x = 3/2 minimum = 1 ]]>

2.83 days ]]>

(7-a)=-1

(7-b)=-2

(7-c)=2

(7-d)=1

(7-e)=7

a=8

b=9

c=5

d=6

e=0

=28 ]]>

LET the remaining areas in cyclic order from left be e,a, d, b.and the shaded area be c.

a+b+c =1/2(a+b+c+d+e+97)

a+b+c = d+e+97

or

d+e+97 = a+b+c -------(1)

similarly,

e+c+d = 1/2(a+b+c+d+e+97)

=>e+c+d= a+b+97 ------(2)

subtract (2) from (1)

c-97 = 97-c

=>c=97

Why have we considered numbers like 7 and such to proceed further?

Could you explain the solution a bit more... ?

]]>(X/2)-1 /x >=0.47 ( for x even)

(X/2 -0.5)/x >=0.47 (x being odd case)

It satisfies at x=17( min)

]]>(x + 8)(x - 3) = 0

x = - 8, x = 3

x^2 + 5x - 36 = 0

(x + 9)(x - 4) = 0

x = -9, x = 4

x varies from -8 to - 9 or 3 to 4

x^2 can vary between 64 to 81 or 9 to 16

23.04 is not within this range.

a. 90%

b. 55%

c. 50%

d. 40%

e. 10% ]]>

[OA : 13C7]

]]>Put x= y,

F(2x) = f(x)^2

Put y=2x

F(3x)= f(x)^3

=>f(nx)=f(x)^n

F(-8)=F(-2×4)=f(4)^-2=1/9

]]>nf(x)= f(x^n)

F(343sqrt3)= 3f(7)+1/2f(3)

F(7) and f(3) can be found using given values.

Let there be "a" meets.

=>[(100a) * 5/(5+3)]=300

Which gives a=4.8

Though it cant be decimal.

But What is wrong with this approach????

As there is a similar question in gyan room-arithmetic shortcuts...and i hv applied the same approach here..

]]>=> xy = nx + ny

=> (x - n)(y - n) = n²

Now, n² must have 5 factors to get 5 ordered pairs

=> n must be of form p², where p is prime number

Hence all primes less than 20

]]>a + b = p and ab = 12

(a + b)^2 = p^2

(a - b)^2 = (a + b)^2 - 4ab

=> (a - b)^2 = p^2 - 12 * 4 = p^2 - 48

If |a - b| ≥ 12 { Difference between the roots is at least 12}

then, (a - b)^2 ≥ 144

p^2 - 48 ≥ 144

p^2 ≥ 192

P ≥ 8√3 or P ≤ -8√3

In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n

[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)

Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)

So the number of distinct integers would be [n/2] + [n/4] + 1

if n = 100,

number of distinct integers would be [100/2] + [100/4] + 1 = 76

if n = 2014,

number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511

if n = 13

number of distinct integers would be [13/2] + [13/4] + 1 = 10

In such scenarios like finding distinct values of [x^2/n] where x can be from 1, 2, 3 ... n

[1^2/n], [2^2/n] ... [(n/2)^2/n] will yield all numbers from 0 to [n/4] (means [n/4] + 1 distinct integers)

Then the next set (from [(n/2 + 1)^2/n] till [n^2/n] will be all different integers (means [n/2] distinct integers)

So the number of distinct integers would be [n/2] + [n/4] + 1

if n = 100,

number of distinct integers would be [100/2] + [100/4] + 1 = 76

if n = 2014,

number of distinct integers would be [2014/2] + [2014/4] + 1 = 1511

if n = 13

number of distinct integers would be [13/2] + [13/4] + 1 = 10

Just trying to generalize a solution shared by Kamal sir (Quant Boosters - Set 1 - Q2).

You can try out with various numbers (may be smaller numbers) so that this can be verified.

]]>number of scores possible = 251 - 15 + 1 ...= 237

Could anyone explain this last step ?

]]>x/m = y + x/v

x/m-2 = 2y + x/v-2

x(v-m) = 2y(m-2)(v-2)

x(v-m) = yvm

vm = 2(vm - 2m -2v + 4)

vm = 2vm - 4m - 4v + 8

4m + 4v - 8 = vm

4m - 8 / (m-4) = v

4 + 8/m-4 = v and m>6

m = 5 , v=12

m = 6 , v = 8

x = 60y/7 and x = 24y

x/m = 12y/7 and 4y

x/v = 5y/7 and 3y

only A satisfies for mini .

Exam Conditions approach : You can take random values of x and y & can try solve

]]>