Let’s say,
I have x coins of 50 paise and (80-x) coins of 100 paise,so the equation is like this ,
50x + (80-x)*100 = 64*100
x = 32
So ,I have 32 coins of 50 paise
Very exited
Let\: the\: total\: price \: be\: Rs. X\: then,
B= 2x/7 & A= \left ( x-2 \right )/7
So, A:B = 5x/7 : 2x/7 = 5 : 2
Let \: B’s \: capital\; be\: Rs.Y\: then\: \left ( 16000\ast 8 \right )/4y= 5/2
=> (16000\ast 8\ast 2)/(4\ast 5) = y
=> y = Rs. 12800
This is more logical …
Let the 1st flag 1 placed at the origin ….
in crossing 8 flags he traveled 7 distances….
s=d/t
=7/8
time for 4 flags t=(d/s)=4/(7/8)=(4*8)/7=4.5714285714285714285714285714286
b
As it is a right angled triangle so we know the hypotenuse must be 13 and 5 & 13 are base and height (in any order).
Area of triangle = 0.5* base * height = 0.5 * 5 * 12 = 30
answer can be 1,2,3,4 can’t be determined exact number
just explaining a case:
there are 10 people in the party.
name of people no. of people with they made handshake list of those people(this can vary but showing the possibility)
1 1 9
2 2 8, 9
3 3 7, 8, 9
4 4 6, 7, 8, 9
5 5 6, 7, 8, 9. 10
6 6 4, 5, 7, 8, 9, 10
7 7 3, 4, 5, 6, 8, 9, 10
8 8 2, 3, 4, 5, 6, 7, 9, 10
9 9 1, 2, 3, 4, 5, 6, 7, 8, 10
10 4 5, 6, 7, 8, 9
jack got 9 different answer so jack can be either 4th number or 10th number and jack’s wife know jack very well so she can’t have handshake with jack so if 4th is jack then she can’t be handshake with 6,7,8,9, in this case she can be 1,2,3, 5, 10 and now depending upon which no is jack’s wife she can have hand shake with- 1- 4 people, and if jack is number 10 then she can’t be 5,6,7,8,9 so again depending upon her number she can handshake with people in range of 1-4
let t = total no of students.. then
students who passed one or both subjects,
n(e U h) = n(e) + n(h) – n(e intersection h)
=> t = 0.8t + 0.7t – 144
=> t = 1.5t – 144
students who failed both subjects is 10% i.e. 0.1t
=>t-n(e U h) = 0.1t,
=>t -(1.5t – 144) = 0.1t
=>t- 1.5t- 0.1t = -144
=> -0.6t = -144
=>t = 240
5 + 6 + 7 + 8 + 9 = 35 so two primes 5 and 7.
Assume Q can do the work alone in Q days:
work of a day by both P and Q =>
(1/15) + (1/Q ) = 1/6
solving this Q = 10.