SNAP 2009 Question Paper

A=P+SI
We get SI =2000
2000 = (6000*R*4)/100
we get R=25/3 %
Now
P =525
A =700
So SI = 175
175 = (525*25/3*T)/100
We get T = 4 years

Let the initial amount with them be
J1 , B1 and T1
Now as per given
J1+B1 =4T1 (1)
T1+B1 = 3J1 (2)
Total amount with them will be 4J1 or 5T1
Now Let final amounts with them be J2 , B2 and T2 respectively .
J2 +B2 = 3T2 (3)
T2 +B2 = 2J2 (4)
Total amount with them will be 4T2 or 3J2
Now 5T1 =4T2 =4J1=3J2
From (1)
B1 = 4T1-J1
= 4T1-5/4 T1
= 11/4 T1
B2 = 3T2 -J2
=3T2 -4/3T2
=5/3 T2
5/3 (5/4) T1
= 25/12 T1
Now B1-B2 = 11/4 T1 -25/12T1 =200
so we get 8/12 T1 = 200
we get T1 = 300
so B1 = 11/4 (300) = 825
B2 = 25/12 (300) = 625
T2 = 5/4(300) = 375
J1 = 375
J2 = 4/3 (375) = 500
Total money = 825+375+300 =1500
Money T had with respect to total = 300/1500 = 1/5

Alternative solution:

It is given that the J and B had 4 times the money as T(in the beginning). Thus, if T had x amount with him, the total amount with them will be 5x.

So, let's assume that the total amount with them is 60x(LCM of 5, 4, and 3).

In the beginning: Now, the amount with T will be 12x, and the sum of the amount with J and B will be 48x.

Since the amount with T and B is 3 times that with J, the amount with J will be 15x. Thus, the amount with B will be 33x.

At the end: Since the amount with J and B is three times that of T, T will have 15x at the end. Also, the amount with T and B is twice that of J; the amount with J will be 20x. Thus, B will have 25x at the end.

Now, the ratio of money with T in the beginning with respect to the total amount = $$\frac{12x}{60x}=\frac{1}{5}$$

Thus, the correct option is D.

Let the initial amount with them be
J1 , B1 and T1
Now as per given
J1+B1 =4T1 (1)
T1+B1 = 3J1 (2)
Total amount with them will be 4J1 or 5T1
Now Let final amounts with them be J2 , B2 and T2 respectively .
J2 +B2 = 3T2 (3)
T2 +B2 = 2J2 (4)
Total amount with them will be 4T2 or 3J2
Now 5T1 =4T2 =4J1=3J2
From (1)
B1 = 4T1-J1
= 4T1-5/4 T1
= 11/4 T1
B2 = 3T2 -J2
=3T2 -4/3T2
=5/3 T2
5/3 (5/4) T1
= 25/12 T1
Now B1-B2 = 11/4 T1 -25/12T1 =200
so we get 8/12 T1 = 200
we get T1 = 300
so B1 = 11/4 (300) = 825
B2 = 25/12 (300) = 625
T2 = 5/4(300) = 375
J1 = 375
J2 = 4/3 (375) = 500
Total money = 825+375+300 = 1500
J won : 500-375 = 125
So fraction = 125/1500 = 1/12

Alternative solution:

It is given that the J and B had 4 times the money as T(in the beginning). Thus, if T had x amount with him, the total amount with them will be 5x.

So, let's assume that the total amount with them is 60x(LCM of 5, 4, and 3).

In the beginning: Now, the amount with T will be 12x, and the sum of the amount with J and B will be 48x.

Since the amount with T and B is 3 times that with J, the amount with J will be 15x. Thus, the amount with B will be 33x.

At the end: Since the amount with J and B is three times that of T, T will have 15x at the end. Also, the amount with T and B is twice that of J; the amount with J will be 20x. Thus, B will have 25x at the end.

Now, the difference between the money with J in the beginning and at the end = $$20x-15x=5x$$

Thus, the required ratio = $$\frac{5x}{60x}=\frac{1}{12}$$

Thus, the correct option is A.

Let the initial amount with them be 
J1 , B1 and T1 
Now as per given 
J1+B1 =4T1   (1)
T1+B1 = 3J1   (2)
Total amount with them will be 4J1 or 5T1
Now Let final amounts with them be J2 , B2 and T2 respectively .
J2 +B2 = 3T2    (3)
T2 +B2 = 2J2 (4)
Total amount with them will be 4T2 or 3J2
Now 5T1 =4T2 =4J1=3J2
From (1)
B1 = 4T1-J1
   = 4T1-5/4 T1
   = 11/4 T1
B2 = 3T2 -J2
    =3T2 -4/3T2
    =5/3 T2
    5/3 (5/4) T1
   = 25/12 T1
Now B1-B2 = 11/4 T1 -25/12T1 =200
so we get 8/12 T1 = 200
we get T1 = 300
so B1 = 11/4 (300) = 825
B2 = 25/12 (300) = 625
T2 = 5/4(300) = 375
J1 = 375
J2 = 4/3 (375) = 500
So B started with Rs 825

Alternative solution:

It is given that the J and B had 4 times the money as T(in the beginning). Thus, if T had x amount with him, the total amount with them will be 5x.

So, let's assume that the total amount with them is 60x(LCM of 5, 4, and 3).

In the beginning: Now, the amount with T will be 12x, and the sum of the amount with J and B will be 48x.

Since the amount with T and B is 3 times that with J, the amount with J will be 15x. Thus, the amount with B will be 33x.

At the end: Since the amount with J and B is three times that of T, T will have 15x at the end. Also, the amount with T and B is twice that of J; the amount with J will be 20x. Thus, B will have 25x at the end.

Now, the difference of money with B in the beginning and in the end = $$33x-25x=8x$$

Since B lost Rs 200 during the game, $$8x=200$$

Thus, $$x=25$$

So, the amount with B, in the beginning = $$33\times\ 25=825$$

Thus, the correct option is C.

I) $$a-c$$ is negative as $$a$$ is negative and $$c$$ is positive. Thus, the value of $$a-c$$ will be smaller than $$a$$(since more negative is added to a negative number). $$a-b$$ can be negative or positive depending upon the value of $$a$$ and $$b$$, but it can be concluded that this number will be greater than $$a$$(since a positive number is added to a negative number). Thus, $$(a-c)<(a-b)$$. Thus, the statement I is false.

II) Given $$a<b$$, $$\frac{a}{c}<\frac{b}{c}$$ as $$c$$ is positive and $$a < b$$. Therefore, statement II is true

III) $$a$$ is negative and $$b$$ is negative, this implies $$\frac{a}{b}$$ is positive. $$a$$ is negative and $$c$$ is positive, this implies $$\frac{a}{c}$$ is negative. Therefore, $$\frac{a}{b}>\frac{a}{c}$$. Statement III is true

Answer is option D

The diagonal of the first square is $$4\sqrt{2}$$
Now we know if side of a square is a
then diagonal is $$a\sqrt{\ 2}$$
Therefore here a = 4
Now area = 16
Now area of second square = 2(16) = 32
Now therefore side of 2nd square = $$\sqrt{\ 32}$$ = $$4\sqrt{\ 2}$$
Now therefore diagonal will be : $$4\sqrt{\ 2}\left(\sqrt{\ 2}\right)=\ 8$$

The maximum length of pencil will be along the longest diagonal of box
so we can say the maximum length will be : $$\sqrt{\ 4^2+6^2+8^2}=\sqrt{\ 104}\ =2\sqrt{\ 26}$$

Now X = 8.7-2.5-0.5-1 =4.7
Y = 9.1-2.5-0.5-1.5 =4.6
Z = 5.6-1-1.5-0.5 =3.1
Now at least 1 will be 
4.7+4.6+3.1+2.5+1+1.5+0.5 = 17.4

Given Readership of newspaper X = 8.7

only X + 2.5 + 0.5 + 1 = 8.7

only X = 4.7

Given Readership of newspaper Y = 9.1

only Y + 2.5 + 1.5 + 0.5 = 9.1

only Y = 4.6

Given Readership of newspaper Z = 5.6

only Z + 0.5 + 1 + 1.5 = 5.6

only Z = 2.6

Number of people who read only one newspaper = 4.7 + 4.6 + 2.6 = 11.9 lakhs

Let Sonali's monthly income be 100x 
Now Expenditure on insurance policies = 15x
Expenditure on shopping and household expenses = 55x
Therefore remaining amount = 100x-55x-15x = 30x 
Which is equal to 12,750 
Now therefore 30x=12,750 
x = 425
Therefore total income = 100x = 42,500