# Free Quant Practice Test - CAT

### Question 1

Triangle PQR is equilateral with sides of length 5 units. O is any point in the interior of triangle PQR. Segment OL, OM, and ON are perpendicular to PQ, PR and QR respectively. Find the sum of segments OL, OM and ON.

10√32

3√32

5√32

#### SOLUTION

Solution :C

Conventional Method Join PO, OQ and OR

There are 3 triangles whose sum of areas = area of the equilateral triangle

Area of a triangle =12× base × height

Thus Area of △POR=12×PR×OM

Area of △QOR=12×QR×ON

Area of △POQ=12×PQ×OL

Sum = Area of △POR+Area of △QOR+Area of △POQ= Area of △PQR

=12×PR×OM+12×QR×ON+12×PQ×OL

Area of an equilateral triangle =√34×a2 , where 'a' is the length of 1 side

√34×52=12×PR×(OM+OL+ON) (Since, PR = QR = PQ)

⟹25√34=12×5×(OM+OL+OM)

⟹OM+ON+OL=5√32Shortcut:- Assumption

Note the most important word in the question-ANY. We can assume ANY point inside the triangle. We will consider a point infinitely close to P.

Thus, the diagram will look as follows.

The answer can now be obtained in a single step.

OL = 0, OM =0 and ON is the height of an equilateral triangle =5√32

Thus, OL+OM+ON=5√32

### Question 2

Let 'x' be the arithmetic mean and y,z be the two geometric means between any two positive numbers. The value of y3+z3xyz is

2

3

12

32

#### SOLUTION

Solution :A

Assumption Technique

This is a question with variables. Thus, we can assume any value which satisfies the basic conditions for x,y and z

Assume any Geometric Progession, say 1 2 4 8 .

This implies that the 2 Geometric Means between 1 and 8 are

y=2 and z=4.

Now, Arithmetic Mean, x = \(\frac {8 + 1}{2} = 4.5 .Substitutethe values for x,y,z in y3+z3xyz

Answer = 23+432×4×4.5 = 2.

Shortcut of a shortcut!You can assume ANY Geometric Progression.

Assume the most basic GP 1 1 1 1 and the most basic AP 1 1 1,

then x=1, y=1 z=1. Answer on substitution in y3+z3xyz=2, which will make the calculation even faster.

Points to note:

1.The moment you get values for x,y,z you are in a position to solve any question on x,y,z and not just the above one. Thus, all variable based questions can be solved in no time at all by assuming simple numbers!

2. You can assume any GP and solve the question. You will get the same answer.

### Question 3

In how many ways can the number 100 be written as a sum of two or more consecutive positive integers?

2

3

5

6

#### SOLUTION

Solution :A

Conventional Approach

To write a number as a sum of two or more consecutive positive integers, we need to form an AP with "n” terms with a common difference 1

Sum to "n” terms of an AP= n2(2a + (n - 1)d) where d=1 (as the numbers are consecutive) n(2a+(n-1))=200

thus, we need to write 200 as a product of 2 numbers, where one number is odd and the other is even

this can be done in 2 ways

200= 5×40

200=25×8

Thus 100 can be expressed in 2 ways as a sum of two or more consecutive positive integers

Shortcut:-

The answer is just one step. The number of ways of writing any number as a sum of two or more

consecutive positive integers= number of odd factors of that number-1

100 = 22×52 Number of odd factors=3

Number of ways of expressing 100 as a sum of two or more consecutive positive integers= 3 - 1=2

### Question 4

The sum of first 3 terms of a G.P is 16 and the sum of next 3 terms is 128. Find the sum of n terms of the G.P?

167(2n + 1)

167(2n - 1)

97(3n - 1)

167(3n + 1)

#### SOLUTION

Solution :B

Conventional Method

From the data given in the question

S3=16=a(r3−1)r−1................................(1)

S6=16+128=144=a(r6−1)r−1=a(r3−1)(r3+1)r−1................................(2)

9 = (r3 + 1)

Hence r =2

Substituting in equation 1, we get

a = 167

We know that,

Sn=a(rn−1)r−1

Substituting the values of a and r found above, we get

Sn=167(2n−1)

Thus, answer is option (b)

Shortcut-

Substitution The answerto this question is in the question itself!

At n=3, the sum should be 16

Hence, in the correct answer option, at n=3, we should get 16

Option (a)= 167(23 + 1)≠16

Option(b)=167(23 - 1) = 16

Option (c)=97(33 - 1)≠16

Option (d)=167(33 - 1)≠16

Options (a), (c) and (d) can never be the answer. Answer is option (b)

### Question 5

Find the remainder when 287 is divided by 100

56

28

76

01

#### SOLUTION

Solution :B

Shortcut of last two digits:

When the question is about finding the remainder when you divide a number by 100, what we actually need to find out is the last two digits, for which we have a direct shortcut. For an indepth explanation, please refer the demo tutorial

287=(210)8×27(210)evev power will always end in..76

27 ends in ...28

Therefore 76x...28=28

### Question 6

ABCDEF is a regular hexagon, with O as the centre. There is another regular hexagon with 2 of the end points PQ inscribed in the above hexagon (again with centre O). Find the area of the unshaded region, given that FPQC lies on a straight line and FP=CQ=2 and FP = 13 FO.

20√3

24√3

18√3

28√3

#### SOLUTION

Solution :B

The hexagon can be graphically divided as follows

Hence, there are totally 54 equilateral triangles out of which 24 are unshaded. Each of the triangles have a side 2. Total unshaded area = √34 x 4 x 24 = 24√3

### Question 7

For the given equation x9 + 5x8 - x3 + 7x+3=0 how many maximum real roots are possible?

3

5

6

7

#### SOLUTION

Solution :B

Let f(x) =x9 + 5x8 - x3 + 7x+3 = 0

We can solve this question easily using Descartes' Rule

According to Descartes' rule maximum number of positive real roots = number of sign

changes in f(x) = 2

Similarly, Maximum number of negative real roots = number of sign changes in f(-x)

Note:To find f(-x) replace "x” by "-x” in each instancetherefore f(-x) = - x9 + 5x8 + x3 - 7x+3

Maximum number of negative real roots = number of sign changes in f(-x) = 3

Zero cannot be a root because constant part is also involved in equation.

So maximum number of real roots = 2 + 3 = 5

DESCARTES' RULE(Points to Remember)

Maximum number of positive real roots = number of sign changes in f(x)

Maximum number of negative real roots = number of sign changes in f(-x)

If a constant term is present, do not consider 0 as a root or else, you need to consider it

### Question 8

At the FIFA World Cup, out of the 2 lakh visiting fans, 90% were fans of Argentina, 85% of Brazil, 70% of Germany, 75% of Spain and 85% of Netherlands. What is the minimum percentage of visiting fans who are fans of all 5 countries if it is given that each of these visiting fans supports at least one country?

5%

10%

0

15%

#### SOLUTION

Solution :A

In this case, the minimum of all is that part of the Venn diagram which has 5 overlapping regions. This can be found using a shortcut technique.

Here X=100,

Step 1- Taking the difference of each from X.

100-90=10

100-85=15

100-70=30

100-75=25

100-85=15

Total (Y) ∑ = 95% (Percentage of fans who do not support atleast 1 country)

Step 2- Take the difference of Y from 100, 100-95 = 5%

Minimum percentage of fans who support all 5 countries = 5%

### Question 9

Suppose the seed of any positive integer n is defined as follows:

Seed (n) = n, if n < 10

= seed(s(n)), otherwise,

Where s(n) indicated the sum of digits of n. For example, seed(7)=7, seed(248) = 2+4+8 = seed(14) = seed(1+4) = seed(5) = 5, etc.

How many positive integers n, such that n< 500, will have seed(n)=5?

39

72

56

108

#### SOLUTION

Solution :C

The correct answer can be marked in 30 seconds if you know that this is an application of "number of terms in an Arithmetic Progression”. If we distribute 500 consecutive numbers with a common difference of 9, number of terms can only be 5009 = 55 or (55+1). You can directly mark option (c)

Why a common difference of 9? Because the first number with a property of its digit sum=5 is 05, the next term is 14 (1+4=5), the next is 23(2+3=5) and so on

The difference between any two consecutive terms with this property is 9.

Thus, the above question is nothing but finding the number of terms in an Arithmetic Progression with the terms as 5,14.... 500.

Notice another interesting thing

When 5 is divided by 9,the remainder is 5

When 14 is divided by 9, the remainder is 5

When 23 is divided by 9, the remainder is 5

3 different ways of asking the same question

1) How many numbers less than or equal to 500 will leave a remainder of 5 when divided by 9?

2) Find the number of terms in the AP 5,14,23,32...........500 ?

3) If B = sum of digits of A, C = sum of digits of B, .........X = X itself. If X = 5 find the different possible values of A given that 0<A <500

### Question 10

A person can only move upwards or towards right. In how many ways can he travel from A to C via B

330

60

55

150

#### SOLUTION

Solution :D

This question can be visualized as a similar to different question in P&C (refer e booklet for indepth explanation of concept)

Number of columns from A to B= 3 and rows= 2. Answer is based on 3 zeros and 2 ones =5C2=10 ways.

Similarly, number of columns from B to C= 4 and number of rows= 2. Answer is based on 4 zeroes and 2 ones= 6C4 =15. Total number of ways = 15×10=150. Option (d)

### Question 11

A man travels from his home to office at 4km/hr and reaches his office 20 min late. If the speed had been 6 km/hr he would have reached 10 min early. Find the distance from his home to office?

4km

6km

8km

9km

#### SOLUTION

Solution :B

Method:Let the distance between home and office =d

Suppose he reaches the office on time, the time taken = x minutes

Case 1:When he reaches office 20 minutes late, time taken = x+20

Case 2:when he reaches office 10 minutes early, time taken = x-10As the distance traveled is the same, the ratio of speed in case 1 to the speed in case 2 will be the inverse of the time taken in both cases.

To understand this better ,Let us call the horizontal distances as H and the vertical distances as V.

The person needs to first go to B.

Number of horizontal distances = Number of rows =2.

Number of vertical distances =Number of columns =3.

i.e.., HHVVV.

Now the question changes to the number of ways of arranging HHVVV=5!3!2!=5C2=10 ways.

Similarly, from B to C, the number of distances =HHVVVV. Number of arrangements=6!4!2!=15 ways .Total number of ways =10×15=150.

Ratio of speed in both cases = 4:6 = 2:3

Ratio of time in both cases = 3:2Therefore,x+20x−10=32

2x+40=30x-30→ x=70 minutes.

Taking case 1:4d=6090(90=x+20)d=36060=6 km.Shortcut- Constant Product Approach

Assume original speed= 4km/hr.

Percentage increase in speed from 4 kmph to 6 kmph= 50% or 12

12 increase in speed will result in 13 decrease in original time (from Constant Product Rule)

=30 minutes.(from given data). Thus, Original time= 90 minutes= 1.5 hours

Answer is Distance = 4 × 1.5=6 km

In case, you want to break down the shortcut in steps, you should proceed like this:

There are 2 cases here,

Case 1 ⇒ speed = 4 km/hr , 20 imn late.

Case 2 ⇒ speed = 6 km/hr , 10 min early.

Step 1:Calculate the increase or decrease in the parameters speed /time.

To calculate increase / decrease we need to consider a case as original case and then calculate the increase or decrease over it.

We will take the first case as the original case,so we can easily observe that the increase in speed s 50% or 12 from case 1 to case 2 (×100=50%)

Step 2:Inverse Proportionality.

Now based on inverse proportionality there will always be a or 33% or decrease in time from case 1 and case 2.

Step 3:Calculating the actual time and distance.

Suppose in the case 1 which we have taken as original, time taken is 'x' mins.

So due,to 12 increase in speed we have decrease in time of 'x' =30 mins.

(form 20 minutes late to 10 mins early there is a decrease of 30 minutes.)

x=90 minutes.

So in the case 1, he is taking 90 mins to travel at a speed of 4 km/hr.So,distance =1.5×4=6 km.

The explanation may be long, but when you actually apply it; you can solve time consuming questions in very little time,helping you save precious minutes which you can use in your weaker areas !

### Question 12

Find the inverse of

#### SOLUTION

Solution :A

Conventional method

Replacing f(x) with y, the given function is

y= (4-(x-7)

^{3})^{1/5}To find the inverse function, we need to write an equation for x in terms of y

Raising both sides to the power 5, we get

y

^{5}= 4 - (x-7)^{3}(x-7)

^{3}= 4- y^{5}Taking cube root on both sides

x-7 = (4 - y

^{5})^{1/3}x = 7 + (4 - y

^{5})^{1/3}We have an expression for x interms of y. We get f

^{-1}(x) if we replace y with x in the above equationHence, f

^{-1}(x) = 7 + (4 - x^{5})^{1/3}. Answer option (a)

Shortcut

What is the basic definition of an inverse function?

An Inverse function is one which can be expressed as f(y) = x when f(x) = y

Using, this very definition, we can use a common sense approach to arrive at the answer In no time at all

1) Substitute a suitable value for x (it can be any value). in this case, for our convenience we will take x=7. Then at x=7, f(x)= 4

^{1//5}2) Now use the definition of inverse function. Put x=4

^{1/5 }in each of the answer options and see where you get f(x)=7. You are just exchanging "x” and "y” values!This is the concept of inverse function after all!

Only option (a) gives 7 at x=4

^{1/5}

All other options can be eliminated as for x=4

^{1/5}, we do not get 7. Answer is option (a)

### Question 13

How much milk should be added to 800 dl of a 15% milk solution to make the strength 32%?

100

150

200

400

#### SOLUTION

Solution :C

Let x dl of pure milk be added. The concentration of pure milk is 100%

Total milk content in x dl of milk = x * 100% = x dl

Total milk content in 800 dl of 15% mixture = 800 * 15% = 120 dl

Therefore, total milk content in final mixture = x + 120

Total volume of mixture = 800 + x

Hence concentration of milk in final mixture = x+120800+x

This is given to be 32%.

Hence x+120800+x = 32 %

Solving, we get x = 200 dl.

Shortcut- Using Alligation

There may be different ways to solve this question, but one of the fastest approaches is using the concept of Alligation. Pure milk has 100% concentration, and it is mixed with a solution with 15% milk to make it 32%. This information can be used to get the ratio of the mixture as follows

For every 4 parts of 15% milk, 1 parts of 100% milk needs to be added.

Hence to 800 dl of the 15% solution, 8004 = 200 dl of 100% milk needs to be added

### Question 14

There are 1000 bulbs in a row numbered 1 to 1000. First person switch on all the 1000 bulbs, next person switch off bulbs 2,4,6,8 and so on to 1000; third person changes the state( switch on bulbs which are OFF, switch off bulbs which are ON) of bulbs 3,6,9,12,15 and so on; fourth person changes the state of bulbs 4,8,12,16 and so on. This goes on till every person in a group of 1000 has had a turn. How many bulbs will be ON at the end?

31

11

49

none of these

#### SOLUTION

Solution :A

Let us take some simple numbers to see if there is a pattern

If you take a number, say 6, In the first round it will be on, 2nd it will be off, 3rd it will be on, and 6th it will be off. Hence its final state will be off.

Consider a number 4-In the first, second and fourth round it will be ON, OFF & finally ON

Consider a number 9. in the first, third and 9th round it will change its state and its final state will be on

From this, you can deduce that the state of the bulb is determined by the property of the number; i.e. all numbers with odd number of factors will remain in an "on” state and all other numbers will be off. Conclusion- All the bulbs which are on are "perfect square” numbered bulbs

Thus, the bulbs which will be on are all in the positions of numbers which are perfect squares.

The question is thus, asking for the number of perfect squares till 1000= 31 (i.e. 12,22...........312)

### Question 15

Find the maximum volume of a right circular cylinder if the sum of the radius and the height of the given cylinder is 6?

32 π

64 π

40 π

#### SOLUTION

Solution :B

Conventional method

Volume of a cylinder, V = π r2h ( r = radius and h = height)

Given that r+h=6. Hence, h = 6-r

Hence, V = π r2(6-r)

V = 6 π r2 - π r3

For maximizing volume, we need to differentiate V with respect to r and equate it to 0.

dvdr = 0

dvdr = 12pr - 3 π r2 = 0

Hence, r = 4

R + h is given as 6, hence h = 2

V = π r2h = π ×42×2 = 32 π

Alternate Method:Volume of a cylinder = π r2h ( r =radius and h= height)

Given that r+h=6

To maximize π r2h, we need to maximize r2h which happens when r2 = h1,

⇒ r = 4 and h =2

Maximum volume = π ×42×2 = 32 π

1. If a+b=constant, ab will be maximum when a=b

Points to remember2. If ab=constant, the minimum value of a+b will be obtained at a=b

3. If a+b+c is a constant, then am. bn .cp is maximum when am = bn = cp (the above question is an example of this)

### Question 16

How many numbers are there between 1 and 100 that are not divisible by 2,3 and 5?

26

36

40

68

48

#### SOLUTION

Solution :A

Numbers divisible by 2 = 50

Numbers divisible by 3 = 33

Numbers divisible by 5 = 20

Numbers divisible by 6 (LCM of 2 & 3) = 16

Numbers divisible by 10 (LCM of 2 & 5) = 10

Numbers divisible by 15 (LCM of 3 & 5) = 6

Numbers divisible by 30 (LCM of 2, 3 & 5) = 3

Numbers divisible by 2, 3 & 5 combined = 50 + 33 + 20 - (16 + 10 + 6) + 3 = 103 - 32 + 3 = 106 - 32 = 74

Numbers not divisible by 2, 3 & 5 = 100 - 74 = 26

### Question 17

For the T-20 World Cup, Indian team of 15 players is to be formed from 20 players. Out of 20 players, 7 are batsmen, 7 are bowlers, 5 all-rounders and 1 wicketkeeper. In how many ways this can be done by taking exactly 5 bowlers, 6 batsmen, 1 wicketkeeper and at most 3 all-rounders.

1470

1890

6280

10

1

#### SOLUTION

Solution :A

To select a team of 15, you have to select almost 3 all rounders.

Therefore the required number of ways

= 7C5 × 7C6 × 1C1 × 5C3

=1470.

### Question 18

If in a ΔABC, cosAa=cosBb=cosCc, then what can be said about the triangle?

Isoceles triangle

Right angled triangle

Equilateral triangle

Acute triangle

#### SOLUTION

Solution :C

Since rule says that = asinA=bsinB=csinC=k

Therefore, a=k(sinA), b=k(sinB) and c=k(sinC)

We can write cosAa=cosBb=cosCc as

cosAksinA=cosBksinB=cosCksinC = cot A=cot B=cot C

⇒ A=B=C(Equilateral Triangle).

### Question 19

Two mutually perpendicular chords AB and CD meet at a point P inside the circle such that AP=6 units PB=4 units and DP=3 units. What is the area of the circle?

125 π4

100 π7

125 π8

52 π43

#### SOLUTION

Solution :A

As AB and CD are two chords that intersect at P, AP ∗ PB = CP ∗ PD

6∗4=CP∗3⇒CP=8

From centre O draw OM ⊥ AB and ON ⊥ CD

We get AM = MB = 5 (perpendicular from center bisects the chord)

Therefore MP = 1, ON = 1, CD = 11

CN = ND = 5.5 (perpendicular from centre bisects the chord)

ON2+CN2+OC2

1+(5.5)2=r2

Area of circle = πr2=π(1+(5.5)2)=125π4

### Question 20

A tank has two pipes A and B. Pipe A is for filling the tank and Pipe B is for emptying the tank. Pipe A can fill the tank in 20 hours and Pipe B can empty the tank in 25 hours. How many hours will it take to completely fill a one-fourth empty tank?

10 hours

25 hours

50 hours

100 hours

5 hours

#### SOLUTION

Solution :B

In one hour pipe A can fill 10020 = 5% of the tank

In one hour pipe B can empty 10025 = 4% of the tank

(A+B)'s work = 5-4 = 1%

To fill a one fourth of empty tank, it will take = 251 = 25 hours

### Question 21

The Karnataka express started from Bangalore to Bhopal at 7 pm at a speed of 60 Kmph. Another train, MP express started from Bhopal to Bangalore at 4 am next morning at a speed of 90 Kmph. Find the time at which the two trains will meet, if the distance between Bangalore and Bhopal is 800 km.

5.30 am

4.48 am

5.44 am

6.02 am

5.12 am

#### SOLUTION

Solution :C

Distance travelled by Karnataka express till 4 am = 9*60 = 540 kmph

Distance remaining = (800-540)km = 260 km

Relative speed = (60+90)kmph = 150 kmph

Time required to travel This distance = 260150 * 60 = 104 minutes = 1 hr 44 mins

Time at which train meets = 4 am + 1 hr 44 mins

= 5.44 am.

### Question 22

Consider three numbers a, b and c Max(a,b,c)+Min(a,b,c)=13. Median (a,b,c)-Mean(a,b,c)=2. Find the median of a, b and c.

11.5

#### SOLUTION

Solution :C

The numbers can be arranged in ascending order min < median < max

We have min + max =13

Mean=min+med+max3=median−12+med3=2

On solving we get median =9.5.

### Question 23

If a 3 digit number 'abc' has 3 factors, how many factors does the 6-digit number 'abcabc' have?

#### SOLUTION

Solution :C

'abc' has exactly 3 factors, so 'abc' should be square of prime number.

Let the square root of 'abc' be 'p' (where 'p' is prime).'abcabc' = abc × 1001 = p2×13×11×7

If 'p' is any prime number other than 13, 11 or 7, then no. of factors = 3 x 2 x 2 x 2 = 24 (no. will be of the form p2×13×11×7)

If 'p' is any prime number among 13, 11 or 7, then no. of factors = 4 x 2 x 2 = 16 (no. will be of the form 73×13×11,if p=7)

∴ No. of factors can be 16 or 24.

### Question 24

Two circles are placed in an equilateral triangle as shown in Fig. What is the ratio of the area of smaller circle to that of equilateral triangle?

#### SOLUTION

Solution :C

### Question 25

Both the H.C.F and the difference of two numbers is 7. If the L.C.M of the two numbers is a three digit number, then what is the maximum possible value of the smaller number?

77

84

70

63

#### SOLUTION

Solution :A

Let the smaller number be 7x, then the larger number will be (7x+7) = 7(x+1)

Since x and (x+1) are co-prime, the LCM would be 7×x×(x+1).

Maximum value of x such that LCM is a three digit number is x=11.

So, smaller number = 77.

### Question 26

2f(x)+3f[(4x+5)(x−4)]=(x+1)2

Find the value of f(5) if x is a real number not equal to 4.

-62

1112

442

None of these

#### SOLUTION

Solution :D

Put x=5, 2f(5)+3f(25)=36.

Put x=25, 2f(25)+3f(5)=676.

f(5) = 391.2

### Question 27

Vikalp, Nilay and Nilabh and visited a shop to buy LED TV and each one of them bought at least one unit of the TV. The shopkeeper gave a discount of 20% to Vikalp. Nilay, a good bargainer got two successive discounts of 20% and 60% . Nilabh, the hardest bargainer of all got three successive discounts of 20% , 40% and 50% respectively. After selling total of 'x' units to three of them, the shopkeeper calculated that he had given an overall discount of 56% . If

10≤x≤20 , then how many values of 'x' are possible? (Assume every one bought at least one unit).

5

6

7

8

9

#### SOLUTION

Solution :B

Discount to Nilay =1−0.8×0.4=68%

Discount to Nilabh =1−0.8×0.6×0.5=76%

Let a,b,c be the number of units sold to Vikalp, Nilay, and Nilabh respectively.

So, 0.2a+0.68b+0.76c=0.56(a+b+c)

So, 5c=3.(3a-b),

So,c is multiple of 3.

If,c=3 then (a,b)= (2,1),(3,4), (4,7)

If, c=6 then (a,b)= (4,2), (5,5), (6,8)

Hence 6 solutions.

### Question 28

How many numbers are there less than 100 that cannot be written as a multiple of a perfect square greater than 1?

61

56

52

65

#### SOLUTION

Solution :A

List all multiples of perfect squares (without repeating any number) and subtract this from 99.

4 - there are 24 multiples of 4 (4,8,12,.....96)

9 - there are 11 multiples, 2 common with 4 (36 and 72) so, add 9 multiples

16 - 0 new multiples

25 - 3 new multiples (25,50,75)

36 - 0 new ones

49 - 2 {49,98}

64 - 0

81 - 0

Total multiples of perfect squares are 38. There are 99 numbers in total. So, there are 61 numbers that are not multiples of perfect squares.

### Question 29

3 friends play a game, rules are as follows whenever there is a contest between any 2 of them, the one who has a higher % alcohol should pour 200ml of his wine into the one having lower % alcohol. The game starts as a contest between A and B, then B and C and then C and A. Post this, the game continues in the same cycle on and on. If a player has emptied all his alcohol, then the remaining two play the game with same rules. If 2 players have alcohol of same percentage level, the younger one pours 200 ml of his alcohol into the elder one's glass. All 3 of them start the game with 600 ml of wine. A's mine has 60% alcohol, B's has 48% alcohol, C's has 50% alcohol. They take 3 minutes to play one round of this game. D, a fourth friend leaves the pub immediately after the game begins, returns after an hour and drinks wine from the person who has the highest alcohol percentage. What is the concentration of alcohol that D had ?

51.5%

52.67%

53%

58.4%

Can't be determined

#### SOLUTION

Solution :B

A v/s B:

A has higher % of alcohol than B, A has to give 200 ml of wine. Considering that A has 60% of strength mixture, B will end up having 800 ml of wine that has 60%×1+48%×34=51% of alcohol.

B v/s C:

B has higher % of alcohol than C, B has to give 200 ml of wine to C now. C will end up having wine that has,

### Question 30

If 3sinx + 4cosx + r is always greater than or equal to 0, what is the smallest value that r can take?

5

-5

4

3

#### SOLUTION

Solution :A

3 sinx +4 cosx + r ≥ 0

3 sinx + 4 cosx ≥ -r

5×(35sinx+45cosx)≥−r

Let 35 = cos A ⇒ sin A ⇒45

5(sin x cos A +sin Acos X)≥ -r

5(sin(X+A)) ≥ -r

We have -1 ≤ sin (angle) ≤ 1

5 sin(X+A)≥ -5

rmin =5.

### Question 31

1 unit of x% alcohol is mixed with 3 units of y% alcohol to give 60% alcohol. If x>y , how many integer values can x take?

19

20

21

13

15

#### SOLUTION

Solution :D

We know that

(x+3y)4=60 or x+3y =240 and x>y

As 0<x<100, let's start with x=99. When x= 99, y=47.

When x=96, y=48 and so on. x has to be greater than y. Now in the limiting case, when x=60, y=60.

So when x=63, y=59. So x can take the values of 63,66,69...... till 99, a total of 13 values.

### Question 32

The sum of all the three digit numbers which when divided by 8 give a remainder of 5 is

#### SOLUTION

Solution :For this condition,

Series is 109,117,125,..........997

a= 109 ,d=8 , n=112

Sn=n2[2a+(n−1)d]

=1122[2×109+(112−1)8]

=61936.

### Question 33

Five balls of different colours are to be placed in three boxes of different sizes. Each box can hold all five. In how many different ways can we place the balls so that no box remains empty?

#### SOLUTION

Solution :Since each box call hold five balls number of ways in which balls; could be distributed so that none is empty are (2. 2, 1) (3, 1, 1)

i.e., (5C2 . 3C2 . 1C1 + 5C3 . 2C1 . 1C1 ) × 3!

=(30+20) × 6 = 300

### Question 34

Two rugby teams (thirty players) have been selected to play for their school. Five of the players speak Spanish, Afrikaans and English. Nine of them speak only Spanish and English. Twenty speak Afrikaans, of which twelve also speak Spanish. Eighteen speak English. No one speaks only Spanish. How many players speak only English?

#### SOLUTION

Solution :From the data given in the question, we can draw a Venn diagram as follows

Since N = 30, the sum of the shaded region is 10. Hence 9 + x = 10 → x = 1.