Let there be x coins of 20 paise and y coins of 50 paise
x+y=90
20x+50y=3690
Solving, you get
x=27 y=63 Option (C)
Let cost price of cow be x
So, that of calf is (1300-x)
Selling price of cow= 1.25x
Selling price of calf= 1.2(1300-x)
Total selling price was 1300+1300*300/1300= 1600
So,
1.25x-1.2x+1560=1600
0.05x=40
x=800 Option (E)
Please post your questions in a single post next time onwards.
Hi Shashank, This "cow-calf" question is from the Alligations chapter in Arun Sharma. Though the solution given here is correct, I think there might be an alternate approach for these types of questions, Some different way of approaching these kind of questions. Kindly suggest.
Thanks, Ashish PS: I am quite comfortable with the current solution, but thought it was more of a very straight and calculation intensive approach .
Points:- 1. Euler number of 5 is 4 nt 1. 2. 5a+2=199b+2^9 .....if b=0,then a=? I mean what is the common term(number) for both the equations??? 3. You have taken 2 common from both denominator n numerator initially but in the last step u have multiplied the term wid 2(numerator wala) but wat abt the 2(denominator wala) wich u also taken as common in the initial step...????
P.S. Clear the Points plz....
(1) Just a typing error.. Thanks for correction.. Euler's number for 5 is 4. I just made a typo there. Solution is correct. (2) We need a number of form 5a+2 = 199b + 2^9 => 5a +2 = 199b + 2*2^8 2^8 will leave remainder by 5 as 1. So, 2^9 will leave remainder by 5 as 2. So, 199b must be divisible by 5. Smallst value to take is b=0 (3) When we take 2 common from number, say x, it will become x/2 Its remainder comes out to be r/2. So, original remainder is (r/2)*2. So here, we just need to multiply by 2.
(1) Just a typing error.. Thanks for correction.. Euler's number for 5 is 4. I just made a typo there. Solution is correct. (2) We need a number of form 5a+2 = 199b + 2^9 => 5a +2 = 199b + 2*2^8 2^8 will leave remainder by 5 as 1. So, 2^9 will leave remainder by 5 as 2. So, 199b must be divisible by 5. Smallst value to take is b=0 (3) When we take 2 common from number, say x, it will become x/2 Its remainder comes out to be r/2. So, original remainder is (r/2)*2. So here, we just need to multiply by 2.
Ok...but what m syng is dat when ur taking 2 as common from both Numerator n Denomenator, u need to bring back both the 2s into picture in the end...But ur nly bringing the 2 taken common frm numerator in last step by multiplying the term wid 2. Where does the second 2 taken from denominator goes? Y its not brought bck in the end???
Ok...but what m syng is dat when ur taking 2 as common from both Numerator n Denomenator, u need to bring back both the 2s into picture in the end...But ur nly bringing the 2 taken common frm numerator in last step by multiplying the term wid 2. Where does the second 2 taken from denominator goes? Y its not brought bck in the end???
P.S. TIA
See.. It is like this. Number (say N) leaves remainder r when divided by n So, N = a*n + r ... where 'a' is some integer
Now, both N and n are multiples of 2 => 'r' must be multiple of 2 .... as (N - a*n) is even.
So, say 2N' = 2a*n' + 2r' .... N = 2N', n = 2n' and r = 2r' => N' = an' + r'
So, r' is remainder of N' with n'. In our case, N = 2^1990 and N' = 2^1989 n = 2*5*199 and n' = 5*199 And we get r' equal to 2^9 which is remainder of N' with n'
Now, we have to find out r. r = 2*r' = 2*2^8 = 2^9
Basically, it has mainly to do with multiplying by 2 for (r') and less to do with numerator and denominator. Hope, this is clear now 😃
See.. It is like this. Number (say N) leaves remainder r when divided by n So, N = a*n + r ... where 'a' is some integer
Now, both N and n are multiples of 2 => 'r' must be multiple of 2 .... as (N - a*n) is even.
So, say 2N' = 2a*n' + 2r' .... N = 2N', n = 2n' and r = 2r' => N' = an' + r'
So, r' is remainder of N' with n'. In our case, N = 2^1990 and N' = 2^1989 n = 2*5*199 and n' = 5*199 And we get r' equal to 2^9 which is remainder of N' with n'
Now, we have to find out r. r = 2*r' = 2*2^8 = 2^9
Basically, it has mainly to do with multiplying by 2 for (r') and less to do with numerator and denominator. Hope, this is clear now :)
Thnks bro,pta ni kaha uljh gaya tha main :-P .....Got it pretty well now....
Please help me with this questions and explaination:
Out of 1500 girls studying in a school, 20% learn only Stitching. One-tenth of the total number of girls learn only Stitching and Pottery, 12% of the total number of girls learn only Farming and 32% learn only Pottery. One-fifteenth of the total number of girls learn only Pottery and Farming and the remaining learn all the three i.e. Stitching, Pottery and Farming together. 81. How many girls learn only any one of the skills? a. 720 b. 840 c. 960 d. 880 e. None of these 82. How many girls learn only Farming? a. 180 b. 160 c. 155 d. 185 e. None of these 83. How many girls do not learn Stitching? a. 680 b. 720 c. 650 d. 760 e. None of these 84. The number of girls learning all the three i.e. Stitching, Pottery and Farming together from approximately what per cent of the total number of girls in the school? a. 11 b. 23 c. 9 d. 27 e. 19 85. What is the total number of girls who learn Pottery? a. 985 b. 1020 c. 1135 d. 998 e. None of these
Please help me with this questions and explaination:
Out of 1500 girls studying in a school, 20% learn only Stitching. One-tenth of the total number of girls learn only Stitching and Pottery, 12% of the total number of girls learn only Farming and 32% learn only Pottery. One-fifteenth of the total number of girls learn only Pottery and Farming and the remaining learn all the three i.e. Stitching, Pottery and Farming together. 81. How many girls learn only any one of the skills? a. 720 b. 840 c. 960 d. 880 e. None of these 82. How many girls learn only Farming? a. 180 b. 160 c. 155 d. 185 e. None of these 83. How many girls do not learn Stitching? a. 680 b. 720 c. 650 d. 760 e. None of these 84. The number of girls learning all the three i.e. Stitching, Pottery and Farming together from approximately what per cent of the total number of girls in the school? a. 11 b. 23 c. 9 d. 27 e. 19 85. What is the total number of girls who learn Pottery? a. 985 b. 1020 c. 1135 d. 998 e. None of these
stitching only= 20% of 1500=300 stitching and pottery=1/10th of 1500=150 farming only=12% of 1500= 180 pottery only=32% of 1500=480 pottery and farming=1/15th of 1500=100 all three= 1500-( 300+150+180+480+100)=290
Now, 81) only one of the skills= 300+180+480= 960 82)only farming= 180 83)not stitching= 180+480+100=760 84) 290 is what percentage of 1500, find out. It will be 19 approx. 85)total pottery=150+480+100+290=1020
The age of shaurya and kauravki is in the ratio 2:6.After 5 years,the ratio of their ages will become 6:8 Find the average of their ages after 10 years?
The age of shaurya and kauravki is in the ratio 2:6.After 5 years,the ratio of their ages will become 6:8 Find the average of their ages after 10 years?
let the age of Shaurya and kauravki be 2x and 6x after 5 years i.e 2x+5/6X+5 = 6/8 16x+40 = 36x+30 20x = 10 x = 1/2 so their ages are 1 and 3 after 10 years their age will be 11 and 13 so average 12
nitya2903 Says
can anyone pls explain the difference between factors and divisor???
Thanks for confirming the solution mate. It is one of the questions in the practice section of the Percentage chapter from Arun Sharma. There, he has mistakenly, given the answer as Rs5.
Also I would like to share one more method, which I though about:-
The method is based on the effect of the change in denominator on the entire fraction.
IF unaware about it::- Refer the Topic "Denominator Effect" in the Illustration section of the Percentage Chapter in Arun Sharma. May be page no. 146
So the solution goes something like this:-
The "price per kg" change, can be demonstrated as:-
400/x -----> 400/(x-20)
Now the question already states that the price has increased by 25%.
The numerator has remained constant, hence the change is only due to the denominator change.
Therefore moving from (x-20) to (x) we must witness a change of 25%
That is:- 20/(x-20) = 25/100 x=100
And hence the initial price = (400/x) = Rs 4 only.
how many terms are identical in the two of this APs 1,3,5,..upto 120 terms and 3,6,9,..upto 80 terms?
i understood the answer given in the book to some extent there is difference of 6 in the series in the first series 1,2,5.. 2 is the common difference and the last term is 239 while the second series has 3 as d common difference and the last term is 240 but from this how can we conclude that the last term of 3,9,15... series as 237 as we dont even know the number of terms?? pls explain how they conculded 237 as the last term thanks in advance..:sneaky:
The series is of the form (1/1*2 + 1/2*3 + 1/3*4 + .....+ 1/11*12 )
Now the shortcut for this is First term: {1/(positive difference between two denominators)}{1/lesser denominator-1/greater denominator} = {1/1}{1/1-1/2}
All together: = {1/1}{1/1-1/2+1/2-1/3+1/3-1/4..........1/11-1/12}
How many numbers of the form xxyy (i.e two digits out of the four distinct and each digit occurs twice) can be formed from the set of whole numbers i,e from 0,1,2,3,4,5,6,7,8,9 ?
I know that the answer is 9C2 x 4!/2!2! = 216 --> selecting the 2 out of the 9 digits for x and y (exclude 0 ) and then arranging them in the four places .
But I thought why not first get the total number of numbers which can be formed if we include zero also , and then substract the numbers which can start with zero .
So total numbers including zero are 10C2 x 4!/2!2! = 270 And the numbers which start with zero are 9 x 3!/2! = 27 The number would be of the form 0 _ _ _ - here one of the three digits in the blanks is 0 and the other two can be anyone of the remaining 9 digits . Subtracting both we get 270-27 = 243
Why the discrepency ?? Am i missing something here ?? Please help me in clearing this doubt !!!!
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Thanks and Regards,