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Reasoning RRB TELUGU
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2. TEST OF DIVISIBILITY
These test help us to know whether a given number however large it maybe is divisible
by a certain division without going into the process of division.
1. Test of divisibility by "2": Any number whose last digit is 0, 2, 4, 6, or 8 is divisible by 2.
Ex: The numbers 72, 908, 1426 are divisible by 2.
2. Test of divisibility by "3": A number is divisible by '3' if the sum of it's digits divisible
by 3.
Ex: 126, 147, 198....
3. Test of divisibility by '"4": A number is divisible by 4 if the last two digits of a number
divisible by 4.
Ex: 124, 184, 864, 1528.....
4. Test of divisibility by "5": If a number ends in '5' (or) '0' the number is divisible by 5.
Ex: 120, 180, 175, 225, 290, 350, 520, 1000, 1550, 1900....
5. Test of divisibility by "6": If a number is divisible by both 3 & 2 the number is also divisible
by '6'.
Ex: 126, 198, 864, 1260....
6. Test of divisibility by "7": If a number is divisible by '7' if its unit digit is multiplied with
2 and sobtracted from the remaining number its result is divisible by '7' the number is divisible
by '7'.
Ex: 1. Find 343 is divisible by '7' (or) not-
Ans: 343 = 342 3 = 36 6 = = 4 So the number 343 is divisible by 7
2. Find 2961 is divisible by 7 (or) not-
Ans: 2961=296 2(1) = 294
294 = 29 2(4) = = 3 So, 2961 is divisible by 7
7. Test of divisibility of "8": If the last three digits of a number is divisible by '8' then the
number is also divisible by '8'.
8. Test of divisibility by "9": If the sum of all the digits of a number is divisible by "9" the
number is also divisible by '9'.
Ex: 387, 549, 657...
217
287
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9. Test of divisibility by "10": Any number which ends with zero it is divisible by '10'.
Ex: 30, 90, 150, 470, 890, 1000, 1500, 1860...
10. Test of divisibility by "11": A number is divisible by 11, if the difference of the sum of its
digits at odd places and the sum of its digits at even places is either "0" (or) a number divisible
by '11'.
Ex: Show that 4832718 is divisible by 11 or not -(sum of digits at odd places) - (sum of digits at even places)
= [ 8 + 7 + 3 + 4] = [8 + 2 + 1] = 22 - 11 = 11, which is divisible by 11.
11. Test of divisibility by "12": Any number which is divisible by both 3 & 4 it is also divisible
by 12.
Ex: 240, 312, 612, 886......
Problems on Test of divisibility:
1. What least value must be given to * so that the number 451*603 is exactly divisible by "9"?
a) 8 b) 9 c) 2 d) 6
Ans: According rule
Sum of the digits must be divisible by "9". Sum of digits = [19 + *]
least value of * is "8".
2. What is the least value of "k" so that the number 6735k1 is divisible by "9".
Ans: 6 + 7 + 3 + 5 + k + 1 = (22 + k)
The least number greater than "22" and divisible by "9" is 27.
27 = 22 + k k = 5
3. For what value of 'k' the number 7236k2 is divisible by "8".
a) 3 b) 7 c) 3&7 d) 3(or)7Ans: According rule.
The last three digits 6k2 is divisible by '8' if k is 3 or 7.
4. 5x2 is a three digit number with 'x' as a missing digit. If the number is divisible by '6' then
the missing digit is-
a) 3 b) 6 c) 7 d) 2
Ans: The given number must be divisible by 2 as well as 3.
5 + x + 2 = 7 + x = must be divisible 3,
7 + 2= x = 2
3
19+0
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Rules on Natural Numbers
Formulae:1) Sum of First 'n' natural numbers : 1 + 2 + 3 +.......n,
2) Sum of squares of 'n' natural numbers: 12 + 22 + 32 + .... n2,
3) Sum of Cubes of 'n' natural numbers: 13 + 23 + 33 + .... nn
4) Sum of first 'n' even natural numbers: 2 + 4 + 6 + .... n,
n (n + 1)
5) Sum of First 'n' odd natural numbers: 1 + 3 + 5 + 7 + .... n.
n2
Problems:1) Find sum of first 100 natural numbers.
a) 5,000 b) 5,050 c) 5,500 d) 6,500
Ans: sum of first 'n' natural numbers
1 + 2 + 3 + ......n
1 + 2 + 3 + .....100
2) The sum of squares of first 10 natural numbers is -
a) 300 b) 385 c) 365 d) 400
Ans: 12 + 22 + 32 +...... 102, n = 10
100(100 1)5,050
2n =
+=
n (n + 1)n =2
2
3n (n + 1)
2n =
2 n (n + 1) (2n + 1)n =6
n (n + 1)n =2
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= 385
3) Find the sum of first twenty multiples of 12.
a) 2,000 b) 2,500 c) 2,520 d) 2,600
Ans: 12 1 + 12 2 + 12 3 + .... 12 20
= 12 [1 + 2 + 3 + ....20] = 12
4) Sum of all odd numbers upto 100 is-
a) 2,000 b) 2,500 c) 2,600 d) 3,000
Ans: From, 1 to 100 there are
50 even numbers
50 odd numbers
There fore, 1 + 3 + 5 + .... 50
Sum of 50 odd numbers = n2
= (50)2 = 2,500
5) Given that (12 + 22 + 32 +..... 102) = 385, then the value of (22 + 42 + 62 +.... 202) is -
a) 770 b) 1,540 c) 1,155 d) (385)2
Ans: 22 + 42 + 62 + ... 202
It can be written as follows
(2 1)2 + (2 2)2 + (2 3)2 + ... (2 10)2
Take 22 common
= 22 [12 + 22 + 32 + ..... 102]
= 4[385] = 1,540
20 (20 + 1) 12 20 21)= =2,520
2 2
n (n + 1) (2n + 1) 10 (10 + 1) (2 10 + 1)2n = =6 6
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6) The sum of [51 +52 + 53 +.... 100] is -
a) 3,700 b) 3,775 c) 3,600 d) 3,800
Ans: [Sum of first 100 natural numbers] [Sum of first 50 natural numbers]
= [1 + 2 + 3 + .... 100] [1 + 2 + .... 50]
= [50 101] 25 (51)
= 25 [2 101 51]
= 25 [2 101 51] 25 [202 51]
= 25
151
= 3,775
7) Sum of cubes of first ten natural numbers is -
a) (50)2 b) (55)2 c) (28)2 d) (52)2
ans: 13 + 23 + 33 + ....103
2 2 2n (n + 1) 10 (10 + 1) 10 113 2n = = = = (55)
2 2 2
100 (100 + 1) 50 (50 + 1)= -
2 2
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Ratio and Proportion
Ratio: "Ratio is a relationship between two numbers of the same kind"
To express the ratio use symbol " : "
Ex: 1) The ratio of two weights of rama to that of krishna is 16 : 23 read of 16 is to 23
2) The ratio a to b is the fraction in a/b it is writen as a:b
In a:b, a = antecedent b = consequent
Rule:
i) The ratio exists b/w Quantities of the same kind
ii) They express in the same units
iii) Ratio being a fraction has no units
iv) A Ratio doesn't alter if its first and second terms are multiplied (or) divided by
the same non-zero number.
Ratio in simplest form:
" A ratio a:b is said to be in simplest form the HCF of a and b divide each of its terms
a&b.
Ex: 14:16
Simplest form 7:8
Inverse Ratio:
"In a given ratio interchanging of the antecedent and consequest is known as inverse
ratio"
Ex: i) The inverse ratio of a:b is b:a (also known as Reciprocal ratio)
a = b denotes Ratio of eaqulity.
For the ratio a:b a2:b2 is called "duplicate ratio"
For the ratio is called "sub-duplicate ratio"
For the ratio a:b a3
:b3
is called Triplicate ratio
For the ratio "sub-triplicate ratio3 3a : b a : b
a : b a : b
14 16:
2 2
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Compounded ratio:Ratio's are compounded by multiplying the fraction which denote
Ex: a:b and c:d are two ratios
Its compounded ratio
Compound Ratio:
Theorem II:
The ratio of first and second quantity = A: B
The ratio of second and third quantity = C: D
The ratio of first, second & third quantity
I : II : III = (A B) : (B C) : (B D)
Ex: If A:B = 3 : 4 and B : C = 8 : 9 find A:B:C
A:B:C = 24 : 32 : 36 = 6 : 8 : 9
Theorem II:
The ratio of first and second quantity = A:B
The ratio of second and third quantity = C:D
The ratio of third and fourth quantity = E:F
a cac :bd
b d
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I Qty : II Qty : III Qty
A B
C D
A : B : C
3 4
8 9
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I:II:III:IV = (ACE) : (BCE) : (BDE) : (BDF)
Ex: If A:B = r : 15, B:C = 5:8 and C:D = 4:5 then A:B:C:D?
A:B:C:D = (854) : (1554) : (1584) : (1585)
= 160 : 300 : 480 : 600
= 8 : 15 : 24 : 30
Theorem III
"If two nembers are in the ratio of a:b and their sum is "x" then these numbers will be"
Theorem IV
If three numbers are in the ratio of a:b:c and the sum of these numbers is 'x' these numbers
will be..a b c
: : r espectivelya b c a b c a b c
x x x + + + + + +
aFirst number
a b
bSecond number
a b
=+
=+
x
x
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I Qty : II Qty : III Qty IV Qty
A B
C D
E F
A : B : C : D
8 15
5 8
4 5
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Examples:
1) a:b = 2:3, b:c = 4:5 then a:b:c=?
Ans: a:b:c = 8:12:15
2) x:y = 3:5, y:z = 2:3 then x:y:z = ?
sol: 3 : 5
2 : 3
6 : 10 : 15
Ans: 6 : 10 : 15
3)
sol:
2 : 3
4 : 5
8 : 12 : 15
Ans: a:b:c = 8:12:15
4) a = 2b, b= 3c, then a:b:c = ?
sol:
2 : 1
3 : 1
6 : 3 : 1 Ans. a:b:c = 6 : 3 : 1
a 2a 2b a : b : 2:1
b 1
b 3b 3c b : c 3 :1
c 1
= = =
= = =
2 a 2a b, b a : b : 2: 3
3 b 3
4 b 4b c b : c 4 : 55 c 5
= = =
= = =
2 4a b, b c, a : b : c ?
3 5= = =
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a b 2 : 3
b c 4 : 5
ab:bb:bc 8 : 12 :15
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5) If 2a = 3b, 4b = 5c, then a:b:c:
sol:
3 : 2
5 : 4
15 : 10 : 8 Ans. 15 : 10 : 8
6) x = 3y = 5z then x : y : z =?
sol:
3 : 1
5 : 3
15 : 5 : 3 Ans. x : y : z = 15 : 5 : 3
7)
sol: 5 : 6 : 7 (answer is same as denominators)
a b c: : then a : b : c
5 6 7=
x 3x 3y x : y 3:1
y 1
y 53y 5z y : z 5 :3
z 3
= = =
= = =
a 32a 3b a : b : 3 : 2
b 2
b 54b 5c b : c 5 : 4
c 4
= = =
= = =
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Problems Based on Quotient, Remainders
Quotient:
A quotient is the result of division. When 'N' is divided by "x" we find a quotient..
Ex: If 6 is divided by 3, the quotient is "2"
This quotient gives number of numbers less than a given number "N"
Problems:
1) Find the number of numbers upto 500 which are divisible by 13.
a) 36 b) 37 c) 38 d) 39
Ans: Divide 500 by 13
13) 500 ( 38
39
110
104
6
The quotient is "38". It means, 38 numbers less than 500 which are divisible by 13.
2) How many numbers upto 100 are divisible by 7?
a) 14 b) 67 c) 93 d) 100
Ans:
7) 100 ( 14
7
30
28
2 The quotient is "14"
3) How many numbers upto 500 are divisible by 23?
a) 23 b) 27 c) 21 d) 19
Ans: 23) 500 ( 21
46
40
23
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17
The quotient is "21"
4) How many numbers upto 200 are divisible by 2 and 3 both?
a) 35 b) 33 c) 29 d) 27
Ans: L.C.M. of 2 & 3 is "6"
6) 200 (33
18
20
18
2
The quotient is 33
5) How many numbers between 100 and 300 are divisible by "11" ?
a) 22 b) 21 c) 20 d) 18
Ans:
11) 100 (9 11) 300 (27
99 22
1 80
77
3
Between 300 and 100, there are 18 numbers (27 - 9 = 18)
6) How many numbers between 300 and 700 are divisible by 2, 3 and 7 together ?
a) 9 b) 8 c) 10 d) 11
Ans: L.C.M of 2, 3 & 7 is "42"
42) 300 (7 42) 700 (16
294 42
6 280
252
28
9 numbers divisible by 42 (16 - 7 = 9)
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7. What is the sum of all the numbers between 3,000 and 5,000 which are divisible by 563?
a) 11,723 b) 11,833 c) 11,823 d) 11,923
Ans: The numbers just more than 3000 and divisible by 563 is
563 6 = 3,378
563 7 = 3,941
563 8 = 4,504
11,823
8) Find the sum of all the numbers upto 10,000 which are divisible by 563.
a) 86,139 b) 78,479 c) 73,671 d) 98,711
Ans: Number of numbers below 10,000 divisible by 563 is "17"
563) 10000 (17
563
4370
3941
329
Sum of such numbers
= 563 (1 + 2 + 3 +....17 )
= 563 17 (17 + 1)
2
= 86,139
9) What is the sum of all the numbers between 300 and 1,000 which are divisible by 179?
a) 2,517 b) 2,527 c) 2,607 d) 2,506
Ans:
179 2 + 179 3 + 179 4 + 179 5
179 (2 + 3 + 4 + 5)
179 14 = 2506
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Remainders
The remainder or residue is the amount "left over" after performing the division of two
integers.
Problems:
1. A certain number 'X' when divided by 51 leaves a remainder 26 what is the remainder if the
number X is divided by 17?
a) 6 b) 7 c) 8 d) 9
Ans: X = 51Q + 26
X = 3 17Q + 17 + 9
X = 17 (3Q + 1) + 9
Remainder = 9
2. A number when divided by 119 leaves 19 as remainder if the same number is divided by 17
the remainder obtained.
a) 2 b) 3 c) 1 d) 5
Ans: Let, the Number = x
Quotient = qDivisor = (d) = 119
Remainder = (r) = 19
x = (d q) + r
x = (119 q) +19
Same number when divided by "17"
x = (17 7 q) + 17 + 2
= 17 (7q + 1) + 2
Remainder is "2"
3. A certain umber when divided by 39 leaves a remainder 20, what is the remainder when the
same number be divided by 13?
a) 7 b) 11 c) 0 d) 5
Ans: Let the number = x
Divisor (d) = 39
Remainder (r) = 20
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x = (d q) + r
x = (39 q) + 20
When divided by "13"
x = 13 3 q + 13 + 7
x =13 (3q + 1) + 7
Remainder = 7
4. When "N" is divided by "4" the remainder is '3' what is the remainder when "2N" is divided
by 4?
a) 1 b) 2 c) 3 d) 4
Ans: When 'N' is divided by 4 the remainder is '3'
Q = Quotient
N = 4Q + r N = 4Q + 3
When '2N' is divided by 4..
2N = 2 (4q + 3) 2N = 8q + 6
2N = 4 (2q + 1) + 2
Remainder = 2
5. What least number must be subtracted from 6,500 to get number exactly divisible by 135?
a) 10 b) 15 c) 20 d) 25
Ans: On dividing 6,500 by 135
135) 6500 (98
540
1100
1050
20 Remainder
If "20" is substracter from 6,500 it is divisible exactly by 135.
6. Find the number which is nearest to 3,105 and exactly divisible by 21.
a) 3,106 b) 3,108 c) 3,110 d) 3,111
Ans: On dividing 3,105 by 21, remainder obtained '18'
Number to be added (21-18) = 3
Required number = (3105 + 3) = 3108
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7. What least number must be added to 3,000 to a number exactly divisible by 19?
a) 1 b) 2 c) 3 d) 4
Ans: 3,000 19 remainder = 17
Number to be added = (19 - 17) = 2
8. A number 'X' when divided by 73 gives a quotient 50 and a remainder one tenth of the
quotient the number 'X' is -
a) 3,500 b) 6,000 c) 3,655 d) 4,050
Ans:
Divisor (d) = 73
Quotiant (Q) = 50
Remainder (r) =
X = [(d q) + r]
X = [73 50) + 5]
X = 3655
Complete Remainder:
"A remainder obtained by dividing a given number by the method of successive division is
called complete remainder"
Ex: Divide 132 by 35
35) 132 (3
105
27
d1
5 132
d2
7 26 2 r1
3 5 r2
Complete remainder = d1 r2 + r1
= (5 5) + 2
= 27
Q 50=
10 10= 5
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Problems:
1. A certain number when successively dividided by "8" and "11" leave remainder 3 and 7
respectivly. Find the complete remainder.
a) 57 b) 58 c) 59 d) 60
Ans:
Let, the number 'X'
d1
8 X
d2
11 - 3 r1
- 7 r2
Complete remainder = d1
r2
+ r1
= (8 7) + 3
= 59
2. A certain number when sucessively divided by 3 and 7 it leaves remainders 1 and 3
respectively. Find the complete remainder.
a) 7 b) 8 c) 9 d) 10Ans:
d1
= 3 d2
= 7
r1
= 1 r2
= 3
Complete remainder = d1
r2
+ r1
= 3 3 + 1 = 10
3. A certain number when sucessively divided by 2, 3 and 5 leave remaindery 1, 2 and 3
respectively. Then what is the complete remainder?
a) 20 b) 21 c) 22 d) 23
Ans:
d1
= 2 d2
= 3 d3= 5
r1
= 1 r2
= 2 r3= 3
Complete remainder = d1 d2 r3 + d1 r2 + r1
= 2 3 3 + 2 2 + 1
= 18 + 4 + 1 = 23
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4. A number when sucessively divided by 3, 4 and 5 leaves remainders 1, 2 and 3 respectively.
if the last quotient is 3, then the number is..
a) 221 b) 222 c) 223 d) 230
Ans:
Let the number be "2"
X = (5 3) + 3 = 18
Y = 4x + 2 = 4 (18) + 2 = 74
Z = 3Y + 1 = 3 (74) +1 = 223
5. Find that number which when successively divided by 7, 5 and 4 leaves remainder 1, 2, 3
respectively the lest quotiant being the sum of the remainders
a) 900 b) 950 c) 960 d) 970
Ans: Let the number be 'Z'
X = 4 6 + 3 = 27
Y = 5x + 2 = 5 (27) + 2 = 137
Z = 7Y + 1 = 7 (137) + 1 = 960
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7 Z
5 Y 1
4 X 2
6 3
3 Z
4 Y-1 1
5 X-2 2
3-3 3
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1. NUMBER SYSTEM
Digit: Collection of certain symbols (or) figures is called Digit. The name "digit" comes from
the fact that the 10 digits (ancient Latin digiti meaning fingers) of the hands correspond to the
10 symbols of the common base 10 number system.
Ex: The ten digits of the numerals are.. 0, 1, 2, 3, 4, 5, 6, 7, 8, 9.
Numbers: A number is a mathematical object used to count, label and measure. Numbers are
formed with the digits. The numbers include such as 0, negative numbers, rational numbers,
irrational numbers, and complex numbers.
Ex: 23, 47, 86, 154, 542, 620, 3540, 8692, 5682, etc....
Facevalue: The value of digit itself is called face value.
Ex: In 8,642 The face value of '2' is 2, face value of '6' is 6
Place value: The place value of a given digit in a given number begins from the extreme
right.
Ex: The number 987654321 is represented as shown below.
10 Crore Crore 10-L Lakh 10-Th Thousands Hundreds Ten's Units place
108 107 106 105 104 103 102 101 100
9 8 7 6 5 4 3 2 1
Ex: Place value of 5 is 510 4 = 50000
Place value of 2 is 2104 = 20000
* 10 lakh = 1 million
* 10 million = 1 crore
* 100 million = 10 crore
* 100 crore = 1 billion
* 1 google = 10100
* 1 google plex = 1010100
* Mahasamudram = 1052
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* Asankhya = 10140
* Tallakshana = 1053
* Anuyoga formula = 296 (number of living things)
* The place value of a number increase 10 times when it moves from right to left.
* The place value of number decreases 10 times when it moves from left to right.
* The largest single digit = 9 * The smallest single digit = 0
* The largest double digit = 99 * The smallest double digit =10
* The largest three digit number = 999 * The smallest three digit number = 100
Various types of numbers:
1. Natural numbers 2. Whole numbers 3. Positive integers
4. Negative integers 5. Integers 6. Even numbers
7. Odd numbers 8. Prime numbers 9. Composite numbers
10. Twin primes 11. Co-prime numbers 12. Perfect numbers
1. Naural numbers (N): Counting numbers starting from '1' are called natural numbers.
Ex. N = {1, 2, 3, 4, 5 .......}
2. Whole numbers (W): All natural numbers together with zero is called whole numbers.
Ex. W = {0, 1, 2, 3, 4 .....}
3. Positive integers (I+): The natural numbers are also called positive integers.
Ex. I+ = {+1, +2, +3, +4 .....}
4. Negative integers (I-
): The set of negative numbers is called negative integers.
Ex. I-= {-1, -2, -3, -4 .....}
5. Integers (I): The set of 'I' of all natural numbers, 0, and negative numbers called set of
integers.
Ex. I = { ...-4, -3, -2, -1, 0 1, 2, 3, 4 ....}
6. Even number: Integers divisible by two are called even numbers.
Ex. {..... -8, -6, -4, -2, 0, 2, 4, 6, 8 ...}
General formula = 2N (N = Natural number)
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7. Odd numbers: Integers not divisible by 2 are called odd numbers.
Ex. { ..... -7, -5, -3, -1, -0, -1, 3, 5, 7 .....}
General formula = 2N + D
8. Prime numbers: A number which has '1' and itself only as factors is called prime number.
Ex: 2, 3, 5, 7, 11, 13, 17.....
Primes in 100 Natural numbers
Code Natural numbers Primes No.of primes
D 1-10 2, 3, 5, 7 4
D 11-20 11, 13, 17, 19 4
B 21-30 23, 29 2
B 31-40 31, 37 2
C 41-50 41, 43, 47 3
B 51-60 53, 59 2
B 61-70 61, 67 2
C 71-80 71, 73, 79 3
B 81-90 83, 89 2
A 91-100 97 1
Total 25
Number of primes can be found with the help of the code "DD BBC BBC BA"
DD Delhi Doordarshan
BBC British Broad Costing Company
BA Bachelor of Arts
Ex: 1. How many primes are there in less than 100 natural numbers? Ans: (b)
a) 20 b) 25 c) 28 d) 30
2. How many numbers of primes between natural numbers 11 to 80? Ans: (c)
a) 16 b) 17 c) 18 d) 19
D + B + B + C + B + B + C = 4 + 2 + 2 + 3 + 2 + 2 + 3 = 18
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3. How many primes between the natural numbers 61 to 80? Ans: (d)
a) 2 b) 3 c) 4 d) 5
C + B 2 + 3 = 5
4. The least prime number is? Ans: (b)
a) 1 b) 2 c) 3 d) 4
5. Which one of the following only prime numbers which is even also? Ans: (b)
a) 1 b) 2 c) 4 d) 6
Test for checking primes greater than "100" natural numbers
Let, x = given number,
"k" be an integer very near to x such that k > xIf 'x' is not divisible by any of the numbers less than k, then 'x' is prime otherwise it is not
a prime.
Ex: Test 191 is prime (or) not..
Ans: let, x = 191
The primes less than 14 are.. 2, 3, 5, 7, 11, 13
191 is not divisible by any of the above. So 191 is a prime number.
Ex: Test 104 is Prime or not..
Ans: Let x = 104
Primes less than 11 are..2, 3, 5, 7. '104' is divisible by '2'. So 104 is not a prime number.
* The largest prime number known so far is (222811) which is a number of about 700 digits.
Formulae related to prime numbers
I. Euclidians formula: If one is added to the product of consecutive primes starting from 2 then
the resultant number is a prime number.
Ex: i. 2 3 + 1 = 7 (prime number)
121 11
11 104
k = =
>
196 14
14 191
k = =
>
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ii. 2 3 5 + 1 = 31 (prime number)
iii. 2 3 5 7 + 1 = 211 (prime number)
II. Leonards formula (or) Eulers formula: If n < 40 then (n2 n + 41) is a prime number.
III. Fermat's formula: The numbers in the form of (22n+1) are prime numbers but it is true for
n = 1, 2, 3, 4
Note: All prime numbers other than 2 are odd numbers but all odd numbers are not prime
numbers.
9. Composite numbers: "The natural numbers which have factors other than "1" and
themselves are called composite number.
Ex: 4, 6, 9, 10, 12, 15, .....
9 = 3 3 other than 1 and 9 10 = 2 5 other than 1 and 10
* "1" is neither prime nor composite number.
* A composite number may be even (or) odd.
* Prime number has two factors.
* Composite number has minimum three factors.
10. Twin primes: If the difference of two prime numbers is 2 then they are called twin primes
Ex: (3, 5); (5, 7); (11, 13); (17, 19); (29, 31), ....
11.Co-primes: "If two numbers have only 1 as common factor then the numbers are called co-
primes. (or) If h.c.f of two numbers is "1" then they are called co-primes.
Ex: (3,15), (4,19)....
12. Perfect numbers: "If the sum of factors of number except itself is equal to the same
number then the number is called perfect number".
Ex: 6, 28, 496...
6 1, 2, 3 1 + 2 + 3 = 6
28 1, 2, 4, 7, 14 1 + 2 + 4 + 7 + 14 = 28
* Successor of a number is - One more than the number
Ex: The successor of 99 is 100
The successor of 999 is 1000
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The successor of 40299 is 40300
* Predecessor of a number: One less than the number
Ex: The predecessor of 40,300 is 40,299
* Two smallest 3-digit number formed by using the digits 6, 3 & 8 is..
Ans: The required number is 3,689
* The smallest 3 digit number formed by using the digits 0, 3 & 5 is..
Ans: The required number is "305"
* Write the greatest three digit number formed by using the digits 7, 6 & 4 is..
Ans: Required numbers is "764"
* The smallest four digit number formed by using the digits 1, 3 & 8 and repeating 8 is twice.
Ans: Required number is 1388
* Write the smallest four digit number of four different digits.
Ans: Required number is 1023
* Write the greatest four digit number of four different digits.
Ans: The greatest four different digits are 9, 8, 7 & 6. The required number = 9876
Ex: 1. What is the differance between the largest numbers and the least number written with
the figures 3, 4, 7, 0, & 3 ?
a) 70983 b) 43893 c) 43983 d) 43883
Ans: Largest number = 74330
Least numbers = 30347
Difference = 74330 -30347 = 43983
2. What is the difference between the largest number and few least number ... with the
figures 3, 4, 1, 7...
a) 6084 b) 6184 c) 5084 d) 6048
Ans: 7431-1347= 6084
3. Find the difference of the place value and face value of 9 in 29735?
a) 8881 b) 8991 c) 9001 d) 8899
Ans: Place value of 9 = 9000 Face value of 9 = 9 Difference = 8991
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4. Evaluate whene 'x' is a prime number
a) 450 b) 600 c) 463 d) 468
Ans: This symbol is called Sigma. It means sum of numbers from n1 to n2
sum of prime numbers between 50 and 80.
Sum = 53 + 59 + 61 + 67 + 71 + 73 + 79 = 463
5. If 'x' is a composite number find the value of
a) 240 b) 245 c) 247 d) 250
Ans: Composite numbers between 30 and 40 are.. 32, 33, 34, 35, 36, 38, 39
Sum = 247
x=40x
x=30
x = n2
x
x = n1
x=80x
x=50
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Least Common Multiple (L.C.M.)
L.C.M.: It is defined as the least number which is exactly divisible each one of the given
numbers is called the L.C.M. of that number.
Method of Finding L.C.M.: There are two methods to find L.C.M.
1. Factorisation method
2. Short Cut method
1. L.C.M. by Factorisation method: Resolve the given numbers into prime factors
then L.C.M. is the product of highest powers of all the factors.
Ex. Find the L.C.M. of 72, 108 & 2100 ?
Ans:
72 = 23 32 = 8 9
108 = 23 33 = 4 27
2100 = 22 3 52 7 = 4 3 25 7 L.C.M. = 23 33 52 7
i) Find the L.C.M. of 23 32 5 11; 24 34 52 7 and 25 33 53 72 11 is
a) 25 34 53 b) 25 34 53 72 11
c) 23 32 5 7 11 d) 23 32 5
Ans: L.C.M. = 25 34 53 72 11
3 2100
7 700
2 100
2 50
5 25
5
2108
2 54
3 27
3 9
3
2 72
2 36
2 18
3 9
3
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2. L.C.M. by Short Cut method:
L.C.M. = 2 2 2 3 3 5 5 3 7 L.C.M.= 23 33 52 7 = 37800
Examples:
i. Find the L.C.M. of 26, 56, 104 and 182.
a) 546 b) 1274 c) 728 d) 784
Ans:
L.C.M = 2 2 2 7 13 = 56 13 = 728
ii. Find the L.C.M. of 148 and 185.
a) 680 b) 740 c) 3700 d) 2960
Ans:
L.C.M = 2 2 5 3 37 = 20 37 = 740.
2 148 185
2 74 185
5 37 185
37 37 37
1 1
2 26 56 104 182
2 13 28 52 91
2 13 14 26 91
13 13 7 13 91
7 1 7 1 7
1 1 1 1
2 72 108 2100
2 36 54 1050
2 18 27 525
3 9 27 5253 3 9 175
5 1 3 175
5 1 3 35
1 3 7
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iii. Find the least number exactly divisible by 12, 15, 20, and 27.
a) 500 b) 540 c) 550 d) 600
Ans:
L.C.M = 2 2 3 3 3 5 = 540
Different Therems on L.C.M.Type-1:
Find the least number which is exactly divisible by x, y and z.
Type-2:Find the least number which when divided by x, y and z leaves the remainder ' r ' in
each case.
Type-3:
Find the least number which when divided by x, y and z leave the remainders p, q and r
(x p) = (y q) = (z r) = k
RequiredL.C.M.of (x, y & z) K
Number
=
RequiredL.C.M.of (x, y &z) r
Number
= +
RequiredL.C.M.of (x, y & z)
Number
=
212 15 20 27
2 6 15 10 27
3 3 15 5 27
3 1 5 5 9
5 1 5 5 3
1 1 1 3
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Examples:
i. Find the least number which when divided by 6, 7, 8, 9 and 12 leaves the semi
remainder 1 in each case.
a) 500 b) 505 c) 506 d) 507
Ans: Required Number = (LCM of 6, 7, 8, 9, 12) + 1
L.C.M. = 23 32 7 = 504Required Number = (504 + 1) = 505
ii. Find the least number which when divided by 48, 60, 72, 108 and 120 leaves the
remainders 38, 50, 62, 98 and 110 respectively.
a) 1950 b) 2050 c) 2120 d) 2150
Ans: Difference between the numbers and the remainder is the same.(48 38) = 10; (60 50) = 10; (72 62)= 10...
= 2160 10 = 2150
iii. What is the least number which when divided by 8, 12, 18 and 24 leaves the
remainders 4, 8, 14 and 20 respectively?
a) 78 b) 68 c) 58 d) None
Ans: Difference between the numbers and the remainder is the same.
8 4 = 4; 12 8 = 4; 18 14 = 4
= 72 4 = 68.
Required(L.C.M.of 8,12,18and24) 4
Number
=
Required(L.C.M.of 48, 60,72,108and120) 10
Number
= =
2 6 7 8 9 12
2 3 7 4 9 6
2 3 7 2 9 3
3 3 7 1 9 3
1 7 1 3 1
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iv. Five bells begins to toll together and toll at interval of 36, 45, 72, 81, and 108 seconds.
After what interval of time they will keep on tolling together?
a) 3240 seconds b) 3080 seconds
c) 3140 seconds d) 3200 seconds
The interval time = L.C.M. of 36, 45, 72, 81, and 108 = 3240 seconds
v. How often will five bells toll together in one hour, if they start together and toll at
interval of interval of 5, 6, 8, 12, 20 seconds respectively?
a) 29 b) 30 c) 31 d) 120
Ans:
= 120 seconds.
The number of times they will toll together in one hour =
One is added because the bells begin tolling together = (30 + 1) = 31 times.
vi. Six bells commence tolling together and toll at intervals of 2, 4, 6, 8 and 12 seconds
respectively. In 30 minutes How many times do they toll together?
a) 4 b) 10 c) 15 d) 16
Ans: L.C.M. of 2, 4, 6, 8, 10, 12 is 120 seconds.
The bell will toll together after every 120 seconds = i.e. 2 min
In 30 minutes they will toll together
301 16 times
2
+ =
360 0
12 030=
The time after whichL.C.M. of 5,6, 8,12 and 20
the bells ring together
=
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vii. An electronic device makes a beep after every 60 seconds. Another device makes a
beep after every 62 seconds. They beeped together at 10 a.m. The time when they will
next make a beep together at the earliest is?
a) 10.30 a.m b) 10.31 a.m. c) 10.59 a.m. d) 11 a.m.
Ans: L.C.M. of 60 & 62 seconds is 1860 seconds = 31 minutes
They will beep together at 10.31 a.m.
viii. The traffic lights at three different road crossings change after every 48, 72 and 108
seconds respectively. If they all change simultaneously at 8.20.00 hours when they will
agian change simultaneously?
a) 8: 27: 12 hrs b) 3: 27: 24 hrs
c) 8 : 27: 36 hrs d) 8 : 27 : 48 hrs
Ans: Interval of change = L.C.M. of (48,72 &108) seconds = 432 seconds
= 7 minutes 12 seconds
Next simultaneous change will take place at
= 8 : 20 : 00
7 : 20
8: 27 : 20 hrs
ix. Find the smallest number which when increased by 11 is divided by 18, 24, 60 and 150.
a) 1700 b) 1780 c) 1789 d) 1790
Ans: L.C.M. of 18, 24, 60 and 150 is
L.C.M = 18, 24, 60 and 150 is 23 32 52 = 1800
Required Number is 1800 11 = 1789
2 18 24 60 150
2 9 12 30 75
2 9 6 15 75
3 9 3 15 75
5 3 1 5 25
3 1 1 5
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x. The least number of five digits when divided by 6, 8, 10, 12 leaves no remainder is?
a) 10080 b) 10070 c) 10065 d) None
Ans:
L.C.M. of 6, 8, 10, 12 is 23 3 5 = 120
Least number of five digits = 10000; now divide 10000 by 120
Least number of five digits = 1000 + (120 40) = 10080
xi. Find the greatest number of four digits which is exactly divisible by 6, 8, 10 and 12.
a) 9990 b) 9960 c) 9950 c) 9860
Ans: L.C.M. 6, 8, 10 and 12 is 120
Greatest number of four digits = 9999
Required number = 9999 39 = 9960
) (120 9999 83960
399
360
39
) (120 10000 83960
400
360
40
2 6 8 10 12
2 3 4 5 62 3 2 5 3
3 1 5 3
1 1 5 1
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Highest Common Factor - HCF
HCF: The HCF of two (or) more numbers is the greatest number that can divide each one of
them exactly.
Method of finding HCF:
There are two methods to find HCF
1. By factorisation
2. By division
1. By factorisation: In this type express the given numbers in to the prime factors.
"The product of least powers of common prime factors gives HCF"
Ex: i. Find the HCF of 108, 288 and 360.
Ans:
2 108 2 288 2 3602 54 2 144 2 180
3 27 2 72 2 90
3 9 2 36 3 45
3 2 18 3 15
3 9 5
3
108 = 22 33 288 = 25 32 360 = 23 32 5
HCF = Primes common with least power
= 22 32 = 36
ii. Find the HCF of 23 32 5 74, 22 35 52 73, 23 53 72 .
a) 2 5 7 b) 22 5 7 c) 22 5 72 d) None
Ans: HCF = Product of common prime numbers with least power
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1056) 1584 (1
1056
528) 1056 (2
1056
0
36) 360 (10
360
0
108) 288 (2
210
72) 108 (1
72
36) 72 (2
72
0
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HCF = 2 5 7 (common primes)
= 22 5 72 (least power)
2. By Division Method:
Ex: i. Find the HCF of 108, 288 and 360.
Ans: Take any two numbers 108 and 288
HCF of 108 & 288 is " 36 "
Divide 360 by 36,
" 36 " is the HCF of 108, 288 and 360.
Problems:
1. HCF of 1056, 1584 and 2178 is -
a) 60 b) 61 c) 65 d) 66
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HCF = " 66 "
2. Find the largest number that can exactly divide by 513, 783 and 1107.
a) 25 b) 26 c) 27 d) 28
Ans: 513) 783 (1
513
270) 513 (1
270
243) 270 (1
243
27) 243 (9
243
0
HCF = " 27 "
Diffrent models on HCF
Type I : Find the greatest number that will exactly divide x, y, z .
Type II : Find the greatest number that will exactly divide x, y & z leaving the remainder p,
q & r respectively.
Type III : Find the greatest number that will divide x, y & z leaving the same remainder in
each case.
Requirednumber= HCFof [(x-y), (y-z), (z-x)]
Required number= HCFof [(x-p), (y-q), (z- r)]
Required number = HCF of (x, y & z)
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528) 21784 (4
2112
66) 528 (8
528
0
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Examples
1. Find the greatest number which can divide 284, 698 and 1618 leaving the same
remainder ' 8 ' in each case.
a) 20 b) 22 c) 23 d) 46
Ans: Required number = HCF [(284 - 8), (698 - 8), (1618 - 8)]
Required. number = HCF of 276, 690 & 1610
276) 690 (2 138) 1610 (11
552 138
138) 276 (2 230
276 138
0 92) 138 (1
92
46) 92 (2
92
0
2. Find the greatest number which will divide 1050, 1250 and 1650 leaving the remainders
43, 31 and 7 respectively.
a) 73 b) 63 c) 59 d) 53
Ans: Required numbers = HCF of [(1050 - 43) : (1250 - 31), (1650 - 7)]
= 1007, 1219 & 1643
1007) 1219 (1 53) 1643 (3
1007 159
212) 1007 (4 53) 53 (1
848 53
159) 212 (1 0
159
53) 159 (3
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159
0
Ans: " 53 "
3. Find the greatest number which will divide 25, 73 and 97 so as to leave the same
remainder in each case.
a) 12 b) 18 c) 24 d) 32
Ans: Required number = HCF of [25 - 73), (73 - 97), (97 - 25)]
= HCF of 48, 24, 72
24) 48 (2 24) 72 (3
48 72
0 0
Ans: " 24 "
4. Find the size of the largest square marble which can be paved on the floor of a room 5
meteres 44 cm long and 3 meters 74 cm broad.
a) 56 b) 42 c) 38 d) 34
Ans: Size of square marble = HCF of 544 cm; 374 cm
374) 544 (1
374
170)374 (2
340
34) 170 (5
170
0
Ans: 34 cm
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4. A room is 4 meteres 37 cm long and 3 metres 23 cm broad. If is required to pave the
floor with minimum square slabs. Find the number of slabs required for this purpose.
a) 485 b) 481 c) 391 d) 381
Ans: Length of room (L) = 437 cm
Breadth (B) = 323 cm
Area of room (A) = l b = 437 323 cm2
Size of marble = HCF of 437 & 323
323) 437 (1
323
114) 323 (2
228
95) 114 (1
95
19) 95(5
95
1
area of marble = 19 19cm2 No. of marbles
5. Three different containers contain 496 litres, 403 liters and 713 litres of mixture of
milk and water respectively. What biggest measure can measure all the different
quantities exactly ?
a) 1 litre b) 7 litre c) 31 litres d) None
Ans: HCF of 496, 403 & 713 is the required measerement
HCF = 31 litres
23437=
17323
19 19391=
Area of room=
Areaof marble
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6. A trader has two varieties of sugar 204 kg and 1190 kg by weights. Find the number
of minimum bags of equal size in which he can store the sugar without mixing.
a) 39 b) 40 c) 41 d) 2
Ans: Size of the bag = HCF of 204 kg & 1190 kg
204) 1190 (5
1020
170) 204 (1
170
34) 170 (5
170
0
HCF = " 34 "
Size of the bag is 34 kg
Number of bags required =
= 6 + 35 = 41
7. A worker was engaged for a certain number of days and was promised to be paid Rs.
1189. He remained absent for some days and was paid Rs.1073 only. What were his
daily wages?
a) 29 b) 30 c) 40 d) 50
Ans: Daily wage is common factor of 1189 & 1073 = HCF of 1189 & 1073 = 29/-
8. Two numbers are in the ratio of 15:11 if their HCF is 13 then the numbers are -
a) 195,140 b) 195,143 c) 190, 80 d) 100,50
Ans: Let, the numbers be 15x & 11x
Their HCF (x) = 13
204 119034 34
+
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The numbers are = [(1513): (1113)]
= (195, 143)
9. Three numbers are in the ratio 1 : 2 : 3 and their HCF is 12, then the numbers are -
a) 10, 20, 30 b) 12, 20, 24 c) 12, 24, 36 d) None
Ans: The numbers be x, 2x & 3x
Their HCF (x) = 12
The numbers are 12, (212), (312)
= 12, 24 & 36
10. If the sum of two numbers is 216 and their HCF is 27, then the numbers are -
a) 27,189 b) 27,100 c) 20,200 d) 27,180
Ans: HCF of two numbers be (x) = 27
Let, the numbers are 279 & 276
sum (27a + 27b) = 216
27 (a + b) = 216
a + b = 216 / 27 a + b = 8
Now, co-primes with sum " 8 " are (1, 7) & (3, 5)
Required numbers are 27a & 27b
(27 1) : (27 7)
27 : 189
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Arithmetic Fundamentals
I. "3x" means 3 multiplied with 'x' (or) 'x' added three times.
II. x3 means 'x' multiplied itself 3 times.
III Addition
'+' + '+' = +
'+' + '' = (or) +
IV. Multiplication
+ + = '+' = '+'
+ = '' + = ''
V. Division
VI. VII. VIII.
IX. X.
XI. XII.
XIII. 0.33 100 = 33(In multiplication decimal point moves right side as many zeros as in
the number)
Ex: i) ii)
XIV. XV.
(During division decimal point moves left as many zeros as in the denominator)
Ex :i) ii) iii) 4 0.0041000
=0.5 0.00051000
=1.232 0.01232100
=
3.20.0032
1000=
0.3 0.3 100 30
0.42 0.42 100 42
= =
4
5320.0.532 10000 =
3
1300.1.3 1000 =
1 1 10 10
0.5 0.5 10
= =
52=
0.25 25 5
0.45 45 9= =
2.4 2.4 10 24
1.2 1.2 10
= =
122=
0.2 0.2 100 5
0.12 0.2 100 3
= =
78 810
7 7=
86 117
5 5=
12618
7=
=
=
+
+=
+= +
+
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XVI. 0.23 0.2 = 0.046
(first multiply 2 23 = 46 and pot decimal point 3 digits towards left)
= 0.046
XVII. first calculate L.C.M of 3, 4,& 6
L.C.M = 2 3 1 2 1 = 12
XVIII.
XIX. 3x 8 = 13 then find 'x' ?
sol: 3x 8 = 13 3x = 13 + 8 3x = 21
XX. 2x + 8 =14, then a = ?
sol: 2x + 8 = 14 2x = 14-8 2x = 6
Rule: i) During addition (or) substraction the sign changes if it moves from left side to other
side of " = "
Ex: i) -5 + x = + 1 x = 1 + 5 x = 6
ii) -5 + x = + 1 -5 + x - 1= 0 -6 + x = 0 x = + 6
iii. Multiplication of one side changes to division in other side of " = "
Ex: i. 5x = 11
35
= =x
6x =
23=
21x =
37
x 7
=
=
3 2 5 53 3 3
6 6 6
++ = + =( )
1 12 3
1 2 ( )+ + +
1 11 2 ?
2 3+ =
2 4 3 3 1 2 8 9 2 19
12 12 12
+ + + += =
4 3 22 3 1
3 4 6+ +
2 3 1
3 4 6+ +
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ii. 2x + 5 = 8 2x = 8 - 5 2x = 3 x = 3/2
XXI. 5x + 8 = 3x + 10 then x = ?
sol : 5x - 3x = 10 - 8
XXII. 7x - 6 = 9x - 18, then x =?
sol : - 6 +18 = 9x -7x
12 = 2x
122 12x= x=
2
6=
22x 2 x= =2
1=
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Finding Unit's Digit
Unit's Digit:
A number is made up from digits in the numeral system is called Unit's Digit. We often use
the decimal system in which we use 10 digits, In writing the any number, many digits are used,
even repitation of digits. when we write any number using the digits, the last digit ( from right
side ) in that number is called unit digit.
For example in the number 48750, here " 0 " is called unit digit.
To find the number in the unit place of Nn, where
N & n = positive numbers
"N" is called base
'n' is called index of power
* Base number "N" can has any of the following digit in the unit's place
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
Case-I:
When 0, 1, 5, 6 are the digits in the unit's place of the base number N, then the number in the
unit's place of Nn will also be 0, 1, 5 & 6 respectively, whatever be the value of "n"
Ex: 1. Number in the unit's place of (370)93 is " 0 "
2. Number in the unit's place (391)775 is " 1 "
3. Number in the unit's place of (75)73 is " 5 "
4. Number in the unit's place (676)99 is " 6 "
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Case-II
When 4 (or) 9 are in the unit's place of N, then
(4)1 = 4 (4)2 = 16
(4)3 = 64 (4)4 = 256
(4)5 = 1024 (4)6 = 4096
From the above figures we observe that..
4n when " n " is odd number [i.e. 1, 3, 5, 7 etc.], it contains " 4 " in the unit's place
4n when " n " even number [i.e. 2, 4, 6, 8 etc.], it contains " 6 " in the unit's place
In the same way..
9n n = odd number " 9 " in the unit's place
9n n = even number " 1 " in the unit's place
Example:
1) (74)99 n = 99 (odd) 4 in the unit's place
2) (84)78 n = 78 (even) 6 in the unit's place
3) (79)33 9 in the unit's place
Case-III
When ' 3 ' is the digit in the unit's place of ' N '
(3)1 = 3 (3)2 = 9
(3)3 = 27 (3)4 = 81
for any value of 3 is always odd
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Problems
1. What is the number in the unit's place of (743)74 ?
Ans: n = 74, it can be written as 74 = (4 18) + 2
'2' is remainder
The number in the unit's place (3) 2 = " 9 "
2. What is the number in the unit's place of (72)75 ?
Ans: n = 75,
75 = (4 18) + 3
Remainder = 3
(72)75 (or) (72)3 23 " 8 "
3. What is the number in the unit's place of (788)94 ?
a) 2 b) 3 c) 4 d) 5
Ans: Divide 94 by 4 ,
Remainder = 2
(8)n = 64
the number is "4" in the unit's place
4. The unit's digit in the product (3127)173 is -
a) 5 b) 6 c) 7 d) 8
Ans: Divide 173 by 4,
Remainder is " 1 "
(7)1 = " 7 " in the unit place
94 234
=
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5. The unit's digit in the product of (256 27 159 182) is -
a) 4 b) 5 c) 6 d) 7
Ans: Product of unit's digits in all the number
(6 7 9 2) = 756
Unit's digit in the given product is " 6 "
6) The unit's digit in the product of (274 318 577 313) is -
a)2 b) 3 c) 4 d)5
Ans: 4 8 7 3 = 672
unit's digit is " 2 "
7. If the unit's digit in the product of (4594628x484) is "2" then the value of "x" is -
a) 1 b) 2 c) 3 d) 4
Ans:
9
6
x
4
216 x
In order to obtain ' 2 ' in unit's place need to multiply by '7'
x = 7
8) The unit's digit in the product (771 659 365) is -
a) 3 b) 4 c) 5 d) 6
Ans: 74 = 1, ((74)4)4
= 764 = 1764 74 73 1 1 73 =
Unit's digit in 771 is " 3 "
659 = every power of " 6 " will give the unit's digit " 6 "
365 = [(34)4]4
31
= 1 3 = 3 (unit's digit of 34 is 1)
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Units digit of 365 is 3
unit's digit in (771 659 365) is
(3 6 3) = 54
Unit's digit is " 4 "
9) The unit's digit in the product of (367 639 753) is -
a) 3 b) 4 c) 5 d) 6
Ans: 367 = [[34]4]4 33 1 33 = 27 7
639 = 6
753 = 7
(7 6 7) = 294
the unit's digit is " 4 "
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DIVISION SUMS
In a division sum we have four quantities namely -
i) Dividend (p)
ii) Divisor (d)iii) Quotient (q)
iv) Remainder (r)
Divisor) Dividend (Quotient
Remainder
Dividend = (Divisor Quotient) + Remainder
Problems:
1. Find a number when divided by a divisor which is five times the quotient, gives a
quotient which is four times the remainder the remainder being 10.
a) 8,000 b) 8,010 c) 8,500 d) 8,110
Ans: Remainder (r) = 10
Quotient (Q) = 4r = 4 10 = 40 = 40
Divisor (d) = 5 40 = 200
Dividend = (200 40) + 10 = 8,010
2. In a division sums the quotient is 120 the divisor 456 and the remainder 333. Find the
dividend.
a) 5,553 b) 50,553 c) 56,053 d) 55,053
Ans: Quotient (Q) = 120,
Divisor (d) = 456
Remainder (r) = 333Dividend (P) = (D Q) + r
= (456 120) + 333 = 55,053
P = (d q) + r
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3. In a division sum the divisor is ten times the quotient and five times the remainder.
what is the dividend if the remainder is 46 ?
a) 5,326 b) 5,306 c) 5,336 d) 5,366
Ans: Remainder (r) = 46
Divisor (d) = 5 r = 5 46 = 230
Divisor (d) = 10 q
10q = 230 q = 23
Dividend - (D Q) + r = (230 23 + 46)
= 5,336
4. The quotient arising from a division of a number by 62 is 463 and the remainder is 60.
what is the number ?
a) 28,666 b) 28,766 c) 28,576 D) 28,676
Ans: Divisor (d) = 62, quotient (Q) = 463
Remainder (r) = 60, find dividend
Dividend = (divisor quotient) + remainder
= (63 463) + 60
= 28,766
Divisors:
Divisor: Each one of the natural numbers that divides a given number exactly is called
a divisor (or)
Factor of the given number is divisor
Ex: Each of the numbers 1, 3 and 5 divides 15 exactly
So, 1, 3 & 5 are divisors of "15"
* Every number has at least two different divisors
* Those numbers which contain only two different divisors are called Prime numbers
* The numbers containing more than two different divisors are called "Composite
numbers"
15 15 15=15, =5, =351 3
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Prime factors:
A factor of a given number is called a prime factor if this factor is a prime number.
Ex: The factors of 30 are 2, 3, & 5 are prime factors.
Prime factorisation
To express a given number as a product of prime factors is called prime factorisation
of the given number.
Ex: Express '90' as a product of prime factors?
Ans: Divide 90 sucessively by primes
starts from least prime...
2 90
3 45
3 15
5 5
1
90 = 2 3 3 5 1
90 = 2 32 5 (product of prime factors of 90)
Problems based on Divisors:
1. Express 108 as a product of prime factors.
a) 22 32 b) 2 32 c) 22 33 d) 2 3
Ans:
108 = 2 2 3 3 3 = 22 33
2. Express "2178" as a product of prime factors.
a) 2 3 11 b) 2 32 11 c) 2 32 112 d) 32 112
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2 108
2 54
3 27
3 9
3
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Ans:
2178 = 2 3 3 11 11
= 2 32 112
3. The prime factors of 15,015 are..
a) 3, 5, 7, 11, 37 b) 3, 5, 11, 13, 37 c) 3, 5, 7, 13 ,37 d) 3, 5, 7, 11, 13
Ans:
15015 = 3 5 11 13 7 1
= 3 5 7 11 13
4. Find the numbers of different prime factors used in (21)17 (15)12 (51)9.
a) 2 b) 3 c) 4 d) 5
Ans:
(21)17 (15)12 (51)9
317 717 312 512 39 179 3, 5, 7, 17
= "4" factors
5. Find the number of different prime factors used in (15)13 (14)21 (62)7.
a) 3 b) 4 c) 5 d) 6
Ans: (15)13 (14)21 (62)7
313 513 221 721 27 317
2, 3, 5, 7, 13 "5" factors
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2 2178
3 1089
3 363
11 121
11
3 15015
5 500511 1001
13 91
7 7
1
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6. What is the number of prime factors contained in the product 307 225 3411?
a) 23 b) 44 c) 46 d) 53
Ans: 307 225 3411
(2 3 5)7 (2 11)5 (2 17)11
27 25 211
37 57 115 1711
223 37 57 115 1711
23 + 7 + 7 + 5 + 11 = 53
7. What is the number of prime factors contained in the product (21)8 (77)11 (26)2 ?
a) 40 b) 41 c) 42 d) 43
Ans: (3 7)8 (7 11)11 (2 13)2
22 38 78 711 1111 132
2 + 8 + 19 + 11 + 2 = 42
Theorem-1
Finding number of division of a composite number
To find the number of divisors of a composite number... "A" be a natural number
ere b, c & d are Distinct Prime Factors
The number of prime factors for "A" = (P + 1) (q + 1) (r + 1).....
Ex: 1. Find the number of factors for 21 .
a) 1 b) 2 c) 3 d) 4
Ans: 21 = 71 31
A = bp
cq
Number of factors = (p + 1) (q + 1)
= (1 + 1) (1 + 1) = 2 2 = 4
Verification:
The set of factors = { 1, 7, 3, 21}
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A = bp
cq
d2
.....
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2. Find the number of factors for "48" ?
a) 7 b) 8 c) 9 d) 10
Ans:
48 = 24 31
A = bq
cr
Number of factors = (4 + 1) (1 + 1)
= 5 2
= 10
Verification:
The set of factors of '48' is {1, 2, 3, 4, 6, 8, 12, 16, 24, 48}
3. The number of divisors of 600 is ?
a) 21 b) 22 c) 23 d) 24
Ans:
600 + 23 31 52
Number of divisors = (3 + 1) (1 + 1) (2 + 1)
= 4 2 3
= 24
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2 48
2 24
2 12
2 6
3
2 600
2 300
2 150
3 75
5 25
5 51