habeas corpus
12, 10
2/3 * 15 miles = 10 miles then
time for 10 miles travel is t1 = 10/40 = 1/4
then remaning distance is 5 miles
time for 5 miles travel is t2 = 5/60 = 1/12
t1+ t2= 1/3 in seconds * 60 = 20 min
Ans- 6,6 litres
Let x Litres from can1 (where, water = x/4 litres and milk = 3x/4 litres)
Now
12-x liters from can 2 ( where, water = 6-x/2, milk = 6-x/2)
Now given
water/milk = 3/5
0.25x+6-0.5x / 0.75 +6-0.5x = 3/5
6-0.25x/6+0.25 = 3/5
Solve it you get
x = 6 litres from can1
Then 12-x= 12-6=6 litres from can2
B
I will first need to understand why he/she behaves so to establish the exact cause and try persuade him to appreciate roles and responsibilities of the various Team members in trying to achieve organizational goals.
Ans 635.5
4000
1 year 5 % 200 rs int
Total 4200
2 year 5 % 210 ra int
Total 4410
4410-2210 = 2200
1 year 5% 110
Total 2310
2 year 5 % 115.5 rs int
Total 2425.5
Total interest paid
200 +210+ 110+115.5. = 635.5
1×2×…100=100!
Number of zeros in product of n numbers =[5n]+[52n]+[53n]+…
Number of zeros in product of 100 numbers =[5100]+[52100]+[53100]
where [.] is greatest integer function
=[20]+[4]+[0.8]=20+4=24
1) time =20mins,speed=15kmph=15*(5/18)=25/6 m/sec
distance=20*(25/6)=250/3 m
then 2) given as time =15min
speed=250/3*15=(50/9)*(18/5)kmph=20kmph
Neice
Let the numbers be x and x + 2.
Then, (x + 2)2 – x2 = 84
⇒ 4x + 4 = 84
⇒ 4x = 80
⇒ x = 20.
∴ The required sum
= x + (x + 2)
= 2x + 2
= 42
73.
one year back father = 72 and son = 36.
4:3
A—–>B (first train)
B——>A(second train)
A/B=Srureroot((time to B)/(time to A))
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
The 3-digit number can be written as the sequence [n, 2n, 3n]
n = 1 → [1, 2, 3] → valid
n = 2 → [2, 4, 6] → valid
n = 3 → [3, 6, 9] → valid
n = 4 → [4, 8, 12] is not valid because this would lead to a 4-digit number
any value of n > 4 would also produce invalid answer
Answer: Three numbers: {123, 246, 369}
The numbers that lie between 100 and 1000 which are divisible by 14 are 112, 126,140 …,994
a = 112; l = 994, d = 14
n= (l−a)/d+1
= (994-112)/14+1
= 64
Sn=n/2(l+a)
= 64/2(994+112)
= 32*1106
= 35392
C. 4