Class 10 RD Sharma Solutions- Chapter 1 Real Numbers - Exercise 1.2 | Set 1
Last Updated : 28 Aug, 2024
In this article, we delve into the solutions for Exercise 1.2 Set 1 from Chapter 1 of RD Sharma's Class 10 Mathematics textbook focusing on "Real Numbers". This chapter is fundamental as it introduces the concept of the real numbers which form the basis for understanding more complex mathematical ideas. The solutions provided will help students grasp the principles and applications of the real numbers ensuring a solid foundation for their studies.
Real Numbers
The Real numbers encompass all the numbers on the number line including both the rational numbers and irrational numbers. They are crucial in various mathematical operations and real-world applications. Understanding real numbers is essential for solving equations, inequalities and other mathematical problems.
Question 1. Define the HCF of two positive integers and find the HCF of the following pairs of numbers:
(i) 32 and 54
Solution:
Applying Euclid’s Division Lemma on 54 , we get,
54 = 32 x 1 + 22
Applying Euclid’s Division Lemma on 32 and remainder 22 , we get,
32 = 22 x 1 + 10
Now, remainder ≠ 0, apply division lemma on 22 and remainder 10, we get,
22 = 10 x 2 + 2
Applying Euclid’s Division Lemma on 10 and remainder 2 , we get,
10 = 2 x 5 + 0
Therefore, the H.C.F. of 32 and 54 is 2.
(ii) 18 and 24
Solution:
Applying Euclid’s Division Lemma on 24 and 18, we get,
24 = 18 x 1 + 6.
Applying Euclid’s Division Lemma on 18 and remainder 6., we get,
18 = 6 x 3 + 0.
Therefore, H.C.F. of 18 and 24 is 6
(iii) 70 and 30
Solution:
Applying Euclid’s Division Lemma on 70 and 30, we get,
70 = 30 x 2 + 10.
Applying Euclid’s Division Lemma on 30 and remainder 10, we get,
30 = 10 x 3 + 0.
Since, remainder is 0 now,
Therefore, H.C.F. of 70 and 30 is 10
(iv) 56 and 88
Solution:
Applying Euclid’s Division Lemma on 56 and remainder 88, we get,
88 = 56 x 1 + 32.
Applying Euclid’s Division Lemma on 56 and remainder 32, we get,
56 = 32 x 1 + 24.
Applying Euclid’s Division Lemma on 32 and remainder 24, we get,
32 = 24 x 1+ 8.
Applying Euclid’s Division Lemma on 24 and remainder 8, we get,
24 = 8 x 3 + 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 56 and 88 is 8
(v) 475 and 495
Solution:
Applying Euclid’s Division Lemma on 475 and 495, we get,
495 = 475 x 1 + 20.
Applying Euclid’s Division Lemma on 475 and remainder 20, we get,
475 = 20 x 23 + 15.
Applying Euclid’s Division Lemma on 20 and remainder 15, we get,
20 = 15 x 1 + 5.
Applying Euclid’s Division Lemma on 15 and remainder 5, we get,
15 = 5 x 3+ 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 475 and 495 is 5
(vi) 75 and 243
Solution:
Applying Euclid’s Division Lemma on 243 and 75, we get,
243 = 75 x 3 + 18.
Applying Euclid’s Division Lemma on 75 and remainder 18, we get,
75 = 18 x 4 + 3.
Applying Euclid’s Division Lemma on 18 and remainder 3, we get,
18 = 3 x 6+ 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 75 and 243 is 3
(vii) 240 and 6552
Solution:
Applying Euclid’s Division Lemma on 6552 and 240, we get,
6552 = 240 x 27 + 72.
Applying Euclid’s Division Lemma on 240 and remainder 72, we get,
240 = 72 x 3+ 24.
Applying Euclid’s Division Lemma on 72 and remainder 24, we get,
72 = 24 x 3 + 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 240 and 6552 is 24
(viii) 155 and 1385
Solution:
Applying Euclid’s Division Lemma on 1385 and 155, we get,
1385 = 155 x 8 + 145.
Applying Euclid’s Division Lemma on 155 and remainder 145, we get,
155 = 145 x 1 + 10.
Applying Euclid’s Division Lemma on 145 and remainder 10, we get,
145 = 10 x 14 + 5.
Applying Euclid’s Division Lemma on 10 and remainder 5, we get,
10 = 5 x 2 + 0.
Since, the remainder is now 0,
Therefore, H.C.F. of 155 and 1385 is 5
(ix) 100 and 190
Solution:
Applying Euclid’s Division Lemma on 190 and 100, we get,
190 = 100 x 1 + 90.
Applying Euclid’s Division Lemma on 100 and remainder 90, we get,
100 = 90 x 1 + 10.
Applying Euclid’s Division Lemma on 90 and remainder 10, we get,
90 = 10 x 9 + 0.
Therefore, H.C.F. of 100 and 190 is 10
(x) 105 and 120
Solution:
Applying Euclid’s Division Lemma on 120 and 105, we get,
120 = 105 x 1 + 15.
Applying Euclid’s Division Lemma on 105 and remainder 15, we get,
105 = 15 x 7 + 0.
Therefore, H.C.F. of 105 and 120 is 15
Question 2. Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
Solution:
On comparing both the integers, we find 225 > 135.
Applying Euclid’s Division Lemma on 225 and 135, we get,
225 = 135 x 1 + 90
Applying Euclid’s Division Lemma on 135 and remainder 90, we get,
⇒ 135 = 90 x 1 + 45
Applying Euclid’s Division Lemma on 90 and remainder 45, we get,
⇒ 90 = 45 x 2 + 0
Since, the remainder is now 0,
Therefore, the H.C.F of 225 and 135 is 45.
(ii) 196 and 38220
Solution:
On comparing both the integers, we find 38220 > 196.
Applying Euclid’s Division Lemma on 38220 and 196, we get,
38220 = 196 x 195 + 0
Since, the remainder is now 0,
Hence, the HCF of 38220 and 196 is 196
(iii) 867 and 255
Solution:
On comparing both the integers, we find 867 > 255.
Applying Euclid’s Division Lemma on 867 and 255 , we get,
867 = 225 x 3 + 192
Applying Euclid’s Division Lemma on 225 and remainder 192 , we get,
225 = 192 x 1 + 33
Applying Euclid’s Division Lemma on 192 and remainder 33 , we get,
192 = 33 x 5 + 27
Applying Euclid’s Division Lemma on 33 and remainder 27 , we get,
33 = 27 x 1 + 6
Applying Euclid’s Division Lemma on 27 and remainder 6 , we get,
27 = 6 x 4 + 3
Applying Euclid’s Division Lemma on 6 and remainder 3 , we get,
6 = 3 x 2 + 0
Since the remainder is now 0,
Hence, the HCF of 867 and 255 is 3.
(iv) 184, 230 and 276
Solution:
We will first choose from 184 and 230 to find the HCF by using Euclid’s division lemma.
Thus, we obtain
230 = 184 x 1 + 46
Applying Euclid’s Division Lemma on 184 and remainder 46 , we get,
184 = 46 x 4 + 0
The HCF is therefore, 230.
Applying Euclid’s Division Lemma on 46 and 276 , we get,
276 = 46 x 6 + 0
Therefore, the HCF of the third number 276 and 46 is 46.
(v) 136, 170 and 255
Solution:
We will first choose from 136 and 170 to find the HCF by using Euclid’s division lemma.
We get,
170 = 136 x 1 + 34
Applying Euclid’s Division Lemma on 136 and remainder 34 , we get,
136 = 34 x 4 + 0
Since, the remainder is now 0, the divisor will be the HCF i.e., 34 for 136 and 170.
Applying Euclid’s Division Lemma on 34 and 255 , we get,
255 = 34 x 7 + 17
Applying Euclid’s Division Lemma on 34 and remainder 17 , we get,
34 = 17 x 2 + 0
Since, the remainder is now 0,
Therefore, the HCF of 136, 170 and 255 is 17.
Question 3. Find the HCF of the following pair of integers and express it as a linear combination of them,
(i) 963 and 657
Solution:
Applying Euclid’s Division Lemma on 963 and 657 , we get,
963 = 657 x 1 + 306………. (1)
Applying Euclid’s Division Lemma on 657 and remainder 306 , we get,
657 = 306 x 2 + 45………… (2)
Applying Euclid’s Division Lemma on 306 and remainder 45 , we get,
306 = 45 x 6 + 36…………. (3)
Applying Euclid’s Division Lemma on 45 and remainder 36 , we get,
45 = 36 x 1 + 9…………… (4)
Applying Euclid’s Division Lemma on 36 and remainder 9 , we get,
36 = 9 x 4 + 0……………. (5)
Therefore, the HCF is 9.
We can express the HCF as a linear combination of 963 and 657, by
9 = 45 – 36 x 1 [from (4)]
= 45 – [306 – 45 x 6] x 1 = 45 – 306 x 1 + 45 x 6 [from (3)]
= 45 x 7 – 306 x 1 = [657 -306 x 2] x 7 – 306 x 1 [from (2)]
= 657 x 7 – 306 x 14 – 306 x 1
= 657 x 7 – 306 x 15
= 657 x 7 – [963 – 657 x 1] x 15 [from (1)]
= 657 x 7 – 963 x 15 + 657 x 15
= 657 x 22 – 963 x 15.
(ii) 592 and 252
Solution:
Applying Euclid’s Division Lemma on 592 and 252 , we get,
592 = 252 x 2 + 88……… (1)
Applying Euclid’s Division Lemma on 252 and remainder 88 , we get,
252 = 88 x 2 + 76………. (2)
Applying Euclid’s Division Lemma on 88 and remainder 76 , we get,
88 = 76 x 1 + 12………… (3)
Applying Euclid’s Division Lemma on 76 and remainder 12 , we get,
76 = 12 x 6 + 4………….. (4)
Applying Euclid’s Division Lemma on 12 and remainder 4 , we get,
12 = 4 x 3 + 0……………. (5)
Therefore, H.C.F. = 4.
We can express the HCF as a linear combination of 592 and 252, by
4 = 76 – 12 x 6 [from (4)]
= 76 – [88 – 76 x 1] x 6 [from (3)]
= 76 – 88 x 6 + 76 x 6
= 76 x 7 – 88 x 6
= [252 – 88 x 2] x 7 – 88 x 6 [from (2)]
= 252 x 7- 88 x 14- 88 x 6
= 252 x 7- 88 x 20
= 252 x 7 – [592 – 252 x 2] x 20 [from (1)]
= 252 x 7 – 592 x 20 + 252 x 40
= 252 x 47 – 592 x 20
= 252 x 47 + 592 x (-20)
(iii) 506 and 1155
Solution:
Applying Euclid’s Division Lemma on 506 and 1155 , we get,
1155 = 506 x 2 + 143…………. (1)
Applying Euclid’s Division Lemma on 506 and remainder 143 , we get,
506 = 143 x 3 + 77…………….. (2)
Applying Euclid’s Division Lemma on 143 and remainder 77 , we get,
143 = 77 x 1 + 66……………… (3)
Applying Euclid’s Division Lemma on 77 and remainder 66 , we get,
77 = 66 x 1 + 11……………….. (4)
Applying Euclid’s Division Lemma on 66 and remainder 11 , we get,
66 = 11 x 6 + 0………………… (5)
Therefore, H.C.F. = 11.
We can express the HCF as a linear combination of 506 and 1155 by,
11 = 77 – 66 x 1 [from (4)]
= 77 – [143 – 77 x 1] x 1 [from (3)]
= 77 – 143 x 1 + 77 x 1
= 77 x 2 – 143 x 1
= [506 – 143 x 3] x 2 – 143 x 1 [from (2)]
= 506 x 2 – 143 x 6 – 143 x 1
= 506 x 2 – 143 x 7
= 506 x 2 – [1155 – 506 x 2] x 7 [from (1)]
= 506 x 2 – 1155 x 7+ 506 x 14
= 506 x 16 – 1155 x 7
(iv) 1288 and 575
Solution:
Applying Euclid’s Division Lemma on 1288 and 575 , we get,
1288 = 575 x 2+ 138………… (1)
Applying Euclid’s Division Lemma on 575 and remainder 138 , we get,
575 = 138 x 4 + 23……………. (2)
Applying Euclid’s Division Lemma on 138 and remainder 23 , we get,
138 = 23 x 6 + 0……………….. (3)
Therefore, H.C.F. = 23.
We can express the found HCF as a linear combination of 1288 and 575, by
23 = 575 – 138 x 4 [from (2)]
= 575 – [1288 – 575 x 2] x 4 [from (1)]
= 575 – 1288 x 4 + 575 x 8
= 575 x 9 – 1288 x 4
Question 4. Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
Solution:
Firstly, since 6 is required as the remainder, we subtract it from both the numbers.
So, the required numbers are 615 – 6 = 609 and 963 – 6 = 957.
The required number is the HCF of newly obtained numbers, 609 and 957.
Applying Euclid’s Division Lemma , we get,
957 = 609 x 1+ 348
609 = 348 x 1 + 261
348 = 261 x 1 + 87
261 = 87 x 3 + 0.
Since, the remainder is 0,
Therefore, the required number is 87
Question 5. If the HCF of 408 and 1032 is expressible in the form 1032m – 408 x 5, find m.
Solution:
Firstly, the HCF of 408 and 1032 is to be found.
Applying Euclid’s Division Lemma on 408 and 1032 , we get,
1032 = 408x 2 + 216.
Applying Euclid’s Division Lemma on 408 and remainder 216 , we get,
408 = 216 x 1 + 192.
Applying Euclid’s Division Lemma on 216 and remainder 192 , we get,
216 = 192 x 1 + 24.
Applying Euclid’s Division Lemma on 192 and remainder 24 , we get,
192 = 24 x 8 + 0.
Since, the remainder is 0 ,
The H.C.F of 408 and 1032 i.e., 24
So, this HCF is expressed as a linear combination that is,
24 = 1032m – 408 x 5
1032m = 24 + 408 x 5
1032m = 24 + 2040
1032m = 2064
m = 2064/1032
We obtain,
∴ m = 2
Question 6. If the HCF of 657 and 963 is expressible in the form 657x + 963 x – 15, find x.
Solution:
Applying Euclid’s Division Lemma on 657 and 963 , we get,
963 = 657 x 1+ 306.
Applying Euclid’s Division Lemma on 657 and remainder 306 , we get,
657 = 306 x 2 + 45.
Applying Euclid’s Division Lemma on 306 and remainder 45 , we get,
306 = 45 x 6 + 36.
Applying Euclid’s Division Lemma on 45 and remainder 36 , we get,
45 = 36 x 1 + 9.
Applying Euclid’s Division Lemma on 36 and remainder 9 , we get,
36 = 9 x 4 + 0.
Now, the remainder = 0.
Hence, the last divisor is the H.C.F of 657 and 963 i.e., 9
by expressing the HCF as a linear combination, we get ,
9 = 657x + 963 (-15).
Finding the value of x, we get
9 = 657x —14445
9 + 14445 = 657x
14454 = 657x
⇒ x = 14454 / 657
∴ x = 22
Question 7. An army contingent of 616 members is to march behind an army band of 32 members in a parade. The two groups are to march in the same number of columns. What is the maximum number of columns in which they can march?
Solution:
We need to compute the max number of columns in which the army band can march, which can be done by finding the HCF of the given two numbers.
Now, this is equal to the H.C.F of 616 and 32.
Applying Euclid’s Division Lemma on 616 and 32 , we get,
616 = 32 x 19 + 8
32 = 8 x 4 + 0.
Therefore, H.C.F. = 8
∴ The maximum number of columns in which the army band can march is 8.
Question 8. A merchant has 120 litres of oil of one kind, 180 litres of another and 240 litres of the third kind. He wants to sell the oil by filling the three kinds of oil in tins of equal capacity. What should be the greatest capacity of such a tin?
Solution:
The greatest capacity of the tin for filling three different types of oil is equivalent to the H.C.F. of the three available quantities 120,180 and 240.
Applying Euclid’s Division Lemma on 180 and 120 , we get,
180 = 120 x 1 + 60
120 = 60 x 2 + 0
Since, the remainder is now 0 ,
The HCF = 60.
Computing the H.C.F of 60 and the third quantity 240.
Applying Euclid’s Division Lemma on 240 and 60 , we get,
240 = 60 x 4 + 0
Since the remainder is 0 ,
Therefore, the HCF = 60
Therefore, the tin should be of 60 litres.
Question 9. During a sale, colour pencils were being sold in packs of 24 each and crayons in packs of 32 each. If you want full packs of both and the same number of pencils and crayons, how many of each would you need to buy?
Solution:
We have,
Number of colour pencils in a pack = 24
Number of crayons in a pack = 32.
The least number of both colour pencils and crayons that needs to be purchased is equivalent to their LCM.
L.C.M of 24 and 32 = 2 x 2 x 2 x 2 x 2 x 3 = 96
Now,
The number of pencil packs to be bought = 96 / 24 = 4 packs
And, the number of crayon packs to be bought = 96 / 32 = 3 packs
Question 10. 144 cartons of Coke Cans and 90 cartons of Pepsi Cans are to be stacked in a Canteen. If each stack is of the same height and is to contain cartons of the same drink, what would be the greatest number of cartons each stack would have?
Solution:
We have,
Number of coke can cartons = 144
Number of Pepsi can cartons = 90.
Therefore, the maximum number of cartons in a stack can be found by computing the H.C.F. of(144, 90).
Applying Euclid’s Division Lemma on 144 and 90 , we get,
144 = 90 x 1 + 54
90 = 54 x 1+ 36
54 = 36 x 1 + 18
36 = 18 x 2 + 0
∴ Since, the remainder is 0 ,
Therefore, the greatest number of cartons together in one stack is 18
Read More:
Summary
Exercise 1.2 in Chapter 1 of RD Sharma's Class 10 mathematics textbook likely covers various aspects of real numbers, including:
- Rational and irrational numbers
- Representing real numbers on a number line
- Operations with real numbers
- Properties of real numbers (closure, commutativity, associativity, distributivity)
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Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.3Problem 1: In a ∆ABC, AD is the bisector of ∠A, meeting side BC at D.(i) If BD = 2.5 cm, AB = 5 cm and AC = 4.2 cm, find DC Solution: Given: Length of side BD = 2.5 cm, AB = 5 cm, and AC = 4.2 cm. To find: Length of side DC In Δ ABC, AD is the bisector of ∠A, meeting side BC at D. Since, AD is ∠A bi
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Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.4Question 1. (i) In fig., if AB || CD, find the value of x. Solution: Given, AB∥ CD. To find the value of x. Now, AO/ CO = BO/ DO [Diagonals of a parallelogram bisect each other] ⇒ 4/ (4x – 2) = (x +1)/ (2x + 4) 4(2x + 4) = (4x – 2)(x +1) 8x + 16 = x(4x – 2) + 1(4x – 2) 8x + 16 = 4x2 – 2x + 4x – 2 -4
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Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 1The Triangles are fundamental geometric shapes characterized by three sides, three angles and three vertices. They are categorized based on their side lengths and angles leading to the various types such as the equilateral, isosceles, and scalene triangles. The study of the triangles includes unders
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Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.5 | Set 2Chapter 4 of the Class 10 RD Sharma Mathematics textbook, "Triangles," covers the properties and theorems related to triangles, including similarity, congruence, and the Pythagorean theorem. Exercise 4.5 focuses on applying these properties to solve problems related to triangles.RD Sharma Solutions
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Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 1This exercise contains questions regarding the similarity of triangles. Questions in this exercise are based on the concepts of properties of two similar triangles. The RD Sharma Solutions Class 10 provides all solutions required for a quick preparation for exams. The following questions and their s
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Class 10 RD Sharma Mathematics Solutions - Chapter 4 Triangles - Exercise 4.6 | Set 2Chapter 4 of RD Sharma's Class 10 mathematics book focuses on "Triangles" which is a fundamental concept in the geometry. This chapter deals with the various properties and theorems related to triangles such as the congruence similarity and Pythagorean theorem. Understanding these concepts is essent
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Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.7 | Set 1Question 1. If the sides of a triangle are 3 cm, 4 cm, and 6 cm long, determine whether the triangle is a right-angled triangle.Solution: According to the question The sides of triangle are: AB = 3 cm BC = 4 cm AC = 6 cm According to Pythagoras Theorem: Â AB2 = 32 = 9 BC2 = 42 = 16 AC2 = 62 = 36 Sinc
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Class 10 RD Sharma Solutions - Chapter 4 Triangles - Exercise 4.7 | Set 2Question 15. Each side of a rhombus is 10 cm. If one of its diagonals is 16 cm, find the length of the other diagonal. Solution: Given, In rhombus ABCD, diagonals AC and BD bisect each other at O at right angles Each side = 10 cm and one diagonal AC = 16 cm ∴ AO = OC = 16/2 = 8 cm Now in ∆AOB, By us
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Chapter 5: Trigonometric Ratios
Chapter 6: Trigonometric Identities
Chapter 7: Statistics
Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 1Problem 1: Calculate the mean for the following distribution:x:56789f:4 814113 Solution: x ffx542068487 1498811889327 N = 40 ∑ fx = 281 We know that, Mean = ∑fx/ N = 281/40 = 7.025 Problem 2: Find the mean of the following data :x:19212325272931f:13151618161513 Solution: x ffx19132472115315231636825
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.1 | Set 2Problem 11: Five coins were simultaneously tossed 1000 times and at each toss, the number of heads was observed. The number of tosses during which 0, 1, 2, 3, 4, and 5 heads were obtained are shown in the table below. Find the mean number of heads per toss.No. of heads per tossNo. of tosses 03811442
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.2Question1: The number of telephone calls received at an exchange per interval for 250 successive one-minute intervals are given in the following frequency table:Number of calls (x) 0 1 2 3 4 5 6 Number of intervals (f) 15 24 29 46 54 43 39 Compute the mean number of calls per interval. Solution: Let
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 1Chapter 7 of RD Sharma’s Class 10 textbook focuses on Statistics a fundamental area in mathematics that deals with data collection, analysis, interpretation, and presentation. Exercise 7.3 | Set 1 of this chapter covers the various problems designed to enhance students' understanding and application
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.3 | Set 2Question 14. Find the mean of the following frequency distribution:Class interval:25 – 2930 – 3435 – 3940 – 4445 – 4950 – 5455 – 59Frequency:1422166534 Solution: Let’s consider the assumed mean (A) = 42 Class intervalMid – value xidi = xi – 42ui = (xi – 42)/5fifiui25 – 2927-15-314-4230 – 3432-10-222
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Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 1In this article, we will explore the solutions to Exercise 7.4 from Chapter 7 of RD Sharma's Class 10 Mathematics textbook which focuses on Statistics. This chapter is crucial as it lays the foundation for understanding how data is collected, organized, analyzed, and interpreted in the various field
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Class 10 RD Sharma Solution - Chapter 7 Statistics - Exercise 7.4 | Set 2Question 11. An incomplete distribution is given below :Variable10-2020-3030-4040-5050-6060-7070-80Frequency1230-65-2518You are given that the median value is 46 and the total number of items is 230. (i) Using the median formula fill up missing frequencies. (ii) Calculate the AM of the completed dis
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 1Statistics is a branch of mathematics dealing with data collection, analysis, interpretation, and presentation. It provides methods for summarizing and drawing conclusions from the data which is essential in various fields such as economics, sociology, and science. Chapter 7 of RD Sharma's Class 10
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.5 | Set 2Question 11. Find the mean, median, and mode of the following data:Classes0-5050-100100-150150-200200-250250-300300-350Frequency2356531 Solution: Let mean (A) = 175 Classes Class Marks (x) Frequency (f) c.f. di = x - A A = 175 fi * di0-502522-150-30050-1007535-100-300100-150125510-50-250150-200175-A
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Class 10 RD Sharma Solutions - Chapter 7 Statistics - Exercise 7.6Chapter 7 of RD Sharma's Class 10 textbook focuses on Statistics a branch of mathematics that deals with the collection, analysis, interpretation, and presentation of data. This chapter is crucial for understanding how to summarize and make sense of numerical data in various contexts. The aim is to
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Chapter 8: Quadratic Equations
Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.1Question 1. Which of the following are quadratic equations?(i) x2 + 6x - 4 = 0 Solution: As we know that, quadratic equation is in form of: ax2 + bx + c = 0 Here, given equation is x2 + 6x - 4 =0 in quadratic equation. So, It is a quadratic equations. (ii) √3x2 − 2x + ½= 0 Solution: As we know that,
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.2Quadratic equations are a fundamental concept in algebra that involves a polynomial of degree two. They play a crucial role in various fields of mathematics, physics, engineering, and many other disciplines. Understanding quadratic equations is essential for solving problems related to motion, area,
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 1Solve the following quadratic equations by factorizationQuestion 1. (x - 4) (x + 2) = 0 Solution: We have equation, (x - 4) (x + 2) = 0 Implies that either x - 4 = 0 or x + 2 = 0 Therefore, roots of the equation are 4 or -2. Question 2. (2x + 3) (3x - 7) = 0 Solution: We have equation, (2x + 3) (3x
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.3 | Set 2The Quadratic equations are fundamental to algebra and appear frequently in various mathematical contexts. They play a crucial role in understanding polynomial functions and solving real-world problems. In this article, we will delve into Exercise 8.3 | Set 2 from RD Sharma’s Class 10 Mathematics bo
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.4Question 1: Find the roots of the following quadratic (if they exist) by the method of completing the square: x^2-4\sqrt{2x}+6=0 . Solution: Given: x^2-4\sqrt{2x}+6=0 We have to make the equation a perfect square. => x^2-4\sqrt{2x} = -6 => x^2 - 2\times2\sqrt{2x} + (2\sqrt{2})^2 = -6 + (2\sqrt
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.5Question 1. Find the discriminant of the following quadratic equations:(i) 2x2 - 5x + 3 = 0 Solution: Given quadratic equation: 2x2 - 5x + 3 = 0 ....(1) As we know that the general form of quadratic equation is ax2 + bx + c = 0 ....(2) On comparing eq(1) and (2), we get Here, a = 2, b = -5 and c = 3
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 1Question 1. Determine the nature of the roots of following quadratic equations : (i) 2x² – 3x + 5 = 0 (ii) 2x² – 6x + 3 = 0 (iii) 3/5 x² – 2/3 x + 1 = 0 (iv) 3x² – 4√3 x + 4 = 0 (v) 3x² – 2√6 x + 2 = 0 Solution: (i) 2x² – 3x + 5 = 0 Here a=2, b=-3, c=5 D=b2-4ac=(-3)2-4*2*5 =-9-40=-31 D<0 Roots ar
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.6 | Set 2Question 11. If – 5 is a root of the quadratic equation 2x² + px – 15 = 0 and the quadratic equation p(x² + x) + k = 0 has equal-roots, find the value of k. Solution: 2x² + px – 15 = 0 -5 is its one root It will satisfy it 2(-5)2+p(-5)-15=0 2*25-5p-15=0 50-15-5p=0 -5p+35=0 -5p=-35 p=-35/-5=7 Now in
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Class 10 RD Sharma Solutions- Chapter 8 Quadratic Equations - Exercise 8.7 | Set 1Question 1: Find two consecutive numbers whose squares have the sum 85. Solution: Let first number is = x ⇒ Second number = (x+1) Now according to given condition— ⇒ Sum of squares of the numbers = 85 ⇒ x2 + (x+1)2 = 85 ⇒ x2 + x2 + 2x + 1 = 85 [ because (a+b)2 = a2 + 2ab + b2] ⇒ 2x2 + 2x + 1 - 85 =
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Class 10 RD Sharma Solutions – Chapter 8 Quadratic Equations – Exercise 8.7 | Set 2Question 11: The sum of a number and its square is 63/4 , find the numbers. Solution: Let the number is = x so it's square is = x2 Now according to condition- ⇒(number)+(number)2=63/4 ⇒ x+x2 = 63/4 ⇒ x2+x-63/4=0 Multiplying by 4- ⇒ 4x2+4x-63=0 ⇒ 4x2 +(18-14)x - 63 =0 [because 63*4=252 so 18*14=252
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.8Chapter 8 of RD Sharma's Class 10 textbook focuses on the Quadratic Equations an essential topic in mathematics that lays the foundation for understanding more advanced algebraic concepts. In this chapter, students will learn how to solve quadratic equations using various methods such as factorizati
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.9The Quadratic equations are a fundamental topic in algebra often introduced at the Class 10 level. They are equations of the form ax2 +bx+c=0 where a, b, and c are constants and x is the variable. Understanding quadratic equations is crucial as they form the basis for many mathematical concepts and
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.10Question 1. The hypotenuse of a right triangle is 25 cm. The difference between the lengths of the other two sides of the triangle is 5 cm. Find the lengths of these sides. Solution: Let the length of other two sides of the triangle be x and y. Therefore, according to the question, x - y = 5 or, x =
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.11Question 1: The perimeter of the rectangular field is 82 m and its area is 400 m2. Find the breadth of the rectangle? Solution: Given: Perimeter = 82 m and its area = 400 m2 Let the breadth of the rectangle be 'b' m. As we know, Perimeter of a rectangle = 2×(length + breadth) 82 = 2×(length + b) 41
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.12Question 1: A takes 10 days less than the time taken by B to finish a piece of work. If both A and B together can finish the work in 12 days, find the time taken by B to finish the work. Solution: Let us assume that B takes 'x' days to complete a piece of work. So, B’s 1 day work = 1/x Now, A takes
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Class 10 RD Sharma Solutions - Chapter 8 Quadratic Equations - Exercise 8.13Question 1. A piece of cloth costs ₹ 35. If the piece were 4 m longer and each meter costs ₹ 1 less, the cost would remain unchanged. How long is the piece? Solution: Let us considered the length of piece of cloth = x m Given: The total cost = ₹ 35 So, the cost of 1 m cloth = ₹ 35/x According to the
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Chapter 9: Arithmetic Progressions
Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progression Exercise 9.1Problem 1: Write the first terms of each of the following sequences whose nth term are:(i) an = 3n + 2 Solution: Given: an = 3n + 2 By putting n = 1, 2, 3, 4, 5 we get first five term of the sequence a1 = (3 × 1) + 2 = 3 + 2 = 5 a2 = (3 × 2) + 2 = 6 + 2 = 8 a3 = (3 × 3) + 2 = 9 + 2 = 11 a4 = (3 × 4)
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.2Problem 1: Show that the sequence defined by an = 5n – 7 is an A.P., find its common difference. Solution: Given: an = 5n – 7 Now putting n = 1, 2, 3, 4,5 we get, a1 = 5.1 – 7 = 5 – 7 = -2 a2 = 5.2 – 7 = 10 – 7 = 3 a3 = 5.3 – 7 = 15 – 7 = 8 a4 = 5.4 – 7 = 20 – 7 = 13 We can see that, a2 – a1 = 3 – (
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.3Problem 1: For the following arithmetic progressions write the first term a and the common difference d:(i) -5, -1, 3, 7, ………… Solution: Given sequence is -5, -1, 3, 7, ………… ∴ First term is -5 And, common difference = a2 - a1 = -1 - (-5) = 4 ∴ Common difference is 4 (ii) 1/5, 3/5, 5/5, 7/5, …… Solut
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 1In arithmetic progressions (AP), the terms follow a consistent pattern where each term is derived from the previous one by adding a fixed value, known as the common difference. To analyze various aspects of APs, such as finding specific terms or determining the number of terms, certain formulas and
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.4 | Set 2Question 14. The 4th term of an A.P. is three times the first and the 7th term exceeds twice the third term by 1. Find the first term and the common difference. Solution: Let’s assume that the first term and the common difference of the A.P to be a and d respectively. Given that, 4th term of an A.P.
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Class 10 RD Sharma Solutions- Chapter 9 Arithmetic Progressions - Exercise 9.5Question 1: Find the value of x for which (8x + 4), (6x – 2) and (2x + 7) are in A.P. Solution: Since (8x + 4), (6x – 2) and (2x + 7) are in A.P. Now we know that condition for three number being in A.P.- ⇒2*(Middle Term)=(First Term)+(Last Term) ⇒2(6x-2)=(8x+4)+(2x+7) ⇒12x-4=10x+11 ⇒(12x-10x)=11+4
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 1Class 10 RD Sharma Solutions are comprehensive guides to the exercises and problems presented in the RD Sharma Mathematics textbook, which is a popular reference for students studying mathematics at the 10th class level. These solutions provide step-by-step explanations and answers to the questions
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 2Question 25. In an A.P. the first term is 22, nth term is –11 and the sum of first n term is 66. Find n and the d, the common difference. Solution: Given A.P. has first term(a) = 22, nth term(an) = –11 and sum(Sn) = 123. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 =>
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Class 10 RD Sharma Solutions - Chapter 9 Arithmetic Progressions - Exercise 9.6 | Set 3Question 49. Find the sum of first n odd natural numbers. Solution: First odd natural numbers are 1, 3, 5, 7, . . .2n – 1. First term(a) = 1, common difference(d) = 3 – 1 = 2 and nth term(an) = 2n – 1. Now by using the formula of sum of n terms of an A.P. Sn = n[a + an] / 2 So, = n[1 + 2n – 1] / 2 =
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