Instructions

For the following questions answer them individually

CAT 1991 - Question 41

Every day Neera's husband meets her at the city railway station at 6.00 p.m. and drives her to their residence. One day she left early from the office and reached the railway station at 5.00 p.m. She started walking towards her home, met her husband coming from their residence on the way and they reached home 10 minutes earlier than the usual time. For how long did she walk?

Video Solution

Solution

Since we know that Neera's husband drives at a uniform speed to and from his residence.
If he saved 10 mins overall travel time, he should have driven 5 mins less towards railway station and 5 mins less while driving towards residence.
If he saved 5 minutes in his return journey, he should have started to return 5 minutes before his actual return time.
When the husband met Neera, he should have met her 5 minutes before the actual meeting time i.e. at 5.55 PM.
So, Neera must have walked for 55 minutes from 5PM.

CAT 1991 - Question 42

In Sivakasi, each boy's quota of match sticks to fill into boxes is not more than 200 per session. If he reduces the number of sticks per box by 25, he can fill 3 more boxes with the total number of sticks assigned to him. Which of the following is the possible number of sticks assigned to each boy?

Video Solution

Solution

Let the number of sticks assigned to each boy be N.

Let the number of boxes be M.

So, number of sticks per box = N/M

Now, if he reduces the number of sticks in each box, the equation becomes N/(M+3) = N/M - 25

So, 25 = N/M - N/(M+3)

From the options, if N = 150, then, we get 25 = 150 [ 1/M - 1/(M+3) ]

=> 1/6 = 1/M - 1/(M+3) => M = 3

So, the number of sticks assigned to each boy = 150

CAT 1991 - Question 43

A sum of money compounded annually becomes Rs.625 in two years and Rs.675 in three years. The rate of interest per annum is

Video Solution

Solution

As we know, formulae of compound interest for 2 years will be:
$$P(1+\frac{r}{100})^{2}$$ = 625 (Where r is rate, P is principal amount)
For 3 years:
$$P(1+\frac{r}{100})^{3}$$ = 675
Dividing above two equations we will get r=8%

CAT 1991 - Question 44

In a six-node network, two nodes are connected to all the other nodes. Of the remaining four, each is connected to four nodes. What is the total number of links in the network?

Video Solution

Solution

Consider the following diagram: In this, the nodes E and F are connected to all the other nodes whereas the other four nodes are connected to four nodes each.

The total number of connections is 13.

CAT 1991 - Question 45

If x is a positive integer such that 2x +12 is perfectly divisible by x, then the number of possible values of x is

Video Solution

Solution

If 2x+12 is perfectly divisible by x, then 12 must be divisible by x.
Hence, there are six possible values of x : (1,2,3,4,6,12)

CAT 1991 - Question 46

A man starting at a point walks one km east, then two km north, then one km east, then one km north, then one km east and then one km north to arrive at the destination. What is the shortest distance from the starting point to the destination?

Video Solution

Solution

In the diagram, A is starting point and B is ending point.

Accordingly, AB = $$\sqrt{(3^2 + 4^2)} = 5$$

CAT 1991 - Question 47

An outgoing batch of students wants to gift PA system worth Rs.4200 to their school. If the teachers offer to pay 50% more than the students, and an external benefactor gives three times teachers contribution. How much should the teachers donate?

Video Solution

Solution

Let's, say students contributed x rs.
So teachers contributed = $$\frac{3x}{2}$$
And external benefactor contributed = $$\frac{9x}{2}$$
Summation of all contribution = 4200 = $$7x$$
or $$x=600$$
Hence, teachers contributed = 900

CAT 1991 - Question 48

A positive integer is said to be a prime number if it is not divisible by any positive integer other than itself and 1. Let $$p$$ be a prime number greater than 5. Then $$(p^2-1)$$ is

Video Solution

Solution

Let the Prime number be 6n+1.

So $$(p^2 - 1)$$ = 6n(6n+2) = 12n(3n+1)

For any value of n , n(3n+1) will have a factor of 2

Hence given equation will be always be divisible by 24

CAT 1991 - Question 49

To decide whether a number of n digits is divisible by 7, we can define a process by which its magnitude is reduced as follows: $$(i_{1}, i_{2}, i_{3}$$,..... are the digits of the number, starting from the most significant digit). $$i_{1} i_{2} ... i_{n} => i_{1}.3^{n-1} + i_{2}.3^{n-2} + ... + i_{n}.3^0$$.
e.g. $$259 => 2.3^2 + 5.3^1 + 9.3^0 = 18 + 15 + 9 = 42$$
Ultimately the resulting number will be seven after repeating the above process a certain number of times. After how many such stages, does the number 203 reduce to 7?

Video Solution

Solution

For 203 :
first step = $$2\times 3^2 + 0 \times 3^1 + 3 \times 3^0$$ = 21
second step = $$2 \times 3^1 + 1 \times 3^0$$ = 7
So two steps needed to reduce it to 7

CAT 1991 - Question 50

If 8 + 12 = 2, 7 + 14 = 3 then 10 + 18 = ?

Video Solution

Solution

8+12 =( 20 = 2+0) = 2
7+14 =( 21 = 2+1) = 3
10+18= (28= 2+8) = 10

CAT Content

CAT 1991 Question Paper

CAT 1991 - Question 41

CAT 1991 - Question 42

CAT 1991 - Question 43

CAT 1991 - Question 44

CAT 1991 - Question 45

CAT 1991 - Question 46

CAT 1991 - Question 47

CAT 1991 - Question 48

CAT 1991 - Question 49

CAT 1991 - Question 50

Related Formulas With Tests

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CAT Content

CAT 1991 Question Paper

CAT 1991 - Question 41

CAT 1991 - Question 42

CAT 1991 - Question 43

CAT 1991 - Question 44

CAT 1991 - Question 45

CAT 1991 - Question 46

CAT 1991 - Question 47

CAT 1991 - Question 48

CAT 1991 - Question 49

CAT 1991 - Question 50

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