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QUESTION: 1

A and B together has Rs. 1210. If 4/15 of A's amount is equal to 2/5 of B's amount, how much amount does B have?

Solution:

QUESTION: 2

Two numbers are 20% and 50% more than "the third number respectively". The ratio of the two numbers is:

Solution:

Let the third number be x.

Then, first number

= 120% of x

= 120x / 100

= 6x / 5

Second number

= 150% of x

= 150x / 100

= 3x / 2

∴ Ratio of first two numbers

= (6x / 5 : 3x / 2)

= 12x : 15x

= 4 : 5.

QUESTION: 3

A sum of money is to be distributed among A, B, C, D in the proportion of 5 : 2 : 4 : 3. If C gets Rs. 1000 more than D, what is B's share?

Solution:

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

⇒ 4*x* - 3*x* = 1000

⇒ *x* = 1000.

∴ B's share :

= Rs. 2*x*

= Rs. (2 x 1000)

= Rs. 2000.

QUESTION: 4

Seats for Mathematics, Physics and Biology in a school are in the ratio 5 : 7 : 8. There is a proposal to increase these seats by 40%, 50% and 75% respectively. What will be the ratio after increased seats?

Solution:

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

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

⇒ (140 / 100 x 5x), (150 / 100 x 7x) and (175 / 100 x 8x)

⇒ 7x, 21x / 2 and 14x.

∴ the required ratio

= 7x : 21x / 2 : 14x

⇒ 14x : 21x : 28x

⇒ 2 : 3 : 4.

QUESTION: 5

In a mixture of 60 litres, the ratio of milk and water is 2 : 1. If this ratio is to be 1 : 2, then the quanity of water to be further added is:

Solution:

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 x litres:

= (40 / 20 + x)

Then, milk : water

Now,

⇒ (40 / 20 + x) = 1 / 2

⇒ 20 + x = 80

⇒ x = 60.

∴ Quantity of water to be added = 60 litres.

QUESTION: 6

Find the ratio A : B : C : D If,

A : B = 3 : 5

B : C = 5 : 7

C : D = 9 : 11

Solution:

A : B = 3 : 5

B : C = 5 : 7

C : D = 9 : 11

► A : B : C : D = 3 x 5 x 9 : 5 x 5 x 9 : 5 x 7 x 9 : 5 x 7 x 11

5 can be taken away as it is common in all.

► 27 : 45 : 63 : 77

The required ratio A : B : C : D is 27 : 45 : 63 : 77.

QUESTION: 7

Find the ratio A : B : C : D : E if,

A : B = 4 : 5

B : C = 6 : 7

C : D = 9 : 10

D : E = 5 : 2

Solution:

A : B = 4 : 5

B : C = 6 : 7

C : D = 9 : 10

D : E = 5 : 2

► A : B : C : D : E = 4 x 6 x 9 x 5 : 5 x 6 x 9 x 5: 5 x 7 x 9 x 5: 5 x 7 x 10 x 5: 5 x 7 x 10 x 2

► 216 : 270 : 315 : 350 : 140

The required ratio A : B : C : D : E is 216 : 270 : 315 : 350 : 140

QUESTION: 8

If 4 : 8 : C are in continued proportion then find C?

Solution:

We know that if a, b and c are in continued proportion, then

► b^{2} = ac

► 8^{2} = 4.C

► C = 64/4 = 16

QUESTION: 9

11 : b : 44 are in continued proportion. Find b.

Solution:

We know that if a, b and c are in continued proportion then,

► b^{2} = ac

► b^{2} = 11 * 44

► b^{2} = 484

► b = 22

QUESTION: 10

The annual income of Victor and Angela are in the ratio 8 : 3 and their annual expenditures are in the ratio 4 : 1. If each save Rs. 2000 per annum. What is the annual expense of Angela?

Solution:

Let the annual income of Victor be 8x. The annual income of Angela will therefore be 3x.

Further, let the expenditure of Victor be 4y. The annual expenditure of Angela will be y.

∴ 8x-4y= 2000 .......(1)and

3x-y=2000.........(2)

Solving above two equations, we get:

► 5x= 3y

By putting this value in equation 2, we get:

► 3x-5x/3=2000

Hence, x=1500

y=2500 (Angela’s Expeniture)

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