where [x] is defined

as integral part of x and f is a fraction,

then x(1–f) equals –

A. 1

B. Less than 1

C. More than 1

D. Between 1 and 2

E. None of the above

Note -

- in R=(a+b * rtc)^k

if k is odd then f=X

fractional part will be f=(a-b * rtc)^n - and if k is even then f+X=1

f=1-(a-b * rtc)^k

here x=(9+4sqrt5)^48

so f=1-(9-4sqrt5)^48

so x * (1-f)=(9+4sqrt5)^48 * (9-4sqrt5)^48 =1

**p, q and r are roots of the equation x^3 - 7x^2 - 6x + 5 = 0. Find the value of (p + 1)(q + 1)(r + 1).**

we need (p+1) (q+1) (r+1)

so need equation whose roots are p+1 ,q+1 and r+1

so apply transformation replace x by x-1

(x-1)^3-7(x-1)^2-6(x-1)+5=0

x^3-1-3x(x-1)-7x^2-7+14x-6x+11=0

x^3-10x^2+11x+3=0

so product of roots will be -3

**p, q, r are roots of the equation x^3 - 7x^2 - 6x + 5 = 0. Find the value of (p + q)(q + r)(r + p).**

p + q + r = 7 so p + q = 7 - r

q + r = 7 - p

and p + r = 7 - q

so we need (7-p)(7-q)(7-r)

means we need equation whose roots are 7-p,7-q,7-r or

x^3-7x^2-6x+5=(x-p)((x-q))(x-r))

put x=7

so 7^3-7 * 7^2-6 * 7+5 =- 37 OA

**If three medians of the triangle are of the lengths 9, 12 and 15, find the area of the triangle.**

Method 1 – Direct formula

Area of triangle using medians, A = 4/3 ( Sqrt(s(s-p)(s-q)(s-r))

p, q and r are medians and s = (p+q+r)/3

Method2- ABC is the triangle, AD, BE, CF are medians and G centroid

extend GD to a point M such that GD = DM, then CM = BG

so, triangle MCG; has sides 6, 8 and 10

area = 24

area of ABC = 24 * 3 = 72.

Method 3

4 * (a^2+b^2+c^2)=3 * (9^2+12^2+15^2)

so a^2+b^2+c^2=3 * (81+225+144)/4

now we have a^2+b^2+c^2

apply a^2+b^2=2*(AD^2+(c^2/4))

**Given 3x^2+x=1, find the value of 6x^3-x^2-3x+2015.**

6x^3+2x^2 = (2x)(3x^2+x) = 2x

and -(3x^2+x) = -1

so 6x^3-x^2-3x+2015 = 2x-1-2x+2015 = 2014.

**Let a, b and c be the roots of x^3 - 2x^2 + 2x + 4 = 0.Find the value of 1/(a^2 + b^2) + 1/(b^2 + c^2) + 1/(c^2 + a^2) + 1/4.**

a + b + c = 2

ab + bc + ac = 2

abc = -4

here a+b+c=ab+bc+ca so a^2+b^2+c^2=0

question reduces to

-(1/a^2 + 1/b^2 + 1/c^2) + 1/4

now we can easily find the value by solving this

Another method : x^3-2x^2+2x+4=0

roots are a,b,c equation whose roots are a^2,b^2,c^2

will be (x^3/2)-2x+2sqrtx+4=0

x*sqrtx-2x+2sqrtx+4=

x^3+4x+4x^2=4x^2+16-16x

so x^3+20x-12=0

now roots of this equation are a^2,b^2,c^2

so equation whose roots are 1/a^2,1/b^2,1/c^2 will be reverse of this

so 1/x^3+20/x-12=0

-12x^3+20x^2+1=0

so 12x^3-20x^2-1=0

so 1/a^2+1/b^2+1/c^2=5/4

so -5/4+1/4=-1

**From 100 litres of milk, 10 litres of water is added and then 20 litres of solution is removed. Next 30 litres of the water is added and 20 litres of solution is removed. Find the amount of milk, in litres, in the solution now.**

Approach- #MULTIPLYING FACTOR

Method to tackle These Type Of Questions

always take initial as those part which is not added,

here milk as it is never added

here in first step 10 litre of water is added so solution=100+10=110

in the 1st time 20 is removed out off 110 so

left-out part is 90/110 so our multiplying factor is 90/110

no 30 is added to 90 so 120 solution

in the 2nd case 20 is removed from 120 so left out is 100/120

so amount of milk= 100 * 90/110 * 100/120=750/11

PS- Always remember Our concentration should be on total solution and removed part .... Added part will only increase the value of solution ..

**A shopkeeper claims to sell his items at a loss of 10% but instead he secretly makes a profit of 20% by using a false weight. what is the effective profit percentage he gets on every 100gm he sells?**

Method1- Approach= He claims to sell 120 when he actually sells 100 (using false weights)

Now, he claims to sell them at a loss of 10%

He makes 120*0.9 = 108

so 8% profit

Method2- = Multiplying factor

100 * 9/10 * 12/10=108 so 8%

Method3- 20-10+(20*(-10)/100=8%

**A dishonest trader, at the time of selling and purchasing, weighs 10% less and 20% more per kilogram respectively. Find the percentage profit earned by treachery. (Assuming he sells at Cost Price)a) 30%b) 20%c) 25%d) 33.33%**

Method1- Let's say original weight is 100 now he is taking 20% extra so total 120 and selling 10% less so 90

So (120-90/90) * 100

= 30/90 * 100= 33.33%

Method2-

multiplying factor

1 * 10/9 * 6/5=1.33 so 33%

**A beaker contains milk and water in the ratio 5 : 3. If a person removes 6 L of this mixture and replaces it with pure water, the ratio of milk to water becomes 1 : 1. What is the amount of milk present in beaker initially**

Always remember one thing

Final =Initial * fraction left 1 * fraction left 2 * fraction left 3 *( so on)

Answer = 75/4

where [x] is defined

as integral part of x and f is a fraction,

then x(1–f) equals –

A. 1

B. Less than 1

C. More than 1

D. Between 1 and 2

E. None of the above

Note -

- in R=(a+b * rtc)^k

if k is odd then f=X

fractional part will be f=(a-b * rtc)^n - and if k is even then f+X=1

f=1-(a-b * rtc)^k

here x=(9+4sqrt5)^48

so f=1-(9-4sqrt5)^48

so x * (1-f)=(9+4sqrt5)^48 * (9-4sqrt5)^48 =1

**p, q and r are roots of the equation x^3 - 7x^2 - 6x + 5 = 0. Find the value of (p + 1)(q + 1)(r + 1).**

we need (p+1) (q+1) (r+1)

so need equation whose roots are p+1 ,q+1 and r+1

so apply transformation replace x by x-1

(x-1)^3-7(x-1)^2-6(x-1)+5=0

x^3-1-3x(x-1)-7x^2-7+14x-6x+11=0

x^3-10x^2+11x+3=0

so product of roots will be -3

**p, q, r are roots of the equation x^3 - 7x^2 - 6x + 5 = 0. Find the value of (p + q)(q + r)(r + p).**

p + q + r = 7 so p + q = 7 - r

q + r = 7 - p

and p + r = 7 - q

so we need (7-p)(7-q)(7-r)

means we need equation whose roots are 7-p,7-q,7-r or

x^3-7x^2-6x+5=(x-p)((x-q))(x-r))

put x=7

so 7^3-7 * 7^2-6 * 7+5 =- 37 OA

**If three medians of the triangle are of the lengths 9, 12 and 15, find the area of the triangle.**

Method 1 – Direct formula

Area of triangle using medians, A = 4/3 ( Sqrt(s(s-p)(s-q)(s-r))

p, q and r are medians and s = (p+q+r)/3

Method2- ABC is the triangle, AD, BE, CF are medians and G centroid

extend GD to a point M such that GD = DM, then CM = BG

so, triangle MCG; has sides 6, 8 and 10

area = 24

area of ABC = 24 * 3 = 72.

Method 3

4 * (a^2+b^2+c^2)=3 * (9^2+12^2+15^2)

so a^2+b^2+c^2=3 * (81+225+144)/4

now we have a^2+b^2+c^2

apply a^2+b^2=2*(AD^2+(c^2/4))

**Given 3x^2+x=1, find the value of 6x^3-x^2-3x+2015.**

6x^3+2x^2 = (2x)(3x^2+x) = 2x

and -(3x^2+x) = -1

so 6x^3-x^2-3x+2015 = 2x-1-2x+2015 = 2014.

**Let a, b and c be the roots of x^3 - 2x^2 + 2x + 4 = 0.Find the value of 1/(a^2 + b^2) + 1/(b^2 + c^2) + 1/(c^2 + a^2) + 1/4.**

a + b + c = 2

ab + bc + ac = 2

abc = -4

here a+b+c=ab+bc+ca so a^2+b^2+c^2=0

question reduces to

-(1/a^2 + 1/b^2 + 1/c^2) + 1/4

now we can easily find the value by solving this

Another method : x^3-2x^2+2x+4=0

roots are a,b,c equation whose roots are a^2,b^2,c^2

will be (x^3/2)-2x+2sqrtx+4=0

x*sqrtx-2x+2sqrtx+4=

x^3+4x+4x^2=4x^2+16-16x

so x^3+20x-12=0

now roots of this equation are a^2,b^2,c^2

so equation whose roots are 1/a^2,1/b^2,1/c^2 will be reverse of this

so 1/x^3+20/x-12=0

-12x^3+20x^2+1=0

so 12x^3-20x^2-1=0

so 1/a^2+1/b^2+1/c^2=5/4

so -5/4+1/4=-1

**From 100 litres of milk, 10 litres of water is added and then 20 litres of solution is removed. Next 30 litres of the water is added and 20 litres of solution is removed. Find the amount of milk, in litres, in the solution now.**

Approach- #MULTIPLYING FACTOR

Method to tackle These Type Of Questions

always take initial as those part which is not added,

here milk as it is never added

here in first step 10 litre of water is added so solution=100+10=110

in the 1st time 20 is removed out off 110 so

left-out part is 90/110 so our multiplying factor is 90/110

no 30 is added to 90 so 120 solution

in the 2nd case 20 is removed from 120 so left out is 100/120

so amount of milk= 100 * 90/110 * 100/120=750/11

PS- Always remember Our concentration should be on total solution and removed part .... Added part will only increase the value of solution ..

**A shopkeeper claims to sell his items at a loss of 10% but instead he secretly makes a profit of 20% by using a false weight. what is the effective profit percentage he gets on every 100gm he sells?**

Method1- Approach= He claims to sell 120 when he actually sells 100 (using false weights)

Now, he claims to sell them at a loss of 10%

He makes 120*0.9 = 108

so 8% profit

Method2- = Multiplying factor

100 * 9/10 * 12/10=108 so 8%

Method3- 20-10+(20*(-10)/100=8%

**A dishonest trader, at the time of selling and purchasing, weighs 10% less and 20% more per kilogram respectively. Find the percentage profit earned by treachery. (Assuming he sells at Cost Price)a) 30%b) 20%c) 25%d) 33.33%**

Method1- Let's say original weight is 100 now he is taking 20% extra so total 120 and selling 10% less so 90

So (120-90/90) * 100

= 30/90 * 100= 33.33%

Method2-

multiplying factor

1 * 10/9 * 6/5=1.33 so 33%

**A beaker contains milk and water in the ratio 5 : 3. If a person removes 6 L of this mixture and replaces it with pure water, the ratio of milk to water becomes 1 : 1. What is the amount of milk present in beaker initially**

Always remember one thing

Final =Initial * fraction left 1 * fraction left 2 * fraction left 3 *( so on)

Answer = 75/4