how to solve these questions..
1)find sum of even factors of 2450?
2)find sum of odd factors of 2450?
3)find the sum and number of factors of 1200 such that factors are(a)divisible by 15 and(b)not divisible by 15??
4)find number of divisors of 1080 excluding the throughout divisors which are perfect square?
@challenger90 said: @AimIIMAC: answer of A is 6.08% annd of B is 660rs.. plz explain how...i tried many times but not able to solve..TIA
See when u perform the given operations in questions, the Cube will be modified into a form of Cuboid. In Cube we have L=B=H. Let it be as 10. Therefore its Surface Area will be 6(a)^2=600. Now when u performed the given operations, the dimensions now will be L=9.5 ,B=9.5 and H=12. Now it transformed into a Cuboid. Try to Imagine the operations in mind, u will get it transformed into a Cuboid(Match Box like). Now the Surface Area of Cuboid is A= 2{(l*b)+(l*h)+(h*b)} =636.25 which is 6.04% more than 600.
The Answer to the B question is Let A having x money, then B=1200-x .
Acc. to question,
.88x + .80(120-x)=.85*1200 and we get x=750. Money Remaining=.88*750=Rs660
I hope u get it :)
The Answer to the B question is Let A having x money, then B=1200-x .
Acc. to question,
.88x + .80(120-x)=.85*1200 and we get x=750. Money Remaining=.88*750=Rs660
I hope u get it :)
@[419968:sidroy09] how could the sum of even factors be odd?????
@[590072:mitali0390] is it 1. 3534,2. 1767,3. a)2790 others i dont know
hi !!
Pls. help with the detailed sol for the below mentioned probs of POM COM:-
Q1) the crew of an 8 member rowing team is to be chosen from 12 men, of which 3 must row on 1 side and 2 must row on the other side only. find the number of ways of arranging the crew with 4 members on each side ??
Options :- A) 40320 B) 30240 C) 60480 D) 10080 E) None of these
Q2) in how many ways 5 MBAs and 6 CA students can be arranged together so that no 2 MBA students are side by side:-
Options:-a) 7!*6!/2! b) 6!*6! c) 6!*5! d) 11P5 e) 11C5
Q3) there are 8 orators A,B,C,D,E,F,G and H. In how many ways can the arrangements be made so that A comes before B and B comes before C alwaz??
Options: -A) 8!/3! B) 8!/6! C) 5!*3! D) 6!*3! E) 8!/(5!*3!) F) None of these
Q4) if we have to make 7 boys and 7 girls sit around a round table which is numbered, then the number of ways in which it can be done are??
Options - A) 2*7!*7! B) 7!*6! C) 7!*7! D) None of these
Q5) in a building there are 12 floors excluding the ground floor. 9 ppl get in the lift at the ground floor. the lift does not stop at the 1st floor. 2,3,4 ppl alight from the lift on its upward journey, then how many ways then can do so?
Options- A) 11C3* 3! B) 11P3*9C4*5C3 C) 10P3*9C4*5C3 D) 12C3
Q6) 7 different objects are divided among 3 ppl. in how many ways it can be done if one or two of them can get no objects??
Options - A) 15120 B) 2187 C) 3003 D) 792
@challenger90 said: A sells his goods 30% cheaper than B and 50% dearer than C. by what %age is cost of C's goods cheaper than B's goods??plz explain in detail solution of this question..answer is 46.15%
take D's sp =100,
so A's sp =150,
now A's sp= 70% of B's sp
so B's sp=214.28
114.28/214.28 *100= 53.XX %
I guess C's sp is 46.15% of B's sp
so A's sp =150,
now A's sp= 70% of B's sp
so B's sp=214.28
114.28/214.28 *100= 53.XX %
I guess C's sp is 46.15% of B's sp
@challenger90 said: A sells his goods 30% cheaper than B and 50% dearer than C. by what %age is cost of C's goods cheaper than B's goods??plz explain in detail solution of this question..answer is 46.15%
Let B's SP be 100
so, A's SP is 70..
A's SP is 50% more than C's SP..
so, C's SP is 33.33 % less than A's SP (check below for detailed calculation)
so, A's SP is 70..
A's SP is 50% more than C's SP..
so, C's SP is 33.33 % less than A's SP (check below for detailed calculation)
33.33 of 70 = 23.31
so C's SP = 70-23.31 = 46.69
hence C is cheaper than B by (100-46.69)% or 53.31%
Please check the answer. I think it wudnt be 46.15 %..
its like 100 ---to--- 150 ---to--- 100
first an increase of 50%.
then we have to calculate % change for 150 to 100.
which is (150-100)/150 or 33.33 %
Q17 . Ranjan purchased a maruti van for rs 1.96000 and the rate of depreciation is 100/7% per annum. Find the value of the van after 2 years?
- 140000
- 144000
- 150000
- 160000
- None of these
Q21 . Harsha makes a fixed deposit of rs 20000 with the bank of India for a period of 3 year. If the rate of interest be 13% SI per annum charged half yearly what amount will he get after 42 months?
- 27800
- 28100
- 29100
- 28500
- None of these
hi guys,
i have trouble understanding one question from progression. page - 97 question-40 and page -98 ques-45.
is there a way to find if the question has been answered in earlier posts?
@kaushikdivya16 said:hi guys,
i have trouble understanding one question from progression. page - 97 question-40 and page -98 ques-45.
is there a way to find if the question has been answered in earlier posts?
1.14400
@hemant89 said:Q17 . Ranjan purchased a maruti van for rs 1.96000 and the rate of depreciation is 100/7% per annum. Find the value of the van after 2 years?
- 140000
- 144000
- 150000
- 160000
- None of these
Q21 . Harsha makes a fixed deposit of rs 20000 with the bank of India for a period of 3 year. If the rate of interest be 13% SI per annum charged half yearly what amount will he get after 42 months?
- 27800
- 28100
- 29100
- 28500
- None of these
@[527148:Subhashdec2] 1) in an infinite GP, each term is equal to 3 times the sum of the terms that follow. If the first term of the series is 8, find the sum of the series?
2) After striking a floor a rubber ball rebounds (7/8)th of the height from which it has fallen. Find the total distance tat it travels before coming to rest, if it is gently dropped from height of 420 m?
@kaushikdivya16 said:@Subhashdec2 1) in an infinite GP, each term is equal to 3 times the sum of the terms that follow. If the first term of the series is 8, find the sum of the series?
2) After striking a floor a rubber ball rebounds (7/8)th of the height from which it has fallen. Find the total distance tat it travels before coming to rest, if it is gently dropped from height of 420 m?
1.32/3
2.6300???
@[527148:Subhashdec2] thanks!!
answer marked in Book for 1) 12 and 2) 6300. can u please post the solution...
@[336473:kaushikdivya16]
For (1), let the GP be : a, ar, ar^2, ar^3 .....
As per the condition given in Qn , every term is 3 times the sum of the terms that follow, so let us take the first term.
a = 3 *[ ar /(1-r) ]
1-r = 3r
or r = 1/4
so the GP is
8 , 8/4 , 8/16 , 8/32 ,........
Sum of GP = [a/(1-r)] = [8/(1-1/4)] = 8*3/(4) = 32 = Ans.
Check : Test if the given conditon is satisfied by the GP we have obtained.
GP: 8, 8/4, 8/16 .....
so the first term should be equal to sum of the terms that follow it
i.e. 8 = 3* [ (8/4) / (1-1/4) ]
Solving RHS, 3*[ 2*4 / 3 ] = 8 = LHS.
So our answer is correct.
@[336473:kaushikdivya16]
For (2), given that the initial height from which the ball falls is 420m.
so first the ball will travel 420m to reach the ground.
Total distance travelled till now : 420
Then, it will rebound to (7/8)*420m after striking the ground.Then again fall thru same distance to strike the ground.
So total distance travelled in this phase : 2[(7/8)*420]
Now for the third time the ball rebounds again, and reaches the height of (7/8) of previous distance rise[ which is (7/8)*420m ].
This is equal to (7/8)*(7/8)*420. after reaching this height the ball will fall for the same distance to reach the ground.
So total distance travelled in this phase : 2[(7/8)*(7/8)*420]
Continuing in this manner, total distance travelled by the ball can be obtained by adding the above distances till infinity.
420 + 2[(7/8)*420] + 2[(7/8)*(7/8)*420] + .............
= 420 + 2*420*[ (7/8) + (7/8)*(7/8) + ............ a G.P. ]
= 420 + 2*420* [ (7/8) / (1-7/8) ]
= 420 + 2*420*7 = 420(1+14) = 420*15 = 6300 = Ans
@kaushikdivya16 said:answer marked in Book for 1) 12 and 2) 6300. can u please post the solution...
Lets see if the answer to (1) can be 12 ,
For sum of GP = 12 and first term = 8
we have 12 = 8 / (1-r)
This gives r = 1/3
So, the GP becomes : 8 , 8/3 , 8/9 , 8/27 .............
Now check the condition given in Qn
any term should be equal to the three times the sum of the terms that follow it,
so, for first term,
8 = 8/3 + 8/ 27 + 8/81............ (a GP)
8 should be equal to 3* [ 8/3 / ( 1-1/3) ]
Solving RHS, 3*[8/2] = 12 ( and not 8 )
Therefore, 12 cannot be the sum of the GP or (1/3) cannot be the common ratio of the GP.
@[527148:Subhashdec2] dude i have the book i know the answer too................i posted the question as i didn't get how to solve it, so can you please explain the method.
Don't take it otherwise, but please it is common sense why would someone post any question of he already has the answer??
Anyone please tell me the solution of this problem:
There are 5 balls of different colours and 5 boxes of the same colour as the balls.The number of ways in which the balls one in each box can be placed,such that a ball does not go in to box of its own colour are ?
@[593895:challenger90]
1) let the length, breadth and height of the cube be 1,1,1 units. which gives a surface area of 6 units.
now,the new l,b,h will be .95, .95 ,1.2. which gives a surface area of 6.365 units.
hence surface area is increased by .365/6 = 6.083%
2) let amount with A = x
then amount with B = 1200-x
A spends 12% of his money that means he has 88% of his share left.
Similarly B, after spending 20% of his money has 80% of his share left.
So,
.88(x) + .80(1200-x) = 85% of 1200
solve the equation and you will obtain the value of x as 750. this was the initial amount with A. so amount left will be 88% of 750 = Rs. 660
Hope this helps!! 