1. what could be the maximum value of q in the following equation ?
5 p 9 + 3 r 7+ 2 q 8 = 1114
(a) 81 (b) 5
(c) 9 (d) 18
Answer: (c) 9
Solution: first add the digits at the unit's place - 9+7+8 = 24. Carry the 2.
Adding digits at the ten's place plus carry 2 from the addition of units place -
we get
p + r+ q+ 2
it is evident from the above equation that the sum should be 11. Now for getting the maximum value of q, r and p should be zero.
=> q+2 = 11
=> q =9
2. Find the value of 1+ 2 + 3 + ........+ 105
(a) 5565 (b) 2782
(c) 4268 (d) 3254
Answer: (a) 5565
Solution: Applying the formula for arithmetic progression, we get
1+ 2 + 3 + ........+ 105 = (105)(1+105)/2=5565
3. Find the value of 1+3+5......+ 49
(a) 298 (b) 625
(c) 525 (d) 400
Answer: (b) 625
Solution:
It form an AP with 1 as the first term and 49 as the last and 2 as the common difference.
Therefore, 1+3+5... ...+ 49= (25)(1+49)/2=(25)(25) = 625
4. Find the value of 2+4+6+8+.......+100
(a) 1125 (b) 625
(d) 2190 (d) 2550
Answer: (d) 2550
Solution: It is an Arithmetic progression with first AP as 2 and the last AP =100
Applying the formula for the sum of the AP to get
2+4+6+8+.......+100= (50)(2+100)/2 = (50)(102)/2 = 25(102) = 2550
5. Find the sum of all 2 digit number divisible by 3
(a) 835 (b) 1665
(c) 1533 (d) 1683
Answer:
2 digits numbers that are divisible by 3 are 3,6,9,12,...99 or 3x1,3x2,3x3,....3x33.
Therefore there are 33 such numbers beginning with 3 till 99.
Sum of 2 digit numbers divisible by 3 = 3+6+9+...99 = (33)(3+99)/2 = 33 x 102/2 = 33x51 = 1683
6. How many number between 11 and 90 are divisible by 7 ?
(a)11 (b) 55
(c) 27 (d) 39
Answer:(a)11
Solution:
After 11, the first number divisible by 7 is 14. And the last number divisible by 7 before the number 90 is 84.
Therefore the numbers between 7 and 90 that are divisible by 7 are
14,21,28,,,,84 or 7x2,7x3,....7x12
Therefore there are 12-2 +1=11 such numbers.
7. Find the sum of all odd number up to 100
(a) 1175 (b) 1250
(c) 2500 (d) 2650
Answer: (c) 2500
Solution: Odd number between 1 and 100 are 50.
First term = 1 and last term =99
therefore, sum = (50)(1+99)/2 = (50)(100)/2 = 50 x 50 = 2500
8. Find the sum of all even number up to 100
(a) 1365 (b) 2550
(c) 1790 (d) 4820
Answer: (b) 2550
Solution: Even numbers between 1 and 100 are 50.
First term = 2 and last term =100
therefore, sum = (50)(2+100)/2 = (50)(102)/2 = 50 x 51 = 2550
9. Find the value 12+ 22+32+....... 102
(a) 570 (b) 450
(c) 132 (d) 582
Answer: (a) 570
Solution
This is an AP with the first term as 12 and the last term as 102 and common difference as 10.
Number of terms is 102 = 12 + (n-1)10
=> 90 = 10n - 10
=>100 = 10n
=> n = 10
Therefore the sum of 12+ 22+32+....... 102 = (10)(12+102)/2 = (10)(114)/2=5x114=570
10. How many terms are there in 2,4,8,16 ........1024?
(a) 8 (b) 10
(c) 12 (d) 6
Answer: (b) 10
Solution:
This is a GP with 2 as the common ratio.
therefore, 2,4,8,16 ........102 is the same as 2^1, 2^2, 2^3, 2^4,.....2^10
1024 = 2^10
Therefore there are 10 terms.
11. How many number up to three digits are divisible by 17 ?
(a) 58 (b) 47
(c) 52 (d) 31
Answer:(a) 58
Solution:
The first number divisible by 17 is 17 itself.
To get the last number before 999 that is divisible by 17 we divide 999 by 17.
we get ,
999/17 = 58x17 + 13
Subtract 13 from 999 to get 986 which is the last three number divisible by 17
Therefore there are 58 numbers that are divisible by 17.
13. Find the no of divisors of 8064
(a) 44 (b) 48
(c) 84 (d)74
Answer: (b) 48
Solution:
8064 = 2^7 x 3^2 x 7^1
The number of divisors of 8064 = (7+1)(2+1)(1+1) = (8)(3)(2)=48
14. Find the sum of the divisors of a number 8064
(a) 26520 (b) 58140
(c) 62195 (d) 29870
Answer: 26520
Solution:
8064 = 2^7 x 3^2 x 7^1
Sum of divisors = (2^0 +2^1 +2^2 +2^3 +2^4 +2^5 +2^6 +2^7 )(3^0 + 3^1+3^2)(7^0+7^1)
= (1+2+4+8+16+32+64+128)(1+3+9)(1+7)
= (26520)
15. The sum of the digits of a two digit number is 8 .If the digits are reversed the number is decreased by 54 find the number .
(a) 97 (b) 71
(c) 33 (d) 49
Answer: (b) 71
Solution: Let the number be 10x +y
then given that x+y =8 ----------------------i
Again if number is reveresed then it is decreases by 54.
=> (10x+y ) - (10y +x ) = 54
=> 10x + y - 10y -x =54
=> 9x - 9y = 54
=> x-y = 6 ----------------------------- ii
Adding i and ii , we have
2x = 14
=> x= 7
And from i the value of y is 8-7 =1
Therefore the number is 71.
16. The ratio of the sum and the difference of two number are 7: 1 find the ratio of those two number
(a) 4:3 (b) 2:3
(c) 3:2 (d) 3:4
Answer:(a) 4:3
Solution:
option (a) 4:3
Sum is 4+3 = 7
Difference is 4-3 =1
And ratio of sum and difference is 7:1.
17. The ratio between a two digit number and the sum of the digits of that number is 4:1 .If the digit in the units place is three more than digit in the ten place. what is the number ?
(a) 44 (b) 36
(c) 24 (d) 18
Answer: (b) 36
Solution:
Let the number be 10x+y with x at the tens place and y at the ones place.
Then according to the question
(10x+y) /(x+y) = 4/1
=> (10x+y) = 4(x+y)
=> 10x + y = 4x +4y
=>6x=3y
=> 2x=y -----------------------------(i)
again given that
y= x+3 --------------------- (ii)
Solving (i) and (ii)
2x = x+3
=> x=3
and y=2x = 2.3 =6
Therefore the number is 36
Shortcut: Lets check the options
Given that the digit in the units place is 3 more than the digit at the tens place. Only option (b) fits the bill and is the answer.
18. A number consists of two digit the sum of digits is 9. If 27 is subtracted from the number ,its digits are interchanged. find the number .
(a) 81 (b) 63
(c) 51 (d) 91
Answer:
Solution:
We will not solve it by forming equations and then solving them.
We will go through the options to zero in on the right answer.
According to questions the sum of the digits is 9. Only option (a)and option (b) has the sum of digits as 9.
Now , if we subtract 27 from 81 we get 54. The digits are not interchanged.
That leaves us the second option. From 63 if we subtract 27 we get 36. Digits are interchanged. Hence option (b) is the correct answer.
19. If three numbers are added in pairs the sum equal 10, 19 and 21 find the number
(a) 2,10,21 (b) 3,9,27
(c) 6,8,3 (d) 6,4,15
Answer:(d) 6,4,15
Solution:
Try each option. only option (d) satisfies the condition.
20 Find the greatest number of 3 digits which is exactly divisible by 35 ?
(a) 980 (b) 220
(c) 880 (d) 1015
Answer: (a) 980
Solution;
Starting by taking the greatest three digit number 999.
Divide this number by 35. we have,
999= 35x28 + 19
Now subtract 19 from the 999 to get 980 as the answer.
21. What least number must be added to 49123 to get a number exactly divisible by 263 ?
(a)10 (b)58
(c)9 (d)18
Answer: (b)58
Solution:
Divide 49123 by 263 to get
49123 = 186.263 + 205
Now we have to add 58 (263-205)a number to 49123 completely divisible by 263.
23. A number when divided by 899 given a remainder of 63. what remainder will obtained by dividing the same number by 29 ?
(a) 13 (b) 3
(c) 5 (d) 21
Answer: (c) 5
Solution:The number could be written as 899xQ + 63 where Q is the quotient of dividing the number by 899. Since 899 is completely divided by 29then the number 899xQ is also divided by 29. what remains is 63. Now if we divide 63 by 29 we get 5 as the remainder,
24. The sum of squares of three consecutive odd number is 2531 find the number
(a) 32,24 12 (b) 27, 29, 31
(c) 8,13,19 (d) 9,15, 25
Answer: (b) 27, 29, 31
Solution:
option(a) : The unit digit in the squares of number 32,24,12 are 4,6and 4 and their sum would have a unit digit of 4. but our sum has 1 in its unit digit. option rejected.
option(b) : The unit digit in the squares of number 27,29,31 are 9,1and 1 and their sum would have a unit digit of 1. and so is our answer. But we would check more options.
option(c) : The unit digit in the squares of number 8,13,19 are 4,9and 1 and their sum would have a unit digit of 4. but our sum has 1 in its unit digit. option rejected.
option(d) :Number are not consecutive.
25. If a number is divided by 84 the remainder is 37. what will be the remainder if it is divided by 21 ?
(a) 16 (b) 18
(c) 12 (d) 9
Answer:(a) 16
Solution:
just divide the remainder by 21 to get the required answer. Here dived 37 by 21 to 16 as the remainder.
26. What is the sum of all odd number between 1 to 50 ?
(a) 180 (b)625
(c) 840 (d) 820
Answer:(b)625
Solution:
There are 25 odd number from 1 to 50 with a common difference as 2.
therefore the sum = (25)(1+49) /2 = 25x25 = 625
27. On converting the binary number 1101101 in to decimal form what will we get ?
(a) 109 (b) 208
(c) 207 (d) -111
Answer: (a) 109
Solution:
1.2^6 +1. 2^5 +0. 2^4+1.2^3+1.2^2+0.2^1+2^0 = 64+32+0+8+4+0+1 = 96+13 = 109
28. What is the unit of digit is the product of 207 X 39 X 94 ?
(a) 9 (b) 1
(c) 7 (d) 2
Answer: (d) 2
Solution:
The unit digit of 207 X 39 is 3 and the unit digit in the 207 X 39 X 94 is 2.
29. How many digits are required to write the number from 1 to 100 ?
(a) 100 (b) 192
(c) 99 (d) 198
Answer: (b) 192
Solution:
From 1 to 9, there are 9 digits.
From 10 to 99 there are 10x2x9 =180 digits.
and to write 100 we need 3 digits.
Therefore, the total numbers of digits are 9+180+3=183+9 = 192
30. Find the largest number of five digits which is divisible by 17 ?
(a) 99999 (b) 99960
(c) 99994 (d) 10013
Answer: (c) 99994
Solution:
First the largest 5 digit number is 99999. Now divide this number by 17 to get the quotient and the remainder.
We get, 99999 = 5882x17 + 5
Therefore the largest number completely divisible by 17 is 99994.
5 p 9 + 3 r 7+ 2 q 8 = 1114
(a) 81 (b) 5
(c) 9 (d) 18
Answer: (c) 9
Solution: first add the digits at the unit's place - 9+7+8 = 24. Carry the 2.
Adding digits at the ten's place plus carry 2 from the addition of units place -
we get
p + r+ q+ 2
it is evident from the above equation that the sum should be 11. Now for getting the maximum value of q, r and p should be zero.
=> q+2 = 11
=> q =9
2. Find the value of 1+ 2 + 3 + ........+ 105
(a) 5565 (b) 2782
(c) 4268 (d) 3254
Answer: (a) 5565
Solution: Applying the formula for arithmetic progression, we get
1+ 2 + 3 + ........+ 105 = (105)(1+105)/2=5565
3. Find the value of 1+3+5......+ 49
(a) 298 (b) 625
(c) 525 (d) 400
Answer: (b) 625
Solution:
It form an AP with 1 as the first term and 49 as the last and 2 as the common difference.
Therefore, 1+3+5... ...+ 49= (25)(1+49)/2=(25)(25) = 625
4. Find the value of 2+4+6+8+.......+100
(a) 1125 (b) 625
(d) 2190 (d) 2550
Answer: (d) 2550
Solution: It is an Arithmetic progression with first AP as 2 and the last AP =100
Applying the formula for the sum of the AP to get
2+4+6+8+.......+100= (50)(2+100)/2 = (50)(102)/2 = 25(102) = 2550
5. Find the sum of all 2 digit number divisible by 3
(a) 835 (b) 1665
(c) 1533 (d) 1683
Answer:
2 digits numbers that are divisible by 3 are 3,6,9,12,...99 or 3x1,3x2,3x3,....3x33.
Therefore there are 33 such numbers beginning with 3 till 99.
Sum of 2 digit numbers divisible by 3 = 3+6+9+...99 = (33)(3+99)/2 = 33 x 102/2 = 33x51 = 1683
6. How many number between 11 and 90 are divisible by 7 ?
(a)11 (b) 55
(c) 27 (d) 39
Answer:(a)11
Solution:
After 11, the first number divisible by 7 is 14. And the last number divisible by 7 before the number 90 is 84.
Therefore the numbers between 7 and 90 that are divisible by 7 are
14,21,28,,,,84 or 7x2,7x3,....7x12
Therefore there are 12-2 +1=11 such numbers.
7. Find the sum of all odd number up to 100
(a) 1175 (b) 1250
(c) 2500 (d) 2650
Answer: (c) 2500
Solution: Odd number between 1 and 100 are 50.
First term = 1 and last term =99
therefore, sum = (50)(1+99)/2 = (50)(100)/2 = 50 x 50 = 2500
8. Find the sum of all even number up to 100
(a) 1365 (b) 2550
(c) 1790 (d) 4820
Answer: (b) 2550
Solution: Even numbers between 1 and 100 are 50.
First term = 2 and last term =100
therefore, sum = (50)(2+100)/2 = (50)(102)/2 = 50 x 51 = 2550
9. Find the value 12+ 22+32+....... 102
(a) 570 (b) 450
(c) 132 (d) 582
Answer: (a) 570
Solution
This is an AP with the first term as 12 and the last term as 102 and common difference as 10.
Number of terms is 102 = 12 + (n-1)10
=> 90 = 10n - 10
=>100 = 10n
=> n = 10
Therefore the sum of 12+ 22+32+....... 102 = (10)(12+102)/2 = (10)(114)/2=5x114=570
10. How many terms are there in 2,4,8,16 ........1024?
(a) 8 (b) 10
(c) 12 (d) 6
Answer: (b) 10
Solution:
This is a GP with 2 as the common ratio.
therefore, 2,4,8,16 ........102 is the same as 2^1, 2^2, 2^3, 2^4,.....2^10
1024 = 2^10
Therefore there are 10 terms.
11. How many number up to three digits are divisible by 17 ?
(a) 58 (b) 47
(c) 52 (d) 31
Answer:(a) 58
Solution:
The first number divisible by 17 is 17 itself.
To get the last number before 999 that is divisible by 17 we divide 999 by 17.
we get ,
999/17 = 58x17 + 13
Subtract 13 from 999 to get 986 which is the last three number divisible by 17
Therefore there are 58 numbers that are divisible by 17.
13. Find the no of divisors of 8064
(a) 44 (b) 48
(c) 84 (d)74
Answer: (b) 48
Solution:
8064 = 2^7 x 3^2 x 7^1
The number of divisors of 8064 = (7+1)(2+1)(1+1) = (8)(3)(2)=48
14. Find the sum of the divisors of a number 8064
(a) 26520 (b) 58140
(c) 62195 (d) 29870
Answer: 26520
Solution:
8064 = 2^7 x 3^2 x 7^1
Sum of divisors = (2^0 +2^1 +2^2 +2^3 +2^4 +2^5 +2^6 +2^7 )(3^0 + 3^1+3^2)(7^0+7^1)
= (1+2+4+8+16+32+64+128)(1+3+9)(1+7)
= (26520)
15. The sum of the digits of a two digit number is 8 .If the digits are reversed the number is decreased by 54 find the number .
(a) 97 (b) 71
(c) 33 (d) 49
Answer: (b) 71
Solution: Let the number be 10x +y
then given that x+y =8 ----------------------i
Again if number is reveresed then it is decreases by 54.
=> (10x+y ) - (10y +x ) = 54
=> 10x + y - 10y -x =54
=> 9x - 9y = 54
=> x-y = 6 ----------------------------- ii
Adding i and ii , we have
2x = 14
=> x= 7
And from i the value of y is 8-7 =1
Therefore the number is 71.
16. The ratio of the sum and the difference of two number are 7: 1 find the ratio of those two number
(a) 4:3 (b) 2:3
(c) 3:2 (d) 3:4
Answer:(a) 4:3
Solution:
option (a) 4:3
Sum is 4+3 = 7
Difference is 4-3 =1
And ratio of sum and difference is 7:1.
17. The ratio between a two digit number and the sum of the digits of that number is 4:1 .If the digit in the units place is three more than digit in the ten place. what is the number ?
(a) 44 (b) 36
(c) 24 (d) 18
Answer: (b) 36
Solution:
Let the number be 10x+y with x at the tens place and y at the ones place.
Then according to the question
(10x+y) /(x+y) = 4/1
=> (10x+y) = 4(x+y)
=> 10x + y = 4x +4y
=>6x=3y
=> 2x=y -----------------------------(i)
again given that
y= x+3 --------------------- (ii)
Solving (i) and (ii)
2x = x+3
=> x=3
and y=2x = 2.3 =6
Therefore the number is 36
Shortcut: Lets check the options
Given that the digit in the units place is 3 more than the digit at the tens place. Only option (b) fits the bill and is the answer.
18. A number consists of two digit the sum of digits is 9. If 27 is subtracted from the number ,its digits are interchanged. find the number .
(a) 81 (b) 63
(c) 51 (d) 91
Answer:
Solution:
We will not solve it by forming equations and then solving them.
We will go through the options to zero in on the right answer.
According to questions the sum of the digits is 9. Only option (a)and option (b) has the sum of digits as 9.
Now , if we subtract 27 from 81 we get 54. The digits are not interchanged.
That leaves us the second option. From 63 if we subtract 27 we get 36. Digits are interchanged. Hence option (b) is the correct answer.
19. If three numbers are added in pairs the sum equal 10, 19 and 21 find the number
(a) 2,10,21 (b) 3,9,27
(c) 6,8,3 (d) 6,4,15
Answer:(d) 6,4,15
Solution:
Try each option. only option (d) satisfies the condition.
20 Find the greatest number of 3 digits which is exactly divisible by 35 ?
(a) 980 (b) 220
(c) 880 (d) 1015
Answer: (a) 980
Solution;
Starting by taking the greatest three digit number 999.
Divide this number by 35. we have,
999= 35x28 + 19
Now subtract 19 from the 999 to get 980 as the answer.
21. What least number must be added to 49123 to get a number exactly divisible by 263 ?
(a)10 (b)58
(c)9 (d)18
Answer: (b)58
Solution:
Divide 49123 by 263 to get
49123 = 186.263 + 205
Now we have to add 58 (263-205)a number to 49123 completely divisible by 263.
23. A number when divided by 899 given a remainder of 63. what remainder will obtained by dividing the same number by 29 ?
(a) 13 (b) 3
(c) 5 (d) 21
Answer: (c) 5
Solution:The number could be written as 899xQ + 63 where Q is the quotient of dividing the number by 899. Since 899 is completely divided by 29then the number 899xQ is also divided by 29. what remains is 63. Now if we divide 63 by 29 we get 5 as the remainder,
24. The sum of squares of three consecutive odd number is 2531 find the number
(a) 32,24 12 (b) 27, 29, 31
(c) 8,13,19 (d) 9,15, 25
Answer: (b) 27, 29, 31
Solution:
option(a) : The unit digit in the squares of number 32,24,12 are 4,6and 4 and their sum would have a unit digit of 4. but our sum has 1 in its unit digit. option rejected.
option(b) : The unit digit in the squares of number 27,29,31 are 9,1and 1 and their sum would have a unit digit of 1. and so is our answer. But we would check more options.
option(c) : The unit digit in the squares of number 8,13,19 are 4,9and 1 and their sum would have a unit digit of 4. but our sum has 1 in its unit digit. option rejected.
option(d) :Number are not consecutive.
25. If a number is divided by 84 the remainder is 37. what will be the remainder if it is divided by 21 ?
(a) 16 (b) 18
(c) 12 (d) 9
Answer:(a) 16
Solution:
just divide the remainder by 21 to get the required answer. Here dived 37 by 21 to 16 as the remainder.
26. What is the sum of all odd number between 1 to 50 ?
(a) 180 (b)625
(c) 840 (d) 820
Answer:(b)625
Solution:
There are 25 odd number from 1 to 50 with a common difference as 2.
therefore the sum = (25)(1+49) /2 = 25x25 = 625
27. On converting the binary number 1101101 in to decimal form what will we get ?
(a) 109 (b) 208
(c) 207 (d) -111
Answer: (a) 109
Solution:
1.2^6 +1. 2^5 +0. 2^4+1.2^3+1.2^2+0.2^1+2^0 = 64+32+0+8+4+0+1 = 96+13 = 109
28. What is the unit of digit is the product of 207 X 39 X 94 ?
(a) 9 (b) 1
(c) 7 (d) 2
Answer: (d) 2
Solution:
The unit digit of 207 X 39 is 3 and the unit digit in the 207 X 39 X 94 is 2.
29. How many digits are required to write the number from 1 to 100 ?
(a) 100 (b) 192
(c) 99 (d) 198
Answer: (b) 192
Solution:
From 1 to 9, there are 9 digits.
From 10 to 99 there are 10x2x9 =180 digits.
and to write 100 we need 3 digits.
Therefore, the total numbers of digits are 9+180+3=183+9 = 192
30. Find the largest number of five digits which is divisible by 17 ?
(a) 99999 (b) 99960
(c) 99994 (d) 10013
Answer: (c) 99994
Solution:
First the largest 5 digit number is 99999. Now divide this number by 17 to get the quotient and the remainder.
We get, 99999 = 5882x17 + 5
Therefore the largest number completely divisible by 17 is 99994.
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