**Question-1****Explanation**

**-2,4,-8,16 ? **

**Options :**

a)-32

b)32

c)-64

d)64

**Answer: ** ** a) -32**

-2*2=4

4*-2=-8

-8*2=16

16*-2=-32

Every number is got by multiplying the previous number by -2. Hence -32 is the answer.

**Question-2****Explanation**

**2,3,5,8,?,21**

**Options :**

a)10

b)11

c)13

d)12

**Answer: ** ** c) 13**

2+3=5

3+5=8

5+8+13

13+8=21

This is a Fibonacci series, where every number is obtained by adding the previous two numbers. Hence answer is 13.

**Question-3****Explanation**

**Sum of seven consecutive odd numbers is 133. Find the sum of the first and third number in the series.**

**Options :**

a)18

b)38

c)48

d)28

**Answer: ** ** b)38**

Since difference between any two successive odd numbers is 2, we can assume 7 consecutive odd numbers to be X-6,X-4,X-2,X,X+2,X+6. ( In our assumed series, difeerence between any two numbers is 2.)

Sum of the numbers in the series =133=X-6+X-4+x-2+x+X+2+x+4+X+6=7X

Therefore, x=133/7=19

Sum of first term and last term=X-6+X+6=2X

Substituting X=19 ( which we found earlier), we can get the answer to be 2X 19=38.

**Question-4****Explanation**

**Sum of six consecutive numbers that are divisible by 3 is 99. Find the sum of the first two numbers.**

**Options :**

a)30

b)21

c)14

d)15

**Answer: ** ** b) 21**

Difference between successive numbers divisible by 3 will be 3. Therefore, the series takes the form as below.

X-6,X-3,X,X+3,X+6,X+9

Sum of all terms in the series=99=x-6+X-3+X+X+3+x+6+X+9=6X+9

6X+9=99

Or 6X=90

Or X=15

Sum of firs two numbers=X-6+X-3=2X-9=2(15)-9=21

**Question-5****Explanation**

**Sum of five consecutive numbers that are divisible by 5 is 275. Find the last number in the series.**

**Options :**

a)65

b)60

c)75

d)70

**Answer: ** ** a) 65 **

Difference between two adjacent numbers divisible by 5 is 5.

Series takes the form X-10,X-5,X,X+5,X+10

Sum =275+X-10+X-5+X+X+5+X+10=5X

5X=275

Or X=55

Last number in the series =X+10=55+10=65

**Question-6****Explanation**

**Sum of 7 consecutive natural numbers is 91. Find the last but one number.**

**Options :**

a)15

b)19

c)21

d)25

**Answer: ** ** a) 15**

Difference between two successive natural numbers is 1.

Therefore series takes the form, x-3,x-2,x-1,x,x+1,x+2,x+3

Sum=91=x-3+x-2+x-1+x+x+1+x2+x+3=7x

7x=91

Or x=13

Last but one number =x+2=13+2=15

**Question-7****Explanation**

**Section a of class VIII had boys and girls in the ratio of 2:1 . The current number of boys in the class is 40. Few girl students where shifted from section B to Section A. This changed the ratio of boys and girls to 2:3 Find the number of girls who got shifted from section B.**

**Options :**

a)30

b)40

c)10

d)5

**Answer: ** ** b) 40**

Let the original number of boys be B and that of girls be g

Then B/G=2/1.

It is given that B =40.

Therefore, 40G=2, Or G=20

Let the number of girls who were shifted from Section B be X. It is given that the new ratio of boys. B to the increased number of girls i.e G+Xis 2/3.

Then B/(G+x)=2/3

Or 40/20+x=2/3

20+X=60

Or X=40

**Question-8****Explanation**

**Number of players in sports namely badminton, tennis and table tennis where in the ratio of 2:3:4. Originally there were 16 badminton players. If 25% of the badminton players withdrew due to some reason, What is the new ratio of the players.**

**Options :**

a)3:6:8

b)3:7:8

c)3:6:4

d)6:6:8

**Answer: ** ** a) 3:6:8**

Let the number of players in Badminton, tennis and table tennis be P,Q and R respectively.

It is given that P:Q:R =2:3:4

From the above ratio, we can easily infer that p/Q=2/3…(1) and P/R=2/4=1/2….(2)

It is given that p=16

From equation 1, P/Q=2/3 or 16/Q=2/3 . Therefore Q=16*3/2=24.

Similarly from equation. 2, P/R=1/2 or 16/R=1/2.

Therefore, R=32.

Let the number of badminton players present after 25% withdrew be P1.

Then p1=75% of p or 75/100*16

Therefore p1=12

New ratio is p1:Q:R or 12:24:32 or 3:6:8

**Question-9****Explanation**

**Let the ratio of water and milk be 1:4 in a filled can of capacity 100 litres. The mixture was added with more water so as to change the ratio of 3:8. Find the quantity of water present originally and the amount of water (in litres) added thereafter.**

**Options :**

a)15

b)8

c)12

d)10

**Answer: ** ** d) 10**

Let the original quantities of water and milk be w and M respectively.

It is given that W/M=1/4….(1)

Also it is given that , total capacity of the can=W+M+100….(2)

If a fraction exists of the from a/b=c/d. Then we can very well write a/a+b=c/c+d.

Applying similar logic to equation 1 we get, W/(W+M)=1/(1+4)=1/5

But we know W+M=100

Therefore, W/100=1/5 or W=20 litres.

Substituting W=20 litres in equation 2, we get

M=100-20=80 litres.

Let X quantity of water be added so that the ratio becomes. 3/8.

Therefore W+X/M=38…(3)

But we know w=20 and M=80

Therefore, equation 3 becomes, 20+X/80=3/8

Or 20+x=30

X=10 litres.

We have found W=20 litres and x=10 litres

**Question-10****Explanation**

** Assume the ratio of monthly wages to team leaders and team members in a factory be 2:1. If salary of team leaders is hiked by 25 % and that of team members is hiked by 30%. What would be the new ratio.**

**Options :**

a)25/13

b)13/25

c)25/26

d)26/25

**Answer: ** ** a) 25/13**

Let monthly wages of team leaders and team members be L and M respectively.

It is given that L/M=2/1

Let the hiked salary ( by 25%) of team leaders be L1.

Therefore L1=125/100*L=1.25L

Let the hiked salary (by 30%) of team members be M1.

Therefore M1=130/100*M=1.3M.

New ratio L1/M1=1.25L/1.3M

But we already know L/M=2/1

Therefore. L1/M1=1/25/1.3*2/1=2.5/1.3=25/13