1*1=1, 1+2=3, 3*3=9, 9+2=11, 11*11=121, 121+2=123
Let S be the sample space.
Then, n(S) = 52C2 = 52 * 51/ (2*1)= 1326.
Let E = event of getting 1 spade and 1 heart. n(E) = number of ways of choosing 1 spade out of 13 and 1 heart out of 13 = (13C1 x 13C1)
= (13 x 13) = 16@P(E) =n (E)/n(S) = 169/ 1326= 13/102
Let the numbers be 3x and 5x.
Then ,(3x-9)/(5x-9)=12/13 23(3x – 9) = 12(5x – 9) 9x = 99 x = 11.
The smaller number = (3 x 11) = 33.
Let A = 2k, B = 3k and C = 5k.
A’s new salary = 115/ 100 of 2k =[115/ 100 x 2k]= 23k/ 10
B’s new salary = 110/ 100 of 3k = [110/ 100x 3k] = 33k/ 10
C’s new salary = 120/ 100 of 5k = [120/ 100 x 5k] = 6k
New ratio = 23k : 33k : 6k = 23 : 33 : 60
CP* (76/100) = 912 => CP = 912 * 100/76
CP= 12 * 100
=> CP = 1200
cost price of article = Rs. 1200
Clearly, there are 52 cards, out of which there are 12 face cards. P (getting a face card)=12/52=3/13
Original fraction = (x - 4)/x
In case II,
8(x - 4 - 2) = x + 1
⇒ 8x - 48 = x + 1
⇒ 7x = 49 ⇒ x = 7
∴Original fraction
= (7 - 4)/7 = 3/7
Given A/p= 4/3 Also A = 26 after 6 years,
so his present age = 20years,
Substituting we get P = 15 years.
speed of the faster train = 2x m/sec.
Relative speed of train = (x + 2x) m/sec = 3x m/sec.
Total distance = (100 + 100)m = 200m
3x = 200/8
=> 24x = 200 => x = 25/3
So speed of the faster train = 2 * 25/3 m/sec
= 50/3 m/sec
= 50/3 * 18/5 = 60 km/hr.
Let S be the sample space. Then, n(S)= number of ways of drawing 3 balls out of 15 = 15C3 = (15 x 14 x 13)/ (3 x 2 x 1) = 455.
Let E = event of getting all the 3 red balls. n(E) = 5C3 = 5C2 = (5 x 4)/ (2 x 1)= 10.
P(E) = n(E) / n(S)= 10/ 455 = 2/ 91
Here, the first divisor (289) is a multiple of second divisor (17)
∴ Required remainder = Remainder obtained on dividing 18 by 17 = 1
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.
X one day work = 1/20
y one day work = 1/12
work done by x in 4 days = 4 * 1/20 = 1/5
left work = (1-1/5) = 4/5
x and y one day work = (1/20 + 1/12) = 8/60 = 2/15
=> time required to do 2/15 part of work by x and y = 1 day
so for whole work = 1/(2/15) = 15/2
so for 4/5 part of work x and y will take =( 4/5*15/2 ) = 6 days.
=> How long did the work last = 4 day + 6 day = 10 days.
As Theft is discovered at 3:00pm but Thief stole the car at 2:30.
This me thief covered some distance in this 30 min gap.
Distance travelled by thief in 30 min = 60 * 1/2 = 30 km
Owner Discovered Car at 3:00pm
Now relative speed = (75-60)km/hr = 15km/hr
Time needed to travel 30km by the speed of 15km/hr.
Time at which owner meets thief = 30/15 = 2 hrs
So After 2 hrs (i.e,at 5:00 pm) the owner will catch/overtake the thief
X’s profit : Y’s profit
= 700 × 3 + 500 × 3 + 620 × 6 : 600 × 12
= 2,100 + 1,500 + 3,720 : 7,200
= 7,320 : 7,200
= 61 : 60
X’s share in the profit = 61/(60+61) × 726 = 366
Let the incomes of Chanda and Kim be 9x and expenditures be 7y and 3y respectively.
Since = Income – Expenditure,
we get 9x – 7y = 2000 and 4x – 3y = 2000.
Solving, we get, x = 8000 and y = 10000.
So Chanda’s expenditure = 7y = 7 × 10000 = Rs. 70,000.
Let the numbers be x and y.
Then, xy = 9375 and x/y=15 Xy/(x/y)=9375/15 y2 = 62@y = 2@x = 15y = (15 x 25) = 375.
Sum of the numbers = x + y = 375 + 25 = 400.
x can do 1/4 of work in = 10 days
so x can do whole work in = (10 x 4) = 40 days.
Y can do (40% or 40/100)of work in = 40 days
so Whole work can be done by Y = (40x100/40)= 100 days.
Z can do 1/3 of work in = 13 days
Whole work will be done by Z in (13 x 3) = 39 days.
so compare x , y ,z work compare = y > x > z
so Z can complete the work first.
Let required speed be x.
So,187.5/{ (x+50)*5/18} =9
(A & B)'s 1 day work = 1/30
(B & C)'s 1 day work = 1/24
(C & A)'s 1 day work = 1/20
so 2 (A + B + C)'s 1 day's work = (1/30+1/24+1/20) = 15/120 = 1/8
=> (A + B + C)'s 1 day's work = 1/16
Work done by A, B and C in 10 days = (10*1/16) = 5/8
so left work = (1?5/8)=3/8
A's 1 day's work (1/16?1/24)=1/48
=> 1/48 part of work is done by A = 1 day.
So, 3/8 part of work will be done by A = (48?3/8) = 6*3 = 18 days.
Let the fourth proportional to 5, 8, 15 be x.
Then, 5 : 8 : 15 : x 5x = (8 x 15) x=(8*15)/5=24
First find the 1 day work of both (A & B)
A 1 day's work = 1/10
and
B 1 day's work = 1/15
So (A + B) 1 day's work = (1/10+1/15)
= (3/30+2/30) = 5/30 = 1/6
So Both (A & B) together can finish work in 6 days
Explanation:
Method: 1
Let's assume Ravi salary = 100
It get reduced by 25% => Salary = 75
75(1 + P/100) = 100
1+ P/100 = 4/3
P = 100/3 = 33 1/3%.
Method: 2
You can use directly formula i.e
[(R*100)/(100-R)]% Where 'R' is decresed %
so put 25 at place of 'R'
=> [(25 * 100)/(100 - 25)]% => [(25 *100)/75]%
=>100/3% = 33 1/3%
Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.
Then, sum of their values=Rs[(25x/100)+(10*2x)/100+(5*3x)/100]=Rs 60x/100 60x/100=30
x=(30*100)/60=50 Hence, the number of 5 p coins = (3 x 50) = 150.
The logic is ×1+1, ×2+4, ×3+9, ×4+16, ×5+25,….
So following the logic we get 178 is wrong instead it should be 172.
(4x - 3)/x + (4y - 3)/y + (4z - 3)/z = 0
=> 4x/x - 3/x + 4y/y - 3/y + 4z/z - 3/z = 0
=> 3/x + 3/y + 3/z = 4 + 4 + 4 = 12
=> 1/x + 1/y + 1/z = 12/3 = 4
@Given fresh fruit has 68% water,
=> Remaining 32% will be fruit content.
@Given Dry fruit has 20% water
=> Remaining 80% is fruit content.
Here assume weight of dry fruit = x kg.
"fruit content in both the fresh fruit and dry fruit is the same"
Fruit % in fresh-fruit = fruit% in dry-fruit
so (32/100) * 100 = (80/100 )* x
=> x = 40 kg.
Find one day work for all three
(A+B)'s 1 day work = 1/20 ----(1)
(B+C)'s 1 day work = 1/30 ----(2)
and (C+A)'s 1 day work = 1/40 ----(3)
2(A+B+C) 1 day work = (1/20 + 1/30 + 1/40)
=> (A+B+C) = (6+4+3)/2*120
=> (A+B+C) = 13/240 -----------(4)
By eq. (2) and (4)
A + 1/30 = 13/240
=> A = 13/240 - 1/30 = (13-8)/240 = 1/48
then A's 1 day work = 1/48
so A alon can finish the job = 48 days