A to Z Exams

5 : 1

Say A : B = 12 : 30

First, we need to find third proportion

A : B = B : C => 12 : 30 = 30 : C

=> C = 75

Next, we need to find mean proportion of 9 and 25

So, Mean Proportion (D) = √ 9 x 25 = 15

We need to find C : D

=> 75 : 15 = 5 : 1

160

Given that, Total amount = Rs.206/-

50p : 20p : 10p = 5 : 9 : 4

As total amount is in rupees convert the paise into rupees.

Say Z is the ratio constant

=> 5Z x 0.5 + 9Z x 0.25 + 4Z x 0.1 = 206

=> Z = 40

The number of 10p coins = 4Z = 4 x 40 = 160

3 : 2

Given that, Total students = 100

Girls = 40

=> Boys = 100 - 40 = 60

Ratio of number of boys to number of girls is

=> 60 : 40 = 3 : 2

196

Ratio of Science and Arts  students = 4:3

Say, the ratio constant is Z, so that the ratio = 4Z:3Z

Given, 14 Science students shifted to Arts

=> (4Z - 14) : (3Z + 14) = 1 :1

=> Z = 28

Total Students = 4Z + 3Z = 196

1.2

Given that 0.75 : Z : : 5 : 8

=> Z x 5 = 0.75 x 8

=> Z = 6/5

=> Z = 1.2

4 : 5

Let the third number be z

First number  =  120% of z = 120z/100 = 6z/5

Second number = 150% of z = 150z/100 = 3z/2

Ratio of these two numbers = 6z/5 : 3z/2 = 12z : 15z = 4 : 5

Rs. 484

(4/15) A = (2/5) B

=> A = [ (2/5) x (15/4) ] B

=> A = (3/2) B

=> A/B = 3/2

=> A : B = 3 : 2

=> B's share = Rs. [ 1210 x (2/5) ] = Rs. 484

Rs. 2000

Let the shares of A, B, C and D be Rs. 5x, Rs. 2x, Rs. 4x and Rs. 3x respectively

Then, 4x - 3x = 1000

=> x = 1000

=> B's share = Rs. 2x = (2 X 1000) = Rs. 2000

2 : 3 : 4

Originally, let the number of seats for Mathematics, Physics and Biology be 5z, 7z and 8z respectively

Number of increased seats are (140% of 5z), (150% of 7z) and (175% of 8z)

=> [ (140/100) x (5z) ] , [ (150/100) x (7z) ] and [ (175/100) x (8z) ]

=> 7z, (21z/2) and 14z

=> The required ratio = 7z : (21z/2) : 14z

=> 2 : 3 : 4

60 litres

Quantity of milk = 60 x (2/3) litres = 40 litres

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

New ratio = 1 : 2

Let quantity of water to be added further be z litres

Then, milk : water = 40/(20+z)

Now, 40/(20+z) = 1/2

=> 20+z = 80

=> z = 60

=> Quantity of water to be added = 60 litres

21 : 22

Originally, let the number of boys and girls in the college be 7z and 8z respectively.

Their increased number is (120% of 7z) and (110% of 8z)

=> [ (120/100) x 7z ] and [ (110/100) x 8z ]

=> 42z/5 and 44z/5

=> The required ratio = ( 42z/5 : 44z/5 ) = 21 : 22

Rs. 38,000

Let the original salaries of Ravi and Sumit be Rs. 2z and Rs. 3z respectively

Then, (2z+4000) / (3z+4000) = 40 / 57

57(2z + 4000) = 40(3z + 4000)

6z = 68,000

3z = 34,000

Sumit's present salary = (3z + 4000) = Rs.(34000 + 4000) = Rs. 38,000

30

Let the three parts be A, B, C.

Then, A : B = 2 : 3 and B : C = 5 : 8 = (5 x (3/5)) : (8 x (3/5)) = 3 : 24/5

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

=> B = 98 x (15/49) = 30

Rs. 204

Given ratio =1/2 : 2/3 : 3/4 = 6 : 8 : 9

=> 1st part = Rs. ( 782 x (6/23) ) = Rs. 204

23 : 33 : 60

Let A = 2k, B = 3k and C = 5k

A's new salary = (115/100) x 2k = 23k/10

B's new salary = (110/100) x 3k = 33k/10

C's new salary = (120/100) x 5k = 6k

=> New Ratio = [ (23k/10) : (33k/10) : 6k ] = 23 : 33 : 60

5 : 3

Let 40% of A = (2/3) B

Then, (40A/100) = (2B/3)

=> 2A/5 = 2B/3

=> A/B = [ (2/3) x (5/2) ] = 5/3

=> A : B = 5 : 3

24

Let the fourth proportional to 5, 8, 15 be z

Then, 5 : 8 : 15 : z

=> 5z = (8 x 15)

=> z = 24

33

Let the numbers be 3z and 5z

Then, (3z-9) / (5z-9) = 12 / 23

=> 23 (3z-9) = 12 (5z-9)

=> z = 11

=> The smaller number = (3 x 11) = 33

150

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

Then, sum of their values = Rs. [ (25z/100) + ((10x2z)/100) + ((5x3z)/100) ] = Rs. (60z/100)

=> (60z/100) = 30

=> z = 50

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

15/19

4 : 15

10 years

Rs. 5,500

2

3 : 4

4 : 3

4 : 15

380.611

10:13

17/24