Assume that the daily wages of man, women and daughter are Rs 5x, Rs. 4x, Rs 3x respectively.
Multiply (no. of days) with (assumed daily wage) of each person to calculate the value of x.
[3 x (5x)] + [2 x (4x)] + [4 x (3x)] = 105
[15x + 8x + 12x] = 105
35x = 105
x = 3
Hence, man’s daily wage = 5x = 5 x 3 = Rs. 15
Wife’s daily wage = 4x = 4 x 3 = Rs. 12
Daughter’s daily wage = 3x = 3 x 3 = Rs. 9

Let the value of one rupee, 50 paise and 25 paise be 11x, 9x, 5x respectively.

No. of 1 rupee coins = (11x / 1) =11x
No. of 50 paise coins = (9x / 0.5) = 18x
No. of 25 paise coins = (5x / 0.25) = 20x

11x + 18x + 9x = 342
38x = 342
x = 9

Therefore, no. of 1 rupee coins = 11 x 9 = 99 coins
No. of 50 paise coins = 18 x 9 = 162 coins
No. of 25 paise coins = 20 x 9 = 180 coins

Let salary of
A = 2k,
B = 3k,
C = 5k

=>A’s new salary = 115/100 of 2k = 115/100 x 2k = 23k/13

=>B’s new salary = 110/100 of 3k = 110/100 x 3k = 33k/10

=>C’s new salary = 120/100 of 5k = 120/100 x 5k = 6k

Hence,new ratio = 23k/13 : 33k/10 : 6k = 23 : 33 : 60

Milk:Water (Initial Ratio) = 5:1 i.e. 5 parts of milk out of 6 (5 + 1) parts of the mixture

Milk:Water (Final Ratio) = 5:2 i.e. 5 parts of milk out of 7 (5 + 2) parts of the mixMilkture

On comparing, 1 part of the mixture has been added (5 litres of water is added). So, 1 part of the mixture = 5 litres

Quantity of milk in the original mixture = 5 parts i.e. 5*5 = 25 litres

Let the share of c be X
then b gets 1/4 of c
so
B = X/4
& a gets two third of B
then
A = 2/3*X/4 = X/6

Now

=> X +X/4 +X/6 = 680

=> take LCM

=>( 12x+3x+2x)/12 = 680

=> 17x = 680×12

=> X = 680*12/15

=> X = 40*12

=> X = 480

so

c got = 480Rs

B got = 480/4 = 120 Rs

A got = 480/6 = 80 Rs

Let the number of 25 p, 10 p and 5 p coins be x, 2x, 3x respectively.

Then, sum of their values = Rs.		25x	+	10 x 2x	+	5 x 3x		= Rs.	60x
		100		100		100			100

	60x	= 30	x =	30 x 100	= 50.
	100			60

Hence, the number of 5 p coins = (3 x 50) = 150.

Let the three parts be A, B, C. Then,

A : B = 2 : 3 and B : C = 5 : 8 =		5 x	3		:		8 x	3		= 3 :	24
			5					5			5

A : B : C = 2 : 3 :	24	= 10 : 15 : 24
	5

B =

98 x

15

= 30.

Quantity of milk =		60 x	2	litres = 40 litres.
			3

Quantity of water in it = (60- 40) litres = 20 litres.

New ratio = 1 : 2

Let quantity of water to be added further be x litres.

Then, milk : water =		40		.
		20 + x

Now,		40		=	1
		20 + x			2

 20 + x = 80

 x = 60.

 Quantity of water to be added = 60 litres.

A                   B                 C

3x                 4x                5x

(3x+x)          2x               (5x+x)         (B Gives 1/4 to A and 1/4 to C)

=4x              2x                 6x

(4x+x)          2x                 5x              ( C Gives 1/6 to A)

=5x              2x                 5x

Therefore, 5 : 2 : 5

Let the two numbers be x and y

Given x : y = 3 : 7 …..(1)

Now, x+6 : y+6 = 5 : 9 …..(2)

From (1), x = 3y/7

From (2), 5y – 9x = 24

=> 5y – 9(3y/7) = 24

=> y = 21

=> From(1), x = 9

Hence, the two numbers be 9 and 21

Angle of triangle = 180 × 3/12

= 45 deg

Other angles of the triangle are = 60 deg, 75 deg

Angle of quadrilateral

= 360 – (45 + 60 + 75)

= 180 deg

Therefore, the four angles of the quadrilateral are 45, 60, 75 & 180 degrees.

Hence, the sum of largest and smallest angle of the quadrilateral are 180 + 45 = 225 deg.

4:5
3:2
——-
=> 12:15:10

12:10

=> 6:5

Given total applicants = 135
Given graduates are G=60
Otherthan graduates=135-60 = 75
Given experienced candidates = 80
1) For maximum number of graduates have experience
Total graduates to have experience = 60

2) For minimum number of graduates have experience
Remaining after taking other than graduates in experience= 80-75 = 5

Given ratio of number of sheep and horses = 4 : 15
Let number of sheep and horses are 4k and 15k
Number of horses = 13200/110 = 120
Now 15K = 120 and k = 8
Number of sheep = 4k = (4 x 8) = 32

The incomes of A and B be 3P and 4P.

Expenditures = Income – Savings

(3P – 100) and (4P – 100)

The ratio of their expenditure = 1:2

(3P – 100):(4P – 100) = 1:2

2P = 100 => P = 50

Their incomes = 150, 200

The ratio of distribution should be:
21 x 15 : 15 x 35 : 35 x 21
= 315 : 525 : 735
= 3 : 5 : 7
The smallest share will be of worth:
3/15 x 52,500 = Rs. 10,500

Only option (B) satisfies the equation

Let the price of required variety = Rs. P/kg

Then, respective amounts were m kg, m kg and 2m kg

= 126m + 135m + 2pm = 153 x 4m

=> 2p = 351

p = 175.5 / kg

Let us say x boys and x girls joined the group.

(64 + x)/(40 + x) = 4/3

192 + 3x = 160 + 4x => x = 32

Number of members in the group = 64 + x + 40 + x

= 104 + 2x = 168.

Frequency of step of A:B:C = 5 : 6 : 7

But in terms of size of step, 6A = 7B = 8C

Therefore, Ratio of speeds of A, B and C = 5/6 : 6/7 : 7/8 = 140 : 144 : 147

If you double the sides of a cube, the ratio of the surface areas of the old and new cubes will be 1: 4. The ratio of the volumes of the old and new cubes will be 1: 8.

Weight is proportional to volume. So, If the first weighs 6 pounds, the second weighs 6×8 pounds =48.

The wages of labourers in a factory increases in the ratio 22:25 and there was a reduction in the number of labourers in the ratio 15:11. Find the original wage bill if the present bill is Rs 5000 ?
Ratio of increase of wages = 22:25
Ratio of decrease of labourers = 15:11
Compound ratio of wages of labourers = 22 x 15 : 25 x 11 = 330:275
Final bill = Rs. 5000
For 275 ratio wages = Rs. 5000
For 1 ratio wages = 5000/275
For 330 ratio wages = 5000/275 x 330 = Rs. 6000

Minimum number of chocolates are possible when he purchases maximum number of costliest chocolates.

Thus,          2 x 5 + 5 x 2 =Rs.20

Now Rs.100 must be spend on 10 chocolates as 100 = 10 x 10.

Thus minumum number of chocolates = 5 + 2 + 10 = 17

Let the number of boys and girls be 8x and 5x.
Total number of students = 13x = 13 * 32 = 416.

Given ratio = 7 : 5 : 3 : 4, Sum of ratio terms = 19.

Smallest part =  (76*3/19)  = 12

3A = 4B = 7C = k,Then A = k/3, B = k/4 and C= k/7.

A : B : C = k/3 : k/4 : k/7 = 28:21 :12.

Cs share = Rs. [427 x (12/61)] = Rs. 84

Step (i): Let x be the number of boys and y be the number of girls.
Given total number of boys and girls = 100
x+y=100 ————– (i)

Step (ii): A boy gets Rs. 3.60 and a girl gets Rs. 2.40
The amount given to 100 boys and girls = Rs. 312
3.60x + 2.40y = 312 ————– (ii)

Step (iii):
Solving (i) and (ii)
3.60x + 3.60y = 360 ——— Multiply (i) by 3.60
3.60x + 2.40y = 312 ——— (ii)
1.20y = 48
y = 48 / 1.20
= 40

 Number of girls = 40

Compounded Ratio :: When we compound two or more ratio’s with each other through product or multiplication, the result is simply a compound ratio.

Thus, the product of two or more ratios; i.e, ab:cd is a ratio compounded of the simple ratios a:c and b:d.

Required compounded ratio = (2/3 x 6/11 x 11/2) = 2/1.