Chapter 1 - Real Numbers, RD Sharma Solutions - (Part - 5) | RD Sharma Solutions for Class 10 Mathematics PDF Download

Page No 1.28

Q.16. Using Euclid's division algorithm , find the largest number that divides 1251, 9377, and  15628 leaving remainders 1, 2 and 3 respectively.
Ans. It is given that 1, 2 and 3 are the remainders of 1251, 9377 and 15628, respectively. Subtracting these remainders from the respective numbers, we get
1251 − 1 = 1250
9377 − 2 = 9375
15628 − 3 = 15625
Now, 1250, 9375 and 15625 are divisible by the required number.
Required number = HCF of 1250, 9375 and 15625
By Euclid's division algorithm a = bq + r, 0 ≤ r < b
For largest number, put a = 15625 and b = 9375
15625 = 9375 × 1 + 6250
⇒ 9375 = 6250 × 1 + 3125
⇒ 6250 = 3125 × 2 + 0
Since remainder is zero, therefore, HCF(15625 and 9375) = 3125
Further, take c = 1250 and d = 3125. Again using Euclid's division algorithm
d = cq + r, 0 ≤ r < c ⇒ 3125 = 1250 × 2 + 625     [∵r ≠ 0]
⇒ 1250 = 625 × 2 + 0
Since remainder is zero, therefore, HCF(1250, 9375 and 15625) = 625 Hence, 625 is the largest number which divides 1251, 9377 and 15628 leaving remainder 1, 2 and 3, respectively.

Q.17. 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?
Ans. We are given that two brands of chocolates are available in packs of 24 and 15 respectively. If he needs to buy an equal number of chocolates of both kinds, then find least number of boxes of each kind he would need to buy.
Given that
Number of chocolates of 1^st brand in one pack = 24
Number of chocolates of 2^nd brand in one pack = 15.
Therefore, the least number of chocolates he need to purchase is
L.C.M. of 24 and 15 = 2 x 2 x 2 x 3 x 5
= 120

Therefore, the number of packet of 1^st brand is
120/24 = 5
And the number of packet of 2^nd brand is
120/15 = 8.

Q.18. 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 cinches of the tile required that has to be cut and how many such tiles are required?
Ans. A mason has to fit a bathroom with square marble tiles of the largest possible size. The size of the bathroom is 10ft. by 8ft. We need to find the size in inches of the tile required that has to be cut and number of such tiles are required.
Size of bathroom = 10ft by 8ft
= (10 x 12) inch. by (8 x 12) inch
= 120 inch. by 96 inch.
The largest size of tile required = H.C.F. of 120 and 96.
By applying Euclid’s division lemma
120 = 96 x 1 + 24
96 = 24 x 4 + 0
Therefore, H.C.F. = 24.
Thus, largest size of tile required = 24 inches.
Therefore,
No. of tiles required = Area of bathroom/Area of 1 tile
= 120 x 96 / 24 x 24
= 5 x 4
= 20 tiles

Q.19. 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?
Ans. We are given that 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. We need to find the number of biscuit packets and number of pastries each box contain.
Given that
Number of pastries = 15
Number of biscuits packets = 12.
Therefore, required number of boxes to contain equal number = H.C.F. of 15 and 12.
By applying Euclid’s division lemma
15 = 12 x 1 + 3
12 = 3 x 4 + 0
Therefore, number of boxes required = 3.
Hence each box will contain 5 pastries and 4 biscuits packets.

Q.20. 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?
Ans. We are given that, 105 goats, 140 donkeys and 175 cows. There is only one boat which will have to make many y 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. We need to tell the number of animals that went in each trip.
Given that
Number of goats = 105
Number of donkeys = 140
Number of cows = 175.
Therefore, the largest number of animals in 1 trip = H.C.F. of 105, 140 and 175.
First we consider 105 and 140.
By applying Euclid’s division lemma
140 = 105 x 1 + 35
105 = 35 x 3 + 0
Therefore, H.C.F. of 105 and 140 = 35
Now, we consider 35 and 175.
By applying Euclid’s division lemma
175 = 35 x 5 + 0
Therefore, H.C.F. of 105, 140 and 175 = 35
Hence, the number of animals went in each trip is 35.

Q.21. The length, breadth and height of a room are 8 m 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.
Ans. We are given the length, breadth and height of a room as 8m 25cm, 6m 75cm and 4m 50cm, respectively. We need to determine the largest room which can measure the three dimensions of the room exactly.
We first convert each dimension in cm
Length of room = 8m 25cm = 825cm
Breadth of room = 6m 75cm = 675cm
Height of room = 4m 50cm = 450cm.
Therefore, the required longest rod = H.C.F. of 825, 675 and 450.
First we consider 675 and 450.
By applying Euclid’s division lemma
675 = 450 x 1 + 225
450 = 225 x 2 + 0
Therefore, H.C.F. of 675 and 450 = 225
Now, we consider 225 and 825.
By applying Euclid’s division lemma
825 = 225 x 3 + 150
225 = 150 x 1 + 75
150 = 75 x 2 + 0
Therefore, H.C.F. of 825, 675 and 450 = 75
Hence, the length of required longest rod is 75cm.

Q.22. Express the HCF of 468 and 222 as 468x + 222y where x, y are integers in two different ways.
Ans. We need to express the H.C.F. of 468 and 222 as 468x + 222y
Where x, y are integers in two different ways.
Given integers are 468 and 222, where 468 > 222
By applying Euclid’s division lemma, we get 468 = 222 x 2 + 24
Since the remainder ≠ 0, so apply division lemma on divisor 222 and remainder 24
222 = 24 x 9 + 6
Since the remainder ≠ 0, so apply division lemma on divisor 24 and remainder 6
24 = 6 x 4 + 0
We observe that remainder is 0. So the last divisor 6 is the H.C.F. of 468 and 222 from we have
6 = 222 - 24 x 9
⇒ 6 = 222 - (468 - 222 x 2) x 9  [substituting 24 = 468 - 222 x 2]
⇒ 6 = 222 - 468 x 9 + 222 x 18
⇒ 6 = 222 x 19 - 468 x 9
⇒ 6 = 222y + 468x, where x = -9 and y = 19.

Page No 1.35

Q.1. Express each of the following integers as a product of its prime factors:
(i) 420
Ans. 420 = 2² x 3 x 5 x 7

(ii) 468 
Ans. 468 = 2² x 3² x 13

(iii) 945 
Ans. 945 = 3³ x 5 x 7

(iv) 7325 
Ans. 7325 = 5² x 293

TO EXPRESS: each of the following numbers as a product of their prime factors  

Q.2. Determine the prime factorisation of each of the following positive integer:
(i) 20570
Ans. 20570 = 2 × 5 × 11² × 17

(ii) 58500
Ans. 58500 = 2² x 3² x 5³ x 13
(iii) 45470971
Ans. 45470971 = 7² x 13² x 17² x 19

TO EXPRESS: each of the following numbers as a product of their prime factors  

Q.3. Explain why 7 × 11 × 13 + 13 and 7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 are composite numbers.
Ans. We can see that both the numbers have common factor 7 and 1.
7 × 11 × 13 + 13 = (77 + 1) x 13
= 78 x 13
7 × 6 × 5 × 4 × 3 × 2 × 1 + 5 = (7 × 6 × 5 × 4 × 3 × 2 + 1) x 5
= 1008 x 5
And we know that composite numbers are those numbers which have at least one more factor other than 1.
Hence after simplification we see that both numbers are even and therefore the given two numbers are composite numbers.

Q.4. Check whether 6ⁿ can end with the digit 0 for any natural number n.
Ans. We know that
6ⁿ = (2 x 3)ⁿ
6ⁿ = 2ⁿ x 3ⁿ
Therefore, prime factorization of 6ⁿ does not contain 5 and 2 as a factor together.
Hence 6ⁿ can never end with the digit 0 for any natural number n.

Q.5. Explain why 3 × 5 × 7 + 7 is a composite number.
Ans. Let a = 3 × 5 × 7 + 7 = 7(3 × 5 + 1) = 7 × 16
It can be seen that a has two more factors 7 and 16 other than 1 and the number itself.
Therefore, it is a composite number.