Warning: Undefined array key "https://nctbsolution.com/brilliants-composite-mathematics-class-7-solutions/" in /home/862143.cloudwaysapps.com/hpawmczmfj/public_html/wp-content/plugins/wpa-seo-auto-linker/wpa-seo-auto-linker.php on line 192
Brilliant’s Composite Mathematics Class 7 Solutions Chapter 7 Direct and Inverse Variations
Welcome to NCTB Solutions. Here with this post we are going to help 7th class students for the Solutions of Brilliant’s Composite Mathematics Class 7 Math Book, Chapter 7, Direct and Inverse Variations. Here students can easily find step by step solutions of all the problems for Direct and Inverse Variations, Exercise 7.1, 7.2, 7.3, 7.4 and 7.5 Also here our mathematics teacher’s are solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily.
Direct and Inverse Variations Exercise 7.1 Solution
Question no – (3)
Solution :
125 steps walking distance = 50 m
∴ 1 steps walking distance = 50/125 m
∴ 80 steps walking distance
= 50/125 × 80 m
= 32 m
Therefore, A will cover 32 meter in 80 steps.
Question no – (4)
Solution :
4 tickets price = 220
∴ 1 tickets price = 220/4
∴ 9 tickets price
= 220/4 × 9
= 495 Rs.
Therefore, the price of 9 such tickets will be 495 Rs.
Question no – (5)
Solution :
1750 oranges can be packed = 50 basket
∴ 1 oranges can be packed = 50/1750
∴ 945 oranges can be packed
= 50 × 945/1750
= 27 baskets
Therefore, there will be 27 baskets required to pack 945 oranges.
Question no – (6)
Solution :
1280 bananas packed = 160 cartons
∴ 1 bananas packed = 160/1280 cartons
∴ 2720 bananas packed,
= 160 × 272/1280
= 340 cartons
Hence, 340 cartons will be required to pack 2720 bananas.
Question no – (7)
Solution :
(i)
| Shelf length | copies |
| 3.6 | 144 |
| 5.5 | ?(x) |
∴ x = 5.5 × 144/3.6
= 55 × 144/36
= 220
(ii)
| Copies | shelf length |
| 144 | 3.6 |
| 200 | ?(x) |
∴ x = 200 × 3.6/144
= 5 m
Question no – (8)
Solution :
| Files | Bundle |
| 21 | 2.8 |
| 135 | ?(x) |
∴ x = 135 × 2.8/21
= 18 cm
Therefore, the thickness of the bundle will be 18 cm.
Question no – (9)
Solution :
| Distance | Bus fare |
| 560 | 560 |
| 364 | ?(x) |
∴ x = 560 × 364/560
= 364 Rs.
Therefore, the journey of the bus fare will be 364 Rs.
Question no – (10)
Solution :
| No of ticket | Ticket fare |
| 2 | 7640 |
| 11 | ?(x) |
∴ x = 7640 × 11/2
= 42020 Rs.
Thus, the amount 42020 Rs will be required to purchase 11 such tickets.
Question no – (11)
Solution :
(i)
| Cloth length | Price |
| 5.6 | 252 |
| 6.4 | ? (x) |
∴ x = 6.4 × 252/5.6
= 288
(ii)
| Price | Cloth length |
| 252 | 5.6 |
| 405 | ?(x) |
∴ x = 405 × 5.6/252
= 9 m
Question no – (12)
Solution :
As we know that, 1 hr = 60 min
| Typing speed | Word |
| 60 | 2520 |
| 1 | ? |
∴ Typing speed = 2520/60
= 42 word
Therefore, Manorma can type 42 words per minute.
Question no – (13)
Solution :
(i)
| Men power | Tunnel length |
| 12 | 5 |
| 12 | ?(x) |
∴ Tunnel length
= 12 × 5/15
= 4 m
(ii)
| Tunnel length | Men power |
| 5 | 5 |
| 18 | ?(x) |
∴ x = 18 × 15/5
= 54 m
Question no – (14)
Solution :
| Day | Wages |
| 16 | 528 |
| 28 | ?(x) |
∴ x = 28 × 528/16
= 924 Rs.
Therefore, his wages for 28 days will be 924 Rs.
Question no – (15)
Solution :
| No of cost | Cost |
| 18 | 42 |
| 28 | ? (x) |
∴ x = 33 × 42/18
= 77
Therefore, the cost of 33 oranges will be 77 Rs.
Question no – (16)
Solution :
| Day | Pounds |
| 30 | 7.8 × 108 |
| 365 | ? (x) |
∴ x = 365 × 7.8 × 108
= 94.84 × 108
Hence, it will pick up 94.84 × 108 dust in 365 days
Question no – (17)
Solution :
(i)
| Height | Length |
| 6 | 7.2 |
| 8.5 | ? (x) |
∴ 7.2 × 8.5/6
= 10.2 metres
Hence, the length will be 10.2 metres
(ii)
| Length | Height |
| 7.2 | 6 |
| 15.3 | ? (x) |
∴ x = 15.3 × 7.2
= 12.75 metres
Therefore, the height of the pole will be 12.75 metres.
Direct and Inverse Variations Exercise 7.2 Solution
Question no – (2)
Solution :
| Time | Distance |
| 15 | 48 |
| 10 | ?(x) |
∴ x = 48 × 10/15
= 32 km/hour
Therefore, the increase of the car speed will be 32 km/hour.
Question no – (3)
Solution :
| Distance | Time |
| 42 | 8 |
| 56 | ? (x) |
∴ x = 8 × 56/42
= 32/3
= 10.6 hours
Hence, it will take 10.6 hours to travel the same distance.
Question no – (5)
Solution :
| Time | No of cow |
| 6 | 66 |
| 9 | ? (x) |
∴ x = 9 × 66/6
= 99 cows
Question no – (6)
Solution :
| Day | Men |
| 20 | 24 |
| 16 | ? (x) |
∴ x = 16 × 24/20
= 19.2
= 20 men
Hence, 20 men will reap the same field in 16 days.
Question no – (7)
Solution :
| Men | Days |
| 32 | 21 |
| 28 | ? (x) |
∴ x = 28 × 21/32
= 13.375 days
Therefore, 28 me will take 13.375 days to dig the tunnel.
Question no – (8)
Solution :
| Men | Days |
| 40 | 21 |
| 30 | ? (x) |
∴ x = 30 × 21/40
= 63/4
= 15 3/4 days
Question no – (9)
Solution :
| Persons | Days |
| 360 | 20 |
| 60 | ? (x) |
∴ x = 60 × 20/360
= 3 1/3 days
Therefore, the provision will last 3 1/3 days.
Question no – (10)
Solution :
| Days | Boys |
| 50 | 315 |
| 30 | ? (x) |
∴ x = 30 × 315/50
= 189 boys
Therefore, 189 boys joined the hostel.
Question no – (11)
Solution :
| Days | Men |
| 30 | 400 |
| 24 | ? (x) |
∴ x = 24 × 400/30
= 120 men
Therefore, in the garrison 120 men arrived.
Direct and Inverse Variations Exercise 7.3 Solution
Question no – (3)
Solution :
Let, d1 = 20, ; d1 = ?, t1
Hence, t1 = 2 seconds ; t1 = 16 seconds
According to questions,
dx + 2
∴ d = kt2
or, d1 = kt12
or, k = d1/t12 = 20/22
= 20/4 = 5
or, k = 5
Now, d2 = kt22
or, d2 = 5 (16)2
= 1280 metres
Therefore, the object will fall 1280 metres in 16 seconds.
Question no – (4)
Solution :
As per the question,
20 m = 2000 cm
1 m = 100 cm
∴ 100 cm stick cast = 55 cm shadow
∴ 1 cm stick cast = 55/100 cm shadow
Let, length of pole = x cm
According to questions,
= 50/100 × x = 200
or, x = 200 × 100/55
= 4000/11
= 3636.36 cm
= 36.36 metres
Therefore, the height of the pole will be 36.36 metres.
Question no – (5)
Solution :
d1 = 40; d2 = ?
F1 = 800 grams;
f2 = (800 + 200) = 100 grams
M1, = 1500 soldier;
M2 = 1500 + 500 = 20 soldiers
We know that,
d1 F1 M1 = d2 F2 M2
or, d2 = d1 F1 M1/F1 M2
= 40 × 800 × 1500/1000 × 2000
= 48 Days
Therefore, the the provisions will last 48 Days.
Question no – (6)
Solution :
As per given question,
M1 = 5 ; M2 = 6
L1 = 200x ; L2 = 160x
D1 = 12 ; D2 = ?
As we know that,
M1 L1 D1 M2L2D2
or, D2 = M1L1D1/M2L2
= 5 × 200 × 12/6 × 160
= 25/4
= 6 1/4 days
Therefore, 6 men will build the similar wall in 6 1/4 days.
Question no – (7)
Solution :
T1 = 4 hr ; 12 minutes = 4.2 hour s1 = 44.8 kh/hour
Distance = 44.8 × 4.2 = 188.16 km
Now, s2 = 57.6 km/hour
T2 = distance/s2
= 188.16/57.6
= 3.266
= 3 hours 12 minutes
Therefore, now to cover the same distance he will take 3 hours and 12 minutes.
Question no – (8)
Solution :
According to the given question,
T1 = 10 hour;
S1 = 48 km/hour;
T2 = 8
Let, speed increased = x km/hour
∴ S2 = 48 + x
According to the questions,
T1 S1 = T2 S2
or, 10 × 48 = 8 (48 + x)
or, 480 = 8x + 384
or, 8x = 480 – 384 = 96
or, x = 96/8
= 12 km/hour
Therefore, its speed should be increased by 12kkm/hour.
Question no – (9)
Solution :
Given, 200 US = 2580.50
∴ 1 US = 2580.50/200 = 12.90
∴ 12.09 = 1 US
∴ 1 = 1/12.90
∴ 14192.75 = 1/12.90 × 14192.75
= 1100 US dollars
Therefore, 1100 US dollars will be obtained.
Direct and Inverse Variations Exercise 7.4 Solution
Question no – (1)
Solution :
25 baskets in 35 days
∴ 1 basket in 35/25 days
∴ 110 baskets in 35 × 110/25 days
= 154 days
Question no – (2)
Solution :
| Days | Time |
| 18 | 8 |
| 12 | ? (x) |
∴ (x) 12 × 8/18
= 16/3
= 5.3 hour
Therefore, to complete the work in 12 days he will need to work 5.3 hour of a day.
Question no – (3)
Solution :
As per the question,
Sham = 56 hours
Paul = 84 hours
Neelkanthan = 280 hours
∴ Together they will work,
= 56 + 84 + 28/3
= 168/3
= 56 hour
Therefore, they will take 56 hours to weave the same number of chairs.
Question no – (4)
Solution :
(A + B) is 1 day polish = 1/25
A alone can do 1/3 of this job in 15 days
A alone can do,
= 3 × 15 = 45 days
∴ A’s 1 day’s polish = 1/45
B’s 1 days polish,
= 1/25 – 1/45
= 9 – 5/225 = 4/225
∴ B 1 days polish,
= 1/4/225 = 225/4
= 56 1/4 days
Hence, in 56 1/4 days ‘B’ alone can polish the floor of the building.
Question no – (5)
Solution :
A’s work done = 1/24
B’s work done = 1/30
∴ A + B work done,
= 1/24 + 1/30
= 5 + 5/120
= 9/120
= 3/40
∴ A + B work can finish work = 40/3
= 13 1/3 days
Therefore, if they work together they will finish the work in 13 1/3 days.
Question no – (6)
Solution :
As per the given question,
1st man’s work done = 1/3 part
2nd man’s work done = 1/6 part
∴ Both man’s done = 1/3 + 1/6
= 2 + 1/6
= 3/6
= 1/2 part
Both complete 1/2 part = 1 day
∴ Both complete 1 part = 1/1/2
= 2 day
Thus, if they work together they will finish the work in 2 days.
Question no – (7)
Solution :
A and B in 10 day complete = 1 part
∴ A and B in 10 day complete = 1/10 part
A and B in 1 day complete = 1/15 part
B and A in 1 day complete = 1/10 – 1/15
= 3 – 2/30
= 1/30 part
B and 1/30 part complete = 1 day
∴ B and 1 part complete = 1/1/30
= 30 days
Therefore, B alone can take 30 days to do the same work.
Question no – (8)
Solution :
Sumon Complete the work = x
Sumon complete work in 1 hr = 1/x
Raju complete work in 1 hr = 1/15
Together complete work in 1 hr = 1/6
According to question,
1/x + 1/15 = 1/6
or, 1/x = 1/6 – 1/15
= 5 – 2/30
= 3/30 = 1/10
or, 1/x = 10 hr
Therefore, Suman alone can do the work in 10 hours.
Question no – (9)
Solution :
A’s 1 day work = 1/3
B’s 1 day work = 1/5
Rest work = 1/3 – 1/5
= 5 – 3/15
= 2/15
∴ B complete work = 15/2
= 7.5 days
Therefore, B will take 7.5 days to complete the work alone.
Question no – (10)
Solution :
1st men work done = 15/2 = 7.5 days
2nd men work done = 1/4 = days
∴ 1st men work x for 2hr = 2/6 = 1/3 part
∴ work remain = 1 – 1/3 = 2/3 work
Both 1 hr both = 1/6 + 1/4 = 4 + 6/24
= 10/24
= 5/12
∴ 1 work = 12/5 hr
∴ 2/3 work = 12/5 × 2/3 = 8/5
= 1.6 hr
Hence, the work will completed in 1.6 hours.
Question no – (11)
Solution :
Together work = 1/A + 1/B + 1/C
= 1/8 + 1/12 + 1/15
= 15 + 1 + 8/120
= 33/120
Together they can work = 120/33
= 40/11
= 3 7/11 days
Therefore, if they work together they will take 3 7/11 days.
Question no – (12)
Solution :
A’s work done = 1/16 part
B’s work done = 1/24 part
(A + B + C) work done = 1/8 part
∴ C work done = 1/8 – (1/16 + 1/24)
= 6 – 2 – 3/48
= 1/48
Therefore, C can finish the work in 48 hours.
Question no – (13)
Solution :
A’s work done = 1/9 hr
B’s work done = 1/18 hr
C’s work done = 1/12 hr
∴ (A + B + C) is work done = 1/9 + 1/185 + 1/12
= 8 + 9 + 6/72 = 23/72
Therefore, (A + B + C) complete the work 72/23 hours.
Question no – (14)
Solution :
As per the question,
A’s work done = 1/10 days
B’s work done = 1/12 days
C’s work done = 1/15 days
∴ (A + B + C) work done = 1/10 + 1/12 + 1/15
= 6 + 5 + 4/60
= 15/16
= 1/4
(A + B + C) Can complete = 4 days
A’s work done in 4 days = 4/10 days
= 2/5 days
B’s work done in 4 days = 4/12 days
= 1/3 days
C’s work done in 4 days
= 4/15 days
Question no – (15)
Solution :
(A + B + C) is work done = 1/8
A’s work done = 1/20
B’s work done = 1/24
∴ C’s work done = 1/8 – (1/20 + 1/24)
= 15 – 6 – 5/20
= 4/120 = 4/120 = 1/30
Therefore, ‘C’ can complete the work in 1/30 days
Question no – (16)
Solution :
Let, C can finish the work x day
A’s work done = 1/40
B’s work = 1/30
C’s work done = 1/x
∴ (A + B + C) is work done (1/40 + 1/30 + 1/x)
Now, according to the question,
= 1/40 + 1/30 + 1/x = 1/10
or, 1/x = 1/10 (1/40 + 1/30)
= 1/10 – [20 + 4/120]
= 1/10 – 7/120
or, 1/x = 12-7/120 = 5/120 = 1/24
or, x = 24
Question no – (17)
Solution :
A’s work done = 1/14 the work
B’s work done = 1/21 the work
∴ (A + B)is work done in 7 days,
= 7 (1/14 + 1/21)
= [3+2/42] × 7
= 5/12 × 7 = 5/6 work
∴ Reaming work = 1 – 5/6 = 6 – 5/6 = 1/6
B can do 1 work = 21 day
∴ B can do 1/6 work
= 21 × 1/6 work = 7/2
= 3.5
∴ B can work do Total no of days,
= 7 + 3.5
= 10.5 days
Therefore, B will complete the work in 10.5 days.
Question no – (18)
Solution :
Time lift = 3 – 1 = 2 day
B takes 3 days
∴ A takes 2 days
Difference of time = 60 days
∴ B takes = (3/2 × 60) = 90 days
A’s work done = 1/30
B’s work done = 1/90
∴ (A + B)’s work done,
= (1/30 + 1/90)
= 4/90 = 2/45
∴ (A + B)’s can do the work in,
= 45/2
= 22 1/2 days
Hence, they will take 22 1/2 dats to complete the work by working together.
Question no – (19)
Solution :
According to the given question,
A’s work done = 1/15 part
B’s work done = 1/12 part
C’s work done = 1/2 part
∴ (A + B + C)’s work done
= 1/15 + 1/12 + 1/20
= 4 + 5 + 3/60
= 12/62 = 1/5
= 5 – 2/5
= 3/5
∴ (A + B)’s work done
= 1/12 + 1/15
= 4 + 5/60
= 9/60
= 3/20
(A + B) complete the remaining work
= 3/5/3/20 = 2/8 × 20/3
= 4 day
Question no – (20)
Solution :
As per the given question,
A’s work done = 1/6
B’s work done = 1/8
(A + B)’s work done = 1/6 + 1/8
= 4 + 3/24 = 7/24
Remaining work of A,
= 1 – 1/6
= 6 – 1/6
= 5/6
Total work done by (A + B) in 1 hr = 7/24
∴ 5/6 WORK DONE BY (A + B) in 1 hr
= 24/7 × 5/6
= 20/7
Total time to finish the work,
= 2 + 120/42
= 84 + 190/42
= 204/42
= 4.86 hr
Therefore, they will take 4.86 hours to complete the work.
Question no – (21)
Solution :
According to the question,
(A + B) work done = 1/12
(B + C)’s work done = 1/15
(B + A)’s work done = 1/20
2 (A + B + C) work done = 1/12 + 1/15 + 1/20 = 5 + 4 + 3/60 = 12/60
∴ A + B + C work done = 1/5 × 2 = 1/10
∴ A’s work done = (A + B + C) – (B + C) = A + B + C – B – C
= 1/10 – 1/15 = 3 – 2/30
= 1/30
Hence, A can do the work in 30 days.
Question no – (22)
Solution :
In the given question,
A’s work done = 1/6
B’s work done = 1/6
C’s work done = 1/12
∴ (A + B + C)‘s work done,
= (1/6 + 1/6 + 1/12)
= 2 + 2 + 1/12
∴ x (A + B + C) take time = 12/5
= 2 2/5 hr
Therefore, three taps will take 2 2/5 hours to fill the empty tank.
Question no – (23)
Solution :
As per the given question,
A’s work done = 1/12
B’s work done = 1/15
C’s work done = – 1/10
∴ (A + B + C)’s work done
= 1/12 + 1/15 – 1/10
= 5 + 4 – 6/60
= 1/20
∴ (A + B + C) to fill done unit
= 1/1/20
= 20 min
Hence, they will take 20 minute to fill the tank completely.
Question no – (24)
Solution :
A’s filling done = 1/10
B’s filling done = 1/15
∴ (A + B)’s filling done
= 1/10 + 1/15
= 3 + 2/30
= 5/30
= 1/6 unit
∴ Tank will filled by (A + B)
= 1/1/6
= 6 hr
Therefore, they will take 6 hours to fill it completely.
Question no – (25)
Solution :
A’s filling done = 1/5
B’s filling done = 1/6
∴ (A + B)’s filling done,
= 1/5 – 1/6
= 6 – 5/30
= 1/30
Therefore, (A + B)’s will take 30 hours to fill the tank completely.
Direct and Inverse Variations Exercise 7.5 Solution
Question no – (1)
Solution :
Speed = 45 km/hr = 45/3600
Time = 6 sec
Distance = S × T
= 45/3600 × 6
= 45 × 1000/600
= 75 m
Therefore, the train will go 75 meter in 6 second.
Question no – (2)
Solution :
Given, Speed = 65 km/hour
And, Time = 3 hr
X = ?
We know, Distance = Speed × Time
= 65 × 3
= 195 km
Therefore, the distance between Agra Cantt. and N. Delhi will be 195 km.
Question no – (3)
Solution :
Given, Distance = 200 m = 0.2 km
Time = 18 sec
= 18/3600
= 3/600
= 1/200 hr
As we know, Speed = D/T
= 0.2/1/200
= 0.2 × 200
= 40 km/hr
Therefore, the car is traveling at 40 km/hours.
Question no – (4)
Solution :
S = 2 1/2 = 5/2 = 2.5 km/hr ;
T1 = 6min = 6/60 = 0.1 hr
S1 = 2.5 × 3 = 7.5
Difference of speed = 3 – 2.5 = 0.5
Distance = 0.26 × 7.5/0.5 = 4 km
T2 = 10 min = 10/60 = 0.16 hr
Total Time = 0.1 + 0.16
= 0.26
Hence, the distance of school will be 4 km.
Question no – (5)
Solution :
Given, Sound travels at 335 metres per second in air
∴ Speed of sound in km/hour,
= 335 × 18/5
= 120 km/hr
Hence, the speed of sound will be 120 km/hr
Question no – (6)
Solution :
A gun is fired at a distance = 1.7 km away from Saurabh.
Saurabh hears the sound = 5 seconds later.
As we know, Speed = D/T
∴ Speed = 17 × 1000/5
= 1700/5
= 240 m/s
Thus, the speed at which sound travels is 240 m/s
Question no – (7)
Solution :
According to the given question,
A gun is fired at a distance of 1.34 km away from Chinka.
She hears the sound 4 seconds later.
∴ Speed = 1.24 × 1000/4
= 336 m/s
Hence, the speed at which sound travels is 336 m/s
Question no – (8)
Solution :
Let, distance = D
For 1st case – T = D/3.5
For 2nd case – T = D/45
Total Time = D/35 + D/45
= 9D + 7D/315
= 16D/315
Now, according to the question,
Speed = Total DIS/Total T
= D + D/16D/315 = 2D/16 × 315
= 39 3/8 km/hr
Therefore, the average speed for the entire journey will be 39 3/8 km/hr.
Question no – (9)
Solution :
Initial speed = S1 = 15 km/hr
Final speed = SF= 10 km/hr
Distance = D = 18 km
Time taken by tonga,
= 18/15
= 1.2 hr
∴ Time taken by cycle,
= 18/10
= 1.8 hr
∴ Average speed,
= 18 + 18/1.2 + 1.8
= 36/3
= 12 km/hr
Question no – (10)
Solution :
Total Dis = 72 km
Total time = 1 hr 30 min
S1 = 54 km/hr = 90 min
For 1st 36 km – T1 = 36/54 = 2/3 hr
Foe remaining 36 km,
= 2 × 60/3
= 40 min
∴ Speed = (72 – 36)/(90 – 40)/60
= 36/50/60
= 36/50 × 60
= 43.2 km/hr
Therefore, it need to travel 43.2 km/hr for the rest of the distance.
Question no – (11)
Solution :
Let, S1 = 4 km/hr
S2 = 12 km/hr
D = 6 km
t1 = 6/4 = 1.5 s
t2 = 6/12
= 0.5 s
Now, the average speed,
= 6 + 6/1.5 + 0.5
= 12/2
= 6 km/hr
Therefore, the man’s average speed for the double journey will be 6 km/hr.
Question no – (12)
Solution :
As per the given question,
8 m/sec = 0.0008/1/3600
= 0.008 × 3600
= 28.8 km/hr
∴ 36 > 28.8 km/hr
∴ 36 km/hr is greater,
36 km/hr = 36000/3600
= 10 m/hr
Now, the difference of distance,
= 10 – 8
= 2 m
Hence, difference in the distances travelled in 1 second will be 2 m.
Question no – (13)
Solution :
According to the given question,
Distance = 6.5 km
Time = 26 min = 26/60 hr
Average speed = 6.5/26/60
= 6.5 × 60/26
= 2390/26
= 15 km/hr
Thus, Priyanka’s average speed will be 15 km/hr.
Question no – (14)
Solution :
Given, Speed = 48 km/hr
and, Time = 8 hr 48 min = 44/5 sec
As we know, Distance = S × T
∴ Distance,
= 48 × 44/5
= 422.4 km
Therefore, the distance of the journey will be 422.4 km.
Question no – (15)
Solution :
First, Speed = 54 km/hr
= 54 × 1000/3600
= 15 m/sec
∴ Distance = 300 m
∴ Time = 300/15
= 20 sec
Therefore, the train will need 20 sec pass the telegraph post.
Question no – (16)
Solution :
Speed = 40.5 km/hr
= 40.5 × 1000/3600
Distance = 270 m = 11.25
∴ Time = D/S
= 270/11.25
= 24 sec
Therefore, it will take 24 sec to cross a tree.
Question no – (17)
Solution :
Speed = 30 km/hr
= 30 × 1000/3600
= 8.3 m/s
Distance = 750 m
∴ Time = 150/8.3
= 18 second
Thus, it will take 18 second to pass a standing man.
Question no – (18)
Solution :
Speed = 36 km/h
= 36 × 1000/3600
= 10 m/s
Time = 25 sec
Distance = 25 × 10
= 250 metres
Therefore, the length of the train will be 250 metres.
Question no – (19)
Solution :
Let, D1 = 180 m ; S = 50 km/h
= 50 × 1000/3600
= 12.5 m/s
and, D2 = 540 m
∴ Total distance = 540 + 180 = 720 m
∴ Time = 720/12.5
= 57.6 sec
Hence, it will take 57.6 second to pass a platform 570 metres long.
Question no – (20)
Solution :
Let, D1 = train length = 80
and, D2 = platform = 220 m
∴ Total distance = 220 + 80 = 300 m
∴ Now, the time,
= 300/12.5
= 24 sec
Therefore, it will take 24 sec to cross a platform 220 metres long.
Question no – (21)
Solution :
Speed of the train relative to number,
= 60 + 6
= 66
= 66 × 100/3600
= 55/3
Distance = 220
∴ Time = 220/55/3
= 660/55
= 12 second
Therefore, the train will cross the man in 12 second.
Question no – (22)
Solution :
Relative speed,
= 62 – 8
= 54 km/h
∴ Distance = 135 m
= 54 × 5/8
= 15 m/second
∴ Time = 135/15
= 9 second
Hence, the train will cross the man in 9 second.
Question no – (23)
Solution :
Let, length of Bridge = D
length of Train = 100 m
Speed = 36 km/hr
= 36 × 5/18
= 10 m/hr
Now, according to questions,
100 + D = 10 × 25 = 250
or, D = 250 – 100
= 150 m
Therefore, the length of the bridge will be 150 meter.
Question no – (24)
Solution :
1st, Speed = 120/5
= 24 m/second
2nd, Traveled distance
= 120 + 180
= 300 m
∴ Now, the time = 300/24
= 50/4
= 12.5 second
Hence, it will take 12.5 second to cross a railway platform 180 metres long.
Question no – (25)
Solution :
L1 = 150
L1 = 180 m
Total = 150 + 180
= 330 m
Total = S = 30 + 24
= 54 km/hr = 54 × 5/18
= 15 m/sec
∴ Now the time,
= 330/15
= 22 second
Therefore, they will cross each other in 22 second.
Question no – (26)
Solution :
Let, L1 = 110
And, L2 = 100 m
Total = 100 + 100 = 210 m
S1 = 66 km/hr
S2 = 39 km/hr
Total = S = 66 – 39
= 27 km/hr
= 27 × 5/8 = 7.5 m/sec
∴ Time = 210/7.85
= 28 second
Therefore, it will take them 28 second to be clear of each other.
Next Chapter Solution :
👉 Chapter 8 👈
