Question

5. Evaluate :

(i) (1.3) ^ 2

(ii) (0.06) ^ 2

(iii) (0.2) ^ 3

(iv) (0.8) ^ 3
6. The cost of one pen is 42.25. Find the cost of one dozen such pens.

7. A car moves at a constant speed of 56.4 km per hour. How much distance does it cover in 3.5 hours?

8. A room is 4.5 m long and 3.8 m broad. Calculate the area of the floor of the room.

9. The cost of 1 litre of refined oil is 124.75. What is the cost of 6.2 litres of this oil?​

Answer 1

Step-by-step explanation:

Solutions:-

5.

(i) (1.3)²

=> (13/10)²

=> (13/10)×(13/10)

=> (13×13)/(10×10)

=> 169/100

=> 1.69

(ii) (0.06)²

=> (6/100)²

=> (6/100)×(6/100)

=> (6×6)/(100×100)

=> 36/10000

=> 0.0036

(iii) (0.2)³

=> (2/10)³

=> (2/10)×(2/10)×(2/10)

=> (2×2×2)/(10×10×10)

=> 8/1000

=> 0.008

(iv) (0.8)³

=> (8/10)³

=> (8/10)×(8/10)×(8/10)

=> (8×8×8)/(10×10×10)

=> 512/1000

=> 0.512

6.

The cost of one pen = Rs. 42.25

The cost of a dozen pens

= 42.25×12

Since dozen = 12 objects

= (4225/100)×12

=(4225×12)/100

= 50700/100

= 507.00

= 507

The cost of dozen pens = Rs. 507

7.

The speed of the car = 56.4 km/hr

Let the distance is to be covered be X km

Time taken for X km = 3.5 hours

We know that

Distance = Speed × Time

=> X = 56.4×3.5

=> X = (564/10)×(35/10)

=> X = (564×35)/(10×10)

=> X = 19740/100

=> X = 197.40 km

The required distance is to be covered by the car is 197.40 km

8.

The measurements of a room = 4.5 m and 3.8 m

Length of the room = 4.5 m

Breadth of the room = 3.8 m

Area of the floor = Length × Breadth sq.units

=> A = 4.5 × 3.8

=> A = (45/10)×(38/10)

=> A = (45×38)/(10×10)

=> A = 1710/100

=> A = 17.10 sq.m

Area of the floor of the room = 17.10 sq.m.

9.

The cost of 1 litre of refined oil = Rs. 124.75.

The cost of 6.2 litres of this oil

=> 6.2 × 124.75

=> (62/10)×(12475/100)

=> (62×12475)/(10×100)

=> 773450/1000

=> 773.450

The cost of 6.2 litres of oil = Rs. 773.45

Answer 2

$\mathtt{EVALUATE} : \\ \\ \sf(i) \: \frak{ {(1.3)}^{2} } \\ \\ \dashrightarrow \frak{(1 + 0.3)^{2} } \\ \\ \dashrightarrow \frak{(1)^{2} +2(1 \times 0.3) + (0.3)^{2}} \\ \\\dashrightarrow \frak{1 + 2(0.3) + 0.09} \\ \\ \dashrightarrow \frak{1 + 0.6 + 0.09} \\ \\\dashrightarrow \frak{1.69} \\ \\ \sf \underline{Hence, \boxed{ \green{1.69}} \: is \: the \: required \: answer} \\ \\ \\ \\ \sf (ii) \: \frak{{(0.06)}^{2}} \\ \\ \hookrightarrow \frak{ {( \frac{6}{100} )}^{2} } \\ \\ \hookrightarrow \frak{ {( \frac{36}{10000} )} } \\ \\ \hookrightarrow \frak{ {0.0036}} \\ \\ \sf \underline{Hence, \boxed{ \green{0.0036}} \: is \: the \: required \: answer} \\ \\ \\ \\ \sf (iii) \: \frak{ {(0.2)}^{3}} \\ \\ \looparrowright \frak{ {( \frac{2}{10} )}^{3}} \\ \\ \looparrowright \frak{ {( \frac{8}{1000} )}} \\ \\ \looparrowright \frak{0.008} \\ \\ \\ \sf \underline{Hence, \boxed{ \green{0.008}} \: is \: the \: required \: answer} \\ \\ \\ \\ \sf (iv) \: \frak{ {(0.8)}^{3}} \\ \\ : \implies \frak{ {( \frac{8}{10} )}^{3}} \\ \\ : \implies \frak{ {( \frac{512}{1000} )}} \\ \\ : \implies \frak{0.512} \\ \\ \\ \sf \underline{Hence, \boxed{ \green{0.512}} \: is \: the \: required \: answer}$

$\text{6) The cost of one pen is 42.25. Find the cost of} \\ \text{one dozen such pens.} \\ \\ \rm{as \: we \: know \: that \: cost \: of \: one \: pen \: is \: 42.25 \: we \: need \: to \: get \: the} \\ \rm{cost \: for \: one \: dozen \: of \: pens} \\ \\ \qquad\boxed{ \frak{\pink{1 \: dozen = 12 \: pens}}} \\ \\ \rightarrow \frak{Cost \: of \: one \: pen = 42.25} \\ \\ \therefore\frak{Cost \: of \: 12 \: pens = 12(42.25) } \rightarrow \frak{ \underline{ \blue{507}}}$

$\text{7. A car moves at a constant speed of 56.4 km per hour.} \\ \text{How much distance does it cover in 3.5 hours?} \\ \\ \sf for \: solving \: this \: question \: we \: need \: to \: apply \: the \: distance \: formula : \\ \\ \qquad \boxed{ \pink{\frak{Distance = Speed \times Time}}} \\ \\ : \implies \frak{Distance = 56.4 \times 3.5 } \\ \\ : \implies \frak{Distance = 197.4 \: km } \\ \\ \therefore \underline{ \red{ \frak{197.4 \: km \: is \: the \: distance \: covered}}}$

$\text{8. A room is 4.5 m long and 3.8 m broad.} \\ \text{Calculate the area of the floor of the room.} \\ \\ \sf {for \: getting \: the \: area \: of \: the \: floor \: we \: need \: to \: apply \: the \: area \: formula :} \\ \\ \qquad \boxed{ \pink{ \frak{Area_{(Floor)} = Length × Breadth}}} \\ \\ \rightarrow \frak{Area_{(Floor)} = 4.5 \times 3.8} \\ \\ \rightarrow \frak{Area_{(Floor)} = 17.1 \: {m}^{2} } \\ \\ \therefore \underline{ \frak{ \red{17.1 \: {m}^{2} \: is \: the \: area \: of \: the \: floor}}}$

$\text{9. The cost of 1 litre of refined oil is 124.75.} \\\text{What is the cost of 6.2 litres of this oil}? \\ \\ \sf for \: getting \: the \: solution \: we \: need \: to \: get \: it \: by \: following \: unitary \: method \\ \\ \rightarrow \frak{Cost \: of \: 1 l \: of \: refined \: oil = 124.75} \\ \\ \rightarrow \frak{Cost \: of \: 6.2 l \: of \: refined \: oil = 6.2(124.75) \rightarrow \underline{ \blue{773.45}}}$