Quant by Arun Sharma

welcome to PG priya

my take::
see u can write the general term as 1/(n*n+4) now in this case
1/4(1/n-1/n+4) now applying values we get
1/4(1-1/5+1/5-1/9.......1/224-1/225)
all the terms will cancel out except
1/4(1-1/225) so the answer is 56/225

am i correct



ya da ans s correct.. thnx shivam

1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%

a) 6kg
b) 3kg
c) 2kg
d) 5kg
e) 4kg

2) a m ixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?

a) 10gals
b) 8.5gals
c) 8gals
d) 6.66gals
e) 8.33gals

1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%

a) 6kg
b) 3kg
c) 2kg
d) 5kg
e) 4kg

2) A mixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?

a) 10gals
b) 8.5gals
c) 8gals
d) 6.66gals
e) 8.33gals

hey divya,

Q1. First of all you need to find the CP of the new mixture which can be got as Rs 4.4/1.1 = Rs 4. (10% profit). Now it becomes a ratio proportion problem where in x kg of 3.6 mix + 8 kg of 4.2 mix gives u (x+kg of
Rs 4 mix. Solve this you get x= 4. Thats your answer.

Q2. 125 gall of mix has 100 gall of wine 25 gall of water. if you add x gall of water to this. Mix becomes 125+x gall . water is 25 +x gall.

(25+x)/(125+x) * 100% =25%

Solve x n you get x=8.33galls.


Let me know if the answers are correct n you r able to follow.

1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%

a) 6kg
b) 3kg
c) 2kg
d) 5kg
e) 4kg

2) A mixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?

a) 10gals
b) 8.5gals
c) 8gals
d) 6.66gals
e) 8.33gals

By mixture and alligation the answer is 4 kg

and for the second one just check in case of option 8.33 is satisfying
1) how many kilograms of sugar worth rs.3.60/kg should b mixed with 8kg of sugar worth rs.4.20/kg such tht by selling d mixture at rs.4.40/kg there may b a gain of 10%

a) 6kg
b) 3kg
c) 2kg
d) 5kg
e) 4kg

2) A mixture of 125 gallon of wine and water contains 20% water.how much water must b added to d mixture in order to increase the percentage of water to 25% of d new mixture?

a) 10gals
b) 8.5gals
c) 8gals
d) 6.66gals
e) 8.33gals


1) use the allegation formula to get the answer i.e qty cheap / qty dear = (CP of dear - Avg CP) / (Avg CP - CP of cheap)

the ans is e) 4kg

2) (25 + x) / (125 + x) = 1 / 4, the ans is e) 8.33gals

Hi All,

I have following questions which needs your immediate attention. All these questions are from LOD2 Permutation and Combinations

1) Seven Different Objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects

a) 15120 b) 2187 c) 3003 d) 792

2) There are 8 orators A,B,C,D,E,F,G and H. In how many ways can the arrangments be made so that A always comes before B and B always comes before C

a) 8!/3! b) 8!/6! c) 5!3! d) 8!/(5!3!)

3) Find the number of ways of selecting n articles out of 3n+1 item of which n items are identical

a) 2^(2n-1) b) (3n+1)Cn / n! c) (3n+1)Pn / n! d) 2^2 n

I am looking for an explanation with formula used (Please give the generic formula)

2) There are 8 orators A,B,C,D,E,F,G and H. In how many ways can the arrangments be made so that A always comes before B and B always comes before C

a) 8!/3! b) 8!/6! c) 5!3! d) 8!/(5!3!)


answer to this one is 8!/6! consider ABC as a single object do total ways of arrangement is this
3) Find the number of ways of selecting n articles out of 3n+1 item of which n items are identical

Option a) 2^(2n-1) its a generic formula i guess
+

1) Seven Different Objects must be divided among three people. In how many ways can this be done if one or two of them can get no objects

a) 15120 b) 2187 c) 3003 d) 792

answer to this is 792

help me to understand Q no.47 of LOD III(number system). the question is very long(it's a passage type) so i m nt posting here.
actually i m not able to get any conclusion frm that passage.

which arun sharma is this i have the book but LOD 3 has only 40 questions

help me to understand Q no.47 of LOD III(number system). the question is very long(it's a passage type) so i m nt posting here.
actually i m not able to get any conclusion frm that passage.

well, i chkd the ques..vil post it in DI thread ...
HI ,
I WANT JUST ASK YOU THAT CAN YOU PROVIDE ME A SOFT COPY OF ARUN SHARMA..
OR THE URL FRM WHICH IT CAN BE DOWNLOADED....

IT WILL BE BETTER IF U FORWARD ME ON MY EMAIL ID:[email protected].
[email protected].


thank you in advance.......:sarcasm:
HI ,
I WANT JUST ASK YOU THAT CAN YOU PROVIDE ME A SOFT COPY OF ARUN SHARMA..
OR THE URL FRM WHICH IT CAN BE DOWNLOADED....

IT WILL BE BETTER IF U FORWARD ME ON MY EMAIL ID:


thank you in advance.......:sarcasm:


well, Its neither a soft copy nor can be downloaded ...

Its a Book to be buyed ..!!! so, i can't send yu by any means ...also don't post ur id's on the thread...its against the rules of PG ..!!

Thanks
:huh:ok...........fine......
any way thanks.........for u r reply..........

thank u...

can sum1 explain how dis sum can b done...
How many batting orders are possible for da indian cricket team if there is a squad of 15 to choose from such that sachin tendulkar is always chosen?
a)1001. 11! b)364. 11!
c)11! d) 15. 11!
thnx in advance....

can sum1 explain how dis sum can b done...
How many batting orders are possible for da indian cricket team if there is a squad of 15 to choose from such that sachin tendulkar is always chosen?
a)1001. 11! b)364. 11!
c)11! d) 15. 11!
thnx in advance....

out of 15 the 1 is fixed so we need to choose the 10 players out of 14 available which can be done in 14C10 ways so the answer is

14C10*11! also they can be arranged in 11! ways
Option 1
1001*11!

can someone help me with the following probs..

Q.67-Lod2-NS-> K is a 3-digit number such tat the ratio of number to the sum of its digits is least. what is the difference between hundreds and the tens digit of K.
(a) 9 (b) 8 (c) 7 (d) none of these.

Q.68 -> In question 67, what can be said about the difference between tens and units digit.
(a) 0 (b) 1 (c) 2 (d) none of these

Q69 -> For the above questions, for how many values of K wil the ratio be highest.
(a) 9 (b) 8 (c) 7 (d) none of these

Q85 -> Find last two digits of the following number
(201*202*203*204*246*247*248*249)^2
(a) 36 (b) 56 (c) 76 (d) 16

Q89 -> The super computer at Ram Mohan Roy Seminary takes an i/p of a no. N & a X where X is a factor of N. In a particular case N is equal to 83p796161q & X is equal to 11 where 0

A triangular number is defined as a number which has the property of being expressed as a sum of consecutive natural numbers starting with 1. How many triangular numbers less than 1000, have the property that they are the difference of squares of 2 consecutive natural numbers?
(a)20 (b)21 (c)22 (d)23

Q93. -> How many integer values of x & y are there such that 4x+7y=3 while
x (a)144 (b)141 (c)143 (d)142

Q95 -> How many 2-digit numbers less than or equal to 50, have product of the factorials of their digits less than or equal to sum of the factorials of their digits.
(a) 17 (b) 16 (c) 15 (d) none of these

Also, questions 97 to 100 from LOD-2(Number systems)

Q.67-Lod2-NS-> K is a 3-digit number such tat the ratio of number to the sum of its digits is least. what is the difference between hundreds and the tens digit of K.
(a) 9 (b) 8 (c) 7 (d) none of these.

is the ans 8 ??

yups the number is 199

nxt ques is also answered..which is 0