## NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2

NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-15-ex-15-2/

 Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 15 Chapter Name Probability Exercise Ex 15.2 Number of Questions Solved 5 Category NCERT Solutions

## NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2

Question 1.
Two customers Shyam and Ekta are visiting a particular shop in the same week (Tuesday to Saturday). Each is equally likely to visit the shop on any day as on another day. What is the probability that both will visit the shop on
(i) the same day?
(ii) consecutive days?
(iii) different days?
Solution: Question 2.
A die is numbered in such a way that its faces show the number 1, 2, 2, 3, 3, 6. It is thrown two times and the total score in two throws is noted. Complete the following table which gives a few values of the total score on the two throws: What is the probability that the total score is at least 6?
(i) even
(ii) 6
(iii) at least 6
Solution: Question 3.
A bag contains 5 red balls and some blue balls. If the probability of drawing a blue ball is doubles that of a red ball, determine the number of blue balls in the bag.
Solution: Question 4.
A box contains 12 balls out of which x are black. If one ball is drawn at random from the box, what is the probability that it will be a black ball? If 6 more black balls are put in the box, the probability of drawing a black ball is now double of what it was before. Find x.
Solution: Question 5.
A jar contains 24 marbles, some are green and others are blue. If a marble is drawn at random from the jar, the probability that it is green is 2/3. Find the number of blue balls in the jar.
Solution: We hope the NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 15 Probability Ex 15.2, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4

NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4 are part of NCERT Solutions for Class 10 Maths. Here we have given NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4. https://mcqquestions.guru/ncert-solutions-for-class-10-maths-chapter-14-ex-14-4/

 Board CBSE Textbook NCERT Class Class 10 Subject Maths Chapter Chapter 14 Chapter Name Statistics Exercise Ex 14.4 Number of Questions Solved 3 Category NCERT Solutions

## NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4

Question 1.
The following distribution gives the daily income of 50 workers of a factory. Convert the distribution above to a less than type cumulative frequency distribution, and draw its ogive.
Solution:  Question 2.
During the medical check-up of 35 students of a class, their weights were recorded as follows: Draw a less than type ogive for the given data. Hence obtain the median weight from the graph and verify the result by using the formula.
Solution:   Question 3.
The following table gives production yield per hectare of wheat of 100 farms of a village. Change the distribution to a more than type distribution, and draw its ogive.
Solution:  We hope the NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4, help you. If you have any query regarding NCERT Solutions for Class 10 Maths Chapter 14 Statistics Ex 14.4, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-2-ex-2-2/

 Board CBSE Textbook NCERT Class Class 6 Subject Maths Chapter Chapter 2 Chapter Name Whole Numbers Exercise Ex 2.2 Number of Questions Solved 7 Category NCERT Solutions

## NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2

Question 1.
Find the sum by suitable rearrangement :
(a) 837 + 208 + 363
(b) 1962 + 453 + 1538 + 647.
Solution :
(a) 837 + 208 + 363
= 837 + 363 + 208 = (837 + 363) + 208 = 1200 + 208 = 1408
(b) 1962+ 453+ 1538 + 647
= 1962 + 1538 + 453 + 647 = (1962 + 1538) + (453 + 647)
= 3500+ 1100 = 4600.

Question 2.
Find the product by a suitable rearrangement :
(a) 2 × 1768 × 50
(b) 4 × 166 × 25
(c) 8 × 291 × 125
(d) 625 × 279 × 16
(e) 285 × 5 × 60
(f) 125 × 40 × 8 × 25.
Solution :
(a) 2 × 1768 × 50
= 2 × 50 × 1768 = (2 × 50) × 1768 = 100 × 1768= 1,76,800
(b) 4 × 166 × 25 = 4 × 25 × 166 = (4 × 25) × 166 = 100 × 166 = 16,600
(c) 8 × 291 × 125 = 8 × 125 × 291
= (8 × 125) × 291 = 1000 × 291 =2,91,000
(d) 625 × 279 × 16 = 625 × 16 × 279
= (625 × 16) × 279 = 10000 ×279 = 27,90,000
(e) 285 × 5 × 60 = 285 × (5 × 60) = 285 × 300
= 85,500
(f) 125 × 40 × 8 × 25 = (125 × 40) × (8 × 25) = 5000 × 200 = 10,00,000.

Question 3.
Find the value of the following :
(a) 297 × 17+297 × 3
(b) 54279 × 92 + 8 × 54279
(c) 81265 × 169 – 81265 × 69
(d) 3845 × 5 × 782 + 769 × 25 × 218.
Solution :
(a) 297 × 17 + 297 × 3 = 297 × (17+ 3)
= 297 × 20 = 5940
(b) 54279 × 92 + 8 × 54279
= 54279 × 92 + 54279 × 8 = 54279 × (92 + 8)
= 54279 × 100 = 54,27,900
(c) 81265 × 169-81265 × 69
= 81265 × (169-69)
= 81265 × 100 = 81,26,500
(d) 3845 × 5 × 782 + 769 × 25 × 218
= 3845 × 5 × 782 + 769 × 5 × 5 × 218 = 3845 × 5 × 782 + (769 × 5) × 5 × 218 = 3845 × 5 × 782 + 3845 × 5 × 218 = 3845 × 5 × (782 + 218)
= 3845 × 5 × 1000 = 19225 × 1000 = 1,92,25,000.

Question 4.
Find the product, using suitable properties :
(a) 738 × 103
(b) 854 × 102
(c) 258 × 1008
(d) 1005 × 168.
Solution :
(a) 738 × 103
= 738 × (100+ 3)
= 738 × 100 + 738 × 3 = 73,800 + 2,214 = 76,014
(b) 854 × 102
= 854 × (100 + 2)
= 854 × 100 + 854 × 2 = 85,400+ 1,708 = 87,108
(c) 258 × 1008
= 258 × (1000+ 8)
= 258 × 1000 + 258 × 8 = 2,58,000 + 2,064 = 2,60,064
(d) 1005 × 168
= 168 × 1005
= 168 × (1000+ 5)
= 168 × 1000+ 168 × 5
= 1,68,000 + 840 = 1,68,840.

Question 5.
A taxi driver filled his car petrol tank with 40 liters of petrol on Monday. The next day he filled the tank with 50 liters of petrol. If the petrol costs ₹ 44 per liter, how much did he spend all on petrol?
Solution :
Petrol filled on Monday = 40 litres
Petrol filled the next day = 50 litres
∴ Total petrol filled on the two days = 40 litres + 50 litres = 90 litres
∴ Cost of petrol per litre = ₹ 44
∴ Cost of 90 litres of 7 petrol = ₹ 44 × 90 = ? 3960.

Question 6.
A vendor’supplies 32 liters of milk to a hotel in the morning and 68 liters of milk in the evening. If the milk costs ₹ 15 per liter, how much money is due to the vendor per day?
Solution :
Milk supplied in the morning = 32 liters
Milk supplied in the evening = 68 liters
∴ Milk supplied per day = 32 litres + 68 litres = 100 litres
Cost of milk per liter = ₹ 15
Money due to the vendor per day = Cost of
100 litres of milk = ₹ 15 × 100 = ₹ 1500.

Question 7.
Match the following : Solution : We hope the NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 2 Whole Numbers Ex 2.2, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-1-ex-1-3/

 Board CBSE Textbook NCERT Class Class 6 Subject Maths Chapter Chapter 1 Chapter Name Knowing Our Numbers Exercise Ex 1.3 Number of Questions Solved 3 Category NCERT Solutions

## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3

Question 1.
Estimate each of the following using general rule:
(a) 730 + 998
(b) 796 – 314
(c) 12,904 + 2,888
(d) 28,292 – 21,496
Make ten more such examples of addition, subtraction and estimation of their outcome.
Solution :
(a) 730 + 998 = 700 + 1000 = 1700
(b) 796 – 314 = 800 – 300 = 500
(c) 12,904 + 2,888 = 13,000 + 3,000 = 16,000
(d) 28,292 – 21,496 = 28,000 – 21,000 = 7,000
Ten more such examples of addition, subtraction and estimation of their outcome are as follows:
Ex. 1.720 + 990
Ex. 2.640 + 880
Ex. 3. 749 + 740
Ex. 4.890 – 420
Ex. 5.680 – 370
Ex. 6.585 – 220
Ex. 7.10803 + 3777
Ex. 8.15663 + 2125
Ex. 9. 30990 – 21660
Ex. 10. 40870-19530
Solution :
1. 720 + 990 = 700 + 1000 = 1700
2. 640 + 880 = 600 + 900 = 1500
3. 749 + 740 = 700 + 700 = 1400
4. 890 – 420 = 900 – 400 = 500
5. 680 – 370 = 700 – 400 = 300
6. 585 – 220 = 600 – 200 = 400
7. 10803 + 3777 = 11000 + 4000 = 15000
8. 15663 + 2125 = 16000 + 2000 = 18000
9. 30990 – 21660 = 31000 – 22000 = 9000
10. 40870 – 19530 = 41000 – 20000 = 21000

Question 2.
Give a rough estimate (by rounding off to nearest hundreds) and also a closer estimate (by rounding off to nearest tens):
(a) 439 + 334 4,317
(b) 1,08,734 – 47,599
(c) 8,325 – 491
(d) 4,89,348-48,365 Make four more such examples.
Solution :
(a)

• Rough estimate (Rounding off to nearest hundreds)
439 + 334 + 4317
= 400 + 300 + 4,300 = 5000
• Closer estimate (Rounding off to nearest tens) 439 + 334 + 4317
= 440 + 330 + 4,320 = 5,090.

(b)

• Rough estimate (Rounding off to nearest hundreds) 1,08,734 – 47,599
= 1,08,700 – 47,600 = 61,100
• Closer estimate (Rounding off to nearest tens) 1,08,734 – 47,599
= 1,08,730 – 47,600 = 61,130.

(c)

• Rough estimate (Rounding off to nearest hundreds) 8325 – 491
= 8300 – 500 = 7800
• Closer estimate (Rounding off to nearest tens)
8325 – 491
= 8330 – 490 = 7840.

(d)

• Rough estimate (Rounding off to nearest hundreds)
4,89,348 – 48,365
= 4,89,300 – 48,400 = 4,40,900
• Closer estimate (Rounding off to nearest tens)
4,89,348 – 48,365
= 4,89,350 – 48,370 = 4,40,980

Four more such examples are as follows:
1. 538 + 432 + 5326
2. 2,09, 849 – 57,698
3. 9426 – 395
4. 5,98,459 – 36,463 Sol.
solution :
1. 538 + 432 + 5326
= 500 + 400 + 5300
(Rough estimate Rounding off to nearest hundreds)
= 6200
538 + 432 + 5326
= 540 + 430 + 5330
(Closer estimate Rounding off to nearest tens) = 6300

2. 2,09,849 – 57.698
= 2.09.800 – 57,700
(Rough estimate Rounding off to nearest hun-dreds)
=152100 2,09, 849 – 57,698
= 2,09, 850 – 57,700
(Closer estimate Rounding off to nearest tens) =152150

3. 9426-395
= 9400 – 400 ..
(Rough estimate Rounding off to nearest hundreds)
= 9000 9426 – 395
= 9430 – 400
(Closer estimate Rounding off to nearest tens) = 9030

4. 5,98,459 – 36,463
= 5,98,500 – 36,500
(Rough estimate Rounding off to nearest hundreds)
= 5,62,000 5,98,459 – 36,463
= 5,98,460 – 36,460
(Closer estimate Rounding off to nearest tens) = 5,62,000

Question 3.
Estimate the following products using the general rule:
(a) 578 ×161
(b) 5281 × 3491
(c) 1291 × 592
(d) 9250 × 29
Make four more such examples.
Solution :
(a) 578 × 161
Estimated product = 600 × 200 = 1,20,000
(b) 5281 × 3491
Estimated product = 5000 × 3500 = 1,75,00,000
(c) 1291 × 592
Estimated product = 1300 × 600 = 7,80,000
(d) 9250 × 29
Estimated product = 10000 × 30 = 3,00,000

Four more such examples are:

1. 678 × 261
2. 4271 × 4391
3. 2391 × 629
4. 8250 × 39

We hope the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.3, drop a comment below and we will get back to you at the earliest.

## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2

NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 are part of NCERT Solutions for Class 6 Maths. Here we have given NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2. https://mcqquestions.guru/ncert-solutions-for-class-6-maths-chapter-1-ex-1-2/

 Board CBSE Textbook NCERT Class Class 6 Subject Maths Chapter Chapter 1 Chapter Name Knowing Our Numbers Exercise Ex 1.2 Number of Questions Solved 12 Category NCERT Solutions

## NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2

Question 1.
A book exhibition was held for four days in a school. The number of tickets sold at the counter on the first, second, third and final day was respectively 1094, 1812, 2050 and 2751. Find the total number of tickets sold on all four days.
Solution :
Number of tickets sold on the first day = 1094
Number of tickets sold on the second day =1812
Number of tickets sold on the third day = 2050
Number of tickets sold on the final day = 2751
Total number of tickets sold on all the four days = 1094 + 1812 + 2050 + 2751 = 7707.

Question 2.
Shekhar is a famous cricket player. He has so far scored 6980 runs in test matches. He wishes to complete 10000 runs. How many more runs does he need?
Solution :
Runs scored so far = 6980
Runs wished to be scored = 10000
Runs needed more = 10000 – 6980 = 3020.

Question 3.
In an election, the successful candidate registered 5,77,500 votes and his nearest rival secured 3,48,700 votes. By what margin did the successful candidate win the election?
Solution :
Votes registered by the successful candidate = 5,77,500
Votes secured by the nearest rival = 3,48.700
Margin by which the successful candidate won the election = 5,77,500 – 3,48,700 = 2,28,800.

Question 4.
Kirti bookstore sold books worth? 2,85,891 in the first week of June and books worth ?. 4,00,768 in the second week of the month. How much was the sale for the two weeks together? In which week was the sale greater and by how much?
Solution :
Sale of books in the first week = ₹ 2,85,891
Sale of books in the second week = ₹ 4,00,768
∴ Sale for the two weeks together = ₹ 2,85,891 + ₹ 4,00,768 = ₹ 6,86,659.
The sale was greater in the second week by ₹ 4,00,768 – ₹ 2,85,891 i.e., by ₹ 1,14,877.

Question 5.
Find the difference between the greatest and least numbers that can be written using the digits 6, 2, 7, 4, 3 each only once.
Sol.
Greatest number that can be written using the digits 6, 2, 7, 4, 3 each only once = 76,432
Least number that can be written using the digits 6, 2, 7, 4, 3 each only once = 23,467
∴ Difference between the greatest and least numbers that can be written using the digits 6,2,7,4, 3 each only once = 76,432 – 23,467 = 52,965.

Question 6.
A machine, on an average, manufactures 2,825 screws a day. How many screws did it produce in the month of January 2006?
Solution :
The number of screws manufactured by the machine a day on an average = 2,825.
Number of days in the month of January 2006 = 31
The number of screws produced by the machine in the month of January 2006 = 2,825 × 31 =87,575.

Question 7.
A merchant had ₹ 78,592 with her. She placed an order for purchasing 40 radio sets at ? 1200 each. How much money will remain with her after the purchase?
Sol.
Money which the merchant had = ₹ 78,592
Cost of 40 radio sets at ? 1200 each = ₹ 1200 × 40 = ₹ 48,000
Money that will remain with the merchant after the purchase = ₹ 78,592 – ₹ 48,000 = ₹ 30,592.

Question 8.
A student multiplied 7236 by 65 instead of multiplying by 56. By how much was his answer greater than the. correct answer?
[Hint: Do you need to do both the multiplications?]
Solution :
= 7236 × 65 – 7236 × 56
= 7236 × (65 – 56)
= 7236 × 9 = 65.124

Question 9.
To stitch a shirt, 2 m 15 cm cloth is needed. Out of 40 m cloth, how many shirts can be stitched and how much cloth will remain ?
[Hint: convert data in cm.]
Solution :
2 m 15 cm = 2 m + 15 cm = 2 × 100 cm + 15 cm = 200 cm + 15 cm = 215 cm
40 m = 40 × 100 cm = 4000 cm Hence, 18 shirts can be stitched and 130 cm,
i. e., 1 m 30 cm cloth will remain.

Question 10.
Medicine is packed in boxes, each weighing 4 kg 500 g. Mow many such boxes can be loaded in a van which cannot carry beyond 800 kg?
Solution :
4 kg 500 g = 4 kg + 500 g
= 4 × 1000 g + 500 g
= 4000 g + 500 g
= 4500 g 800 kg
= 800 × 1000 g = 800000 g Hence, 177 such boxes can be loaded.

Question 11.
The distance between the school and the house of a student’s house is 1 km 875 m. Every day she walks both ways. Find the total distance covered by her in six days.
Solution :
1 km 875 m = 1 km + 875 m
= 1 × 1000 m + 875 m
= 1000 m + 875 m = 1875 m
Distance covered by her in a day in walking both ways between school and home = 1875 × 2 m = 3750 m
∴ Total distance covered by her in six days in walking both ways between school and home = 3750 m × 6 = 22500 m
= 22000 m + 500 m = $$\frac { 22000 }{ 1000 }$$ km + 500 m
= 22 km + 500 m = 22 km 500 m.

Question 12.
A vessel has 4 liters and 500 ml of curd. In how many glasses, each of 25 ml capacity, can it be filled?
Solution :
41500 ml = 41 + 500 ml
= 4 × 1000 ml + 500 ml
= 4000 ml + 500 ml = 4500 ml Hence, it can be filled in 180 glasses.

We hope the NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2 help you. If you have any query regarding NCERT Solutions for Class 6 Maths Chapter 1 Knowing Our Numbers Ex 1.2, drop a comment below and we will get back to you at the earliest.