S.chand class 6 Mathematics Chapter 17 Exercise 17B

Exercise 17 B

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Question 1

In the given fig. Q1 , which setting will the pointer face after it turns 

(i) 45º clockwise from normal ?

Sol :

A circle consist of total angle of 360º  . Now lets find the angle between two options .

And to do so , we take $=\dfrac{360}{\text{total options}}=\dfrac{360^º}{8}$

= 45º

According to question as we move from normal in clockwise direction 45º , then pointer face normal .

 

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(ii) 135º anti-clockwise from off ?

Sol :

As we already know angle between two options is $=\dfrac{360º}{8}= 45º$ .

So , if we divide 135º by 45º $=\dfrac{135^º}{45^º}=3$

Which means from off position pointer have to move 3 steps in anti-clockwise direction .

First step = off→very hot

Second step = very hot→hot

Third step = hot→warm

So , at last pointer stop at warm position .

 

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(iii) 360º anti-clockwise from hot ?

Sol :

We know angle between two options is $=\dfrac{360º}{8}= 45º$

So , if we divide 360º by 45º $=\dfrac{360º}{45º}=8$

Which means from hot position pointer have to move 8 steps in anti-clockwise direction .

First step = hot→warm

Second step = warm→normal

Third step = normal→cool

Fourth step = cool→cold

Fifth step = cold →very cold

Sixth step = very cold →off

Seventh step = off →very hot

Eighth step = very hot →hot

So , at last pointer stop at hot position .

 

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(iv) 270º clockwise from cold ?

Sol :

We know angle between two options is $=\dfrac{360º}{8}= 45º$

So , if we divide 270º by 45º $=\dfrac{270º}{45º}=6$

Which means from cold position pointer have to move 8 steps in anti-clockwise direction .

First step = cold→cool

Second step = cool→normal

Third step = normal→warm

Fourth step = warm→hot

Fifth step = hot →very hot

Sixth step = very hot → off

So , at last pointer stop at off position .

 

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Question 2

There are markings as shown on a domestic appliance . Through how many degrees does the pointer turn when it move in the fig.Q1

(i) clockwise from off to cold

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from off to cold option in clockwise direction there are total 2 steps i.e.

First step = off→very cold

Second step = very cold→cold

And single step covers 45º , then 2 steps cover 2×45º = 90º

 

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(ii) anti-clockwise from hot to warm

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from hot to warm option in anti-clockwise direction there are total 1 steps i.e.

hot→warm

And single step covers 45º

 

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(iii) anti-clockwise from very cold to very hot

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from very cold to very hot option in anti-clockwise direction there are total 2 steps i.e.

First step = very cold →off

Second step = off→ very hot

And single step covers 45º , then 2 steps cover 2×45º = 90º

 

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(iv) clockwise from warm to cool

Sol :

Between two options , angle is $=\dfrac{360º}{8}=45º$

And moving from warm to cool option in clockwise direction there are total 6 steps i.e.

First step = warm→hot

Second step = hot→very hot

Third step = very hot→off

Fourth step = off→very cold

Fifth step = very cold →cold

Sixth step = cold → cool

And single step covers 45º , then 2 steps cover 6×45º = 270º

 

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Question 3

Answer the following questions related to  a clock in the fig. Q3.

(i) When the seconds hand has moved from 12 to 6 , how many degrees has it turned through ?

Sol :

As we know $=\dfrac{360^º}{12}=30º$ which means 30º per hours  .

And from 12 to 6 we have 6 hours ,= 30º×6

= 180º

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(ii) When the seconds hand has moved from 6 to 8 , how many degrees has it turned through ?

Sol :

As we know $=\dfrac{360^º}{12}=30º$ which means 30º per hours  .

And from 6 to 8 we have 2 hours ,= 30º×2

= 60º

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(iii) What is the time on the clock when hour hand moves clockwise ?

(a) 60º from 6 O' clock ?

Sol :

As we know hour hand moves 30º per hour . So , on dividing 60º by 30º we get $=\dfrac{60^º}{30^º}=2$ which means 2 hours

Which means 6 + 2 = 8 O' clock is the answer

 

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(b) 180º from 10 O' clock ?

Sol :

As we know hour hand moves 30º per hour . So , on dividing 180º by 30º we get $=\dfrac{180^º}{30^º}=2$ which means 6 hours

Which means 10 + 6 = 16 = 12 + 4  i.e at first hour hand moves from 10→12 then 12→4

4 O' clock is the answer

 

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(c) 270º from 12 O' clock ?

Sol :

As we know hour hand moves 30º per hour . So , on dividing 270º by 30º we get $=\dfrac{270^º}{30^º}=9$ which means 9 hours

Which means hour hand moves from 12 → 9

9 O' clock is the answer

 

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(iv) Through how many degrees does the minute hand of a clock turn in 

(a) 1 minute

Sol :

As we know $=\dfrac{360^º}{60^º}=6º$ which means minute hand moves 6º per minutes .

So , in 1 minute it covers 6º

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(b) 8 minutes 

Sol :

As we know minute hand moves 6º per minutes . Then , in 8 minutes it covers 6×8 = 48º

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(c) $\dfrac{3}{4}\text{ hours}$ 

Sol :

Firstly , lets convert hours to minutes . We know 1 hour = 60 minutes

on multiplying both sides by 3/4 we get

$\dfrac{3}{4}\text{ hour}=\dfrac{3}{4}\times 60 \text{ minutes}$

$\dfrac{3}{4}\text{ hour}=\dfrac{3\times 15}{4}\text{ minutes}$

$\dfrac{3}{4}\text{ hour}=45 \text{ minutes}$

and as we know minute hand moves 6º per minutes . Then , in 45 minutes it covers 6×45 = 270º

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(d) $1\dfrac{1}{2}\text{ hours}$

Sol :

Firstly , we convert mixed fraction to improper fraction

$=1\dfrac{1}{2}=\dfrac{3}{2}$

Lets convert hours to minutes . We know 1 hour = 60 minutes

On multiplying both sides by 3/2 we get

$\dfrac{3}{2}\text{ hour}=\dfrac{3}{2}\times 60 \text{ minutes}$

$\dfrac{3}{2}\text{ hour}= 90 \text{ minutes}$

and as we know minute hand moves 6º per minutes . Then , in 90 minutes it covers 6×90 = 540º

 

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(v) Through how many degrees does the hour hand of a clock turn in 

(a) 1 minutes

Sol :

$=\dfrac{360^º}{720}=0.5$ which means hour hand moves 0.5º per minutes

So , in 1 minute it turns only 0.5º or $\dfrac{1}{2}$

 

(b) 10 minutes

Sol :

$=\dfrac{360^º}{720}=0.5$ which means hour hand moves 0.5º per minutes

So , in 10 minute it turns 10×0.5º = 5º

 

(c) 30 minutes

Sol :

$=\dfrac{360^º}{720}=0.5$ which means hour hand moves 0.5º per minutes

So , in 30 minute it turns 30×0.5º = 15º

 

(d) 2 hours 

Sol :

$=\dfrac{360^º}{12}=30^º$ which means hour hand moves 30º per hour

So in 2 hour , the hour hand moves 2×30 = 60º

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Question 4

In the given Fig. Q.4 , find the number of degrees , when we turn in a clockwise directions

(i) From NW to SW

Sol :

 

(ii) From N to S

Sol :

 

(iii) From N to E

Sol :

 

(iv) From NE to W

Sol :

 

 

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Question 5

Look at Fig. Q.5 . The circle is divided into 16 equal parts . Find

(i)  ∠A₁OA₂

(ii) ∠A₁OA3

(iii) ∠A₁OA5

(iv) How many times is the angle ∠A₁OA12 of the angle ∠A₁OA₂ ?

(v) How many times is the angle ∠A₁OA4 of the angle ∠A₁OA2 ?

(vi) What fraction of the whole revolution is the angle ∠A₅OA₇ ?

Sol :

 

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Question 6

Add the following .

(i) 36º 45' and 56º 87'

(ii) 47º 28' 55'' and 27º 35' 49''

 

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Question 7

Subtracting the following

(i) 39º 39' from 50º 15'

(ii) 26º 31'  from 50º 8'

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Multiple Choice Questions (MCQs)

Tick (✔) the correct option.

8. A boat is sailing North East. A little later it is found sailing south. Through what angle has it turned?

(a) $150^{\circ}$

(b) $180^{\circ}$

(c) $135^{\circ}$

(d) $45^{\circ}$

9. The minute hand when it moves $330^{\circ}$ from $110^{\prime}$ Clock is now at

(a) 9 O clock $^{\prime}$

(b) $110^{\prime}$ Clock

(c) 12 O' Clock

(d) 10 O' Clock