Free Quant Practice Test - CAT

Question 1

Anil has bought a new game, which is in the form of 6 concentric circles.The 6th circle has a radius of 6 and it has 6 openings. A ball is positioned on the circumference of the innermost circle and it finally exits from the circumference of the outermost circle. Thus, the ball starts travelling from the smallest circle 1 and leaves it at the point of opening on the normal to reach the next circle 2, then it moves in a similar manner to 3 and so on.The total number of different paths in which the ball can take to come out of the 6th circle is?

2 $^{6}$ *6!

2 $^{5}$ *6!

2 $^{5}$ (5)!

SOLUTION

Solution : A

There are n openings to come out of the nth circle and it can travel in

the clockwise or anticlockwise direction to enter the (n+1)th circle.

Thus the number of paths is

(1*2)*(2*2)*(3*2)......(6*2) = 6!*2 $^{6}$

Shortcut

This is a numbers to numbers question.

Use the technique of double substitution

Rewrite the answer options in terms of n as follows,

a) 2 $^{n}$ n! B) 2 $^{n - 1}$ *n! c) n! d) 2 $^{n - 1}$ (n -1)! e) 2 $^{n + 1}$ (n-1)!

Now at n=1, there are 2 ways in which they can exit from the innermost

circle. At n=1 only option (a) gives 2. Answer is option (a)

Question 2

If a,b,c are in Geometric Progression, then the equation $a x^{2} - 2 b x + c = 0$ and $d x^{2} - 2 e x + f = 0$ have a common root if $\frac{d}{a}, \frac{e}{b}, \frac{f}{c}$ are in?

Arithmetic Progession

Geometric Progression

Harmonic Progression

GP or HP

SOLUTION

Solution : A

Solve it using Assumption

Put a=1,b=1,c=1.

Equation (1) is $x^{2} - 2 x + 1 = 0$ . Sum of the roots= 2 and product of roots= 1

Thus the roots are 1,1

Let the common root be 1

Let us assume the other root =2

The equation changes to

$x^{2} + 3 x + 2 = 0$

Thus, $\frac{2 e}{d} = 3$ and $\frac{f}{d} = 2$

When d=1, f=2 and $e = \frac{3}{2}$

Thus $\frac{d}{a}$ ; $\frac{e}{b}$ ; $\frac{f}{c}$ = 1, 1.5, 2 are in AP. Answer is option (a)

Question 3

Let g(x)= |x+1|. The number of values that 'x' can take (given that its range is between -2 $&$ + 2 (both inclusive) for which g(x-3), g(x-1), g(x+1) are in AP is?

SOLUTION

Solution : B

g(x-3) = |x-2|

g(x-1)=|x|

g(x+1)= |x+2|

These will be in AP if

2|x| = |x-2|+|x+2|

Only x= -2 and x=2 satisfy the above equation in the given range.

Question 4

There are 3 types of Coffee powder available at La Cafe. They are graded based on the chicory content. The three grades of coffee contain $1 %$ chicory, $2 %$ chicory and $3 %$ chicory in the coffee blend. If p kgs of the $1 %$ grade and q kgs of the $2 %$ grade and r kgs of the $3 %$ grade are mixed to give p+q+r kgs of $1.5 %$ grade, what is p in terms of q and r?

q+3r

$\frac{q + r}{4}$

2q + 3r

4.5q+3r

3q+ r

SOLUTION

Solution : A

Put r=0, then if 1 kg of $1 %$ grade is mixed with 1 kg of $2 %$ grade, we get

2 kg of $1.5 %$ Grade. Hence p=1, q=1 and r=0. Substitute in answer

Options to get answer as option (a)

Question 5

There are 300 people in a locality. Every person in the locality reads 5 different magazines and every magazine is read by 60 people. How many different magazines are there?

atleast 30

exactly 25

atleast 35

between 30 & 35

SOLUTION

Solution : B

Let the number of magazines = n

The total number of times the magazines are read = 300*5 = 1500 = 60*n

Hence n=25.

Question 6

Which of the following values of x would give a number such that $(1073)^{71}$ -x is divisible by 100?

SOLUTION

Solution : B

Conventional Approach

Using the technique of last two digits:

The question is to find the last two digits of $(1073)^{71}$ (which is nothing but the remainder when the number is divided by 100).

We will solve this using a technique as explained below and also in our demo tutorial in arithmetic.

$(1073)^{71} = (. .73)^{71} = (. .73)^{68} \times (. .73)^{3}$
$= (. {.73}^{2} \times . {.73}^{2})^{17} \times (. .73)^{3}$
$= (. .29 \times . .29)^{17} \times (. .73)^{3}$
$= (. .41)^{17} \times (. .73)^{3}$
$= 81 \times 29 \times 73 = . . .77$

Shortcut of a shortcut!

Using answer options:

From the power cycle of 3 (3,9,7,1) -

71= 4k+3

Thus, the last digit should be 7. The only answer option ending with 7 is option (b).

Question 7

Let $y = \frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + . . . . .}}}}$ What is the value of y?

$\frac{\sqrt{13} + 3}{2}$

$\frac{\sqrt{13} - 3}{2}$

$\frac{\sqrt{15} + 3}{2}$

$\frac{\sqrt{15} - 3}{2}$

SOLUTION

Solution : D

Given : $y = \frac{1}{2 + \frac{1}{3 + \frac{1}{2 + \frac{1}{3 + . . . . . . . .}}}}$

Thus $y = \frac{1}{2 + \frac{1}{3 + y}}$

$y = \frac{3 + y}{6 + 2 y + 1}; \Rightarrow 2 y^{2} + 7 y = 3 + y;$ On solving, $y = \frac{- 3 \pm \sqrt{15}}{2};$ since the given fraction is positive, the value of y is $\frac{\sqrt{15} - 3}{2}$

You can also go from elimination of answer options. Answer should

$<$ 0.5 and close to it. This happens only at option (d)

Question 8

In the given figure, MN is the common tangent to both the inner circles with M and N lying on the circumference of the biggest circle. MN=16. The centers of the smaller circles lie on the diameter of the biggest circle. Find the area of the shaded region

16 $π$

18 $π$

20 $π$

32 $π$

SOLUTION

Solution : D

e. 32 $π$

If the radii of the three circles be a, b and c decreasing order of size,then

2b+2c=2a; $\Rightarrow$ a= b+c.

$Δ$ AMC and $Δ$ ANC are congruent. $Δ$ ABM and $Δ$ ABN are similar; so $Δ$ BMC and $Δ$ BNC are congruent. Thus MB=BN=8.

Thus, AB.BC = MB.BN=64.

AB.BC= 2c.2b=64. (since triangle AMB=triangle MCB)

The required area is π ( (b+c) $^{2}$ - b $^{2}$ - c $^{2}$ ); = 2 $π$ bc = 2 $π$ .16=32 $π$ .

Question 9

Find the remainder when $7^{55}$ is divided by 29.

SOLUTION

Solution : B

We need to use reverse of Euler's theorem as the Euler's number for 29=28.

And the question reduces to $(\frac{7^{27}}{29})$ (Rem).

The right answer multiplied by 7, should give a remainder of 1 when divided by 29, as $(\frac{7^{28}}{29})$ (Rem) = 1.

Thus, the answer is option (b).

Verification -

25 or $(- 4) \times 7$ = -28 $⟹$ remainder of 1.

Question 10

If $2 g (x) + 3 g (\frac{1}{x}) = x - 1$ , then find the value of g(2).

$- \frac{7}{10}$

$\frac{12}{10}$

$\frac{7}{11}$

$\frac{1}{2}$

SOLUTION

Solution : A

Shortcut: Direct substitution

First put x=2

$2 g (2) + 3 g (\frac{1}{2}) = 2 - 1 = 1$

Put, $x = \frac{1}{2}$

$2 g (\frac{1}{2}) + 3 g (2) = \frac{1}{2} - 1 = - \frac{1}{2}$

Solving (1) and (2), $g (2) = - \frac{7}{10}$

Question 11

An eight digit number divisible by 9 is to be formed by using 8 digits out of the digits 0,1,2,3,4,5,6,7,8,9 without replacement.
The number of ways in which can be done is

2(7!)

4(7!)

(36)(7!)

(7!)(33)

SOLUTION

Solution : C

From the digits 0 1 2 3 4 5 6 7 8 9

The sum is 45

Thus, every time we remove 2 digits which add up to 9, we will have a new set of 8 numbers which are divisible by 9

Case 1

Removing 0 and 9

Number of cases possible = 8!

Case 2

Removing

(1)8 and 1

(2)7 and 2

(3)6 and 3

(4)5 and 4

In each case, total number of cases = 8!-7!

(we subtract 7! Because those are the cases where 0 is the first digit)

Hence, total number of ways = 4(8!-7!)+8!= 36x7!.

Question 12

Four concentric circles share the same centre. The smallest circle has a radius of 1 inch. For n ≥ 1, the area of the nth smallest circle in square inches, A_n, is given by $A_{n + 1} = A_{n} + (2 n + 1) π$ . What is the ratio of the sum of the areas of the four circles to the sum of their circumferences?

$\frac{3}{2}$

SOLUTION

Solution : B

Following the pattern

A $_{1}$ = $π$
A $_{2}$ =4 $π$
A $_{3}$ =9 $π$
A $_{4}$ =16 $π$
Hence r = 1,2,3 4
Thus
C $_{1}$ = 2 $π$
C $_{2}$ = 4 $π$
C $_{3}$ = 6 $π$
C $_{4}$ = 8 $π$
Ratio = 1.5

Question 13

Given that a_i $>$ 0 and i belongs to a set of natural numbers. If $a_{1}, a_{2}, a_{3} . . . . . a_{2 n}$ are in AP, then find the value of $\frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{2}}} + \frac{a_{2} + a_{2 n} - 1}{{\sqrt{a}}_{2} + \sqrt{a_{3}}} . . . . . . . . . . . . + \frac{a_{n} + a_{n + 1}}{\sqrt{a_{n}} + \sqrt{a_{n + 1}}}$

n-1

$n \frac{a_{1} + a_{2 n}}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}$

$\frac{n - 1}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}$

$\frac{n + 1}{\sqrt{a_{1}} + \sqrt{a_{n + 1}}}$

SOLUTION

Solution : B

Let the AP be 1,2,3,4,5

At n=1, the AP = 1,2

$a_{1} = 1$ and $a_{2} = 2$

The value of the expression $= \frac{3}{2.4}$

Only option (b) gives this value

Question 14

Find the number of values of x for which $\frac{8^{x} + 27^{x}}{12^{x} + 18^{x}} = \frac{7}{6}$

$>$ 3

SOLUTION

Solution : B

Let $2^{x} = a$ and $3^{x} = b,$ the equation becomes

$\frac{a^{3} + b^{3}}{a^{2 b} + b^{2 a}} = \frac{7}{6}$

$o r \frac{a^{2} - a b + b^{2}}{a b} = \frac{7}{6}$

$o r 6 a^{2} - 13 a b + 6 b^{2} = 0$

$(2 a - 3 b) (3 a - 2 b) = 0$

Therefore $2^{x + 1} = 3^{x + 1}$ or $2^{x - 1} = 3^{x - 1}$ which implies that x= -1 and x= 1. Hence only two solutions exist.

Question 15

Two guys Abhinav and Bineesh are walking downward and upward respectively on a descending escalator. Abhinav takes three steps in the same time when Bineesh takes two steps. When Abhinav covers 90 steps he gets out of the escalator while Bineesh takes 80 steps to get out of the escalator. If they start from opposite ends using the same escalator find the difference in steps covered by them when they meet.

SOLUTION

Solution : D

Let the escalator move x steps when A walks down 90 steps. Total number of steps on a stationary escalator = x + 90.

When A takes 90 steps, B should have taken 60 steps and the escalator x steps. So when B takes 80 steps, the escalator should have taken $\frac{4}{3} \times x$ steps.

So, $\frac{4}{3} \times x + 80 = x + 90$

Total number of steps in the escalator when it is stationary. So x = 30. Hence, total number of steps = 120.

By the time they meet, together they will 120 steps in the ratio 3:2. i.e 72 and 48 steps. So difference is 24. Option (d).

Question 16

If $N = 70^{2730} + 30^{2780}$ is divisible by $10^{x}$ (x is a natural number), find the rightmost non-zero integer in the remainder when N is divided by $10^{x + 1}$ (if x is maximum).

A. 7

B. 1

C. 9

D. 0

SOLUTION

Solution : C

The question is based on the last non-zero digit (i.e. remainder when the number is divided by 10), as when x is maximum, $10^{x}$ will cancel out all the zeroes in the end.

It now becomes a question based on the rightmost non-zero integer in $7^{2730}$ ( $30^{2780}$ will end with more zeroes).

$7^{2730} ⟹ 2730 = 4 K + 2$
Thus, the rightmost non-zero integer is 9.

Question 17

When Arpita was going through her address book, she glanced upon the telephone number of her old friend, Gita. However, the last two digits were erased. But being a math freak, Arpita remembered that Gita had a 6 digit telephone number starting with 2345_ _ whose digital sum =7 (digital sum is the sum of the digits of a number till a single digit number is obtained). Find the maximum number of trials till she dials the right number.

SOLUTION

Solution : D

As of now the sum is 14. Let the last two digits be a and b.

With an upper limit of 9 for a and b:
a+b = 2 and a+b = 11

Number of solutions for a+b = 2 is 3 [(0,2), (2, 0), (1, 1)].

Number of solutions for a+b = 11 is 8 ( $^{12} C_{1} - 4 = 8$ )

Total number of ways = 11

Shortcut:

The question is about an AP with common difference = 9 in a span of 100 numbers. Answer is $\frac{100}{9} = 11$ .

Question 18

A 50m long platoon is marching ahead. The last person in the platoon wants to give a letter to the first person leading the platoon. So while the platoon is marching he runs ahead, reaches the first person and hands over the letter to him and without stopping he runs and comes back to his original position. In the mean time the whole platoon has moved ahead by 50m. How much distance (approximately) did the last person cover in that time. Assuming that he ran the whole distance with uniform speed

120m

100m

102m

97m

111m

SOLUTION

Solution : A

The last person covered 120.71 meters.

It is given that the platoon and the last person moved with uniform speed. Also, they both moved for the identical amount of time. Hence, the ratio of the distance they covered - while person moving forward and backward - are equal. Let's assume that when the last person reached the first person, the platoon moved X meters forward.

Thus, while moving forward the last person moved (50+X) meters whereas the platoon moved X meters.

Similarly, while moving back the last person moved [50-(50-X)] = X meters whereas the platoon moved (50-X) meters.

Now, as the ratios are equal,

$\frac{[50 + X]}{X} = \frac{X}{[50 - X]}$
$[50 + X] \times [50 - X] = X \times X$

Solving, X=35.355 meters

Thus, total distance covered by the last person
$= [50 + X] + X$
$= 2 \times X + 50$
$= 2 \times [35.355] + 50$
=120.71 metres.
=120 m (Approximately)

Question 19

A number is said to be a 'zeroth number' if the sum of the squares of its digits ends in a zero. How many two-digit 'zeroth numbers' are there?

SOLUTION

Solution : D

Let the number be 10x + y.

Then, $1 \leq x \leq 9, 0, \leq y \leq 9$ and $x^{2} + y^{2} = 10 k .$

$\begin{matrix} x & x^{2} & y^{2} m u s t e n d i n & P o s s i b l e v a l u e o f y 1 & 1 & 9 & 3, 7 2 & 4 & 6 & 4, 6 3 & 9 & 1 & 1, 9 4 & 16 & 4 & 2, 8 5 & 25 & 5 & 5 6 & 36 & 4 & 2, 8 7 & 49 & 1 & 1, 9 8 & 64 & 6 & 4, 6 9 & 81 & 9 & 3, 7 \end{matrix}$

The required number = 17, hence option (d).

Question 20

Let x and y be consecutive integers. Suppose m be the number of solutions of $x^{3} - y^{3} = 3 k^{2}$ and n be the number of solutions of $x^{3} - y^{3} = 2 t^{2}$ , where k, t are integers. Then m + n equals:

D. None of these

SOLUTION

Solution : D

Let x=y+1 so, $x^{3} - y^{3} = 1 + 3 y (y + 1)$ clearly this is not divisible by either 2 or 3 so m = 0 and n = 0 so m+n=0.
Hence option (d)

Question 21

In the given equilateral triangle, the smaller circle and the bigger semi circle are inscribed inside. What is the value of $\frac{X}{Y}$

4.5

SOLUTION

Solution : A

Line PV bisects line QR. $Δ$ PST and $Δ$ TUV are similar.

Hence $\frac{Z}{Y + X} = \frac{Y}{X = Y}; \Rightarrow \frac{Y}{Z} = \frac{X - Y}{X + Y};$

but in $Δ$ PST, $\frac{Y}{Z}$ = sin 30 = $\frac{1}{2}$

Hence $(\frac{1}{2})$ = $\frac{X - Y}{X + Y}; \Rightarrow \frac{X}{Y} = 3$

Question 22

An engine without wagon can go at the rate of 24 kmph and its speed diminishes by a quantity which varies as the square root of the
number of wagons attached. With four wagons its speed 20 kmph. The greatest number of wagons with which the wagon can move is :

143

144

145

122

123

SOLUTION

Solution : A

Decrease in speed = 24 - 20 = 4

$4 = k \sqrt{4} \Rightarrow k = 2$

Now, for the speed to decrease to zero

$24 = k \sqrt{x},$ where x is the number of wagons which

will stop the engine

X= 144

The greatest number of wagons with which the wagon can move is 143

Question 23

Find the minimum value of a+b+c+d in the equation $x^{5} - a x^{4} + b x^{3} - c x^{2} + d x - 243 = 0,$ if it is given that the roots are positive real numbers ?

780

660

720

SOLUTION

Solution : A

There are 5 roots in this equation.

Here the product of the roots is given as 243. Hence the sum will be minimum when the roots are equal. In fact a,b,c and d will be minimum when the roots are equal.

243 = $3^{5}$

Hence, the roots will be 3,3,3,3 and 3

a= sum of the roots = 15

b=Sum of product of roots taken 2 at a time = (3 $\times$ 3)+(3 $\times$ 3)...10 times or = $^{5} C_{2} \times 9$ = 90

c= Sum of product of roots taken 3 at a time = (3 $\times$ 3 $\times$ 3)+(3 $\times$ 3 $\times$ 3)...5 times or = $^{5} C_{3} \times 27$ = 270

d=Sum of product of roots taken 4 at a time = (3 $\times$ 3 $\times$ 3 $\times$ 3)....5 times= 405

Hence, a+b+c+d = 780

Question 24

In a class, there are 100 students. They are asked to vote for the teachers they like among five teachers namely Vanita Sehgal, Rashmi Ojha, Dharmendra Kumar, Satish M and L Maraditya. Any student can vote for any number of teachers but they have to vote for at least one. Below is the number of votes achieved by the five teachers. Vanita Sehgal: 47, Rashmi Ojha: 15, Dharmendra Kumar: 41, Satish M: 28, L Maraditya: 25 Determine the least number of students who have voted for a minimum of two of the five teachers.

SOLUTION

Solution : B

X= I + II + III + IV + V = 100

S= I + 2II + 3III + 4IV +5V = 47 + 15 + 41 + 28 + 25 = 156 (each circle added)

S-X= 56= II +2III+3IV+4V

To minimize (II+III+IV+V), we have to maximize 1

To do this make II, III and IV=0

Thus, 4V= 56 $\Rightarrow$ V= 14

Thus minimum of (II+III+IV+V)= 14

Question 25

Suppose $f (x) = (x + 1)^{2}$ for $x \geq 1.$ If g(x) is the function whose graph is reflection of the graph of f(x) with respect to the line y = x, then g(x) equals:

$- \sqrt{x} - 1$

$\sqrt{x + 1}$

$\sqrt{x} - 1$

$\sqrt{x}$

SOLUTION

Solution : C

Reflection of a function about the line y=x is nothing but finding the inverse of the function. So the question basically is to find the inverse of f(x)

$y = (x + 1)^{2}$
$\sqrt{y} - 1 = x$
Replace y with x gives inverse function of f(x)
$f^{- 1} (x) = \sqrt{x} - 1$

Question 26

If $f (x) = \frac{x}{x + 1},$ determine the value of g(2) if, $g (x) = f [\frac{1}{x + 1}] - \frac{1}{f (x + 1)} .$

$\frac{1}{12}$

$\frac{13}{12}$

$\frac{- 13}{12}$

$\frac{- 1}{12}$

SOLUTION

Solution : C

Method :1 Conventional Approach

$f [\frac{1}{x + 1}] = \frac{[\frac{1}{x + 1}]}{[{\frac{1}{x + 1}} + 1]} = \frac{1}{x + 2}$

$\frac{1}{f (x + 1)} = \frac{1}{[\frac{x + 1}{x + 2}]} = \frac{x + 2}{x + 1}$

$g (x) = f [\frac{1}{x + 1}] - \frac{1}{f (x + 1)} = [\frac{1}{x + 2}] - [\frac{x + 2}{x + 1}]$

$g (2) = (\frac{1}{4}) - (\frac{4}{3}) = - \frac{13}{12}$

Hence option (c)

SHORTCUT : Direct Substitution

When x=2
$G (2) = f (\frac{1}{3}) - \frac{1}{f (3)} = (\frac{1}{4}) - \frac{4}{3} = - \frac{13}{12} .$

Question 27

Find the remainder when 4001000 is divided by 17?

-1

SOLUTION

Solution : B

This needs the application of basic remainder theorem, fermat theorem and frequency method

First find the remainder when 400 is divided by 17= 9

The problem changes to $\frac{9^{1000}}{17}$

From fermat theorem the Euler's number of 17 is 16.(for all prime numbers, euler's number=N-1)

1000 = 16k +8

Problem now changes to $\frac{9^{8}}{17}$

Going by frequency method . $\frac{9^{4}}{17}$ gives remainder -1 so $\frac{9^{8}}{17}$ will give remainder 1.

Question 28

If $\frac{1}{6} t h$ of a number is subtracted from 290, the result is equal to 50% of the number. Find the number.___

SOLUTION

Solution :
Let the number be x.

According to the statement,

$290 - \frac{x}{6} = \frac{x}{2}$

or $\frac{x}{2} + \frac{x}{6} = 290$

or $\frac{4 x}{6} = 290$

or x = 435.

Therefore, the number is 435.

Question 29

The greatest perfect square less than 20000 which is divisible by each of 2,5,7 and 8 is :___

SOLUTION

Solution :
The least number that is exactly divisible by 2, 5, 7 and 8 is its LCM which is 280.

Since 280 is not a perfect square, the multiples of 280 are checked for perfect square.

Now, $280 = 2^{3} \times 5^{1} \times 7^{1}$ . So, 280 must be multiplied with a minimum of $2^{1} \times 5^{1} \times 7^{1}$ i.e. 70 to make it a perfect square.
So, the required number is: $280 \times 70 = 19600$ which is the greatest perfect square less than 20000 satisfying this condition.

Question 30

How many positive integers $\leq$ 1260 are relatively prime to 1260?

723

360

900

288

SOLUTION

Solution : D

Method 1: Conventional Approach

$1260 = 2^{2} \times 3^{2} \times 5 \times 7$

We need to find the number of positive integers 1260 which are not divisible by 2, 3, 5 or 7.

Let A be the set consisting of multiples of $2 \leq 1260$ ;
B be the set consisting of multiples of 3;
C be the set consisting of multiples of 5 & D of multiples of 7.

$A = \frac{1260}{2} = 630,$
$B = \frac{1260}{3} = 420,$
$C = \frac{1260}{5} = 252,$
$D = \frac{1260}{7} = 180$
$A Π B = \frac{1260}{2 \times 3} = 210 A Π C = \frac{1260}{2 \times 5} = 126 A Π D = \frac{1260}{2 \times 7} = 90$

$B Π C = \frac{1260}{3 \times 5} = 84$
$B Π D = \frac{1260}{3 \times 7} = 60 C Π D = \frac{1260}{5 \times 7} = 36$
$A Π B Π C = \frac{1260}{2 \times 3 \times 5} = 42$
$A Π B Π D = \frac{1260}{2 \times 3 \times 7} = 30$
$A Π C Π D = \frac{1260}{2 \times 5 \times 7} = 18$
$B Π C Π D = \frac{1260}{3 \times 5 \times 7} = 12$
$A Π B Π C Π D = \frac{1260}{2 \times 3 \times 5 \times 7} = 6$

So, $A \cup B \cup C \cup D = 630 + 420 + 252 + 180 - 210 - 126 - 90 - 84 - 60 - 36 + 42 + 30 + 18 + 12 - 6 = 972$

So, 1260 - 972 = 288 integers are relatively prime to 1260.

Method 2: Shortcut: Using the concept of Euler's number

The number of positive integers $\leq 1260$ which are relatively prime to 1260 can be directly found by finding the Euler's number of 1260
$= N (1 - \frac{1}{2}) \times (1 - \frac{1}{3}) \times (1 - \frac{1}{5}) \times (1 - \frac{1}{7})$
$= 1260 \times \frac{1}{2} \times \frac{2}{3} \times \frac{4}{5} \times \frac{6}{7} = 288$

(where 2,3,5,7 are the prime factors of 1260)

Question 31

How many positive integers $x \leq 10000$ are there such that the difference $2^{x} - x^{2}$ is not divisible by 7?

6662

7142

3687

7562

SOLUTION

Solution : B

$2^{x} - x^{2}$ is divisible by 7 if and only if $2^{x}$ and $x^{2}$ both leave the same remainder when divided by 7.

$\begin{matrix} x & 2^{x} & R e m a i n d e r w i t h 7 1 & 2 & 2 2 & 4 & 4 3 & 8 & 1 4 & 16 & 2 5 & 32 & 4 6 & 64 & 1 7 & 128 & 2 8 & 256 & 4 \end{matrix}$

$\begin{matrix} x & x^{2} & R e m a i n d e r w i t h 7 1 & 1 & 1 2 & 4 & 4 3 & 9 & 2 4 & 16 & 2 5 & 25 & 4 6 & 36 & 1 7 & 49 & 0 8 & 64 & 1 9 & 81 & 4 10 & 100 & 2 \end{matrix}$

Remainder of $2^{x}$ repeats after every 3 and remainder of $x^{2}$ repeats after 7. The remainders of both $2^{x}$ and $x^{2}$ will repeat after a period of length 3 $\times$ 7 = 21

$\begin{matrix} x & 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & \frac{1}{0} & \frac{1}{1} & \frac{1}{2} & \frac{1}{3} & \frac{1}{4} & \frac{1}{5} & \frac{1}{6} & \frac{1}{7} & \frac{1}{8} & \frac{1}{9} & \frac{2}{0} & \frac{2}{1} 2 x & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 & 2 & 4 & 1 x^{2} & 1 & 4 & 2 & 2 & 4 & 1 & 0 & 1 & 4 & 2 & 2 & 4 & 1 & 0 & 1 & 4 & 2 & 2 & 4 & 1 & 0 \end{matrix}$

In this set of 21 values, we see that there are 6 values of x for which the remainders are same.

As per the periodicity, within every consecutive 21 values, 6 will repeat.

10,000 = 21 $\times$ 476 + 4

The 476 groups contribute 476 $\times$ 6 = 2856 values and the remaining 4 will give 2 values.
So, total 2858 integers $\leq 10000$ are divisible. Thus, 10000- 2858 = 7142 integers are not divisible.

Question 32

The integers from 1 to 1000 are written in order around a circle. Starting at 1, every fifteenth number is marked (that is 1, 16, 31, etc.). This process is continued until a number is reached which has already been marked. How many unmarked numbers remain?

732

136

635

800

SOLUTION

Solution : D

Consider the first round: - All the numbers which leave a remainder of 1 when divided by 15 will be marked. (1, 16, 31, 46 ... 991)

Consider the second round: - The first number to be marked is 991 + 15 - 1000 = 6. Thereafter all the numbers which leave a remainder 6 when divided by 15 will be marked. (6, 21, 36, 51 ... 996)

Consider the third round: - The first number to be marked is 996 + 15 - 1000 = 11. And all the numbers which leave a remainder 11 when divided by 15 will be marked. (11, 26, 41, 56 ... 986)

The first number to be marked in the fourth round is 986 + 15 - 1000 = 1. Now the cycle will repeat.

If we see the numbers which are getting marked are: 1, 6, 11, 16, 21, 26, 31, 36 ... 996

i.e. all the numbers which divided by 5 leave a remainder 1. So there are $\frac{1000}{5} = 200$ numbers. So, 1000 - 200 = 800 numbers remain unmarked.

Question 33

In a 4 person race, medals are awarded to the fastest 3 sprinters. The first-place sprinter receives a gold medal, the second-place sprinter receives a silver medal, and the third-place sprinter receives a bronze medal. In the event of a tie, the tied sprinters receive the same color medal. (For example, if there is a two-way tie for first-place, the top two sprinters receive gold medals, the next-fastest sprinter receives a silver medal, and no bronze medal is awarded). Assuming that exactly three medals are awarded, and that the three medal winners stand together with their medals to form a Winner's triangle, how many different Winners' triangles are possible?

144

SOLUTION

Solution : B

First, let's consider the different medal combinations that can be awarded to the 3 winners:

(1) If there are NO TIES then the three medals awarded are: GOLD, SILVER, BRONZE.

(2) What if there is a 2-WAY tie?
--If there is a 2-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, SILVER.
--If there is a 2-WAY tie for SECOND, then the medals awarded are: GOLD, SILVER, SILVER.
--There cannot be a 2-WAY tie for THIRD (because exactly three medals are awarded in total).

(3) What if there is a 3-WAY tie?
--If there is a 3-WAY tie for FIRST, then the medals awarded are: GOLD, GOLD, GOLD.
--There are no other possible 3-WAY ties.
Thus, there are 4 possible medal combinations:

(1) G, S, B (2) G, G, S (3) G, S, S (4) G, G, G

COMBINATION 1: Gold, Silver, Bronze

Number of possibilities = $4 \times 3 \times 2$ =24 different Winner's triangles that will contain 1 GOLD, 1 SILVER, and 1 BRONZE medalist.

COMBINATION 2: Gold, Gold, Silver.

Selection = $^{4} C_{3}$ and arrangement = $\frac{4!}{3! (4 - 3)!}$ = $4 \times 3$ =4 ways

COMBINATION 3: Gold, Silver, Silver.
Using the same reasoning as for Combination 2, we see that there are 24 possible Winner's triangles, but only 12 unique Winner's triangles that contain 1 GOLD medalist and 2 SILVER medalists.

COMBINATION 4: Gold, Gold, Gold
Finally , then we have the following:
Selection = $^{4} C_{3}$ and arrangement $= \frac{3!}{2!} =$ 4 $\times$ 1=4 ways

Thus, there are 24 + 12 + 12 + 4 = 52 unique Winner's triangles.

The correct answer is B.

Question 34

Consider two points from which two persons start walking towards each other. They cross each other after time T1. They went further to the opposite points and return instantaneously to cross each other again after time T2. Then T2 =

2T1

3T1

$\frac{3}{2} \times T_{1}$

Data Insufficient

SOLUTION

Solution : A

Suppose the distance between the two points is 'd'.

When the persons meet for the first time, total distance covered = d

When persons meet for the second time, total distance covered = d + 2d

Distance = d, time taken = T1; distance = 2d, time taken = 2T1 = T2

Hence Option (a)

Alternative Approach: Using Assumption

You can also solve this using numbers. If the distance is 100 km, and the speed is 50 kmph for each person, they will meet after $\frac{1}{2}$ an hour. They will reach the opposite corners in 1 hour and will meet again in $1 \frac{1}{2}$ hours; thus T₂=1 hour = 2T1. Option (a)

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