Balbharati Solutions Class 5 Mathematics Problems on Measurement

Welcome to NCTB Solutions. Here with this post we are going to help 5th class students for the Solutions of Balbharati Class 5 Math Book, Problem Set 46 and 47, Problems on Measurement. Here students can easily find step by step solutions of all the problems for Problems on Measurement. Also our Expert Mathematic Teacher’s solved all the problems with easily understandable methods with proper guidance so that all the students can understand easily. Here in this post students will get Problems on Measurement solutions. Here all the solutions are based on Maharashtra State Board latest syllabus.

Problems on Measurement all Question Solutions :

Problem Set 46 :

Question no – (1)

Solution :

(1) Rs 9, 50 paise + Rs 14, 60 paise

Rupee :

= 9 + 14

= 23

Paise :

50 + 60

= 110

110 paise = 100 paise + 10 paise

We know, 1 rupee = 100 paise

Add 1₹ in 23₹ is 24₹

So, The answer is ₹24, 10 paise.

(2) 6 cm 5 mm + 7 cm 9 mm

cm = (6 + 7) = 13

mm = (5 + 9) = 14

Therefore,

14 mm = 10 mm + 4 mm

We know, 1 cm = 10 mm

Add 1 cm in 13 cm is 14 cm

So, The answer is 14 cm 4m

(3) 22 m 50 cm + 25 m 75 cm

Meter = (22 + 25)

= 47

Centimetre = (50 + 75)

= 125

Therefore,

125 cm = 100 cm + 25 cm

We know, 1 m = 100 cm

Add 1m in 47 m is 48 m

So, The answer is 48 m 25 cm.

(4) 15 km 740 m + 13 km 950 m

Kilo meter :

= (15 + 13) = 28

Metre :

= (740 + 950) = 1690

Therefore,

1690 m = 1000m + 690 m

We know, 1 km = 1000m

Add 1 km in 28 km is 29 km

So , The answer is 29 km 690 m

(5) 25 kg 650 g + 29 kg 770 g

Kilo gram :

= (25 + 29) = 54

Gram :

= (650 + 770) = 1420

Therefore,

1520 g = 1000 g + 420 g

We know, 1 kg = 1000 g

Add 1 kg in 54 kg is 55 kg

So, The answer is 55 kg 420 g

(6) 19 l 840 ml + 25 l 250 ml

Liter :

= (19 + 25) = 44

Milliliter :

= (840 + 250) = 1090

Therefore,

1090 = 1000 + 90

We know, 1 l = 1000 ml

Add 1l in 44 l is 45 l

Thus, the answer is 45 l 90 ml.

Question no – (2)

Solution :

(1) ₹19, 50 paise – ₹12, 60 paise

Rupee :

= (19 -12) = 06

Paise :

= (50 + 60) = 90

Therefore,

We cannot subtract 60 paise into 50 paise

We know, ₹ 1 = 100 paise

So, the answer is

₹6.90 paise.

(2) 24 cm 2 mm – 3 cm 8 mm

Centimetre :

24 – 3 = 20

Millimetre :

(2 – 8) = 4

Therefore,

We cannot subtract 2 mm into 8 mm

We know,

1 cm = 10 mm

So, The answer is

20 cm 4 mm

(3) 20 m 30 cm – 17 m 60 cm

Miter :

= (20 – 17) = 02

Centimetre :

(30 – 60) = 70

Therefore,

We cannot subtract 30 cm into 60 cm

We know,

1m = 100 cm

So , The ans is

2m 70 cm

(4) 40 km 255 m – 17 km 960 m

Kilometre :

(40 – 17) = 22

Miter :

(255 – 960) = 265

Therefore,

We cannot subtract 255 m into 960 m

We know,

1 km = 1000m

So, The answer is

22 km 265 m

(5) 35 kg 150 g – 26 kg 470 g

Kilo gram :

(35 – 26) = 08

Gram :

(150 – 470) = 680

Therefore,

We cannot subtract 150 g into 470 g

We know,

1 kg = 1000 g

So, The answer is 8 kg 680 g

(6) 46l 200ml – 38l 750ml

Litre :

(46 – 38) = 07

Millilitre :

(200 – 750) = 450

Therefore,

We cannot subtract 200 ml into 750 ml

We know,

1l = 1000 ml

So, The answer is 7l 450 ml

Problem Set 47 :

Question no – (1)

Solution :

20 l 450 ml of milk to the children in an Ashramshala

28 l 800 ml to the children in an orphanage

We add, Milk donate to the children in an Ashramshala and Milk donate to the children in an orphanage

Liter :

= (20 + 28) = 49

Millilitre :

= (450 + 800) = 250

So, the answer is 49 l 250 ml milk Ajay donate.

Question no – (2)

Solution :

College students cleaned 1 km 750 m of a village road

2 km 575 m long of a village road.

Road remained to be cleaned = road length – road to be cleaned

Kilometre :

-1
____
0

Miter:

575

– 750
_____
825

Therefore, the answer is 825 m Road remain to be cleaned.

Question no – (3)

Solution :

Babhulgaon used 21,250 litres of treated waste water in the fields

Samvatsar used 31,350 litres of similar water

Treated waste water used in all = Babhulgaon used treated waste water + Samvatsar used treated waste water.

Therefore,

= (21250 + 3135)

= 52600

Hence, the answer is 52600 litres treated waste water used in all.

Question no – (4)

Solution :

1/2 a litre of milk costs 22 rupees

We know, 1/2 + 1/2 = 1

22 rupees + 22 rupees = 44 rupees

So, 1 litre of milk costs 44 rupees

For 7 liters,

We multiply 7 into 1 litre of milk costs

7 × 44 = 308 rupees

7 litres costs 308 rupees.

Question no – (5)

Solution :

The speed of a motorcycle is 40 km per hour

hour and a quarter means 1 + 1/4

1 hour = 40 km

And 1/4 = 40 × 1/4 = 10 km

1 + 1/4 = 40 km + 10 km

1 + 1/4 = 50 km

So, 50 km travel in an hour and a quarter.

Question no – (6)

Solution :

A man walks at a speed of 4 kmph

In 1 hour Aman walks 4 km

For same unit

Convert hour into minutes

1 hour = 60 minutes

It means

man walks 2 km in 30 minutes

And man walks 1 km in 15 minutes

If man walks 1 km in 15 minutes

Then in 3 km man walks

3 × 15 = 45 minutes

Therefore, Man walks 3 km in 45 minutes.

Question no – (7)

Solution :

A rickshaw travels at a speed of 30 kmph

A rickshaw travels 30 km in 1 hours

Convert the hour in minutes

1 hour = 60 minutes

So that,

60 minutes/2 = 30 minutes

30 km/2 = 15 km

in 30 minutes travels 15 km

30 minutes/2 = 15 minutes

15 km/2 = 7.5 km

In 15 minutes travels 15 km

three quarters of an hour means 45 minutes

If we add 30 minutes and 15 minutes it gives 45 minutes

It means in 45 minutes travels

15 km + 7.5 km = 22.5 km

22.5 km travel in three quarters of an hour.

Question no – (8)

Solution :

1 Kg = 1000 gm

So, 1/4 Kg = 1000/4 = 200 gm

We know

Three quarters = 1/4 + 1/4 + 1/4

Three quarters = 250 gm + 250 gm + 250 gm

Three quarters = 750 gm

Kg :

+ 5 = 5

Gm :

= (750 + 500) = 1250

Therefore,

1250 gm = 1000 gm + 250 gm

1250 gm = 1 kg + 250 gm

We add 1 kg in 5 kg is 6 kg

So, 6 Kg 250 gm garbage they collect in all.

Question no – (9)

Solution :

Given, One shirt needs 2 m 50 cm of cloth

For 5 shirts = 5 × 2 m 50 cm.

Miter :

= (2 × 5) = 10

Centimetre :

= (50 × 5) = 250

Therefore, 12 m 50 cm cloth need.

Question no – (10)

Solution :

Given, A car travels 60 km in an hour

(1) 2 hours ?

We know,

60 km = 1 hour

So,In 2 hour

2 × 60 km = 120 km

In 2 hour car travels 120 km

(2) 15 minutes?

We know,

1 hour = 60 minutes

60 km = 60 minutes

The half of 60 min = 30 min

The half of 60 km = 30 km

The half of 30 min =15 min

The half of 30 km =15 km

In 30 min travels 60 km

And In 15 travels 15 km

In 15 minutes car travels 15 km

(3) Half an hour?

We know,

1 hour = 60 minutes

60 km = 60 minutes

The half of 60 min = 30 min

The half of 60 km = 30 km

In half an hour travels 30 km

(4) Three and a half hours?

We know,

60 km = 1 hour

So, In 3 hour

3 × 60 km = 180 km

In 3 hour travels 180 km

1 hour = 60 minutes

The half of 60 min = 30 min

The half of 60 km = 30 km

In half an hour travels 30 km

In Three and a half hours = 180 km + 30 km

In Three and a half hours = 210 km

So, In Three and a half hours travels 210 km.

Question no – (11)

Solution :

Gram :

= (12 × 8)

= 96

Milligram :

= (250 × 8) = 2000

= (96 + 2) = 98 (A s we know 1000 mg = 1 gm)

Explanation :

We know, 1 g =1000 mg

2000 mg = 1000 mg + 1000 mg

2000 mg = 2 g

So, Add 2 g in 96 g

Therefore, 98 gram gold needed to make 8 such bangles.

Question no – (12 )

Solution :

As we know, 1 kg = 1000 g

1 kg 240 g = 1000 g + 240 g

1 kg 240 g = 1240 g

Therefore,

1240 ÷ 20 = 62

∴ 62 pouches made from 1 kg 240 g of clovers.

Question no – (13)

Solution :

Seema’s mother bought 2m 70 cm of cloth for a kurta

Seema’s mother bought 2m 40 cm for a shirt

Add the bought cloth for a kurta and shirt

Miter :

= (2 + 2) = 4

Centimetre :

= (70 + 40) = 110

= 5 m 10 cm

Explanation :

We know,

1m = 100 cm

110 cm = 100 cm + 10 cm

Add 1m in 4m is 5m

Therefore, 5m 10 cm cloth she buy in all.

Question no – (14)

Solution :

A water tank holds 125 l of water

97 l 500ml of the water is used

Subtracted 97 l 500ml from 125 l

Liter :

= (125 – 97)

= 27

Millilitre :

= (000 – 500)

= 500

Explanation :

We cannot subtract 000 from 500

We know,

1 l = 1000 ml

27 l 500ml water remains in the tank.

Question no – (15)

Solution :

57 kg 500 g of wheat from one shop

36 kg 800 g of wheat from another shop

Add them

Kilogram :

= (57 + 36)

= 93

Gram :

= (500 + 800) = 1300

= 94 kg 300 g

Explanation :

We know, 1 kg = 1000 g

1300 g = 1000 g + 300 g

Add 1 kg in 93 kg is 94 Kg

Therefore, 94 kg 300 g wheat he buy altogether.

Question no – (16)

Solution :

Given in the question :

Renu took part in a 100 m race

Renu tripped and fell after running 80 m 50 cm.

Miter :

= (100 – 80)

= 19

Centimetre :

= (00 – 50)

= 50

Explanation :

We cannot subtracted 50 cm from 00

We know, 1 m = 100 cm

Therefore, 19 m 50 cm distance Renu Have left to run.

Question no – (17)

Solution :

A sack had 40 kg 300 grams of vegetables

17 kg 700 g potatoes

13 kg 400 g cabbage

= (17 + 13)

= 30

gram :

= (700 + 400)

= 1100

= 31 kg 100 g

Explanation :

We know,

1 kg = 1000 g

1100 g = 1000 g + 100 g

Add 1 kg in 30 kg is 31 kg

So, total weight of potatoes and cabbage is 31 kg 100 g

Now, Onions weight = Total weight of vegetables – Total weight of potatoes and cabbage

Onions weight = 40 kg 300 g – 31 kg 100 g

Kg :

(40 – 31) = 8

gram :

(300 – 100) = 200

The weight of the onions is 8 kg 200 g.

Question no – (18)

Solution :

Gurminder Singh walked 3km 750m

Parminder Singh walked 2 km 825m

Km :

= (2 – 3 – 2) = 0

Metre :

= 1000 + 750 = 1750

= (750 – 825) = 925

Explanation :

We cannot subtracted 750 m from 925 m

We know,

1 km = 1000 m

Therefore, Gurminder Singh walked 925 m more.

Question no – (19)

Solution :

3 kg 250 g of tomatoes

2 kg 500 g of peas

1 kg 750 g of cauliflower

Total weight = weight of tomatoes + weight of peas + weight of cauliflower

Kg :

= (3 + 2 + 1) = 6

Gram :

= (250 + 500 + 750) = 1500

= 7 kg 500 g

Explanation :

We know,

1 kg = 100 g

1500 g = 1000 g + 500 g

Thus, total weight of the vegetables Suresh bought is 7 kg 500 g.

Question no – (20)

Solution :

Given,

Amravati – 95 km, Bhusawal – 154 km,

Nagpur – 249 km, Jalgaon – 181 km

(i) What is the distance between jalgaon and Amravati if Akola to jalgaon distance is 181 and Akola to Amravati distance is 95?

Akola to jalgaon distance is 181 km

Akola to Amravati distance is 95 km

distance between jalgaon and Amravati = Akola to jalgaon distance – Akola to Amravati distance

distance between jalgaon and Amravati =181 km – 95 km

distance between jalgaon and Amravati = 86 km

distance between jalgaon and Amravati is 86 km

(ii) What is the distance between Bhusawal and Amravati if Akola to Amravati distance is 95km and Akola to bhusawal distance is 154km?

Akola to Amravati distance is 95 km

Akola to Bhusawal distance is 154 km

distance between Bhusawal and Amravati = Akola to Bhusawal distance + Akola to Amravati distance

distance between Bhusawal and Amravati = 154 km – 95 km

distance between Bhusawal and Amravati = 59 km

distance between Bhusawal and Amravati is 59 km.

Question no – (21)

Solution :

Sugar rate is 32 ₹ per kg

Rate of 2.5kg sugar is = 2.5 × 32

Rate of 2.5kg sugar is = 80.00 ₹

Rice rate is 35 ₹ per kg

Rate of 4.0kg Rice is = 4.0 × 35

Rate of 4.0kg Rice is = 140.00 ₹

Chana dal rate is 60 ₹ per kg

Rate of 1.5kg chana dal is = 1.5 × 60

Rate of 1.5kg chana dal is = 90.00 ₹

Toor dal rate is 70₹ per kg

Rate of 3.0kg Toor dal is = 3.0×70

Rate of 3.0kg Toor dal is =210.00 ₹

Wheat rate is 21₹ per kg

Rate of 7.0kg wheat is = 7.0 × 21

Rate of 7.0kg wheat is = 147.00 ₹

Oil rate is 110₹ per kg

Rate of 1.5kg oil is = 1.5 × 110

Rate of 1.5kg oil is = 165.00 ₹

Total cost = sugar cost + rice cost + Chana Dal cost + toor Dal cost +wheat cost +oil cost

Total cost = 80.00 + 140.00 + 90.00 + 210.00 + 147.00 + 165.00

Total cost = 832.00 ₹

Foodstuff :	Weight (kg) :	Rate(₹ per kg) :	Cost :
Sugar	2.5	32	80.00
Rice	4.0	35	140.00
Chana Dal	1.5	60	90.00
Toor Dal	3.0	70	210.00
Wheat	7.0	21	147.00
Oil	1.5	110	165.00
			Total = Rs, 832.00