Q.1. How many threedigit numbers are greater than 100 and increase by 198 when the three digits are arranged in the reverse order? [2021]
Ans: 70
Let the numbers be of the form 100a + 10b + c, where a, b, and c represent single digits.
Then (100c + 10b + a)  (100a+ 10b + c) = 198
99c  99a = 198
c  a = 2.
Now, a can take the values 17. a cannot be zero as the initial number has 3 digits and cannot be 8 or 9 as then c would not be a singledigit number.
Thus, there can be 7 cases.
B can take the value of any digit from 09, as it does not affect the answer. Hence, the total cases will be 7× 10 = 70.
Q.2. For a 4digit number, the sum of its digits in the thousands, hundreds and tens places is 14, the sum of its digits in the hundreds, tens and units places is 15, and the tens place digit is 4 more than the units place digit. Then the highest possible 4digit number satisfying the above conditions is [2021]
Ans: 4195
Given the 4 digit number :
Considering the number in thousands digit is a number in the hundredth digit is b, number in tens digit is c, number in the units digit is d.
Let the number be abcd.
Given that a + b + c = 14. (1)
b + c + d = 15. (2)
c = d + 4. (3).
In order to find the maximum number which satisfies the condition, we need to have abcd such that a is maximum which is the digit in thousands place in order to maximize the value of the number. b, c, and d are less than 9 each as they are singledigit numbers.
Substituting (3) in (2) we have b + d + 4 + d = 15, b + 2 x d = 11. (4)
Subtracting (2) and (1) : (2)  (1) = d = a+1. (5)
Since c cannot be greater than 9 considering c to be the maximum value 9 the value of d is 5.
If d = 5, using d = a + 1, a = 4.
Hence the maximum value of a = 4 when c = 9, d = 5.
Substituting b + 2 x d = 11. b = 1.
The highest fourdigit number satisfying the condition is 4195
Q.3. Suppose one of the roots of the equation ax^{2}  bx + c = 0 is 2+ √3 , Where a,b and c are rational numbers and a ≠ 0. If b = c^{3 } then  a  equals. [2021]
(a) 1
(b) 2
(c) 3
(d) 4
Correct Option is B
Given a, b, c are rational numbers.
Hence a, b, c are three numbers that can be written in the form of p/q.
Hence if one both the root is 2 + √3 and considering the other root to be x.
The sum of the roots and the product of the two roots must be rational numbers.
For this to happen the other root must be the conjugate of 2 + √3 so the product and the sum of the roots are rational numbers which are represented by: b/a, c/a
Hence the sum of the roots is 2 + √3 + 2  √3 = 4
The product of the roots is 2 + √3 x 2  √3 =1
b/a = 4, c/a = 1.
b = 4 x a, c= a.
Since b = c^{3}
4 x a = a^{3}
a^{2} = 4.
a = 2 or 2.
a = 2
Q.4. For a sequence of real numbers x_{1} ,x_{2} ,...x_{n} , If x_{1} −x_{2} +x_{3} −....+(−1)^{n+1} x_{n} = n^{2} +2n for all natural numbers n, then the sum x_{49} + x_{50} equals [2021]
(a) 200
(b) 2
(c) 200
(d) 2
Correct Option is D
Now as per the given series :
we get x_{1} =1 + 2 =3
Now x_{1} −x_{2} = 8
So, x_{2} = 5
Now x_{1} − x_{2} + x_{3} = 15
So, x_{3} = 7
So, we get
So x_{49 }= 99 and x_{50} = −101
Therefore x_{49} + x_{50} = −2
Q.5. For a real number x the condition ∣3x−20∣+∣3x−40∣ = 20 necessarily holds if [2021]
(a) 10 < x < 15
(b) 9 < x < 14
(c) 7 < x < 12
(d) 6 < x < 11
Correct Option is C
Case 1 : x ≥ 40 / 3
we get 3x  20 + 3x  40 = 20
6x = 80
Case 2:
we get 3x  20 + 40  3x = 20
we get 20 = 20
So, we get
Case 3:
we get 20  3x + 40  3x = 20
40 = 6x
x = 20/3
but this is not possible
so we get from case 1,2 and 3
Now looking at options
we can say only option C satisfies for all x .
Hence 7 < x < 12.
Q.6. If n is a positive integer such that , then the smallest value of n is [2021]
Ans: 6
For minimum value of n,
1 + 2 + ... + n = 21
We can see that if n = 6, 1 + 2 + 3 + ... + 6 = 21.
Q.7. If 3x + 2∣y∣ + y=7 and x + ∣x∣ + 3y = 1 then x + 2yx + 2y is: [2021]
(a) 4/3
(b) 8/3
(c) 0
(d) 1
Correct Option is C
We need to check for all regions:
x >= 0, y >= 0
x >= 0, y < 0
x < 0, y >= 0
x < 0, y < 0
However, once we find out the answer for any one of the regions, we do not need to calculate for other regions since the options suggest that there will be a single answer.
Let us start with x >= 0, y >= 0,
3x + 3y = 7
2x + 3y = 1
Hence, x = 6 and y = 11/3
Since y > = 0, this is not satisfying the set of rules.
Next, let us test x >= 0, y < 0,
3x  y = 7
2x + 3y = 1
Hence, y = 1
x = 2.
This satisfies both the conditions. Hence, this is the correct point.
We need the value of x + 2y
x + 2y = 2 + 2(1) = 2  2 = 0.
Q.8. If a, b and c are positive integers such that ab = 432, bc = 96 and c < 9, then the smallest possible value of a + b + c is [2020]
(a) 56
(b) 59
(c) 49
(d) 46
Correct Option is D
Given ab = 432, bc = 96 and c < 9
To find the minimum value for a + b + c, the two larger numbers should be as close as possible.
The closest combination whose product is 432 is 18 x 24 . For b = 24, we get c = 4 and a = 18 .
Hence the least value for a + b + c = 46 .
Q.9. How many 3digit numbers are there, for which the product of their digits is more than 2 but less than 7? [2020]
Ans: 21
The product of the digits of the threedigit numbers should be more than 2 and less than 7 .
Hence the possible numbers are as follows.
Q.10. The mean of all 4 digit even natural numbers of the form 'aabb', where a >0, is [2020]
(a) 5050
(b) 4466
(c) 5544
(d) 4864
Correct Option is C
The sum of possible even digit numbers in the form aabb is 1100 + 1122 + 1144 + 1166 + 1188 + 2200 + 2222 + 2288 + ......9900 + 9922 + 9988 i.e. (45 numbers)
⇒ 1100 + 1100 +1100+ 1100 + 1100 + 22 + 44 + 66 + 88 + 2200 + 2200 + 2200 + 2200 + 2200 +22 + 44 + 66 + 88 +… + 9900 + 9900 + 9900 + 990 + 9900 + 22 + 44 + 66 + 88
⇒ 5(1100 + 2200 + .....9900) + 9(22 + 44 + 66 + 88)5 x 1100(1 + 2 + .... 9) + 9 x 22(1 + 2 + 3 + 4)
⇒ 5500(45) + 45 ´ 44 = 45(5544)
Hence mean =5544
Q.11. Let A, B and C be three positive integers such that the sum of A and the mean of B and C is 5. In addition, the sum of B and the mean of A and C is 7. Then the sum of A and B is [2020]
(a) 6
(b) 5
(c) 7
(d) 4
Correct Option is A
Given A, B and C are positive integers such that
The least value for A=1 in which case B=5.
Hence A + B = 6
Q.12. If x and y are positive real numbers satisfying x + y=102, then the minimum possible value of is [2020]
Ans: 2704
AM ≥ GM ≥ HM
Given x + y = 102
⇒
⇒
The minimum value of
=
= 2704
Q.13. How many 4digit numbers, each greater than 1000 and each having all four digits distinct, are there with 7 coming before 3? [2020]
Ans: 315
Consider four blanks
7 is in thousand place, then 3 can be placed in any of the 3 places in 3 ways. Remaining two blanks can be filled with remaining eight digits in ^{8}P_{2} ways. The number of numbers that have 7 is in thousand place is 3 x ^{8}P_{2} = 168
Thousand place cannot be 0,7 and 3, it can be filled with remaining 7 digits in 7 ways. In remaining 3 blanks, 7 and 3 can be arranged in 3 ways. Fourth blank can be filled in 7 ways. The number of numbers that are formed where 7 and 3 is not in thousand place is 7 x 3 x 7 = 147 . Hence total required numbers = 168 + 147 = 315 .
Q.14. In how many ways can a pair of integers (x , a) be chosen such that x^{2}  2 x +  a  2 = 0? [2020]
(a) 4
(b) 5
(c) 6
(d) 7
Correct Option is D
x^{2}  2 x +  a 2 = 0
If a > 2;  a  2 = a  2
since x is integer 3  a ≥ 0
a ≤ 3
The possible values of a is = 3
Then x = ± 1;
If a = 2,  x = 1 ± 1 , ⇒ x = ± 2, 0
If a < 2,  a  2 = 2  a
Since x is integer a 1 ≥ 0 ⇒ a ≥ 1
∴ The possible values of a is 1
If a = 1,  x = 1 ⇒ x = ± 1
∴ The possible pairs =(1,3), (1,3), (1,1), (1,1), (2,2),(2,2), (0,2) i.e., 7
Q.15. The number of integers that satisfy the equality ( x^{2}  5x + 7 )^{ x+1} = 1 is [2020]
(a) 2
(b) 3
(c) 5
(d) 4
Correct Option is B
( x^{2}  5x + 7 )^{ x+1 } = 1
We know, for ab = 1,
if a = 1 then b is even.
a = 1 then b is any number
a > 0 then b = 0
Case 1: x + 1 = 0 Þ x = 1
Case 2: x^{2} 5x+7 =1 ⇒ x2 5x + 6 = 0 ⇒ x = 2 or 3
Case 3: x^{2 }5x + 7 = 1 ⇒ x^{2} 5x + 8 = 0
but x is not an integer
∴ The number of integers satisfies the equation is 3
Q.16. The number of pairs of integers (x , y) satisfying x ≥ y ≥ 20 and 2 x + 5 y = 99 is [2020]
Ans: 17
2 x + 5 y = 99
When y =  19, x = 97; since x ≥ y ; the maximum value of y is 13 and corresponding value of x is 17.
We know that the solutions of y are in arithmetic progression with common difference 2.
Here a =  19, d = 2, tn = 13
t_{n} = a + (n 1)d
19 + (n  1)(2) = 13
(n  1)2 = 32 ⇒ n = 17
Hence number of pairs integers is 17
Q.17. If x and y are nonnegative integers such that x + 9 = z, y + 1 = z and x + y < z + 5, then the maximum possible value of 2x + y equals [2020]
Ans: 23
Given x + 9 = z = y + 1 and x + y < z + 5
⇒ (z 9) + (z  1) < z + 5
⇒ z < 15
Hence the maximum value of z = 14, max of x = 5 and max of y = 13
Required answer, 2 x + y = 2 x 5 + 13 = 23
Q.18. How many pairs (a, b) of positive integers are there such that a ≤ b and ab = 4^{2017} ? [2020]
(a) 2017
(b) 2019
(c) 2020
(d) 2018
Correct Option is B
Given, a × b = 4^{2017} = 2^{4034}
Since a × b = 4^{2017} , is a perfect square the number of factors of 2^{4034} is odd.
Required answer, the number of values of
= 2018
Q.19. If x_{1} = 1 and x_{m} = x_{m +1} + (m + 1) for every positive integer $m,$ then x_{100} equals [2020]
(a) 5151
(b) 5150
(c) 5051
(d) 5050
Correct Option is D
x_{m + 1 }= x_{m}  (m + 1)
x_{2 }= x_{1 } 2 = 1  2 = 3
x_{3} = x_{2}  3 = 1  2  3 =  6
Similarly,
Hence,
= 5050
Q.20. Let m and n be natural numbers such that n is even and . Then m  2n equals [2020]
(a) 3
(b) 4
(c) 1
(d) 2
Correct Option is C
Given,
⇒ n = 4
since 0.2 < n/m < 0.5 m and n = 4, m = 9
m  2n = 9  2 x 4 = 1
Q.21. Let N, x and y be positive integers such that N = x + y, 2 < x < 10 and 14 < y < 23. If N > 25, then how many distinct values are possible for N? [2020]
Ans: 6
Given, 2 < x < 10 and 14 < y < 23 Þ 17 < ( x + y ) < 32 i.e. 17 < N < 32
But N > 25 hence 25 < N < 32
N can take 6 distinct values.
Q.22. How many of the integers 1, 2, … , 120, are divisible by none of 2, 5 and 7? [2020]
(a) 40
(b) 42
(c) 43
(d) 41
Correct Option is D
The required answer
=
= 41.14
Required answer is the integral part of 41.14 = 41
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