Alligations page no. 133 quest 9??? explaination plsssssss
a man purchased two qualities of pulses at the rate of rs. 220 per quintal and rs. 260 per quintal. In 52 quintals of the second quantity, how much pulse of the first quality should be mixed so that by selling the resulting mixture at rs. 300 per quintal, he gains a profit of 25%?
1) 100 quintals
2) 104
3)26 quintals
4)none of these
explanation plsss
Kushwaha Sayswat is the ans buddy is that 4%.
======================
@ Kushwaha: plz can u explain hw did u get 4%.
a man purchased two qualities of pulses at the rate of rs. 220 per quintal and rs. 260 per quintal. In 52 quintals of the second quantity, how much pulse of the first quality should be mixed so that by selling the resulting mixture at rs. 300 per quintal, he gains a profit of 25%?
1) 100 quintals
2) 104
3)26 quintals
4)none of these
explanation plsss
CP will be Rs 240
220 260
240
x 52
20/20 = 52/x
x = 52 quintals
Hey puys! I am having hardtime solving inequalities LOD1 ques from Arun sharma...
Can someone help me with this.
Chapter : Inequalities
LOD1 Q 27
sqrt >1
a) 0
P.S : I'm sure this can be solved by substituting values from options. But I would like to know the process of obtaining at an answer w/o options. Because if we have None of these as one of the answer, I would like to arrive at it rather than checking with random values. Hope you get the point :)
======================
@ Kushwaha: plz can u explain hw did u get 4%.
=====================
@khushwaha, dear alwayz mention the question by coping and pasting it on the top so that everybdy can cm to knw the question of the soln wch u provide. It will add the value to solution posted by you .
hi puys..i have some query's.. I could not solve some problems, kindly give me the solutions and explanations. the problems are from arun sharma quant book.
1. anand and kasparov plays chess, they played 5 matches. P(anand wins)=2/5, P( kaspar wins)= 3/5, P (match draw)= zero..find the probability that Anand wins?
options : 992/3125 273/625 | 1021/3125 1081/3125
2. find the probability that a year chosen at random has 53 mondays? option: 5/28....3/28.....1/28.....3/28
3. probability that four S comes sonsecutively in MISSISSIPPI? option: 4/165 .......2/165....3/165...1/165
P.S: if possible, please PM me the solutions, as I do not get time to go through every page of this thread. thanks in advance to all the seniors!
Some Questions puys.
Q1. The sum of series
1/(sqrt2+sqrt1) + 1/(sqrt2+qrt3) +.......................+ 1/(sqrt120+sqrt121)
Ans = 10, 11 , 12, None of these
Q2. Find Infinite sum of the series 1/1+1/3+1/6+1/10+1/15.............
Ans = 2, 2.25, 3, 4
Q3. How many 3 digit nos have the property that their digits taken from left to right form an AP or a GP?
Ans = 15, 36, 20, 42
Q4. An AP P consists of n erms.
From the progressions three different progressions P1, P2, & P3 re created such that P1 is obtained by the 1st,4th,7th....terms of P, P2 has the 2nd,5th,8th .......terms of P and P3 hs the 3rd,6t,9th....terms of P. It is found that of P1,P2 and P3 two progression have the propertythat their average is ielfa term of the original Progression P. Find possible values of n.
Ans = 20, 26, 36, both 1 & 2
Waiting for the Explanation puys.
The Questions is of Progression from Arun Sharma in LOD2 & LOD 3
Q1) Rationalizing the Dr. for each term in the series, we have,
1/(sqrt2+sqrt1) = sqrt(2) - sqrt(1)
1/(sqrt2+sqrt3) = sqrt(3) - sqrt(2)
and so on...
Finally, you end up with,
(sqrt2-sqrt1) + (sqrt3-sqrt2) + (sqrt4 - sqrt3) + ... + (sqrt120 - sqrt119) + (sqrt121 - sqrt120)
= sqrt121 - sqrt1
= 10
Some Questions puys.
Q1. The sum of series
1/(sqrt2+sqrt1) + 1/(sqrt2+qrt3) +.......................+ 1/(sqrt120+sqrt121)
Ans = 10, 11 , 12, None of these
Q2. Find Infinite sum of the series 1/1+1/3+1/6+1/10+1/15.............
Ans = 2, 2.25, 3, 4
Q3. How many 3 digit nos have the property that their digits taken from left to right form an AP or a GP?
Ans = 15, 36, 20, 42
Q4. An AP P consists of n erms.
From the progressions three different progressions P1, P2, & P3 re created such that P1 is obtained by the 1st,4th,7th....terms of P, P2 has the 2nd,5th,8th .......terms of P and P3 hs the 3rd,6t,9th....terms of P. It is found that of P1,P2 and P3 two progression have the propertythat their average is ielfa term of the original Progression P. Find possible values of n.
Ans = 20, 26, 36, both 1 & 2
Waiting for the Explanation puys.
The Questions is of Progression from Arun Sharma in LOD2 & LOD 3
Q1 ) If the median PT and RS of a triangle with vertices P(o,b), Q(0,0) and R(a,0) are perpendicular to each other , which of the following hold?
1) 4b^2=a^2
2) 2b^2=a
3) a=-2b
4) a^2+b^2=0
--------------------------
Q2) How many pts on x+y=4are their lie at a unit distance from the line 4x+3y=10?
1) 1 (2) 2 (3) 3 (d) none of these
hw these can be done, i mean solutions..
If the median PT and RS of a triangle with vertices P(o,b), Q(0,0) and R(a,0) are perpendicular to each other , which of the following hold?
1) 4b^2=a^2
2) 2b^2=a
3) a=-2b
4) a^2+b^2=0
solution plz?
it wil b option 4..
si da co-ordinate of T wud (a/2,0) and co-ordinate of S wud b (,b/2)..
if PT and RS r perpendicular..then their product of slopes should b equal to -1
so slope of PT = -2b/a
slope of RS= -b/2a
so -2b/a * -b/2a = -1
b^2= -a^2
so b^2+a^2=0
Hi frndz i have a question no. 40 from chapter 18 set theory.
waiting for explanation puys......................
Hi frndz i have a question no. 40 from chapter 18 set theory.
waiting for explanation puys......................
a=zee
b=sp
c=sony
d=z and son
e=z and sp
f=sony and sp
g= all three
a+b+c+d+e+f+g=78
a+d+e+g=36
b+e+f+g=48
c+d+f+g=32
e+g=14
f+g=20
d+g=12
b+f=34
b+f+c=42
c=8
d+f+g=24
d=4
a=18
b=28-e
e-d=2
e=6
a/c=18/8=9/4
Hey puys! I am having hardtime solving inequalities LOD1 ques from Arun sharma...
Can someone help me with this.
Chapter : Inequalities
LOD1 Q 27
sqrt >1
a) 0b)0.75 c)0.75 d)0 e) None of the above.
P.S : I'm sure this can be solved by substituting values from options. But I would like to know the process of obtaining at an answer w/o options. Because if we have None of these as one of the answer, I would like to arrive at it rather than checking with random values. Hope you get the point :)
if you solve the inequation entirely,you will get x>0.75
now check for consistency of the solution.
terms inside the root cant be negetive.
therefore you get xis the ans 0.75
answer is 2 for question 2.....will give explanation if rite...
Some Questions puys.
Q1. The sum of series
1/(sqrt2+sqrt1) + 1/(sqrt2+qrt3) +.......................+ 1/(sqrt120+sqrt121)
Ans = 10, 11 , 12, None of these
Q2. Find Infinite sum of the series 1/1+1/3+1/6+1/10+1/15.............
Ans = 2, 2.25, 3, 4
Q3. How many 3 digit nos have the property that their digits taken from left to right form an AP or a GP?
Ans = 15, 36, 20, 42
Q4. An AP P consists of n erms.
From the progressions three different progressions P1, P2, & P3 re created such that P1 is obtained by the 1st,4th,7th....terms of P, P2 has the 2nd,5th,8th .......terms of P and P3 hs the 3rd,6t,9th....terms of P. It is found that of P1,P2 and P3 two progression have the propertythat their average is ielfa term of the original Progression P. Find possible values of n.
Ans = 20, 26, 36, both 1 & 2
Waiting for the Explanation puys.
The Questions is of Progression from Arun Sharma in LOD2 & LOD 3
answer for question 2 is 2...will xplain if i am rite....
hi puys..i have some query's.. I could not solve some problems, kindly give me the solutions and explanations. the problems are from arun sharma quant book.
1. anand and kasparov plays chess, they played 5 matches. P(anand wins)=2/5, P( kaspar wins)= 3/5, P (match draw)= zero..find the probability that Anand wins?
options : 992/3125 273/625 | 1021/3125 1081/3125
2. find the probability that a year chosen at random has 53 mondays? option: 5/28....3/28.....1/28.....3/28
3. probability that four S comes sonsecutively in MISSISSIPPI? option: 4/165 .......2/165....3/165...1/165
P.S: if possible, please PM me the solutions, as I do not get time to go through every page of this thread. thanks in advance to all the seniors!
I am posting it again, please solve it.
thanks in advance!
Please help me... got stuck..
Q. A shrewd milkman mixes water and milk in the ratio of 2:3. What part of this mixture should be removed and replaced with water so that the solution contains water and milk in the ratio 1:1?
1)1/6
2) 1/4
3) 1/3
4) 5/6
Q. A shrewd milkman mixes water and milk in the ratio of 2:3. What part of this mixture should be removed and replaced with water so that the solution contains water and milk in the ratio 1:1?
1)1/6
2) 1/4
3) 1/3
4) 5/6
Find the max. Value of n such that
570*60*30*90*100*500*700*343*720*81
is perfectly divisible by 30^n.
A. 12
b. 11
c. 10
d. 13
Please help me... got stuck..
Q. A shrewd milkman mixes water and milk in the ratio of 2:3. What part of this mixture should be removed and replaced with water so that the solution contains water and milk in the ratio 1:1?
1)1/6
2) 1/4
3) 1/3
4) 5/6
Ans: (1)1/6
let water =20 ltr and milk =50 ltr (2:3 ratio)
let x ltr of mixture is replaced and it is replaced by x ltr of water.
(concept: ratio of water: milk in removed mixture is also 2:3 ,)
so water removed=2/5x and milk removed=3/5x .
Finally both are equal
20-2/5x+x=30-3/5x
x=50/6 hence 1/6th
Ans is (1) right ?
