doubts on -> Ratio, proportion and variation
q1> A few factory employs skilled workers,unskilled workers and clerks in the proportion 8:5:1 and the wages of a skilled worker,an unskilled worker and a clerk are in the ratio 5:2:3 .when 20 unskilled workers are employed ,the total daily wages of all amont to Rs 318 .The wages paid to each ategory of workers are
(a)rs 240,rs 60,rs 18
(b)rs 200 ,rs 90 ,rs 28
(c)rs 150 ,rs 108 ,rs 60
(d) non of these
q2>total expenses of running the hostel at harvard business school are partly fixed and partly varying linearly with the number of boarders . the average expense per boarder is $70 when there are 25 boarders and $60 when there are 50 boarders. what is the average expense per boarder when there are 100 boarders ?
(a)55
(b)56
(c)54
(d)50
q3> Mr am , the magnanimous chashier at xyz ltd. , while distributing salary ,adds whatever money is needed to make the sum a multiple of 50. he adds rs 10 and rs 40 to A's and B's salary respectively and then he realises that the salaries of A,B and C are now in the ratio 4:5:7 . The salary of C could be
(a)2300
(b)2150
(c)1800
(d)2100
plz post the method also...... 😃
LOD Q.1. 3x^2 -7x+4 getting ans which is none of the option
LOD Q.12 x^2-3x+ x-2 getting (1+ 3 , 2+2) as ans which is none of the option.
LOD Q.13 x^2-7x+12
(a) x4 (c)2
Sol.
x^2-7x+12
x^2-7x+12
Taking intersection of the interval , I got none of the options are tru.
What say guys ?? Am i goin wrong ????
Inequalities
LOD Q.1. 3x^2 -7x+4 getting ans which is none of the option
LOD Q.12 x^2-3x+ x-2 getting (1+ 3 , 2+2) as ans which is none of the option.
LOD Q.13 x^2-7x+12
(a) x4 (c)2frm the Q. we get 2 Inequalities (IE)
Sol.
x^2-7x+12
x^2-7x+12
Taking intersection of the interval , I got none of the options are tru.
What say guys ?? Am i goin wrong ????
Q1. You are correct
Q2. Answer is option (c) (1- 3 , 2-2)
Q3. Answer is again option (c)
When you solve two cases, then finally you take the union of two solutions as final solution not the intersection. Put in some values and try out, you will understand.
doubts on -> Ratio, proportion and variation
q1> A few factory employs skilled workers,unskilled workers and clerks in the proportion 8:5:1 and the wages of a skilled worker,an unskilled worker and a clerk are in the ratio 5:2:3 .when 20 unskilled workers are employed ,the total daily wages of all amont to Rs 318 .The wages paid to each ategory of workers are
(a)rs 240,rs 60,rs 18
(b)rs 200 ,rs 90 ,rs 28
(c)rs 150 ,rs 108 ,rs 60
(d) non of these
q2>total expenses of running the hostel at harvard business school are partly fixed and partly varying linearly with the number of boarders . the average expense per boarder is $70 when there are 25 boarders and $60 when there are 50 boarders. what is the average expense per boarder when there are 100 boarders ?
(a)55
(b)56
(c)54
(d)50
q3> Mr am , the magnanimous chashier at xyz ltd. , while distributing salary ,adds whatever money is needed to make the sum a multiple of 50. he adds rs 10 and rs 40 to A's and B's salary respectively and then he realises that the salaries of A,B and C are now in the ratio 4:5:7 . The salary of C could be
(a)2300
(b)2150
(c)1800
(d)2100
plz post the method also...... :)
soln. 1.
unskilled worker=20
hence, skilled=32, clerk=4(ratio 8:5:1)
let the wages are 5x,2x & 3x.
32*5x+20*2x+4*3x=318
solving x=1.5
total wages to skilled worker=32*5*1.5=240 hence (a)
hence 60 for unskilled, clekr=18
2)Total expenses T=ax+b where x= no. of boarders
70*25=a*25+b
60*50=a*50+b
on solving a=50, b=500
when x= 100
ax+b=5000+500=5500
average=55 (a)
3) The salary C should be multiple of 7 hence 2100.
Q1. You are correct
Q2. Answer is option (c) (1- 3 , 2-2)
Q3. Answer is again option (c)
When you solve two cases, then finally you take the union of two solutions as final solution not the intersection. Put in some values and try out, you will understand.
Sorry,for Q2 option (c) is tru. I had done a silly mistake.
For,Q3 I don't understand why I sud take union. But putting some values as u said I got (c) as tru. But it is not clear Why I sud take union.
Fro ex. x^2 -2x
Solving got 2 inequalities so 2 intervals
x1....1
0
Taking intersection 1
Plz explain yaar !!!
why we sud do Union for particular this Q !!!!
mejogi
LOD I Q 17 x-2
(a) x> (4-2)/2 (b) x3)/2 (c) Both a and b (d) None of these
LOD I Q 29 (x^2-2x-3)
(a) (-1-5
I have got x2)/2 and x>= (5+3)/2 of the 1st Q, which is none of the option.
and for the 2nd Q.(Which is printed wrong in the book, I have written without doing any change) have got 1-5
Am i wrong ???
If so then plz post the solution !!!!
regards
mejogi
Plz solve the above Q. Mathematically without directly putting the option and checking.
Hi please help me out with this percentage proble
1>Price of sugar reduced by 25% but inspite of the decrease Ayush ends up increasing his expenditure on sugar by 25% whats the percentage change in his monthly consumption on sugar?
a)+60%
b)-10%
c)33.33%
d)50%
e)none.
Ans is option a.
Hi please help me out with this percentage proble
1>Price of sugar reduced by 25% but inspite of the decrease Ayush ends up increasing his expenditure on sugar by 25% whats the percentage change in his monthly consumption on sugar?
a)+60%
b)-10%
c)33.33%
d)50%
e)none.
Ans is option a.
answer should be e 66.66%
New cosumption=1.25/0.75 of old consumption
Hi please help me out with this percentage proble
1>Price of sugar reduced by 25% but inspite of the decrease Ayush ends up increasing his expenditure on sugar by 25% whats the percentage change in his monthly consumption on sugar?
a)+60%
b)-10%
c)33.33%
d)50%
e)none.
Ans is option a.
Ans is : 66.66%(125/75)
Hi please help me out with this percentage proble
1>Price of sugar reduced by 25% but inspite of the decrease Ayush ends up increasing his expenditure on sugar by 25% whats the percentage change in his monthly consumption on sugar?
a)+60%
b)-10%
c)33.33%
d)50%
e)none.
Ans is option a.
current expenditure = (1.25/.75) * (previous expenditure) = 1.667 times the previous expenditure => 66.7% increase
Q.How many squares are possible if two of the vertices of a quadrilateral are (1,0) and (2,0) ?
a.1 b.2 c.3 d.4
I think 2 if we consider as 2D figure , bcoz can't consider 3 D as coordinates given refers to 2D.
What say ?
Have any one done the Q of Inequalities Q.17 and Q.29 Mathematically.I have posted to the following link. http://www.pagalguy.com/discussions/quant-by-arun-sharma-25023813 I tried to solve all the Question Mathematically but after Q. 28 gave up and started solving through the option, which was the easiest way.And solved few of Q Mathematically...Now at LOD Q 60 R u guys doing the same way or solving each and every Q mathematically, without going through the option ?????
Q.How many squares are possible if two of the vertices of a quadrilateral are (1,0) and (2,0) ?
a.1 b.2 c.3 d.4
I think 2 if we consider as 2D figure , bcoz can't consider 3 D as coordinates given refers to 2D.
What say ?
its should be 3 (two considering two points as adjacent vertices & 3 rd one considering them opposite)
Hello All,
I am facing a lot of problems in the Inequalities chapter. What I get as answer by solving the questions, is completely opposite of the correct answer (although because of this I am able to guess the correct answer always :grin:).
I think that I am doing something wrong somewhere fundamentally... I searched the threads but couldn't find a correct and comprehensive solution... so putting one example of how I solved it. Please tell me my mistake...
LOD 1 :- Q 13 :- x^2-7x+12
Ans a) x4 c) 2
Solution: 1st : ------> x^2-7x+12 ------> x^2-8x+16 ------> (x-4)^2 ------> x-4 0
------> x4
2nd :------> x^2-7x+12 > -(x-4)
------> x^2-7x+12 > -x+4
------> x^2-6x+08 > 0
------> (x-4) (x-2) > 0
------> (x-4) > 0 and (x-2) > 0
or
(x-4) ------> x > 4 and x > 2 which gives x>4
or
x which gives x
Now please tell me what I did wrong in all these steps... I am not able to reach the answer... although I guessed it correctly that answer would be (c) 
LOD 1 :- Q 13 :- x^2-7x+12
Ans a) x4 c) 2
Solution: 1st : ------> x^2-7x+12 ------> x^2-8x+16 ------> (x-4)^2 ------> x-4 0
------> x4
2nd :------> x^2-7x+12 > -(x-4)
------> x^2-7x+12 > -x+4
------> x^2-6x+08 > 0
------> (x-4) (x-2) > 0
------> (x-4) > 0 and (x-2) > 0
or
(x-4) ------> x > 4 and x > 2 which gives x>4
or
x which gives x
x^2-7x+12
= (x-4)(x-3)
Case 1: (x-4) > 0 is same as (x-4)
hence comparison operator doesnt change
(x-4)(x-3) x-3 x
Case 2: (x-4) hance comparison operator changes
(x-4)(x-3) > -(x-4)
x-3 > -1
x > 2
hence (c) 2>x>4
x^2-7x+12
= (x-4)(x-3)
Case 1: (x-4) > 0; hence x-4| is same as (x-4)
hence comparison operator doesnt change
(x-4)(x-3) x-3 x
Case 2: (x-4) hance comparison operator changes
(x-4)(x-3) > -(x-4)
x-3 > -1
x > 2
hence (c) 2>x>4
Was there something wrong in my method? The reasons I am asking this are:
1. I also used the same condition operators as you have used. I just didn't cut the common part from both sides in both cases.
2. The reason I didn't cut the common part is that in some questions, (Solved examples especially) I noticed that you dont get the correct answer if you cut the common terms on both sides.
So how to find out when to cancel the common terms and when not to do that???
??:
You can ALWAYS cut out common sections.... just remember to consider two cases, and REVERSE the comparison operator in the appropriate one...
2. Let me clarify a concept
Assume
a this means two cases
Case1
a 0
Case 2
a > b if (a-b)>0
*what you did wrong was not change the direction of the comparison operator in case 2 and thats why your answer came up wrong...
gaurishankar Saysits should be 3 (two considering two points as adjacent vertices & 3 rd one considering them opposite)
I didn't understand ur explanation. Better write the co ordinates of the 3 squares considering a 2 dimensional figure.
regards
mejogi
Hello All,
I am facing a lot of problems in the Inequalities chapter. What I get as answer by solving the questions, is completely opposite of the correct answer (although because of this I am able to guess the correct answer always :grin:).
I think that I am doing something wrong somewhere fundamentally... I searched the threads but couldn't find a correct and comprehensive solution... so putting one example of how I solved it. Please tell me my mistake...
LOD 1 :- Q 13 :- x^2-7x+12
Ans a) x4 c) 2
Solution: 1st : ------> x^2-7x+12 ------> x^2-8x+16 ------> (x-4)^2 .....It is not possible for any real value of x.So no solution
------> x-4 0
------> x4
or
2nd............
For the 2nd case it is explained by Nikolai seems right.Then We have to take union of 2 cases, as or is used in between case 1 and case 2.
so nothing union 2
What u say ?
mejogi