Hey hi...:)
Please can you solve these problems for me...i am confused on how to solve these 3 problems....
1)Hursh wanted to subtract 5 from a number. Unfortunately, he added 5 instead of subtracting.Find the percentage change in the result.
(a) 300% (b)66.66% (c)50% (d)33.33%
2)Hans and Bhaskar have salaries that jointly amount to Rs.10000 per month.They spend the same amount monthly and it is found that the ratio of their savings is 6:1. Which of the following can be Hans's salary?
1)Rs.6000 2)Rs.5000 3)Rs.4000 4)Rs.3000
3)The population of a village is 5500. If the number of males increases by 11% and the number of females increases by 20%, then the population becomes 6330. Find the population of females in the town.
1) 2500 2)3000 3)2000 4)3500
4)During winters, an athelete can run 'x' metres on one bottle of Glucose. But in a summer, he can only run 0.5x metres on one bottle of Glucose. How many bottles of Glucose are required to run 400 meters during summer?
1)800/x 2)890/x 3)96 4)454/x
Thank you!:-o
Hey hi...:)
Please can you solve these problems for me...i am confused on how to solve these 3 problems....
3)The population of a village is 5500. If the number of males increases by 11% and the number of females increases by 20%, then the population becomes 6330. Find the population of females in the town.
1) 2500 2)3000 3)2000 4)3500
Thank you!:-o
The answer for this is A. 2500
Thanks
Ashish
Hey hi...:)
2)Hans and Bhaskar have salaries that jointly amount to Rs.10000 per month.They spend the same amount monthly and it is found that the ratio of their savings is 6:1. Which of the following can be Hans's salary?
1)Rs.6000 2)Rs.5000 3)Rs.4000 4)Rs.3000
Thank you!:-o
let hans salary is x.
harish salary is y.
x+y=10000
1) 6000=x
4000=y
x-r/y-r=6/1
6000-r=24000-6r
5r=18000
r=3600
2) x=5000 y=5000
5000-r=30000-6r
5r=25000
r=5000
here no saving for both.
3) x=4000 y=6000
4000-r=24000-6r
5r=20000
r= 4000
so no saving for hans.
4) x= 3000, y=7000
3000-r=28000-6r
5r=25000
r=5000
not possible
So only 1st choice is correct.
Thanks
Ashish
hi
please can someone solve this ques?
It is ques 97-99(LOD 2) chapter1 of quant by arun sharma.
hi
please can someone solve this ques?
It is ques 97-99(LOD 2) chapter1 of quant by arun sharma.
Hi
Can you Please Post the question as it might happen we all are having different edition of the book?
It would be great if you post the questions..
Thanks
Ashish
In a game, the 1st person tosses a coin. If the outcome is head, he throws a die and the result is recorded as his score and he gets one more chance to toss the coin. If the outcome is tail then the next person gets a chace to toss the coin. What is theprobability that a person scores 35 points after 6 tosses in this game?
1) (1/32)(1/6)^5
2) (1/32)(1/6)^6
3) (1/64) (1/6)^6
4) (1/64) (1/6)^5
only 666665 is feasible in the above secenario...so answer is 4
((1/2)*(1/6))^6 * 6!/5!
A mock test is taken at AMS System.The test paper consists if ques in three levels of difficulty-LOD1,LOD2,LOD3.
The following table gives the details of the positive and negative marks attached to each ques type.
LOD1- POSITIVE MARKS 4,NEGATIVE MARKS 2
LOD2-POSITIVE MARKS 3,NEGATIVE MARKS 1.5
OLD 3-POSITIVE MARKS 2,NEGATIVE MARKS 1
The test had 200 ques with 80 on LOD 1 and 60each on LOD2 & 3.
Q1)if a student has solved 100 ques exactly and scored 120 marks, the maximum number of incorrect questions he might have marked
a)44 b)56
c)60 d)none of these
Q2)if amit attempted the least number of questions and got a total of 130 marks&he; has attempted at lest one of each type,then the no of ques he must have attempted
a)34 b)35
c)36 d)none of these
Q3)in the above ques what is the least number of ques he might have got incorrect?
a)0 b)1
c)2 d)none of these
My answers and explanation
1) b
Here we have to maximize the no. of incorrect attempts, thereby minimising the no. of correct attempts
so allot the least no. of negative marks for each question i.e 1 and the highest marks for a correct answer i.e 4.
44x4=176-(56x1)=120.
2) b
Finding the maximum multiple of 4( going by the above logic)
124- 31 q
1 each of LOD 2,3- 33 q
LOD 3 1 correct 1 wrong
So, 35 q
3) b
1 wrong
pls solve my this problem:
Mr X wants to establish his own unit.For this he needs an instant loan of 500,000 and every five years hence he requires an additional loan of Rs 100,000.If he had to clear all his outstandings in 20 years and he repays the principal of the first loan equally over the 20 years , find what amount he would have to pay as interest on his initial borrowing if the rate of interest is 10% p.a. Simple Interest .
a. Rs 560,000 b.Rs 540,000 c.Rs 525,000 d.Rs 500,000
2222's power 7777 + 7777's 2222
is divisible by
?
2222's power 7777 + 7777's 2222
is divisible by
?
is it 11........
options are
1)999
2)99
3)88
4)77
5)44
My take
2) 99
My take
2) 99
it's ryt. bt cn u xplain it 4 me?
plz rahulmn.............
2222^7777 + 7777^2222
=(2222^7)^1111 + (7777^2)^1111
The number has to be divisible by 2222^7 + 7777^2( a^n+b^n is divisible by a+b if n is odd)
2222^7 + 7777^2
=1111^2( 2^7*1111^5 + 7^2 )
If we check for 88 and get it right then 44 should also be right, hence 88 cannot be the answer.
We are left with 44, 77 and 99. .
11 divides 1111^2.
So the remaining term should be divisible by either 7,4 or 9
2^3 = -1(mod 9) => 2^7 = 2(mod 9)
1111 = 4(mod 9) => 1111^5 = 4^5(mod 9) = 2^10(mod 9) = -1*2
= -2(mod 9)
2^7*1111^5 = -4(mod 9) and 7^2 = +4( mod 9 )
9 divides ( 2^7*1111^5 + 7^2 )
Hence,99 divides the given number
Let me know if you have any problem
2222^7777 + 7777^2222
=(2222^7)^1111 + (7777^2)^1111
The number has to be divisible by 2222^7 + 7777^2( a^n+b^n is divisible by a+b if n is odd)
2222^7 + 7777^2
=1111^2( 2^7*1111^5 + 7^2 )
If we check for 88 and get it right then 44 should also be right, hence 88 cannot be the answer.
We are left with 44, 77 and 99. .
11 divides 1111^2.
So the remaining term should be divisible by either 7,4 or 9
2^3 = -1(mod 9) => 2^7 = 2(mod 9)
1111 = 4(mod 9) => 1111^5 = 4^5(mod 9) = 2^10(mod 9) = -1*2
= -2(mod 9)
2^7*1111^5 = -4(mod 9) and 7^2 = +4( mod 9 )
9 divides ( 2^7*1111^5 + 7^2 )
Hence,99 divides the given number
Let me know if you have any problem
did not get this step.
2222^7777 + 7777^2222
this question cumes to me ... i would first look at the option
suppose it is 22,99
now 11 is there in both ohk
when u divide 2222 by 11 remainder is zero and if u divide 7777 by 11 it is also zero so
in sum number is divisible by 11
now we need to check whether it is divisible by 2
2222 / 2= remainder 0
7777/2 remainder 1
so we need to find a number that when divided by 2 gives 0 &1 NEVER POSSIBLE
then cum to 9
2222/9 = remainder 8
7777/9= remainder 1
now equation turns into 8^7777 + 1^2222
k
1^2222 divide by 9 always leave 1 remainder
and 8^7777 by cyclicity u can find leave 8 only becoz (cyclicity is 8 &1)
so it will leave 8
and remainder sum would be 8+1 =9
that is divisible by 9 ( forexample if u need to find remainder of 40 by 9 u can write it as 36+4 ohk and 36 is divisible so remainder will be nothing but 0+4 =4)
hence number is divisible by 9
& 99
2222^7= 2222*2222*.......2222( 7 times)
= (1111*2)*(1111*2).......(1111*2)( 7 times)
7777^2= 7777*7777
= (1111*7)*(1111*7)
As you can see 1111*2 is common in both eqns
2222^7 + 7777^2
= 1111^2(1111^5*2^7 + 7^2)
Hope it is clear 
Find the remainder when 37^(1257) *28^3512 is divided by 5.
2
6
8
4