1. If 2
2. In a famous Bel Air Apartments in Ranchi, there are three watchmen meant to protect the precious fruits in the campus. However, one day a theif got in without being noticed and stole some precious mangoes. On the way out however, he was confronted by the three watchmen, the first two of whom asked him to part with 1/3rd of the fruits and one more. The last asked him to part with 1/5th of the mangoes and 4 more. As a result he had no mangoes left. What was the number of mangoes he has stolen?
a)12 b)13 c)15 d)None of these.
1. If 2a)6 b)7 c)8 d)4 e)None of these.
upper limit of x+y tends to 7 and lower limit of x-y tends to -1
ratio is -7..
1000 has 16 factors ( 1000= 2^3*5^3)..but product of consecutive 3 integers or more is a multiple of 3...and 1000 is not a multiple of 3...so how can we have 1000 as product of consecutive integers..
or am i missing a trick here..
I think there must have been a printing error in the question in the book,anyways thanks for the explanation
2. In a famous Bel Air Apartments in Ranchi, there are three watchmen meant to protect the precious fruits in the campus. However, one day a theif got in without being noticed and stole some precious mangoes. On the way out however, he was confronted by the three watchmen, the first two of whom asked him to part with 1/3rd of the fruits and one more. The last asked him to part with 1/5th of the mangoes and 4 more. As a result he had no mangoes left. What was the number of mangoes he has stolen?
a)12 b)13 c)15 d)None of these.
From the question we can conclude that initial no. of mangoes must be a multiple of 3,hence we can rule out option b,13.
If we look at first option after giving mangoes to 1st watchman,the thief would be left with 12 -(12/3 + 1) = 7 mangoes,hence ruled out.
Now if he had 15 mangoes,after giving to first watchman 15-(15/3 +1) =9 mangoes remain,so after giving to 2nd watchman,9 - (9/3 +1) =5 mangoes remain.Now if he gives 5/5 + 4 mangoes,no mangoes are left
Hence option c =15 is the ans
Hi puys ,
Another problem from number systems
Find the remainder when 2^2 + 22^2 +222^2 +.....+222....49twos^2 is divided by 9
a.2 b.5 c.6 d.7
Hi puys ,
Another problem from number systems
Find the remainder when 2^2 + 22^2 +222^2 +.....+222....49twos^2 is divided by 9
a.2 b.5 c.6 d.7
is d ans 6??...approach not worth mentioning ...looking for shorter method..
Hi puys ,
Another problem from number systems
Find the remainder when 2^2 + 22^2 +222^2 +.....+222....49twos^2 is divided by 9
a.2 b.5 c.6 d.7
Ans is c.6 .
Solution
--------
(2^2+22^2+222^2.................+2222....49timestwo^2)
First divide by 9 and put the remainders in place
ex: 2/9 = 2 , 22/9 = 4 and so on
thus (2^2 + 4^2 + 8^2 + ...............98^2)
2^2( 1+2^2+3^2...........49^2) = 4 * (49*50*99/6) Mod 9 Comes out to be 6
:)
Hi puys ,
Another problem from number systems
Find the remainder when 2^2 + 22^2 +222^2 +.....+222....49twos^2 is divided by 9
a.2 b.5 c.6 d.7
2%9 = 2
22 % 9 = 4
222 % 9 = 6
:
:
:222..... 9 times % 9 = 0
The first 9 nos from 2 to 222.....9 times will have remainder from 1,2,3....8. After 222.....9 digits the remainder will start repeating. i.e.-
2 and 222... 10 times(9n+1) have same remainder 2
2 and 222.... 11 times(9n+2) have same remainder 4
2 and 222.... 12 times(9n+3) have same remainder 6
2 and 222.... 13 times(9n+4) have same remainder 8
Hence, using binomial theorem-
2^2 + 22^2 +222^2 +.....+222....49twos^2 % 9 = % 9 = (1020 +120)%9 = 6
(6) is the answer!!!
Cheers!!!
Cheers!!!
please let me know how to solve the questions of LOD 2, number systems question no. 60 onwards.
please help me with as many questions possible..........most of the questions are of similar type so let me know the concept of how to solve them..........
thnxx in advance.............
please let me know how to solve the questions of LOD 2, number systems question no. 60 onwards.
please help me with as many questions possible..........most of the questions are of similar type so let me know the concept of how to solve them..........
thnxx in advance.............
dear post the Qns
they will all be solved........
How many 3-digit even numbers can you form such that if one of the digits is 5, the following digit must be 7?
a) 5 b) 405 c)365 d)495
a) 5 b) 405 c)365 d)495
hey 
plz help me .....
Assume that the trains between delhi and karnal leave each city every hour on the hour. It takes exactly six hours for the trains to travel from one city to another. On its run from delhi to karnal , a train will meet N trains going in the opposite direction. What is the value of N?
a) 13
b) 6
c) 11
d) 7
plz help me .....Assume that the trains between delhi and karnal leave each city every hour on the hour. It takes exactly six hours for the trains to travel from one city to another. On its run from delhi to karnal , a train will meet N trains going in the opposite direction. What is the value of N?
a) 13
b) 6
c) 11
d) 7
plz help me wid these questions.......:nervous:
Q1) a train travels 8 km in 1st quarter of an hour, 6 km in second quarter and 40 km in 3rd quarter. find the avg speed of train per hr over the entire journey.
ans- 72km/hr
Q2)the avg age of 8 person is inc. by 2 yrs when 2 men aged 35 & 45 yrs are replaced by 2 women.find avg age of 2 women
ans - 48
Q3)in a hotel where rooms are numbered from 101 to 130 each room give an earning of 3000 for 1st 15 days and 2000 for the later half,per room. find the avg earning per room per day(assume 30 days per month)
ans - 2500
thnxx in advance..............
@nivedita
1)
8km in 1/4 hr
6km in 1/4 hr
40km in 1/4 hr
avg speed= total distance/total time
so; (8+6+40)/(3*1/4)= 72kmph
2)
let average = x
so; total age =8x
after 2 men have left... total age=8x-(35+45)
let the age of the 2 women be y
so; new total age = 8x-(35+45)+y
now; as per question new average is = x+2
therefore; /8 = x+2
which yields y =96
now y =total age of 2 women
so average = 96/2= 48
3)
for one room total for a month: 15*3000+15*2000
average=/30= 2500
1) the weight of the body as calculated by avg. of 7 diff. experiment is 53.735 gm.The avg of 1st 3 exp is 54.005gm, of the 4th exp. is .oo4 gm greater than 5th, while avg of 6th n 7th was .010 gm less than avg of 1st 3. Find the wt. of the body obtd. by 4th exp.
hint: take 53 as the base to reduce ur calculations.
ans : 53.072
2) one collective farm got an avg harvest of 21 tonnes of wheat & another collective farm had 12 acres of land less given to wheat, got 25 tonnes from a hectare. As a result, the second farm harvested 300 tonnes of wheat more than the first. How many tonnes of wheat did each farm harvest?
ans : 3150, 3450
thnxx in advance..............:wink:
plz help me wid these questions.......:nervous:
Q1) a train travels 8 km in 1st quarter of an hour, 6 km in second quarter and 40 km in 3rd quarter. find the avg speed of train per hr over the entire journey.
ans- 72km/hr
Q2)the avg age of 8 person is inc. by 2 yrs when 2 men aged 35 & 45 yrs are replaced by 2 women.find avg age of 2 women
ans - 48
Q3)in a hotel where rooms are numbered from 101 to 130 each room give an earning of 3000 for 1st 15 days and 2000 for the later half,per room. find the avg earning per room per day(assume 30 days per month)
ans - 2500
thnxx in advance..............
For the second problem you can solve it by the conventional method which is already posted. Also you can treat this problem as:
The two women who are replacing the two women should have the weight such that they can make up for the weights of the two men as well as increase the avg of the group by 2. Thus the weights of the two women are
35+45+2*8=96. So the average is 96/2=48
hey
Assume that the trains between delhi and karnal leave each city every hour on the hour. It takes exactly six hours for the trains to travel from one city to another. On its run from delhi to karnal , a train will meet N trains going in the opposite direction. What is the value of N?
a) 13
b) 6
c) 11
d) 7
Is the answer 6?
hey
Assume that the trains between delhi and karnal leave each city every hour on the hour. It takes exactly six hours for the trains to travel from one city to another. On its run from delhi to karnal , a train will meet N trains going in the opposite direction. What is the value of N?
a) 13
b) 6
c) 11
d) 7
I think its 7
Logic.
Suppose a train just starts from Delhi , assume that somne train which had started 6 hours back from karnal is just reaching delhi , in that case JUST when the train leaves DELHI , it will meet the furst train , ditto it will meet another 5 trains on the way and JUST when it is reaching Karnal , it will meet the last train ==> 1+5+1 = 7...
Please let me know if its correct 😃
1) the weight of the body as calculated by avg. of 7 diff. experiment is 53.735 gm.The avg of 1st 3 exp is 54.005gm, of the 4th exp. is .oo4 gm greater than 5th, while avg of 6th n 7th was .010 gm less than avg of 1st 3. Find the wt. of the body obtd. by 4th exp.
hint: take 53 as the base to reduce ur calculations.
ans : 53.072
2) one collective farm got an avg harvest of 21 tonnes of wheat & another collective farm had 12 acres of land less given to wheat, got 25 tonnes from a hectare. As a result, the second farm harvested 300 tonnes of wheat more than the first. How many tonnes of wheat did each farm harvest?
ans : 3150, 3450
thnxx in advance..............:wink:
1) I will like to use hint for solving.
Avg of all weights = 53.735
Hence each weight is 0.735 more than 53.
Net Deviation = 7*(0.735)= 5.145
Avg of 1st 3 = 54.005
Hence net Deviation for 1st 3 from 53= (1.005)*3 = 3.015
Avg of 6 and 7th = 54.005 - 0.010 = 53.995
Hence net Deviation for 6th and 7th = 2(0.995) = 1.990
Since 4th is 0.004gm more than 5th. Hence its net deviation will be 0.004 greater than 5.
Let,
Deviation of 5th be a
So, deviation of 4th = a+0.004
Now, sum of deviation of all experiments= net deviation of the average weight
3.015 + 1.990 + a + (a+0.004) = 5.145
a= 0.068
Hence, Weight of 4th experiment = 53 + 0.068 + 0.003 = 53.072
2)
Let Area of 1st farm be x hectares.
Hence area of 2nd farm = (x-12) hectars.
Harvets from first farm = 21x tonnes
Harvest from 2nd farm= 25(x-12) tonnes
25(x-12) - 21x = 300
x= 150
Hence harvest from 1st farm= 21*150 = 3150 tonnes
harvest from second farm = 3150 +300 = 3450 tonnes