Let's assume the sample space = S
and Event of selecting 1 girl and 2 boys = E
So, n(S) = Number ways of selecting 3 students out of 25 = 25C3
=> (25 * 24 * 23)/(3 * 2 * 1) = 2300.
n(E) = (10C1* 15C2)
= 10 * [(15 * 14)/(2 * 1)] = 1050.
P(E) = n(E)/n(S) = 1050/2300 = 21/46.
A's 1 day's work = x
and B's 1 day's work = y
So (A & B) 1 day work = 1/30 => x+y =1/30
=> 30x + 30y = 1 -------- (1)
So 16x + 44y = 1 -------- (2)
By Solving above two equations,
x = 1/60 and y = 1/60
B's 1 day's work = 1/60
Hence, B alone shall finish the whole work in 60 days.
Let total distance be S
total time=1hr24min
A to T :: speed=4kmph
diistance=2/3S
T to S :: speed=5km
distance=1-2/3S=1/3S
21/15 hr=2/3 S/4 + 1/3s /5
84=14/3S*3
S=84*3/14*3
= 6km.
Let they meet x hours after 7 a.m.
Distance covered by A in x hours = 20x km.
Distance covered by B in (x - 1) hours = 25(x - 1) km.
So Total Distance
=> 20x + 25(x - 1) = 110
=> 45x - 25 = 110 => 45x = 135
=> x = 3.
As They meet x hrs after 7 a.m. so they meet at 10 a.m.
Let C capital = x. so B capital = 4x
2 *(A capital) = 3 * 4x
=> A's capital = 6x
So A : B : C = 6x : 4x : x = 6 : 4 : 1
So, B's capital = Rs. [16500 * 4/11] = Rs. 6000.
Let the three parts be A, B, C. Then,
A : B = 2 : 3 and B : C = 5 : 8 = 5 * 3/5 : 8 * 3/5 = 3 : 24/5
=> A : B : C = 2 : 3 : 24/5 = 10 : 15 : 24
=> B = 98 x 15/49 = 30.
LCM of 11 and 17 = 187.
When divided by 11 remainder 7,so difference 4.
When devided by 17 remainder 13,so difference 4.
Largest no exactly devide by 11 & 17=9911
The no's = 9911-4 = 9907.
Let the length of the train be x meters and its speed by y m/sec.
Then, x/y = 8 => x = 8y ----------- (1)
As per the question total distance = (x + 264) meters.
(x + 264)/20 = y
Put the value of x from equation 1.
=> 8y + 264 = 20y
=> y = 22.
Therefore Speed = 22 m/sec = ( 22 x 18 /5) km/hr = 79.2 km/hr.
Average price of a sheep = Rs. 74
:Total price of 13 sheeps = (74*13) = Rs. 962
But, total price of 13 sheeps and 9 pigs
= Rs. 1291.85
Total price of 9 pigs
= Rs. (1291.85-962) = Rs. 329.85
Hence, average price of a pig
= (329.85/9) = Rs. 36.65.
(2x + 2/x) = 1
=> (x + 1/x) = 1/2
(x^3+1/x^3) = (x+1/x)^3-3x*1/x(x+1/x)
= (1/2)^3-1/2 = (1/8 - 3/2) = (1-12)/8= -11/8.
Time = Distance/Speed
Time taken for each 6 km can be given by
6/25, 6/50, 6/75 and 6/150
Total time = (6/25) + (6/50) + (6/75) + (6/150) = (36 + 18 + 12 + 6)/150 = 72/150
Average speed = Distance/time = (24/72) x 150 = 50 kmph.
x^3 + 3x^2 + 3x = 7
Adding both side 1
=> x^3 + 3x^2 + 3x + 1= 7 +1
=> (x+1)^3=2^3
=> x+1 = 2 => x =1.
Tank part filled by pipes (A+B+C) in 1 hrs = 1/6 ------- (1)
so tank part filled by (A+B+C) in 2 hrs = 2*1/6 = 1/3
Now find the remaining part = (1-1/3) = 2/3
=> (A+B) 7 hs work = 2/3
so (A+B) 1 hrs work = 2/21 ------ (2)
To find the C 1 hrs work use eq. 1 & 2
=> 1/6-2/21 = 1/14
so C alone can fill the tank in 14 hrs.
Let's assume mother age = x years. ----- (1)
sum of mother and her son age = 45
so son age will be = (45-x) years. ------- (2)
Five year ago:
mother age will be = (x-5) years
son age will be = (45-x-x) years = (40-x) year.
As per question
(x-5) * (40-x) = 3*(x-5)
=> (40-x) = 3
=> x = 37 year.
so son age will be (45-37) = 8 years.
Let the period is 'T' and Sum= 'P'.
As given money become 2.5.
=> 2.5 * P = P + S.I
=> S.I = 1.5 * P -------------- (1)
=> S.I = (P * T * 12.5)/100 -----------(2)
By eq. (1) and (2)
=> 1.5 * P = (P * T * 12.5)/100
=> T=12 years.
Let total income is X
X=20,000+(X-20,000/2)+6000
X-X/2=20,000-10,000+6000
X/2=16,000
X=32,000.
(A + B)'s 1 day's work = 1/30
so (A&B) 20 days work = (20*1/30) = 2/3
so left work = (1?2/3)=1/3
1/3 work is done by A = 20 days.
So whole work will be done by A = (20 x 3) = 60 days.
Quick Approach
for x =1 given eq. will be satisfy. (1+1/1)=2
so (x^100 + 1/x^100) = (1^100 + 1/1^100) = 2.
A alone one day work = 1/8
B alone one day work = 1/12
Both A and B one day work = (1/8 + 1/12) = (3/24 + 2/24)
= 5/24
so A and B together finish the work in 24/5 day
or 4 4/5 days.
Time taken by tap A to fill the cistern=4 hrs
so work done by tap A in 1 hour = 1/4th
Time taken by tap B to empty the full cistern = 9 hours
so work done by tap B in 1 hour = 1/9th
=> Work done by (A + B) in 1 hour=(1/4 - 1/9)=5/36
Therefore, the tank will fill the cistern = 36/5 hours=7.2 hours.
Let's Assume the distance travelled Albert = x km.
Formula Used: Time = Distance/Speed
so (x/10 - x/15) = 2 hrs
=> (3x - 2x)/30 = 2 hrs.
=> (3x - 2x) = 60
x = 60 km.
Time taken to travel 60 km at 10 km/hr = 60/10 hrs = 6 hrs.
Formula Used: Speed = Distance/Time
So, Albert started 6 hours before 2 P.M. i.e @ 8 A.M.
Required speed = 60/5 kmph. = 12 kmph.
Lets assume time required by slower pipe alone to fill the tank = x minutes.
Then, faster pipe will fill it in x/3 minutes.
=> 1/x+3/x = 1/36
=>4/x = 1/36 => x = 144 min.
Speed = (200/24) m/sec = 25/3 m/sec
convert m/sec to km/hr
(25/3 * 18/5) km/hr = 30 km/hr.
Lets assume the required distance = x km.
Difference in the times taken at two speeds=12mins=1/5 hr.
Therefore (x/5-x/6)=1/5 or (6x-5x) = 6 or x = 6km.
So required distance = 6 km.
Product of two no. = H.C.F*L.C.M
So,x*y=135*9=1215 -----(1)
and x^2+y^2=2754
So,(x+y)^2=x^2+y^2+2*x*y = 2754+2*1215=5184
So,x+y=72 ----------- (2)
By solving eq. (1) & (2)
nos. are 45 and 27.
Total distance travelled in 1 hour
= 50 + (&frac; x 32) km
= 66 km
? Average speed = Total distance/Total time
= 66/(3/2) = 44 kmph.
7 years ago, A's age=4x years
and B's age=5xyears
so (4x+14)/(5x+l4)=5/6
=> 25x + 70 = 24x + 84
x = (84 - 70) = 14
B's present age = 5x + 7 = 5*14 + 7 = 77 years.
Man's downstream speed = 72/ 9 kmph => 8kmph
Man's up stream speed = 45/ 9 => 5 kmph
So speed of current = (8 - 5)/2 = 1.5 kmph.
3/4 of his share = 75000
so his share = 100000.
2/3 of business value = 100000
so total value = 150000.
Formula Used: Time = ( Distance / Speed)
As all the option given in sec., so convert the train speed (Kmph) in to mps multiply by 5/18
speed (mps) = 73 * 5/18
Time = 600 / (73 * 5/18)
= (600 * 18 )/(73 * 5) sec
= (10800 / 365)
Time take by Train = 29.58Sec.
Tank part filled by pipe A in 1 hour =1/9
Tank part filled by pipe B in 1 hour =1/3
Given Pipe A and B are opened alternatively.
So Part filled in every 2 hours =(1/9+1/3)=4/9
Tank Part will be filled in 4 hour =2*4/9=8/9
Remaining part = (1-8/9)=1/9
So next is A turn.
So Pipe A will fill remaining 1/9 part in next 1 hour.
Total Time = (4 hrs + 1 hrs) = 5 hrs.
Man's row in downstream by speed = 6 kmph
and upstream by speed = 4 kmph
so man rate = (6 - 4)/2 = 1 kmph.
Trains are running in opposite Direction:
So need to find Length of two Trains = 300m + 400m = 700m
and Total Speed = 40 Kmph + 50 Kmph (Opposite Direction)
= 90 Kmph
so speed (m/sec) = 90 * 5/18 m/sec = 25 m/sec
Formula Used: Time = Distance/Speed
Time = 700/ 25 sec
Time = 28 Sec.
As train are running in same direction
so Relative speed = (40 - 20) km/hr = 20 km/hr
= ( 20 x 5/18 ) m/sec = 50/9 m/sec.
Formula Used: Distance = Speed * Time
Now Length of Faster Train = ( 50/9 x 5 ) m = 250/9 m
= 27 7/9 m.
First find the 1 day work of both (A & B)
A's 1 day's work = 1/18
and
B's 1 day's work = 1/9 (B can do work in half time)
(A + B)'s 1 day's work = (1/18+1/9)
= (1+2)/18 = 3/18 = 1/6
so A & B together can do 1/6 of work in 1 day.
Split the series as below
4,2,5 | 9,5,11 | 13,7,16 | 17,9,?4,2,5 -> diff b/w 4,5 = 1
9,5,11 -> diff b/w 9,11 = 2
13,7,16 -> diff b/w 13,16 = 3
17,9,21 -> diff b/w 17,21 = 4.
Let the average for 17 innings is x runs
Total runs in 17 innings = 17x
Total runs in 18 innings = 17x + 150
Average of 18 innings = 17x + 150/18
17x + 150/18 = x + 6 -- > x = 42
Thus, average after 18 innings = 42.
man's upstream speed = 25 kmph
Man's downstream speed = 35 kmph
so Speed of current = (35 - 25)/2 = 5 kmph.
Time taken by one tap to fill half tank = 3hrs.
Part filled by the four taps in 1 hrs = (4*1/6) = 2/3.
Remaining part = (1-1/2) = 1/2.
so 2/3 : 1/2 :: 1:x => x = 1/2*1*3/2) = 3/4 hrs. => 45 min.
=> Total time taken = 3 hrs 45 min.
Let's assume time required by Pipe A to fill the cistern = X hours
So Time required by Pipe B to fill the cistern = (X + 6) hours
? Both Pipes (A+B) can fill cistern in 1 hour = [1/X + 1/(X + 6)]
Given Both pipe fill the cistern in 4 hours
=> [1/X + 1/(X + 6)] = 1/4 => [(X+6) + X]/(X+6)*x = 1/4
4X + 24 + 4X = X2 + 6x
X2 - 2X - 24 = 0
(X-6)(X+4) = 0
=> A can fill cistern in 6 hours.
Given sum of ages is 360 years.
The ratio of children and parents ages is 2:1.
So total age of parents = 360 x 1 / 3 = 120 years
Given Ratio of wife and husband age is 5:7.
So the age of husband = 120 x 7
Let's assume Monit's present age = X years.
So father's present age = (X + 3X) years = 4X years.
After 8 years.
(4X + 8) =5/2 * (X + 8)
=> 8X + 16 = 5X + 40
=> 3X = 24 so, X=8
Hence, required ratio = (4X + 16) / (X + 16) = 48/24 = 2.
A & B one day work = 1/8
A alone one day work = 1/12
B alone one day work = (1/8 - 1/12) = ( 3/24 - 2/24)
=> B one day work = 1/24
so B can complete the work in 24 days.
Formula Used: Time = ( Distance / Speed)
As all the option given in sec., so convert the train speed (Kmph) in to mps multiply by 5/18
speed (mps) = 72 * 5/18 = 20 mps
Time = (120 / 20) sec = 6 sec
Time take by Train = 6 Sec.