official quant thread for cat08

[naga25french]

Originally Posted by jigar_p_civil

the total number of dogs in a colony is a composite number less than 200. the ratio of two number of dogs which are high breed to those who are not could be

100:49
102:55
87:76
94:85
86:101
49:100


no ans with me so plz post approach

the answer is 86:101... as 187 is the composite number (17*11)

when we see other number in the options , we get a prime number

[naga25french]

Originally Posted by naiquevin

yes u are right...how did u arrive at 162 ?

the question is atmost one digit repeat..

so lets inverse of this statement

there is no repetition of number ..

divisibility rule of 11:

sum of odd places number ~ sum of even place number must be 0 or multiple of 11

here there can be 0 or 11 as its a two digit subtraction

case 1:

when diff = 0

possible values are 3-3,4-4,5-5,6-6,7-7,8-8,9-9

for 3-3 , we have pair of number 12 , 30

by this we can form 6 numbers 1023,2013,3102,3201,2310,1320..

like wise working on various combinations , we will arrive at number close 160...

now using the options provided in the question

eliminate the option to get desired result...

hope its clear...

[shivam_01]

Originally Posted by jigar_p_civil
the total number of dogs in a colony is a composite number less than 200. the ratio of two number of dogs which are high breed to those who are not could be

100:49
102:55
87:76
94:85
86:101
49:100

answer is 86:101
Others will give u a prime number on addition

[bishweshwar1234]

Originally Posted by singh_kanwar

some questions making my head go round n round
Q1: in the above fig.<QYR = 120 Degrees and < PXQ = 150 degrees.if <QPS = 105 degrees and d area of d quad PQRS is 2( root 3 +2) sq.cm find the area of the circle.
a) 3 pi
b) 4 pi
c) 6pi
d) none
ps: fig attached

Q2: in d above fig. ABCD & BPRS r parallelograms,PR intersects DC at Q and RS passes thru the point C.how many of the following statements r always true?
A) the sum of the areas of triangles QRC and BCS is half d area of ||gm BPRS.
B) the sum of the areas of triangles APB and PDQ is half d area of ||gm ABCD
C) the sum of the areas of triangles QRC and BCS is half d area of d triangles PDQ & APB
D) d areas of ||gms ABCD & BPRS r equal

1) 4
2) 1
3) 2
4) 3
ps: fig attached

Q3: an AP is such that the ratio of the sum of d 1st n terms to that of d next n terms is equal to d sum of d 1st m terms to that of next m terms,for any two distinct natural nos m & n.if the common diff is nt 0,wat is the ratio of the sum of 1st 2n terms to dat of next 3n terms?
a) 4:21
b) 9:16
c) 2:3
d) cannot b determined.

Q4: let (PQRST) base 12 be a 5 digit no.,in base 12,with distinct,non-zero digits.al d possible 3 digit nos r now formed using d digits of d above no widout repetition.wat is d quotient wen d sum of all such 3 digits nos is divided by the sum of the digits of the original no.?(assume all d nos to b in base 12)
a) (1110)base 12
b) (333) base 12
c) (3330)base 12
d) (111)base 12

soory . was out of touch for long. Could not see reply to these questions. If any one gets, please mention the URL for the same.

[jigar_p_civil]

Originally Posted by Unhappy_Demon

Attached again.

ABGF is a square . All values are given in the figure .

How to calculate the Arc length of BF ?

here ans is simple

total circumference=2pie *r

so arc AF=2 pie *r(90/360)

Little Star

[Niki_10]

The absolute term i.e the term independent of x in the expression of(3x^2/2-1/3x)^9

Please explain me the approach to solve this ..

[ankaj]

Originally Posted by Niki_10

The absolute term i.e the term independent of x in the expression of(3x^2/2-1/3x)^9

Please explain me the approach to solve this ..

is it 7/36...
u can use binomial expression to expand this....

[Ralfi]

Originally Posted by Niki_10

The absolute term i.e the term independent of x in the expression of(3x^2/2-1/3x)^9

Please explain me the approach to solve this ..

ncr * x^n-r * y^r

9cr * (3x^2/2)^9-r * (-1/3x)^r

now, x^18-2r * x^-r = x^0
=> r = 6

now substitute u get ...7/18

[bishweshwar1234]

Originally Posted by ankaj

how to approach this one...i am getting set of two numbers...but third one is difficult???

Would you post your approach, for the two numbers you found.
i'm in lost way

[ankaj]

Originally Posted by bishweshwar1234

Would you post your approach, for the two numbers you found.
i'm in lost way

was using trial and error method...e.g 63, 64...these are divided by 9 (square) and 4 (cube)...but 65 is not divisible by fourth power???