1. An error 2% in excess is made while measuring the side of a square. What is the percentage of error in the calculated area of the square? | |

A. 4.04 % | B. 2.02 % |

C. 4 % | D. 2 % |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

Error = 2% while measuring the side of a square.

Let the correct value of the side of the square = 100

$MF#%\text{Then the measured value = }100 \times \dfrac{(100 + 2)}{100} = 102 \text{ (∵ error 2% in excess)}$MF#%

Correct Value of the area of the square = 100 × 100 = 10000

Calculated Value of the area of the square = 102 × 102 = 10404

Error = 10404 - 10000 = 404

$MF#%\text{Percentage Error = }\dfrac{\text{Error}}{\text{Actual Value}}\times 100 = \dfrac{404}{10000}\times 100 = 4.04\%$MF#%

2. A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the park and rest of the park has been used as a lawn. The area of the lawn is 2109 sq. m. what is the width of the road? | |

A. 5 m | B. 4 m |

C. 2 m | D. 3 m |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

Please refer the diagram given above.

Area of the park = 60 × 40 = 2400 m^{2}

Given that area of the lawn = 2109 m^{2}

∴ Area of the cross roads = 2400 - 2109 = 291 m^{2}

Assume that the width of the cross roads = x

Then total area of the cross roads

= Area of road 1 + area of road 2 - (Common Area of the cross roads)

= 60x + 40x - x^{2}

(Let's look in detail how we got the total area of the cross roads as 60x + 40x - x^{2}

As shown in the diagram, area of the road 1 = 60x. This has the areas of the

parts 1,2 and 3 given in the diagram

Area of the road 2 = 40x. This has the parts 4, 5 and 6

You can see that there is an area which is intersecting (i.e. part 2 and part 5)

and the intersection area = x^{2}.

Since 60x + 40x covers the intersecting area (x^{2}) two times ( part 2 and part 5)

,we need to subtract the intersecting area of (x^{2}) once time to get the total area.

. Hence total area of the cross roads = 60x + 40x - x^{2})

Now, we have

Total areas of cross roads = 60x + 40x - x^{2}

But area of the cross roads = 291 m^{2}

Hence 60x + 40x - x^{2} = 291

=> 100x - x^{2} = 291

=> x^{2} - 100x + 291 = 0

=> (x - 97)(x - 3) = 0

=> x = 3 * (x can not be 97 as the park is only 60 m long and 40 m wide)*

3. A towel, when bleached, lost 20% of its length and 10% of its breadth. What is the percentage of decrease in area? | |

A. 30 % | B. 28 % |

C. 32 % | D. 26 % |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

---------------------------------------------------------

Solution 1

---------------------------------------------------------

Let original length = 100 and original breadth = 100

Then original area = 100 × 100 = 10000

$MF#%\begin{align}&\text{Lost 20% of length}\\

&\Rightarrow\text{New length = Original length}\times \dfrac{(100 - 20)}{100} = 100 \times \dfrac{80}{100} = 80\\\\\\

&\text{Lost 10% of breadth}\\

&\Rightarrow \text{New breadth= Original breadth}\times \dfrac{(100 - 10)}{100} = 100 \times \dfrac{90}{100} = 90\\\\

\end{align} $MF#%

New area = 80 × 90 = 7200

Decrease in area = Original Area - New Area = 10000 - 7200 = 2800

$MF#%\text{Percentage of decrease in area = }\dfrac{\text{Decrease in Area}}{\text{Original Area}} \times 100 = \dfrac{2800}{10000} \times 100 = 28\% $MF#%

--------------------------------------------------------- Solution 2 ---------------------------------------------------------

Let original length = l and original breadth = b

Then original area = lb

$MF#%\begin{align}&\text{Lost 20% of length}\\

&\Rightarrow\text{New length = Original length}\times \dfrac{(100 - 20)}{100} = l\times \dfrac{80}{100} = \dfrac{80l}{100}\\\\\\\\\\

&\text{Lost 10% of breadth}\\

&\Rightarrow \text{New breadth= Original breadth}\times \dfrac{(100 - 10)}{100} = b \times \dfrac{90}{100} = \dfrac{90b}{100}\\\\\\\\\\

&\text{New area =}\dfrac{80l}{100} \times \dfrac{90b}{100} = \dfrac{7200lb}{10000}= \dfrac{72lb}{100}\\\\

&\text{Decrease in area = Original Area - New Area = }lb - \dfrac{72lb}{100} = \dfrac{28lb}{100}\\\\\\

&\text{Percentage of decrease in area = }\dfrac{\text{Decrease in Area}}{\text{Original Area}} \times 100 \\

&= \dfrac{\left(\dfrac{28lb}{100}\right)}{lb} \times 100 = \dfrac{28lb \times 100}{100lb} = 28\%

\end{align} $MF#%

4. If the length of a rectangle is halved and its breadth is tripled, what is the percentage change in its area? | |

A. 25 % Increase | B. 25 % Decrease |

C. 50 % Decrease | D. 50 % Increase |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

---------------------------------------------------------

Solution 1

---------------------------------------------------------

Let original length = 100 and original breadth = 100

Then original area = 100 × 100 = 10000

$MF#%\begin{align}&\text{Length of the rectangle is halved}\\

&\Rightarrow\text{New length = }\dfrac{\text{Original length}}{2} = \dfrac{100}{2} = 50\\\\\\

&\text{breadth is tripled}\\

&\Rightarrow \text{New breadth= Original breadth}\times 3 = 100 \times 3 = 300\\\\

\end{align} $MF#%

New area = 50 × 300 = 15000

Increase in area = New Area - Original Area = 15000 - 10000= 5000

$MF#%\text{Percentage of Increase in area = }\dfrac{\text{Increase in Area}}{\text{Original Area}} \times 100 = \dfrac{5000}{10000} \times 100 = 50\% $MF#%

--------------------------------------------------------- Solution 2 ---------------------------------------------------------

Let original length = l and original breadth = b

Then original area = lb

$MF#%\begin{align}

&\text{Length of the rectangle is halved}\\

&\Rightarrow \text{New length = }\dfrac{\text{Original length}}{2} = \dfrac{l}{2}\\\\\\

&\text{breadth is tripled}\\

&\Rightarrow \text{New breadth = Original breadth}\times 3 = 3b \\\\

&\text{New area = }\dfrac{l}{2}\times 3b = \dfrac{3lb}{2}\\\\

&\text{Increase in area = New Area - Original Area = } \dfrac{3lb}{2} - lb = \dfrac{lb}{2}\\\\

&\text{Percentage of Increase in area = }\dfrac{\text{Increase in Area}}{\text{Original Area}} \times 100 \\

&= \dfrac{\left(\dfrac{lb}{2}\right)}{lb} \times 100 = \dfrac{lb \times 100}{2lb} = 50\% \end{align} $MF#%

5. A person walked diagonally across a square plot. Approximately, what was the percent saved by not walking along the edges? | |

A. 35% | B. 30 % |

C. 20 % | D. 25% |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

---------------------------------------------------------

Solution 1

---------------------------------------------------------

Consider a square plot as shown above and let the length of each side = 1

$MF#%\text{Then length of the diagonal = }\sqrt{(1+1)} = \sqrt{2}$MF#%

Distance travelled if walked along the edges = BC + CD = 1 + 1 = 2

$MF#%\text{Distance travelled if walked diagonally = BD = }\sqrt{2} = 1.41$MF#%

Distance Saved = 2 - 1.41 = .59

$MF#%\text{Percent distance saved = }\dfrac{.59}{2} \times 100 = .59 \times 50 \approx 30 \%$MF#%

--------------------------------------------------------- Solution 2 ---------------------------------------------------------

Consider a square plot as shown above and let the length of each side = x

$MF#%\text{Then length of the diagonal = }\sqrt{(x+x)} = \sqrt{2x}$MF#%

Distance travelled if walked along the edges = BC + CD = x + x = 2x

$MF#%\text{Distance travelled if walked diagonally = BD = }\sqrt{2x} = 1.41x$MF#%

Distance Saved = 2x - 1.41x = .59x

$MF#%\text{Percent distance saved = }\dfrac{.59x}{2x} \times 100 = .59 \times 50 \approx 30 \%$MF#%

6. A rectangular field has to be fenced on three sides leaving a side of 20 feet uncovered. If the area of the field is 680 sq. feet, how many feet of fencing will be required? | |

A. 95 | B. 92 |

C. 88 | D. 82 |

| Discuss |

Here is the answer and explanation

Answer : Option C

Explanation :

Given that area of the field = 680 sq. feet

=> lb = 680 sq. feet

Length(l) = 20 feet

=> 20 × b = 680

$MF#%\Rightarrow b = \dfrac{680}{20} = 34\text{ feet}$MF#%

Required length of the fencing = l + 2b = 20 + (2 × 34) = 88 feet

7. A rectangular parking space is marked out by painting three of its sides. If the length of the unpainted side is 9 feet, and the sum of the lengths of the painted sides is 37 feet, find out the area of the parking space in square feet? | |

A. 126 sq. ft. | B. 64 sq. ft. |

C. 100 sq. ft. | D. 102 sq. ft. |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

Let l = 9 ft.

Then l + 2b = 37

=> 2b = 37 - l = 37 - 9 = 28

=> b = ^{28}/_{2} = 14 ft.

Area = lb = 9 × 14 = 126 sq. ft.

8. The area of a rectangle plot is 460 square metres. If the length is 15% more than the breadth, what is the breadth of the plot? | |

A. 14 metres | B. 20 metres |

C. 18 metres | D. 12 metres |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

lb = 460 m^{2} ------(Equation 1)

Let the breadth = b

$MF#%\text{Then length, l = }b \times \dfrac{(100 + 15)}{100} = \dfrac{115b}{100}\text{------(Equation 2)}$MF#%

From Equation 1 and Equation 2,

$MF#%\begin{align}

&\dfrac{115b}{100} \times b = 460\\\\

&b^2 = \dfrac{46000}{115} = 400\\\\

&\Rightarrow b = \sqrt{400} = 20\text{ m}

\end{align} $MF#%

9. A large field of 700 hectares is divided into two parts. The difference of the areas of the two parts is one-fifth of the average of the two areas. What is the area of the smaller part in hectares? | |

A. 400 | B. 365 |

C. 385 | D. 315 |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

Let the areas of the parts be x hectares and (700 - x) hectares.

$MF#%\begin{align}&\text{Difference of the areas of the two parts = x - (700 - x) = 2x - 700}\\\\

&\text{one-fifth of the Average of the two areas = }\dfrac{1}{5}\dfrac{[ x + (700 - x)]}{2}\\

&= \dfrac{1}{5} \times \dfrac{700}{2}= \dfrac{350}{5} = 70\end{align} $MF#%

Given that difference of the areas of the two parts = one-fifth of the Average of the

two areas

=> 2x - 700 = 70

=> 2x = 770

$MF#%\Rightarrow x = \dfrac{770}{2}= 385$MF#%

Hence, Area of smaller part = (700 - x) = (700 – 385) = 315 hectares.

10. The length of a room is 5.5 m and width is 3.75 m. What is the cost of paying the floor by slabs at the rate of Rs. 800 per sq. metre. | |

A. Rs.12000 | B. Rs.19500 |

C. Rs.18000 | D. Rs.16500. |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

Area = 5.5 × 3.75 sq. metre.

Cost for 1 sq. metre. = Rs. 800

Hence total cost = 5.5 × 3.75 × 800 = 5.5 × 3000 = Rs. 16500

11. The length of a rectangle is twice its breadth. If its length is decreased by 5 cm and breadth is increased by 5 cm, the area of the rectangle is increased by 75 sq.cm. What is the length of the rectangle? | |

A. 18 cm | B. 16 cm |

C. 40 cm | D. 20 cm |

| Discuss |

Here is the answer and explanation

Answer : Option C

Explanation :

Let breadth = x cm

Then length = 2x cm

Area = lb = x × 2x = 2x^{2}

New length = (2x - 5)

New breadth = (x + 5)

New Area = lb = (2x - 5)(x + 5)

But given that new area = initial area + 75 sq.cm.

=> (2x - 5)(x + 5) = 2x^{2} + 75

=> 2x^{2} + 10x - 5x - 25 = 2x^{2} + 75

=> 5x - 25 = 75

=> 5x = 75 + 25 = 100

=> x = ^{100}/_{5} = 20 cm

Length = 2x = 2 × 20 = 40cm

12. If a square and a rhombus stand on the same base, then what is the ratio of the areas of the square and the rhombus? | |

A. equal to ½ | B. equal to ¾ |

C. greater than 1 | D. equal to 1 |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

**Hence ratio of the areas of the square and the rhombus will be equal to 1 since
they stand on the same base**

================================================================

Note : Please find the proof of the formula given below which you may like to go through

Let ABCD be the square and ABEF be the rhombus

Consider the right-angled triangles ADF and BCE

We know that AD = BC (∵ sides of a square)

AF = BE (∵ sides of a rhombus)

∴ DF = CE [∵ DF

^{2}= AF

^{2}- AD

^{2}and CE

^{2}= BE

^{2}- BC

^{2}]

Hence Δ ADF = Δ BCE

=> Δ ADF + Trapezium ABCF= Δ BCE + Trapezium ABCF

=> Area of square ABCD = Area of rhombus ABEF

13. The breadth of a rectangular field is 60% of its length. If the perimeter of the field is 800 m, find out the area of the field. | |

A. 37500 m | B. 30500 m |

C. 32500 m | D. 40000 m |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

Given that breadth of a rectangular field is 60% of its length

$MF#%\Rightarrow b = \dfrac{60l}{100} = \dfrac{3l}{5}$MF#%

perimeter of the field = 800 m

=> 2 (l + b) = 800

$MF#%\begin{align}

&\Rightarrow 2\left(l + \dfrac{3l}{5} \right) = 800\\\\

&\Rightarrow l + \dfrac{3l}{5} = 400\\\\

&\Rightarrow \dfrac{8l}{5} = 400\\\\

&\Rightarrow \dfrac{l}{5} = 50\\\\

&\Rightarrow l = 5 \times 50 = 250\text{ m}\\\\\\\\

&\text{b = }\dfrac{3l}{5} = \dfrac{3 \times 250}{5} = 2 \times 50 = 150\text{ m}\\\\

&\text{Area = lb = }250 \times 150 = 37500\text{ m}^2

\end{align} $MF#%

14. A room 5m 44cm long and 3m 74cm broad needs to be paved with square tiles. What will be the least number of square tiles required to cover the floor? | |

A. 176 | B. 124 |

C. 224 | D. 186 |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

l = 5 m 44 cm = 544 cm

b = 3 m 74 cm = 374 cm

Area = 544 × 374 cm^{2}

Now we need to find out HCF(Highest Common Factor) of 544 and 374.

Let's find out the HCF using long division method for quicker results)

374) 544 (1374170) 374 (234034) 170 (51700

Hence, HCF of 544 and 374 = 34

Hence, side length of largest square tile we can take = 34 cm

Area of each square tile = 34 × 34 cm^{2}

$MF#%\text{Number of tiles required = }\dfrac{544 \times 374}{34 \times 34} = 16 \times 11 = 176

$MF#%

15. The length of a rectangular plot is 20 metres more than its breadth. If the cost of fencing the plot @ Rs. 26.50 per metre is Rs. 5300, what is the length of the plot in metres? | |

A. 60 m | B. 100 m |

C. 75 m | D. 50 m |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

Length of the plot is 20 metres more than its breadth.

Hence, let's take the length as l metres and breadth as (l - 20) metres

Length of the fence = perimeter = 2(length + breadth)= 2[ l + (l - 20) ] = 2(2l - 20) metres

Cost per meter = Rs. 26.50

Total cost = 2(2l - 20) × 26.50

Total cost is given as Rs. 5300

=> 2(2l - 20) × 26.50 = 5300

=> (2l - 20) × 26.50 = 2650

=> (l - 10) × 26.50 = 1325

=> (l - 10) = ^{1325}/_{26.50} = 50

=> l = 50 + 10 = 60 metres

16. The ratio between the length and the breadth of a rectangular park is 3 : 2. If a man cycling along the boundary of the park at the speed of 12 km/hr completes one round in 8 minutes, then what is the area of the park (in sq. m)? | |

A. 142000 | B. 112800 |

C. 142500 | D. 153600 |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

l : b = 3 : 2 ------------------------------------------**(Equation 1)**

Perimeter of the rectangular park

= Distance travelled by the man at the speed of 12 km/hr in 8 minutes

= speed × time = 12 × ^{8}/_{60} (∵ 8 minute = 8/_{60} hour)

= ^{8}/_{5} km = ^{8}/_{5} ×1000 m = 1600 m

Perimeter = 2(l + b)

=> 2(l + b) = 1600

=> l + b = ^{1600}/_{2} = 800 m ---------------------------**(Equation 2)**

From (Equation 1) and (Equation 2)

l = 800 × ^{3}/_{5} = 480 m

b = 800 × ^{2}/_{5} = 320 m *(Or b = 800 - 480 = 320m)*

Area = lb = 480 × 320 = 153600 m^{2}

17. What is the percentage increase in the area of a rectangle, if each of its sides is increased by 20%? | |

A. 45% | B. 44% |

C. 40% | D. 42% |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

---------------------------------------------------------

Solution 1

---------------------------------------------------------

Let original length = 100 and original breadth = 100

Then original area = 100 × 100 = 10000

$MF#%\begin{align}&\text{Increase in 20% of length}\\

&\Rightarrow\text{New length = Original length}\times \dfrac{(100 + 20)}{100} = 100 \times \dfrac{120}{100} = 120\\\\\\

&\text{Increase in 20% of breadth}\\

&\Rightarrow \text{New breadth= Original breadth}\times \dfrac{(100 + 20)}{100} = 100 \times \dfrac{120}{100} = 120\\\\\\

\end{align} $MF#%

New area = 120 × 120 = 14400

Increase in area = New Area - Original Area = 14400 - 10000 = 4400

$MF#%\text{Percentage increase in area = }\dfrac{\text{Increase in Area}}{\text{Original Area}} \times 100 = \dfrac{4400}{10000} \times 100 = 44\% $MF#%

--------------------------------------------------------- Solution 2 ---------------------------------------------------------

Let original length = l and original breadth = b

Then original area = lb

$MF#%\begin{align}&\text{Increase in 20% of length}\\

&\Rightarrow\text{New length = Original length}\times \dfrac{(100 + 20)}{100} = l\times \dfrac{120}{100} = \dfrac{120l}{100}\\\\\\\\\\

&\text{Increase in 20% of breadth}\\

&\Rightarrow \text{New breadth= Original breadth}\times \dfrac{(100 + 20)}{100} = b \times \dfrac{120}{100} = \dfrac{120b}{100}\\\\\\\\\\

&\text{New area =}\dfrac{120l}{100} \times \dfrac{120b}{100} = \dfrac{14400lb}{10000}= \dfrac{144lb}{100}\\\\

&\text{Increase in area = New Area - Original Area = }\dfrac{144lb}{100} - lb = \dfrac{44lb}{100}\\\\\\

&\text{Percentage of increase in area = }\dfrac{\text{Increase in Area}}{\text{Original Area}} \times 100 \\

&= \dfrac{\left(\dfrac{44lb}{100}\right)}{lb} \times 100 = \dfrac{44lb \times 100}{100lb} = 44\%

\end{align} $MF#%

18. If the difference between the length and breadth of a rectangle is 23 m and its perimeter is 206 m, what is its area? | |

A. 2800 m | B. 2740 m |

C. 2520 m | D. 2200 m |

| Discuss |

Here is the answer and explanation

Answer : Option C

Explanation :

l - b = 23 ...................**(Equation 1)**

perimeter = 2(l + b) = 206

=> l + b = 103.............**(Equation 2)**

(Equation 1) + (Equation 2) => 2l = 23 + 103 = 126

=> l = ^{126}/_{2} = 63 metre

Substituting this value of l in (Equation 1), we get

63 - b = 23

=> b = 63 - 23 = 40 metre

Area = lb = 63 × 40 = 2520 m^{2}

19. The ratio between the perimeter and the breadth of a rectangle is 5 : 1. If the area of the rectangle is 216 sq. cm, what is the length of the rectangle? | |

A. 16 cm | B. 18 cm |

C. 14 cm | D. 20 cm |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

$MF#%\text{Given that }\dfrac{2(l + b)}{b} = 5$MF#%

=> 2l + 2b = 5b

=> 2l = 3b

$MF#%=> b = \dfrac{2l}{3}$MF#%

Also given that area = 216 cm^{2}

=> lb = 216 cm^{2}

$MF#%\begin{align}

&\text{Substituting the value of b, we get, }l \times \dfrac{2l}{3} = 216\\\\

&\Rightarrow l^2 = \dfrac{3 \times 216}{2} = 3 \times 108 = (3 \times 3) \times 36\\\\

&\Rightarrow l = 3 \times 6 = 18\text{ cm}

\end{align} $MF#%

20. What is the least number of squares tiles required to pave the floor of a room 15 m 17 cm long and 9 m 2 cm broad? | |

A. 814 | B. 802 |

C. 836 | D. 900 |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

l = 15 m 17 cm = 1517 cm

b = 9 m 2 cm = 902 cm

Area = 1517 × 902 cm^{2}

Now we need to find out HCF(Highest Common Factor) of 1517 and 902.

Let's find out the HCF using long division method for quicker results)

902) 1517 (1902615) 902 (1615287) 615 (257441) 287 (72870

Hence, HCF of 1517 and 902 = 41

Hence, side length of largest square tile we can take = 41 cm

Area of each square tile = 41 × 41 cm^{2}

$MF#%\text{Number of tiles required = }\dfrac{1517 \times 902}{41 \times 41} = 37 \times 22 = 407 \times 2 = 814

$MF#%

21. The diagonal of the floor of a rectangular room is $MF#%7\dfrac{1}{2}$MF#% feet. The shorter side of the room is 4 $MF#%\dfrac{1}{2}$MF#% feet. What is the area of the room? | |

A. 27 square feet | B. 22 square feet |

C. 24 square feet | D. 20 square feet |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

$MF#%\begin{align}

&\text{Diagonal, d = 7}\dfrac{1}{2}\text{ feet } = \dfrac{15}{2}\text{ feet}\\

&\text{Breadth, b = 4}\dfrac{1}{2}\text{ feet} = \dfrac{9}{2}\text{ feet}\\\\\\\\

&\text{In the right-angled triangle PQR,}\\

&l^2 = \left(\dfrac{15}{2}\right)^2 - \left(\dfrac{9}{2}\right)^2 \\\\

&= \dfrac{225}{4} - \dfrac{81}{4} = \dfrac{144}{4}\\\\

&l = \sqrt{\dfrac{144}{4}} = \dfrac{12}{2}\text{ feet = 6 feet}\\\\\\\\\\

&\text{Area = lb = }6 \times \dfrac{9}{2} = 27 \text{ feet}^2

\end{align} $MF#%

22. The diagonal of a rectangle is $MF#%\sqrt{41}$MF#% cm and its area is 20 sq. cm. What is the perimeter of the rectangle? | |

A. 16 cm | B. 10 cm |

C. 12 cm | D. 18 cm |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

$MF#%\boxed{\text{For a rectangle, }d^2 = l^2 + b^2 \\\\

\text{where l = length , b = breadth and d = diagonal of the of the rectangle}}$MF#%

$MF#%\begin{align}

&d = \sqrt{41}\text{ cm}\\\\

&d^2 = l^2 + b^2\\\\

&\Rightarrow l^2 + b^2 = \left(\sqrt{41}\right)^2 = 41\text{........(Equation 1)}\\\\\\\\\\\\

&\text{Area = lb = 20 cm}^2\text{............(Equation 2)}\\\\\\\\\\\\

&\text{Solving (Equation 1) and (Equation 2)}\\

&\end{align} $MF#%

$MF#%\boxed{(a + b)^2 = a^2 + 2ab + b^2}$MF#%

$MF#%\begin{align}

&\text{using the above formula, we have}\\\\

&(l + b)^2 = l^2 + 2lb + b^2 = (l^2 + b^2) + 2lb = 41 + (2 \times 20) = 81\\\\

&\Rightarrow (l + b) = \sqrt{81} = 9 \text{ cm}\\\\

&\text{perimeter = }2(l + b) = 2 \times 9 = 18\text{ cm}

\end{align} $MF#%

23. A tank is 25 m long, 12 m wide and 6 m deep. What is the cost of plastering of its walls and bottom at the rate of 75 paise per sq. m? | |

A. Rs. 558 | B. Rs. 502 |

C. Rs. 516 | D. Rs. 612 |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

**Consider a rectangular solid of length l, width w and height h. Then**

1. Total Surface area of a rectangular solid, S = 2lw + 2lh + 2wh = 2(lw + lh + wh)

2. Volume of a rectangular solid, V = lwh

1. Total Surface area of a rectangular solid, S = 2lw + 2lh + 2wh = 2(lw + lh + wh)

2. Volume of a rectangular solid, V = lwh

In this case, l = 25 m, w = 12 m, h = 6 m

and all surface needs to be plastered except the top

Hence total area needs to be plastered

= Total Surface Area - Area of the Top face

= (2lw + 2lh + 2wh) - lw

= lw + 2lh + 2wh

= (25 × 12) + (2 × 25 × 6) + (2 × 12 × 6)

= 300 + 300 + 144

= 744 m^{2}

Cost of plastering = 744 × 75 = 55800 paise = Rs.558

24. It is decided to construct a 2 metre broad pathway around a rectangular plot on the inside. If the area of the plots is 96 sq.m. and the rate of construction is Rs. 50 per square metre., what will be the total cost of the construction? | |

A. Rs.3500 | B. Rs. 4200 |

C. Insufficient Data | D. Rs. 4400 |

| Discuss |

Here is the answer and explanation

Answer : Option C

Explanation :

Let length and width of the rectangular plot be l and b respectively

Total Area of the rectangular plot = 96 sq.m.

Width of the pathway = 2 m

Length of the remaining area in the plot = (l - 4)

breadth of the remaining area in the plot = (b - 4)

Area of the remaining area in the plot = (l - 4)(b - 4)

Area of the pathway

= Total Area of the rectangular plot - remaining area in the plot

= 96 - [(l - 4)(b - 4)]

= 96 - [lb - 4l - 4b + 16]

= 96 - [96 - 4l - 4b + 16]

= 96 - 96 + 4l + 4b - 16]

= 4l + 4b - 16

= 4(l + b) - 16

We do not know the values of l and b and hence total area of the rectangular plot

can not be found out. So we can not find out total cost of the construction.

25. The area of a parallelogram is 72 cm | |

A. 6 cm | B. 7 cm |

C. 8 cm | D. 12 cm |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

**Area of a parallelogram , A = bh**

where b is the base and h is the height of the parallelogram

where b is the base and h is the height of the parallelogram

Let the base = x cm.

Then the height = 2x cm (∵ altitude is twice the base)

Area = x × 2x = 2x^{2}

But the area is given as 72 cm^{2}

=> 2x^{2} = 72

=> x^{2} = 36

=> x = 6 cm

26. Two diagonals of a rhombus are 72 cm and 30 cm respectively. What is its perimeter? | |

A. 136 cm | B. 156 cm |

C. 144 cm | D. 121 cm |

| Discuss |

Here is the answer and explanation

Answer : Option B

Explanation :

Remember the following two properties of a rhombus which will be useful in solving this question

1. The sides of a rhombus are congruent.

2. The diagonals of a rhombus are unequal and bisect each other at right angles.

Let the diagonals be PR and SQ such that PR = 72 cm and SQ = 30 cm

$MF#%\begin{align}

&\text{PO = OR = }\dfrac{72}{2} = 36\text{ cm }\\\\
&\text{SO = OQ = }\dfrac{30}{2} = 15\text{ cm }\\\\
&\text{PQ = QR = RS = SP = }\sqrt{36^2 + 15^2}=\sqrt{1296 + 225}=\sqrt{1521}\text{ = 39 cm}\\\\

&\text{perimeter = 4 × 39 =156 cm}

\end{align} $MF#%

27. The base of a parallelogram is (p + 4), altitude to the base is (p - 3) and the area is (p | |

A. 40 sq. units | B. 54 sq. units |

C. 36 sq. units | D. 60 sq. units |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

**Area of a parallelogram , A = bh**

where b is the base and h is the height of the parallelogram

where b is the base and h is the height of the parallelogram

Hence, we have

p^{2} - 4 = (p + 4)(p - 3)

=> p^{2} - 4 = p^{2} - 3p + 4p - 12

=> -4 = p - 12

=> p = 12 - 4 = 8

Hence, actual area = (p^{2} - 4) = 8^{2} - 4 = 64 - 4 = 60 sq. units

28. A circle is inscribed in an equilateral triangle of side 24 cm, touching its sides. What is the area of the remaining portion of the triangle? | |

A. $MF#%144\sqrt{3} - 48\pi$MF#% cm | B. $MF#%121\sqrt{3} - 36\pi$MF#% cm |

C. $MF#%144\sqrt{3} - 36\pi$MF#% cm | D. $MF#%121\sqrt{3} - 48\pi$MF#% cm |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

$MF#%\boxed{\text{Area of an equilateral triangle = }\dfrac{\sqrt{3}}{4}a^2 \\

\text{where a is length of one side of the equilateral triangle}}$MF#%

$MF#%\begin{align}

&\text{Area of the equilateral Δ ABC = }\dfrac{\sqrt{3}}{4}a^2 =\dfrac{\sqrt{3}}{4}24^2 = 144\sqrt{3}\text{ cm}^2\text{.............(1)}

\end{align} $MF#%

$MF#%\boxed{\text{Area of a triangle = }\dfrac{1}{2}\text{bh}\\

\text{where b is the base and h is the height of the triangle}}$MF#%

Let r = radius of the inscribed circle. Then

Area of Δ ABC

= Area of Δ OBC + Area of Δ OCA + area of Δ OAB

= (½ × r × BC) + (½ × r × CA) + (½ × r × AB)

= ½ × r × (BC + CA + AB)

= ½ x r x (24 + 24 + 24)

= ½ x r x 72 = 36r cm2 ------------------------------------------ (2)

From (1) and (2),

$MF#%\begin{align}

&144\sqrt{3} = 36r\\\\

&\Rightarrow r = \dfrac{144}{36}\sqrt{3}= 4\sqrt{3}------------(3)

\end{align} $MF#%

$MF#%\boxed{\text{Area of a circle = }\pi r^2 \\ \text{ where = radius of the circle}}$MF#%

$MF#%\begin{align}

&\text{From (3), the area of the inscribed circle = }\pi r^2 = \pi \left(4\sqrt{3}\right)^2 = 48 \pi ------------(4)\\\\\\\\

&\text{Hence , Area of the remaining portion of the triangle }\\\\

&\text{Area of Δ ABC – Area of inscribed circle}\\\\

&144\sqrt{3} - 48\pi \text{ cm}^2

\end{align} $MF#%

29. A rectangular plot measuring 90 metres by 50 metres needs to be enclosed by wire fencing such that poles of the fence will be kept 5 metres apart. How many poles will be needed? | |

A. 30 | B. 44 |

C. 56 | D. 60 |

| Discuss |

Here is the answer and explanation

Answer : Option C

Explanation :

**Perimeter of a rectangle = 2(l + b)**

where l is the length and b is the breadth of the rectangle

where l is the length and b is the breadth of the rectangle

Length of the wire fencing = perimeter = 2(90 + 50) = 280 metres

Two poles will be kept 5 metres apart. Also remember that the poles will be placed

along the perimeter of the rectangular plot, not in a single straight line which is

very important.

Hence number of poles required = ^{280}⁄_{5} = 56

30. If the diagonals of a rhombus are 24 cm and 10 cm, what will be its perimeter | |

A. 42 cm | B. 64 cm |

C. 56 cm | D. 52 cm |

| Discuss |

Here is the answer and explanation

Answer : Option D

Explanation :

Let the diagonals be PR and SQ such that PR = 24 cm and SQ = 10 cm

$MF#%\begin{align}

&\text{PO = OR = }\dfrac{24}{2} = 12\text{ cm }\\\\

&\text{SO = OQ = }\dfrac{10}{2} = 5\text{ cm }\\\\

&\text{PQ = QR = RS = SP = }\sqrt{12^2 + 5^2}=\sqrt{144 + 25}=\sqrt{169}\text{ = 13 cm}\\\\

&\text{perimeter = 4 × 13 = 52 cm}

\end{align} $MF#%

31. What will be the length of the longest rod which can be placed in a box of 80 cm length, 40 cm breadth and 60 cm height? | |

A. $MF#%\sqrt{11600}$MF#% cm | B. $MF#%\sqrt{14400}$MF#% cm |

C. $MF#%\sqrt{10000}$MF#% cm | D. $MF#%\sqrt{12040}$MF#% cm |

| Discuss |

Here is the answer and explanation

Answer : Option A

Explanation :

The longest road which can fit into the box will have one end at A and

other end at G (or any other similar diagonal)

Hence the length of the longest rod = AG

Initially let's find out AC. Consider the right angled triangle ABC

AC^{2} = AB^{2} + BC^{2} = 40^{2} + 80^{2} = 1600 + 6400 = 8000

$MF#%\Rightarrow \text{AC = }\sqrt{8000}\text{ cm}$MF#%

Consider the right angled triangle ACG

AG^{2}= AC^{2}+ CG^{2}

$MF#%\begin{align}&= \left(\sqrt{8000}\right)^2 + 60^2 = 8000 + 3600 = 11600\\\\

&\Rightarrow \text{AG = }\sqrt{11600}\text{ cm}\\\\

&\Rightarrow \text{The length of the longest rod = }\sqrt{11600}\text{ cm}

\end{align} $MF#%

Comments(95)

Area of the larger square = 2root(2) * 2root(2) = 8

Area of smaller square = 8/2 = 4

Side of the smaller square = root(4) = 2

Diagonal of the smaller square = 2root(2)

^{2}= 2464

22/7 * r

^{2}= 2464

r

^{2}=2464 * 7/22= 784

r = 28

diameter = 2r = 56 m

root(30^2+24^2+18^2)

= root(1800)

= 30 * root(2) m

2 * pi * r = 44

2 * (22/7) * r = 44

r = 7

Area = pi * r

^{2}= (22/7) * 7

^{2}= 154

Consider the triangle ABC in which three identical circles are inscribed. Let length of each circle be r

Let circle with centre O is close to the vertex A.

Let circle with centre O touches one side of triangle ABC at D

Since each angle of an equilateral is 60 degree, angle DAO is 30 degree

Also, ADO will be a right angled triangle with angle ADO is 90 degree

$MF#% \tan 30 = \dfrac{\text{OD}}{\text{AD}}$MF#%

$MF#%\dfrac{1}{\sqrt{3}} = \dfrac{\text{r}}{\text{AD}}$MF#%

$MF#%\text{AD = r } \sqrt{3}$MF#%

One side of the triangle = distance between two centres of the circles + 2(AD)

24 = 2r + 2(AD)

24 = 2r + 2r $MF#%\sqrt{3}$MF#%

r= $MF#%\dfrac{12}{1+ \sqrt{3}}$MF#%

Area of each circle = $MF#%\pi*r^2 = \pi \left(\dfrac{12}{1+ \sqrt{3}}\right)^2$MF#%

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