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. Solution
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? Solution
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. Solution
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. Solution
HCF * LCM = product of number
So answer is 435
Question 5
If HCF (26, 169) = 13, then LCM (26, 169) =? Solution
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. Solution
3
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. Solution
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. Solution
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. Solution
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. Solution
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. Solution
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. Solution
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) $x^2y^2$ Solution
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 Solution
LCM is Product of the greatest power of each prime factor involved in the number
So LCM=a3b2 .Hence Option (C)
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 Solution
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.
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