Solving Quadratic Equations, Assignments of Mathematics

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CBSE 2025

MATHS STANDARD

Including Competency Based Questions

C L A S S 1 0C L A S S 1 0

Chapter-wise Question Bank

Based on Previous 20 Years 63 Papers

NODIA AND COMPANY

CONTENTS

• CHAP 1. Real Numbers 5-
• CHAP 2. Polynomials 27-
• CHAP 3. Pair of Linear Equation in Two Variables 57-
• CHAP 4. Quadratic Equations 103-
• CHAP 5. Arithmetic Progression 154-
• CHAP 6. Triangles 204-
• CHAP 7. Co-ordinate Geometry 237-
• CHAP 8. Introduction of Trigonometry 275-
• CHAP 9. Some Applications of Trigonometry 313-
• CHAP 10. Circle 367-
• CHAP 11. Areas Related to Circles 408-
• CHAP 12. Surface Areas and Volumes 438-
• CHAP 13. Statistics 480-
• CHAP 14. Probability 538-

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Page 6 Real Numbers Chap 1

1. The HCF and the LCM of 12, 21, 15 respectively are (a) 3, 140 (b) 12, 420 (c) 3, 420 (d) 420, 3

Sol : [Board 2020 Delhi Standard] We have 12 = (^2) # 2 # 3 21 = (^3) # 7 15 = (^3) # 5 HCF(12, 21, 15) = 3 LCM (12, 21, 15) = (^2) # 2 # 3 # 5 # 7 = 420 Thus (c) is correct option.

1. The LCM of smallest two digit composite number and smallest composite number is (a) 12 (b) 4 (c) 20 (d) 44

Sol : [Board 2020 SQP Standard] Smallest two digit composite number is 10 and smallest composite number is 4. LCM ( 10 4, ) = 20 Thus (c) is correct option.

1. HCF of two numbers is 27 and their LCM is 162. If one of the numbers is 54, then the other number is (a) 36 (b) 35 (c) 9 (d) 81

Sol : [Board 2020 OD Basic] Let y be the second number. Since, product of two numbers is equal to product of LCM and HCM, (^54) # y =LCM (^) #HCF (^54) # y = (^162) # 27

y = 16254 #^27 = 81

Thus (b) is correct option.

1. HCF of 144 and 198 is

(a) 9 (b) 18 (c) 6 (d) 12

Sol : [Board 2020 Delhi Basic] Using prime factorization method, 144 = (^2) # 2 # 2 # 2 # 3 # 3 = (^2 4) # 32 and 198 = (^2) # 3 # 3 # 11

= (^2) # 3 2 # 11 HCF(144, 198) = (^2) # 3 2 = (^2) # 9 = 18 Thus (b) is correct option.

1. 225 can be expressed as (a) 5 (^) # 3 2 (b) 5 (^2) # 3 (c) (^5 2) # 32 (d) 5 (^3) # 3

Sol : [Board 2020 Delhi Basic] By prime factorization of 225, we have 225 = (^3) # 3 # 5 # 5 = (^3 2) # 52 or 5 (^2) # 32 Thus (c) is correct option.

1. 2 3 is (a) an integer (b) a rational number (c) an irrational number (d) a whole number

Sol : [Board 2020 OD Basic] Let us assume that 2 3 is a rational number. Now 2 3 = r where r is rational number

or 3 = r 2

Now, we know that 3 is an irrational number, So, r 2 has to be irrational to make the equation true. This is a contradiction to our assumption. Thus, our assumption is wrong and 2 3 is an irrational number. Thus (c) is correct option.

1. The product of a non-zero rational and an irrational number is (a) always irrational (b) always rational (c) rational or irrational (d) one Sol : Product of a non-zero rational and an irrational number is always irrational i.e., (^43) # 2 = 3 42 which is irrational. Thus (a) is correct option.
2. If two positive integers a and b are written as a = x y^3 and b = xy^3 , where x , y are prime numbers, then HCF (^) ^ a b , h is (a) (^) xy (b) (^) xy^2 (c) x y^3 3 (d) x y^2

Sol : [Board Term-1 2014, 2011] We have a = x y^3 2 = x (^) # x (^) # x (^) # y (^) # y b = xy^3 = x (^) # y (^) # y (^) # y

Chap 1 Real Numbers Page 7

HCF( a , b ) = HCF ^ x y 3^3 ,^ xy^3 h = x (^) # y (^) # y = xy^2 HCF is the product of the smallest power of each common prime factor involved in the numbers. Thus (b) is correct option.

1. If two positive integers p and q can be expressed as p = ab^2 and q = a b^3 ; where a , b being prime numbers, then LCM (^) ^ p q , h is equal to (a) ab (b) a b^2 (c) a b^3 2 (d) a b^3

Sol : [Board 2010] We have p = ab^2 = a (^) # b (^) # b and q = a b^3 = a (^) # a (^) # a (^) # b LCM( p , q ) = LCM (^) ^ ab^2 ,^ a b^3 h = a (^) # b (^) # b (^) # a (^) # a = a b^3 LCM is the product of the greatest power of each prime factor involved in the numbers. Thus (c) is correct option.

1. The least number that is divisible by all the numbers from 1 to 10 (both inclusive) (a) 10 (b) 100 (c) 504 (d) 2520

Sol : [Board Term-1 2016, 2015] Factor of 1 to 10 numbers 1 = 1 2 = (^1) # 2 3 = (^1) # 3 4 = (^1) # 2 # 2 5 = (^1) # 5 6 = (^1) # 2 # 3 7 = (^1) # 7 8 = (^1) # 2 # 2 # 2 9 = (^1) # 3 # 3 10 = (^1) # 2 # 5 LCM(1 to 10) = LCM 1 2 3 4 5 6 7 8 9 10^ , , , , , , , , , h = (^1) # 2 # 2 # 3 # 3 # 5 # 7 = 2520 Thus (d) is correct option.

1. If p 1 and p 2 are two odd prime numbers such that p (^) 1 > p 2 , then p (^) 12 - p 22 is (a) an even number (b) an odd number (c) an odd prime number (d) a prime number

Sol : p (^) 12 - p 22 is an even number. Let us take p 1 = 5 and p 2 = 3 Then, p (^) 12 - p 22 = 25 - 9 = 16 16 is an even number. Thus (a) is correct option.

1. The number 3 13 - 310 is divisible by (a) 2 and 3 (b) 3 and 10 (c) 2, 3 and 10 (d) 2, 3 and 13

Sol : 3 13 - 310 = 3 10 (^3 3 - 1 ) = 3 10 ( 26 ) = (^2) # 13 # 310 Hence, 3 13 - 310 is divisible by 2, 3 and 13. Thus (d) is correct option.

1. 1. The L.C.M. of x and 18 is 36.
   2. The H.C.F. of x and 18 is 2. What is the number (^) x? (a) 1 (b) 2 (c) 3 (d) 4

Sol : [Board Term-1 2013] LCM (^) # HCF (^) = First number (^) # second number

Hence, required number = 3618 #^2 = 4 Thus (d) is correct option.

1. If a = (^2 3) # 3 , b = (^2) # 3 # 5 , c = 3 n^ # 5 and LCM ( , , ) a b c = (^2 3) # 3 2 # 5 ,then n is (a) 1 (b) 2 (c) 3 (d) 4

Sol : Value of n must be 2. Thus (b) is correct option.

1. The least number which is a perfect square and is divisible by each of 16, 20 and 24 is (a) 240 (b) 1600 (c) 2400 (d) 3600

Chap 1 Real Numbers Page 9

1. Find the HCF and LCM of 90 and 144 by the method of prime factorization. Sol : [Board Term-1 2012] We have 90 = (^9) # 10 = (^9) # 2 # 5 = (^2) # 3 2 # 5 and 144 = (^16) # 9 = (^2 4) # 32 HCF = (^2) # 3 2 = 18 LCM = (^2 4) # 3 2 # 5 = 720

PRACTICE  Find HCF of 144 and 198. [Board 2020 Delhi Basic] Ans : 18

1. If two positive integers a and b are written as a = x y^3 and b = xy^3 , where x , y are prime numbers, then find HCF (^) ^ a b , h. Sol : [Board Term -1 2014] We have a = x y^3 2 = x (^) # x (^) # x (^) # y (^) # y b = xy^3 = x (^) # y (^) # y (^) # y HCF( a , b ) = HCF ^ x y 3^2 ,^ xy^3 h = x (^) # y (^) # y = xy^2 HCF is the product of the smallest power of each common prime factor involved in the numbers.

PRACTICE  If two positive integers p and q can be expressed as p = ab^2 and q = a b^3 ; where a , b being prime numbers, then what is the LCM of ^ p q , h? [Board Term -1 2014] Ans : a b^3

1. What are the values of x and y in the given figure?

Sol : [Board Term -1 2012] We have 1001 = x (^) # 143 & x = 7 143 = y (^) # 11 & y = 13 Hence x = 7 , y = 13

1. If HCF(336, 54) = 6 , find LCM(336, 54). Sol : [Board 2019 OD] HCF (^) # LCM = Product of number (^6) # LCM = (^336) # 54

LCM =^336 6 #^54

= (^56) # 54 = 3024 Thus LCM of 336 and 54 is 3024.

1. a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. Then calculate the least prime factor of ( a + b ). Sol : [Board Term-1 2014] Here a and b are two positive integers such that the least prime factor of a is 3 and the least prime factor of b is 5. The least prime factor of ( a + b )would be 2.
2. What is the HCF of the smallest composite number and the smallest prime number? Sol : [Board Term-1 OD 2018] The smallest prime number is 2 and the smallest composite number is 4 = 2 2. Hence, required HCF is ( 2 2 ,^2 ) = 2.
3. Calculate the HCF of 3 3 × 5 and 3 2 × 52. Sol : [Board 2007] We have 3 3 × 5 = 3 2 × (^5) # 3 3 2 ×^52 = 3 2 × (^5) # 5 HCF ( 3 3 × , 5 3 2 ×^52 )^ = 3 2 × 5 = 9 × 5 = 45
4. If HCF ( , ) a b = 12 and a (^) # b = 1 800, , then find LCM ( , ) a b. Sol : We know that HCF ( , ) a b (^) # LCM( , ) a b = a × b Substituting the values we have (^12) # LCM ( , ) a b = 1800

or, LCM ( , a b ) = 1 800, 12 = 150

Page 10 Real Numbers Chap 1

1. Find the smallest natural number by which 1200 should be multiplied so that the square root of the product is a rational number. Sol : [Board Term-1 2016, 2015] We have 1200 = (^12) # 100 = (^4) # 3 # 4 # 25 = (^4 2) # 3 # 52 Here if we multiply by 3, then its square root will be 4 (^) # 3 # 5 which is a rational number. Thus the required smallest natural number is 3.
2. Can two numbers have 15 as their HCF and 175 as their LCM? Give reasons. Sol : [Board Term-1 2012] LCM of two numbers should be exactly divisible by their HCF. Since, 15 does not divide 175, two numbers cannot have their HCF as 15 and LCM as 175.
3. Find the least number that is divisible by all numbers between 1 and 10 (both inclusive). Sol : [Board 2010] The required number is the LCM of 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, LCM = (^2) # 2 # 3 # 2 # 3 # 5 # 7 = 2520
4. Find HCF of the numbers given below: k , 2 k , 3 k , 4 k and 5 k ,, where k is a positive integer.

Sol : [Board Term-1 2015] Here we can see easily that k is common factor between all and this is highest factor Thus HCF of k , 2 k , 3 k , 4 k and 5 k ,is k.

1. Complete the following factor tree and find the composite number x.

Sol : [Board Term-1 2015] We have y = (^5) # 13 = 65 and x = (^3) # 195 = 585

TWO MARKS QUESTIONS

1. Prove that 5 - 2 3 is an irrational number. It is given that 3 is an irrational number. Sol : [Board 2024 OD Standard] Assume that 5 - 2 3 is a rational number. Therefore, we can write it in the form of (^) qp^ where p and q are co-prime integers and q! 0.

Now 5 - 2 3 = qp where q! 0 and p and q are integers. Rewriting the above expression as,

2 3 = 5 - qp

3 = 5 q 2^ - qp

Here 5 q^2^ - qp^ is rational because p and q are co-prime integers, thus 3 should be a rational number. But 3 is irrational. This contradicts the given fact that 3 is irrational. Hence 5 - 2 3 is an irrational number.

1. Show that the number (^5) # 11 # 17 + 3 # 11 is a composite number. Sol : [Board 2024 OD Standard] (^5) # 11 # 17 + (^3) # 11 = 11 5^ (^) # 17 + 3 h = (^11) #^ 85 + (^3) h = (^11) # 88 = (^11) # 8 # 11 = (^2 3) # 112 As we can see 5 (^) # 11 # 17 + (^3) # 11 can be factorised as 2 (^3) # 112. It means it has factors other than 1. Thus (^5) # 11 # 17 + (^3) # 11 is a composite number
2. Show that 6 n^ cannot end with digit 0 for any natural number n. Sol : [Board 2023 OD Standard] If the number 6 n^ for any n , were to end with the digit five, then it would be divisible by 5. That is, the prime factorization of 6 n^ would contain the prime 5. This is not possible because the only prime in the factorization of 6 n^ = ( (^2) # 3 ) n are 2 and
   1. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 6 n^. Since there is no prime factor 5, 6 n^ cannot end with the digit five.

Page 12 Real Numbers Chap 1

Sol : [Board Term-1 2015, 2014] We complete the given factor tree writing variable y and z as following.

We have z = 1617 = 23

y = (^7) # 161 = 1127 Composite number, (^) x = 2 # 3381 = 6762 PRACTICE  Complete the factor-tree and find the composite number M.

[Board Term-1 2013] Ans : M = 32760 , N = 16380 , O = 3 , P = 7 and Q = 13

 Complete the following factor tree and find the composite number x.

[Board Term-1 2015] Ans : x =^585 and y = 65

 Complete the following factor tree and find the composite number x

[Board Term-1 2015] Ans : x =^11130 , y =^5565 and z = 53

 Find the missing numbers a b c , , and d in the given factor tree:

Chap 1 Real Numbers Page 13

[Board Term-1 2012] Ans : c = 11 , b = 7 , c = 13 and d = 11.

1. Check whether 4 n^ can end with the digit 0 for any natural number n. Sol : [Board Term-1 2015] If the number 4 n ,^ for any n , were to end with the digit zero, then it would be divisible by 5 and 2. That is, the prime factorization of 4 n^ would contain the prime 5 and 2. This is not possible because the only prime in the factorization of 4 n^ = 22 n is 2. So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4 n^. So, there is no natural number n for which 4 n^ ends with the digit zero. Hence 4 n cannot end with the digit zero.

PRACTICE  Check whether ( 15 ) n^ can end with digit 0 for any n! N. [Board Term-1 2012] Ans : No

 Show that 7 n^ cannot end with the digit zero, for any natural number n. [Board Term-1 2012] Ans : Proof

 Show that numbers 8 n^ can never end with digit 0 of any natural number n. [Board Term-1 2015] Ans : Proof

1. Show that 5 6 is an irrational number. Sol : [Board Term-1 2015] Let 5 6 be a rational number, which can be expressed as (^) ba^ ,where b! 0 ; a and b are co-primes.

Now 5 6 = ba

6 = 5 ab

or, 6 = rational But, 6 is an irrational number. Thus, our assumption is wrong. Hence, 5 6 is an irrational number.

1. Write a rational number between 2 and 3. Sol : [Board Term-1 2013] We have 2 = 100200 and 3 = 100300 We need to find a rational number x such that

101 200 <^^ x < 101 300 Choosing any perfect square such as 225 or 256 in between 200 and 300, we have

x = 100225 = 1015 = 35

Similarly if we choose 256, then we have

x = 100256 = 1016 = 58

1. Show that 571 is a prime number. Sol : Let x = 571 x = 571 Now 571 lies between the perfect squares of (^) ^ (^23) h^2 = 529 and (^) ^ (^24) h^2 = 576. Prime numbers less than 24 are 2, 3, 5, 7, 11, 13, 17, 19, 23. Here 571 is not divisible by any of the above numbers, thus 571 is a prime number.
2. If two positive integers p and q are written as p = a b^2 and q = a b^3 ,where a and b are prime numbers than verify LCM ( , ) p q (^) # HCF( , ) q q = pq Sol : [Sample Paper 2017] We have p = a b^2 3 = a (^) # a (^) # b (^) # b (^) # b and q = a b^3 = a (^) # a (^) # a (^) # b Now LCM ( , ) p q = a (^) # a (^) # a (^) # b (^) # b (^) # b = a b^3 and HCF ( , ) p q = a (^) # a (^) # b = a b^2

Chap 1 Real Numbers Page 15

where q! 0 and p and q are co-prime integers. Rewriting the above expression as, 2 5 = (^) qp^ + 3

5 = p^ + 2 q^3 q

Here p^ 2 + q^3 q is rational because p and q are co-prime integers, thus 5 should be a rational number. But 5 is irrational. This contradicts the given fact that 5 is irrational. Hence 2 5 - 3 is an irrational number. PRACTICE  Given that 3 is irrational, prove that 2 3 - 5 is an irrational number. [Board 2008, 2015] Ans : Proof

1. Given that 2 is irrational, prove that ( 5 + 3 2 )is an irrational number. Sol : [Board 2018] Assume that ( 5 + 3 2 ) is a rational number. Therefore, we can write it in the form of (^) qp^ where p and q are co-prime integers and q! 0.

Now 5 + 3 2 = qp

where q! 0 and p and q are integers. Rewriting the above expression as,

3 2 = (^) qp^ - 5

2 = p^ - 3 q^5 q

Here p^ 3 - q^5 q is rational because p and q are co-prime integers, thus 2 should be a rational number. But 2 is irrational. This contradicts the given fact that 2 is irrational. Hence (^) ^ 5 + (^3 2) h is an irrational number.

PRACTICE  Given that 3 is irrational, prove that ( 5 + 2 3 ) is an irrational number. [Board 2020 Delhi Basic] Ans : Proof

 Prove that 2 + 5 3 is an irrational number, given that 3 is an irrational number. [Board 2019 OD] Ans : Proof

 Prove that 2 + 5 3 is an irrational number, given that 3 is an irrational number. [Board 2019 Delhi] Ans : Proof

1. Write the smallest number which is divisible by both 306 and 657. Sol : [Board 2019 OD] The smallest number that is divisible by two numbers is obtained by finding the LCM of these numbers Here, the given numbers are 306 and 657. 306 = (^6) # 51 = (^3) # 2 # 3 # 17 657 = (^9) # 73 = (^3) # 3 # 73 LCM(306, 657) = (^2) # 3 # 3 # 17 # 73 = 22338 Hence, the smallest number which is divisible by 306 and 657 is 22338.
2. 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 if it equal contain cartons of the same drink, what would be the greatest number of cartons each stack would have? Sol : [Board Term-1 2011] The required answer will be HCF of 144 and 90. 144 = (^2 4) # 32 90 = (^2) # 3 2 # 5 HCF(144, 90) = (^2) # 3 2 = 18 Thus each stack would have 18 cartons.
3. Three bells toll at intervals of 9, 12, 15 minutes respectively. If they start tolling together, after what time will they next toll together? Sol : [Board Term-1 2011] The required answer is the LCM of 9, 12, and 15 minutes. Finding prime factor of given number we have, 9 = (^3) # 3 = 3 2 12 = (^2) # 2 # 3 = (^2 2) # 3 15 = (^3) # 5 LCM(9, 12, 15) = (^2 2) # 3 2 # 5 = 150 minutes The bells will toll next together after 180 minutes.
4. Find HCF and LCM of 16 and 36 by prime factorization and check your answer. Sol : [Board 2009] Finding prime factor of given number we have, 16 = (^2) # 2 # 2 # 2 = 2 4

Page 16 Real Numbers Chap 1

36 = (^2) # 2 # 3 # 3 = (^2 2) # 32 HCF(16, 36) = (^2) # 2 = 4 LCM (16, 36) = (^2 4) # 32 = (^16) # 9 = 144 Check : HCF( , ) a b ×LCM( , ) a b = a (^) # b or, 4 × 144 = 16 × 36 576 = 576 Thus LHS = RHS

PRACTICE  Find the HCF and LCM of 510 and 92 and verify that HCF × LCM = Product of two given numbers. [Board Term-1 2011] Ans : 2, 23460

 Find HCF and LCM of 404 and 96 and verify that HCF (^) # LCM = Product of the two given numbers. [Board 2018] Ans : 4, 9696

1. The length, breadth and height of a room are 8 m 50 cm, 6 m 25 cm and 4 m 75 cm respectively. Find the length of the longest rod that can measure the dimensions of the room exactly. Sol : [Board Term-1 2016] Here we have to determine the HCF of all length which can measure all dimension.

Length, l = 8 m 50 cm = 850 cm

= (^50) # 17 = (^2) # 5 2 # 17 Breadth, b = 6 m 25 cm = 625 cm = (^25) # 25 = (^5 2) # 52 Height, h = 4 m 75 cm = 475 cm = (^25) # 19 = (^5 2) # 19 HCF( , , l b h ) = HCF ( 850 625 475, , ) = HCF ( (^2) # 5 2 # 17 , 5 2 ,^5 2 # 19 ) = 5 2 = 25 cm Thus 25 cm rod can measure the dimensions of the room exactly. This is longest rod that can measure exactly.

FIVE MARKS QUESTIONS

1. If p is prime number, then prove that p is an irrational. Sol : [Board Term-1 2013] Let p be a prime number and if possible, let p be rational Thus p = mn , where m and n are co-primes and n! 0. Squaring on both sides, we get

p n

m 2

2

or, pn^2 = m^2 ...(1) Here p divides pn^2. Thus p divides m^2 and in result p also divides m. Let m = pq for some integer q and putting m = pq in eq. (1), we have pn^2 = p q^2 or, n^2 = pq^2 Here p divides pq^2 .Thus p divides n^2 and in result p also divides n. [ p is prime and p divides n^2 & p divides n ] Thus p is a common factor of m and n but this contradicts the fact that m and n are primes. The contradiction arises by assuming that p is rational. Hence, p is irrational.

1. Prove that 3 is an irrational number. Sol : [Board 2020 OD Basic] Assume that 3 is a rational number. Therefore, we can write it in the form of ab^ where a and b are co- prime integers and q! 0. Assume that 3 be a rational number then we have 3 = (^) ba , where (^) a and (^) b are co-primes and (^) b! 0. Now a = b 3 Squaring both the sides, we have a^2 = 3 b 2 Thus 3 is a factor of a^2 and in result 3 is also a factor of a. Let a = 3 c where c is some integer, then we have a^2 = 9 c 2 Substituting a^2 = 3 b 2 we have 3 b 2 = 9 c 2 b^2 = 3 c 2

Page 18 Real Numbers Chap 1

the prime 5. This is not possible because the only prime in the factorization of 6 n^ = ( (^2) # 3 ) n are 2 and

1. The uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 6 n^. Since there is no prime factor 5, 6 n^ cannot end with the digit five.
2. State Fundamental theorem of Arithmetic. Is it possible that HCF and LCM of two numbers be 24 and 540 respectively. Justify your answer. Sol : [Board Term-1 2015] Fundamental theorem of Arithmetic : Every integer greater than one ither is prime itself or is the product of prime numbers and that this product is unique. Up to the order of the factors. LCM of two numbers should be exactly divisible by their HCF. In other words LCM is always a multiple of HCF. Since, 24 does not divide 540 two numbers cannot have their HCF as 24 and LCM as 540. HCF = 24 LCM = 540

HCF

LCM (^). 24 = 540 = 22 5not an integer

COMPETENCEY BASED QUESTIONS

1. You have a piece of construction paper that measures 32 cm by 48 cm. You want to cut it into squares of equal size. (i) What will be the dimensions of the largest possible square? (ii) How many squares will you have?

Sol : Here we have to find HCF of 32 and 48. 32 = 2 # 2 # 2 # 2 # 2 48 = 2 # 2 # 2 # 2 # 3 HCF (32, 48) =^2 #^2 #^2 # 2 = 16 Area of construction paper =^32 # 48 cm^2 Area of square =^16 # 16 cm 2

1. When the marbles in a bag are divided evenly between two friends, there is one marble left over. When the same marbles are divided evenly among three friends, there is one marble left over. When the marbles are divided evenly among five friends, there is one marble left over. (i) What is the least possible number of marbles in the bag? (ii) What is another possible number of marbles in the bag?

Sol : (i) In all three case one marble is left after division. Thus total marble will be one more than LCM of numbers. LCM(2, 3, 5) =^2 #^3 # 5 = 30 Thus 31 marbles are in bag. (ii) If we add one in multiple of 30, we will get another possible number of marble. These are 61, 91, 121....

1. An online shopping website sells 10 types of items which are packed into various sizes of cartons which are given below. Carton type Inner Dimensions (L # W)cm^2 Small 6 # 8 Medium 12 # 24 Large 24 # 36 Extra large 36 # 48 XXL 48 # 96

Chap 1 Real Numbers Page 19

The company places supporting thermocol sheets inside every package along the edges. The company thought of procuring same sized sheets for all types of cartons. (i) What should be the maximum size of the sheet that fits into all type of cartons? (ii) How many such sheet sizes are possible? (iii) The company later introduced a new size of carton called semi large whose measurements are 14 #

1. Whether the existing maximum size sheet fits this shape? (iv) What should have been the size of the semi large carton (which is larger than medium carton but smaller than large carton) so that the maximum sized sheet remains same?

Sol : (i) To find dimension of maximum size sheet which can be fitted in all carton, it is required that we should find HCF of length of all different sized cartons that is HCF(6, 12, 24, 36, 48) = 6 and HCF of width of all cartons that is HCF(8, 24, 36, 48, 96) = 4. Thus maximum size of the sheet is 6 by 4. (ii) Because HCF of certain numbers is always unique so only one sized sheet is possible. (iii) 14 is not multiple of 6 and 15 is not multiple of 4 so it is not possible to have a carton with dimension 14 # 15. (iv) 18 is the only multiple of 6 between 12 and 24 for length of semi sized carton and there are choices for width of semi sized cartons from 28 and 32, so possible answers are 18 #^ 28 and 18 #^ 32.

1. Two oil tankers contain 825 litres and 675 litres of kerosene oil respectively. (i) Find the maximum capacity of a container which can measure the kerosene oil of both the tankers when used an exact number of times.

(ii) How many times we have to use container for both tanker to fill?

Sol : The maximum capacity of the required container has to measure both the tankers in a way that the count is an exact number of times. So its capacity is exactly divisible by both the tankers. So we have to find the HCF of 825 and 675. First we find prime factorization of 825 and 675. 675 = 3 #^3 #^3 #^5 # 5 825 = 3 #^5 #^5 # 11 HCF(675, 825) = 3 #^5 # 5 = 75 (i) Thus the maximum capacity of the required container is 75 litres. (ii) Therefore, the first tanker will require 82575 = 11 times to fill it and 2nd tanker will require 67575 = 9 times to fill it.

1. Last year my grand mother was admitted to Fortis hospital due to a small accident. She was prescribed a pain medication to be given every 4 hr and an antibiotic to be given every 5 hr. Bandages applied to the my grand mother’s external injuries needed changing every 12 hr. The nurse changed the bandages and gave my grand mother both medications at 6: AM Monday morning. (i) How many hours will pass before the grand mother is given both medications and has her bandages changed at the same time? (ii) What day and time will this be?