1. Find out the number of all even numbers from 1 to 300 ?
Solution :
Since 300 is an even number so total number of even numbers wil be (n/20) = (300/2)
= 150 even numbers
2. What is the sum of all the even numbers from 1 to 38
Solution :
Even numbers will be
=(381-1)/2= 190
Sum of even numbers =n*(n+1)
= 190(190+1)
= 36,290
3. Find out of sum of all the odd numbers from 50 to 200
Solution :
Required Sum
= Sum of all odd numbers from 1 to 200 - Sum of all odd numbers from 1 to 50
=100^2-25^2
=9375
4. What least number must be added to 7963 to make it exactly divisible by 65 ?
Solution :
On dividing 7963 by 65 we get 33 as remainder,
So the number to be added will be
65 - 33 = 32
5. What least number must be subtracted from 7963 to make it exactly divisible by 65 ?
Solution :
On dividing 7963 by 65 we get 33 as remainder,
So the number to be subtracted will be 33
6. Find the least number of five digits which is exactly divisible by 73 ?
Solution:
Least number of five digits will be 10000,
on dividing 10000 by 73 we get 72 as remainder,
so the number will be
= 10000 + 72 = 10072
7. find the greatest number of five digits which is exactly divisible by 147 ?
Solution :
The greatest number of five digit will be 99999,
on dividing it by 147 we get 39 as remainder,
so the required number will be
99999 - 39 = 99960
8. What is the unit digit in 7^105 ?
Solution :
Since 300 is an even number so total number of even numbers wil be (n/20) = (300/2)
= 150 even numbers
2. What is the sum of all the even numbers from 1 to 38
Solution :
Even numbers will be
=(381-1)/2= 190
Sum of even numbers =n*(n+1)
= 190(190+1)
= 36,290
3. Find out of sum of all the odd numbers from 50 to 200
Solution :
Required Sum
= Sum of all odd numbers from 1 to 200 - Sum of all odd numbers from 1 to 50
=100^2-25^2
=9375
4. What least number must be added to 7963 to make it exactly divisible by 65 ?
Solution :
On dividing 7963 by 65 we get 33 as remainder,
So the number to be added will be
65 - 33 = 32
5. What least number must be subtracted from 7963 to make it exactly divisible by 65 ?
Solution :
On dividing 7963 by 65 we get 33 as remainder,
So the number to be subtracted will be 33
6. Find the least number of five digits which is exactly divisible by 73 ?
Solution:
Least number of five digits will be 10000,
on dividing 10000 by 73 we get 72 as remainder,
so the number will be
= 10000 + 72 = 10072
7. find the greatest number of five digits which is exactly divisible by 147 ?
Solution :
The greatest number of five digit will be 99999,
on dividing it by 147 we get 39 as remainder,
so the required number will be
99999 - 39 = 99960
8. What is the unit digit in 7^105 ?
Solution :
Unit digit in 7105
= Unit digit in [(74)26 x 7]
But, unit digit in (74)26 = 1
Unit digit in 7105
=(1x7) = 7
9. What is the unit digit in the product (3^65 x 6^59 x 7^71)?
Solution :
Unit digit in 3^4 = 1 Unit digit in (3^4)^16 = 1
Unit digit in 3^65 = Unit digit in [ (3^4)^16 x 3 ] = (1 x 3) = 3
Unit digit in 6^59 = 6
Unit digit in 7^4
Unit digit in (7^4)^17 is 1.
Unit digit in 7^71 = Unit digit in [(7^4)^17 x 7^3] = (1 x 3) = 3
Required digit = Unit digit in (3 x 6 x 3) = 4.
10. Find the number at the unit place in (257)^61
Solution:
(257)^61=(257)^60(257)
=(257)^4*15(257)
=(...1)(...7)
=7 is the unit digit
11. Find the number at the unit place in (142)^65
Solution
(142)^65=(142)^64(142)
=(142)^4*16(142)
=(..6)(..12)
=12 so unit digit is 2
12. How many prime numbers exist in exist in
6^5 * 21^4 * 13^6
Solution:
=(2*3)^5 (7*3)^4 (13)^6
=2^5 * 3^9 * 7^4 *13^6
sum of the power of the prime number
=5+9+4+6
=24
13. A watch ticks 90 times in 95 seconds and another watch ticks 315 times in 323 seconds.
If both the watches are started together, how many times will they tick together in first hour?
Solution:
The first watch ticks every 95 / 90 seconds,
second watch ticks every 323 / 315 seconds.
They will tick together after ( LCM of 95 / 90 and 323 / 315 ) seconds.
= LCM of (95, 323) / HCF of (90, 315) = (19 * 5 * 17 ) / 45
The number of times they will tick in the first 3600 seconds
= 3600 / (1615 / 45 )
= 100.30 = 100 + 1 ,
as it has already ticked first time.
14. Find out the number of zeroes in 8 × 15 × 23 × 17 × 25 × 22 ?
Solution : (2^3 ) × (3×5) × (23) × (17) × (5^2 ) × (11×2)
Zeroes are formed by combination of 2 * 5,
here number of pairs of (2, 5) is 3 so the numbers of zeros will be three
15. The product of two numbers is 60480 and their HCF is 12 . Find the numbers ?
Solution :
Since HCF s 12, so the two numbers will be multiple of their HCF
let the first number is 12P and the second number be 12Q
∴12P × 12Q = 60480
∴ P × Q = 420
Now pair of numbers whose product is 420 is
(420, 1) (210, 2) (140, 3) (105, 4) (60, 7) (20, 21)
Out of these ( 210, 2 ) is not prime so neglected
Now the required numbers will be
(420×12,1×12) (140×12,3×12) (105×12,4×12) (60×12 ,7×12)
(20×12,21×12) (5040,12) (1680, 36) ( 1260, 48) (720,84)
(240,252) be the required numbers
16. Find the greatest number that will divide 37, 109 and 157 so as to leave the same remainder in each case ?
Solution :
Let the remainder be x, then the numbers :
(37 - x) (109 - x) (157 - x) must be divisible by the required number.
Also if two numbers are divisible by the certain number then their difference is also divisible by that number
(109 - x) - (37 - x) = 72
(157 - x) - (109 - x) = 48
(157 - x) - (37 - x) = 120
So, the numbers 72, 48, 120 will also be divisible by that number,
So HCF of 72, 48, 120 is 24, therefore required number will be 24
17. Ex. Find out the number of zeros at the end of products 20×15×16×44×72×95×25
Solution :
Note : Zeroes can be produced by two ways
(1) If there is any zero at the end of any multiplicand.
(2) If 5 or its multiple are multiplied by any even number.
Now 20×15×16×44×72×95×25 =
(2^2*5)(3*5)(2^4)(2^2*11)(2^3*3^2)(5*9)(5^2)
(2^11)(5^5).....
So total number of zeros are 5
18. The number of prime factors of (3 x 5)^12 (2 x 7)^10 (10)^25 is?
a) 47
b) 60
c) 72
d) None of these
Solution
3^12*5^38*2^35*7^10
=12+38+35+10=95
Ans: None of these
19. The smallest number, which is a perfect square and contains 7936 as a factor is:
a) 251664
b) 231564
c) 246016
d) 346016
e) None of these
Solution;
7936 => 2^2 * 2^2 * 2^2 * 2^2 * 31^1
To make it as a perfect square, we have to multiply 7936 with 31...
Hence the reqd no. is 7936*31 = 246016
Ans : 246016
20. There are four prime numbers written in ascending order of magnitude.
The product of first three is 385 and that of last three is 1001. Find the first number.
a) 5
b) 7
c) 11
d) 17
Solution:
The prime nos are 5,7,11,13
1st three no product is 385 so 5*7*11 = 385
Last three no product is 1001 so 7*11*13 =1001
four prime numbers are 5,7,11,13
the largest of the given prime number is 13
21. Every time x is increased by a given constant number, y doubles and z becomes three times.
How will log(y) and log(z) behave as x is increased by the same constant number?
a) Both will grow linearly with different slopes
b) Both will grow linearly with same slopes
c) y will grow linearly, while z will not
d) z will grow linearly, while y will not
Solution:
y₂/y₁ = 2
log(y₂/y₁) = log(2)
log(y₂) - log(y₁) = log(2) = constant
log(y) increases linearly ---> ∆log(y)/∆x = log(2)/k
z₂/z₁ = 3
log(z₂/z₁) = log(3)
log(z₂) - log(z₁) = log(3) = constant
log(z) increases linearly ---> ∆log(z)/∆x = log(3)/k
Ans: Both will grow linearly with different slopes
log(y₂/y₁) = log(2)
log(y₂) - log(y₁) = log(2) = constant
log(y) increases linearly ---> ∆log(y)/∆x = log(2)/k
z₂/z₁ = 3
log(z₂/z₁) = log(3)
log(z₂) - log(z₁) = log(3) = constant
log(z) increases linearly ---> ∆log(z)/∆x = log(3)/k
Ans: Both will grow linearly with different slopes
22. Rajeev multiplies a number by 10, the log (to base 10) of this number will change in what way?
a) Increase by 10
b) Increase by 1
c) Multiplied by 10
d) None of these
Solution:
log x^x=1
base=10 , multiplies number=10
log10^10=1
Ans:Increase by 1
23. .How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated ?
a) 5
b) 10
c) 15
d) 20
Solution:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
24. The number obtained by interchanging the two digits of a two-digit number is more than the original number by 27. If the sum of the two digits is 13, what is the original number?
1) 63
2) 74
3) 85
4) 58
5) None of these
Solution:
Lets say the 2 digit number is xy
tens digit is x and units is y
based on given condition
10y+x=10x+y+27 --> (1)
x+y=13 -->(2)
On Solving 1 and 2
x=5 y=8
number is 58
25. The number obtained by interchanging the two digits of a two-digit number is less than the original number by 18. The sum of the two digits of the number is 16. What is the original number?
1) 97
2)87
3)79
4) Cannot be determined
Solution:
original number =10x+y
tens digit is x and units is y
x+y=16--> (1)
10y+x=10x+y-18--> (2)
On Solving 1 and 2
x=9, y=7
Applying on original number
10(9)+7=97
26. When the digits of a two-digit number are interchanged, the number obtained is less than the original number by 36. What is the original number if the difference of the two digits is 4?
1) 84
2) 51
3) 73
4) Cannot be determined
27. If the positions of the digits of a two-digit number are interchanged, the number obtained is smaller than the original number by 27. If the digits of the number are in the ratio of 1:2, what is the original number?
1) 36
2) 63
3) 48
4) Cannot be determined
Solution:
x=10a+b,
y=10b+a,
x-y=27,
b/a=1/2
63-36=27
a = 6,
b = 3,
x = 63,
y = 36
63 is the original number.
28. If the digits of a two-digit number are interchanged, the number formed is greater than the original number by 45. If the difference between the digits is 5, what is the original number?
1) 16
2) 27
3) 38
4) Cannot be determined
5) None of these
29. The sum of four consecutive even numbers is 44. What is the sum of the original squares of these numbers?
1) 288
2) 502
3) 696
4) 920
5) None of these
Solution:
let the first no. be x so next will be x+2,next one x+4 so on,
So sum of no. is 4x+12=44,
so 4x=44-12=32, so x=8.
So no. (s) are 8, 10,12 ,14.
So sum of the original squares of these numbers is 504
30. A, B, C, D and E are five consecutive odd numbers. The sum of A and C is 146. What is the value of E?
1) 75
2) 81
3) 71
4) 79
5) None of these
Solution:
A,B,C,D,E are odd numbers
A=A, B=A+2, C=A+4
D=A+6, E=A+8
given, A+C=146----(1)
C=A+4 apply on equ 1
A+A+4=146
2A+4=146
A=71
Find E
E=A+8
=71+8=79
31. The product of two successive numbers is 4692. Which is the smaller of the two numbers?
1) 69
2) 62
3) 68
4) 67
5) None of these
Solution:
68
as 68*69=4692
32. The product of two successive numbers is 9506. Which is the smaller of the two numbers?
1) 96
2) 97
3) 98
4) 99
5) None of these
Solution:
97*98=9506
Smallest num is 97
33. .The product of two consecutive even numbers is 3248. Which is the larger number?
1) 58
2) 62
3) 56
4) 60
5) None of these
Solution:
58*56=3248
Largest number is 58
34. The sum of five consecutive even numbers is 200. What is the sum of the next set of the consecutive even numbers?
1) 215
2) 235
3) 240
4) 250
5) None of these
Solution:
nos will be x-4 ,x-2 , x , x+2 , x+4 = 200, which is x=40 ,
36+38+40+42+44 =200
so next set of even no
= 46+48+50+52+54= 250
35. .The sum of five consecutive odd numbers is 575. What is the sum of the next set of the consecutive odd numbers?
1) 615
2) 635
3) 595
4) Cannot be determined
5) None of these
Solution:
x+(x+2)+(x+4)+(x+6)+(x+8)=575
x=111
5 Numbers
111,113,115,117,119
Next copnsequte
121,123,125,127,129=625
a) Increase by 10
b) Increase by 1
c) Multiplied by 10
d) None of these
Solution:
log x^x=1
base=10 , multiplies number=10
log10^10=1
Ans:Increase by 1
23. .How many 3 digit numbers can be formed from the digits 2, 3, 5, 6, 7 and 9, which are divisible by 5 and none of the digits is repeated ?
a) 5
b) 10
c) 15
d) 20
Solution:
Since each desired number is divisible by 5, so we must have 5 at the unit place. So, there is 1 way of doing it.
The tens place can now be filled by any of the remaining 5 digits (2, 3, 6, 7, 9). So, there are 5 ways of filling the tens place.
The hundreds place can now be filled by any of the remaining 4 digits. So, there are 4 ways of filling it.
Required number of numbers = (1 x 5 x 4) = 20.
24. The number obtained by interchanging the two digits of a two-digit number is more than the original number by 27. If the sum of the two digits is 13, what is the original number?
1) 63
2) 74
3) 85
4) 58
5) None of these
Solution:
Lets say the 2 digit number is xy
tens digit is x and units is y
based on given condition
10y+x=10x+y+27 --> (1)
x+y=13 -->(2)
On Solving 1 and 2
x=5 y=8
number is 58
25. The number obtained by interchanging the two digits of a two-digit number is less than the original number by 18. The sum of the two digits of the number is 16. What is the original number?
1) 97
2)87
3)79
4) Cannot be determined
Solution:
original number =10x+y
tens digit is x and units is y
x+y=16--> (1)
10y+x=10x+y-18--> (2)
On Solving 1 and 2
x=9, y=7
Applying on original number
10(9)+7=97
26. When the digits of a two-digit number are interchanged, the number obtained is less than the original number by 36. What is the original number if the difference of the two digits is 4?
1) 84
2) 51
3) 73
4) Cannot be determined
27. If the positions of the digits of a two-digit number are interchanged, the number obtained is smaller than the original number by 27. If the digits of the number are in the ratio of 1:2, what is the original number?
1) 36
2) 63
3) 48
4) Cannot be determined
Solution:
x=10a+b,
y=10b+a,
x-y=27,
b/a=1/2
63-36=27
a = 6,
b = 3,
x = 63,
y = 36
63 is the original number.
28. If the digits of a two-digit number are interchanged, the number formed is greater than the original number by 45. If the difference between the digits is 5, what is the original number?
1) 16
2) 27
3) 38
4) Cannot be determined
5) None of these
29. The sum of four consecutive even numbers is 44. What is the sum of the original squares of these numbers?
1) 288
2) 502
3) 696
4) 920
5) None of these
Solution:
let the first no. be x so next will be x+2,next one x+4 so on,
So sum of no. is 4x+12=44,
so 4x=44-12=32, so x=8.
So no. (s) are 8, 10,12 ,14.
So sum of the original squares of these numbers is 504
30. A, B, C, D and E are five consecutive odd numbers. The sum of A and C is 146. What is the value of E?
1) 75
2) 81
3) 71
4) 79
5) None of these
Solution:
A,B,C,D,E are odd numbers
A=A, B=A+2, C=A+4
D=A+6, E=A+8
given, A+C=146----(1)
C=A+4 apply on equ 1
A+A+4=146
2A+4=146
A=71
Find E
E=A+8
=71+8=79
31. The product of two successive numbers is 4692. Which is the smaller of the two numbers?
1) 69
2) 62
3) 68
4) 67
5) None of these
Solution:
68
as 68*69=4692
32. The product of two successive numbers is 9506. Which is the smaller of the two numbers?
1) 96
2) 97
3) 98
4) 99
5) None of these
Solution:
97*98=9506
Smallest num is 97
33. .The product of two consecutive even numbers is 3248. Which is the larger number?
1) 58
2) 62
3) 56
4) 60
5) None of these
Solution:
58*56=3248
Largest number is 58
34. The sum of five consecutive even numbers is 200. What is the sum of the next set of the consecutive even numbers?
1) 215
2) 235
3) 240
4) 250
5) None of these
Solution:
nos will be x-4 ,x-2 , x , x+2 , x+4 = 200, which is x=40 ,
36+38+40+42+44 =200
so next set of even no
= 46+48+50+52+54= 250
35. .The sum of five consecutive odd numbers is 575. What is the sum of the next set of the consecutive odd numbers?
1) 615
2) 635
3) 595
4) Cannot be determined
5) None of these
Solution:
x+(x+2)+(x+4)+(x+6)+(x+8)=575
x=111
5 Numbers
111,113,115,117,119
Next copnsequte
121,123,125,127,129=625
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