Q 1 – A Mathematics Exhibition is being conducted in your School and one of your friends is making a model of a factor tree. He has some difficulty and asks for your help in completing a quiz for the audience.
Observe the following factor tree and answer the following:
(a) What will be the value of x?
a) 15005
b) 13915
c) 56920
d) 1742)
(b) 13915
(b) What will be the value of z?
a) 22
b) 23
c) 17
d) 19
(b) 23
(c) The product of HCF and LCM of 60, 84 and 108 is
a) 55360
b) 35360
c) 45500
d) 45360
d) 45360
Q 2 – The HCF and the LCM of 12, 21, 15 respectively are
(a) 3, 140
(b) 12, 420
(c) 3, 420
(d) 4,203
(c) 3, 420
Q 3 – The LCM of smallest two digit composite number and smallest composite number is
(a) 12
(b) 4
(c) 20
(d) 4
(c) 20
Q 4 – 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
(d) 81
Q 5 – HCF of 144 and 198 is
(a) 9
(b) 18
(c) 6
(d) 1
(b) 18
Q 6 – The decimal expansion of the rational number will 14587/1250 terminate after
(a) one decimal place
(b) two decimal places
(c) three decimal places
(d) four decimal place
(d) four decimal place
Q 7 – The values of x and y in the given figure are
(a) 7, 13
(b) 13, 7
(c) 9, 12
(d) 12,
(a) 7, 13
Q 8 – 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
(d) 3600
Q 9 – The decimal expansion of the rational number 
will terminate after how many
Q 10 – A trader has 612 A type soaps and 342 B type soaps. He packs them in boxes and each box contains exactly one type of soap. If every box contains the same number of soaps, then find the number of soaps in each box such that number of boxes is the least
The required number is HCF of 612 and 342.
This gives the maximum number of soaps in each box and the number of boxes with them be the least.
By using Euclid’s division algorithm, we have
612 = 342 × 1 + 270
342 = 270 × 1 + 72
270 = 72 × 3 + 54
72 = 54 ×1 + 18
54 = 18 × 3 + 0
Here we notice that the remainder is zero, and the divisor at this stage is 18.
Therefore. HCF of 612 and 342 is 18.
So, the trader can pack 18 soaps per box.
Q 11 – The HCF of 65 and 117 is expressible in the form 65m – 117. Find the value of m.
(a) 4
(b) 2
(c) 1
(d) 34
(b) 2
Q 12 – Consider the number 4n where n is a natural number. Check whether there is any value of n for which 4n ends with the digit zero
If the number 4n, for any n, were to end with the digit zero, then it would be divisible by 5.
That is, the prime factorization of 4n would contain the prime 5.
But, 4n=(2×2)n
⇒ The only prime in the factorization of 4n is 2.
So, the uniqueness of the Fundamental Theorem of Arithmetic guarantees that there are no other primes in the factorization of 4n.
So, there is no natural number n for which 4n ends with the digit zero
Q 13 – For positive integers a and 3, there exist unique integers q and r such that a = 3q + r, where r must satisfy: Marks)
(a) 0 < r < 3
(b) 1 < r < 3
(c) 0 < r < 3
(d) 0 < r < 3
(a) 0 < r < 3
Q 14 – The LCM of two numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then find the other number
GIVEN: HCF + LCM= 600, ONE NUMBER= 280
Let the HCF= x
LCM= 14 x
HCF + LCM= 600
x +14 x = 600
15x = 600
x= 600/15
x= 40
HCF = 40
LCM = 14x = 14 × 40= 560
HCF×LCM= 1ST NUMBER × 2ND NUMBER
40× 560= 280× 2ND NUMBER
2nd Number = (40 × 560) /280
= 40×2
= 80
Q 15 – If the HCF of 408 and 1032 is expressible in the form 1032 × 2 + 408 × p, then find the value of p.
To determine the value of p we have to determine the HCF between 408 and 1032.
Then we have to equalise the HCF with the expression.
The HCF of 408 and 1032 is-
408 = 4 × 6 × 17
1032 = 4 × 6 × 43
4 × 6 = 24
Equalising 24 with the expression we get-
24 = 1032 × 2 480p
1 = 43 × 2 + 17 p
17p = – 86
p = –5.05
Thus the value of p is -5.05.
Q 16 – Show that any positive odd integer is of the form 4q + 1 or 4q + 3, where q is some integer.
Here we have to prove that for any positive integer q, the positive odd integer will be form of 4q+1 or 4q+3.
Now let us suppose that the positive odd integer is a then by Euclid’s division rule
a = 4q + r ……(1 )
Where q (quotient) and r (remainder) are positive integers, and 0 r 4
We are putting the values of r from 0 to 3 in equation (1), we get
But we can easily see that 4q and 4q+2 are both even numbers.
Therefore for any positive value q, the positive odd integer will be the form of 4q+1 and 4q+3.
Q 17 – Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.
We need to find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11 and 15 respectively. Hence, we need to find the HCF of 3 numbers (398-7), (436-11) and (542-15).
Hence,
398 – 7 = 391
436 – 11 = 425
542 – 15 = 527
Now, you can find the HCF(391, 425, 527) = 17
Hence, the largest number that will divide 398,436 and 542 leaving remainders 7, 11, and 15 respectively is 17.
Q 18 – Which of the following rational numbers have a terminating decimal expansion?
(a) 125/441
(b) 77/210
(c) 15/1600
(d) 
(c) 15/1600
Q 19 – HCF and LCM of two numbers is 9 and 459 respectively. If one of the numbers is 27, find the other number.
Q 20 – 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.
Length of the room = 8m 25cm = 825 cm
Breadth of the room = 6m 75cm = 675 cm
Height of the room = 4m 50cm = 450 cm
Prime factorization of 825, 675 and 450 are as follows:
825 = 3×5×5×11
675 = 3×3×3×5×5
450 = 2×3×3×5×5
Common factors = 3×5×5
H.C.F. = 75
Hence, H.C.F. of 825, 675, and 450 is 75
Q 21 – If two positive integers x and y are expressible in terms of primes as x = p2 q3 and y = p3q, what can you say about their LCM and HCF. Is LCM a multiple of HCF?
Q 22 – Find the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the
He given numbers are 306 and 657
The LCM and HCF of these numbers can be found out using the method of prime factorization
The prime factorization of 306 and 657 are:
306 = 2 × 3² × 17
657 = 3² × 73
Now, HCF of these numbers are given by multiplying the common factors of these numbers
The common factors of 306 and 657 is 3²
Therefore, HCF = 3² = 9
Now, the LCM is given by multiplying the HCF of 306 and 657 with the non-common prime factors of 306 and 657
Therefore, LCM = 9 × 2 × 17 × 73 = 22,338
Now, LCM × HCF = 9 × 22,338 = 201042
a × b = 306 × 657 = 201042
Hence verified that LCM × HCF = product of the two numbers
Q 23 – Find whether decimal expansion of 13/64 is a terminating or non-terminating decimal. If it terminates, find the number of decimal places its decimal expansion has.
For any integer to have a terminating decimal operation, it
must be in the form of a/b = a/22m 52m
⇒ a/102m
13/64
13/22 6 × 520
So, 13/64 is a terminating decimal expression and it terminates after six places of decimal.
Q 24 – Floor of a room is to be fitted with square marble tiles of the largest possible size. The size of the room is 10 m 7 m. What should be the size of tiles required that has to be cut and how many such tiles are required?
Least number of square tiles are required if the side of the square tile is the H.C.F. of the length and breadth.
length of the room = 10 m
Breadth of the room = 7 m
H.C.F. of 10 and 7
10 = (7*1) + 3
7 = (3*2) + 1
3 = (1*3) + 0
So, H.C.F. of 10 and 7 is 1
Therefore, the length of the square tile is 1 m
Area of 1 square tile = 1*1 = 1 sq m
Area of the room = 10*7 = 70 sq m
Number of square tile required each measuring 1m × 1m = 70/1
= 70 tiles
Least number of square tiles are required if the side of the square tile is the H.C.F. of the length and breadth.
length of the room = 10 m
Breadth of the room = 7 m
H.C.F. of 10 and 7
10 = (7*1) + 3
7 = (3*2) + 1
3 = (1*3) + 0
So, H.C.F. of 10 and 7 is 1
Therefore, the length of the square tile is 1 m
Area of 1 square tile = 1*1 = 1 sq m
Area of the room = 10*7 = 70 sq m
Number of square tile required each measuring 1m × 1m = 70/1
= 70 tiles
Hence, the size of each square tile is 1m × 1m and 70 square tiles will be required.
Q 25 – The LCM of 2 numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then find the other number.
GIVEN: HCF + LCM= 600, ONE NUMBER= 280
Let the HCF= x
LCM = 14x
HCF + LCM= 600
X+ 14X=600
15X = 600
X= 600/15
X= 40
HCF= 40
LCM = 14x = 14 × 40= 560
HCF×LCM= 1ST NUMBER × 2ND NUMBER
40× 560= 280× 2ND NUMBER
2nd Number = (40×560) /280
= 40×2
= 80
Q 26 – Use Euclid’s division algorithm to find the HCF of 210 and 55.
Apply Euclid’s division lemma to given numbers c and d to find whole numbers q and r such that
c= dq + r, 0 ≤ r < d
Here, c=210, d = 55
210 = 55 × 3 + 45
– The remainder is not equal to 0. Therefore, we apply the same process again on 55 and 45.
55 = 45 × 1 + 10
Again the remainder is not equal to 0. Therefore, we apply the same process again on 45 and 10.
45=10×4+5
Now the third step Again ….The remainder is not equal to 0 . Therefore, we apply the same process again on 10 and 5.
10 = 5 ×2 + 0
The remainder is equal to 0.
Therefore, HCF of 210 and 55 is 5
Q 27 – Assertion The HCF of two numbers is 5 and their product is 150, then their LCM is 30
Reason For any two positive integers a and b, HCF (a, b) + LCM (a,b) = a b.
a) Both assertion (A) and reason (R) are true and reason
(R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason
(R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
(c) Assertion (A) is true but reason (R) is false.
Q 28 – Assertion: 34. 12345 is a terminating decimal fraction.
Reason : Denominator of 34.12345, when expressed in the form p/q , q≠ 0 , is of the form 2m × 5n ,Where m and n is of the form
(a) Both assertion (A) and reason (R) are true and reason
(R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason
(R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true
(c) Assertion (A) is true but reason (R) is false.
Q 29 – Assertion When a positive integer a is divided by 3, the values of remainder can be 0, 1 or 2.
Reason: According to Euclid’s Division Lemma a= bq + r ,where 0 ≤ r < b and r is an integer.
(a) Both assertion (A) and reason (R) are true and reason
(R) is the correct explanation of assertion (A).
(b) Both assertion (A) and reason (R) are true but reason
(R) is not the correct explanation of assertion (A).
(c) Assertion (A) is true but reason (R) is false.
(d) Assertion (A) is false but reason (R) is true.
Q 30 – Express (15/4 + 5/40) as a decimal fraction without actual division. a/b
Q 31 – The prime factorisation of 13915 is
a) 5 × 113 × 132
b) 5 × 113 × 232
c) 5 × 112 × 23
d) 5 × 112 × 132
c) 5 × 112 × 23
Q 32 – For any two positive integers a and b, there exist (unique) whole numbers q and r such that
(a) q = ar + b , 0 ⩽ r < b.
(b) a = bq + r , 0 ⩽ r < b.
(c) b = aq + r , 0 ⩽ r < b.
(d) none of these
(b) a = bq + r , 0 ⩽ r < b.
Q 33 – The least number that is divisible by all the numbers from 1 to 8 (both inclusive) is
(a) 840
(b) 2520
(c) 8
(d) 420
(a) 840
Q 34 – For positive integers a and 3, there exist unique integers q and r such that a = 3q + r, where r must satisfy:
(a) 0 < r < 3
(b) 1 < r < 3
(c) 0 < r < 3
(d) 0 < r < 3
(a) 0 < r < 3
Q 34 – If the HCF of 408 and 1032 is expressible in the form 1032 * 2 + 408 p, then find the value of p.
HCF of 408 and 1032 is 24.
1032 × 2 + 408 × (p) = 24
408p = 24 – 2064
p = – 5
Q 35 – The decimal expansion of π is:
(a) Terminating
(b) Non-terminating and Non-recurring
(c) Non-terminating and recurring
(d) Doesn’t exist
(b) Non-terminating and recurring
Q 36 – Find the largest number that will divide 398, 436 and 542 leaving remainders 7, 11, and 15 respectively.
Algorithm
398 – 7 = 391, 436 – 11 = 425, 542 – 15 = 527
HCF of 391, 425, 527 = 17
Q 37 – HCF and LCM of two numbers are 9 and 459 respectively. If one of the numbers is 27, find the other number.
We know,
1st number × 2nd number = HCF × LCM
⇒ 27 × 2nd number = 9 × 459
⇒ 2nd number = 153
Q 38 – Find LCM of numbers whose prime factorisation is expressible as 3 × 52 and 32 × 72.
LCM is the multiplication of the highest powers of prime factors of the numbers.
(3 x 52 , 32 x 72) = 32 x 52 x 72 = 9 x 25 x 49 = 11025
Q 39 – Check whether 4n can end with the digit 0 for any natural number n.
4n = (22)n = 22n
The only prime in the factorization of 4n is 2.
There is no other prime in the factorization of 4n = 2n
(By uniqueness of the Fundamental Theorem of Arithmetic).
5 does not occur in the prime factorization of 4n for any n.
Therefore, 4n does not end with the digit zero for any natural number n.
Q 40 – Prove that √5 is irrational and hence show that 3 + √5 is also irrational.
Step-by-step explanation:
To prove :√5 is irrational number.
Proof :
Let us assume that √5 is rational
Then √5 = a/b
(a and b are co primes, with only 1 common factor and b≠0)
⇒ √5 = a/b
(cross multiply)
⇒ a = √5b , Taking squares on both sides.
a² = 5b² …………………. eq.(1)
5 is a factor of a² from eq.(1)
(by theorem if p divides q then p can also divide q²)
⇒ 5 is a factor of a (Theorem) eq.(2)
a = 5c (where c is an integer )
(squaring on both sides)
⇒ a² = 25c² …………………. eq.(3)
From equations eq.(1) and eq.(3)
⇒ 5b² = 25c²
b² = 5c²
5 is a factor of b² from
(by theorem if p divides q then p can also divide q²)
⇒ 5 is a factor of b (Theorem) eq.(4)
We know that a and b are co-primes having only 1 common factor but from eq.(2) and eq.(4) we can say that it is wrong and 5 is also common factor of a & b .
This contradiction arises because we assumed that √5 is a rational number
∴ Our assumption is wrong
∴ √5 is irrational number
Part-b) Let us assume that 3 + √5 is a rational number.
Now,
3 + √5 = p/q
[Here p and q are co-prime numbers]
√5 = [p/q] – 3
√5 = [p/q] – 3
Here, [p/q] – 3 is a rational number.
But we know that √5 is a irrational number but [p/q] – 3 is a rational number. We know that LHS = RHS.
So, [p/q] – 3 is also a irrational number.
So, our assumption is wrong.
3 + √5 is a irrational number.
Hence, proved.
Q 41 – By using Euclid’s algorithm, find the largest number which divides 650 and 1170.
Given numbers are 650 and 1170.
1170 > 650
1170 = 650 × 1 + 520
650 = 520 × 1 + 130
520 = 130 × 4 + 0
HCF = 130
The required largest number is 130.
Q 42 – 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.
Length of room = 8m 25cm = 825 cm
Breadth of room = 6m 75m = 675 cm
Height of room = 4m 50m = 450 cm
∴ The required longest rod
= HCF of 825, 675 and 450
First consider 675 and 450
By applying Euclid’s division lemma
675 = 450 × 1 + 225
450 = 225 × 2 + 0
∴ HCF of 675 and 450 = 825
By applying Euclid’s division lemma
825 = 225 × 3 + 150
225 = 150 × 1 + 75
150 = 75 × 2 + 0
HCF of 825, 675 and 450 = 75
Q 43 – If two positive integers x and y are expressible in terms of primes as x = p2 q3 and y = p3q, what can you say about their LCM and HCF. Is LCM a multiple of HCF? Explain.
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
= 6 × 6
LCM = 6 (HCF) …[From (ii)]
Here LCM is 6 times HCF.
Q 44 – Find the HCF and LCM of 306 and 657 and verify that LCM × HCF = Product of the two numbers.
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
Q 45 – Floor of a room is to be fitted with square marble tiles of the largest possible size. The size of the room is 10 m by 7 m. What should be the size of tiles required that has to be cut and how many such tiles are required?
Least number of square tiles are required if the side of the square tile is the H.C.F. of the length and breadth.
length of the room = 10 m
Breadth of the room = 7 m
H.C.F. of 10 and 7
10 = (7 x 1) + 3
7 = (3 x 2) + 1
3 = (1 x 3) + 0
So, H.C.F. of 10 and 7 is 1
Therefore, the length of the square tile is 1 m
Area of 1 square tile = 1 x 1 = 1 sq m
Area of the room = 10 x 7 = 70 sq m
Number of square tile required each measuring 1m × 1m = 70/1
= 70 tiles
Hence, the size of each square tile is 1m x 1m and 70 square tiles will be required.
Q 47 – The LCM of 2 numbers is 14 times their HCF. The sum of LCM and HCF is 600. If one number is 280, then find the other number.
Given, One no. is 280
Let, HCF be x
so LCM = 14x
Sum of HCF and LCM = 600
x + 14x = 600
15x = 600
So, HCF = 40
LCM = 14 x 40 = 560
To Find Other no.
first no. = 280
second no. = y
We Know that
HCF x LCM = product of two numbers
40 × 560 = 280 × y
22400 = 280y
y = 22400 / 280