Instructions

For the following questions answer them individually

Question 11

In 4 years, Rs. 6000 amounts to Rs. 8000. In what time at the same rate will Rs. 525 amount to Rs. 700?

Video Solution

Solution

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

Instructions

At the start of a game of cards, J and B together had four times as much money as T, while T and B together had three times as much as J. At the end of the evening, J and B together had three times as much money as T, while T and B together had twice as much as J. B lost Rs. 200

Question 12

What fraction of the total money did T have at the beginning of the game?

Video Solution

Solution

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.

Question 13

What fraction of the total money did J win/lose?

Video Solution

Solution

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.

Question 14

What amount did B start with?

Video Solution

Solution

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.

Instructions

For the following questions answer them individually

Question 15

If a and b are negative, and c is positive, which of the following statement/s is/are true?
I) $$a - b < a - c$$
II) if $$a < b$$, then $$\frac{a}{c} < \frac{b}{c}$$
III) $$\frac{a}{b} > \frac{a}{c}$$

Video Solution

Solution

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

Question 16

The diagonal of a square is $$4\sqrt{2}$$ cm. The diagonal of another square whose area is double that of the first square is

Video Solution

Solution

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 : $$4\sqrt{\ 2}\left(\sqrt{\ 2}\right)=\ 8$$

Question 17

The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is

Video Solution

Solution

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

Instructions

Answer the questions based on the information given below
The Venn diagram given below shows the estimated readership of 3 daily newspapers (X, Y & Z) in a city. The total readership and advertising cost for each of these papers is as below

The total population of the city is estimated to be 14 million. The common readership (in lakhs) is indicated in the given Venn diagram

Question 18

The number of people (in lakhs) who read at least one newspaper is

Video Solution

Solution

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

Question 19

The number of people (in lakhs) who read only one newspaper is

Video Solution

Solution

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

Instructions

For the following questions answer them individually

Question 20

Sonali invests 15% of her monthly salary in insurance policies. She spends 55% of her monthly salary in shopping and on household expenses. She saves the remaining amount of Rs. 12,750. What is Sonali’s monthly income?

Video Solution

Solution

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

CAT Content

SNAP 2009 Question Paper

In 4 years, Rs. 6000 amounts to Rs. 8000. In what time at the same rate will Rs. 525 amount to Rs. 700?

What fraction of the total money did T have at the beginning of the game?

What fraction of the total money did J win/lose?

What amount did B start with?

If a and b are negative, and c is positive, which of the following statement/s is/are true?
I) $$a - b < a - c$$
II) if $$a < b$$, then $$\frac{a}{c} < \frac{b}{c}$$
III) $$\frac{a}{b} > \frac{a}{c}$$

The diagonal of a square is $$4\sqrt{2}$$ cm. The diagonal of another square whose area is double that of the first square is

The maximum length of a pencil that can be kept in a rectangular box of dimensions 8 cm x 6 cm x 2 cm, is

The number of people (in lakhs) who read at least one newspaper is

The number of people (in lakhs) who read only one newspaper is

Sonali invests 15% of her monthly salary in insurance policies. She spends 55% of her monthly salary in shopping and on household expenses. She saves the remaining amount of Rs. 12,750. What is Sonali’s monthly income?

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