Chapter 1 - Real Numbers, RD Sharma Solutions - (Part - 4) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 1.27

Q.4. Find the largest number which divides 615 and 963 leaving remainder 6 in each case.
Ans. We need to find the largest number which divides 615 and 963 leaving remainder 6 in each case.
The required number when divides 615 and 963, leaves remainder 6, this means 615 - 6 = 609 and 963 - 6 = 957 are completely divisible by the number.
Therefore,
The required number = H.C.F. of 609 and 957.
By applying Euclid’s division lemma
957 = 609 x 1 + 348
609 = 348 x 1 + 261
348 = 216 x 1 + 87
261 = 87 x 3 + 0.
Therefore, H.C.F. = 87.
Hence, the required number is 87.

Q.5. If the HCF of 408 and 1032 is expressible in the form 1032 m – 408 × 5, find m.
Ans. We need to find m if the H.C.F of 408 and 1032 is expressible in the form
1032 m - 408 x 5.
Given integers are 408 and 1032 where 408 < 1032.
By applying Euclid’s division lemma, we get 1032 = 408 x 2 + 216.
Since the remainder ≠ 0, so apply division lemma on divisor 408 and remainder 216
408 = 216 x 1 + 192.
Since the remainder  ≠ 0, so apply division lemma on divisor 216 and remainder 192
216 = 192 x 1 + 24.
Since the remainder  ≠ 0, so apply division lemma on divisor 192 and remainder 24
192 = 24 x 8 + 0.
We observe that remainder is 0. So the last divisor is the H.C.F of 408 and 1032.
Therefore,
24 = 1032 m - 408 x 5
⇒ 1032 m = 24 + 408 x 5
⇒ 1032 m = 24 + 2040
⇒ 1032 m = 2064
⇒ m = 2064/1032
⇒ m = 2

Q.6. If the HCF of 657 and 963 is expressible in the form 657x + 963 × – 15, find x.
Ans. 
We need to find x if the H.C.F of 657 and 963 is expressible in the form
657x + 963y(-15).
Given integers are 657 and 963.
By applying Euclid’s division lemma, we get
963 = 657 x 1 + 306
Since the remainder ≠ 0, so apply division lemma on divisor 657 and remainder 306
657 = 306 x 2 + 45.
Since the remainder ≠ 0, so apply division lemma on divisor 306 and remainder 45
306 = 45 x 6 + 36.
Since the remainder ≠ 0, so apply division lemma on divisor 45 and remainder 36
45 = 36 x 1 + 9.
Since the remainder ≠ 0, so apply division lemma on divisor 36 and remainder 9
36 = 9 x 4 + 0.
Therefore, H.C.F. = 9.
Given H.C.F = 657x + 936(-15).
Therefore,
⇒ 9 = 657x - 14445
⇒ 9 + 14445 = 657x
⇒ 14454 = 657x
⇒ x = 14454 / 657
⇒ x = 22

Q.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?
Ans. We are given that 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. We need to find the maximum number of columns in which they can march.
Members in army = 616
Members in band = 32.
Therefore,
Maximum number of columns = H.C.F of 616 and 32.
By applying Euclid’s division lemma
616 = 32 x 19 + 8
32 = 8 x 4 + 0.
Therefore, H.C.F. = 8
Hence, the maximum number of columns in which they can march is 8.

Q.8. 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?
Ans. The merchant has 3 different oils of 120 liters, 180 liters and 240 liters respectively.
So the greatest capacity of the tin for filling three different types of oil is given by the H.C.F. of 120,180 and 240.
So first we will calculate H.C.F of 120 and 180 by Euclid’s division lemma.
180 = (120)(1) + 60
120 = (60)(2) + 0
The divisor at the last step is 60. So the H.C.F of 120 and 180 is 60.
Now we will find the H.C.F. of 60 and 240,
240 = (60)(4) + 0
The divisor at the last step is 60. So the H.C.F of 240 and 60 is 60.
Therefore, the tin should be of 60 litre.

Q.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?
Ans. We are given that during a sale, color pencils were being sold in packs of 24 each and crayons in packs of 32 each. If we want full packs of both and the same number of pencils and crayons, we need to find the number of each we need to buy.
Given that
Number of color pencils in one pack = 24
Number of crayons in pack = 32.
Therefore, the least number of both colors to be purchased
L.C.M. of 24 and 32 = 2 x 2 x 2 x 2 x 2 x3
= 96

Hence, number of packs of pencils to bought
96/24 = 4,
And number of packs of crayon to be bought
96/32 = 3.

Page No 1.28

Q.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?
Ans. Given that 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 contains cartons of the same drink We need to find the greatest number of cartons, each stack would have
Given that
Number of cartons of coke cans = 144
Number of cartons of Pepsi cans = 90.
Therefore, the greatest number of cartons in one stack = H.C.F. of 144 and 90.
By applying Euclid’s division lemma
144 = 90 x 1 + 54
90 = 54 x 1 + 36
54 = 36 x 1 + 18
36 = 18 x 2 + 0
H.C.F. = 18.
Hence, the greatest number cartons in one stack 18.

Q.11. Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.
Ans. We need to find the greatest number which divides 285 and 1249 leaving remainder 9 and 7 respectively.
The required 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.
Therefore, the required number = H.C.F. of 276 and 1242.
By applying Euclid’s division lemma
1242 = 276 x 4 + 138
276 = 138 x 2 + 0.
Therefore, H.C.F. = 138
Hence, required number is 138.

Q.12. Find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
Ans. We need to find the largest number which exactly divides 280 and 1245 leaving remainders 4 and 3, respectively.
The required number when divides 280 and 1245, leaves remainder 4 and 3, this means 280 - 4 = 276 and 1245 - 3 = 1242 are completely divisible by the number.
Therefore, the required number = H.C.F. of 276 and 1242.
By applying Euclid’s division lemma
1242 = 276 x 4 + 138
276 =138 x 2 + 0.
Therefore, H.C.F. = 138.
Hence, the required number is 138.

Q.13. What is the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.
Ans. We need to find the largest number that divides 626, 3127 and 15628 and leaves remainders of 1, 2 and 3 respectively.
The required number when divides 626, 3127 and 15628 leaves remainders 1, 2 and 3 this means 626 - 1 = 625, 3127 - 2 = 15625 are completely divisible by the number.
Therefore, the required number = H.C.F. of 625, 3125 and 15625.
First we consider 625 and 3125.
By applying Euclid’s division lemma
3125 = 625 x 5 + 0.
H.C.F. of 625 and 3125 = 625
Now, consider 625 and 15625.
By applying Euclid’s division lemma
15625 = 625 x 25 + 0.
Therefore, H.C.F. of 625, 3125 and 15625 = 625
Hence, the required number is 625.

Q.14. Find the greatest number that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.
Ans. Find the greatest number that divides 445, 572 and 699 and leaves remainders of 4, 5 and 6 respectively.
The required number when divides 445, 572 and 699 leaves remainders 4, 5 and 6 this means 445 - 4 = 441, 572 - 5 = 567 and 699 - 6 = 693  are completely divisible by the number.
Therefore, the required number = H.C.F. of 441, 567 and 693.
First consider 441 and 567.
By applying Euclid’s division lemma
567 = 441 x 1 + 126
441 = 126 x 3 + 63
126 = 63 x 2 + 0.
Therefore, H.C.F. of 441 and 567 = 63
Now, consider 63 and 693
By applying Euclid’s division lemma
693 = 63 x 11 + 0.
Therefore, H.C.F. of 441, 567 and 693 = 63
Hence, the required number is 63.

Q.15. Find the greatest number which divides 2011 and 2623 leaving remainders 9 and 5 respectively
Ans. Find the greatest number which divides 2011 and 2623 leaving remainder 9 and 5 respectively.
The required number when divides 2011 and 2623 leaves remainders 9 and 5 this means 2011 - 9 = 2002 and 2623 - 5 = 2618 are completely divisible by the number.
Therefore, the required number = H.C.F. of 2002 and 2618
By applying Euclid’s division lemma
2618 = 2002 x 1 + 616
2002 = 616 x 3 + 154
616 = 154 x 4 + 0.
H.C.F. of 2002 and 2618 = 154
Hence, the required number is 154.