?(a) 40 kg (b) 45 kg (c) 48 kg (d) 53 kg
45??
@grkkrg said:There were 4 parcels all of whose weights were integers (in kg). The weights of all the possible pairsof parcels were noted down and amongst these the distinct values observed were 94 kg, 97 kg,101 kg and 104 kg. Which of the following can be the weight of one of the parcels?(a) 40 kg (b) 45 kg (c) 48 kg (d) 53 kg
45 kg??
@Zedai said:thanks bhai
Just liking a post is enough 😃 writing this creates unnecessary posts that make thread large
Raj invests Rs.2400 partly in 3% stock at 75 and partly in 4% stock at 96. If the total
income from both is Rs.97.50, find the sum invested in each.
(a) Rs.1500 in 3%, Rs.900 in 4%
(b) Rs.900 in 3%, Rs.1500 in 4%
(c) Rs.1200 in each
(d) Rs.2000 in 3%, Rs.400 in 4%
income from both is Rs.97.50, find the sum invested in each.
(a) Rs.1500 in 3%, Rs.900 in 4%
(b) Rs.900 in 3%, Rs.1500 in 4%
(c) Rs.1200 in each
(d) Rs.2000 in 3%, Rs.400 in 4%