Official Quant thread for CAT 2013

@Highway66 said:
In a regular octagon1.how many pairs of parallel diagonals are there? 1. 4 2. 8 3. 12 4. 162. if there is a regular polygon of 20 sides, how many diagonals originate from one vertex?1.17 2. 18 3. 19 4. 20p.s. second question ke liye approach plz
for second 20c2-20/10 = 17

for 1st in total 8c2-8 = 20 diagnols
4 will not have any parallel line counterpart
16 /2 8 pairs i guess

guys if you find time please update your answers with little explanation. it will be useful for fellow / future aspirants

@albiesriram said:
4pi-3_/3/12

option d
@Highway66 said:
This one is a nice ques:A pencil, eraser and notebook together cost Rs 100. the notebook cost more than two pencils. three pencils cost more than four erasers and three erasers cost more than one notebook. Answer the questions given that cost are positive, non-zero, integers.1) The cost of pencil isa) 27 b) 26 c) 28 d) data insufficient2) if u paid the shopkeeper Rs 210 for 2 notebooks, 3 pencils and 1 eraser, how many change would he returna) 3 b) 5 c) 7 d) data insufficient
pehle wale ka data insufficient hai kya??
@albiesriram said:
D. 1/3rd of circle of r = 1 - eqtr of side rt3.

Aur oral solution ko likhna bahut painful hota hai 😞 I know it is bad but...

regards
scrabbler

@albiesriram said:
d? 4pie-3rt(3)/12
@albiesriram said:


area=pi/3-sin120/2=pi/3-sqrt(3)/4 =4pi-3sqrt(3)/12 =option d

ďťż
@Dexian nahi
@Highway66 said:
@Dexian nahi
1)b.26?
@albiesriram said:
is it d?

area of sector - area of equilateral/3

area of sector =pi*1*1*120/360
side of equilateral= rt3
area of equil. triangle = rt3 *rt3 *rt3 /(4*3)


@albiesriram said:
option A:
@Highway66 said:
This one is a nice ques:A pencil, eraser and notebook together cost Rs 100. the notebook cost more than two pencils. three pencils cost more than four erasers and three erasers cost more than one notebook. Answer the questions given that cost are positive, non-zero, integers.1) The cost of pencil isa) 27 b) 26 c) 28 d) data insufficient



a) P=27
N(min)=55
E=18

it doesn't satisfy the 3rd quation(3E>N)

u cannt increase N as it will decrease E and again 3rd equation

b)P=26
N(min)=53
E=21
EDIT:
it satisfy the 3rd quation(3E>N)

c)P=28
N(min)=57
E=15

it doesn't satisfy the 3rd quation(3E>N)


to D hua na....
@Highway66 said:
This one is a nice ques:A pencil, eraser and notebook together cost Rs 100. the notebook cost more than two pencils. three pencils cost more than four erasers and three erasers cost more than one notebook. Answer the questions given that cost are positive, non-zero, integers.1) The cost of pencil isa) 27 b) 26 c) 28 d) data insufficient2) if u paid the shopkeeper Rs 210 for 2 notebooks, 3 pencils and 1 eraser, how many change would he returna) 3 b) 5 c) 7 d) data insufficient
1> 26?
2 > 3??

@Dexian Let the cost of Pencil, Eraser and Notebook be P, E and N respectively.
4P+4E+4N=400
4P+(3P-K1)+(8P+K2)=400
15P+(K2-K1)=400
P={400-(K2-K1)}/15 ~~~~eqn -1
Thus,the numerator should be less than 400.
From the given options only 26 is the value which when divided by 15 gives 390(less than 400) where 27 and 28 gives value more than 400. Thus they are not the possible values.
Hence 1)-option b)-26
4E=3P-K1
E=(78-K1)/4
4N=8P+K2
N=(208+K2)/4
K2-K1=10(can be obtianed from eqn -1)
only 1 pair of value satisfies this condition i.e. K1=2 and K2=12
hence P=26
E=19
N=55
2)
2N+3P+1E=207
210-207=3(change returned)
Option-a)-3
@heylady said:
option A:
OA
@Dexian said:
a) P=27N(min)=55E=18it doesn't satisfy the 3rd quation(3E>N)u cannt increase N as it will decrease E and again 3rd equationb)P=26N(min)=53E=21it doesn't satisfy the 3rd quation(3E>N)c)P=28N(min)=57E=15it doesn't satisfy the 3rd quation(3E>N)to D hua na....
E=19 aayeha.. 26*3=78 but E can be got only by dividing by 4 hence E=76/4=19
@Subhashdec2 said:
area=pi/3-sin120/2=pi/3-sqrt(3)/4 =4pi-3sqrt(3)/12 =option d
[Image] !!
yeh formula yaad hi nhi aa rha tha..directly nikal jata :splat:
#Me n my memory! :banghead:
@heylady said:
option A:
bhai explain kar do