Free Mixed Bag 5 Practice Test - CAT

Question 1

In a triangle r is the radius of the incircle, R is the radius of the circumcircle and A is the area of the triangle. If h $_{1}$ , h $_{2}$ and h $_{3}$ are the 3 altitudes of the triangle, then $\frac{1}{h_{1}} + \frac{1}{h_{2}} + \frac{1}{h_{3}} = ?$

\frac{r}{A}

\frac{1}{A}

\frac{1}{r}

\frac{R}{A}

SOLUTION

Solution : C

Option (c)

Let a = side, h = height,

Since no conditions are mentioned and only variables are used, we can assume an equilateral triangle.

Let r =1. Then R = 2., h = 3. In an equilateral triangle r : R : h = 1 : 2 : 3

Side a = 2 $\sqrt{3}$ ; Area= $\frac{\sqrt{3}}{4}$ a $^{2}$ = 3 $\sqrt{3}$ .

$\frac{1}{h_{1}} = \frac{1}{h_{2}} = \frac{1}{h_{3}} = \frac{1}{3}$ . Sum = 1

Only option c gives 1.

Question 2

In how many ways can 1000 be written as sum of 2 or more consecutive natural numbers?

A. 4

B. 10

C. 3

D. 16

SOLUTION

Solution : C

Shortcut - Number of ways of writing a number as a sum of 2 or more consecutive natural numbers = number of odd factors of that number-1
$1000 = 2^{3} \times 5^{3}$ . Number of odd factors = 4

Answer = 4-1 = 3

Question 3

Two guys A and B are walking down an escalator in the direction of the motion of the escalator. A takes two steps on the same time when B takes one step. When A covers 60 steps he gets out of the escalator while B takes 40 steps to get out of the escalator. The number of steps in the escalator when it is stationary___

SOLUTION

Solution :
Let the escalator moves x steps when A walks down 60 steps. Total number of steps on a stationary escalator = x + 60.

When A takes 60 steps, B should have taken 30 steps and the escalator x steps. So when B takes 40 steps, the escalator should have taken $\frac{4}{3} \times x$ steps.

So, $\frac{4}{3} \times x + 40$ = x + 60 = Total number of steps in the escalator when it is stationary. So x = 60. Hence, total number of steps = 120

Question 4

ABCDEF is a regular hexagon. PR= $\frac{1}{3}$ AF, QS= $\frac{1}{3}$ BC and PQ is parallel to RS. Also TE= $\frac{1}{3}$ FE and UD= $\frac{1}{3}$ CD and TU is parallel to ED. Find the ratio of shaded regions to the hexagon ABCDEF.

A. 1:4

B. 8:27

C. 17:54

D. 1:3

SOLUTION

Solution : B

Solution: Option (b)

A regular hexagon can be divided into 54 equal triangles as follows

Of which the shaded part constitutes $\frac{16}{54}$ part.

Hence, the answer is option (b)

Question 5

The sum of 3^rd and 15^th elements of an arithmetic progression is equal to the sum of 6^th, 11^th and 13^th elements of the same progression. Then which element of the series should necessarily be equal to zero?

A. 12^th

B. 11^th

C. 9^th

D. 8^th

E. 7^th

SOLUTION

Solution : A

Soln: Let the first term be a
Given that a+2d+a+14d=a+5d+a+10d+a+12d
=> a+11d=0. The 12^th term is necessarily zero. Answer is option (a)

Question 6

A,B,C and D are contesting for the district elections. Of the total population, 50% like A, 25% like B, 30% like C. All of those who don’t like any of these like D. All those who like B, like both A and C. All those who like C, like A as well. Number of people who like C and not B is what percent of those who like D?

A. 0

B. 10

C. 15

D. 25

E. 20

SOLUTION

Solution : B

option (b)From the Venn diagram given, it's clear that no. of people who like C and not B are 30 - 25 = 5. People who like D are 100 - 50 = 50. Therefore, $\frac{5}{50} \times 100 = 10$ .

Question 7

If x, y, z are distinct positive real numbers then the expression below would be

A. Greater than 4

B. Greater than 5

C. Greater than 6

D. Less than 6

E. Less than 2.

SOLUTION

Solution : C

Alternatively:- Use assumption

At x=1, y=2, z=3 (these are the 3 minimum distinct positive integers)

Value of the expression is 48/6 = 8

Try for one more value x=3, y=2 and z=4

Again the value >6. Answer is option (C)

Question 8

The value of

A. 2√2

B. 3

C. 3√3

D. 9+4√5

E. 3+2√5

SOLUTION

Solution : C

Question 9

Given that x,y,z are all distinct. Then find the value of

A. x²

B. x+5

C. (x+2)²

D. (x-6)²

E. (x-4)²

SOLUTION

Solution : A

Option (a) Assume values. Take x=1,y=2,z=3 and a=1

Substitute in equation and the result you get is 1. look in the answer options for 1. answer a

Question 10

In a game of twist, there are 10 indistinguishable coins which need to be arranged in the boxes below such that no row is empty and each box contains no more than one coin. In how ways can the coins be arranged in the following layout

A. 64

B. 63

C. 61

D. 60

E. none of these

SOLUTION

Solution : A

Total number of possible selection of boxes for the 10 coins = $^{12} C_{10} = 66$

Cases in which rows will be empty = 2, when row 2, 3 and 4 and completely filled and when row 1,2, and 3 are completely filled. In all other cases, atleast one coin will be in each row.

Answer = 66 - 2 = 64

Question 11

ABCD is a square. O is the point of intersection of the diagonals of the square. Given that RS= $\frac{1}{4}$ BC and QC= $\frac{1}{4}$ DC. Find ratios of areas of Shaded region and unshaded region ?

A. 1:8

B. 2:7

C. 1:6

D. 1:7

SOLUTION

Solution : D

Option (d)
We can graphically divide the square into 16 equal parts as follows

The shaded region forms 2 out of the 16 parts. Hence, ratio of areas = 2:14 = 1:7

Question 12

If a and b are the roots of the equation x²+px+1=0 and ; c and d are the roots of the equation x²+qx+1=0, then (a-c)(b-c)(a+d)(b+d)=?

A. p²-q²

B. a²+b²

C. q²-p²

D. q²+p²

E. p²

SOLUTION

Solution : C

Shortcut: Assumption
Option (c)
product of roots=1 is the only constraint.
Assume values that satisfy the constraint and substitute, take a= 1 and b= 1, then p= -2
Assume c=1/2 and d= 2; q= -5/2
Question is = (a-c)(b-c)(a+d)(b+d)=9/4
Only option (c) satisfies this => q² –p² = 25/4 – 4 =9/4

Question 13

Find the last two digits of the expression $84^{289}$ ?

A. 24

B. 44

C. 64

D. 84

SOLUTION

Solution : C

$84 = 21 \times 4$
The question changes to $(21)^{289} \times 4^{289}$

Using the last two digits technique for 1 -

For all numbers ending in 1, last digit is always 1.

Second last digit is the product of $10^{t h}$ digit of the base and the unit digit of the power.

E.g.

In this question $21^{289} =__81$

Using the last two digits technique based on the pattern of 2, we can arrive at the answer very fast.

$4^{289} = (2^{2})^{289} = 2^{578} = (2^{10})^{57} \times 2^{8} =__24 \times__56$

Since, $(2^{10})^{o d d}$ power always ends in 24.

Answer $=__81 \times__24 \times__56 =__64$

Question 14

If the graph of the function is symmetric about the y axis then determine the possible value of ‘n.’

A. 2

B. 4

C. 3

D. 8

E. 6

SOLUTION

Solution : C

Soln:

f(y) = (3^y – 1)/[(yⁿ)(3^y + 1)] = [1/(yⁿ)][ (3^y – 1)/(3^y + 1)] = h(y) g(y)

Consider g(y) = (3^y – 1)/(3^y + 1)

g(-y)= (3^-y – 1)/(3^-y + 1)= (1 – 3^y)/(1 + 3^y) = -g(y)

So, g(y) is an odd function.

Now, as f(y) is symmetric about the y axis, f(y) is an even function.

h(y)*odd function = even function

→ h(y) should be an odd function

h(y) = 1/yⁿ

for h(y) to be odd, n has to be odd.

Hence option (c)

Shortcut

Put y=1 and y=-1, since it is mentioned that it is symmetric about the y axis. The values of both expressions should be equal. This will happen only if n is odd. Hence the answer is option (c)

Question 15

What is the number of distinct terms in the expansion of $(p + q + r + s)^{10}$ ?

A. 245

B. 105

C. 378

D. none of these

SOLUTION

Solution : D

option (d)

The number of distinct terms = arrangement of 10 zeroes and 3 ones = $^{13} C_{3} = 286$

Free Mixed Bag 5 Practice Test - CAT

Question 1

SOLUTION

Question 2

SOLUTION

Question 3

SOLUTION

Question 4

SOLUTION

Question 5

SOLUTION

Question 6

SOLUTION

Question 7

SOLUTION

Question 8

SOLUTION

Question 9

SOLUTION

Question 10

SOLUTION

Question 11

SOLUTION

Question 12

SOLUTION

Question 13

SOLUTION

Question 14

SOLUTION

Question 15

SOLUTION

Related ExamsMixed Bag 6Mixed Bag 7Mixed Bag 8

Related Exams
Mixed Bag 6
Mixed Bag 7
Mixed Bag 8