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TECHGURU CLASSES for SSC-JE/RRB/TTA/ORDINANCE/RAILWAYS
CHAPTER -1 (NUMBER SYSTEM) : VERBAL ABILITY & NUMERICAL
CHAPTER-1 : NUMEBER SYSTEMS

PERSONAL REMARK
 Introduction to Numbers
 The numbers written by the symbols, 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9
are known as digits.
 The numbers which are divisible by 2 much as 2, 4, 6 etc. are known
as even numbers.
 The numbers which are not divisible by 2 such as 1, 3, 5, etc. are
known as odd numbers.
 The numbers 1, 2, 3……. Which are used for counting are known as
the natural numbers.
Thus 528 is a natural number but 0, -3, 14.6 and ¾ etc. are not natural
numbers.
 If zero is included in natural numbers. The set so obtained is known
as whole numbers. For example 0, 1, 2, 3… etc. Are whole
numbers. But -1, -2, 32, 1/9 are not whole numbers.
 The whole numbers positives as well as negative are known as
integers. For example -7, -6, -5,……..0. 1, 2, 3, 4 etc are integers but
5 , 0.32, -1.2 and 1/3 etc. are not integers.
 The numbers, which can be expressed in the form of p/q, where p
and q are integers but q≠0, are known as rational numbers. For
37 2
example 25, -62,
, - , 0 etc. are rational numbers.
125 3
 The number, which cannot be expressed in terminating decimal, is
known as an irrational number. For example 5 , 3 + 2 , etc. are
irrational numbers.
 The numbers, which are either rational or irrational, are known as
real numbers. For example 3, -2, 7 etc. are real numbers.
 A natural number other than 1, is a prime number if and only if it is
divisible by 1 and the number itself. For example 2 a prime number
but 6 is not a prime number.
 The natural number other than 1 is a composite number which is
divisible by 1, the number itself and at least one other number
more. For example 6 is a composite number because it is divisible
by 1, 2, 3 and 6.
Note: 1 is neither prime number nor composite. 2 is only the number
which is prime as well as even number.
 Characteristics of Rational and Irrational Number The sum or product of two rational numbers is always a rational
number.
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CHAPTER -1 (NUMBER SYSTEM) : VERBAL ABILITY & NUMERICAL
 The sum of product of two irrational numbers is some times a
rational number and sometimes an irrational number.

PERSONAL REMARK
 The sum or product of a rational number and an irrational number
is always an irrational number.
 There are infinite of a rational number and an irrational numbers
between two rational numbers or two irrational numbers.
 Test of Divisibility
 A number is divisible by 2 if it ends in zero or in a digit which is a
multiple of 2.
 A number is divisible by 3 if the sum of its digits is divisible by 3.
 A number is divisible by 4 if the last two digits i.e. tens and units are
divisible by 4.
 A number is divisible by 5 if it ends in zero or 5.
 A number is divisible by 6 if it divisible by 2 as well as by 3.
 A number is divisible by 8 if its last three digits i.e., hundreds, tens
and units, are divisible by 8.
 A number is divisible by 9 if the sum of its digits is divisible by 9.
 A number is divisible by 10 if it ends in zero.
 A number is divisible by 11 if, the difference between the sums of
the digits in the even and odd places is zero or a multiple of 11.
SOLVED EXAMPLES
Ex.1. What is the unit digit in the product of 237 × 72 ×
28 × 15?
Sol. The unit digit of the numbers 237, 72, 28 and 15 are
7, 2, 8 and 5 respectively.
And the product of 7, 2, 8 and 5 = 560.
∴ the unit digit in the required product is 0.
Ex. 2 If a number is divided by 84 the remainder is 37. What will be
the remainder if it is divided by 21?
Sol. Since 84 is divisible by 21, therefore, (The given number) will
also
be perfectly divisible by 21. So it is the remainder i.e. 37 which is
to be considered when divided by 21. But we know if 37 is
divided by 21, the remainder will be 16.
By short-cut method
If the first divisor is exactly divisible by the second divisor.
Then required remainder is obtained by dividing the given
remainder by the second divisor.
Here 84 is divisible by 21.
Required remainder will be same as remainder obtained by
dividing 37 by 21. But the remainder 16 is obtained when 37
is divided by 21. Required remainder = 16.
Ex. 3 what will be the sum of all odd numbers from 1 to 50?
Sol. The odd numbers from 1 to 50 are as following:
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CHAPTER -1 (NUMBER SYSTEM) : VERBAL ABILITY & NUMERICAL
1, 3, 5,7, 9, 11, 13, 15, 17, 19, 21, 23, 25, 27, 29, 31, 33,
35, 37, 39, 41, 43, 45, 47, and 49.
Thus, there are 25 odd numbers from 1 to 50 and the sum of all
these odd numbers = n2, where n is the number of odd
numbers = (25)2 = 625

PERSONAL REMARK
Note: In order to find out the number of odd numbers, the last even
number is halved. If the last number is odd. Then next even
numbers from 1 to 50, half of the last even number i.e. 50 is
halved.
Hence there are 25 odd numbers from 1 to 50.
In order of find out the sum of all these odd numbers, the total
number of odd numbers is squared.
Ex. 4 what will be the sum of all even numbers from 1 to 40?
Sol. The even numbers from 1 to 40 are as following:
2, 4, 6, 8, 10, 12, 14, 16, 18, 20, 22, 24, 26, 28, 30, 32, 34,
36, 38 and 40.
Thus, there are 20 even numbers from 1 to 40 and the sum of all
these even numbers = n(n+1), where n is the number of even
Numbers = 20 × 21 = 420
Note: In order to find out the number of even numbers the last
even number is halved.
In order to find out the sum of even numbers the number of
even numbers is multiplied by its next number.
Ex. 5 which of the digits is possible in unit place of the perfect square
number?
Sol.
Since
Square
of
0
=
0
‘’
“
“
1
=
1
“
“
“
2
=
4
”
“
“
3
=
9
“
“
“
4
=
16
“
“
“
5
=
25
“
“
“
6
=
36
“
“
“
7
=
49
“
“
“
8
=
64
“
“
“
9
=
81
Hence on the unit place of a perfect square one of the
following numbers may be possible 0, 1, 4, 5, 6 and 9.
Ex. 6
Sol.
what will be unit digit in the product (3127)173?
173  1 172
∴

 43
4
4
173  1
∴ The remainder in
is zero.
4
∴ The unit digit in the product (3127)173 = (7)0 +1 = 7
Ex. 7
what will be the unit digit in the product 1723 × 2246 ×
339?
23  1
Sol. ∴ The remainder in
is 2, hence the unit digit in 2246 =
4
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21 +1 = 4 and the remainder in
9 1
is 0, hence the unit
4

PERSONAL REMARK
digit in 339 = 30+1 = 3.
∴ The unit digit in 1723 × 2246× 339 = unit digit in 3 × 4 ×
3 is 6.
Ex. 8 In a party each member shakes hand with other members. If
they shook hand 210 times. What is the number of the
members?
Sol. Let the number of members be x
∴ Each member shakes hand (x -1) times
∵ x(x - 1) = 210 or x2- x – 210= 0
or (x - 15)( x + 14) = 0 or x = 15
By Short – cut Method
Since the larger nearest whole squared number to
210 is 225 and
225  15
∴ The number of members = 15.
Ex. 9 How many numbers between 200 and 300 are such which are
divisible by 13?
Sol. The numbers which are divisible by 13 are.
208, 221, 234, 247, 260, 273, 286 and 299.
Thus, there are 8 numbers between 200 and 300,
which are exactly divisible by 13.
By Short – cut Method
Required Number = 300/100 – 200/13
= 23
1
5
- 15 = 23 – 15 = 8
13
13
CLASSROOM PRACTICE SHEET: NUMBER SYSTEM
1. Which of the following numbers will not unit place if any number
is squared? 0, 4 , 2, 1,
(a) 0
(b) 1
(c) 4
(d) 2
2. Sum of the numbers from 1 to 20 is –
(a) 210
(b) 110
(c) 220
(d) 105
3. The difference between the greatest and the least numbers of
eight digits, which begin with 8 and end with 6 is(a) 99999999
(b) 9999990
(c) 10000000
(d) 8000006
4. The difference between the greatest and the least numbers of 5
digits, 0, 2, 3, 6 and 7 using all but once is –
(a) 50953
(b) 95821
(c) 35905
(d) 55953
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CHAPTER -1 (NUMBER SYSTEM) : VERBAL ABILITY & NUMERICAL
5. Which of the following is equal to 115 × 15 ?
(a) 105 × 10 + 115 × 5
(b) 11 × 515
(c) 1151 × 5
(d) 110 × 15 +5 × 15

PERSONAL REMARK
6. The least number of four digits exactly divisible by 235 (a) 1060
(b) 9980
(c) 1935
(d) 1175
7. A number when divided by 296 given a remainder 75. The
remainder when the same number is divided by 37 is(a) 2
(b) 8
(c) 1
(d) 11
8. If 232 +1 is exactly divisible by 641, which one of the following is
also divisible by 641?
(a) 296 + 1
(b) 216 +1
(c) 216 -1
(d) 7.233
9. The square of an odd number is –
(a) always an even number
(b) sometimes even and sometimes odd
(c) always an odd number
(d) always an irrational number
10. A number when divided by 221 leaves remainder 64. What is
the remainder if the number is divided by13?
(a) 0
(b) 1
(c) 11
(d) 12
11. Which is the greatest one of the following numbers?
(a) (2 +2 +2)2
(b) (2 × 2 × 2 ×)2
(b) [(2 + 2)2]2
(d) 43
12. The number of prime numbers between 20 and 60 is–
(a) 7
(b) 8
(c) 10
(d) 9
13. The greatest number of 5 digits divisible exactly by 777 is –
(a) 99654
(b) 99645
(c) 99456
(d) 99465
14. The digit in the blank space of the number 34*7 so that the
number is divisible by 11 will be(a)7
(b) 6
(c) 8
(d) 3
15. A number when divided by 3, 5 and 7 successively in order
gives remainder 2, 3 and 4 respectively. If it is divided by 105,
the remainder will be(a) 15
(b) 38
(c) 59
(d) 71
16. If a*b = a2+b2, then 3 *5 is equal to –
(a) 16
(b) 8
(c) 34
(d) 15
17. If the sum of two numbers is 97 ,if the sum of two numbers is
97 and their difference is 37, their product is(a) 8040
(b) 2010
(c) 2378
(d) 1736
18. The unit’s digit in the sum (264)102 + (264)103 is –
(a) 0
(b) 4
(c) 6
(d) 8
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CHAPTER -1 (NUMBER SYSTEM) : VERBAL ABILITY & NUMERICAL
19. 461 + 462 + 463 + 464 is divisible by –
(a) 3
(b) 11
(c) 13
(d) 17

PERSONAL REMARK
20. How many numbers between 1000 and 5000 are exactly
divisible by 225?
(a) 16
(b) 18
(c) 19
(d) 12
21. If the first 400 numbers are written down and those divisible by
2 are deleted and again those divisible by 5 are struck off, how
many numbers are left out?
(a) 150
(b) 170
(c) 200
(d) 160
22.The smallest natural number by which the product of 3
consecutive even numbers is always divisible is –
(a) 16
(b) 32
(c) 48
(d) 96
23. Let m and n be the two whole numbers. If mn= 25, the value of
nm is –
(a) 4
(b) 10
(c) 25
(d) 32
24. The product 459 ×4* × 787 × 483 has 7 at its unit place. What
should be placed at *?
(a) 5
(b) 1
(c) 7
(d) 3
25. How many digits are required to write numbers between 1 and
100?
(a) 196
(b) 188
(c) 192
(d) 200
26. How many of prime factors of (6)10 × (7)17 × (55)27 is –
(a) 54
(b) 64
(c) 81
(d) 91
27. The unit digit in (2467)153 is –
(a) 7
(b) 3
(c) 9
(d) 1
28. which one of the following is always odd?
(a) Sum of two odd numbers
(b) Difference of two odd
(c) Product of two odd numbers
(d) None of these
29. Which of the following can be a product of two three digit
numbers × × 3 and × × 8 ?
(a) 1010024
(b) 991014
(c) 9124
(d) can not be determined
30. A prime number is the number which is divisible by?
(a) 1
(b) 2
(c) itself
(d) 1 and itself only
31. Mohan engaged a servant on the condition that he would pay
him Rs. 200 and a set of clothes after 10 days of his service. The
servant left the job after 5 days only. Mohan paid him Rs. 20 in
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CHAPTER -1 (NUMBER SYSTEM) : VERBAL ABILITY & NUMERICAL
cash and a set of clothes as his proportionate wages. The price
of the set of clothes is(a) Rs. 80
(b) Rs. 120
(c) Rs. 140
(d) Rs. 160

PERSONAL REMARK
32. A number being successively divided by 9, 11 and 13 leaves 8, 9
and 8 remainders respectively. If the order of divisors is reversed,
then remainders will be(a) 8, 9, 8
(b) 9, 8, 8,
(c) 10, 1, 6
(d) 10, 8, 9
33. Which one of the following numbers is multiple of 11?
(a) 978626
(b) 112144
(c) 447355
(d) 869756
34. Four prime numbers are written in ascending order of their
magnitudes the product of the first three is 385 and that of the
last three is 1001. The largest given prime number is(a) 11
(b) 13
(c) 17
(d) 19
35. A number divided by 19, gives the quotient 19 and remainder 9.
The number is(a) 370
(b) 352
(c) 361
(d) 371
ANSWERS WITH EXPLANATORY NOTES
1.
2.
(d)
(a) Sum of numbers from 1 to 20
= 20 × 21/2 = 210
n  (n  1)
2
Where n is known number of first natural numbers.
Note: Sum of first natural numbers =
4.
(b)Greatest number of the given digits.
= 89999996 – 800000006 = 9999990
(d) Greatest number of the given digits
= 76320
And the least
” “ = 20367
5.
(d) 115 × 15 = (110 + 5) × 15= 110 × 15 +5 × 15
6.
(d) The least four digit number is 1000
∴ on dividing 1000 by 235
235)1000(4
940
60 is the remainder
∴ Reqd. number = 235 × 5= 1175
(c) Let the number be A,
A = 296 k + 75 = 37 × 8k + 37 × 2 + 1
= 37 (8k + 2) +1
Hence remainder is 1, when the number is divided by 37.
3.
7.
8.
9.
(a) ∴ 296 +1 = (232)3 + (1)3 = (232 + 1)[264 – 232 + 1]
If 232 +1 is exactly divisible by 641, then 296 +1 will also
be divisible by 641.
(c)
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10.
(d) Since the first divisor 221 is exactly divisible by the second
divisor 13, therefore, the reqd. remainder will be same as the
remainder obtained after dividing the remainder 64 by the
second divisor 13.
∴ required remainder = 12
11.
(c) ∵ ( 2 + 2 + 2)2 = 36
(2 × 2 × 2)2 = 64
[(2 + 2)2]2 = (16)2 + 256
43 = 64
Hence [(2 + 2)2]2 is the greatest.

PERSONAL REMARK
12. (d) The prime numbers between 20 and 60 are 23, 29, 31, 37,
41, 43, 47, 53 and 59.
Hence, the no. of prime numbers between 20 and 60 = 9.
13. (c) The greatest five digit number is 99999
∴ On dividing
777)99999(128
777
2229
1554
6759
6216
543
∴ The required number = 99999 - 543
= 99456
14. (c) In order to make the number divisible by 11, the sum of 7
and 4 must be the same as the sum of 3 and missing digit.
Hence the missing digit is 8.
15. (d) Required remainder = First remainder + (second remainder
× Fist divisor) + (Third remainder × Fist divisor × Second
divisor)
= 2 + (3 × 3) + (4 × 3 × 5)
= 2 + 9 + 60
= 71
16. (c) ∵ a*b = a2 + b2
∴ – 3*5 = (–3)2 + (5)2 = 34
17. (b) ∵ 4 xy = (x +y)2 – (x – y)2
∴
(x  y) 2  (x  y)2
4
2
(97  37)(97  37)
(97)  (37)2
=
=
4
4
134  60
=
= 134 × 15 = 2010
4
xy =
18. (a) The remailer in 102– 1/4 is 1.
∴ The unit digit in (264)102 is the same as the unit digit in
(4)1 +1 is 6.
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The remainder in 103 – 1/4 = 2.
∴ The unit digit in (246)103 is the same as the unit digit in
(4)1+2 = 4.
Hence unit digit in sum of (264)102 + (264)103 →6+4 → 0

PERSONAL REMARK
19. (d) 461 + 462 +463+464 = 461 (1 + 4 + 42 +43)
= 461 × 85 = 461 × 5 × 17
∴ The given expression is divisible by 17.
20. (b) The number which are divisible by 225 and are between
1000 and 5000, are as
50
50
(5000/225)-(1000/225) = 22
= 22 – 4 = 18
4
225
225
21. (d) In numbers from 1 to 400 the numbers divisible by 2
= 200
And “ “
by 5
= 80
And “ “
10 = 40
Since the numbers divisible by 10, are common in those
numbers which are divisible by 2 and 5.
∴ Numbers actually divisible by 2 and 5.
= 200 +80 – 40 = 240
∴Remaining number which are not divisible by 2 and 5
= 400 – 240= 160
22. (a) Let the three consecutive even numbers be 2, 4 and 6.
∴ Product of these numbers = 2 × 4 × 6 = 48
Now this product can be divisible by 16 and 48 among the
given numbers. But among these two 16 is the smallest
natural number.
∴ Product of three even consecutive no. always divisible by
16.
Note: You can take any three even consecutive numbers.
Hence, the smallest reqd. number = 48.
23. (d)
mn
= 25
= (5)2
∴m = 5 and n = 2
∴ nm = 25 =32
24. (d) Since the unit digit of the product 459 × 4 * × 787 × 483 is
the same as of the product 9 × * × 7 × 3 and same as of the
product 189 × * and same as k of the product of 9 × *. But it
is 7.
Hence, * should be replaced by 3.
25. (c)No. of digits required to write from
1 to 9 = 9
“
“
“ from 10 to 99 = 180
And
“
“
100
=3
∴ Total no. of digits required
= 9 + 180 + 3 = 192
26. (d) (6)10 × (7)17 × (55)27
= 210 × 310 × 710 × 5 × 1127
∴ The number of prime factors
= 10+ 10+17 +27 = 91
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27. (a) The unit digit in
(2467)2 = 9
“
“
“
(2467)3 = 3
“
“
“
(2467)4 = 1
“
“
“
(2467)5 = 7
“
“
“
(2467) 6 = 9
Thus, the unit’s digit is repeated, then where is a difference
of 4 in power.
∴ The unit’s digit in (2467)150 = 9
And “
“
“(2467) = 7

PERSONAL REMARK
By Short –cut Method
The unit digit in (2467)153
153  1 152

 38
4
4
There for remainder = 0
There units digit = 70 +1 = 7
28. (c)
29.(b)
30. (d)
31. (d) Let the price of the set of clothes be Rs. x.
∴ salary for 10 days = Rs. (x + 200)
(x  200)  5
x  200
And Salary for 5 days =
=
10
2
x  200
∴
= x + 20
2
or x + 200 = 2x + 40 ∴ X = Rs. 160
32. (c) True remainder = 8 + (9 × 9) + (8 × 9 × 11)= 8 + 81 + 792
= 881
13 881
11 67 – 10
∴ New remainders are 10, 1, 6
9 6–1
0-6
33. (a) In 978626
The sum of the digits in the odd places
= 6 + 6 + 7 =19
And the sum of the digits in the even places
= 2 + 8 +9 = 19
∴ 978626 is multiple of 11
34. (b) In the product of first three numbers and the product of last
three prime numbers. The product of second third number is
common which the H.C.F. of the given products is.
∴ H.C.F. of 385 and 1001 = 77
∴ Largest number = 1001/ 77
35. (a) The number = 19 × 19 + 9 = 361 + 9 = 370
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