Are you searching for solutions for RD Sharma Class 10th Ex 2.1? In this post, we have provided solutions to RD Sharma 10th grade math exercise 2.1.
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RD Sharma Class 10 Ex 1.2
Question 1. Define HCF of two positive integers and find the HCF of the following pairs of numbers:
- 32 and 54
- 18 and 24
- 70 and 30
- 56 and 88
- 475 and 495
- 75 and 243
- 240 and 6552
- 155 and 1385
- 100 and 190
- 105 and 120
Answer – Definition of HCF (Highest Common Factor): The largest positive integer which divides two or more integers without any remainder is called Highest Common Factor (HCF) or Greatest Common Divisor or Greatest Common Factor (GCF).
(i) Prime factorization of 32 and 54 are:
- 32 = 2 × 2 × 2 × 2 × 2
- 54 = 2 x 3 x 3 x 3
From above prime factorization we got that the highest common factor of 32 and 54 is 2
(ii) Prime factorization of 18 and 24 are:
- 18 = 2 x 3 x 3
- 24 = 2 x 2 x 2 x 3
From above prime factorization we got that the highest common factor of 18 and 24 is 3×2 = 6
(iii) Prime factorization of 30 and 70 are:
- 30 = 2 × 3 × 5
- 70 = 2 x 5 x 7
From above prime factorization we got that the highest common factor of 30 and 70 is 2×5 = 10
(iv) Prime factorization of 56 and 88 are:
- 56 = 2 x 2 x 2 x 7
- 88 = 2 x 2 x 2 x 11
From above prime factorization we got that the highest common factor of 56 and 88 is 2x2x2 = 8
(v) Prime factorization of 475 and 495 are:
- 475 = 5 x 5 x 19
- 495 = 3 x 3 x 5 x 11
From above prime factorization we got that the highest common factor of 475 and 495 is 5
(vi) Prime factorization of 75 and 243 are:
- 75 = 3 x 5 x 5
- 243 = 3 x 3 x 3 x 3 x 3
From above prime factorization we got that the highest common factor of 75 and 243 is 3
(vii) Prime factorization of 240 and 6552 are:
- 240 = 2 x 2 x2 x 2 x 3 x 5
- 6552 = 2 x 2 x 2 x 3 x 3 x 7 x 13
From above prime factorization we got that the highest common factor of 240 and 6552 is 2x2x2x3 ⇒ 24
(viii) Prime factorization of 155 and 1385 are:
- 155 = 5 x 31
- 1385 = 5 x 277
From above prime factorization we got that the highest common factor of 155 and 1385 is 5
(ix) Prime factorization of 100 and 190 are:
100 = 2 x 2 x 5 x 5
190 = 2 x 5 x19
From above prime factorization we got that the highest common factor of 155 and 1385 is 2×5 = 10
(x) Prime factorization of 105 and 120 are:
105 = 3 x 5 x 7
120 = 2 x 2 x 2 x 3 x 5
From above prime factorization we got that the highest common factor of 105 and 120 is 3×5 = 15
2. Question – Use Euclid’s division algorithm to find the HCF of
- 135 and 225
- 196 and 38220
- 867 and 255.
Answer
(i) Concept used: To obtain the HCF of two positive integers, say c and d, with c > d,we followthe steps below: Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c=dq + r, 0 < r < d.Step 2: If r = 0, d is the HCF of c and d. If r 0, apply the division lemma to d and r.Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.Now, We know that,
= 225>135
Applying Euclid’s division algorithm: (Dividend = Divisor x Quotient + Remainder)
225 = 135 x1+90
Here remainder = 90,
So, Again Applying Euclid’s division algorithm
135 = 90×1+45
Here remainder = 45,
So, Again Applying Euclid’s division algorithm
90 = 45×2+0
Remainder = 0,
Hence,
HCF of (135, 225) = 45
(ii) Concept used: To obtain the HCF of two positive integers, say c and d, with c > d,we followthe steps below: Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 < r < d.Step 2: If r = 0, d is the HCF of c and d. If r 0, apply the division lemma to d and r.Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF. Now, We know that,
38220>196
So, Applying Euclid’s division algorithm
38220 = 196×195+0 (Dividend = Divisor x Quotient + Remainder)
Remainder = 0
Hence,
HCF of (196, 38220) = 196
(iii) Concept used: To obtain the HCF of two positive integers, say c and d, with c > d, we followthe steps below: Step 1: Apply Euclid’s division lemma, to c and d. So, we find whole numbers, q and r such that c = dq + r, 0 ≤ r < d.Step 2: If r = 0, d is the HCF of c and d. If r 0, apply the division lemma to d and r.Step 3: Continue the process till the remainder is zero. The divisor at this stage will be the required HCF.Now, We know that,
867>255
So, Applying Euclid’s division algorithm
867 = 255×3+102 (Dividend = Divisor x Quotient + Remainder)
Remainder = 102
So, Again Applying Euclid’s division algorithm
255 = 102×2+51
Remainder = 51
So, Again Applying Euclid’s division algorithm
102 = 51×2+0
Remainder = 0
Hence,
(HCF Of 867 and 255) = 51
3. Question – Find the HCF of the following pairs of integers and express it as a linear combination of them
(i) 963 & 657 (ii) 592 & 252
(iii) 506 & 1155 (iv) 1288 & 575
Answer (i) – Using Euclid’s Division Lemma
a = bq+r, (o <r<b)
963 = 657×1 + 306
657 = 306×2 + 45
306 = 45×6+36
45 = 36×1+9
36 = 9×4+0
HCF (657, 963) = 9
linear form: 657a+ 306b = 9 The above equation have many solutions, one of them is a = -15, b = 22i.e. 9657(-15) + 306(22)
(ii) Using Euclid’s Division Lemma
a = bq + r, (o <r<b)
- 592 = 252×2+88
- 252 = 88×2+76
- 88 = 76×1+12
- 76 = 12×6+4
- 12 = 4×3+0
HCF (592, 252) = 4
linear form:592a + 252b = 4, The above equation have many solutions, one of them is a = 77, b = -201.e. 4592(77) + 252(20)
(iii) Solve Using Euclid’s Division Lemma
a = bq + r, (o <r<b)
- 1155 = 506×2+143
- 506 = 143×3+77
- 143 = 77×1+66
- 77 = 66×1+11
- 66 = 11×6+0
.. HCF (506, 1155) = 11
linear form: 506a+ 1155b 11The above equation have many solutions, one of them is a = 16, b = -7i.e. 11 506(16) + 1155(-7)
(iv) Using Euclid’s Division Lemma of 1288
a = bq + r, (o <r<b)
- 1288 = 575×2+138
- 575 = 138×4+23
- 138 = 23×6+0
::. HCF (1288, 575) = 23
linear form:1288a+ 575b = 23
The above equation have many solutions, one of them is a = 4, b = 9
i.e. 23 1288(4) + 575(9)
4. Question – Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways. Answer
HCF of 468 and 222 is found by division mehtod:
Therefore, HCF (468,222) = 6
Now, we need to express the HCF of 468 and 222 as 468x + 222y where x and y are any two integers.
Now, HCF i.e. 6 can be written as,
HCF = 222 216 222 (24 × 9)
Writing 468 222 x 2 + 24, we get,
HCF = 222 {(468 222 x 2) × 9}
⇒ HCF = 222 – {(468 ×9) – (222 × 2 × 9)} HCF 222 (468 × 9) + (222 x 18)
⇒ HCF = 222 + (222 x 18) – (468 × 9)
Taking 222 common from the first two terms, we get,
⇒ HCF = 222[1 +18] – 468 x 9
HCF = 222 x 19 468 x 9
HCF 468 x(-9) + 222 ×(19)
Let, say, x = -9 and y =19
Then, HCF = 468 x(x) + 222 x(y)
Therefore the HCF of 468 and 222 is written in the form of 468x + 222y where, -9 and 19 are the two integers.
5. Question – If the HCF of 408 and 1032 is expressible in the form 1032 m – 408 x 5, find m.
Answer –
- Prime factors of 408: 2 x 2 x 2 x 3 x 17
- Prime factors of 1032: 2 x 2 x2 x 3 x 43
HCF of 408 and 1032 = 2 x 2 x 2 x 3 = 24
According to question:
- 24 = 1032 m – 408 x 5
- 24 = 1032 m – 2040
- 2064 = 1032 m
So, m = 2064 ÷1032 = 2
Therefore m = 2
6. Question – If the HCF of 657 and 963 is expressible in the form 657 x + 963 x -15, find x.
Answer –
- Prime factors of 657: 3 x 3 x 73
- Prime factors of 963: 3 x 3 x 107
HCF of 657 and 963 = 3 x3=9
According to question: the HCF of 657 and 963 is expressible in the form 657 x + 963 x -15
9 = 657x+963 x -15
9 = 657x – 14445
9+14445 = 657x
So, X = 14454/657 = 22
Therefore x = 22
7. Question – Find the he largest number which divides 615 and 963 leaving remainder 6 in each case.
Answer
To find: the largest number which divides 615 and 963 leaving remainder 6 in each case.
Solution: Let the HCF of 615 and 963 be x.
Since it is given remainder is 6 in each case, Therefore for the numbers to be completely divisible, 6 should be subtracted from both the numbers.
Therefore new numbers are:
615-6 = 609
963-6 = 957
- Prime factors of 609 = 3 × 3 × 29
- Prime factors of 957= 3 x 11 x 29
Therefore HCF of 609 and 957 is: 3 x 29 = 87
Hence the largest number which divides 615 and 963 leaving remainder 6 in each case is 87. 8. Question
Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.
Answer – The new numbers after subtracting remainders are:
285-9 = 276
1249-7 = 1242
- Prime factors of 276 = 23 × 3 × 2
- Prime factors of 1242 = 2 x 3 x 3 x 3 x 23
Therefore HCF of 276 and 1242 is: 2 x 3 x 23 = 138
Hence the greatest number which divides 285 and 1249 leaving remainder 9 and 7 respectively is138 9. Question
Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively. Answer
The new numbers after subtracting remainders are:
280-4 = 276
1245-3 = 1242
- Prime factors of 276 = 23 x 3 x 2
- Prime factors of 1242 = 2 x 3 x 3 x 3 x 23
Therefore HCF of 276 and 1242 is: 2 x 3 x 23 = 138
Hence the greatest number which divides 280 and 1245 leaving remainder 4 and 3 respectively is138 10. Question
What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively?
Answer
The new numbers after subtracting remainders are:
626-1 = 625
3127-2 = 3125
15628-3 = 15625
- Prime factors of 625 = 5 x 5 x 5 x 5
- Prime factors of 3125 = 5 x 5 x 5 x 5 x 5
- Prime factors of 15625 = 5 x 5 x 5 x 5 x 5 x 5
Therefore HCF of 625, 3125 and 15625 is: 5 x 5 x 5 x 5= 625
Hence the largest number which divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively is 625
11. Question – Find the greatest number that will divide 445, 572 and 699 leaving remainder 4, 5 and 6 respectively. Answer
The new numbers after subtracting remainders are:
445-4 = 441
572-5 = 567
699-6 = 693
Prime factors of 441= 3 x 3 x 7 x 7
- Prime factors of 567 = 3 x 3 x 3 x 3 x 7
- Prime factors of 693 = 3 x 3 x 7 x 11
Therefore HCF of 441, 567 and 693 is: 3 x 3 x7 = 63
Hence the greatest number that will divide 445, 572 and 699 leaving remainder 4, 5 and 6 respectively is 63
12. Question – Find the greatest number which divides 2011 and 2623 leaving remainder 9 and 5 respectively. Answer
The new numbers after subtracting remainders are:
2011-9 = 2002
2623-5 = 2618
- Prime factors of 2002= 2 x 7 x 11 x 13
- Prime factors of 2618 = 2 x 7 x 11 x 17
Therefore HCF of 2002 and 2618 is: 2 x 7 x 11 = 154
Hence the greatest number which divides 2011 and 2623 leaving remainder 9 and 5 respectively is 154.
13. Question – 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?
Answer – To find maximum number of columns we should find HCF of 616 and 32
Using Euclid’s algorithms:
Let a = 616 and b = 32
a = bq+r, (o <r<b)
616 = 32×19+8
32 = 8×4+0
.. HCF of 616 and 32 is 8
Therefore the maximum number of columns in which army contingent to march is 8
14. Question – A merchant has 120 litres of oil of one kind, 180 litres of another kind and 240 litres of 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?
Answer – To find greatest capacity of tin we should find HCF of 120 and 180 and 240
- Prime factors of 120= 2 x 2 x 2 x 3 x 5
- Prime factors of 180 = 2 x 2 x 3 x 3 x 5
- Prime factors of 240 = 2 x 2 x 2 x 2 x 3 x 5
Therefore HCF of 120, 180 and 240 is: 2 x 2 x 3 x 5 = 60
Therefore the greatest capacity of a tin is 60 Liters
15. Question – 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?
Answer – For having the full packs of both and the same number of pencils and crayons, we need to find LCM of 24 and 32
- Prime factors of 24 = 2 x 2 x 2 x 3
- Prime factors of 32 = 2 x 2 x 2 x 2 x 2
Therefore LCM of 24 and 32 is: 2 x 2 x 2 x 2 x 2 x 3 = 96
Number of colour pencil packs = 96÷24 = 4
Number of crayons packs = 96 ÷ 32 = 3
Hence 4 packets of colour pencils and 3 packets of crayons would be bought.
16. Question – 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?
Answer – To find greatest number of cartons each stack would have, we should find HCF of 144 and 90.
- Prime factors of 144 = 2 x 2 x 2 x 2 x 3 x 3
- Prime factors of 90 = 2 × 3 × 3 × 5
Therefore HCF of 144 and 90 is: 2 x 3 x 3 = 18
Therefore the greatest number of cartons each stack would have is: 18
17. Question – Two brands of chocolates are available in packs of 24 and 15 respectively. If I need to buy an equal number of chocolates of both kinds, what is the least number of boxes of each kind I would need to buy?
Answer – Given: Two brands of chocolates are available in packs of 24 and 15 respectively.
To find the least number of boxes of each kind.
Solution: To find the least number of boxes of each kind.we need to find LCM of 24 and 15
- Prime factors of 24 = 2 × 2 × 2 × 3
- Prime factors of 15 = 3 x 5
Therefore LCM of 24 and 15 is: 2 x 2 x 2 x 3 x 5 = 120
Number of boxes for first chocolate kind 120÷ 24 =5
Number of boxes for second chocolate kind = 120÷15 =8
Hence 5 boxes of first kind and 8 boxes of second kind needed to buy.
18. Question – A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10 ft. by 8 ft. What would be the size in inches of the tile required that has to be cut and how many such tiles are required?
Answer – Given: The size of the bathroom is 10 ft. by 8 ft.
To find: the size in inches of the tile required that has to be cut and how many such tiles are required?
Solution:To find the largest size of tile, we should find HCF of 10 and 8
Prime factors of 10 = 2 x 5
Prime factors of 8 = 2 x 2 x 2
Therefore HCF of 10 and 8 is: 2 ftSince i ft 12 inches
Therefore the largest size of tile is: 2 x 12 inches 24 inches
Area of bathroom length of bathroom x breadth of bathroom
= 10 × 8
= 80 sq. ftArea of 1 tile = length of tile x breadth of tile
= 2 × 2
= 4 sq. ft
Number of tiles = area of bathroom ÷ area of one tile = 80÷4 = 20 tiles
NOTE: Always find the HCF of the given values to find their maximum.
19. Question – 15 pastries and 12 biscuit packets have been donated for a school fete. These are to be packed in several smaller identical boxes with the same number of pastries and biscuit packets in each. How many biscuit packets and how many pastries will each box contain?
Answer – To find the number of biscuit packets and pastries, we should find HCF of 15 and 12
- Prime factors of 15 = 3 x 5
- Prime factors of 12 = 2 x 2 x 3
Therefore HCF of 15 and 12 is: 3
Number of pastries = 15÷3 =5
Number of biscuit packets = 12÷3 = 4
20. Question – 105 goats, 140 donkeys and 175 cow have to be taken across a river. There is only one boat which will have to make many trips in order to do so. The lazy boatman has his own conditions for transporting them. He insists that he will take the same number of animals in every trip and they have to be of the same kind. He will naturally like to take the largest possible number each time. Can you tell how many animals went in each trip?
Answer – To find the largest number of animals, we should find HCF of 105, 140 and 175
- Prime factors of 105 = 3 x 5 x 7
- Prime factors of 140 = 2 × 2 × 5 × 7
- Prime factors of 175 = 5 x 5 x 7
Therefore HCF of 105, 140 and 175 is: 5 x 7 = 35
Hence, 35 animals went in each trip
21. Question – The length, breadth and height of a room are 8 m and 25 cm, 6 m 75 cm and 4 m 50 cm, respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Answer – Given: The length, breadth and height of a room are 8 m and 25 cm, 6 m 75 cm and 4 m 50 cm. To find: the longest rod which can measure the three dimensions of the room exactly.
Solution:To find the length of largest rod, we should find HCF of 8m and 25 cm, 6 m 75 cm and 4 m 50 cmAs 1 m = 100 cm⇒ 8m and 25 cm = 8 x 100 cm + 25 cm
- Prime factors of 825 = 3 x 5 x 5 x 11
- Prime factors of 675 = 3 x 3 x 3 x 5 x 5
- Prime factors of 450 = 2 x 3 x 3 x 5 x 5
Therefore HCF of 825, 675 and 450 is: 3 x 5 x 5 = 75
Hence, the length of rod is: 75 cm
NOTE: Always find the HCF of the given values to find their maximum.