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Welcome to ICOME Quiz Corner

Q1. Find the greatest number that will divide 360 and 650 and will leave as remainder 6 and 1 respectively .

Solution.
The greatest number will be the HCF of (360 - 6) and (650 - 1).
So, HCF of 354 and 649 is 59 .
Solution

Q2. Three bells sound together and they tell after 0.1, 0.25 and 0.125 seconds. After what intervals will they again toll together .

Solution.
In this case find the LCM of 0.1, 0.25, 0.125 = 0.5 sec.
Solution

Q3. What is the smallest sum of money which contains Rs.15, Rs.12, Rs.18 and Rs.21.

Solution.
In this case find the LCM of 15,12,18,21 = 1260
Solution

Q4. Find the LCM of
4/5
,
6/11
,
7/15

Solution.
LCM of fraction =
LCM of Numerators/ HCF of Denominators


LCM of 4,6,7 = 84

HCF of 5,11,15 = 1

LCM of fraction = 84
Solution

Q5. Three sportsman run on a circular track, time taken by them to complete one revolution is 4 hrs, 3 hrs and 9 hrs respectively . When they will meet if start at the same point.

Solution.
LCM of 4,3,9 = 36

After 36 hrs they will meet
Solution

Q6. Three sportsman A,B and C run on a circular track , having speed of 4 hrs, 3 hrs and 9 hrs respectively . Circumference of circular track is 27 km. When they will at the starting point.

Solution.
First find time to complete one revolution .

Time =
distance/ speed


LCM of
27/4
,
27/3
and
27/9
=
27/4
,
27/3
,
3/1


LCM of
27/4
,
27/3
,
3/1
=
LCM of 27,27,3/HCF of 4,3,1


=
27/1
= 27

All will meet after 27 hrs.
Solution

Q7. Find the LCM and HCF of 96 and 404.

Solution.
96 = 25 x 3 and 404 = 22 x 101

Least exponents common prime factorisation

HCF = 22 = 4

As we know that .

Product of HCF and LCM = Product of two numbers

4 x (LCM) = 96 x 404

LCM =
404 x 96/4
= 101 x 96 = 9696

HCF = 4 , LCM = 9696
Solution

Q8. Find the largest positive integer that will divide 398, 436 and 503 leaving remainders 7, 11 and 10 respectively .

Solution.
In this case find HCF of (398 - 7), (436 - 11) and (503 - 10)

Using prime factorisation HCF of 391, 425,493 = 17
Solution

Q9. In a conference , the number of participants in statistics, maths and english are 78, 90 and 98 respectively. Find the minimum number of rooms required if in each room the same number of participants are to be seated and all of them belong to same subject

Solution.
In this case find HCF of 78, 90 and 98

78 = 2 x 3 x 13

90 = 22 x 3 2 x 5

96 = 25 x 3

rooms required =
total participants/6
=
264/6
= 44

rooms required = 44
Solution

Q10. Find the highest positive integer that will divide 278, 182 and 145 and leaving remainder 8,2 and 10 respectively.

Solution.
In this case we have to find HCF of 278 - 8 = 270, 180 and 135 .

270 = 2 x 33 x 5

180 = 22 x 32 x 5

135 = 33 x 5

HCF is 32 x 5 = 45
Solution