@Pradeep_Rss said:No of solutions for x+y+z = 60 if xyz are (i)natural numbers and (ii)whole numbers .post the approach in solving these kind of questions ...
whole no. solutions = 62c2..
natural no. 59c2..
natural no. 59c2..
@Pradeep_Rss said:No of solutions for x+y+z = 60 if xyz are (i)natural numbers and (ii)whole numbers .post the approach in solving these kind of questions ...
For +ive solutions , 59C2
For non-negative solutions, 62C2
Refer point 23 here http://quantexpert.co.in/qenotes/quantnotes/modern-maths/65-permutation-and-combination.html
For non-negative solutions, 62C2
Refer point 23 here http://quantexpert.co.in/qenotes/quantnotes/modern-maths/65-permutation-and-combination.html
Divide 3903 b/w A and B such that A's share at the end of 7 yrs is equal to B's share at the end of 9 yrs at 4%p.a rate of compound interest
A-2008 B-1000
A-2902 B-1001
A-2028 B-1875
A-2600 B-1303
@vbhvgupta said:Divide 3903 b/w A and B such that A's share at the end of 7 yrs is equal to B's share at the end of 9 yrs at 4%p.a rate of compound interestA-2008 B-1000A-2902 B-1001A-2028 B-1875A-2600 B-1303
Let the parts be a and 3903-a
Now, a(1+4/100)^7 = (3903-a)(1+4/100)^9
=> a / (3903-a) = 26^2 / 25^2
=> a / 3903 = 26^2 / (25^2+26^2)
=> a / 3903 = 26^2 / 1301
=> a / 3 = 26^2
=> a = 2028
So (c) 2028 and 1875
Now, a(1+4/100)^7 = (3903-a)(1+4/100)^9
=> a / (3903-a) = 26^2 / 25^2
=> a / 3903 = 26^2 / (25^2+26^2)
=> a / 3903 = 26^2 / 1301
=> a / 3 = 26^2
=> a = 2028
So (c) 2028 and 1875
A shopkeeper increases the price of an article by x % and then decreases it by x % as a result the price of article reduced by rs 180 after one more such change the price further reduced by 153 rs .find the original price of article.
@vbhvgupta said:Divide 3903 b/w A and B such that A's share at the end of 7 yrs is equal to B's share at the end of 9 yrs at 4%p.a rate of compound interestA-2008 B-1000A-2902 B-1001A-2028 B-1875A-2600 B-1303
x(1.04)^7= (3903-x)1.04^9;
x=(3903-x)*1.04^2
x=1875 aprox..
3rd option..
x=(3903-x)*1.04^2
x=1875 aprox..
3rd option..
@IIM-A2013 said:@dragster approach
It's recursive , form an equation and solve it;
or just take 6-7 terms and approximate , you will get the same.
or just take 6-7 terms and approximate , you will get the same.
@Pradeep_Rss said:A shopkeeper increases the price of an article by x % and then decreases it by x % as a result the price of article reduced by rs 180 after one more such change the price further reduced by 153 rs .find the original price of article.
Let the cost price be CP
We have successive changes of +x% and then -x%
So effective change = - x^2/100 %
So CP decreses by x^2/100
i.e. (x^2/100)CP = 180--(i)
So Cost price now is (1-x^2/100)CP
Now it is done once again
So x^2/100((1-x^2/100)CP = 153
Now, Using (i) we get
(1-x^2/100)*180 = 153
=> x^2 = 15
From (i) we get CP=1200
We have successive changes of +x% and then -x%
So effective change = - x^2/100 %
So CP decreses by x^2/100
i.e. (x^2/100)CP = 180--(i)
So Cost price now is (1-x^2/100)CP
Now it is done once again
So x^2/100((1-x^2/100)CP = 153
Now, Using (i) we get
(1-x^2/100)*180 = 153
=> x^2 = 15
From (i) we get CP=1200
@vishcat said:@naveenkrs 17
according to you one lap is just going from one end to the other? I had taken 1 lap to mean come to the initial point.
@Pradeep_Rss said:A shopkeeper increases the price of an article by x % and then decreases it by x % as a result the price of article reduced by rs 180 after one more such change the price further reduced by 153 rs .find the original price of article.
m getting 1200.... (edited)
1st iteration difference
A - A ( 1 - x^2/100^2) = 180
Ax^2/100^2 = 180
2nd iteration
A(1 - x^2/100^2) - A(1-x^2/100^2)^2 = 153
A(1 - x^2/100^2)x^2/100^2 = 153
let x^2/100^2 = y
so 1/1-y = 180/153
y = 27/180
x^2/100 = 27/180
so A = 180*180/27 = 1200
@catahead said:Let the cost price be CPWe have successive changes of +x% and then -x% So effective change = - x^2/100 %So CP decreses by x^2/100i.e. (x^2/100)CP = 180--(i)So Cost price now is (1-x^2/100)CPNow it is done once againSo x^2/100((1-x^2/100)CP = 153Now, Using (i) we get (1-x^2/100)*180 = 153=> x^2 = 15From (i) we get CP=1200
1st iteration
C.P(1 + x/100)(1 - x/100) = CP (1 - x^2/100^2)
so (x^2/100^2)C.P = 180
am i wrong?
If the diff between the CI and SI on a certain sum of money for 3 years at 2% p.a is Rs 604, what is the sum?
@vbhvgupta said:If the diff between the CI and SI on a certain sum of money for 3 years at 2% p.a is Rs 604, what is the sum?
75500000?
formula lagaya: diffrnce= p (r/100)^n
fir calculator use kiya...
@vbhvgupta said:no, it is 5,00,000. I applied the formula but not getting it.
mera ans formula mein dalke dekhlo.....
ans to aa raha hai...