# Brilliant’s Composite Mathematics Class 7 Solutions Chapter 7

## 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

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 × 10dust 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, D= 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 :

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

= 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

L= 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 :

Updated: June 10, 2023 — 6:15 am