Exercise 17 B
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 .
(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 .
(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 .
(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 .
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
(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
(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
(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
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
(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
(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
(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
(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
(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
(b) 8 minutes
Sol :
As we know minute hand moves 6º per minutes . Then , in 8 minutes it covers 6×8 = 48º
(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
(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
(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
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 :
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 :
Question 6
Add the following .
(i) 36º 45' and 56º 87'
(ii) 47º 28' 55'' and 27º 35' 49''
Question 7
Subtracting the following
(i) 39º 39' from 50º 15'
(ii) 26º 31' from 50º 8'
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
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