Quant by Arun Sharma

devananth Says
*In an AP Sp=q, and Sq=p (Sn being the sum of the first n terms of the progression). find Sp+q.


My take is -(p+q).

Let the first term be a and the common difference be d.
=> p/2*{2a + (p-1)*d} = q.
=> 2ap + pd - pd = 2q ---- (1).
Similarly, 2aq + qd - qd = 2p --- (2).
(1) - (2).
=> 2(q - p) = 2a(p - q) + d(p - q) - d(p - q).
Dividing by (p - q) on both sides.
=> -2 = 2a + d(p + q) - d.
=> 2a + {p+q-1}*d = -2.
Multiplying by (p+q)/2 on both sides.
=> (p+q)/2*{2a + (p+q-1)*d} = -2*(p+q)/2.
=> S(p+q) = -(p+q).
my take is -(p+q).


i think u r right.:)

Hi puys,
can u share the approach to solve this problem of interest.

A sum of money is borrowed and paid back in two equal instalments of Rs. 882, allowing 5% compound interest. Find the sum borrowed??

Hi puys,
can u share the approach to solve this problem of interest.

A sum of money is borrowed and paid back in two equal instalments of Rs. 882, allowing 5% compound interest. Find the sum borrowed??


sol:
882*2 = p(1+r/100)^2

P = 1600 :)

..............................................................................................................

bt the answer to the problem in the buk is 1640.. thats y i thght i must be doing sumthing wrng..

Solve this:
Manas, Mirza, Shorty and Jaipal bought a motorbike for $60,000. Manas paid 50% of the amounts paid by the other 3 boys,Mirza paid one third of the sum of the amounts paid by the other boys, and Shorty paid one fourth of the sum of amounts paid by the other boys. How much did Jaipal have to pay?

Solve this:
Manas, Mirza, Shorty and Jaipal bought a motorbike for $60,000. Manas paid 50% of the amounts paid by the other 3 boys,Mirza paid one third of the sum of the amounts paid by the other boys, and Shorty paid one fourth of the sum of amounts paid by the other boys. How much did Jaipal have to pay?


SOL:
manas - x mirza - y shorty - z jaipal - t (let)

y+z+t =2x

now if we add x to the above equation both sides

x+y+z+t = 2x+x

60000 = 3x

x = 20000

x+z+t = 3y

x+y+z+t = 4y

60000 = 4y

y = 15000


x+y+t = 4z

x+y+z+t = 5z

z = 12000

so amount paid by JAIPAL = 60000-(20000+15000+12000) = 13000:)

Hi puys,
can u share the approach to solve this problem of interest.

A sum of money is borrowed and paid back in two equal instalments of Rs. 882, allowing 5% compound interest. Find the sum borrowed??


Let X be the annual installment

((882*(105/100) - X)(105/100) - X) = 0


solve this for X and u will get the answer.
Let X be the annual installment

((882*(105/100) - X)(105/100) - X) = 0


solve this for X and u will get the answer.




can u elaborate ur approach.. how did u get this equation?
dtha Says
can u elaborate ur approach.. how did u get this equation?


Let X be the annual installment.
Now first year compound interest will be 852(105/100).

X will be deducted from it and then CI will be calculated on that remaining amount after deduction.

Again from that amount X will be deducted to be debt free therefore debt will be 0.

ps:iam travelling ryt now so if u didn't get the above explanation..will explain it later...
Let X be the annual installment.
Now first year compound interest will be 852(105/100).

X will be deducted from it and then CI will be calculated on that remaining amount after deduction.

Again from that amount X will be deducted to be debt free therefore debt will be 0.

ps:iam travelling ryt now so if u didn't get the above explanation..will explain it later...


hey can u solve the equation

am little confused with brackets:|
Hi puys,
can u share the approach to solve this problem of interest.

A sum of money is borrowed and paid back in two equal instalments of Rs. 882, allowing 5% compound interest. Find the sum borrowed??


Let X be the amount borrowed.
The sum would become X*1.05 at the end of first year.
Deducting the amount of 882 as the installment, it would become X*1.05 - 882
At the end of second year the sum would be equal to ((X*1.05) - 882)*1.05
This would be equal to 882 since the debt is paid off by paying that much amount.
So, the equation would become
((X*1.05) - 882)*1.05 = 882
=> X*1.05 - 882 = 882/1.05
=> X = (882/1.05 +882) / 1.05

Solve to get X = 1640 (I used a calculator you will have to solve without it )

Hope this helps. :)

Let X be the annual installment

((882*(105/100) - X)(105/100) - X) = 0


solve this for X and u will get the answer.


Ladke tune 882 ko borrowed money samajh liya and X ko installment. The question expect the opposite 😃

Ladke tune 882 ko borrowed money samajh liya and X ko installment. The question expect the opposite :)


OOhhh bhai kya karu us time train main tha toh jaldi jaldi main kar diya post...neways thanx for correction
Q. 4/5th of the voters in Barrely promised to vote for Sonia gandhi and the rest promised to vote for Sushma Swaraj . Of these voters 10% of the voters who had promised to vote for Sonia Gandhi , did not vote on the election day.What is the total no. of votes polled if Sonia Gandhi got 216 votes ?

(a) 200 (b)300 (c) 264 (d) 100

Let the total no of votes be 100.
4/5th votes out of 100 is 80
out of these 10% people were not voted means 10% of 80 is 72,
so If total votes are 100 sonia got 72 votes,
if total are 300 then sonia will get 216 votes(cross method)

hope it was correct.





hey bro , i am getting the same ans, but in arun sharma solution it has given 264 as answer.
which one do u think is correct
hey bro , i am getting the same ans, but in arun sharma solution it has given 264 as answer.
which one do u think is correct

Q. 4/5th of the voters in Barrely promised to vote for Sonia gandhi and the rest promised to vote for Sushma Swaraj . Of these voters 10% of the voters who had promised to vote for Sonia Gandhi , did not vote on the election day.What is the total no. of votes polled if Sonia Gandhi got 216 votes ?

(a) 200 (b)300 (c) 264 (d) 100

Let the total no of votes be 100.
4/5th votes out of 100 is 80
out of these 10% people were not voted means 10% of 80 is 72,
so If total votes are 100 sonia got 72 votes,
if total are 300 then sonia will get 216 votes(cross method)

hope it was correct.



TOTAL NO OF VOTES=300
BUT THE QUESTION IS ABOUT TOTAL NO. OF VOTES POLLED ON THAT DAY,SO WE HAV TO SUBTRCT 24(4/5*1/10*300).

SO AM GETTING 276 WHICH IS NONE OF THE ABOVE.
Q. 4/5th of the voters in Barrely promised to vote for Sonia gandhi and the rest promised to vote for Sushma Swaraj . Of these voters 10% of the voters who had promised to vote for Sonia Gandhi , did not vote on the election day.What is the total no. of votes polled if Sonia Gandhi got 216 votes ?

(a) 200 (b)300 (c) 264 (d) 100

Let the total no of votes be 100.
4/5th votes out of 100 is 80
out of these 10% people were not voted means 10% of 80 is 72,
so If total votes are 100 sonia got 72 votes,
if total are 300 then sonia will get 216 votes(cross method)

hope it was correct.



TOTAL NO OF VOTES=300
BUT THE QUESTION IS ABOUT TOTAL NO. OF VOTES POLLED ON THAT DAY,SO WE HAV TO SUBTRCT 24(4/5*1/10*300).

SO AM GETTING 276 WHICH IS NONE OF THE ABOVE.




the question is incomplete here it also states:: 20% of the voters who promised to vote for sushma swaraj did not vote on the election day.

So taking this into consideration the total votes polled comes out to be 264..

I have two doubts.

Q:1 There are two mixtures of honey and water, the quantity of honey in them being 25 and 75% of the mixture. If 2 gallons of the first are mixed with 3 gallons of the second, what will be the ratio of honey to water in the new mixture?

Q:2 A vessel is full of mixture of kerosene and petrol in which there is 18% kerosene. Eight litrs are drawn off and then the vessel is filled with petrol. I kerosene is now 15%, how much does the vessel hold?

Thanks in advance.



Q:1 There are two mixtures of honey
and water, the quantity of honey in
them being 25 and 75% of the
mixture. If 2 gallons of the first are
mixed with 3 gallons of the second,
what will be the ratio of honey to
water in the new mixture?

Q:2 A vessel is full of mixture of kerosene and petrol in which there is 18% kerosene. Eight litrs are drawn off and then the vessel is filled with petrol. I kerosene is now 15%, how much does the vessel hold?

Thanks in advance.

1)11/9

2)48 litres?

ques 1
N= 202 x 20002 x 200000002 x 20000000000000002 x 2000...02 (31 times 0)
find sum of digits of this multiplication..
options 112,160,144, cant be determined

ques 2
25 sets of problems on DI- one for each of di sections of 25 tests were prepared by a team. di section of each paper had 50 questions each. out of the 50 ques 35 ques were unique in each test (not repeated in any other 24 tests). find max numbers of questions prepared for DI sections of all tests put together..
options: 1100, 975, 1070, 1055

please share the method. both from number system.

ques 1
N= 202 x 20002 x 200000002 x 20000000000000002 x 2000...02 (31 times 0)
find sum of digits of this multiplication..
options 112,160,144, cant be determined

answer is option 1 i.e.112 for 1st question?????
not sure if this is correct then i will share my approach