We know,

(a + b + c)^{2 =} a²+b²+c² + 2(ab + bc + ca)

= 20 + 2×8 = 36

a + b + c = 6

Therefore,

(1/2)(a + b + c) [(a − b)² + (b − c)² + (c − a)²] = (1/2)(a + b + c)[2(a²+b²+c²) − 2(ab + bc + ca)]

= (1/2)×6[2(20−8)]

= 72

25% of CP = Rs. 44.80

100% of CP

Since we know that, 



Since sec θ – tan θ = P       .....................(i)

then       .....................(ii)

Adding eq.(i) and (ii):


Subtracting eq.(ii) from (i):

It can be written as:


Thus, 

Area of a triangle = Inradius × Semi Perimeter 
 
15 = 3 × Semi Perimeter 

Semi Perimeter = 5 cm
Perimeter = 10 cm

Let the efficiencies of A, B and C be 5, 3, 8 respectively.

Efficiencies of A , B and C together =16

Together A, B and C can complete the work in 30 days

So the total work= 30 x 16 = 480 units

Efficiency of A and B together = 5+3 = 8

Work done by A+B in 20 days = 8 x 20 =160 units

Remaining work = 480-160 =320 units

Remaining work will be done by C alone = 320/8 =40 days.

Relative speed of train X w.r.t. Y = 100 km/hr

Time take to cross each other = 800/100 = 8 hr

Distance covered by train X = 40 x 8 = 320 km = Distance from A at which X and Y meet.

Let the present ages of A and B is 8x and 9x

After 9 years ratio

(8x+9)/(9x+9)=19/21

(8x+9)/(3x+3)=19/7

56x+63=57x+57

X=6

So age of B=9x=9 x 6=54

C is 3 year younger than B so age of C

=54-3=51 years

Let the number of articles is 100

Let the cost price of one article is 100 Rs.

MRP is increased by 40% from cost price so MRP

=140

70% of articles are sold at MRP so selling price of 70% articles =70 x 140=9800 Rs.

30% of articles are sold at 40% discount at MRP (at 84 Rs. Per article)

So selling price of 30% articles=30 x 84=2520 Rs.

Selling price of 100% articles=9800+2500=12320 Rs.

Total cost price=100 x 100=10000 Rs.

Profit= 12320-10000

=2320 Rs.

Percentage profit=2320/10000 x 100

=23.2%

For divisibility by 88, number should be divisible by 11 and 8 both.

For divisible by 8 last 3 digit of the number (5y2) must be divisible by 8.

So, the value of y can be 1,5 and 9.

For divisibility by 11, difference of sum of digits at odd place and sum of digits at even place must be zero or divisible by 11.

So, (11+x) – (7+y) = 0 or multiple
For the smallest value of y, y=1
11+x-7-1=0
3+x=0
so, only one value of x that is x=8 will satisfy the equation.

By putting y=1 and x=8

(4x-y)= 31