Free Mixed Bag 4 Practice Test - CAT

Question 1

$21^{643} = 25 A + B .$ Find the least value that B can take?

A. 11

B. 1

C. 6

D. 16

SOLUTION

Solution : A

The question is asking for the remainder when $21^{643}$ is divided by 25.
We can use the last two digits technique to get the answer, since the remainder when a number is divided by 25, depends only on the last two digits.

For numbers ending in 1, the last two digits of a power expression $a^{b}$ can be determined as -

The last digit is always 1.

The $2^{n d}$ last digit = product of tens digit of base * unit digit of the power i.e. $21^{643} = . . . . . . 61 (6 = 2 \times 3)$

When 61 is divided by 25, the remainder is 11.

Question 2

A square is inscribed in a triangle with the two shorter sides p and q such that the square has a common right angle with the triangle. What is the perimeter of the square?

A. 4

(\frac{1}{p} + \frac{1}{q})

(p q)^{3}

\frac{p^{2} q^{2}}{(p + q)^{2}}

\frac{4 p q}{p + q}

SOLUTION

Solution : D

Option (d)

Assume a 45-45-90 triangle. p=1 and q=1

Since only variables are used, the conditions should satisfy any right angled triangle. The side of the square becomes 0.5 or $\frac{1}{2}$ ,. The perimeter = 4x $\frac{1}{2}$ =2

Look in the answer options, where on the substitution of p=1 and q=1, you get 2. only option (e) gives 2

Question 3

How many integers satisfy the inequality ?

A. 4

B. 5

C. 7

D. 10

E. infinitely many

SOLUTION

Solution : A

option (a)

The critical points are 1, -4, -2, 3, -8, -3, which can be plotted as

Hence, the regions <0 are -8 to -4 and 1 to 3= totally 4 points

Question 4

An Arithmetic Progression of positive integers has its n^th term as 66541⁸⁷⁶⁸. Find the first term of this Arithmetic Progression if the common difference = 9.

A. 0

B. 1

C. 2

D. 6

E. 7

SOLUTION

Solution : E

option (e)
From constant remainder theorem, if any of the terms in an AP is divided by the common difference, the remainder will be constant in each case.
Let us find the remainder when the n^th term of the AP is divided by 9
66541⁸⁷⁶⁸/ 9rem = 4⁸⁷⁶⁸/9rem
Euler’s number of 9 = 6.
8768 = 6k + 2
The number changes to 4²/9 = 16/9 (Rem) = 7.
answer is option (e)

Question 5

If f(x – 1) = 3x² + 5x + 6, find the value of f(x + 1).

A. 3x² + 17x + 28

B. 3x² + 17x – 28

C. 3x² + 16x + 25

D. 3x² - 16x + 25

E. 3x² + 17x – 32

SOLUTION

Solution : A

option (a)
f(x-1) = 3x² + 5x + 6
put x = x + 2, f(x + 2 – 1) = 3(x+2)² + 5(x+2) + 6
f( x + 1)= 3(x² + 4 + 4x) + 5x + 16
= 3x² + 12 + 12x + 5x + 16
= 3x² + 17x + 28
Hence option (a)

Shortcut:-Assumption method
Assume x = 1
f(0) = 14, f(2) = 48
Among the answer options, Sub x=1 and check where you are getting 48,
This is obtained only at option (a)

Shortcut of a shortcut!
Put x=1 in the question and x=-1 in the option, both should give 14

Question 6

How many numbers less than 10000 are there whose sum of the digits is less than or equal to 5?

A. 125

B. 100

C. 126

D. 189

SOLUTION

Solution : C

Let the required number be represented as ABCD.
$A + B + C + D \leq 5$

Introducing a dummy variable E,

$A + B + C + D + E = 5$

This is the arrangement of 5 zeroes and 4 ones (the number of plus signs). Answer is $^{9} C_{4} = 126$ .

Question 7

A boy had to distribute some chocolates at school,

A. 32

B. 48

C. 40

D. 5

SOLUTION

Solution : C

Method 1: conventional
Let the boy have x chocolates initially and let him have y amount of money to buy these chocolates
So with y amount of money he can buy x-8 chocolates and with 1.5y he can buy x+8 chocolates
So =
Hence x= 40
Method 2: Assumption
If we assume that the boy had 100 rs with him initially
We can now go from the answer option and pick the number of chocolates he had initially. Put in the conditions given and the price of each chocolate has to be the same:
If we assume 40 is the right option
In condition 1 we will have to buy 32 chocolates with 100 rs, where as we will have to buy 48 chocolates with 150 rs. If each of these chocolates cost the same then the assumption is correct.
i.e ,100/32 =150/48 . hence 40 is the right option.

Question 8

In a survey conducted among 100 college students in Bangalore, 100 students prefer jeans of Brand A, 80 prefer brand B, 75 prefer brand C, 70 prefer brand D and 80 prefer brand E. The minimum possible number of students who prefer all the five brands is___

SOLUTION

Solution :
option (b)

The minimum value can be calculated using a short cut technique as given below.

Step 1 :Take the difference of all from 100( total number in question is 100)

100-100 = 0

100-80= 20

100-75= 25

100-70= 30

100-80= 20

$\sum$ 95

Find the sum of all the differences = 95

Take the difference again from 100 = 100-95= 5

This is the minimum possible number of students preferring all 5 brands. Answer is option (b)

Question 9

Pipe X and Y take 60 minutes and 90 minutes respectively to fill a cistern. Pipe Z can empty a cistern in 60 minutes. Pipe X and Y are opened when the tank is empty. Pipe Z is opened by an attender when the tank is half full. However, one rainy day, the attender comes late and delays the opening of the Pipe Z by 9 minutes. What is the time difference of the tank overflow on that day as compared to the other days?

A. 13 min 50 secs

B. 13 min 30 secs

C. 22 min 30 secs

D. 14 minutes

SOLUTION

Solution : B

Pipe X and Y in one minute fill $\frac{1}{60} + \frac{1}{90} = \frac{1}{36}$ . this means that to fill half the tank i.e. $\frac{18}{36}$ , it will take 18 minutes

After half the tank in filled, all 3 taps are kept open. Pipe X and Z cancel out each other’s actions, hence only Pipe Y will be considered in filling the tank. To fill the tank Y takes 90 minutes. To fill half the tank, it will take 45 minutes. Total time taken to fill a tank on a normal day = 18+45 = 63

On the rainy day, Pipe X and Y fill for 9 minutes extra = $\frac{3}{4}$ of the tank is filled .this takes 27 minutes.

Pipe Z is opened, and Pipe Y fills for the remaining $(\frac{1}{4})^{t h}$ of the tank which will take $\frac{90}{4}$ = 22.5 minutes. Total time taken on a rainy day = 27+22.5 = 49.5

Time difference = 63- 49.5 = 13 min 30 secs. Option (b)

Question 10

The numbers 44997 and 43080, leave the same remainder when divided by x. Given that $100 \leq x \leq 999$ , how many solutions are possible for x?

A. 1

B. 3

C. 2

>

SOLUTION

Solution : C

This is based on constant remainder theorem.
Take the difference of 44997 and 43080 = 1917.

Number of 3 digit factors $= 2 (\frac{1917}{3} = 639$ and $\frac{1917}{9} = 213)$

Question 11

Find the $7^{t h}$ digit from the right in 25!

A. 0

B. 4

C. 8

D. 2

SOLUTION

Solution : B

First find the number of zeros in $25! = (\frac{25}{5} = 5) + (\frac{25}{25} = 1) = 6$
Now, we need to find the last non-zero digit at the end of 25!

1! to10! ends with 8 (non-zero).

11! to 20! also ends with 8 (non-zero).

21! to 25! ends with 6 (non-zero).

Hence $25! = 8 \times 8 \times 6 = . . . . .4$

Question 12

In an examination, the maximum marks a student can score is 20 and the minimum is 12. The students score an average of 16 marks. 3 students score 18 or more. If the sum of the scores of all the students is 240, what is the maximum number of students who can score exactly 16 marks?

A. 9

B. 11

C. 10

D. 12

E. 8

SOLUTION

Solution : C

solution: option (c)
there are totally 15 students (240/16)
3 students score 18 or more marks.
We need to maximize the number of students scoring 16, hence, assume that 3 students have scored 18 marks.
The average is 16, to balance the +6 surplus, there is one student who scores 14 and one student who scores 12.
All the remaining students score exactly 16. Maximum = 15-3-2 =10

Approach 2-Reverse Gear
Go from answer options.
option (c)
If 11 students score 16, then 176 marks are distributed. We next need to distribute 18x3 =54 marks. Total
= 230. Only 10 marks are left. Since minimum possible score is 12 for a student, 11 cannot be the answer

Question 13

Given that |x|<1 and |y|<1, then find the sum of the series x(x+y) + x²(x²+y²) + x³(x³+y³)….

C. 2xy

E. xy

SOLUTION

Solution : B

option (b)

Start out by assuming values for x and y

Say x=1/2 and y=1/2

Find the sum of the first two terms of the series = ½ + 1/8 = 5/8 = 0.63 (Approx)

The sum of the series will be slightly more than 0.63 as the denominator will keep increasing and the value of each term will reduce.

Look at the answer options

Substitute values of x=1/2 and y=1/2

Option a= xy²/(1-xy²) = 1/7 = 0.16

Option b = (x²)/(1-x²) + (xy) /( 1-xy)

. Answer = 2/3 = 0.667 (Approx)

option c = 2xy = 2/4 = ½ = 0.5

option d = 2xy + 1/x = 0.5 + 2 = 2.5

option e = xy = ¼ = 0.25

answer can never be (a), option (c) and option (e), and option (d) is too high. Answer = option (b)

Question 14

Find the common difference of an Arithmetic Progression in which the ratio of the sum of the first half of any even number of terms to the second half of the same number of terms is constant.

A. 1

B. 3

C. 4

D. -1

E. none of these

SOLUTION

Solution : E

Option (e)
Go by answer options.
For eg) for option a
1 2 3 4 5 6 7 8 is an AP
Sum of 1^st 2 terms = 3 Sum of next two terms = 7 ratio = 2/7
Sum of 1^st 4 terms = 10 Sum of next 4 terms= 26. ratio = 10/26
This cannot be the answer. Similarly, you can eliminate all other answer options as well

Question 15

A person has to move from A to B. He can only move upwards or towards the right. Find the probability that he passes through C?

\frac{1}{2}

\frac{10}{11}

\frac{13}{24}

\frac{1}{3}

SOLUTION

Solution : B

Option (b)

Totally there are 4 vertical distances and 5 horizontal distances VVVVHHHHH.

Number of ways in which a person can go from A to B via C= Arrangement of 9 things out of which 4 are identical and of one type and 5 are identical and of another type. = $\frac{9!}{4! \times 5!} = 126$ .

Number of ways of reaching C from A= HHVV= $^{4} C_{2} = 6$ . Number of ways of reaching B from C= HHHVV = $^{5} C_{3} = 10$ .
Required Probability = $\frac{(10 \times 6)}{126} = \frac{10}{21}$

Free Mixed Bag 4 Practice Test - CAT

Question 1

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Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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