# Free Quant Practice Test - CAT

### Question 1

A circle of radius 3 crosses the centre of a square of side length 2. Find the approximate positive difference between the areas of the non-overlapping portions of the figures.

26

24

22

cannot be determined

#### SOLUTION

Solution :B

Let the area of the square= s, the area of the circle= c, and the area of the overlapping portion= x. The area of the circle not overlapped by the square is "c -x” and the area of the square not overlapped by the circle is "s - x”, so the difference between these two is (c -x) -(s -x) = c -s = 9π2 -4 .(approximately = 24.26). Hence the correct answer is option (b).

### Question 2

How many pairs of positive integers (a, b) are there such that a and b have no common factor greater than 1 and ab+14b9a is an integer?

1

2

3

4

#### SOLUTION

Solution :D

Let u=ab. Then the problem is equivalent to finding all possible rational numbers 'u' such that

u+149u=K, for some integer K.

This equation is equivalent to 9u2−9uk+14=0

9u2−9uk+14=0 whose solutions are,

u=9k±√81k2−50418

=K2±16√9k2−56

Hence u is rational if and onlly if

√9k2−56 is rational, Which is true if and only if 9k2−56 is a perfect square. Suppose that 9k2−56=S2 for some positive integer S.

(3k+s)(3k-S)=56

The only factors of 56 are 1,2,4,7,8,14,28 and 56. So, (3k-S) and (3k+s) is one of the

ordered pairs (1,56), (2,28), (4,14) and (7,8). The cases (1,56) and (7,8) yield no (4,14) yield k=5 and k=3 respectively.

If k=5 then u=13 or u=143

If k=3 then u=23 of u=73.

Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3)

### Question 3

Let a, b, c, d and e be distinct integers such that (6 - a) (6 - b) (6 - c) (6 - d) (6 - e) = 45. What is a+b+c+d+e?

25

28

27

16

#### SOLUTION

Solution :A

If 45 is expressed as a product of five distinct integral factors, the unique way is 45 =(+1) (-1) (+3) (-3) 5 (no factor has its absolute value greater than 5)

⇒ (6 - a)(6 - b) (6 - c) (6 - d) (6 - e) =(1) (-1) (3) (-3) 5The corresponding values of a, b, c, d and e are 5, 7, 3, 9 and 1 and their sum is 5 + 7 + 3 + 9 + 1= 25.

### Question 4

Sum of the first 30 terms of an arithmetic progression is 0. If the first term is -29, then find the sum of the 28th, 29th and 30th terms of this arithmetic progression.

81

84

-84

-81

#### SOLUTION

Solution :A

Let the common difference of the arithmetic progression be 'd'.

Sum of first 30 terms of the arithmetic progression

= 30/2*[2(-29) + (30-1)d]

Hence, 15(−58+29d) = 0

Hence, d=2

Sum of 28th, 29th and 30th term of this arithmetic progression= 3(-29) + (27 + 28 +29) × 2 = 81

### Question 5

If f(x) = 9x9x+3,find the value of f(150) + f(250) + ...... + f(4850) + f(4950)

24

24.5

25

25.5

#### SOLUTION

Solution :B

f(x) = 9x9x+3

f(1−x) = 91−x91−x+3=99x[(99x)+3]=33+9x

f(x) + f(1 - x) = 1

So, f(150) = 1 - f(1−150) = 1 - f(4950) → f(150) + f(4950) = 1

Similarly f(250) + f(4850) = 1

So, the given expression becomes:1 + 1 + ..... 24 times + f(2550) = 24 + f(12)

We have f(x) + f(1 - x) = 1.

Put x = 12, 2f(12) = 1 → f(12) = 12

so, the answer is 24.5. Hence option (b)

### Question 6

Tiles are packed into packets of 6, 9 and 50. What is the largest number of tiles that cannot be purchased with any combination of above packets. Tiles cannot be sold loose.

449

451

103

49

#### SOLUTION

Solution :C

After 6 all multiples of 3 would be covered. After 50+6 all 3x+2 tiles would be covered. After 50+50+6 all 3x+1 tiles would be covered. So for 106 and above all tile combinations are possible. 105 again is possible (it is a multiple of 3), 104 is also possible as it is of the form 3x+2. The largest possible value is 103 which can't be obtained.

### Question 7

A supplier provides imported TV'S to a retailer each year. Each TV costs Rs 10000 for the supplier. 5% of the TVs are defective and they are replaced

without extra cost. If the supplier still makes a profit of 20% , what is the SP of each TV to the retailer?

12000

12600

13000

13500

#### SOLUTION

Solution :B

To make calculation simpler, take cost of each TV as 1000

5% defective means, out of every 20, 1 is defective

cost of 20 TVs = Rs. 20000

To make a profit of 20000 = 1.2 × 20000 = 24000Selling 19 TVs at 24000, entails a SP of (2400029)=12600 ( approx)

### Question 8

Find the number of factors of 3150.

36

52

26

None of these

#### SOLUTION

Solution :A

3150 can be prime factorized as follows:

3150=52×2×7×32Number of factors = (2+1)(1+1)(1+1)(2+1)=3×2×2×3=36.

### Question 9

How many three- digit numbers can be formed with at least 2 distinct digits such that the product of the digits is the cube of a positive integer?

33

#### SOLUTION

Solution :A

The combination of three digits, whose product can give a perfect cube are (1,2,4) (1,3,9) (4,6,9) (3,8,9) and (2,4,8) when all three digits are distinct.With each of these combinations we can have 3! three digit numbers with distinct digits. Therefore, the number of such 3-digit numbers is 5(3!) = 5*6 = 30. There is another case of (8,8,1) and the number of such 3 digit numbers is 3. answer = 30+3= 33.

### Question 10

2 cows and 7 calves can graze a field in 14 days. 3 cows and 8 calves can graze the same field in 11 days. Then 8 cows and 6 calves can graze 4 times the same field in how many days? It is known that, once grazed; the grass does not grow back

21

281

392

18

#### SOLUTION

Solution :C

Aàcows

Bàcalves

2 cows and 7 calves are needed to do a piece of work in 14 days. This means to complete the work we need 2 cows to graze for 14 days and 7 calves to graze for 14 days. So if we want to complete the work in one day we need: 2×(14)=28 cows and 7×(14)=98 calves ⇒ 28A + 98B.

Also, 3 cows and 8 calves take 11 days, so if they need to complete the work in 1 day, we need 3(11) =33 cows and 8 (11) =88 calves ⇒ 33A + 88B

We can now equate the two cases as 28A + 98B = 33A + 88B

Solving we get 5A=10B or 1A=2B

i.e. one cow is equivalent to two calves.

We need to calculate for 8 cows & 6 calves or 8+3=11 cows

3 cows and 8 calves can graze the field in 11 days ⇒ 7 cows can graze the field in 11 days

⇒ 11 cows can graze the field in 7 daysThus, 11 cows can graze 4 fields in 7 × 4=28 days. Option (d)

### Question 11

Find the number of different possible points scored by Peter, Thomas and Ram given that they score a total of 32 points and Peter scores a minimum of 4 points. (Note:- All points scored are whole numbers)

561

445

485

435

#### SOLUTION

Solution :D

It is an Sà D question

P+T+R = 32, where P≥4

Question changes to P+T+R= 28.

Number of different possible points scored = 30C2= 435This question is not considered for evaluation

### Question 12

If the (p+1)th, (q+1)th, (r+1)th terms of an AP are in GP when p, q, r are in HP, then the ratio of the first term to the common difference of the AP is

2q

-2q

−q2

q2

#### SOLUTION

Solution :C

Let the terms be a+pd, a+qd, a+rd where a is the first term of AP and d is common difference of AP. Now p,q,r are in HP, So, q=2pr(p+r) and also the terms are in GP. So (a+qd)2=(a+pd)×(a+rd). Solving this leaves us with (ad)=(q2−rp)(p+r−2q). Substituting p+r=2prq, we get ad=−q2 .Hence, choice (c) is the right option

### Question 13

In a survey on a fraction of the teenage population on their preference for the "in” brands this season, 78% of these asked were in favour of at least one of the brands: 1; 2 and 3. 50% of those asked favoured Brand 1, 30% favoured Brand 2, and 20% Brand 3. If 5% of those asked favoured all three Brands, what percentage of those asked favoured more than one of the three Brands?

10

12

17

22

#### SOLUTION

Solution :C

We will use the S-X technique to solve this question

X (those favouringatleast 1)=78

S (I+2II+3III+4IV)=50+30+20=100

All 3 (III) = 5

S-X = II + 2III

22=II + 2(5) ⇒ II = 22-10=12More than one = II + III = 12+5=17

### Question 14

The area of regular hexagon ABCDEF is 144. Then, the area of shaded region is

#### SOLUTION

Solution :

A Regular hexagon is the combination of 12 identical triangles, out of 12 three triangles are shaded. Thus the area of shaded region =312 × 144 = 36. Option (b).

### Question 15

A palindrome is a positive integer which is unchanged if you reverse the order of its digits. If all palindromes are written in increasing order, how many possible prime values can the difference between successive palindromes take?

1

2

3

none of these

#### SOLUTION

Solution :B

Let x be a palindrome and x' the next highest palindrome. If x<101, then it is easy to see by inspection that x' - x = 1, 2 or 11, so the only prime differences are 2 and 11.

So assume x>100. If x and x' have the same final digit, then their difference is divisible by 10 and hence not prime. So they must have different digits. Thus either x = d9...9d and x' = d'0...0d', where d<9 and d' = d+1, or x' has one more digit than x and d = 9, d' = 1. In the first case x' - x = 11. In the second case x' - x = 2. So again the only prime differences are 2 and 11.Hence, choice (b) is the right answer

### Question 16

Find the remainder when 4001000 is divided by 17?

0

1

-1

2

#### SOLUTION

Solution :B

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

Step 1 - Remainder when 400 is divided by 17 is 9

The problem changes to 9100017

From fermat theorem, the Euler's number of 17 is 16.

1000=16k +8

Problem now changes to 9817

Going by frequency method, 9417 gives remainder -1 so 9817 will give remainder 1.

Altenative solution :

We know tha 92=81=>(81)500=>(−4)500=>16250=>(−1)250=1

### Question 17

Find the number of 6 letter words that can be formed using all letters of the English alphabet so that the words contain atleast 3 consonants and each letter of the word is unique in every case?

(21C3 × 5C4 + 21C4 × 5C3 + 21C5 × 5C2 + 21C6) × 6!

(21C3 × 5C3 + 21C4 × 5C2 + 21C5 × 5C1 + 21C6) × 6!

21C3 × 5C3 + 21C4 × 5C2 + 21C5 × 5C1 + 21C6

None of these

#### SOLUTION

Solution :B

TOTAL = 21 consonants and 5 vowels

Three cases

3 consonants & 3 vowels= 21C3 × 5C3

4 consonants & 2 vowels= 21C4 × 5C2

5 consonants & 1 vowel= 21C5 × 5C1

6 consonants & 0 vowels= 21C6

Arrangement = 6! In each case(21C3 × 5C3 + 21C4 × 5C2 + 21C5 × 5C1 + 21C6) × 6!

### Question 18

A boy has to climb 10 steps. He tosses one coin. If head appears, he climbs a single step. If tail appears, he climbs 2 steps. In how many ways can he climb 10 steps?

90

88

92

89

#### SOLUTION

Solution :D

Single StepDouble StepNumber of ways1001819!8!=9628!6!×2!=28437!4!×3!=35246!2!×4!=15051

Total number of ways = 1 + 9 + 28 + 35 + 15 + 1 = 89. Option(d)

### Question 19

If this pattern continues, where would the number 289 appear?.

Row 17

Row 18

Row 19

Row 20

#### SOLUTION

Solution :A

Perfect squares are the middle term of odd rows. (17)2=289 should be middle term of 17th row. Option (a).

Alternate Solution:

There is 1 odd number in the 1st row.

There are 2 odd numbers in the 2nd row.

There are 3 odd numbers in the 3rd row. ...

Hence this can be considered as a∑n question where, n is the number of odd numbers.

As 289 will be 2892=144145th term in the sequence, find a∑n value which is close to the number. which is equal to 153 and 17 numbers before 153 will be in the 17th row. Hence the answer is 17.

### Question 20

A park has the shape of a regular hexagon of sides 2 m each. Ram walks a distance of 5 m around the perimeter. What is the direct distance between the start point and the end point?

√12

√13

√17

√16

#### SOLUTION

Solution :B

Vertex angle for regular hexagon = 180∘(6−2)6=30∘×4=120∘

In the ΔABC, ∠ACB = ∠CAB = 30∘

We know 30∘ - 60∘ - 90∘ = 1 : √3 : 2 ; AC = 2√3

Now, from ΔACD:-

12 + (2√3)2 = AD2;AD = √13.Option (b)

The starting point of Ram is A and end point is D.

Alternate Solution:

The answer option will have to be between 2√3 and 4 i.e. between √12 and √16.Only option (b) fulfils the given condition and hence option (b) is the correct answer choice.

### Question 21

Every year Aman consumes 250 kg wheat and 90 kg rice. The price of wheat is 20% of the price of rice and he spends a total of Rs. 3500 on wheat and rice per year. If the price of rice is increased by 20% then what is the percentage reduction of wheat consumption for the same expenditure of Rs. 3500? (Given that the price of wheat and consumption of rice is constant)

24%

40%

25%

36%

#### SOLUTION

Solution :D

Let the price of Wheat =X per kg. Then, the price of Rice =5X per kg (As it is given that the price of Wheat is 20% the price of Rice.)

Wheat Rice

250 90

x250 x 5x450 x

700 X = 3500

X=5

Hence the price of wheat is = Rs.5 per kg

Let the new amount of wheat be M kg,then M×5+90×30=3500;M=160

(As the price of Rice increases by 20% i.e., from Rs.25 to Rs.30)

Hence decrease (in %) of amount of wheat =250−160250×100=36%

Alternate Method:

As wheat is 20% of rice, instead of consuming 250 kg of wheat, Aman can consume 50 kg of rice. Hence,

(50+90)x=3500 ⇒ X=25.

Now,x(25)+90(30)=3500.⇒ x=32.

Therefore,change in consumption =50 - 32=18kg.

Thus,percentage increase will be 1850 or 36%

### Question 22

The sum of first n terms of a A.P. is 100 and sum of first n terms of an other A.P. is 300. Find the ratio of common difference:

1:6

1:3

1:1

cannot be determined

#### SOLUTION

Solution :D

If sums of the series are in ratio 1:3, then the ratio of common difference will remain same.For example: - Let an A.P. 18, 19, 20, 21, 22 , an other one 58, 59, 60, 61, 62 and an other A.P. 54, 57, 60, 63, 66 sum of first A.P. =100 and sum of others A.P. = 300. So, the ratio of common difference = 1: 1 and 1 : 3.Option (d).

### Question 23

Four equal jugs are filled with the mixture of sprit and water. The ratio of sprit to water in four jugs is 1:3, 3:5, 5:11 and 7:9. The mixture of the four jugs is emptied into a single vessel. What is the proportion of sprit to water in the vessel?

11:21

11:30

12:24

3:5

#### SOLUTION

Solution :A

Let x (in ml) be the volume of each jug.

Jug1st2nd3rd4thS:W1:33:55:117:9S14x38x516x716xW34x58x1116x916x

(14x+38x+516x+716x)(34x+58x+1116x+916)=(2216)(4216)=2242=1121

### Question 24

Given that n∑x=1(x+1)(2x+3) = pn4+qn3+rn2+73n+s, then

i)p=12,q=83 ii)p=13,q=14 iii)r=92,s=0 iv)r=56,s=0

I and IV only

II and III only

I and III only

II and IV only

#### SOLUTION

Solution :C

Put n = 1, x = 1

Then LHS = (1+1)× (2+3) = 10

RHS = p+q+r+73+s should = 10

Plug in all possible answer values

Options I and III give the right answer as 12+83+92+73=10

### Question 25

A college has 10 tennis players. A 4-member team has to be selected and a non-playing captain will be selected out of the remaining players. How many different selections can be made?

1260

210

10C4 × 6!

10C4 × 6

#### SOLUTION

Solution :D

A 4-member team can be selected in 10C4 ways. Now as the captain is non-playing he needs to be selected from the remaining 6 players.

Therefore final answer is 10C4 × 6.

### Question 26

How many 3 digit numbers contain 5 as one of their digits?

271

252

300

243

#### SOLUTION

Solution :B

Total number of 3 digit numbers = 999-99= 990

Those not having 5 s any of their digits = 8×9×9=648

Those having 5 as one of their digits = 990-648 = 252

Alternate Solution:

This can also be manually counted.

5 can come either in hundreds place or tens place or units place.

In hundreds place: 500 to 599 : 100 numbers.

In tens place: x 5y where y can take values from 0 to 9 and x can take values from 1 to 9 (except 5) = 80 numbers.

In units place: xy5 where y can take values from 0 to 9 (except 5) and x can take values from 1 to 9 (except 5) = 72 numbers.Total = 100 + 80 + 72 numbers = 252 numbers.

### Question 27

The quadratic function f(x)=ax2+bx+c is known to pass through the points (-1, 6), (7, 6), and (1, - 6). Find the smallest value of the function.

-6

-10

-20

-26

#### SOLUTION

Solution :B

If the points (-1,6), (7,6),and (1,-6) lie oh the graph of y=ax2+bx+c,

Then 6 = a - b + c, 6 = 49a + 7b + c, -6 = a + b + c.

Solving this system, we get a = 1, b = -6, c = -1.

The graph of f(x) = x2−6x−1 is a parabola that opens upward and the vertex is (−b2a,−D4a)=(3,-10).

Option (b).

### Question 28

Ahmed and Sahil set out together on bicycle traveling at 15 and 12 kilometers per hour, respectively. After 40 minutes, Ahmed stops to fix a flat tire. If it takes Ahmed one hour to fix the flat tire and Sahil continues to ride during this time, how many hours will it take Ahmed to catch up to Sahil assuming he resumes his ride at 15 kilometers per hour? (consider Ahmed's deceleration/acceleration before/after the flat to be negligible)

4.5

3.33

3.5

4

#### SOLUTION

Solution :B

A travels 10 km in 40 min

B travels 8 km in 40 min

After one hour, A would have still traveled only 10 km and B would have traveled 20 kmRepresenting this on a timeline:

Time in hours123455:20DistanceA101025405560DistanceB82032445660

A gains this 10 km in 103=3.33 hours.

Answer is Option (b).

### Question 29

In an automated plant assembly line, the rate of rejection of components was 10% on July 1st and 6% on July 2nd . The combined rate of rejection for the two days was 9%. The ratio of production volumes on July 1st and on July 2nd is

2:1

3 : 2

3 : 1

None of these

#### SOLUTION

Solution :C

Approach 1:Conventional Approach

Ans. (c) Let the production on July 1st and on July 2nd be X and Y units respectively.

Therefore, Rejection on July 1st = 0.1 X

Rejection on July 2nd = 0.06 YTherefore, Total rejection = 0.1 X + 0.06 Y = 0.09 (X + Y). Hence, X : Y = 3 : 1

Required ratio is 3:1.

Option (c)

### Question 30

The sum to 2n terms of the series 53+4×63+73+4×83+93.....is ?

n2(10n2+96n+103)

n(12n3+12n2+187)

n(10n2+96n2+343n+540)

None of these

#### SOLUTION

Solution :C

At n=1

2n= 2

Sum to 2n terms = 125 + 864 = 989Substitute n=1 in answer options. Only option c gives 989

### Question 31

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

0

g(1)3

#### SOLUTION

Solution :A

Put a = 1

Then the equation changes to g(2) + g(0) = 2g(1) ⇒ 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 32

If 2a^{2}
- 3a + 15 < 20, then the solution set of 'a' for the inequality is

#### SOLUTION

Solution :B

Substitute a value for "a” which is there in 2 answer options and not there in the other two

Take a=2

At a=2 LHS<RHS. Hence, options (c) & (d) can be eliminated

Answer can be option (a) or (b)

Now substitute a value that is there in option (b) and not in option (a). put a =0LHS<RHS. Hence, option (a) can also be eliminated. Answer is option (b)

### Question 33

For how many integer values of m does the equation x2−10x+(m+3)(m−4)=0 have roots of opposite signs?

#### SOLUTION

Solution :If the roots are of opposite signs, product of roots must be negative.

i.e.(m+3)(m−4)<0⟹−3<m<4Number of integer values satisfying the above inequality is 6 (-2, -1, 0, 1, 2, 3).

### Question 34

If 2x+3y=12; x, y>0, then the maximum possible value of xy2 is

#### SOLUTION

Solution :Power of x is 1 and that of y is 2 in xy2.

So, we split the term with y into two.

2x+2y+y=12

He know that, AM≥ GM.

2x+2y+y3≥(2x×2y×y)13