Mix Test 4

The diameter of the silver coin is 2r = 1.75 cm

Therefore, the radius of the silver coin is r = $\frac{1.75}{2}$ = 0.875 cm.

The area of the silver coin is $\pi r^2$ = $0.875\times 0.875 \pi$ = $0.765625 \pi \,cm^2$ = $2.40625\,cm^2$

Since, given that the thickness of the coin is d = 2 mm = 0.2 cm

Therefore, the volume of silver coin = Area of the coin $\times$ thickness of the coin

= 2.40625 $\times$ 0.2 = $0.48125\, cm^3.$

Since, we are forming a cuboid of dimensions 5.5 cm$\times$10 cm $\times$ 3.5 cm from n number of silver coins.

Therefore, the volume of the cuboid = n $\times$ the volume of a silver coin.

$\Rightarrow$ $5.5\times 10\times 3.5$ = 0.48125 n ($\because$ volume of the cuboid = lbh)

$\Rightarrow$ n = $\frac{5.5\times 10\times 3.5}{0.48125}$ = $\frac{192.5}{0.48125}$ = 400.

Hence, total 400 silver coins are required to forming cuboid of given dimension.

Let one son is x years old and other son is y years old,

Given that man’s age is 3 times the sum of ages of his two sons.

Therefore, the man’s age = 3(x + y). ..........(1)

After 5 years the man’s age is 3(x + y) + 5.

But given that after 5 years, the man’s age will be twice of the ages of his two sons. After 5 years the age of first son is x + 5 and the age of second son is y + 5.

Therefore, 3(x + y) + 5 = 2((x+5) + (y+5))

$\Rightarrow$ 3(x + y) + 5 =2(x + y + 10)= 2(x + y) +20

$\Rightarrow$ x + y = 20 – 5 = 15.

$\Rightarrow$ x + y = 15.

Therefore, the man’s age = 3(x + y) = 3 × 15 = 45 years. [From equation (1)]

$\frac{\mathrm x+2}{y+2}=\frac{1}{3}$

$\Rightarrow$ 3(x +2) = y + 2

$\Rightarrow$ 3x + 6 = y + 2

$\Rightarrow$ 3x – y + 4 = 0

$\Rightarrow$ 3x – y = – 4.

$\frac{\mathrm x+3}{y+3}$ = $\frac{2}{3}$

$\Rightarrow$ 5(x+3) = 2(y+3)

$\Rightarrow$ 5x + 15 = 2y + 6

$\Rightarrow$ 5x – 2y + 9 = 0

$\Rightarrow$ 5x – 2y = – 9.

Given that if 2 is added to both of numerator and denominator of a fraction, it becomes $\frac{1}{3}$.

$\therefore$ $\frac{\mathrm x+2}{y+2}=\frac{1}{3}$

$\Rightarrow$ 3(x +2) = y + 2

$\Rightarrow$ 3x + 6 = y + 2

$\Rightarrow$ 3x – y + 4 = 0

$\Rightarrow$ 3x – y = – 4. ..........(1)

Given that when 3 is added to both numerator and denominator of same fraction, it becomes $\frac{2}{5}$.

$\therefore$ $\frac{\mathrm x+3}{y+3}=\frac{2}{5}$

$\Rightarrow$ 5(x+3) = 2(y+3)

$\Rightarrow$ 5x + 15 = 2y + 6

$\Rightarrow$ 5x – 2y + 9 = 0

$\Rightarrow$ 5x – 2y = – 9.   .........(2)

Now, multiplying equation (1) by 2, we get

6x – 2y = – 8. ............(3)

Now, subtracting equation (2) from equation (3), we get (6x – 2y) – (5x – 2y) = – 8 –(– 9)

$\Rightarrow$ 6x – 5x – 2y + 2y = – 8 + 9

$\Rightarrow$ x = 1

Hence, the value of x is 1.

Given that if 2 is added to both of numerator and denominator of a fraction, it becomes $\frac{1}{3}$.

$\therefore$ $\frac{\mathrm x+2}{y+2}=\frac{1}{3}$

$\Rightarrow$ 3(x +2) = y + 2

$\Rightarrow$ 3x + 6 = y + 2

$\Rightarrow$ 3x – y + 4 = 0

$\Rightarrow$ 3x – y = – 4. ..........(1)

Given that when 3 is added to both numerator and denominator of same fraction, it becomes $\frac{2}{5}$

$\therefore$ $\frac{\mathrm x+3}{y+3}=\frac{2}{5}$

$\Rightarrow$ 5(x+3) = 2(y+3)

$\Rightarrow$ 5x + 15 = 2y + 6

$\Rightarrow$ 5x – 2y + 9 = 0

$\Rightarrow$ 5x – 2y = – 9. ...............(2)

Now, multiplying equation (1) by 2, we get

6x – 2y = – 8. ...........(3)

Now, subtracting equation (2) from equation (3), we get (6x – 2y) – (5x – 2y) = – 8 –(– 9)

$\Rightarrow$ 6x – 5x – 2y + 2y = – 8 + 9

$\Rightarrow$ x = 1.

By putting x = 1 in equation (1), we get 3 × 1 – y = – 4

$\Rightarrow$ 3 – y = – 4

$\Rightarrow$ y = 3 + 4 = 7.

Hence, the value of y is 7

Given that if 2 is added to both of numerator and denominator of a fraction, it becomes $\frac{1}{3}$.

$\therefore$ $\frac{\mathrm x+2}{y+2}$ = $\frac{1}{3}$

$\Rightarrow$ 3(x +2) = y + 2

$\Rightarrow$ 3x + 6 = y + 2

$\Rightarrow$ 3x – y + 4 = 0

$\Rightarrow$ 3x – y = – 4..... ... (1)

Given that when 3 is added to both numerator and denominator of same fraction, it becomes $\frac{2}{5}$.

$\therefore$ $\frac{\mathrm x+3}{y+3}=\frac{2}{5}$

$\Rightarrow$ 5(x+3) = 2(y+3)

$\Rightarrow$ 5x + 15 = 2y + 6

$\Rightarrow$ 5x – 2y + 9 = 0

$\Rightarrow$ 5x – 2y = – 9.  ...........(2)

Now, multiplying equation (1) by 2, we get

6x – 2y = – 8. ...........(3)

Now, subtracting equation (2) from equation (3), we get (6x – 2y) – (5x – 2y) = – 8 –(– 9)

$\Rightarrow$ 6x – 5x – 2y + 2y = – 8 + 9

$\Rightarrow$ x = 1.

By putting x = 1 in equation (1), we get 3 $\times$ 1 – y = – 4

$\Rightarrow$ 3 – y = – 4

$\Rightarrow$ y = 3 + 4 = 7.

Hence, the required (original) fraction is $\frac{\mathrm x}{y}=\frac{1}{7}$.

We know that the ratio of areas of two similar triangles is the square of the ratio of corresponding sides of these similar triangles.

Let the area of the larger triangle is $A\, cm^2.$

Given that the area of the smaller triangle is $48 \,cm^2$ and their corresponding sides are in the ratio 2 : 3.

Therefore, $\frac{48}{A}=\left(\frac{2}{3}\right)^2$

(The ratio of Areas of similar triangles is the square of the ratio of their corresponding sides.)

$\Rightarrow$ A = $48\times \left(\frac{3}{2}\right)^2$

$\Rightarrow$ A = $48\times \frac{9}{4}$ = $12\times 9$ = 108.

Hence, the area of the larger triangle is $108\,cm^2$.