Free Quant Practice Test - CAT

Question 1

In the following figure, PQ is the diameter of the circle with centre O and QR is the tangent to the circle at Q.
PR intersects the circle at S and the tangent to the circle as S intersects QR at T. If the diameter of the circle is 20 cm and TR=6 cm, find the approximate length of the line segment PR.

33.2 cm

26.16 cm

24.57 cm

22.5 cm

SOLUTION

Solution : D

Clearly. $∠$ PSQ=90 $^{\circ}$ [Angle in a semi circle]
Also, TQ=TS [Tangents from an external point are equal ] ... (1)
Let $∠$ TQS= $∠$ TSQ=x
Clearly, $∠$ TSR=90 $^{\circ}$ - x
Also in $Δ$ QSR, $∠$ QRS=90 $^{\circ}$ -x
:.TR=TS ... (2)
From (1) and (2), QT=TR
Given that TR=6 cm.
$\Rightarrow$ QR=QR+TR =6+6=12cm
Now, the length of the line segment PR = $\sqrt{P Q^{2} + Q R^{2}}$
= $\sqrt{20^{2} + 12^{2}} = \sqrt{544} = \sqrt{16 \times 34}$
=4 $\sqrt{34}$ cm=22.5 cm

Question 2

Given that N= abcdefghij is a ten-digit number . All of the digits (a, b, c, d, ...) are different from one another. If 1 1 1 1 1 divides it evenly, how many different possibilities are there for abcdefghij?

3024

3456

5076

1692

SOLUTION

Solution : B

If all the digits are different, and there are 10 of them, then all the digits from 0 to 9 must appear.
That implies that the digits of abcdefghij add up to 45 =0+1+2+3+...+9.
That implies that abcdefgh is definitely divisible by 9.
Since 9 and 1 1 1 1 1 are relatively prime, their least common multiple is 9 $\times$ 11111=99999,
Thus abcdefgh must be divisible by 99999.
We have to find the condition such that abcdefghij is divisible by 99999.
abcdefghij =abcde $\times$ 10 $^{5}$ + fghij and, this number is exactly divisible by (10 $^{5}$ -1).
Hence by remainder theorem, (abcde $\times$ 10 $^{5}$ + fghij) must be 99999.
Then the following equations must all hold:
f =9-a, g =9-b, h =9-c, I =9-d, j =9-c
Now there are 9 Options for a (it can't be 0) and then f is known.That leaves 8 options for b, and
then'g' is known. That leaves 6 Options for c, and then h is known.
That leaves 4 Options for d and then I is known.
That leaves 2 Options for c, and then j is known.
Thus, the total number of such number abcdefghij is 9 $\times$ 8 $\times$ 6 $\times$ 4 $\times$ 2=3456.

Question 3

N = abcdefghij is a ten-digit number. All of the digits (a, b, c, d, …) are different from one another.
If 11111 divides N evenly, find the remainder if the greatest possible abcdefghij is divided by 8?

SOLUTION

Solution : A

The greatest possible value of N, satisfying the condition is N=9876501234.

The divisibility rule for 8 is to check only the last three digits - $\frac{234}{8}$ , remainder = 2.

Question 4

S consists of a set of numbers from -50 to 50 (zero inclusive), we have to pick a set K such that K consists of {A, B, C} such that the sum of numbers in set K is equal to zero. How many such sets are possible?

1250

650

1300

1220

SOLUTION

Solution : C

We can have either 2 +ve numbers and 1 -ve number or 2 -ve numbers and 1 +ve number.
Let's consider the case when we have 1 -ve and 2 +ve numbers.
For the -ve number we can assume it to be -50, the +ve numbers have to be chosen in pairs such that 2 of them add upto 50. i.e. (0,50),(1,49).. We see that 25 such combinations are possible. (Note: Here only 25 will not have a pair)
Similarly for -49, 25 such combinations will be possible. (Note: Here -50 will be left out)
Based on same logic we can extrapolate: Hence the total number of combinations possible will be 2(25 + 24 + 23 +.... +1).
As we can also have 2 -ve numbers and 1 +ve number, the same number of combinations will again be possible.
Hence, the total number of solutions will be $2 \times 2 \times \frac{n (n + 1)}{2}$ ; where n = 25. Thus, the total number of ways will be 1300.

Question 5

A certain clock marks every hour by striking a number of times equal to the hour and the time required for a stroke is exactly equal to the time interval between strokes. At 9 : 00 a.m. the time lapse between the beginning of the first stroke and the end of the last stroke is 34 seconds. At 18 : 00 hrs. how many seconds elapse between the beginning of the first stroke and the end of the last stroke:

SOLUTION

Solution : D

At 9:00 a.m. there are 9 strokes and 8 intervals between strokes.
Thus for 17 equal time intervals, Time = 34 sec.

For 1 interval time = 2 seconds. At 18:00 hours i.e. (6:00 p.m.) there are 6 strokes and 5 intervals i.e. total 11 time intervals of 2 seconds.

Answer = 22 seconds.

Question 6

In the figure find b in terms of a c & d.

b = c + d - a

B. b =

\frac{(c + d - a)}{2}

b = a +c + d

b = $\frac{a + c + d}{2}$

SOLUTION

Solution : A

The right answer is b = c + d - a.

x = 360 - b

w = 360 - a

m + w + x = 180

$\Rightarrow$ m = a + b - 540

Y = 360 - c and z = 360 - d

Therefore a + b - 540 + 360 - c + 360 - d = 180

$\Rightarrow$ a + b = c + d. answer is option (a)

Question 7

If g(a+1)+g(a-1)=2g(a) and g(0)=0, then g(n), where n is a natural number is?

ng(1)

[g(1)] $^{n}$

g(1) $^{3}$

g(n)

SOLUTION

Solution : A

Solve by assumption
Put a=1
Then the equation changes to g(2)+g(0)=2g(1) $\Rightarrow$ g(2)=2g(1)
Look in the answer options on substituting n=2
Eliminate answer options where you are not getting 2g(1)
Answer is option (a)

Question 8

How many zeros does the expression: $1^{1} \times 2^{2} \times 3^{3} . . . . . . {.99}^{99} \times 100^{100}$ end in?

1050

1300

1250

None of these

SOLUTION

Solution : B

The number of 5's present in the expression will determine the number of 0's in the expression.
There are numbers which have single 5's as their factors, like 5,10,15.... each of these contributes that many 5's to the expression.
This gives a sum of 1050 5's (=5+10+15+ ...100).
But there are numbers like 25, 50, 75 and 100 which contribute two 5's.
So add 25+50+75+100 = 250 more to it.
It gives 1300 5's.
Hence, the answer is 1300.

Question 9

L and M make an appointment to meet on 20th Nov. 2005 at the CAT examination centre, but without fixing anything other than that the appointment is between 10 a.m. to 11 a.m. They decide to wait no longer than 10 minutes for each other. Assuming that each is independently likely to arrive at any time during the hour, find the probability that they will meet.

$\frac{5}{6}$

$\frac{25}{36}$

$\frac{11}{36}$

None of these

SOLUTION

Solution : C

There can be 2 approaches to this problem:
Approach 1
We can consider the entire time to be represented by the x-y plot: where on the x-axis we represent the time of one of them coming and on the y-axis the other person coming. L&M will meet provided |x-y|<=10 (x, y) lies on or inside the square with vertices 0(0, 0) A(60, 0) B(60, 60) and C (0, 60).
The region lying inside the square OABC and satisfying the inequality |x - y| > 0 consists two triangles DAE and GFC.
Sum of the areas of two triangles = $\frac{1}{2}$ (50) $^{2}$ + $\frac{1}{2}$ (50) $^{2}$ = (50) $^{2}$
Probability = 1 - $\frac{50^{2}}{60^{2}}$ = 1 - $\frac{25}{36}$ = $\frac{11}{36}$

Alternate approach:

The probability that M will not meet L is $(\frac{50}{60})$ (as he cannot meet him any way outside the 10 min he is present at the venue)
Similarly the probability that L will not meet M is also $(\frac{50}{60})$
Hence the combined probability that they will not meet is $\frac{50^{2}}{60^{2}}$
Hence the probability that they will meet will be = 1 - $\frac{50^{2}}{60^{2}}$ = 1 - $\frac{25}{36}$ = $\frac{11}{36}$

Question 10

The digits p, q and r of a three-digit natural number pqr satisfies the condition 49p + 7q + r = 286. Find the sum of the digits of pqr.

SOLUTION

Solution : A

Since 49p+7q+r = 286
Or $7^{2} p + 7^{1} q + 7^{0} r = 286$
In other words, this question is the conversion of 286 in base 10 to base 7.
pqr in the equivalent of 286 in base $7 = (p q r)_{7} = (556)_{7}$
Thus, the sum of the digits = 16.

Question 11

The largest set of real values of x for which $f (x) = \sqrt{(x + 2) (5 - x)} - \frac{1}{\sqrt{x^{2} - 4}}$ is real is:

(1,2) $\cup$ (2,5)

(2,5)

(3,4)

(5,8)

SOLUTION

Solution : B

For f(x) to be real, the least value x can take is something slightly greater than 2 so that the denominator does not go to zero. Also, the maximum value x can take is 5, because beyond that the first part of the expression will be an imaginary number.

Question 12

In a town which has a barter system , one kg of sugar is equivalent to one dozen mangoes. A gives 1.5 kg sugar to B, using a false weight of 800 gm as 1kg and takes 15 mangoes. Who is in profit and what is the percentage profit?

$8.34 % t o A$

8.34 to B

4.17 to A

4 to A

Cannot be determined.

SOLUTION

Solution : C

Let the cost for 1kg sugar = Rs 1000
For 1.5 kg sugar, A is supposed to get 18 mangoes, but he takes only 15
However, instead of 1.5 kg, he is giving only 1.2 kg
His cost is therefore, 1200 rupees
For 15 mangoes, the cost is 1250 (this is the value which he gets)
Hence, he makes a 50 rupees profit on 1200, which is $4.17 %$ to A.

Question 13

If a, b, c >0, a + b + c =6 and f(x) = $(\frac{6}{x})$ - 1, find the minimum value of f(a).f(b).f(c)?

SOLUTION

Solution : D

f(a).f(b).f(c) = $\frac{(6 - a) (6 - b) (6 - c)}{a b c}$ = $\frac{216 - 36 (a + b + c) + 6 (a b + b c + c a) - a b c}{a b c}$ = $\frac{(a b + b c + c a)}{a b c}$ - 1 = 6 $(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$ - 1
As AM $\geq$ HM
$\frac{(\frac{1}{a} + \frac{1}{b} + \frac{1}{c})}{3}$ = $\frac{\frac{6}{a} + \frac{6}{b} + \frac{6}{c}}{3}$ > 3
$\frac{6}{a} + \frac{6}{b} + \frac{6}{c}$ > 9
f(a).f(b).f(c)> 9 - 1 = 8

Question 14

ABCD and CFEG are squares. HF = 5 cm, AH: HF = 3:1

Which of the following could be the value of area of ABCD? (All the sides are integers)

144

324

112

357

SOLUTION

Solution : A

METHOD 1

triangles AHD and HCF are similar. So $\frac{A H}{H F} = \frac{A D}{C F} = \frac{D H}{C H}$

If CH is x, DH = 3x so, DC = DH + CH = 4x

If CF = y, AD = 3y

Area of ABCD = 3y $\times$ 4x = 12xy i.e. the area has to be a multiple of 12, hence 144 is the answer.

METHOD 2

The diagonal has to be between 15 and 20. Thus the area will be between 113 and 200. Only option (a) satisfies this

Question 15

A, B and C can do a work together in a certain number of days. If A leaves after half the work is done, then the work takes 4 more days for completion, but if B leaves after half the work is done, the work takes 5 more days for completion. If A takes 10 more days than B to do the work alone, then in how many days can C alone do the work?

52.5

SOLUTION

Solution : A

Let A alone can do the work in a days, B alone in b days and C alone in c days.
Let A, B and C together completes the work in x days.

So, $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{x} . . . . . . . . . . . (1)$

Then half of the work they will complete together in $\frac{x}{2}$ days.
When A leaves, Rest half of the work is done by B and C together,

$(\frac{2}{b} + \frac{2}{c}) = \frac{1}{{(\frac{x}{2}) + 4}}$

Using (1), $2 \times (\frac{1}{x} - \frac{1}{a}) = \frac{1}{{(\frac{x}{2}) + 4}}$

Solving we get, a $= \frac{[x \times {(\frac{x}{2}) + 4}]}{4} . . . . . . . . . . . . . . . . (2)$

On similar lines b $= \frac{[x \times {(\frac{x}{2}) + 5}]}{5} . . . . . . . . . . . . . . . . (3)$

Given that a=b+10 ..........................(4)

Thus we have $\frac{[x \times {(\frac{x}{2}) + 4}]}{4} = \frac{[x \times {(\frac{x}{2}) + 5}]}{5} + 10$

Solving we get x = 20.
From (2) and (3) we get
a = 70 and b = 60
Put these values in (1) we get c = 52.5 days.
Hence, choice (a) is the right option

Question 16

Find the 50th term of the series 1,0,3,10,21,36,55

4568

4656

3467

D. 4135

SOLUTION

Solution : B

Method 1:
First we calculate the differences -1,3,7,11,15..
Again we calculate differences of the above series 4,4,4,4
We see that differences are constant, therefore it's a quadratic expression..
$a x^{2} + b x + c$ .. Put x=0 we get c=1.. Then x=1 and 2.. We get a= 2 and b= -3..
Therefore the equation is $2 x^{2} - 3 x + 1$ . Put x= 49. Answer is 4656

Method 2:
First we calculate the differences -1,3,7,11,15..
$49^{t h}$ term in the above sequence = 191.
Sum of 49 terms in the above sequence = 4655
$50^{t h}$ term in the given sequence = 1+4655 = 4656.

Question 17

For any natural number $x, (18)^{8 x} - (7)^{8 x}$ is divisible by:

1) 25 2) 373 3) 11 4) 275

1,2,3

1,2,4

Only 1,2

1,2,3 and 4

SOLUTION

Solution : D

8x is even. Hence it will be divisible by $18 - 7, 18 + 7, 18^{2} + 7^{2}, 18^{2} - 7^{2}$ . Answer is option (d).

Question 18

Let X be the set of first 100 natural numbers. Sets S1 and S2 are subsets of X such that each of them has more than zero elements and no common element. If the maximum number of elements in S1 = x1, in S2 = y1 and minimum no. of elements in S1 = x2, in S2 = y2 then:
Suppose the union of sets S1 and S2 is X and S1, S2 have the same number of elements. If elements from S1 and S2 are chosen randomly, it was found that by exchanging two elements of S1 with two elements of S2, every element of S1 was greater than that of S2. Find the no. of sets S1 that can be formed.

^{50} C_{2}

^{50} C_{49}

100

$^{50} C_{48} \times^{50} C_{2}$

SOLUTION

Solution : D

In set S1, 48 elements can be selected randomly in $^{50} C_{48}$ ways (from 51 to 100) and the other two elements in S2 can be selected in $^{50} C_{2}$ ways (from 1 to 50). So, set S1 can be formed in $^{50} C_{48} \times^{50} C_{2}$ ways. Hence, option (d).

Question 19

Let X be the set of first 100 natural numbers. Sets S1 and S2 are subsets of X such that each of them has more than zero elements and no common element. If the maximum number of elements in S1 = x1, in S2 = y1 and minimum no. of elements in S1 = x2, in S2 = y2 then:
If |x1 - y1| and |x2 - y2| are at their minimum possible values then which of the following is true?

|x1 - x2| $>$ |y1 - y2|

|x1 - x2| = |y1 - y2|

|x1 - x2| $<$ |y1 - y2|

|x1 - x2| $\leq$ |y1 - y2|

SOLUTION

Solution : B

The minimum possible value of |y1 - y2| = 1 - 1 = 0. The minimum possible value of |x1 - x2| = 99 - 99 = 0. So, |x1 - x2| = |y1 - y2|.

Hence, option (b) is correct.

Question 20

If $x^{2} - 3 x + 2$ is a factor of $x^{4} - p x^{4} + q$ , then the values of p and q are

5, -4

5, 4

-5, 4

-5, -4

SOLUTION

Solution : B

Roots of $x^{2} - 3 x + 2 = 1, 2$
Put x =1 and x=2 in the $2^{n d}$ equation $x^{4} - p x^{2} + q$
At x=1, 1-p+q=0 $\Rightarrow$ p-q=1
And at x=2, 16-4p+q=0 $\Rightarrow$ 4p-q=16
Thus, p= 5 and q= 4. Option (b)

Question 21

The number of 7-letter sequences which use only P, Q, R (and not necessarily all of these) and with P never immediately followed by Q, Q never immediately followed by R and R never immediately followed by P is

384

192

288

none of these

SOLUTION

Solution : B

We have 3 choices for the first letter, and 2 for each subsequent letter. So 3 $\times$ 2 $^{6}$ = 192

Question 22

The sum of 6 natural numbers (not necessarily distinct) is 23. Let M denote the positive difference of the maximum and minimum of the LCM of these 6 numbers. Then total number of divisors that divide $M^{2}$ but doesn't divide M are:

SOLUTION

Solution : B

Minimum occurs when the numbers are 6, 6, 6, 2, 2, 1 (LCM = 6). Maximum occurs when the numbers are 7, 5, 4, 3, 2, 2 (LCM = 420).
To achieve minimum LCM, we have to search for maximum number of GCDs among the $^{6} C_{2}$ pairs. To look for maximum LCM, we need the numbers as co-prime.
Thus, M = 414. Now, 414 can be written as $2 \times 3^{2} \times 23 ⟹$ M has $2 \times 3 \times 2$ = 12 divisors.
$M^{2}$ = $2^{2} \times 3^{4} \times 23^{2} ⟹$ $M^{2}$ has $3 \times 5 \times 3$ = 45 divisors.
Hence, option (b) (45-12 = 33) is the correct answer.

Question 23

a, b, c and d are real positive numbers such that a + b = 5, c + d = 2 and ad = bc = 1. Then $a c + b d + (\frac{a}{c})$ is:

11.5

12.5

10.5

SOLUTION

Solution : C

Multiply both equations - (a+b)(c +d) = 10

$⟹$ ac+bd=8 (ad=bc=1)...(i)

a(c+d)=2a (c+d=2)

$⟹$ ac+ad=ac+bc=c(a+b)=2a

$⟹ \frac{a}{c} = \frac{5}{2}$ ...(ii)
Hence (i)+(ii) = $a c + b d + (\frac{a}{c})$ = $8 + \frac{5}{2} = 10.5$

Question 24

Consider a circular track of circumference = 1400 m. There are 2 bikes which start from point A and move in the opposite direction. Once they meet, they start moving in the opposite directions. The one that moves in the anticlockwise direction to the other one has a speed in the ratio of 36:48 kmph. Find the distance of the two bikes from A when they meet for the 15th time?

1200

250

350

450

SOLUTION

Solution : C

1400 m track is divided into 7 parts of 200 m each. The speeds of the two bikes are in a ratio of 3:4, each time they meet, they will cover a distance in the ratio of 3:4
They will thus meet for the first time at ${\frac{3}{4}}^{t h}$ the distance which will happen at point 4 à 1050 m. Second time they will meet at the starting point.
3 $^{r d}$ time at point 4... and so on.so the 15 $^{t h}$ time, they meet at 1050 cm which is 350 m from A

Question 25

A two-digit number N is 'q' times the sum of its digits. The two-digit number formed by reversing the digits of N is 'p' times the sum of the digits.
Which of the following is equal to 'p'?

(9 - q)

(q + 1)

(11 - q)

(10 - q)

SOLUTION

Solution : C

Let the number be xy.

Therefore, 10x + y = q(x + y)

Or (10 - q)x = y(q - 1)

Or $\frac{x}{y} = \frac{q - 1}{10 - q}$ ...(i)

If number xy is reversed, it is equal to 'p' times the sum of digits.

Then 10y + x = p(x + y)

Therefore, x(p - 1) = y(10 - p)

Or $\frac{x}{y} = \frac{10 - p}{p - 1}$ ...(ii)

Comparing, $\frac{10 - p}{p - 1} = \frac{q - 1}{10 - q}$

Solving, we get p = 11 - q.

Question 26

Which of the two is smaller - $5^{\frac{1}{3}}$ or $6^{\frac{1}{4}}$ ?

$5^{\frac{1}{3}}$

$6^{\frac{1}{4}}$

Both are equal

Cannot be determined

SOLUTION

Solution : B

Take a common power. 12 is the LCM of 3 and 4.

So $[5^{\frac{1}{3}}]^{12}$ and $[6^{\frac{1}{4}}]^{12}$

This gives $5^{4} = 625$ and $6^{3} = 216.$ Therefore $5^{\frac{1}{3}}$ > $6^{\frac{1}{4}}$ .

Question 27

Find S1+S2+S3.....S10 given that Sn= 1+2+3+4.....n

260

240

340

220

SOLUTION

Solution : D

S1= 1

S2= 1+2 =3

S3=1+2+3= 6

This is of the form $\frac{n (n + 1)}{2}$

We need to find $\sum \frac{n (n + 1)}{2}$ where n=10

Expanding we get

$\frac{1}{2} [\sum n^{2} + \sum n]$

$\frac{1}{2} [\frac{n (n + 1) (2 n + 1)}{6} + \frac{n (n + 1)}{2}]$

Substitute n=10 to get the answer as $\frac{440}{2}$ =220

Question 28

In a multiplication problem, instead of taking 45 as one of the multipliers, Vineet took 54. As a result, the product went up by 540. What is the correct product?

2700

5400

1440

1590

SOLUTION

Solution : A

Let the number that Vineet wanted to multiply be 'X'.
He is expected to find the value of 45X.
Instead, he found the value of 54X. The difference between the value that he got (54X) and what he was expected to get (45X), according to the question, is 540.
i.e., 54X - 45X = 540
or (54 - 45) $\times$ X = 540
$⟹$ X = 60.
$∴$ 45X = 2700.

Question 29

When a number is divided by 48, it leaves a remainder of 17. What will be the remainder when the number is divided by 12?

SOLUTION

Solution : C

We can write the number as 48x+17. Therefore, when this is divided by 12, we have $\frac{48 x}{12} + \frac{17}{12}$ . Now we can see that the remainder is 5.
Answer is option (c).

Question 30

The sum of first two terms of a GP is $\frac{14}{5}$ and the sum of infinite terms is $\frac{10}{3}$ . The first term of the GP if the common ratio is positive is___.

SOLUTION

Solution :
Let a and r be the first term and common ratio respectively.
Given, a + ar= $\frac{14}{5}$ ......(1)
and     $\frac{a}{(1 - r)} = \frac{10}{3}$
             a= $\frac{10 (1 - r)}{3}$ .......(2)
Substituting (2) in (1),
             $\frac{10 (1 - r^{2})}{3} = \frac{14}{5}$
                        r²= $\frac{4}{25}$
                        r= $\frac{2}{5}$ (since r is positive)
From (2), a=2
Therefore, first term = a = 2

Question 31

What is the maximum number of times 10 squares of same side length can intersect each other?
___

SOLUTION

Solution :
Number of ways of selecting two squares from 10 squares= $^{10}$ C $_{2}$
Each pair of squares can meet at maximum of 8 points.
If there are no common points of intersection, maximum number intersections= $^{10}$ C $_{2}$ * 8=360

Question 32

The roots of the equation (x+10)(x+1)+19 is given by a and b. The greater ratio of roots of (x+a)(x+b)-11 is ___. (The answer will be of p/q form where p and q are integers, then input 'pq' in the provided space.)

SOLUTION

Solution :
(x+10)(x+1)+19 = $x^{2} + 11 x + 29$
Since a and b are the roots, a+b= -11 and ab=29
$∴$ (x+a)(x+b)-11 =
$x^{2} + (a + b) x + a b - 11$
= $x^{2} - 11 x + 18$
=(x-9)(x-2)
ie., the roots are 9 and 2.
$T h e g r e a t e r r a t i o i s \frac{9}{2}$ .

Question 33

Anil bought a 10 kg water-melon in Bangalore that had $99 %$ water. After the water-melon was left outdoors for a day, it was $95 %$ water. What was the weight of the dehydrated water-melon?

9.6 kg

9.5 kg

2 kg

D. None of these

SOLUTION

Solution : C

Initially, there was 0.1 kg of dry part + 9.9 kg of water. After dehydration, 0.1 kg of dry part is $5 %$ of total water-melon's weight. Hence, weight of dehydrated water-melon is 2 kg.

Question 34

Three circles C1, C2 and C3 having diameters 3 cm, 4 cm and 5 cm are intersecting each other such that their diameters form a triangle. If the area common to circles C1 and C2, C2 and C3, C3 and C1 are a, b and c respectively then find the area common to all the three circles.

a + b - c

a - b - c

a - b + c

None of these

SOLUTION

Solution : D

Notice that the diameters form a Pythagoras triplet. The common area to all three circles will be the one between C1 and C2 i.e. "a”.

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