official quant thread for cat08

[naga25french]

Originally Posted by getintoiimb

puys pls help me with these qns


3)what will be the value of x^1/2.x^1/4.x^1/8 to infinity
a)x^2 b)x c)x^3/2 d)x^3

pls let me know ur approach as well

required value = x ^(1/2 + 1/4 + 1/8 .... infinity)
so sum can be calculated as

s = x^[g/(1-r)]

g = power of first term
r = common ratio

s = x ^ [(1/2)]/[1- 1/2]

s = x

hope i am clear...

[Iota]

Originally Posted by getintoiimb

puys pls help me with these qns

2)the sum of an infinite GP whose common ratio is < 1is 32 and the sum of the first two terms is 24.what will be the 3rd term?
a)2 b)16 c)8 d)12 e)4

3)what will be the value of x^1/2.x^1/4.x^1/8 to infinity
a)x^2 b)x c)x^3/2 d)x^3

2) a/(1-r) = 32 , so a = 32(1-r) --(1)
again a+ar = 24
32(1-r)(1+r) = 24 ; so r = 1/2.
and a= 16 so the 3rd term = 4

3) S = x^1/2.x^1/4.x^1/8 to infinity
= X^(1/2+1/4+...to infinity)
= X^(0.5/(1-0.5)
=X

[ramsolutions]

Originally Posted by guneet_01

Shaurya and Ajit take a straight route to the same terminal point and travel with constant speeds. At the initial moment, the positions of the two and the terminal point form an equilateral triangle. When Ajit covers a distance of 80 km, the triangle becomes right-angled. When Arjit was at a distance of 120 km from the terminal point, Shaurya arrived at the point. Find the distance between them at the initial moment assuming that there are integral distances throughout the movements described.
(a) 300 km
(b) 240 km
(c) 200 km
(d) 225 km
(e) 275 km

Is the answer is b)240.

[guneet_01]

Originally Posted by apatchofsnow

Working from the options ,
I get the answer as option B 240 km...

did u use trignometry (sin 30 and stuff) for working thru options?

[guneet_01]

Originally Posted by ramsolutions

Is the answer is b)240.

yes it is 240.. how did u solve it?

[getintoiimb]

few more qns

1)the sum of the 2nd and 5th term of an AP is 8 and that of the 3rd term and the 7th termis 14.find the 11th term.
a)6 b)9 c)7 d)8 e)5

2)find the 2nd term of an AP if the sum of its 1st five even terms is equal to 15 and the sum of the 1st 3 terms is equal to -3
a)-3 b)-2 c)-1 d)0 e)1

3)find the sum of all 3 digit natural numbers which on beinf divided by 5 leave a remainder of 4
a)57270 b)96780 c)49680 d)99270 e)90270

[apatchofsnow]

Originally Posted by guneet_01

yes it is 240.. how did u solve it?

Hi

I used the options and tried the pythagoras theorem..
only one option satisfied..................

[naga25french]

Originally Posted by getintoiimb

few more qns

3)find the sum of all 3 digit natural numbers which on beinf divided by 5 leave a remainder of 4
a)57270 b)96780 c)49680 d)99270 e)90270

simple approach

for number to be divided by 5 and remainder is 4 means , number should have unit place value of 4 or 9

so three digit numbers goes as follow
series = 104,109,114,..............994,999 which is AP
with a = 104
l = 999
d = 5

now n should be found
l = a + (n - 1)* d
999 = 104 + ( n - 1) *5

so n = 180

now sum = n/2[ a+ l]

= 90 [104 + 999 ]

= 99270 which is option d)

hope i am clear..

[apatchofsnow]

Originally Posted by getintoiimb

few more qns

1)the sum of the 2nd and 5th term of an AP is 8 and that of the 3rd term and the 7th termis 14.find the 11th term.
a)6 b)9 c)7 d)8 e)5

2)find the 2nd term of an AP if the sum of its 1st five even terms is equal to 15 and the sum of the 1st 3 terms is equal to -3
a)-3 b)-2 c)-1 d)0 e)1

3)find the sum of all 3 digit natural numbers which on beinf divided by 5 leave a remainder of 4
a)57270 b)96780 c)49680 d)99270 e)90270

Answer for 1 : I am getting 19..
for 2 option c -1...........
3) option d 99270

[naga25french]

Originally Posted by getintoiimb

few more qns

1)the sum of the 2nd and 5th term of an AP is 8 and that of the 3rd term and the 7th termis 14.find the 11th term.
a)6 b)9 c)7 d)8 e)5

let the terms in ap be a , a+d ,a+ 2d , a + 3d

so now according to condition given

2a + 5d = 8

2a + 8d = 14

solvind two eqn we get

a = -1 and d = 2

so term 11 is a + 10d = 19

@getintoiimb

pls check the options again...