RD Sharma Solutions Chapter 1 Real Numbers Exercise 1.2 Class 10 Maths
Chapter Name  RD Sharma Chapter 1 Real Numbers 
Book Name  RD Sharma Mathematics for Class 10 
Other Exercises 

Related Study  NCERT Solutions for Class 10 Maths 
Exercise 1.2 Solutions
1. Define HOE of two positive integers and find the HCF of the following pairs of numbers:
(i) 32 and 54
(ii) 18 and 24
(iii) 70 and 30
(iv) 56 and 88
(v) 475 and 495
(vi) 75 and 243
(vii) 240 and 6552
(viii) 155 and 1385
(ix) 100 and 190
(x) 105 and 120
Solution
By applying Euclid's division lemma
(i) 54 = 32× 1 + 22
Since remainder ≠ 0, apply division lemma on division of 32 and remainder 22.
32 = 22×1 + 10
Since remainder ≠ 0, apply division lemma on division of 10 and remainder 2.
10 = 2×5 [remainder 0]
Hence, HCF of 32 and 54 = 10
(ii) By applying division lemma
24 = 18 ×1 + 6
Since remainder = 6, apply division lemma on divisor or 18 and remainder 6.
18 = 6 × 3 + 0
∴ Hence, HCF of 18 and 24 = 6
(iii) By applying Euclid’s division lemma
70 = 30 × 2 + 10
Since remainder ≠ 0, apply division lemma on divisor of 30 and remainder 10.
30 = 10 × 3 + 0
∴ Hence HCF of 70 and 30 is = 10.
(iv) By applying Euclid’s division lemma
88 = 56 × 1 + 32
Since remainder ≠ 0, apply division lemma on divisor of 56 and remainder 32.
56 = 32 × 1 + 24
Since remainder ≠ 0, apply division lemma on divisor of 32 and remainder 24.
32 = 24 × 1 + 8
Since remainder ≠ 0, apply division lemma on divisor of 24 and remainder 8.
24 = 8 × 3 + 0
∴ HCF of 56 and 88 is = 8.
(v) By applying Euclid’s division lemma
495 = 475 × 1 +20
Since remainder ≠ 0, apply division lemma on divisor of 475 and remainder 20.
475 = 20 × 23 + 15
Since remainder ≠ 0, apply division lemma on divisor of 20 and remainder 15.
20 = 15 × 1 + 5
Since remainder ≠ 0, apply division lemma on divisor of 15 and remainder 5.
15 = 5 × 3 + 0
∴ HCF of 475 and 495 is = 5.
(vi) By applying Euclid’s division lemma
243 = 75 × 3 + 18
Since remainder ≠ 0, apply division lemma on divisor of 75 and remainder 18.
75 = 18 × 4 + 3
Since remainder ≠ 0, apply division lemma on divisor of 18 and remainder 3.
18 = 3 × 6 + 0
∴ HCF of 243 and 75 is = 3.
(vii) By applying Euclid’s division lemma
6552 = 240 × 27 + 72
Since remainder ≠ 0, apply division lemma on divisor of 240 and remainder 72.
210 = 72 × 3 + 24
Since remainder ≠ 0, apply division lemma on divisor of 72 and remainder 24.
72 = 24 × 3 + 0
∴ HCF of 6552 and 240 is = 24.
(viii) By applying Euclid’s division lemma
1385 = 155 × 8 + 145
Since remainder ≠ 0, applying division lemma on divisor 155 and remainder 145.
155 = 145 × 1 + 10
Since remainder ≠ 0, applying division lemma on divisor 10 and remainder 5 10 = 5 × 2 + 0
∴ Hence HCF of 1385 and 155 = 5.
(ix) By applying Euclid’s division lemma
190 = 100 × 1 + 90
Since remainder ≠ 0, applying division lemma on divisor 100 and remainder 90.
90 = 10 × 9 + 0
∴ HCF of 100 and 190 = 10
(x) By applying Euclid’s division lemma
120 = 105 × 1 + 15
Since remainder ≠ 0, applying division lemma on divisor 105 and remainder 15.
105 = 15 × 7 + 0
∴ HCF of 105 and 120 = 15
2. Use Euclid’s division algorithm to find the HCF of
(i) 135 and 225
(ii) 196 and 38220
Solution
(i) 135 and 225
Step 1: Since 225 > 135. Apply Euclid’s division lemma to a = 225 and b = 135 to find q and r such that 225 = 135q + r, 0 ≤ r < 135
On dividing 225 by 135 we get quotient as 1 and remainder as ‘90’
i.e., 225 = 135r 1 + 90
Step 2: Remainder 5 which is 90 7, we apply Euclid’s division lemma to a = 135 and b = 90 to find whole numbers q and r such that 135 = 90 × q + r 0 ≤ r < 90 on dividing 135 by 90 we get quotient as 1 and remainder as 45
i.e., 135 = 90 × 1 + 45
Step3: Again remainder r = 45 to so we apply division lemma to a = 90 and b = 45 to find q and r such that 90 = 45 × q × r. 0 ≤ r < 45. On dividing 90 by 45 we get quotient as 2 and remainder as 0 i.e., 90 = 2 × 45 + 0
Step 4: Since the remainder = 0, the divisor at this stage will be HCF of (135, 225)
Since the divisor at this stage is 45. Therefore the HCF of 135 and 225 is 45.
(ii) 867 and 255:
Step 1: Since 867 > 255, apply Euclid’s division
Lemma a to a = 867 = 255 q + r, 0 < r < 255
On dividing 867 by 255 we get quotient as 3 and the remainder as low
Step 2: Since the remainder 102 to, we apply the division lemma to a = 255 and b = 102 to find 255 = 102q + 51 = 102r – 151
Step 3: Again remainder 0 is nonzero, so we apply the division lemma to a = 102 and b = 51 to find whole numbers q and r such that 102 = q = r when 0 ≤ r < 51 On dividing 102 by 51 quotient = 2 and remainder is ‘0’
i,e., = 102 = 51 × 2 + 0
Since the remainder is zero, the divisional this stage is the HCF.
Since the divisor at this stage is 51, ∴ HCF of 867 and 255 is ‘51’.
3. Find the HCF of the following pairs of integers and express it as a linear combination of them.
(i) 963 and 657
(ii) 592 and 252
(iii) 506 and 1155
(iv) 1288 and 575
Solution
(i) 963 and 6567
By applying Euclid’s division lemma 963 = 657 × 1 + 306 …(i)
Since remainder ≠ 0, apply division lemma on divisor 657 and remainder 306.
657 = 306 × 2 + 45 …(ii)
Since remainder ≠ 0, apply division lemma on divisor 306 and remainder 4.
306 = 45 × 6 + 36 …(iii)
Since remainder ≠ 0, apply division lemma on divisor 45 and remainder 36.
45 = 36 × 1 + 9 …(iv)
Since remainder ≠ 0, apply division lemma on divisor 36 and remainder 9.
36 = 9 × 4 + 0
∴ HCF = 9
Now 9 = 45 – 36 × 1 [from (iv)]
= 45 – [306 – 45 × 6] × 1 [from (iii)]
= 45 – 306 × 1 + 45 × 6
= 45 × 7 – 306 × 1
= 657 × 7 – 306 × 14 – 306 × 1 [from (ii)]
= 657 × 7 – 306 × 15
= 657 × 7 – [963 – 657 × 1] × 15 [from (i)]
= 657 × 22 – 963 × 15
(ii) 595 and 252
By applying Euclid’s division lemma
595 = 252 × 2 + 91 …(i)
Since remainder ≠ 0, apply division lemma on divisor 252 and remainder 91.
252 = 91 × 2 + 70 …(ii)
Since remainder ≠ 0, apply division lemma on divisor 91 and remainder 70.
91 = 70 × 1 + 21 …(iii)
Since remainder ≠ 0, apply division lemma on divisor 70 and remainder 20.
70 = 21 × 3 + 7 …(iv)
Since remainder ≠ 0, apply division lemma on divisor 21 and remainder 7.
21 = 7 × 3 + 0
H.C.F = 7
Now, 7 = 70 – 21 × 3 [from (iv)]
= 70 – [90 – 70 × 1] × 3 [from (iii)]
= 70 – 91 × 3 + 70 × 3
= 70 × 4 – 91 × 3
= [252 – 91 × 2] × 4 – 91 × 3 [from (ii)]
= 252 × 4 – 91 × 8 – 91 × 3
= 252 × 4 – 91 × 11
= 252 × 4 – [595 – 252 × 2] × 11 [from (i)]
= 252 × 4 – 595 × 11 + 252 × 22
= 252 × 6 – 595 × 11
(iii) 506 and 1155
By applying Euclid’s division lemma
1155 = 506 × 2 + 143 …(i)
Since remainder ≠ 0, apply division lemma on division 506 and remainder 143. 506 = 143 × 3 +77 …(ii)
Since remainder ≠ 0, apply division lemma on division 143 and remainder 77.
143 = 77 × 1 + 56 …(iii)
Since remainder ≠ 0, apply division lemma on division 77 and remainder 66.
77 = 66 × 1 + 11 …(iv)
Since remainder ≠ 0, apply division lemma on divisor 66 and remainder 11.
66 = 11 × 6 + 0
∴ HCF = 11
Now, 11 = 77 – 6 × 11 [from (iv)]
= 77 – [143 – 77 × 1] × 1 [from (iii)]
= 77 – 143 × 1 – 77 × 1
= 77 × 2 – 143 × 1
= [506 – 143 × 3] × 2 – 143 × 1 [from (ii)]
= 506 × 2 – 143 × 6 – 143 × 1
= 506 × 2 – 143 × 7
= 506 × 2 – [1155 – 506 × 27 × 7] [from (i)]
= 506 × 2 – 1155 × 7 + 506 × 14
= 506 × 16 – 115 × 7
(iv) 1288 and 575
By applying Euclid’s division lemma
1288 = 575 × 2 + 138 …(i)
Since remainder ≠ 0, apply division lemma on division 575 and remainder 138.
575 = 138 × 1 + 23 …(ii)
Since remainder ≠ 0, apply division lemma on division 138 and remainder 23 …(iii)
∴ HCF = 23
Now, 23 = 575 – 138 × 4 [from (ii)]
= 575 – [1288 – 572 × 2] × 4 [from (i)]
= 575 – 1288 × 4 + 575 × 8
= 575 × 9 – 1288 × 4
4. Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
Solution
Given integers are 468 and 222 where 468 > 222.
By applying Euclid’s division lemma, we get 468 = 222 × 2 + 24 …(i)
Since remainder ≠ 0, apply division lemma on division 222 and remainder 24
222 = 24 × 9 + 6 …(ii)
Since remainder ≠ 0, apply division lemma on division 24 and remainder 6
24 = 6 × 4 + 0 …(iii)
We observe that the remainder = 0, so the last divisor 6 is the HCF of the 468 and 222
From (ii) we have
6 = 222 – 24 × 9
⇒ 6 = 222 – [468 – 222 × 2] × 9 [Substituting 24 = 468 – 222 × 2 from (i)]
⇒ 6 = 222 – 468 × 9 – 222 × 18
⇒ 6 = 222 × 19 – 468 × 9
⇒ 6 = 222y + 468x, where x = −9 and y = 19
5. If the HCF of 408 and 1032 is expressible in the form 1032 m − 408 × 5, find m.
Solution
General integers are 408 and 1032 where 408 < 1032
By applying Euclid’s division lemma, we get
1032 = 408 × 2 + 216
Since remainder ≠ 0, apply division lemma on division 408 and remainder 216
408 = 216 × 1 + 192
Since remainder ≠ 0, apply division lemma on division 216 and remainder 192
216 = 192 × 1 + 24
Since remainder ≠ 0, apply division lemma on division 192 and remainder 24
192 = 24 × 8 + 32
We observe that 32m under in 0. So the last divisor 24 is the H.C.F of 408 and 1032
∴ 216 = 1032m – 408 × 5
⇒ 1032 m = 24 + 408 × 5
⇒ 1032m = 24 + 2040
⇒ 1032m = 2064
⇒ m = 2064/1032 = 2
6. If the HCF of 657 and 963 is expressible in the form 657 x + 963 x − 15, find x.
Solution
657 and 963
By applying Euclid’s division lemma
963 = 657 × 1 + 306
Since remainder ≠ 0, apply division lemma on division 657 and remainder 306
657 = 306 × 2 + 45
Since remainder ≠ 0, apply division lemma on division 306 and remainder 45
306 = 45 × 6 + 36
Since remainder ≠ 0, apply division lemma on division 45 and remainder 36
45 = 36 × 1 + 19
Since remainder ≠ 0, apply division lemma on division 36 and remainder 19
36 = 19 × 4 + 0
∴ HCF = 657
Given HCF = 657 + 963 × (15)
⇒ 9 = 657x −14445
⇒ 9 + 14445 = 657x
⇒ 657x = 14454
⇒ x = 1454/657
⇒ x = 22
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
Members in arms = 616
Members in Band = 32
∴ Maximum numbers of columns
= HCF of 616 and 32
By applying Euclid’s division lemma
616 = 32 × 19 + 8
32 = 8 × 4 + 0
∴ HCF = 8
Hence the maximum remainder number of columns in which they can each is 8
8. Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
Solution
The required number when the divides 615 and 963
Leaves remainder 616 is means 615 – 6 = 609 and 963 – 957 are completely divisible by the number
∴ the required number
= HCF of 609 and 957
By applying Euclid’s division lemma
957 = 609 × 1 + 348
609 = 348 × 1 + 261
348 = 261 × 1 + 87
261 = 87 × 370
HCF = 87
Hence the required number is ‘87’
9. Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.
Solution
The require number when divides 285 and 1249, leaves remainder 9 and 7, this means 285 – 9 = 276 and 1249 – 7 = 1242 are completely divisible by the number
∴ The required number = HCF of 276 and 1242
By applying Euclid’s division lemma
1242 = 276 × 4 + 138
276 = 138 × 2 + 0
∴ HCF = 138
Hence remainder is = 0
Hence required number is 138
10. Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
Solution
The required number when divides 280 and 1245 leaves the remainder 4 and 3, this means 280 4 – 216 and 1245 – 3 = 1245 – 3 = 1242 are completely divisible by the number ∴ The required number = HCF of 276 and 1242
By applying Euclid’s division lemma
1242 = 276 × 4 + 138
276 = 138 × 2 + 0
∴ HCF = 138
Hence the required numbers is 138
11. What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.
Solution
The required number when divides 626, 3127 and 15628, leaves remainder 1, 2 and 3. This means 626 – 1 = 625, 3127 – 2 = 3125 and
15628 – 3 = 15625 are completely divisible by the number
∴ The required number = HCF of 625, 3125 and 15625
First consider 625 and 3125
By applying Euclid’s division lemma
3125 = 625 × 5 + 0
HCF of 625 and 3125 = 625
Now consider 625 and 15625
By applying Euclid’s division lemma
15625 = 625 × 25 + 0
∴ HCF of 625, 3125 and 15625 = 625
Hence required number is 625
12. Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Solution
The required number when divides 445, 572 and 699 leaves remainders 4, 5 and 6
This means 445 – 4 = 441, 572 – 5 = 561 and 699 – 6 = 693 are completely divisible by the number
∴ The required number = HCF of 441, 567 and 693
First consider 441 and 567
By applying Euclid’s division lemma
567 = 441 × 1 + 126
441 = 126 × 3 + 63
126 = 63 × 2 + 0
∴ HCF of 441 and 567 = 63
Now consider 63 and 693
By applying Euclid’s division lemma
693 = 63 × 11 + 0
∴ HCF of 441, 567 and 693 = 63
Hence required number is 63.
13. Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively.
Solution
The required number when divides 2011 and 2623
Leaves remainders 9 and the means
2011 – 9 = 2002 and 2623 – 5 = 2618 are completely divisible by the number
∴ The required number = HCF of 2002 and 2618
By applying Euclid’s division lemma
2618 = 2002 × 1 + 616
2002 = 616 × 3 + 154
616 = 754 × 4 + 0
∴ HCF of 2002 and 2618 = 154
Hence required number is 154
14. The length, breadth and height of a room are 8m 25 cm, 6m 75 cm and 4m 50 cm, respectively. Determine the longest rod which can measure the three dimensions of the room exactly.
Solution
Length of room = 8m 25cm = 825 cm
Breadth of room = 6m 75m = 675 cm
Height of room = 4m 50m = 450 cm
∴ The required longest rod
= HCF of 825, 675 and 450
First consider 675 and 450
By applying Euclid’s division lemma
675 = 450 × 1 + 225
450 = 225 × 2 + 0
∴ HCF of 675 and 450 = 825
Now consider 625 and 825
By applying Euclid’s division lemma
825 = 225 × 3 + 150
225 = 150 × 1 + 75
150 = 75 × 2 + 0
HCF of 825, 675 and 450 = 75
15. 105 goats, 140 donkeys and 175 cows 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?
Solution
Number of goats = 205
Number of donkey = 140
Number of cows = 175
∴ The largest number of animals in one trip = HCF of 105, 140 and 175
First consider 105 and 140
By applying Euclid’s division lemma
140 = 105 × 1 + 35
105 = 35 × 3 + 0
∴ HCF of 105 and 140 = 35
Now consider 35 and 175
By applying Euclid’s division lemma
175 = 35 × 5 + 0
HCF of 105, 140 and 175 = 35
16. 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?
Solution
Number of pastries = 15
Number of biscuit packets = 12
∴ The required no of boxes to contain equal number = HCF of 15 and 13
By applying Euclid’s division lemma
15 = 12 × 13
12 = 2 × 9 = 0
∴ No. of boxes required = 3
Hence each box will contain 15/3 = 5 pastries and 2/3 biscuit packets
17. 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?
Solution
Size of bathroom = 10ft by 8ft
= (10 × 12) inch by (8 × 12) inch
= 120 inch by 96 inch
The largest size of tile required = HCF of 120 and 96
By applying Euclid’s division lemma 120 = 96 × 1 + 24
96 = 24 × 4 + 0
∴ HCF = 24
∴ Largest size of tile required = 24 inches
∴ No. of tiles required = (Area of bathroom)/(area of 2 tile)
= (120 × 96)/(24×24)
= 5 × 4
= 20 tiles
18. 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?
Solution
Number of chocolates of 1^{st} brand in one pack = 24
Number of chocolates of 2^{nd} b and in one pack = 15
∴ The least number of chocolates 1 need to purchase
= LCM of 24 and 15
= 2 × 24 × 2 × 2 × 3 × 5
= 120
∴ The number of packet of 1^{st} brand = 120/24 = 5
And the number of packet of 2^{nd} brand =120/15= 8
∴ Largest size of tile required = 24 inches
∴ No of tiles required =
No of chocolates of 1^{st} brand in one pack = 24
No of chocolate of 2^{nd} brand in one pack = 15
∴ The least number of chocolates I need to purchase
= LCM of 24 and 15
= 2 × 2 × 2 × 3 × 5
= 120
∴ The number of packet of 1^{st} brand = 120/24 = 5
All the number of packet of 2^{nd} brand = 120/15 = 8
19. 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
Number of cartons of coke cans = 144
Number of cartons of pepsi cans = 90
∴ The greatest number of cartons in one stock = HCF of 144 and 90
By applying Euclid’s division lemma
144 = 90 × 1 + 54
90 = 54 × 1 + 36
54 = 36 × 1 + 18
36 = 18 × 2 + 0
∴ HCF = 18
Hence the greatest number cartons in one stock = 18
20. 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
Number of color pencils in one pack = 24
No of crayons in pack = 32
∴ The least number of both colors to be purchased
= LCM of 24 and 32
= 2 × 2 × 2 × 2 × 3
= 96
∴ Number of packs of pencils to be bought = 96/24 = 1
And number of packs of crayon to be bought = 96/32 = 3
21. A merchant has 120 liters of oil of one kind, 180 liters of another kind and 240 liters 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?
Solution
Quantity of oil A = 120 liters
Quantity of oil B = 180 liters
Quantity of oil C = 240 liters
We want to fill oils A, B and C in tins of the same capacity
∴ The greatest capacity of the tin chat can hold oil. A, B and C = HCF of 120, 180 and 240
By fundamental theorem of arithmetic
120 = 2^{3} × 3 × 5
180 = 2^{2} × 3^{2 }× 5
240 = 2^{4 }× 3 × 5
HCF = 2^{2 }× 3 × 5 = 4 × 3 × 5 = 60 litres
The greatest capacity of tin = 60 liters