Question 1:
Which of the following sets of numbers can be expressed in the form 4q + 1 or 4q + 3,
where q is a whole number?
Solution :
Let a be any positive odd integer and let b = 4.
Then, by Euclid’s division lemma, whole numbers q and r exist such that
a = 4q + r 0 ≤ r < 4
Then, r = 0 or 1 or 2 or 3.
Thus, a can be 4q or 4q + 1 or 4q + 2 or 4q + 3.
Since a is odd, a cannot be 4q or 4q + 2, as they are divisible by 2.
Thus, a can be expressed in the form 4q + 1 or 4q + 3.
Thus, every positive odd integer can be expressed in the form 4q + 1 or 4q + 3,
where q is a whole number.
The correct answer is C.
Option1 :
Option2 :
Option3 :
Option4 :
Question 2:
If k is any whole number, then any perfect square can be written in which of the following forms?
Solution :
Let a be any positive integer. On applying division algorithm on a and b = 2,
we obtain a = 2q + r; 0 ≤ r < 2
Hence, the possible remainders are 0 and 1.
This means a is of the form 2q or 2q + 1.
Thus, all positive integers can be written in the form 2q or 2q + 1.
On squaring 2q and 2q + 1, we obtain 4q2 and (4q2 + 4q + 1).
We can write them as 4(q2) and [4(q2 + q) +1] or, 4k and 4k +1, where k is a constant.
Thus, any perfect square can be written in the form 4k or 4k + 1.
The correct answer is A.
Option1 :
Option2 :
Option3 :
Option4 :
Question 3:
Use the following information to answer the next question.
Three statements are given as
I. All positive integers can be expressed in the form 3m or 3m + 1 or 3m +2, where m is a whole number. II. The square of any positive integer can be expressed in the form 3m or 3m + 2, where m is a whole number. III. The cube of any positive integercan be expressed in the form 3m or 3m + 1,where m is a whole number. |
Among the given statements,
Solution :
Statement I
Let a be any positive integer and let b = 3
Then, by Euclid’s division lemma, whole numbers q and r exist such that
a = 3q + r 0 ≤ r < 3
Then, r = 0 or 1 or 2.
Thus, a can be 3q or 3q + 1 or 3q + 2.
Thus, every positive integer can be expressed in the form 3m or 3m + 1 or 3m + 2,
where m is a whole number.
Thus, statement I is correct.
Statement II
As proved in statement I, any positive integer can be expressed in
the form 3m or 3m + 1 or 3m + 2, where m is a whole number.
Consider their squares i.e., (3q)2, (3q + 1)2, and (3q + 2)2.
Case (i)
(3q)2 = 9q2 = 3(3q2) = 3m, where m = (3q2) is a whole number.
Case (ii)
(3q + 1)2 = 9q2 + 6q + 1 = 3(3q2 + 2q) + 1 = 3m + 1,
where m = (3q2 + 2q) is a whole number.
Case (iii)
(3q + 2)2 = 9q2 + 12q + 4 = 9q2 + 12q + 3 + 1 = 3(3q2 + 4q + 1) + 1
= 3m + 1, where m = (3q2 + 4q + 1) is a whole number.
Thus, the square of every positive integer can be expressed in the form 3m or 3m + 1,
where m is a whole number.
Thus, statement II is incorrect.
Statement III
As proved in statement I, any positive integer can be expressed in the
form 3m or 3m + 1 or 3m + 2, where m is a whole number. Consider their cubes i.e., (3q)3,
(3q + 1)3, and (3q + 2)3.
Case (i)
(3q)3 = 27q3 = 3(9q3) = 3m, where m = (9q3) is a whole number.
Case (ii)
(3q + 1)3 = 27q3 + 27q2 + 9q + 1 = 3(9q3 + 9q2 + 3q) + 1
= 3m + 1, where m = (9q3 + 9q2 + 3q) is a whole number.
Case (iii)
(3q + 2)3 = 27q3 + 54q2 + 36q + 8 = 3(9q3 + 18q2 + 12q + 2) + 2
= 3m + 2, where m = (9q3 + 18q2 + 12q + 2) is a whole number.
Thus, the cube of any positive integer can be expressed in the form 3m or 3m +1 or 3m + 2.
Thus, statement III is incorrect.
Hence, among the given statements, only statement I is correct.
The correct answer is B.
Option1 :
statements I and II are correct
Option2 :
only statement I is correct
Option3 :
statements I and III are correct
Option4 :
onlystatement II is correct
Question 4:
1051 bottles of soft drink have to be kept in crates. Each crate can hold a maximum of 28 bottles.
How many empty spaces will be left in the last crate?
Solution :
We first need to find the number of bottles that have to be kept in the last crate.
Using Euclid’s division lemma, we can write 1051 as
1051 = 28 × 37 + 15
Hence, 15 bottles have to be kept in the last crate.
It is given that one crate can hold a maximum of 28 bottles.
Therefore, number of empty spaces in the last crate = (28 − 15) = 13
Hence, the correct answer is option A.
Option1 :
Option2 :
Option3 :
Option4 :
Option5 :
Question 5:
Use the following information to answer the next question.
The H.C.F. of the numbers 8008 and 700 is calculated by Euclid’s division lemma. The first two steps of the method are given as: 8008 = A × 11 + B A = B × 2 + C |
What are the respective values of A, B, and C?
Solution :
The H.C.F of 8008 and 700 can be found by Euclid’s division lemma as:
Since 8008 > 700, applying Euclid’s division lemma to 8008 and 700:
8008 = 700 × 11 + 308 ...(1)
Since remainder 308 ≠ 0, on applying Euclid’s lemma again:
700 = 308 × 2 + 84 ...(2)
Comparing equations (1) and (2) with the given steps, the values of A, B, and C are obtained as:
A = 700, B = 308, and C = 84
Thus, the values of A, B, and C are 700, 308, and 84 respectively.
The correct answer is C.
Option1 :
Option2 :
Option3 :
Option4 :
Question 6:
Let x and y be two consecutive positive integers such that
x<y.
Then the difference between their squares is always equal to
Solution :
Since x and y are two consecutive positive integers,
Therefore, one of them is even and other is odd.
By Euclid's algorithm, we know that every positive even integer is of form 2k,
and that every positive odd integer is of form 2k + 1, where k is some integer.
∴
x=2k, y=2k+1 ∀k∈I
Then
y2−x2=(2k+1)2−(2k)2=4k+1 ∀k∈I.
Now
4k+1=(2k)+(2k+1)=x+y
Hence, the correct answer is option A.
Option1 :
the sum of the two consecutive positive integers
Option2 :
the difference of the two consecutive positive integers
Option3 :
the product of the two consecutive positive integers
Option4 :
the quotient of the two consecutive positive integers
Option5 :
Question 7:
The square of every positive odd integer can be expressed in terms of m (where m is a whole number) as
Solution :
Let a be any positive odd integer and let b = 6.
Then, by Euclid’s algorithm, whole numbers q and r exist such that
a = 6q + r 0 ≤ r < 6
Thus, the possible values of remainder r are 0, 1, 2, 3, 4, and 5.
Thus, a can be 6q, 6q + 1, 6q + 2, 6q + 3, 6q + 4, or 6q + 5.
Since the squares of positive odd integers are being considered, a will always be odd.
Thus, a cannot be 6q, 6q + 2, or 6q + 4.
∴ a = 6q + 1 or 6q + 3 or 6q + 5
Consider the squares of 6q + 1, 6q + 3, and 6q + 5.
Case (i)
(6q + 1)2 = 36q2 + 12q + 1
= 6(6q2 + 2q) + 1
= 6m + 1, where m is a whole number and m = (6q2 + 2q)
Case (ii)
(6q + 3)2 = 36q2 + 36q + 9
= 6(6q2 + 6q + 1) + 3
= 6m + 3, where m is a whole number and m = (6q2 + 6q + 1)
Case (iii)
(6q + 5)2 = 36q2 + 60q + 25
= 6(6q2 + 10q + 4) + 1
= 6m + 1, where m is a whole number and m = (6q2 + 10q + 4)
Thus, the square of every positive odd integer can be written in the form
6m + 1 or 6m + 3, where m is a a whole number.
The correct answer is C.
Option1 :
Option2 :
Option3 :
Option4 :
Question 8:
Use the following information to answer the next question.
Vijay is finding the H.C.F. of (33976, 18495). He starts the process correctly but commits a mistake in the third step. However, he manages to get the correct answer. Ist step: 33976 = 18495 × 1 + 15481 IInd step : 18495 = 15481 × 1 + 3014 IIIrd step: 15481 = 3014 × 7 + 1233 IVth step: 3014 = 1233 × 2 + 137 Vth step: 1233 = 137 × 9 + 0 |
After the IInd step, the correct steps are
Solution :
It is given that the first two steps are correct.
Ist step: 33976 = 18495 × 1 + 15481
IInd step: 18495 = 15481 × 1 + 3014
On applying Euclid’s division algorithm, we get
15481= 3014 × 5 + 411 |
3014 = 411 × 7 + 137 |
411 = 137 × 3 + 0 |
Hence, the correct steps are shown in table B.
The correct answer is B.
Option1 :
15481= 3014 × 5 + 548 |
3014 = 548 × 5 + 137 |
548 = 137 × 4 + 0 |
Option2 :
15481= 3014 × 5 + 411 |
3014 = 411 × 7 + 137 |
411 = 137 × 3 + 0 |
Option3 :
15481= 3014 × 6 + 274 |
3014 = 274 × 11 + 137 |
274 = 137 × 2 + 0 |
Option4 :
15481= 3014 × 4 + 685 |
3014 = 685 × 4 + 137 |
685 = 137 × 5 + 0 |
Option5 :
Question 9:
Use the following information to answer the next question.
The given steps show the procedure for finding the H.C.F. of the numbers 133763 and 10185 by using Euclid’s division algorithm. 133763 = 10185 × a + b 10185 = b × 7 + c b =c× 2 + 0 |
The values of a, b, and c are respectively
Solution :
The H.C.F. of two positive integers 133763 and 10185 can be found by using Euclid’s division
algorithm. This can be done through the following steps.
Step 1
As 133763 > 10185, by applying Euclid’s division lemma to the numbers 133763 and 10185,
we obtain 133763 = 10185 × 13 + 1358
Step 2
As the remainder 1358 ≠ 0, by applying Euclid’s division lemma to the numbers 10185 and 1358,
we obtain 10185 = 1358 × 7 + 679
Step 3
Again as the remainder 679 ≠ 0, by applying Euclid’s division lemma to the numbers 1358 and
679, we obtain 1358 = 679 × 2 + 0
Thus, H.C.F. (133763, 10185) = 679
Hence, the values a, b, and c are 13, 1358, and 679 respectively.
The correct answer is C.
Option1 :
Option2 :
Option3 :
Option4 :
Question 10:
If LCM(70, 125) = 17
(k+1)−1. The value of k is
Solution :
LCM(70, 125) = 1750 =
17×103−1
Comparing it with 17
(k+1)−1, we get
k+1=103k=102
Hence, the correct answer is option C.
Option1 :
100
Option2 :
101
Option3 :
102
Option4 :
1
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