Real Number Problems And Solutions

Given below are the Class 10 Maths Word Problems for Real Numbers
a. HCF and LCM Problems
b. Prime Factorisation Problem
c. Division Problems
e. Word Problems

Short Answer Type

Question 1
If 3 is the least prime factor of number a and 7 is the least prime factor of numbers b, then the least prime factor of a + b, is.

Answer

Since least prime factor of number a =3, It must be odd number
Since least prime factor of number b =7, It must be odd number
Odd Number + Odd number= even number.
Least prime factor of any even number is 2. So Answer is 2

---

Question 2
A circular field has a circumference of 360km. Three cyclists start together and can cycle 48, 60 and 72 km a day, round the field. When will they meet again?

Answer

Circumference of field =360 km
Cycle 1 speed is 48 km/day
Cyclist 1 can cover 360 km in = 360/48​=7.5 days=180 hours
Cycle 2 speed is 60 km/day
Cyclist 2 can cover 360 km in=360/60=6 days =114 hours
Cyclist 3 speed is 72 km/day
Cyclist 3 can cover 360 km in=360/72=5 days =120 hours
Now LCM of 180,144 & 120 is 720
which is equal to 30 days
​So,The cyclist will meet after 30 days

---

Question 3
If the sum of LCM and HCF of two numbers is 1260 and their LCM is 900 more than their HCF, then the product of two numbers is.

Answer

Let a be HCM and b be LCM
a+b=1260
b=900+a
Therefore
a+ 900+a=1260
So, a=180
Therefore b =1080
Product of Numbers=a * b= 194400

---

Question 4
The HCF of two numbers is 145 and their LCM is 2175. If one number is 725, find the other.

Answer

HCF * LCM = product of number So answer is 435

---

Question 5
If HCF (26, 169) = 13, then LCM (26, 169) =?

Answer

HCF * LCM = product of number
338

---

Long Asnwer type

Question 6
In a seminar, the number of participants in Hindi, English and Mathematics are 60, 84 and 108, respectively. Find the minimum number of rooms required if in each room the same numbers of participants are to be seated and all of them being in the same subject.

Answer

We need to find the HCF of 60,84 and 108
By prime factorisation
60=2*2*3*5
84=2*2*3*7
108=2*2*3*3*3
HCF =12
Rooms required= 60/12 + 84/12+108/12=5+7+9=21

---

Question 7
If the HCF of 657 and 963 is expressible in the form $657x + 963 \times (-15)$, find x.

Answer

First Lets find out the HCF of 657 and 963
By prime factorisation
$657 = 9 \times 63$
$963 = 9 \times 107$
HCF(657, 963) = 9
Now it is given that HCF is expressed in the form of $657x + 963 \times (-15)$
Then, $657x + 963 \times (-15) = 9$
$657x - 963 \times 15 = 9$
$657x = 14454$
$x = \frac {14454}{657}$
x = 22

---

Question 8
Find the greatest number which divides 285 and 1249 leaving remainders 9 and 7 respectively.

Answer

We have to find the HCF of 285-9 and 1249-7 i.e 276 and 1242
By Euclid division
1242=276X4+138
276=2X138 + 0
So 138 is the HCF

---

Question 9
Find the greatest numbers that will divide 445, 572 and 699 leaving remainders 4, 5 and 6 respectively.

Answer

We have to find the HCF of 445-4,572-5 and 699-6 i.e 441,567 and 693
We can find using the Euclid division or prime factorization method HCF =63

---

Question 10
Find the greatest number which divides 2011 and 2623 leaving remainder 9 and 5 respectively.

Answer

This is same as earliar questions Answer is 154

---

Question 11
Find the smallest number which leaves remainders 8 and 12 when divided by 28 and 32 respectively.

Answer

First we need to find the LCM of 28 and 32. Using prime factorization 28=2 x 2 x7 32=2 x 2 x 2 x 2 x2 LCM = 224 The smallest number which when divided by 28 and 32 leaves remainders 8 and 12 is 224 -(8 +12) = 204 Hence the answer is 204

---

Question 12
Find the smallest number which when increased by 17 is exactly divisible by both 520 and 468.

Answer

520 = 2* 2 * 2 * 5* 13 468 = 2 * 2 * 3 *3 * 13 LCM of 520 and 469 = 4680 Hence the required number is =4680-17=4663

---

Multiple Choice Questions

Question 13
If two positive integers a and b are written as
$a = x^3y^2$ and $b = xy^3$ . x, y are prime numbers, then HCF (a, b) is
(A) $xy$
(B) $xy^2$
(C)$x^3y^3$
(D) $x^2y^2$

Answer

HCF is common factor between the number
So HCF=xy² .Hence Option (B)

---

Question 14
If two positive integers p and q can be expressed as
p = ab² and q = a³b   a, b being prime numbers, then LCM (p, q) is
(A) ab
(B)a²b²
(C)  a³b²
(D) a³b³

Answer

LCM is Product of the greatest power of each prime factor involved in the number
So LCM=a³b² .Hence Option (C)

---

Question 15

Answer

a

---

Question 16
Two tankers contain 583 litres and 242 litres of petrol respectively. A container with maximum capacity is used which can measure the petrol of either tanker in exact number of litres. How many containers of petrol are there in the first tanker.
(A)53
(B)50
(C)54
(D)11

Answer

HCF of 583 and 242=11
So Container Capacity is 11 litres
Number of container in first tanker=583/11=53

---

Summary

This Class 10 Maths word Problems And Solutions for Real Number is prepared keeping in mind the latest syllabus of CBSE . This has been designed in a way to improve the academic performance of the students. If you find mistakes , please do provide the feedback on the mail.

Also Read

---

• Notes
  • Real Numbers Class 10 Notes
  • Real Number Formula Cheat sheet
• NCERT Solutions & Assignments
  • NCERT solutions Real Numbers Exercise 1.1
  • NCERT solutions Real Numbers Exercise 1.2,1.3,1.4
  • Real numbers Class 10 Important questions
  • Real number Class 10 Worksheet
  • Real numbers Class 10 problem and Solutions

---

---

Go back to Class 10 Main Page using below links
Class 10 Maths Class 10 Science

Latest Updates Sound Class 8 Science Quiz

---

Limits Class 11 MCQ

---

Circles in Conic Sections Class 11 MCQ

---

Plant Kingdom free NEET mock tests

---

The Living World free NEET mock tests

---