### Question 1:

A survey was conducted on 100 children about their liking for 3 fruits i.e. Apple , Banana and Cherry .It was found that 20 children did not like any of these fruits. There were 30 children who liked apple , 30 liked banana and 40 liked cherry. Also, the number of children who liked all the three fruits were 5 .Find the number of children who liked exactly two fruits.a)10

b) 20

c) 12

d) cannot be determined

a) 6

### Question 2:

From among 6 couples , a committee of five members is to be formed. If the selected committee has no couple , then in how many ways can the committee be formed ?a) 6

b) 792

c)40

d) 192

a) 4

### Question 3:

Arun invested a sum of money at a certain rate of simple interest for a period of four years . Had he invested the same sum for a period of six years the total interest earned by him would have been fifty per cent more than the earlier interest amount . What was the rate of interest per annum ?a) 4

b) 8

c)5

d) cannot be determined

e) no one of these

### Question 4:

55% of the employees in a certain office have a tv and a refrigerator . What percentage of employees in the office have a tv, but no refrigerator ?**(i)**65% of the employees have a tv.

**(ii)**75% of the employees have a refrigerator .

### Question 5:

There are 180 blue and red balls in a cylinder and a square box together. What is the number of blue balls in the square box ?**(i)**There are70 balls in the square box

**(ii)**There are30 blue balls in the cylinder

## Solution

#### 1) a

The given data can be represented through a venn diagram ,as shown below:

IA =Number of children who liked only apple

IB = Number of children who liked only banana

IC = Number of children who liked only cherry

IIAB = Number of children who liked apple and banana only

IIBC= Number of children who liked banana and cherry only

IIAC= Number of children who liked apple and cherry only

III= Number of children who liked all the three fruits

Where , U = total number of children surveyed = 100

A=Number of children who liked apple = (IA + IIAB + IIAC + III ) = 30 ……. (i)

B= Number of children who liked banana =( IB + IIBC + IIAB + III ) =30 …… (ii)

C = Number of children who liked cherry = (IC + IIBC+ IIAC +III ) = 40 …….. (iii)

Let,

Y= number of children who did not like any of these fruits = 20

X=Number of children who like at least one of the fruits = U - Y =80

S= sum of all the three sets = A+B+C = 30+30+40 = 100

THEN , number of children who liked exactly one fruit (either apple or banana or cherry ) = I = IA + IB + IC

number of children who liked two fruits = IIBC + IIAB + IIAC = II

X = I+ II+III

Looking at (i) (ii) and (iii) and rearranging the terms we get

S = IA + IB + IC + IIAB + IIAC + IIBC + IIAB + IIBC+ IIAC + III + III+ III

=I + 2 II + 3 III

So S-X = (I + 2II + 3III ) – (I + II + III)

= II + 2 III

Putting the values in above result we get 100 – 80 = II + 2(5)

So number of children who liked exactly two fruits = II = 10

IB = Number of children who liked only banana

IC = Number of children who liked only cherry

IIAB = Number of children who liked apple and banana only

IIBC= Number of children who liked banana and cherry only

IIAC= Number of children who liked apple and cherry only

III= Number of children who liked all the three fruits

Where , U = total number of children surveyed = 100

A=Number of children who liked apple = (IA + IIAB + IIAC + III ) = 30 ……. (i)

B= Number of children who liked banana =( IB + IIBC + IIAB + III ) =30 …… (ii)

C = Number of children who liked cherry = (IC + IIBC+ IIAC +III ) = 40 …….. (iii)

Let,

Y= number of children who did not like any of these fruits = 20

X=Number of children who like at least one of the fruits = U - Y =80

S= sum of all the three sets = A+B+C = 30+30+40 = 100

THEN , number of children who liked exactly one fruit (either apple or banana or cherry ) = I = IA + IB + IC

number of children who liked two fruits = IIBC + IIAB + IIAC = II

X = I+ II+III

Looking at (i) (ii) and (iii) and rearranging the terms we get

S = IA + IB + IC + IIAB + IIAC + IIBC + IIAB + IIBC+ IIAC + III + III+ III

=I + 2 II + 3 III

So S-X = (I + 2II + 3III ) – (I + II + III)

= II + 2 III

Putting the values in above result we get 100 – 80 = II + 2(5)

So number of children who liked exactly two fruits = II = 10

#### 2) d

(representing cases where no males are selected ; 1 male is selected while his partner is not ; two males are selected while their partners are not ; and so on , through to ; no female is selected ) = 192 Hence (d)

Alternatively , the number of ways the committee can be formed = 6c5 * 25 = 192 , hence (d)

#### 3) d

cannot be determined . (d)

#### 4) c

(c) Using any one statement you can find the answer .

#### 5) a

total number of balls = 180

Statement 1 : number of balls in the square box is 70 .threfore , number of balls in the cylinder is 110 but the breakup as blue or red is not known .

Statement II : number of blue balls I nthe cylinder is 30 .But the total number of blue balls is not known . Hence , number of blue balls in the square box cannot be found out.

After combining statements I and II too answer cannot be found . Hence (a)

Statement 1 : number of balls in the square box is 70 .threfore , number of balls in the cylinder is 110 but the breakup as blue or red is not known .

Statement II : number of blue balls I nthe cylinder is 30 .But the total number of blue balls is not known . Hence , number of blue balls in the square box cannot be found out.

After combining statements I and II too answer cannot be found . Hence (a)

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