# Free Quant Practice Test - CAT

How many numbers are there from 1 to 1000 which are perfect squares as well as perfect cubes?

A.

1

B.

2

C.

3

D.

> 3

#### SOLUTION

Solution : C

The number should be of the form a6, where 6 is the LCM of 2 and 3.

The required numbers are 16, 26  and 36.

The least number which gives a remainder of 2, 3 and 4 when successively divided by 3,x and 8 respectively is 71. Find x.

A.

4

B.

6

C.

5

D.

7

#### SOLUTION

Solution : C Solving, we get (4x + 3)×3+2 = 71 x=5.

Taps to a tank are switched on in the following manner. Tap 1 for the first hour. Tap 2 for the next two hrs and so on.The number indicated on each tap (1 for Tap number 1, 2 for tap number 2...) indicates its flow rate in litre/hr. These taps are filling a tank whose capacity is more than 1000 litre. Moreover the capacity cannot be expressed as a difference of squares of two numbers. Find the tap no that fills the tank assuming that the tank's capacity is minimum.

A.

11

B.

12

C.

13

D.

14

#### SOLUTION

Solution : D

The smallest number possible is 1002 (this is a 4k+2 number, it cannot be expressed as difference of squares) .

Therefore our equation becomes. 1×1+2×2+........N×n=1002.

N=13.93

i.e. when the 14th tap is open the tank gets filled.

A group of 100 people plays carrom, snooker and chess. 90 people play chess, 80 people play snooker and 80 people play carom. Find the maximum number of people who play all 3 games, if each person plays at least one game.

A.

80

B.

70

C.

75

D.

65

#### SOLUTION

Solution : C

S= 90+80+80=250

X= 100

Using the Maxima of all shortcut
SXN1=1502=75

The milk and water in two vessels A and B are in the ratio 4:5 and 7:2 respectively. In what ratio the liquids in both the vessels be mixed to obtain a new mixture in vessel C consisting half milk and half water?

A.

2: 1

B.

5: 1

C.

4: 1

D.

7: 2

#### SOLUTION

Solution : B

Suppose in the vessels A & B, one litre of liquid is present.

So, Amount of milk in vessel A =49        ;         Amount of milk in vessel B = 79

The required mixture in vessel C should be 12 milk.

So, let the ratio be x: 1. Applying alligation: (518)(118)=x:1    x=5.

Which of the following is the factor of f(x) = x5 - 2x4 + 3x3 - 6x2 - 4x + 8.

A.

x - 2

B.

x + 2

C.

x - 3

D.

x + 3

#### SOLUTION

Solution : A

Use the reverse gear approach

Option (a) f(2) = 32 - 32 + 24 - 24 - 8 + 8 = 0

Option (b) f(-2)  0

Option (c) f(3) 0

Option (d) f(-3) 0

If x, y, z are positive numbers such that  x + [y] + {z} = 3.8, [x] + {y} + z = 3.2, {x} + y + [z] = 2.2, where [p] denotes the greatest integer less than or equal to p and {p} denotes the fractional part of p. E.g. [1.23] = 1, {1.23} = 0.23 = 23100. The numerical value of [x2+y2+z2] is -

A.

7

B.

8

C.

9

D.

None of these

#### SOLUTION

Solution : C

Adding the three equations we get,  2(x+y+z)=9.2x+y+z=4.6.
Subtracting first eq. from the above eq. we get {y}+[z]=0.8{y}=0.8 and [z]=0.
Subtracting second eq. from the above eq. {x}+[y]=1.4{x}=0.4 and [y]=1.
Subtracting third eq. from the above eq. [x]+{z}=2.4[x]=2 and {z}=0.4.
x=2.4,y=1.8 and z=0.4.
Hence, choice (c) is the right answer.

Anil looked up at the top of a lighthouse from his boat, and found the angle of elevation to be 30. After sailing in a straight line 50 m towards the lighthouse, he found that the angle of elevation changed to 45. Find the height of the lighthouse.

A.

25

B.

253

C.

25(3 - 1)

D.

25(3 + 1)

#### SOLUTION

Solution : D Now since tan 45 = 1,BC=DB=x
Also, tan 30 = 13=x50+x
Thus x(3 - 1) = 50
or, x = 25(3 + 1)m

Find the equation with roots 2, 3, 4, 5 and 6.

A. x5+72x4+140x3580x2+1024x+720=0
B.

x520x4+155x3580x2+1044x720=0

C. x5+20x4+165x3+585x21044x720=0
D. x530x4165x3680x621024x+720=0

#### SOLUTION

Solution : B

Let the equation be Ax5+Bx4+Cx3+Dx4+Ex+F=0
Sum of the product of the roots taken 1 at a time =BA
=BA

Sum of the product of the roots taken 2 at a time =CA
(2×3)+(2×4)+(2×5)+(2×6)+(3×4)+(3×5)+(3×6)+(4×5)+(4×6)+(5×6)=CA
155=CA

Sum of the product of the roots taken 3 at a time =DA
(2×3×4)+(2×3×5)+(2×3×6)+(2×4×5)+(2×4×6)+(2×5×6)+(3×4×5)+(3×4×6)+(3×5×6)+(4×5×6)=DA
580=DA

Sum of the product of the roots taken 4 at a time =EA
(2×3×4×5)+(2×3×4×6)+(2×3×5×6)+(2×4×5×6)+(3×4×5×6)=1044
1044=EA

Sum of the product of the roots taken 5 at a time=FA
(2×3×4×5×6)=FA
720=FA

Take A=1. Then B= -20,C=155, D= -580, E=1044, F=-720

In the given figure, two circles are inscribed inside two adjoining triangles.

Given: PR=12, RV=18, PV=10. Find TV. A.

8

B.

12

C.

16

D.

10

#### SOLUTION

Solution : A

Let ST=TM=TU=x; UV=VW=y.

Then RS= 18-(2x+y); PW=10-y=PM=PQ.

PR=RQ+PQ= 18-(2x+y)+ (10-y)=12; x+y=8. Now, TV=TU+UV=x+y. Hence TV=8.

A rectangle is drawn so that none of its sides has a length greater than 'x'. all lengths lesser than 'x' are equally likely. The chance that the rectangle has its diagonal greater than 'x' is (in %)?

A.

29.3%

B.

21.5%

C.

66.66%

D.

33.33%

#### SOLUTION

Solution : B

Draw a square of side x and an arc of radius x. All rectangles with diagonal less than or equal to x will lie within or on the quadrant of the circle. The unshaded region is the favorable area.
Hence, required probability = x2(1π4)x2×100 = 0.215 = 21.5% chance

84n74n is divisible by which of the following?

A.

15

B.

113

C.

23

D.

More than one of these

#### SOLUTION

Solution : D

(82)2n(72)2n is of the form anbn where the power is always even because 2n is always even.

That means it is divisible by 82+72 as well as 8272.

Check for n=1, it becomes 8474, which is divisible by 82+72 and 8272.

Thus. the given number is divisible by both 15 and 113.

Find the rightmost non-zero integer of the expression 1430343+1470367?

A.

3

B.

9

C.

7

D.

1

#### SOLUTION

Solution : C

It can be easily observed that 1470367 has more number of zeroes as compared to

1430343. hence, the rightmost non-zero integer will depend on the unit digit of only 1430343, which in turn will depend on the unit digit of 3343

343 can written as 4k+3

In the power cycle of 3 (3,9,7,1), the third digit is 7. Hence, the last non-zero integer in the expression is 7.

If a1+2a2+3a3=1, where a1,a2&a3 are all positive numbers; then what is the minimum value of 2a1+4a2+8a3?

A.

413

B.

(3×2)13

C.

213

D.

(27×2)13

#### SOLUTION

Solution : D

AM=(2a1+4a2+8a3)3 and GM =(2a1×4a2×8a3)(13)

Using condition AMGM, we get (2a1+4a2+8a3)3(2a1×4a2×8a3)(13)

2a1+4a2+8a33(2a1×22a2×23a3)(13)

3(2a1+2a2+3a3)(13)

But a1+2a2+3a3=1

Find the remainder when 9400+9401+9402..........9410is divided by 6?

A.

0

B.

1

C.

3

D.

2

#### SOLUTION

Solution : C

Any power of 9 when divided by 6, gives a remainder 3. hence answer =11×36 ((As there are 11 terms)=336= remainder =3

The tens digit of 617 is -

A.

1

B.

3

C.

5

D.

7

#### SOLUTION

Solution : B

We can use the last two digits technique to solve this question.

617=317×217
317=(34)4×3=(32×32)4×3=(09×09)4×3=814×3=21×3=63
217=(210)1×27=24×......28=.....72
Answer is the tens digit of the product of 72 and 63 = ...36.

A government land consists of 16 square plots of equal size. If I start from the SW corner and move towards the NE corner (movement is always in the positive direction), in how many different ways can I reach the NE corner?

A.

252

B.

74

C.

70

D.

87

#### SOLUTION

Solution : C

Method 1- Unitary method

Consider a 2x2 grid

No. of different paths = (1+ 2x2 +1)

For a 3x3 = (1 + 3x3 + 3x3 + 1) ..

For a 4x4 =( 1+ 4x4 + 6x6 + 4x4 +1) ..

Method 2- Using SàD Shortcut:- No of rows=4. number of columns=4.

i.e. using the one zero method, total number of paths= 4+4C4= 70

There are 256 players in a chess tournament (singles). Two players play a match. Matches are played on a knockout basis, the loser is eliminated after each match. How many matches need to be played to declare a winner? There is no draw.___

#### SOLUTION

Solution :

For two players, we need one match. The first round, having 256 players will need 2562= 128 matches.

Now there are 128 players left. To ensure that all players play a match, we need 1282= 64 matches

Continuing in the same way

Total number of matches = 128+64+ 32+ 16+8+4+2+1= 255

OR

In 1 match 1 player is eliminated

For tournament to conclude there should be only 1 winner and hence 255 players need to lose.
Hence we need 255 matches to eliminate 255 players.

Given that the ath, bth and cth term of an Arithmetic Progression are in Geometric Progression, then common ratio of that GP is?

A.

b-a

B.

a+b

C.

cbba

D.

c-1

#### SOLUTION

Solution : C

Assume a GP, 1, 2, 4 with common ratio =2

Assume an AP with common difference 1 = 1, 2, 3, 4

Here ath,bth and cth term is 1,2 and 4

Look in the answer options where you get answer 2 after substitution

If p, q, r and s are four positive real numbers such that pqrs = 1, what is the minimum value of (1+p)(1+q)(1+r)(1+s)?___

#### SOLUTION

Solution :

Take p=q=r=s=1.

(1+1)(1+1)(1+1)(1+1) = 16

The number of different 12-letter arrangements that can be made from the letters of the word 'BYJUSCLASSES' so that all vowels do not occur together is 10!4! × A  Find the value of 'A'. Given 'A' is a three digit natural number.
Ans: ___.

#### SOLUTION

Solution :

There are 12 letters, in which 'S' appears 4 times and 3 vowels are A,E,U.

Take all vowels (AUE) as a single object

So number of permutations when vowels occur together = 10!4! × 3!
Arrangements of 12 letters take all at atime = 12!4!
Required arrangements when al vowels do not occur together
12!4! - 10!4! × 3!
10!4! (12 × 11 -6)
10!4! × 126

A = 126

At a cricket match, out of the 2 lakh spectators, 90% were fans of team A, 85% of Team B, 70% of Team C, 75% of Team D and 80% of Team E. What is the minimum number of spectators who are fans of all 5 teams?

A.

500

B.

10000

C.

2000

D.

None of these

#### SOLUTION

Solution : D

To find the minimum number of spectators who are fans of all 5 teams, take the difference from 100 from each, and the difference from 100 again for the total

100-90=10

100-85=15

100-70=30

100-75=25

100-80=20

total = 100

100-100=0

Minimum no. of spectators who are all 5 teams' fans= 0

Find the sum to n terms of the series 312×22+522×33+732×42...... is

A.

n2+1(n+1)2

B.

n2+2n(n+1)2

C.

n2+2(n+1)3

D.

(n+1)3

#### SOLUTION

Solution : B

At n=1, we should get 34 (this is the sum to 1 term of the series)

Option (a) gives =24=12

Option (b) gives 34

Option (c) gives 38

Option(d) gives 8

When N is divided by X, the remainder is 28. When N is divided by 10X, the remainder is 125. Find the remainder when N is divided by 5X.

A.

100

B.

125

C.

112

D.

None of these

#### SOLUTION

Solution : B

Number = Quotient × Divisor + Remainder

Given that:

N = rX+ 28 and N = s*10X+125

rX+28 = s*10X + 125

Solving the above two equations, we get X(r-10s) = 97.

Therefore X is a factor of 97. So possible values are 1 and 97.

Now X leaves a remainder of 28. Therefore its greater than 28. Thus X=97. 5X=485.

5X will also leave a remainder of 125 (since 5X is a factor of 10X and 5X>125).

Find the sum of areas of POQ and ORS A.

pq2

B.

p2q22

C.

p2+q22

D.

p2+q28

#### SOLUTION

Solution : A

Since the only condition we need to maintain is that the diagonals need to intersect at right angles, assume the simplest quadrilateral which satisfies this condition- a square. Now the question can be solved by simple graphical division.

Take the side of the square =p=q=2. Area = 4. We need the area of the shaded region, which is half of that of the square= 2. Look in the answer options for 2 on substituting p=2 and q=2. Answer is option (a)

The diameter of the circle on which the vertices of the quadrilateral lie is?

A.

2p2q

B.

p2+q22

C.

p2+q2

D.

pq2

#### SOLUTION

Solution : C

The diameter of the circle = diagonal of the square = 22

A man , a woman and a child are doing a piece of work. A man can do it in 5 days, a woman can do it alone in 10 days and the child will take 20 days to complete it. During the first day the man is not feeling well and does not work properly, and from the next day onward he does not come for work. If the child got paid Rs.5100 out of a total of 18,000 for its share of work, find the efficiency of the man on the first day.

A.

63%

B.

65%

C.

75%

D.

None of these

#### SOLUTION

Solution : C

The woman is twice as efficient as the child hence she gets paid twice the money= 10200. Therefore the man gets paid 18000 - (5100+10200)= 2700. The man did

270018,000  of the work.

=960th in one day.Hence his efficiency is (15)(960)=75%.

A rice trader sells one type of rice at Rs 2.70 per kg and loses 10%. He sells another type of rice at 4.5/kg and gains 1212%. His objective is to mix these two types and sell the mixture at 3.95/kg and get a profit of 25%. What should be the ratio of quantities of rice mixed?

A.

23:6

B.

84:16

C.

6:1

D.

None of these

#### SOLUTION

Solution : B

This question can be solved using alligation. For alligation; remember to always consider the cost price only.

Selling Price (SP)

Cost Price (CP)

SP of rice 1 = 2.70/kg                           CP (rice 1) = 2.70.9 = 3/kg

SP of rice 2 = 4.5/kg                             CP (rice 2) = 4.51.125 = 4/kg

SP of mixture = 3.95/kg                      CP (mixture) = 3.951.25 = 3.16/kg Required Ratio = 84:16

On June 1, 2004 two clubs, S1 and S2, are formed, each with n members. Everyday S1 adds b members while S2 multiplies its current number of members by a constant factor r. Both the societies have the same number of members on June 7, 2004 (end of day). If b =10.5n, what is the value of r?

A.

3

B.

2

C.

1.5

D.

Cannot be determined

#### SOLUTION

Solution : B

S1 club members increase in an arithmetic progression.

Hence on June 7th ,the number of members in club S1 is  n+6(10.5n)=64n.

S2 club members increase in an geometric progression.

Hence on June 7th  ,the number of members in club S2 is  nr6=64n....(As given in the question).

Therefore r=2.

Let g be a function satisfying g(a+b)= g(a)+g(b) for all a,b which is real. If

g(1)=k, then find g(n), where n is a natural number

A.

kn

B.

nk

C.

nk

D.

1n

#### SOLUTION

Solution : B

Assumption

Assume n=2

Then

g(2) = g(1) +g(1) = 2g(1)= 2k

substitute n=2 in answer options to eliminate wrong options.

Find  p12q2+p3q3+p8q3 if p and q are the roots of the equation mx22nx+v=0

A.

mnv[(n2+v2)m4]

B.

v2(v+2n)m3

C.

v2m3

D.

nv2m3

#### SOLUTION

Solution : B

Assume the roots p and q = 1 and 2 respectively.

Sum of roots =(2nm)=3

Product of roots =vm=2

Write all the terms is the form of variable m,

2nm=3n=3m2

vm=2v=2m

p12q2+p3q3+p8q3=20

Substitute values in terms of m in answer options. Only option b gives 20

There are nine distinct numbers of which five numbers are positive and four numbers are negative. Three numbers are chosen at random and the product of these numbers is found. How many of these products are positive?

A.

40

B.

25

C.

50

D.

30

#### SOLUTION

Solution : A

For product of three numbers to be positive it must have either 0 or 2 negative numbers.

Case 1: 0 negative numbers.

3 positive numbers can be selected out of 5 in 5C3 ways = 10 ways.

Case 2: 2 negative numbers

2 negative numbers can be selected out of 4 in 4C2 ways = 6 ways.

1 positive number can be selected out of 5 in 5C1 ways = 5 ways.

Hence this combination can be selected in a total of (6×5) = 30 ways.

In the figure given below ABCD is a square and ΔPDC is isosceles with PD = PC. If the area of ΔPDC is twice the area of the square, then find the ratio of the area of ΔPDC that lies outside the square to the area of the square. A.

32

B.

45

C.

98

D.

107

#### SOLUTION

Solution : C Drop a perpendicular PM on the side DC.
Let PN be 'x' and side of square be 2a
12×2a× (2a + x) = 8a2
x = 6a
ΔPQRΔPDC=(6a8a)2=916ΔPQR=916ΔPDC
ΔPQR=916(2×ABCD)
Required Ratio = 98
Hence, option (c)

Given that p 1. What is the minimum value of p2+625p2?

A.

5

B.

10

C.

25

D.

50

#### SOLUTION

Solution : D

The minimum value will occur when p2=625p2

p4=625p=5

Putting the value of p in the equation, min value =p4=625p=5=25+62525=25+25=50