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

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