Free Quant Practice Test - CAT 

Question 1

In the unit square, find the distance from E to AD in terms of a and b, the lengths of DF and AG, respectively

A.

aba+b

B.

ba+b

C.

aba+b

D.

aa+b

SOLUTION

Solution : A

Let x be the desired distance and y be the distance from E to DF as shown, Since ΔDEFΔGEA,y1y=ab;y=aa+b Since HE is parallel to side AG of ΔADG,
xb=y1;x=aba+b

 

 

Question 2

Find the remainder when 1373+185 is divided by 11.

A.

8

B.

7

C.

10

D.

4

SOLUTION

Solution : B

137311r=27311r=3214×811r=(1)14×8=8
Similarly , 18511r=7511r=343×4911r=2×5=10
So, the remainder =8+10=1811r=7.

Question 3

What is the smallest number that has exactly 8 factors?

A.

27

B.

24

C.

54

D.

30

SOLUTION

Solution : B

Number of the form paqbrc (where p, q and r are primes) has (a + 1) (b + 1) (c + 1) factors.
So, if some number has 8 factors (a + 1)(b + 1)(c + 1) should be 8.
8 can be written as 8 or 4×2 or 2×2×2.
If a number has to have 8 factors, it should be in one of the following forms,
p77+1=8 factors
p3q14×2=8 factors
pqr2×2×2=8 factors
Let us deduce the smallest possible number in each form.
p727
p3q123×3=24
pqr2×3×5=30

Therefore, smallest number that has exactly 8 factors = 24.

Question 4

In how many ways can the number 286 be written as a sum of two or more consecutive integers?

A.

3

B.

1

C.

2

D.

4

SOLUTION

Solution : A

Shortcut - Number of ways in which a no. N can be written as a sum of two or more consecutive integers = (No. of odd factors of that number-1)
286=2×11×13. So, it has 4 odd factors - 1, 11, 13, 143.

(No. of odd factors - 1) = 4 - 1 = 3.

Question 5

Find the number of solutions of the equation xx1+x1x=137.

A.

0

B.

1

C.

2

D.

More than 2

SOLUTION

Solution : A

If we suppose xx1=a.
Then, xx1=1a
And the equation becomes a+1a=137
Now we know that for a real number a :
a+1a is greater than or equal to 2 "OR" less than or equal to -2.
Therefore a+1a can never be 137 (less than 2 but greater than -2).
Hence the given equation has no solution i.e., the number of solutions is zero.

Question 6

If the integers m and n are chosen at random from integers 1 to 100 with replacement, then the probability that a number of the form 7m ? is divisible by 5 equals

A.

14

B.

17

C.

18

D.

None of these

SOLUTION

Solution : A

Option A is the correct answer.

Question 7

A rubber ball is dropped from a height of 4 meters onto a concrete floor.  It bounces to 75% of its original height and drops back down, bouncing many times in ever-smaller bounces.  If it did this forever how much total distance did the ball cover?

A.

30m

B.

26m

C.

28m

D.

24m

SOLUTION

Solution : C

Distance travelled before first hit = 4 meters.
Distance travelled before second hit = 4+0.75 ×4×2
Distance travelled before third hit =4+0.75×4×2+(0.75)2×4×2. And so on.
Hence required distance
=4+8×0.75+8×0.752+8×0.753+............=4+8×0.750.25=28m




 

 

Question 8

Before his last Physics test, Nitish's average for Physics was 33%. In his last test he scored 40% which increased his average to 34%. What must he score in his next test to increase his average to 35%?

A.

45%

B.

40%

C.

44%

D.

None of these

SOLUTION

Solution : D

Method 1: Equation Approach.

Suppose that Nitish wrote 'n' Physics tests before his last test.
Since his average is 33% in those n tests, the sum of his scores in those tests equal 33n percent.
He then scored 40% in his next test to increase his average to 34%.
The sum of all his scores is then equal to 33n+40 percent,giving an average of 33n+40n+1=34%. Solving this equation for n gives n=6.
Thus,if Nitish wants an average of 35 after 6+2 =8 tests,the sum of his scores should equal  8×35%=280%.
After 7 tests the sum of his scores equal 7×34%=238%,so he needs to score 280%238%=42%in his eighth test.

Method 2: Alligation Approach



So, we can easily determine X=42%

 

Question 9

Two coins are kept in a square of side 1 as shown in the diagram. The sum of the radii of coins is

A.

(2 - 2

B.

(2 - 1) 

C.

132

D.

122

SOLUTION

Solution : C


Suppose the coins have radii R and r.
The centres of the coins and the point where they touch divide the diagonal into four line segments with lengths R2 , R, r, r2
Since the diagonal has length 2, it follows that (R + r) (2 + 1) = 2,
So R + r = 22+1 = (2 - 2

Question 10

A car is moving towards a building. When it is at a distance of 500 metres from the building, another car starts moving away from the building. The two cars meet at a distance of 100 metres from the building, at an angle of elevation of 600Find the height of the building?

A.

58 metres

B.

100 metres

C.

173 metres

D.

220 metres

SOLUTION

Solution : C

The given information can be drawn as the adjoining figure.
So, tan(60) = h100 3 = h100
h = 1003 = 173.14 metres 

Question 11

The diagram shows a sequence of squares, decreasing in size and alternating between white and grey. Each new square is drawn by connecting the midpoints of the sides of the larger square that precedes it. If this process is continued indefinitely, then the fraction of the area that will ultimately be grey is

A.

14

B.

13

C.

516

D.

25

SOLUTION

Solution : B

Each square is half the area of the square just bigger than it (which is the other colour), so it is one-quarter of the area of the next larger square of the same colour. Taking the area of the whole figure to be 1, we see that the area of the four largest grey triangles is 14, and each set of four grey triangles is % of the area of the previous one. Thus the total grey area is 14 + 142143+ ........=14(114) = 13

Question 12

What number should be subtracted from the expression P=x3+4x27x+12 for it to be perfectly divisible by x + 3?

A.

42

B.

39

C.

13

D.

None of these

SOLUTION

Solution : A

Method 1 :-
According to remainder theorem, the remainder for f(x)x+a is f(-a).

Let the number to be subtracted be 'k'.
Since, x+3 divides x3+4x27x+12k perfectly the remainder when x is substituted by -3 is 0f(3)=0.
(3)3+4(3)27(3)+12k=0
27+36+21+12=k
k=42

Method 2 :-
If we want x3+4x27x+12 to be perfectly divisible by x+3,
x3+4x27x+12 should be of the form of (x+3)×g(x).
(x3+4x27x+12) 
=(x+3)(x2+x10)+12+30
=(x+3)(x2+x10)+42
So, the number that needs to be subtracted = 42.

Method 3 :- Shortcut!
Since this is a variable-number question, assume x=1; then the expression P=10.
Now we need to check which answer option satisfies the following condition.
10(answer option)4(Remainder)=0
10424(Remainder)=324(Remainder)=0

Question 13

Aman and Bahadur are walking in the same direction beside a railroad track, and Aman is far behind Bahadur. Both walk at constant speeds, and Aman walks faster than Bahadur. A long train traveling at a constant speed in the same direction will take 10 seconds to pass Aman (from the front to the end) and will take 9 seconds to pass Bahadur. If it will take twenty minutes for the front of the train to travel from Aman to Bahadur, how many minutes will it take for Aman to catch up to Bahadur?

A.

200

B.

220

C.

240

D.

260

SOLUTION

Solution : A

Denote by VA,VB, and VT the speeds of Aman, Bahadhur, and the train,respectively .The relative speeds of the train with respect ti Aman and Bahadur are than VTVA and VTVB. It follows from the condition that the train passes Aman in 10 seconds and Bahadur in 9 seconds that

VTVAVTVB=910.

The relative speed of Aman with respect to Bahadur is VAVB. We have

VAVBVTVB=VTVBVTVBVTVAVTVB=1910=110

Thus, Aman's speed with respect to Bahadur is ten times smaller than the train's, so, in order to reach Bahadur, it would take Aman ten times the time it would take the train, (i.e.,)200 minutes.

Question 14

What is the remainder when 15987797 is divided by 10?

A.

3

B.

1

C.

7

D.

9

SOLUTION

Solution : C

Express the power in the form of 4K + X. In this case 797 = 4K + 1.
159871 = Unit digit is 7. Hence, remainder is 7 when divided by 10.

Question 15

"A” and "B” have to clear their respective loans by paying 3 equal annual installments of Rs.4,00,000 each. "A” pays at 10% per annum of simple interest while "B” pays 10% per annum compound interest. What is the difference in their payments?

A.

4333.33

B.

5250

C.

3450

D.

4000

SOLUTION

Solution : D

- 4, 00, 000{1 + 1.1 + (1.1)2} - 4, 00, 000{1 + 1.1 + 1.2} = 4000. Option(d).

Question 16

The number assigned to each of the 26 alphabets namely A, B, C... Y and Z is 1, 2, 3... 25 and 26 respectively. A selection of certain alphabets is made such that the selection includes all the vowels and all those alphabets which have been assigned a prime number and a five-letter word using the alphabets from the selection is formed. Which of the following is definitely not a word that can be formed?

A.

IGCAM

B.

SCEKA

C.

SUOEM

D.

OGVUE

SOLUTION

Solution : D

The alphabets which are selected are: A,E,I,O,U,B,C,G,K,M,Q,S and W.

Since, V is not selected the correct answer will be option (d).

Question 17

A person starts multiplying consecutive positive integers starting from 20. How many numbers should he multiply before he will have a result that will end with 3 zeroes?

A.

11

B.

10

C.

6

D.

5

SOLUTION

Solution : C

A number will end in 3 zeroes when it is multiplied by three 10s.
To get a 10, one needs a 5 and a 2.
20=2×1020 will give one zero and it has one extra 2.
So, to get three zeroes he should multiply till he encounters two more 5s and one more 2.
Since 25 has two 5s and 22 has one 2 he should multiply till 25 i.e. 20, 21, 22, 23, 24, 25 = 6 numbers.

Question 18

Three numbers are chosen at random from 1,2,3....10 without replacement. The probability that the minimum of the chosen numbers in 4 or the maximum is 8 is?

A.

1140

B.

410

C.

112

D.

35

E.

23

SOLUTION

Solution : A

Probability of 4 being the minimum number of the three numbers selected = 6C210C3 
Probability of 8 being the maximum number of the three numbers selected = 7C210C3 
The probability of 4 being the minimum number and 8 being the maximum number = 310C3 
Therefore, required probability is 6C210C3  + 7C210C3310C31140 

Question 19

What is the remainder when (37)2012 is divided by 25?

A.

6

B.

21

C.

11

D.

1

SOLUTION

Solution : A

Last two digits of (37)2012=(69)1006=(61)503=8125r=6.
Note - The ten's digit of 61n has a cyclicity of 5 (61-21-81-41-01-61) and 503=5k+3.

Question 20

Which of the following graphs represent the function- f(x) = x4 + x3- 2x2

A.

B

B.

A

C.

D

D.

C

SOLUTION

Solution : D

Shortcut- Go by elimination
At x=1, f(x)=0, option (b) and option (d) are eliminated

At x= -1, f(x)= -2.

Question 21

Let two adjacent sides of a parallelogram be a and b while the diagonals be x and y. Then x2+y2 is ___

A.

2(a2+b2)  

B.

12(1a2+1b2)

C.

2ab

D.

2ab

SOLUTION

Solution : A

Method 1 

From the figure,
x2=BD2=(ap)2+n2  
y2=AC2=(a+p)2+n2 
b2=p2+n2 
x2+y2=2(a2+p2+n2)=2(a2+b2) 

Method 2- Assumption
A rectangle is a type of parallelogram. Assume a rectangle with length = 4 and breadth = 3
Then the diagonal = 5 (Pythagoras triplet) 

Thus x=y=5 and a=4 and b=3
Substitute in answer options. Only option (a) is satisfied. Hence, answer is option (a)

Question 22

A positive integer is called "Panache” if it can be written as a sum of distinct positive powers of 4, and "Elan" if it can be written as a sum of distinct positive powers of 6. From the given number, which of the following can we write as a sum of a "Panache” number and an "Elan" number?

A.

2012

B.

2013

C.

2011

D.

None of these

SOLUTION

Solution : D

Suppose that 2013 = a + b, where a is "Panache” and b is an "Elan". Then 'a' must be a sum of some of the powers of 4 less than 2013, namely 1, 4, 16, 64, 256, and 1024. Similarly, b must be a sum of some of the numbers 1, 6, 36, 216, and 1296. So, a + b must be a sum of some distinct entries from the list 1, 1, 4, 6, 16, 36, 64, 216, 256, 1024, 1296. If we use both 1024 and 1296, we get at least 1024+1296 = 2320 which is too big. But if we omit one of them, the most we can get 1 + 1 + 4 + 6 + 16 + 36 + 64 + 216 + 256 + 1296 = 1896; which is too small. So there is no way to achieve a sum of 2011, 2012 and 2013.

Question 23

If n is a positive integer, what is the smallest value of n such that

(n+10) + (n+11) + (n+12) + ......+ (n+58) is a perfect square?

A.

2

B.

4

C.

3

D.

1

SOLUTION

Solution : A

This is an arithmetic sequence with 49 terms, and middle term (n + 34), so the sum is equal to 49(n + 34). Since 49 is 72, which is already a perfect square, we need the smallest value of n such that (n + 34) is a perfect square. We see that for n=2, n+34 is a perfect square (since 36=62).

Question 24

There are 30 students in a class whose average age is 20 years. The average age of the boys in the class is 3 less than the average age of the girls in it. If the age of their class teacher, which is between 49 years and 54 years is included with boys, the average age of the boys equals to that of the girls. If the average age of boys as well as the average age of girls are integers, then how many boys are there in the class?

A.

11

B.

12

C.

10

D.

Cannot be determined.

SOLUTION

Solution : C

The total of ages of the students of the class = 600 years.Let the number of boys in the class be x.Let there average be y.Let the age of teacher be z.
xy+(30-x)(y+3)=600
30y+90-3x=600 ; x=10y-170.
When teacher is also included xy+zx+1=y+3 
xy+z=xy+3x+y+3.
z=x+17010+3x+3=31x10+20;
Now use answer options.Only possibility is x=10.Option (c).

Question 25

Find the value of a and b if 1×5+2×52+3×53+.....x×5x=(4x1)5a+1+b16

A.

 a=x,b=2

B.

a=x, b=5

C.

a=0,b=1

D.

None of these

SOLUTION

Solution : B

Conventional Approach :
S=1×5+2×52+3×53+....+x×5x.....(i)
5S=1×52+2×53+.....+x×5(x+1).........(ii)
(S5S)=1×5+52(21)+53(32)+....+5x(x(x1))x×5(x+1)
4S=5+52+53+....+5xx×5(x+1)
4S=x×5(x+1)(5+52+53+.......+5x)
S=x×5(x+1)(5+52+53+,,,+5x)4
(4x1)5(a+1)+b16=x×5(x+1)(5+52+53+,,,+5x)4
(4x1)5(a+1)+b16=x×5(x+1)(5(x+1))544
(4x1)5(a+1)+b16=(4x1)5(x+1)+516,
Now we can compare both sides and we can determine a=x and b=5.
Go from answer options.
For answer option b.
When a=x and b=5. Take x=1 , then a=1 and b=5.
LHS =5.
RHS =[3(25)+5]16=5

Question 26

An unlimited number of coupons bearing the letters A, B and C are available. What is the possible number of ways of choosing 3 of these coupons so that they cannot be used to spell BAC?

A.

15

B.

18

C.

21

D.

27

SOLUTION

Solution : C

Each of the 3 places can take 3 letters 27. But we don't want the combination (A, B, C) 3! = 6 are out 27-6 = 21. Hence, choice (c) is the right answer.

Question 27

A boat leaves it's dock and travels upstream at a speed of 40 km/hr and then returns to the dock. What is the speed of the stream if they average speed of the entire trip is 60 km/hr?

A.

20 km/hr

B.

30 km/hr

C.

50 km/hr

D.

None of these

SOLUTION

Solution : D

Speed of stream = x.
Speed of boat = b.
Given, b - x = 40.
Also,

2[(b+x)(bx)](b+x+bx)=602(b+x)(40)2b=60.

Solving we get  b= 2x.
2x - x = 40 , x = 40 (d).

Question 28

A firm has tractors of four models A, B, C and D. Four tractors (two of model B and one each of models C and D) plough a field in 2 days. Two model A tractors and one model C tractor take 3 days to do this job. Three tractors one each of models A, B and C take 4 days to do the same task. How long will it take to do the job if a team is made up of four tractors of different models?

A.

107 days

B.

147 days 

C.

137 days 

D.

127 days

SOLUTION

Solution : D

Method 1:
2B+1C+1D=12    .............(i)

2A+1C=13         ...................(ii)

And 1A+1B+1C=14  ..........(iii)

On subtracting (i) and (ii) we get,

2A2B1D=16  ...................(iv)

On subtracting (iii) from (ii) we get 

1A1B=112

Therefore ,1B=1A112   .................(v)

From (ii), 1C=132A  .......................(vi)

Using (i), (v) and (vi), we get 

 1D=122(1A112)(132A)=13

Therefore, 1A+1B+1C+1D=14+13=712

Hence time taken to complete the job =127 days.

Method 2: Using percentage approach (Recommanded)
Given that 
2B+C+D=50 ............(1)
2A+C=33.33.............(2)
A+B+C=25...............(3)

(3)×2(2)
2B+C=16.66
Substituting in (1)
D=33.33%
From (3) and adding D to it we get 
A+B+C+D=68.88%=127 days.

Question 29

How many natural numbers less than 2500 will have odd number of factors?

A. 50
B.

16

C.

15

D. None of these

SOLUTION

Solution : D

Only perfect squares have odd number of factors. The numbers range from 22 to 492 = 48 such cases.

Question 30

The function f(x) = |x - 2| + |2.5 - x| + |3.6 - x|, where x is a real number, attains a minimum at

A.

x = 2.3

B.

x = 2.5

C.

x = 2.7

D.

None of the above

SOLUTION

Solution : B

When a minimum question is asked for a modulus function we should always try to make one of the components zero ( as that is the minimum the modulus values can take). Hence we take x=2, x= 2.5 and x=3.6 and check where the value of y is minimum.

Case 1: If x < 2, then y = 2 - x + 2.5 - x + 3.6 - x = 8.1 - 3x.
This will be least if x is highest i.e. just less than 2.
In this case y will be just more than 2.1

Case 2: If 2  x< 2.5 , then y = x - 2 + 2.5 - x 3.6 - x = 4.1 - x
Again, this will be least if x is the highest case y will be just more than 1.6.

Case 3: If 2.5 x< 3.6 , then y = x - 2 + x - 2.5 + 3.6 - x = x - 0.9
This will be least if x is least i.e. X = 2.5.

Case 4: If In this case y = 1.6 X  3.6 , then 
y = x - 2 + x - 2.5 + x - 3.6 = 3x - 8.1
y = x - 2 + x - 2.5 + x - 3.6 = 3x - 8.1
The minimum value of this will be at x = 3.6 = 27
Hence the minimum value of y is attained at x = 2.5

Shortcut:
Put x=2, f(x) = 2.1
When x=2.5, f(x) = 1.6
When x=3.6, f(x) = 2.7

Therefore, minimum value is attained at x=2.5. 

 

Question 31

P is represented as P=xaya+xa1ya1+.....x1y+x0 where 0<x0..xay1. If x and y are positive integers and y=4, then which of the following could be a representation of an integer with y as a base?

A.

12343

B.

132421

C.

132321

D.

None of these

SOLUTION

Solution : C

When the base is 4, the maximum digit the number can contain is 3, it cannot contain 4.

Question 32

How many five digit number can be formed using the digits 2,3,8,7,5 exactly once such that they are divisible by 125?

___.

SOLUTION

Solution : Any number is divisible by 125, if the number formed by the hundred's place, ten's place and unit's place is also divisible by 125.
In the digits 2,3,8,7,5. There will be two combinations for hundred, tens and units place i.e. _ _ 375 and_ _ 875.
Hence five digits numbers, which can be formed using

2,3,8,7,5 are 23875,32875,82375 and 28375.

 

Question 33

What is the sum of digits of the least multiple of 13, which when divided by 6,8 and 12 leaves 5,7 and 11 as remainder?___

SOLUTION

Solution : 6-5 = 8-7 = 12-11 = 1 (Difference is constant)
Find the LCM of 6, 8, 12 = 24
The smallest number which when divided by 6,8 and 12 leaving 5,7 and 11 as remainder = 24 -1 = 23
The required number will be of the form 23 + 24k (LCM of 6, 8, 24), which is also a multiple of 13.
This means (-3) + (-2)k should give a remainder of 0 when divided by 13.
The first integral value of k which satisfies this condition is k = 5.
Number = 23+120 = 143, Sum of digits = 8.

Question 34

A man has 14 friends out of which 5 are married couples and 4 are single people. He goes out for dinner with 3 of his friends every night. If he invites one member of a married couple, then he also invites the other. He stops going out for dinner when he has gone out with all possible combinations of his friends. How many times does he go out for dinner?
___

SOLUTION

Solution : The man can only go out to dinner with one couple and one single person or with three single people. 

The number of ways in which he can go out for a dinner with one couple (out of 5) and one single person (out of 4) = 5C1 × 4C1 = 20

The number of ways in which he can go out for a dinner with three single people (out of 4) = 4C3 = 4
 

The total number of times he goes out for a dinner = 20 + 4 = 24