### Question 1:

A sum of money was distributed equally among a certain number of children . If there were 5children less , each would have received a rupee more .But if there were 15children more , each would have received Rs.2 less .Find the sum of money distributed.a)120

b)125

c)100

d) 360

a) 15

### Question 2:

In a class test, A scored 51 marks. The test had 4 sections. In the second section, A scored 7 marks more than in the first section, and in the third section A Scored 7 marks more than in the second section. If A did not score at all in the fourth section , how much did A score in the second section ?a) 15

b) 10

c) 24

d) 17

a) 3

### Question 3:

A three digit number is such that , when divided by 13 , it yields twice the digit in the unit’s place as quotient and zero as remainder .We also know that the digit in the hundreds’place is not a prime number and that the digit in the tens’place is zero .Find the digit in the untis’place.a) 3

b)4

c)1

d)9

## DATA SUFFICIENCY :

### Question 4:

Is the product of two given consecutive natural numbers greater than 92 ?**(i)**The difference between the square of the larger number and the larger number is equal to 90.

**(ii)**The smaller of the two numbers , when added be a squared even integer and then divided by 2 , gives the result 6.5

### Question 5:

Certain quantities of two mixtures X and Y are mixed to get a new mixture containing 20% alcohol. Find the percentage of alcohol. Find the percentage of alcohol in mixture X .**(i)**Mixture Y contains twice as much alcohol as mixture X

**(II)**Qunatity of mixture X is twice that of mixture Y

## Explanation

#### 1) d

Let the amount each child has be Rs.y and the number of children be x

(x-5)(y+1) = xy ….. (i)

(x+15) (y-2) = xy (ii)

From (i) xy+x-5y-5 = xy

X – 5y = 5 …… (iii)

From (ii) xy -2x +15y – 30 = xy

-2x +15y = 30 …… (iv)

Multiplying (iii) by 2

2x-10y = 10 ….. (v)

Adding (iv) and (v), we get 5y = 40

so y =8

substituting the value of ‘y’in equation(iii)

x-40 = 5

x=45

amount distributed = xy = 45* 8 = 360

Hence , (d)360

(x-5)(y+1) = xy ….. (i)

(x+15) (y-2) = xy (ii)

From (i) xy+x-5y-5 = xy

X – 5y = 5 …… (iii)

From (ii) xy -2x +15y – 30 = xy

-2x +15y = 30 …… (iv)

Multiplying (iii) by 2

2x-10y = 10 ….. (v)

Adding (iv) and (v), we get 5y = 40

so y =8

substituting the value of ‘y’in equation(iii)

x-40 = 5

x=45

amount distributed = xy = 45* 8 = 360

Hence , (d)360

#### 2) d

If A Scored x marks in section 2 , he scored (x-7) marks I nsection 1 and (x+7) marks in section 3

Therefore , (x-7) + x + (x+7) + 0 =51

3x=51

X=17

Therefore , (x-7) + x + (x+7) + 0 =51

3x=51

X=17

#### 3) b

Let the digit in the units place be x .and that in the hundreds’place be y.

So the number be 100y + x

(100y + x ) = 13*2x +0

i.e. 100y+x = 26X so x=4y

As x and y are single digits , y can only take the values of 1 and 2 . but as y is not prime y=1 and hence x=4 , Hence (b) 4

So the number be 100y + x

(100y + x ) = 13*2x +0

i.e. 100y+x = 26X so x=4y

As x and y are single digits , y can only take the values of 1 and 2 . but as y is not prime y=1 and hence x=4 , Hence (b) 4

#### 4) e

Let the smaller and larger numbers be n and n+1 respectively …

Statement (I ) : (n+1)² – (n-1) = 90

Or n² +2n +1 –n-1 =90

Or n² +n = 90

Or n(n+1) = 90

Or (smaller number ) * ( larger number ) = 90

So , I alone is not sufficient .

Statement (II) : Let n bethe smaller of the two numbers then

N+2a² / 2 = 6.5

N+(2a)² = 13

The only squared even integer less than 13 is 4

N=9

Hence , II alone is sufficient .

Statement (I ) : (n+1)² – (n-1) = 90

Or n² +2n +1 –n-1 =90

Or n² +n = 90

Or n(n+1) = 90

Or (smaller number ) * ( larger number ) = 90

So , I alone is not sufficient .

Statement (II) : Let n bethe smaller of the two numbers then

N+2a² / 2 = 6.5

N+(2a)² = 13

The only squared even integer less than 13 is 4

N=9

Hence , II alone is sufficient .

#### 5)

Any of the statements is individually insufficient to answer the question ..

Combining statements I and ii

Let y and 2y be the quantities of mixtures Y and X respectively , and X and 2x be the quantities of alcohol in mixture X and Y , respectively .

By allegation :

x/2y * 100 2x/y *100

20

(100x/2y – 20 ) / (20-200x/y) = ½

Thus ,we can find

x/2y *100 i.e. the required value ,Hence (b) both statements needed .

Combining statements I and ii

Let y and 2y be the quantities of mixtures Y and X respectively , and X and 2x be the quantities of alcohol in mixture X and Y , respectively .

By allegation :

x/2y * 100 2x/y *100

20

(100x/2y – 20 ) / (20-200x/y) = ½

Thus ,we can find

x/2y *100 i.e. the required value ,Hence (b) both statements needed .

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