CAT 2007: QQAD, Practice Test Discussions

[amit_vce]

Solution Of 5 easiest ones : Rest will follow

4. Time Difference = BC+CA /3 - CB+AB /2
sOLVING THIS BC+20 / 6
Solving for BC
COS(120) = 12^2 + 8^2 - BC^2 / 2 * 96
BC = root ( 304 ).
After putting BC , time difference = 2(root(19) + 10 ) /3
So answer = b.
3. Going by options : c
6. f(x,1-x) = f(1,1) + f(1,1) = 2f(1,1)
f(x,-x) = f(0,1) + f(1,0) = f(1,1)+f(1,1)+f(1,1)+f(1,1) = 4f(1,1).
so f(x,1-x) / f(x,-x) = 1/2 ,answer = a.
8. trying some values like a= 2, b =2 , c =2 .
Its always greater then 1.

and its always less then 2.
14. putting i = 1 in equation ,
ax - y = a-1
putting i = 2 in equation
ax - 2y = 2a -1
solving 2 equations , y = -a ,
solving further x = -1/a
so xy =1. answer = b.

[amit_vce]

Originally Posted by arnabsadhu

Arav & Vineet,

Q7: 253 = 23*11.

Possible no. of terms 11 & 23.

Difference = 12. But where is the option.

Series corresponding 23-> 0, 1, 2,......, 11, ......, 20, 21, 22.
Series corresponding 11-> 18, 19, 20,..., 23, ........., 26, 27, 28.

Today in stipulated 50 minutes, I've attempted first 17 problem correctly, I think. What can be my standing?

I'll post all my approaches tomorrow.

The terms can be 46 and 11 .

46 starting from -17 and 11 starting from 18 . So the difference can be 35.

[sunit_chotu]

I attempted 11 ; 10 correct and 1 wrong i am posting the detailed solution as under.

[sunit_chotu]

1) seven digit binary number hence left most digit cannot be 0, putting 1 there rest 6 digits can be arranged in (6!/(3!*3!))=20 ways that means there are 20 numbers where 1 comes at this digit place .

also when 1 comes at 2nd digit from left there are (5!/(3!*2!)) = 10 such numbers.

when 0 is at a particular position it will not contribute to the final sum.

now for the sum of all the numbers when 1 comes at different positions is:
20*(2^6)+10*[2^5+2^4+2^3+2^2+2^1+2^0] = 1910.

hence ans is (e).

[sunit_chotu]

2) we are making cone by alligning the straight sides hence the slant height will be 10.

now the circumference of the cone and the length of the arc from which this cone is made should be equal,

hence, (252/360)*(2*pi)*(10) = 2*pi* r
=> r =7.
hence (c) is the correct answer.

[sunit_chotu]

4) by cosine law in triangle ABC,

a^2 = b^2+c^2 - 2*b*c*cosA
= 144 + 64 - 2*12*8*(-1/2)
= 304
a = 4(v19)

now the time interval between their return = [(12+(4*v19))/2 - (8+4*v19)/3]

= (10+2*v19)/3

hence option (b)

[sunit_chotu]

6) f(x,y) = f(x+y,1) + f(1,x+y) --------------------------(1)

now using this f(x,1-x) = f(x+1-x,1) + f(1,x+1-x) = 2* f(1,1).

similarly f(x,-x) = f(0,1) + f(1,0) = f(0+1,1) +f(1,0+1) + f(1+0,1) +f(1,1+0)
= 4*f(1,1)

f(x,1-x) / f(x,-x) = 2* f(1,1) / 4*f(1,1) = 1/2.

hence option (a)

[sunit_chotu]

8. this i did by trial and error by putting some values and got (a) and (b) both as true.

looking forward for the exact solution.

[sunit_chotu]

11) using the formula (M1*D1*H1)/W1 = (M2*D2*H2)/W2

we get ,

(100*1)/(300x+200y) = (60*2)/(240x+300y) = (50*3)/(150x+my)


=> 1/(3x+2y) = 2/ (4x+5y) = 150/(150x+my)
solving first two we get y = 2x.

now take first and third,
we get, 1/7x = 150/(150x+2mx)
=> m = 450.

hence option (d)

[sunit_chotu]

14) system of equations are y-1 = a* (x-1) -----------------(1)
2y-1 = a*(x-2) ------------------(2)
3y-1 = a*(x-3) ------------------(3)

solving 1 and 2 we get y = -a,
put the value of y in 3 we get x = -1/a,
hence xy = 1.
hence option (b)