Class 10 Maths Chapter 1 Question Answers - Real Numbers

Q1: Kerosene, paraffin, or lamp oil is a combustible hydrocarbon liquid which is derivative from petroleum. Kerosen’s uses vary from fuel for oil lamps to cleaning agents , jet fuel , heating oil or fuel for cooking  Two oil tankers contain 825 litres and 675 litres of kerosene oil respectively.
(a) Find the maximum capacity of a container which can measure the Kerosene oil of both the tankers when used an exact number of times.
(b) How many times we have to use container for both the tanker to fill?

Sol: 
(a) HCF of 825 and 625
825 = 3 x 5 x 5 x 11
675 = 3 x 3 x 3 x5 x 5
HCF = 3 x 5 x 5 = 75
Maximum capacity reqired is 75 litres
(b) The first tanker will require 875/75 = 11 times to fill
The second tanker will require 675/75 = 9 times to fill

Q2: The sum of LCM and HCF of two numbers is 7380.If the LCM of these numbers is 7340 more than their HCF. Find the product of the two numbers.

Sol: LCM + HCF = 7380
LCM – HCF = 7340
2LCM = 14720
LCM = 14720/2
LCM = 7360
LCM + HCF = 7380
7360+ HCF = 7380
HCF = 7380 – 7360
HCF = 20 (1)
HCF x LCM = product of numbers
20 x 7360= product of numbers
147200 = product of numbers 

Q3: If two positive integers x and y are expressible in terms of primes as x = p²q³ and y = p³q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain.

Sol: x = p²q³ and y = p³q
LCM = p³q³
HCF = p²q …..(i)
Now, LCM = p³q³
⇒ LCM = pq² (p²q)
⇒ LCM = pq² (HCF)
Yes, LCM is a multiple of HCF.
Explanation:
Let a = 12 = 2² × 3
b = 18 = 2 × 3²
HCF = 2 × 3 = 6 …(ii)
LCM = 2² × 3² = 36
LCM = 6 × 6
LCM = 6 (HCF) …[From (ii)]
Here LCM is 6 times HCF. 

Q4: Find HCF of numbers 134791, 6341 and 6339 by Euclid’s division algorithm. 

Sol: First, we find HCF of 6339 and 6341 by Euclid’s division method.
6341 > 6339
6341 = 6339 × 1 + 2
6339 = 2 × 3169 + 1
2 = 1 × 2 + 0
HCF of 6341 and 6339 is 1.
Now, we find the HCF of 134791 and 1
134791 = 1 × 134791 + 0
HCF of 134791 and 1 is 1.
Hence, the HCF of the given three numbers is 1. 

Q5: A woman wants to organise her birthday party. She was happy on her birthday but there was a problem that she does not want to serve fast food to her guests because she is very health conscious. She as 15 apples and 40 bananas at home and decided to serve them. She want to distribute fruits among guests. She does not want to discriminate among guests so she decided to distribute equally among all. So
(a) How many guests she can invite?
(b) How many apples and banana will each guest get?

Sol: 
(i)HCF of(15,40) =5

Fruits will be distributed equally among 5 guests 

(ii) Out of 15 apples each guest will get 15 /5 = 3 apples

Out of 40 banana each guest will get 40/5 =8 bananas

Q6: A charitable trust donates 28 different books of Maths,16 different books of science and 12 different books of Social Science to the poor students. Each student is given maximum number of books of only one subject of his interest and each student got equal number of books
(a) Find the number of books each student got.
(b) Find the total number of students who got books.

Sol:
(i) HCF of 28,16 and 12 is 4

Therefore maximum number of books each student get is 4 

(ii) Number of maths books 28/4 = 7

Number of science books 16/4 = 4

Number of social science = 12/4 = 3

Total books = 7 + 4 + 3 =14 

Q7: Show that any positive odd integer is of the form 4q + 1 or 4q + 3 where q is a positive integer.

Sol: Let a be a positive odd integer
By Euclid’s Division algorithm:
a = 4q + r …[where q, r are positive integers and 0 ≤ r < 4]
a = 4q
or 4q + 1
or 4q + 2
or 4q + 3
But 4q and 4q + 2 are both even
a is of the form 4q + 1 or 4q + 3.

Q8: Find the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the two numbers.

Sol: 306 = 2 × 3² × 17
657 = 3² × 73
HCF = 3² = 9
LCM = 2 × 3² × 17 × 73 = 22338
L.H.S. = LCM × HCF = 22338 × 9 = 201042
R.H.S. = Product of two numbers = 306 × 657 = 201042
L.H.S. = R.H.S.

Q9: Show that one and only one out of n, (n + 1) and (n + 2) is divisible by 3, where n is any positive integer.

Sol:
Let n, n + 1, n + 2 be three consecutive positive integers.
We know that n is of the form 3q, 3q + 1, or 3q + 2.
Case I. When n = 3q,
In this case, n is divisible by 3,
but n + 1 and n + 2 are not divisible by 3.
Case II. When n = 3q + 1,
In this case n + 2 = (3q + 1) + 2
= 3q + 3
= 3(q + 1 ), (n + 2) is divisible by 3,
but n and n + 1 are not divisible by 3.
Case III. When n = 3q + 2, in this case,
n + 1 = (3q + 2) + 1
= 3q + 3 = 3 (q + 1 ), (n + 1) is divisible by 3,
But n and n + 2 are not divisible by 3.
Hence, one and only one out of n, n + 1 and n + 2 is divisible by 3. 

Q10: Amita, Sneha, and Raghav start preparing cards for all persons of an old age home. In order to complete one card, they take 10, 16 and 20 minutes respectively. If all of them started together, after what time will they start preparing a new card together?

Sol: To find the earliest (least) time, they will start preparing a new card together, we find the LCM of 10, 16 and 20.
10 = 2 × 5
16 = 2⁴
20 = 2² × 5
LCM = 2⁴ × 5 = 16 × 5 = 80 minutes
They will start preparing a new card together after 80 minutes. 

Q11: Dudhnath has two vessels containing 720 ml and 405 ml of milk respectively. Milk from these containers is poured into glasses of equal capacity to their brim. Find the minimum number of glasses that can be filled.

Sol: 1st vessel = 720 ml; 2nd vessel = 405 ml
We find the HCF of 720 and 405 to find the maximum quantity of milk to be filled in one glass.
405 = 3⁴ × 5
720 = 2⁴ × 3² × 5
HCF = 3² × 5 = 45 ml = Capacity of glass
No. of glasses filled from 1st vessel = 720/45 = 16
No. of glasses filled from 2nd vessel = 405/45 = 9
Total number of glasses = 25 

Q12: There are 104 students in class X and 96 students in class IX in a school. In a house examination, the students are to be evenly seated in parallel rows such that no two adjacent rows are of the same class. 
(a) Find the maximum number of parallel rows of each class for the seating arrange¬ment.
(b) Also, find the number of students of class IX and also of class X in a row.
(c) What is the objective of the school administration behind such an arrangement?

Sol:
104 = 2³ × 13
96 = 2⁵ × 3
HCF = 2³ = 8
(a) Number of rows of students of class X = 104/8 = 13
Number maximum of rows class IX = 96/8 = 12
Total number of rows = 13 + 12 = 25
(b) No. of students of class IX in a row = 8
No. of students of class X in a row = 8
(c) The objective of school administration behind such an arrangement is fair and clean examination, so that no student can take help from any other student of his/her class. 

Q13: Prove that 3 + 2√5 is irrational.

Sol:
Let us assume, to the contrary, that 3 + 2√5 is rational
So that we can find integers a and b (b ≠ 0), such that
3 + 2 √5 = ab, where a and b are coprime.
Rearranging this equation, we get

⇒ 2√5 = ab - 3   ⇒     √5 = a - 3b2b

⇒ √5 =  a2b - 3b 2b  ⇒  √5 =  a2b - 3 2
Since a and b are integers, we get that  a2b - 3  2 is rational and so √5 is rational. But this contradicts the fact that √5 is irrational.
So we conclude that 3 + 2√5 is irrational.

Q14: Prove that √5 is irrational.

Sol:
Consider that √5 is a rational number.
i.e. √5 = x/y (as well as, x and y are co-primes)
y√5= x
Squaring the both the sides, we observe,
(y√5)2 = x2
⇒5y2 = x2……………………………….. (1)
Therefore, x2 is divided by 5, so x is also divided by 5.
Consider, x = 5k, for some value of k and putting the value of x in equation (1), we observe,
5y2 = (5k)2
⇒y2 = 5k2
is divisible by 5 it means y is divisible by 5.
Clearly, x and y are not co-primes. Thus, our assumption about √5 is rational is incorrect.
Hence, √5 is an irrational number.

Q15: A positive integer is of the form 3q + 1, q being a natural number. Can you write its square in any form other than 3m + 1, i.e., 3m or 3m + 2 for some integer m? Justify your answer.

Sol:
No.
Let the positive integer 3q + 1, where q is the natural number.
(3q + 1)2 = 9q2 + 6q + 1
= 3(3q2 + 2q) + 1
= 3m + 1, (here, m is an integer which is equal to 3q2 + 2q.
Hence (3q + 1)2 cannot be expressed in any other form apart from 3m + 1.