**1) Find the number of zeroes at the end of the product: 25 × 40 × 50 × 55 × 65 × 125?**

Solution:

The number of zeroes at the end of the product depends upon the combination of 2 × 5. Each pair of 2 and 5 will give us one zero. So we just have to calculate the number of pairs of 2 and 5 exist in the multiplication.

Prime factorization of each number:

25 = 5^{2}

40 = 2^{3} × 5

50 = 2 × 5^{2}

55 = 5 × 11

65 = 5 × 13

125 = 5^{3}

**25 × 40 × 50 × 55 × 65 × 125**

= 5^{2 }× 2^{3} × 5 x 2 × 5^{2} × 5 × 11 × 5 × 13 × 5^{3}

= 5^{10} × 2^{4} × 11 × 13 = (2 × 5)^{4} × 5^{6} × 11 × 13.

∴ The number of zeroes at the end of the product **=**** 4.**

**2) The profit earned when an article is sold for Rs 800 is 20 times the loss incurred when it is sold for Rs 275. At what price should the article be sold if it is desired to make a profit of 25%.**

**(a) Rs 300 (b) Rs 350 (c) Rs 375 (d) Rs 400**

Solution:

Let the Cost Price (C.P) of an article is Rs X.

Profit earned when it is sold for Rs 800 = Rs (800 – X)

Loss incurred when it is sold for Rs 275 = Rs (X – 275)

800 – X = 20 × (X – 275)

21 X = 6300

X = Rs 300

To make a profit of 25% the article must be sold at Rs (1.25 × 300) = **Rs 375.**

*Another Approach (With the help of options)*

If the article was sold at Rs 375 then to earn a profit of 25%, the cost price of the article must be Rs (100 × 375/125) = Rs 300

Profit earned when it is sold for Rs 800 = Rs (800 – 300) = Rs 500

Loss incurred when it is sold for Rs 275 = Rs (300 – 275) = Rs 25.

So, profit earned when it is sold for Rs 800 is 20 times the loss incurred when it is sold for Rs 275.

Therefore, **option (C)** is the correct answer.

**3) A person travels three equal distances at a speed of X km/hr, Y km/hr and Z km/hr respectively. What will be the average speed during the whole journey?**

Solution:

Let the total distance be “3D” km in which the 1^{st} D km person travels at a speed of X km/hr, next D km at a speed of Y km/hr & the remaining D km is with Z km/hr.

Time taken to cover 1^{st} D km is D/X hr.

Time taken to cover next D km is D/Y hr.

The remaining D km is covered in D/Z hr.

Average speed during the whole journey:

= Total Distance ÷ Total Time Taken

= 3D / (D/X + D/Y + D/Z)

= **3XYZ / (XY + YZ + ZX)**.

**4) The length of a rectangle is 12 cm more than its breadth. If the area of the rectangle is 448 sq cm. What is the perimeter of the rectangle?**

Solution:

Let ‘L’ & ‘B’ are the length & breadth of the rectangle (refer * Figure 1*)

L = 12 + B

Area of rectangle = L × B = 448

(12 + B) × B = 448

B^{2} + 12B – 448 = 0

(B + 28) (B – 16) = 0 or, B = −28, 16.

Breadth can’t be negative, therefore** B** = 16 cm & Length **L** = 12 + 16 = 28 cm.

Perimeter of the rectangle = 2 × (L + B) = 2 × (28 + 16) = **88 cm.**

**5) If two cylinders of equal volume have their heights in the ratio 2:3, then the ratio of their radii is ___________?**

Solution:

Let the height of 1^{st }cylinder is ‘h’ & its radius is ‘r’.

Whereas, the height of 2^{nd} cylinder is ‘H’ & its radius is ‘R’.

Volume of 1^{st} Cylinder = Volume of 2^{nd} Cylinder.

⇒ Πr²h = ΠR²H

⇒ r²/R² = H/h

⇒ r/R = (H/h)½ (Given h: H = 2 : 3)

⇒ r/R = (3/2)½ = √3/√2

∴ Ratio of their radii is **√3 : √2.**

6**) The radius of circle is so increased that its circumference increased by 5%. The area of the circle then increases by ______________?**

**(a) 12.5% (b) 10.25% (c) 10.5% (d) 11.25%**

Solution:

Let the radius of the circle be ‘r’.

Circumference of the circle = 2Πr.

**Initial Area = Πr²**

So, when circumference is increased by 5%, it implies that the radius increases by 5%.

New radius = r + 5% × r = 1.05r.

**New area = Π × (1.05r) ^{ 2} = 1.1025Πr²**

Percentage increase in Area = {(1.1025Πr² − Πr²)/Πr²} × 100 = **10.25%.**

**7) The HCF & LCM of two numbers are 12 and 144 respectively. If one of the numbers is 36, the other number is?**

Solution:

Let the other number be X.

HCF of 36 & X is 12.

LCM of 36 & X is 144.

Product of numbers = HCF × LCM

36 × X = 12 × 144

X = 48

Therefore, the other number is **48.**

**8) A boat goes 40 km upstream in 8 hrs and a distance of 49 km downstream in 7 hrs. The speed of the boat in still water is _____________?**

Solution:

Let the speed of the boat in still water is u km/hr & the speed of the stream (or current) be v km/hr.

In water, the direction along the stream is called downstream. And, the direction against the stream is called upstream.

Speed Upstream = (u − v) km/hr = 40/8 = 5 km/hr.

Speed Downstream = (u + v) km/hr = 49/7 = 7 km/hr.

Speed of the boat in still water:

u = (Speed Upstream + Speed Downstream)/2

= (5 + 7)/2 = **6 km/hr**

**9) The speed of three motorbikes is in the ratio 2:3:4. The ratio between the times taken by these cars to travel the same distance is _______________?**

Solution:

Let the speed of motorbikes be 2x, 3x and 4x km/hr.

Suppose the distance travelled by each motorbike is ‘D’ km.

Ratio of time taken = D/2x : D/3x : D/4x = **6 : 4 : 3 **(Take LCM of 2, 3, 4 = 12)

**10) The length of a rope, to which a goat is tied, is increased from 19m to 30m. How much additional ground will it be able to graze? Assume that the goat is able to move on all sides with equal ease. (Use π = 22/7)**

Solution:

As the goat is able to move on all sides with equal ease, therefore the initial area that goat can graze = Πr² = Π × 19^{2 }m^{2}.

When the length of the rope is increased to 30m, the new grazing area = ΠR² = Π × 30² m²

Additional area the goat will able to graze (shaded region refer * Figure 2*)

= Π × (30² − 19²) = 22/7 × 11 × 49 = 154 × 11 = **1694 m2.**

If you have any doubt on the above questions. Please feel free to ask your doubt in the comment section.

Thank You, Readers!!

Believe You Can and You’re Halfway There.~Theodore Roosevelt

[…] Note: If you want to cross-check just put the value of X & Y in the above equation (1) & (2) to see whether L.H.S = R.H.S or not. And, if left-hand side value is equal to the right-hand side then the above obtained solution is correct. […]