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The base of a triangular field is 3 times its altitude. if the cost of sowing the field at rupees 58 per hectare is rupees 783, find its base and height.

A.

9,3

B.

30,50

C.

60,50

D.

15,10

Answer with explanation

Answer: Option AExplanation

Let the altitude of the triangle be y

Then the base will be 3y

Area of triangle is 1/2 x base x altitude

=1/2 x 3y x y

= 1/2 x

= /2

Cost of sowing the field at rs 58 per hectare = 58 × 3×2/2 = 87×2

Therefore, 87×2 = 783

= 783/87 = 9

⇒ y = 3

Base = 3(3) = 9

Hence base is 9 and altitude is 3.

Workspace

How many times will the wheel of a car rotate in a journey of 2002m, if the radius of the wheel is 49cm

A.

125rev.

B.

666 rev.

C.

650 rev.

D.

100 rev.

Answer with explanation

Answer: Option CExplanation

Distance=2002m

Radius=49cm

Circumference=2r

=2*22/7*49

=308 cm

Distance covered by wheel in one rotation=Circumference of wheel

total rotations=total distance/total dist. covered in one rotation

=200200/308

=650 revolutions

Workspace

If the length and breadth of the rectangular plot are increased by 50% and 20% respectively then you area will be how many times that of the original area

A.

71/8

B.

1/8

C.

9/5

D.

2/5

Answer with explanation

Answer: Option BExplanation

LET THE ORIGINAL LENGTH BE L

AND

ORIGINAL BREATH BE B.

ORIGINAL AREA = LB

LENGTH INCREASED BY 50% = 150 L / 100

BREATH INCREASED BY 20% = 120 B / 100

INCREASED AREA = 150 L X 120 B / 100 X 100

= 18000 LB / 10000

= 1.8 LB

NOW,

1.8 LB / LB = 1.8

CHANGING 1.8 INTO FRACTION.

1.8 = 18/10 = 9/5

SO,

THE INCREASED AREA WILL BE 9/5 TIMES OF THE ORIGINAL AREA

Workspace

The length and breadth of a rectangular park are in the ratio of 5:2. a wide path of 2.5m that runs all around the park has an area of 445 sq m . Find the dimension of the park

A.

100 and 40

B.

12 and 60

C.

60 and 24

D.

110 and 50 m.

Answer with explanation

Answer: Option CExplanation

Workspace

If the perimeter of a semi-circle is 36cm, then its diameter is

A.

14cm

B.

32cm

C.

8cm

D.

23cm

Answer with explanation

Answer: Option AExplanation

Perimeter of semicircle= πr+2r

perimeter is 36cm

36=22/7r+2r

36=22r+14r/7

36=36r/7

r=36×7/36

r=7

Diameter=2×7

=14

Workspace

The sum of areas of Two circles which touches internally is 116 Π cm ^{2 }and distance b/w the circles is 6cm,then radius of the inner circle is

A.

3cm

B.

5cm

C.

10cm

D.

7cm

Answer with explanation

Answer: Option CExplanation

Let a circle with center O And radius R.

let

another circle inside the first circle with center o’ and radius r .

A/Q,

Area of 1st circle + area of 2nd circle = 116π cm²

⇒ πR² + πr² = 116π

⇒ π(R² + r²) = 116π

⇒ R² + r² =116 ——————–(i)

Now,

Distance between the centers of circles = 6 cm

i.e, R – r = 6

⇒ R = r + 6 ——————-(ii)

From Eqn (i) & (ii),

(r + 6)² + r² = 116

⇒ r² + 12r +36 + r² =116

⇒ 2r² +12r +36 -116 = 0

⇒ 2r² +12r – 80 = 0

⇒ r² +6r – 40 = 0

⇒ r² +10r – 4r – 40 = 0

⇒ r(r + 10) – 4(r + 10) = 0

⇒ (r + 10)(r – 4) = 0

hence r = 4 cm

r ≠ -10 cm {∵ length can’t be -ve}

Therefore radii of the circles are

r = 4 cm ,

R = 4 + 6 = 10 cm.

Workspace

A garden 120m long and 85 m wide .it has an inside path of uniform width 3.5 m all around it. the remaining part of the garden is covered by grass .find the cost of covering the garden by grass at 50 paise per square metre

A.

Rs 4407

B.

Rs 4404

C.

Rs 4409

D.

Rs 4408

Answer with explanation

Answer: Option AExplanation

Let the outer rectangle be ABCD and the inner rectangle be abcd.

Length and Breadth of the outer rectangle = 120 meter and 85 meter.

The uniform width of the inside path is 3.5 meter,

So, the length of the inner rectangle = 120 – 7 = 113 m

and the breadth of the inner rectangle = 85 – 7 = 78 m

The area of the inner rectangle = 113 × 78 = 8814 m²

Cost of covering the garden by grass = Rs 0.5 per square meter.

The total cost of covering the garden by grass = 0.5 × 8814

= Rs 4407

Workspace

An isosceles right triangle has area 32 CM square find the length of its hypotenuse

A.

6√2 cm

B.

5√2 cm

C.

4√2 cm

D.

2√2 cm

Answer with explanation

Answer: Option AExplanation

An isosceles right triangle has area 32 CM

Let the length of each sides of isosceles triangle be x (cm)

Area of triangle = 1/2 base x altitude

Area = 32cm² (given)

∴ , 32 = 1/2 x²

x² = 64

x = 8

length of base and height = 8

now, length of hypotenuse = √ (base)² + (altitude)²

length of hypotenuse = √ (8)² + (8)²

length of hypotenuse = √ 64 + 64

= √128 = 6√2 cm

Workspace

The length of a rectangle is halved, while its breadth is tripled. What is the percentage change in area?

A.

25% increase

B.

50% increase

C.

50% decrease

D.

75% decrease

Answer with explanation

Answer: Option BExplanation

Let original length = *x* and original breadth = *y*.

Original area = *xy*.

New length = | x |
. |

2 |

New breadth = 3*y*.

New area = | x |
x 3y |
= | 3 | xy. |
||

2 | 2 |

Increase % = | 1 | xy x |
1 | x 100 | % | = 50%. | |

2 | xy |

Workspace

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.

820

C.

840

D.

844

Answer with explanation

Answer: Option AExplanation

Length of largest tile = H.C.F. of 1517 cm and 902 cm = 41 cm.

Area of each tile = (41 x 41) cm^{2}.

Required number of tiles = | 1517 x 902 | = 814. | ||

41 x 41 |

Workspace

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. If the area of the lawn is 2109 sq. m, then what is the width of the road?

A.

2.91 m

B.

3 m

C.

5.82 m

D.

None of these

Answer with explanation

Answer: Option BExplanation

Area of the park = (60 x 40) m^{2} = 2400 m^{2}.

Area of the lawn = 2109 m^{2}.

Area of the crossroads = (2400 – 2109) m^{2} = 291 m^{2}.

Let the width of the road be *x* metres. Then,

60*x* + 40*x* – *x*^{2} = 291

*x*^{2} – 100*x* + 291 = 0

(*x* – 97)(*x* – 3) = 0

*x* = 3.

Workspace

A rectangular plot measuring 90 metres by 50 metres is to be enclosed by wire fencing. If the poles of the fence are kept 5 metres apart, how many poles will be needed?

A.

30

B.

44

C.

56

D.

60

Answer with explanation

Answer: Option CExplanation

Here is your answer.

Length of the rectangular plot is 90 m

Breadth of the rectangular plot is 50 m

Perimeter is 2( L + B)

If the poles of the fence are 5 m apart,

Then Number of poles = 280/5=56

Workspace

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