Concepts Examples

How To Solve Allegations

Suppose the average weight of a group of objects O₁ is A₁ and average weight of group of objects O₂ is A₂ and A₁ < A₂, then we can write the data in the manner as shown in the figure. (A_w−A₂) : (A_w−A₁) gives the ratio of O₁ : O₂.

Allegations does not give the actual volume to be mixed but only the ratio in which volumes are to be mixed.

Question 1 A shopkeeper sells two types of rice A and B. If he sold 30% of type A rice at a profit of 50% and sold 90% of the rice of type B at a profit of 10%. What is the averge profit percent of the shopkeeper, if he sells only these two types of rice.

35%	45%
40%	55%

C

Since the profit on rice of type B = 10% ∴ A₁ = 10

Since the profit on rice of type A = 50% ∴ A₂ = 50

Let the average profit be Rs x ∴ A_w = x

$\frac{50-x}{x-10}=\frac{3}{1}\Rightarrow50-x=3x-30$50−xx−10 =31 ⇒50−x=3x−30

⇒ x = 20

∴ Average profit of the shopkeeper is 40%

Question 2 How much water should be mixed with 6L pure milk to make the concentration of milk 60%.

4L	2L
3L	5L

A

Since the concentration of milk in water = 0 ∴ A₁ = 0

Since the concentration of pure milk = 100 ∴ A₂ = 100

Since the concentration of milk in milk-water solution = 60 ∴ A_w = 60

∴ Ratio in which water and milk should be added = 2:3

∴ Volume of water to be mixed = 4L

Question 3 A dealer buys 11 kg of wheat at Rs 275 and mixes it with another quantity of wheat in the ratio of 4:5. The price of the resulting mixture is Rs 30 per kg. Finf the price of this other quantity of wheat.

Rs 37 per kg	Rs 31 per kg
Rs 24 per kg	Rs 34 per kg

D

Price per kg of wheat = $\frac{275}{11}=25$27511 =25 Rs

Since the of wheat = Rs 25 ∴ A₁ = 25

Let the price of rice of other type be Rs x ∴ A₂ = x

The average price of wheat = Rs 30 ∴ A_w = 30

∴$\frac{5}{x-30}=\frac{5}{4}\Rightarrow20=5x-150$5x−30 =54 ⇒20=5x−150

⇒ x = 34

∴ Price of wheat of other type = Rs 34 per kg

Question 4 If the average weight of a class of 60 students is 50 and the average weight of a class of 40 students is 45 , then what will be the average weight of both the classes combined.

24	26
45	48

D

Since the average weight of class of 40 students = 45 ∴ A₁ = 45

Since the average weight of class of 60 students = 50 ∴ A₂ = 50

Let the average weight of both the classes combined be x kg.

O₁:O₂ = 40:60 = 2:3

$\frac{50-x}{x-45}=\frac{2}{3}$50−xx−45 =23 

$150-3x=2x-90$150−3x=2x−90

$x=48$x=48

∴ Average weight of both the classes combined = 48 kg

Question 5 A person travels at a speed of 25 km/hr for 30min and then travels at 40km/hr for next 20 min to reach his destination. Find his average speed in the whole journey.

21 km/hr	31 km/hr
33 km/hr	None of these

B

Since the person travels for 30 min at speed of 25 km/hr ∴ A₁ = 25

Since the person travels for 20 min at speed of 40 km/hr ∴ A₂ = 40

Let the average speed of the person be x km/hr ∴ A_w = 500

$\frac{40-x}{x-25}=\frac{3}{2}\Rightarrow80-2x=3x-75$40−xx−25 =32 ⇒80−2x=3x−75

⇒ x = 31

∴ Average speed of the person in the whole journey was 31 Km/hr

Question 6 In what ratio should a 65% milk solution and 85% milk solution should be mixed so that the new solution is a 80% milk solution.

1:2	1:3
1:4	2:3

B

Since the concentration of milk in water = 65 ∴ A₁ = 65

Since the concentration of pure milk = 85 ∴ A₂ = 85

Since the concentration of milk in milk-water solution = 80 ∴ A_w = 80

∴ Ratio in which the two solutions should be mixed = 1:3

Concepts Examples

When we have to calculate the average of some groups of different things, having different number of elements is being calculated, then it is called the weighted average.

If the number of elements in n different groups be N₁,N₂,N₃ .. N_n and their respective averages be A₁,A₂,A₃...A_n ,

then Weighted average = $\frac{\mathbb{N}_1\cdot A_1+\mathbb{N}_2\cdot A_2+\mathbb{N}_3\cdot A_3...\mathbb{N}_n\cdot A_n}{\mathbb{N}_1+\mathbb{N}_2+\mathbb{N}_3...\mathbb{N}_n}$N₁·A₁+N₂·A₂+N₃·A₃...N_n·A_nN₁+N₂+N₃...N_n 

Question 1 Average cost of buying 5 quality A crackers is 60 and cost of buying 10 quality B crackers is 80. WHhat will be average cost of buying all the crackers.

71.33	72.33
73.33	74.33

C

Total number of employees = 5 + 10 = 15

Average cost of all the crackers = $\frac{5\cdot60+10\cdot80}{5+10}=\frac{300+800}{15}=\frac{1100}{15}=73.33$5·60+10·805+10 =300+80015 =110015 =73.33

Question 2 Average weight of a class of 40 students is 30 and average weight of a class of 20 students is 15. Find the average weight of both the classes combined.

21	22
24	25

D

Average weight = $\frac{40\times30+20\times15}{40+20}=25$40×30+20×1540+20 =25

Question 3 There are two sections A and B of a class, consisting of 36 and 44 students' respectively. If the average weight of section A is 40kg and that of section B is 35kg, find the average weight of the whole class.

37.25	74.5
36.25	36

A

Total weight of (36+44) Students = (36×40 + 44×35) kg = 2980 kg

Therefore average weight of the whole class = $\frac{2980}{80}$298080  kg

Therefore average weight = 37.25 kg

Question 4 Average salary of 12 employees in company A is 15000 pm and 15 employees is company B is 16000 pm. What will be the average salary of all the employees.

17000	16888
15000	14888

B

Total number of employees = 15 + 12 = 27

Average salary = $\frac{12\cdot18000+15\cdot16000}{12+15}$12·18000+15·1600012+15 

= $\frac{456000}{27}=16888.88$45600027 =16888.88

Concepts Examples

If a number x divides another number y completely then x is a divisor of y. If x is a prime number then x is called the prime factor of number x.

e.g 1,2,3,6 are the factors of 6 and 2,3 are the prime factors of 6.

If any composite number N which can be expressed as N=$a^x\cdot b^y\cdot c^z...$a^x·b^y·c^z... , where a,b,c are the prime numbers and x,y,z are the powers of a,b,c respectively and so on. then the number of factors of N =(x+1)×(y+1)×(z+1)...

e.g Find the number of factors of 24

24=2²×3¹

∴ number of factors of 24 = (3+1)(1+1) = 8

Question 1 Find the total number of factors of 1024 except 1 and 1024 itself.

8	9
10	11

B

1024= 2¹⁰

∴ total number of factors except 1 and 1024 = (10+1) - 2 = 9

Question 2 Find number of factors of 1001

8	3
16	24

A

1001 = 7¹×11¹×13¹

∴ total number of factors = (1+1)(1+1)(1+1) = 2×2×2 = 8

Question 3 Find the total number of factors of 144

15	16
18	25

A

144= 2⁴ × 3²

∴ Number of factors of 144 = (4+1)(2+1) = 15

Question 4 Find the number of factors of 970299

64	32
24	28

D

970299 = 3⁶×11³

∴ total number of factors = (6+1)(3+1) = 7×4 = 28

Question 5 Find the total number of factors of 56 × 324

40	50
60	70

C

56 × 324 = (2³×7¹) × (2²×3⁴)

56 × 324 = 2⁵×3⁴×7¹

∴ total number of factors = (5+1)(4+1)(1+1) = 6 × 5 × 2= 60

Concepts Examples

Properties of Averages

1. Average of N things/quantities is equal to the sum of all the things/quantities divided by number of things/quantities.

2. Average of N number of things/quantities always lies between the lowest and the highest quantities.

3. If each quantity is increased by a certain value K, then the new average is increased by K.

4. If each quantity is decreased by a certain value K, then the new average is decreased by K.

5. If each quantity is multiplied by a certain value K, then the new average becomes K times the the original average.

6. If each quantity is divided by a certain value K, then the new average becomes $\frac{1}{K}$1K  times the the original average.

Question 1 Average number of chocolates with 10 children is 5. If a person distributes all 20 chocolates he has to the children then what is the average number of chocolates with the children.

7	8
9	10

A

Total number of chocolates with the students = 10 × 5 = 50

New Total number of chocolates = 50 + 20 = 70

∴ new average number of chocolates = $\frac{70}{10}=7$7010 =7

Question 2 The average score of a cricketer for ten matches is 38.9 runs. If the average for the first six matches is 42, then find the average for the last four matches.

35	36
35.25	36.25

C

Total sum of last 4 matches = (10 × 38.9) - (6 × 42)

= 389 - 252 = 137

Average = $\frac{137}{4}=35.25$1374 =35.25

Question 3 Average of a class of 50 students is 50 kg. If 2 guys of average weight 45 leaves the class. What is the new average of the class.

50	50.21
51.2	55.2

B

Total weight of the class = 50 × 50 =2500 kg

Total weight of the two students who leave the class = 45 × 2 = 90 kg

∴new average weight of the class = $\frac{2500-900}{48}=50.21$2500-90048 =50.21

Question 4 The number of sweets in a box is 15. If the shop keeper $\frac{2}{3}$23 ^rd's the number of sweets in each box, then what will be the new average of number of sweets in the box.

5	15
98	10

D

Its simple application of the rule, Since the number of sweets is $\frac{2}{3}$23  of the original quantity in each box

∴ the average number of sweets in each box will be 15 × $\frac{2}{3}$23  = 10

Question 5 If the average weight of a family of 5 people is 42 kg and one member gets married and add to the family. Weight of this member is 48 kg. What will be the new average weight of the family.

42	43
44	45

B

Total weight of the family of 5 people = 5 × 42 = 240

Total weight of family after new member joins the family = 210 + 48 = 258

∴ new average of the family = $\frac{258}{6}=43$2586 =43

Question 6 The number of sweets in a box is 15. If the shop keeper triples the number of sweets in each box, then what will be the new average of number of sweets in the box.

3	15
45	55

C

Its simple application of the rule, Since the number of sweets is tripled in each box

∴ the average number of sweets in each box will be 15 × 3 = 45

Concepts Examples

There are three types of questions that are based on LCM with remainders. They are as follows.

1. When the remainders are same for all the divisors
   
   In this case the required number will be the LCM × N + remainder, where N is any natural number
   
   N=1, will give smallest such number
   
   N=2, will give second smallest such number and so on

2. When the remainders are different for different divisors but the respective difference between the divisors and the remainders remain constant.
   
   In this case the required number will be LCM × N +difference of any (divisor - remainder)
   
   refer to example 2.

3. When neither the divisors are same nor the respective difference between the divisors and the remaniders remain constant.
   
   In this case solve the question by forming equations and solving them.

Question 1 What is the least possible number which when divided by 4,5,6 leaves the remainder as 3,4,5 respectively.

59	60
61	119

A

Since the difference is same in all i.e (4-3)=(5-4)=(6-5)=1

∴ the required number = LCM (4,5,6) -1

= 60 - 1 = 59

Question 2 Find the least possible 5 digit number which when divided by 2,4,6,8 it leaves the remainder 1,3,5,7 respectively.

10006	10007
10008	None of these

B

i) LCM of 2,4,6,8 is 24



∴ all possible values = 24N - 1 , where N is a natural number

Least possible 5 digit number will be for N=417, 24 × 417 -1 = 10007

Question 3 What is the least possible number which when divided by 10,12,14 leaves the remainder as 2. How many such numbers are there between 5000 and 6000.

1	2
3	4

C

i) LCM of 10,12,14 is 420

∴ the least possible number that leaves the remainder as 2 when divided by 10,12,14 = 420 × 1 + 2 = 422

ii) Number of such numbers between 5000 and 6000 can be found by putting different values of N,

for N=12, 420N +2 = 5042

for N=14, 420N +2 = 5882

∴ there are 3 such numbers between 5000 and 6000.(for N=12,13,14 420N+2 lies between 5000and 6000)

Question 4 What is the least possible number which when divided by 11 leaves the remainder 3 and when divided by 5 leaves the remainder as 2.

46	47
48	49

B

Let the required number be N,

Since N divided by 5 gives remainder as 2 ∴ N = 5 × m + 2

Since N divided by 11 gives remainder as 3 ∴ N = 11 × n + 3

∴ 5m+2=11n+3

$m=\frac{11n+1}{5}$m=11n+15 

For n = 4, m = 9 that is the smallest possible value of m for any n.

∴ the smallest number is 5×9+2=47

Next such number will be LCM of (11 and 5) + 47

Question 5 What will be the least possible number which when divided by 4,5,6 always leaves the remainder 3. Find the following such numbers which satisfy the given condition.

i) Smallest

ii) Second Smallest

iii) Greatest number smaller than 1000

1. The smallest number which when divided by 4,5,6 leaves the remainder 3 in each case is LCM (4,5,6) + 3 = 63

2. To find all the set of such numbers which gives remainder as 3 when divided by 4,5,6 will be
   
   60N + 3, where N is a natural number.
   
   ∴ Second largest such number will be 60 × 2 + 3 = 123

3. To find the largest number of such form smaller than 1000, we will have to find the largest number of form 60N + 3 , that is smaller than 1000.
   
   Taking N=16, 60N +3 = 963, that is the largest such number

Concepts Examples

There can be direct questions on finding last two digits of any power of a natural number and this is very important for eliminating options in questions. Following are the four simple cases with which we will be able to solve almost every problem.

Rules for Finding Last Two Digits of square of a number

In order to find the last two digits of square of any two digit number number, we can write it as difference or sum from 50 or 100, whichever is closer. In general the last two digits of a number (50±k)² or (100-k)² will be determined by square of k.

e.g 47² can be written as (50-3)²,

Now, the last two digits will be determined by square of 3 that is 09

Rules for Finding Last Two Digits of any power of a number

1. For a natural number N^k, where N is a natural number ending with 0 and K is a natural number greater than 2,
   
   Last two digits will always be 00.

2. For a natural number N^k, where N is a natural number ending with 5 and K is a natural number greater than 2,
   
   Last two digits will always be 25.

3. For a natural number (2×m)^40k+1, where m is a odd number not ending with 5 and K is a natural number,
   
   Last two digits will always be (2×m +50).

4. For all the remaining cases, the last two digits of N^40k+x,
   
   Last two digits will always be equal to the last 2 digits of N^x

Question 1 Find the last two digits of 12345⁹⁸⁷⁴⁵.

05	00
25	50

C

Since the number ends with 5, thus rule 2 applies here

∴ the last two digits will be 25.

Question 2 Find the last two digits of 39².

21	81
79	67

A

Since 39² can be written as (50-11)²

∴ last two digits will be determined by 11²=21

Question 3 Find the last two digits of 22⁴⁸².

84	48
92	64

A

Since no rule applies to this, thus we will apply rule 4 here

22⁴⁸² can be written as 22^40×12+2

∴ the last two digits can be determined by 22² and will be 84

Question 4 Find the last two digits of 42⁸⁰¹.

29	92
45	54

B

Since the number can be written as (2×21)^40×20+1, thus rule 3 applies

∴ the last two digits will be (2×21+50)=92

Question 5 Find the last two digits of 39⁹⁰⁰.

01	81
27	34

A

Since no rule applies to this, thus we will apply rule 4 here

39⁹⁰⁰ can be written as 39^40×22+20

∴ the last two digits can be determined by 39²⁰



Now we will keep squaring 39 to reach closet to 39²⁰, we will consider only the last two digits.

39²- Last two digits will be 21, (using rule to find last two digits of square of a number (50-11)²)

39⁴- Last two digits will be 41, squaring 21

39⁸- Last two digits will be 81, squaring 41 (using rule to find last two digits of square of a number (50-9)²)

39¹⁶- Last two digits will be 61, squaring 81 (using rule to find last two digits of square of a number (100-19)²)



Now, last two digits of 39²⁰=Last two digits of (39⁴×39¹⁶)

=Last two digits of (41×61)

=01

∴ last two digits of 39⁹⁰⁰=01

Question 6 Find the last two digits of 53².

01	02
03	04

D

Since 81² can be written as (50+2)²

∴ last two digits will be determined by 2²=04

Question 7 Find the last two digits of 81².

21	61
27	67

B

Since 81² can be written as (100-19)²

∴ last two digits will be determined by 19²=61

Question 8 Find the last two digits of 80⁷⁵⁴.

64	16
00	08

C

Since the number ends with 0, thus rule 1 applies here

∴ the last two digits will be 00.

Concepts Examples

To find the last digit i.e the unit digit you need to be aware of the cyclicity of the numbers and their powers

21=2	31=3	41=4
22=4	32=9	42=16
23=8	33=27	43=64
24=16	34=81	44=256
25=32	35=243	45=1024
26=64	36=729	46=4096
27=128	37=2187
28=256	38=6561

We can see that unit's digit of 2¹,2⁵,2⁹ is 2 and so on. Therefore, after every four powers of 2, the units digit of the number starts repeating.Thus we can say that cyclicity of unit's digit of higher powers of 2 is 4.

Similarily unit digit of power of 4 starts reeating after 2, thus its cyclicity is 2.

Unit digit follows a periodic pattern that is after a particular period it repeats in a cyclic form, this is called cyclicity.

• Unit digits of numbers ending with 0,1,5,6 is always the same irrespective of their powers raised on them.

• Unit digit of numbers ending with 4,9 follows the pattern with cyclicity of 2

• Unit digit of numbers ending with 2,3,7,8 follows the pattern with cyclicity of 4

In order to find the last digit of any number xyz^abc whose who's last digit 'z' has the cyclicity of 4, then to find the last digit write the number as z^{4k + m} where 4k+m = abc, and m is smaller than equal to z and not 0.

In order to find the last digit of any number xyz^abc whose who's last digit 'z' has the cyclicity of 2, then to find the last digit write the number as z^{2k + m} where 2k+m = abc, and m is smaller than or equal to z ans is not equal to 0.

Question 1 What will be the unit digit of 7²³×8¹³

2	3
4	6

C

Since the cyclicity of both 7 and 8 is 4,

∴ unit digit of 7²³ and 8¹³=7^4×5+3 × 8^4×3+1 =unit digit of 7³ and 8¹

∴ unit digit of 7²³×8¹³=unit digit of 3×8=4

Question 2 What will be the unit digit of 888⁴⁴⁴+444⁸⁸⁸

2	4
5	6

A

Since the cyclicity of 8 is 4 and cyclicity of 4 is 2,

∴ 888⁴⁴⁴+444⁸⁸⁸=888^4×111+444^2×444

∴ last digit of 888⁴⁴⁴ and 444⁸⁸⁸=last digit of 8⁴ and 4² = 6 and 6

Thus unit digit =6+6=12 ∴ unit digit is 2

Question 3 Find the last digit of 3⁵⁷

3	6
1	7

A

3 has the cyclicity of 4. 3⁵⁷= 3^4×14+1

∴ last digit of 3⁵⁷=last digit of 3¹ = 3

Question 4 What will be the unit digit of 4567¹⁵⁷⁸

7	1
9	3

C

Since the cyclicity of 7 is 4, 4567¹⁵⁷⁸= 4567^4×394+2

∴ last digit of 4567¹⁵⁷⁸=last digit of 7² = 9

Question 5 What will be the unit digit of 199!

1	2
9	0

D

Factorial 199 = 1×2×3×4×5×..9×10×11..198×199 Thus the unit digit will be 0 (a number multiplied by 10 always end with 0)