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1. Find the sum of first 25 natural numbers.
(a) 432 (b) 315 (c) 325 (d) 335 (e) None of the above
2. Find the sum of the squares of first 35 natural numbers.
(a) 14910 (b) 15510
(c) 14510 (d) 16510
(e) None of the above
3. Find the sum of the cubes of first 15 natural numbers.
(a) 15400 (b) 14400
(c) 16800 (d) 13300
4. Find the sum of first 37 odd numbers.
[Hotel Mgmt. 2010]
(a) 1369 (b) 1295 (c) 1388(d) 1875 (e) None of the above
5. Find the sum of first 84 even numbers.
[Bank Clerks 2008] (a) 7140 (b) 7540 (c) 6720 (d) 8832 (e) None of the above
6. Sum of first 15 multiples of 8 is
(a) 960 (b) 660
(c) 1200 (d) 1060
7. The product of four consecutive natural numbers plus one is [CDS 2014]
(a) a non-square
(b) always sum of two square numbers
(c) a square
(d) None of the above
Find the unit digit in the product of (268 x 539 x 826 x 102). [MBA 2009]
(a) 5 (b) 3 (c) 4 (d) 2
(e) None of the above
9. Find the unit digit in the product of (4326 x 5321). [Hotel Mgmt. 2010]
(a) 6 (b) 8 (c) 1 (d) 3 (e) None of the above
10. What is the unit digit in 6817754) ?
(a) 8 (b) 4 (c) 2 (d) 9 (e) None of the above
11. What is the unit digit in
(365 X659X771
(a) 6 (b) 4 (C) 2 (d) 1 (e) None of the above
12. Find the last two-digits of 15x37x63x51x97x17 [IBAC1O2012]
(a) 35 (b) 45 (c) 55 (d) 85
13. How many rational numbers are there between 1 and 1000? [CDS 2012]
(a) 998 (b) 999
(c) 1000 (d) Infinite
14. The sum of 5 consecutive even numbers A,B,C,D andE is 130. What is the product of A and E ? [Bank Clerks 2009]
(a) 720 (b) 616 (c) 660 (d) 672 (e) None of the above
15. The sum of the five consecutive numbers is equal to 170. What is the product of largest and the smallest numbers?
[Bank Clerks 2011]
(a) 1512 (b) 1102 (c) 1152 (d) 1210 (e) None of the above
16. Which of the following numbers always divides the difference between the squares of two consecutive odd integers?
[Bank Clerks 2009]
(a) 7 (b) 3 (c) 8 (d) 6 (e) None of the above
17. A number divided by 56 gives 29 as remainder. If the same number is divided by 8, the remainder will be …
[SSC CCL 2007]
(a) 4 (b) 5
(c) 6 (d) 7
18. On dividing a certain number by 357, the remainder is 39. On dividing the same number by 17, what will be the
remainder?
(a) 5 (b) 3 (c) 7 (d) 6 (e) None of the above
19. A number when divided by 5, leaves 3 as remainder. What will be the remainder when the square of this number is
divided by 5?
(a) 3 (b) 4 (c) 5 (d) 0 (e) None of the above
20. In a question on division with zero remainder, a candidate took 12 as divisor instead of 21. The quotient obtained by
him was 35. Find the correct quotient.
(a) 10 (b) 12 (c) 20 (d) 15 (e) None of the above
21. A number when divided by a divisor leaves a remainder of 24. When twice the original number is divided by the
same divisor, the remainder is 11. What is the value of the divisor? [IB ACIO 2013]
(a) 13 (b) 59
(c) 35 (d) 37
22. The number 58129745812974 is divisible by [CDS 2012]
(a) 11 (b)9
(c) 4 (d) None of these
23. How many numbers between – Hand 11 are multiples of 2 or 3? [CDS 2012]
(a) 11 (b) 14
yoursmahboob.wordpress.comyoursmahboob.wordpress.com
(c) 15 (d) None of these
24. Which one of the following numbers is divisible by 11? [CDS 2013]
(a) 45678940 (b) 54857266
(c) 87524398 (d) 93455120
25. When 17200 is divided by 18, find the remainder.
(a) 1 (b) 4 (c) 5 (d) 3 (e) None of the above
26. What is the remainder when 41000 is divisible by 7? [CDS 2014]
(a) 1 (b) 2
(c) 4 (d) None of these
27. A common factor of (4143 + 4343) and (4143 + 4341)is…
(a) (43 – 41) (b) (4141 + 4341)
(c) (4143 + 4343) (d) (41 + 43) (e) None of the above
28. The remainder when 9 + 6 is divided by 8 is [SSC CGL (Main) 2012]
(a) 2 (b) 3
(c) 5 (d) 7
29. What will be the remainder when 19100 is divided by 20? [SSC CGL (Main) 2012]
(a) 19 (b) 20
(c) 3 (d) 1
30. It is given that (2S2 + 1) is exactly divisible by a certain number. Which of the following is also definitely divisible
by the same number?
(a) (216 + 1) (b) (216-1)
(c)7×213 (d) (296 + 1)
(e) None of the above
31. The number (6x + &x) for natural number x is always divisible by …
(a) 6 and 12 (b) 12 only
(c) 6 only (d) 3 only
(e) None of the above
32. 195
+ 215
is divisible by [CDS 2013]
(a) Only 10 (b) Only 20
(c) Both 10 and 20
(d) Neither 10 nor 20
33. If ‘a’ is a natural number, then the largest number dividing (a – a) is
(a) A (b) 5
(c) 6 (d) 7
(e) None of the above
34. 7 – 4 is exactly divisible by which of the following number? [SSC FCI 2012]
(a) 34 (b) 33
(c) 36 fdj 35
35. If N, (N + 2) and (N + 4) are prime numbers, then the number of possible solutions for N are [CDS 2013]
(a) 1 (b) 2
(c) 3 (d) None of these
36. The smallest positive prime (say p) such that 2
P
– lis not a prime is ^DS 2013]
raj 5 (b) 11
(c) 17 fdj 29
37. If b is the largest square divisor of c and a divides c, then which one of the following is correct? (where, a, b and c
are integers) [CDS 2013]
(a) b divides a
(b) a does not divide b
(c) a divides b
(d) a and b are coprime
38. If re is a whole number greater than 1, then re (re – 1) is always divisible by
[CDS 2014]
(a) 12 (b) 24 (c) 48 (d) 60
39. What is the sum of all positive integers lying between 200 and 400 that are multiples of 7? [IB ACIO 2013]
(a) 8729 (b) 8700 (c) 8428 (d) 8278
40. Consider the following statements
I. To obtain prime numbers less than 121, we are to reject all the multiples of 2, 3, 5 and 7. II. Every composite number
less than 121 is divisible by a prime number less than 11.
Which of the statements given above is/are correct? [CDS 2013]
(a) Only I (b) Only II
(c) Both I and II (d) Neither I nor II
41. Consider the following statements
I. 7710312401 is divisible by 11. II. 173 is a prime number. Which of the statements given above is/are correct? [CDS
2013]
(a) Only I (b) Only II
(c) Both I and II (d) Neither I nor II
42. If & is a positive integer, then every square integer is of the form [CDS 2013]
(a) Only 4k (b) Ak or Ak + 3
(c) 4/c + 1 or 4/c + 3 (d) Ak or Ak + 1
43. Every prime number of the form 3k + 1 can be represented in the form 6m + 1 (where k, m are integers), when
[CDS 2013]
(a) k is odd
(b) k is even
(c) k can be both odd and even
(d) No such form is possible
1. (c)2. (a) 3. (6) 4. (a) We know that,
2 Sum of first n odd numbers = n
Given,
n = 37
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∴ Required sum = (3 if =37×37
= 1369
5. (a) We know that,
Sum of first n even numbers = n {n + 1)
Given, n = 84
∴ Required sum = 84 (84 + 1)
= 84 X 85= 7140
6. (a) first 15 multiple of 8 are
8, 16, 24…..120
So, 8(1,2,3,4…..15
7. (c) Product of four consecutive numbers plus one is always a square Illustration 1 Let four consecutive numbers be 3,
4, 5 and 6.
= (3 x 4 x 5 x 6) + 1 = 361 = (19)z
Illustration 2 Let four consecutive numbers be 9, 10, 11 and 12.
= (9 x 10 x 11 x 12) + 1
= 11881 = (109)2
8. (c) Product of unit digits
= 8X9X6X2 = 864 /. Required digit = 4
9. (a) Product of unit digits =6×1=6 /. Required digit = 6
10. (d) Required digit = Unit digit in (7)754
= Unit digit in {(74
)
188 x 72
} = Unit digit in (1 x 49) = 9
11. (6) Unit digit in 34
= 1 ∴ Unit digit in (34
)
16 = 1
Unit digit in 385 = 3
Unit digit in 659 = 6
Unit digit in 74
= 1
∴ Unit digit in (74
)
17 =1
77l = (74jl7 x f
As,
Unit digit in 73
=3 Unit digit in 771 =3 .”. Required unit digit = Unit digit in
(3x6x3) = Unit digit in 54 =4
12. (a) 15 X 37 X 63 X 51 X 97 X 17
= 255X37X63X51 X 97 = 35X21 [last two digits]
= 735 = 35 [last two digits]
Hence, last two digits of the product is 35.
13. {d) There can be infinite number of rational numbers between 1 and 1000.
14. (c) Let five consecutive even number be A = x, .B = x + 2, C = x+4,
D = x + 6 and E = x + 8 According to the question, X+X+2+X+4+ x + 6+ x + 8 = 130 => 5x+20 = 130
=> 5x = 130-20 = 110
15. (c) Let the five consecutive numbers are x, (x + 1) and (x + 2), (x + 3) and (x + 4). According to the question, x+x+l
+ x+2+x+3+x + 4 = 170
5x+ 10 = 170 => 5x = 160
= 32 Largest number = (x + 4)
= 32 + 4 = 36 /. Required product
= 32×36 = 1152
16. (c) Let the two consecutive odd numbers be (2x + 1) and (2x + 3).
∴ Difference = (2x + 3)2
– (2x + l)2
= (2x + 3 + 2x + 1) (2x + 3 – 2x- 1) = (4x + 4) X2 = 8 (x+ 1),
which is exactly divisible by 8.
17. (6) Let the number be x. According to the question,
x = 56A- + 29 Then,
x = (8 X 7 k) + (8 X 3) + 5 = 8 X (7i + 3) + 5 Therefore, when x is divided by 8, the required remainder = 5
18. (a) Let the given number be (357Jr + 39).
Then,
(357* + 39) = (17 X 21k) + (17 X 2) + 5
= 17x(21i + 2) + 5 .•. Required remainder = 5
19. (6) Let the number be x. According to the question,
x = (5k + 3) On squaring both sides, we get => x
2
= (5k + 3)2
= (25£2
+ 30Jc + 9)
= 5(5£2
+ 6k + 1) + 4 .•. On dividing x by 5, the remainder is 4.
20. (C) Number = 35 X 12 = 420
21. (d) Let the divisor be x and quotient be y. Then, number = xy + 24
Twice the number = 2xy + 48 Now, 2xy is completely divisible by x. On dividing 48 by x remainder is x. x = 48 -11 =
37
22. (a) We know that, a number is divisible by 11 when the difference between the sum of its digit at even places and
sum of digit at odd places is either 0 or the difference is divisible by 11.
So, number is 58129745812974 Sum of digits at odd places
= 4 + 9 + 1 + 5+7+2 + 8 = 36 Sum of digit at even places
7+2 + 8 + 4 + 9 + 1 + 5 = 36 So, the required difference = 36 – 36 = 0 /. The number is divisible by 11.
23. (c) Method I
Following are the numbers between – 11
and 11 which are multiples of 2 or 3?
-10,-9,-8,-6,-4,
-3,-2, 0, 2, 3, 4, 6, 8, 9, 10.
/. The numbers of multiples 2 or 3, between
– 11 and 11 are 15. Method II
Numbers between 0 and 11 which are multiples of 2 or 3
/. Number be 15, including “0”.
24. (d) we know that, if the difference between the sum of digits at even places and sum of digits at odd places is (0),
then the number is divisible by 11. From options.
(a) 45678940
Sum of even places =5+7+9 + 0 = 21 Sum of odd places =4 + 6 + 8 + 4 = 22 Their difference = 22-21*0
(b) 54857266
Sum of even places =4 + 5+2+6 = 17 Sum of odd places = 5+8 + 7+6 = 26 Their difference = 26-17 = 9*0
(c) 87524398
Sum of even places =7+2+3+8= 20 Sum of odd places =8 + 5 + 4 + 9= 26 Their difference = 26-20=6*0
(d) 93455120
Sum of even places =3+5+1 + 0=9 Sum of odd places =9 + 4 + 5+2 = 20 Their difference = 20-9 = 11+11 = 1 So, it is
divisible by 11. 25. (a) We know that, (xm – a
m) is divisible by (x + a), for even values of m. .: (17200
– l
200) is divisible
by (17 + 1). => (17200 – 1) is divisible by 18.
When 17200 is divisible by 18, then the remainder is 1.
26.(C)
27. (d) We know that, when m is odd (xm + am) is divisible by (x + a).
.•. Each one is divisible by (41 + 43). ∴ Common factor = (41 + 43)
28. (d) Required remainder =9 +6
= (l)19 + 6=4
[v8 = 9 – 1, so replaced by 1]
29.(D)
30. (d)
31. (a)
32. (c)
33. (C)
35. (a)
36. (b)
37. (c) Since, b is largest square divisor of c. So, c = bx
(where, x is not a whole square number)
Also, a divides c.
So, a will divide bx.
or a will divide b.
(since, it cannot divide x as it is not a whole
square)
38. (a) If n greater than 1, then n
z
(nz
– 1) is always divisible by 12
39. (a) Least number divisible by 7 and above 200 is 203.
Greatest number divisible of 7 and below 400 is 399.
Total numbers divisible by 7 between 200 to 400 are 29
Now, sum of n terms of AP = — (a + 1) 2
where, a = 203, 1 = 399 and n = 29
40. (c) Both the statements given are correct. As 121 is the square of 11. So, to obtain prime numbers less than 121, we
reject all the multiples of prime numbers less than 11 i.e., 2, 3, 5 and 7. Similarly, every composite number less than 121
is divisible by a prime number less than 11 i.e., 2, 3, 5 or 7.
41. (c) I. Any number in order to get completely divided by 11 must have the difference between the sum of even place
digits and the sum of odd place digits equal to 0 or the multiple of 11. In 7710312401, difference between sum of even
place digits and the sum of odd place digit 0.
So, it is divisible by 11
42. (d) If k is a positive integer, then every square integer is of the form 4i or 4£ + 1, as every square number is either a
multiple of 4 or exceeds multiple of 4 by unity.
43. (b) Every prime number of the form 3k + 1 can be represented in the form 6m + 1 only, when k is even.

## 2 thoughts on “DAILY MATH PRACTICE SET- Number System-HIGHLEVEL-18/01/2022”

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