Free Quant Practice Test - CAT

Question 1

Digit sum (DG) is defined as the sum of the digits of a number till you get a single digit number. Eg) DG(40)=4+0=4. DG(345)=DG(12)=DG(3)=3. How many positive integers (n) are there between 500 and 1500 such that DG(N)=7?

110

111

SOLUTION

Solution : D

Approach 1: Logic
The first number which satisfies this is 502. The next number will be 511 and so on. The terms will fall in an Arithmetic Progression with common difference = 9. Last term = 1492.
Number of terms $= \frac{990}{9} + 1 = 111.$

Approach 2: Shortcut in a Shortcut!
Digit sum of numbers will fall in an Arithmetic Progression with a common difference of 9. Thus, there can be $\frac{1500 - 500}{9} = 111$ or 112 such numbers.
Since only 111 is there in the answer options, you can mark it directly.

Question 2

If |x-1| - |x| + |2x+3| $>$ 2x + 4 then ?

$x < \frac{- 3}{2}$

$(- \infty, - \frac{3}{2}) \cup (1, \infty)$

$- \frac{3}{2} < x < 0, x \geq 1$

$(- \infty, - \frac{3}{2}) \cup (0, 1)$

SOLUTION

Solution : A

Again use the technique of going from answer options. We need to find the range of x for which LHS $>$ RHS
Choose a value for x which is there is 2 answer options and not there in 2 answer options.
Take x = 2
LHS = |x-1| - |x| + |2x+3| = 1-2+7 = 6
RHS = 2x + 4 = 8
LHS $<$ RHS.
Hence, option (b) and (c) can be ruled out.
Next put x = $\frac{1}{2}$ .
LHS = $\frac{1}{2} - \frac{1}{2} + 4 = 4$
RHS = 1 + 4 = 5
LHS $<$ RHS. Hence, option (d) is ruled out. Answer is option (a)

Question 3

Two friends A and B are stationed at two different points P and Q respectively along the same bank of a straight stretch of a flowing stream. The ratio of the speeds of A and B in still water is 1: 3. When A and B swim in the opposite directions from P and Q respectively, they meet at a point 30 meters from Q. If the time taken by A to reach point Q and B to reach point P from their original positions P and Q is the same, then the distance (in meters) between P and Q is

120

SOLUTION

Solution : D

Let the speeds of A and B be x m/s and 3x m/s respectively and let the speed of the stream br y m/s.
According to the question:

$\frac{P Q - 30}{x + y} = \frac{30}{3 x - y} . . . . . . (1)$
And,
$\frac{P Q}{x + y} = \frac{P Q}{3 x - y} . . . . . . . . . (2)$
(If we assume the water flows in the direction Q to P,then 3x+y=x-y, which is not possible.)
From (2),we get:

3x-y=x+y
$\Rightarrow$ x=y
$\Rightarrow$ Substituting in (1),we get : PQ=60 m.

Shortcut Method :
As time taken by A and B is the same to cover equal distances in opposite direction,means effective speeds of both the friends are the same . Hence, distances between P and Q =30+30 = 60 metres.

Question 4

Let $f (x) = a x^{2} + b x + c$ where a,b, and c are constants. If f(x) takes it maximum value at $x = \frac{1}{3},$ then which of the following is necessarily true?

$f (0) = f (\frac{2}{3})$

$f (0) = f (- \frac{2}{3})$

$f (0) = - f (\frac{2}{3})$

$f (0) = - f (\frac{- 2}{3})$

SOLUTION

Solution : A

$f (x) = a x^{2} + b x + c$
Since maxima of f(x) occurs at $x = \frac{1}{3},$ f(x) can be rewritten as $f (x) = a {(x - \frac{1}{3})}^{2} + k$ (where $f (\frac{1}{3}) = k$ )
It is obvious that a $<$ 0.
Assuming f(n)=f(0) where n is any constant.
$f (0) = 9 (\frac{1}{9}) + k = f (n) = a {(n - \frac{1}{3})}^{2} + k$
$\Rightarrow {(n - \frac{1}{3})}^{2} = \frac{1}{9}$
$\Rightarrow n = \frac{2}{3}$
Therefore,f(0)= $f (\frac{2}{3})$

Alternative Method:
The maximum value of $f (x) = a x^{2} + b x + c$ occurs at $x = - \frac{b}{2 a} .$
$- \frac{b}{2 a} = \frac{1}{3} \Rightarrow b = \frac{- 2}{3 a}$
Also,f(0)=c. Substituting $b = \frac{- 2}{3 a}$ in f(x),we can easily observe that $f (0) = f (\frac{2}{3}) .$

Question 5

The figure below shows the plan of a town. A person "X" rides his bicycle from house at A to office at B, taking the shortest route.
The number of shortest paths that he can choose is:

168

336

208

308

SOLUTION

Solution : B

Number of ways "X” can go from A to C= $^{8}$ C $_{3}$ and from C to B= $^{4}$ C $_{2}$ . Then the total number of shortest paths that he can choose is $^{8}$ C $_{3}$ $\times$ $^{4}$ C $_{2}$ = 336

Question 6

Points (1, 2), (2, 4), (7, 5) and (6, x) are the vertices of a parallelogram. What is the value of x?

SOLUTION

Solution : C

On plotting these points on the X-Y plane, we get:

Let m $_{1}$ be the slope of the line joining the points (1, 2) and (2, 4).
So, m $_{1}$ = $\frac{4 - 2}{2 - 1}$ = $\frac{2}{1}$
As slope of two parallel lines are equal, slope m $_{2}$ of the lines joining the points (6, x) and (7, 5) should also be $\frac{3}{2}$ .
m $_{2} = \frac{5 - x}{7 - 6} = \frac{2}{1} \Rightarrow$ x = 3

Shortcut Method:

We know that in a paralellogram, the diagonals bisect each other. Therefore the point of intersection shall be the midpoint of both the diagonals.
$\Rightarrow$ Coordinates of point of intersection = $(\frac{1 + 7}{2}, \frac{2 + 5}{2}) = (4, \frac{7}{2})$
$\Rightarrow$ $\frac{x + 4}{2} = \frac{7}{2}$
$\Rightarrow$ x = 3

Question 7

In a Degree college there are 3 branches and it is mandatory to opt for at least one branch. 70% students opted for Science, 60% students opted for commerce and 92% students opted for Arts. A student can opt more than one branch. What is the maximum and minimum number of students that can opt all three streams?

61% and 20%

60% and 22%

61% and 22%

60% and 20%

SOLUTION

Solution : C

Here, S= 222% and X = 100%, S - X = 122% = II + 2 III, Maximum Possible value of all three= $\frac{122}{2}$ = 61%
For minimum, III = 100 - {(100 - 70) + (100 - 60) + (100 - 92)} = 22%

Question 8

A sum of Rs. 12000 is borrowed at $12 %$ p.a. compounded annually which is paid back in 3 equal annual installments. What is the amount of each installment?

4997

5990

4599

3896

SOLUTION

Solution : A

$1^{s t} m e t h o d : 12000 = x \times [\frac{25}{28} + (\frac{25}{28})^{2} + (\frac{25}{28})^{3}]$

$= x \times \frac{25}{28} [1 + \frac{25}{28} + \frac{625}{784}]$

$= \frac{25 x}{28} [\frac{2109}{784}]$
x=4996.1878

$2^{n d} m e t h o d :$
$12000 (1.12)^{3} = x [1 + (1.12) + (1.12)^{2}]$
x= 4996.187

Question 9

What is the number of digits in $5^{40}$ ? (Given $l o g_{10} 2$ = 0.3010)

SOLUTION

Solution : A

$l o g 5^{40} = 40 l o g 5 = 40 \times [l o g (\frac{10}{2})] = 40 (l o g 10 - l o g 2)$
$= 40 (1 - 0.3010) = 40 \times 0.6990 = 27.96.$
Characteristic = 27. Hence, the number of digits in $5^{40}$ is 28.

Question 10

What are the last two digits of $2^{253}$ ?

SOLUTION

Solution : B

$2^{253} = (2^{10})^{25} \times 2^{3} = 24 \times 8 = 92$
Note - $(2^{10})^{o d d n u m b e r}$ always ends in 24.

Question 11

If n! and (6n)! end with 22 zeros and 144 zeros respectively, then which of the following is a divisor of n?

SOLUTION

Solution : A

Possible values of n = 95, 96, 97, 98 and 99. Possible values of 6n = 570, 576, 582, 588 and 594. Only 588! ends with 144 zeros. So, n = 98. Only 14 is a divisor of 98.

Question 12

Nupur and Sheetal play a game "eight time coin throw”. If the fair coin shows heads, Nupur gets a point & if it shows tails, Sheetal gets a point. After eight throws, they compare their scores. What is the probability that the game will be a tie?

$\frac{1}{8}$

$\frac{1}{9}$

$\frac{35}{128}$

$\frac{7}{128}$

$\frac{33}{128}$

SOLUTION

Solution : C

For a tie, there would be 4 heads and 4 tails.
Probability of "4” heads in 8 throws can be found using $^{n}$ C $_{r}$ p $^{r}$ q $^{n - r}$ = $^{8}$ C $_{4}$ ${(\frac{1}{2})}^{4}$ ${(\frac{1}{2})}^{4}$ = $\frac{35}{128}$

Question 13

In how many ways can the number $(9!)^{2}$ be written as a sum of two or more consecutive positive integers?

SOLUTION

Solution : B

No. of ways of writing a number as a sum of two or more consecutive positive integers = No. of odd factors of the number - 1

$(9!)^{2} = 2^{14} \times 3^{8} \times 5^{2} \times 7^{2}$

(No. of odd factors - 1) = 81 - 1 = 80.

Question 14

S={1,2,3,4,5,6,7,8,9}. How many distinct two-digit numbers can be formed using numbers from the set S such that when it is divided by 13, it gives a remainder of 1.
(Repetition is not allowed i.e. two-digit number should be formed using distinct numbers from the set)

SOLUTION

Solution : C

Only 14, 27, 53, 79 and 92 are of the form of 13K + 1.

Question 15

In an exciting finish to a T - 20 cricket match between MI and RCB, the MIs require 10 runs in the last 3 balls to win. If any one of the scores 0, 1, 2, 3, 4, 6 can be made from a ball and no wides or no-balls are bowled then in how many different sequences can batsmen make exactly 10?

SOLUTION

Solution : B

A + B + C = 10, Where 0 ≤ A, B, C ≤ 6
Number of solution of given equation is ( $^{13} C_{2} - 3 \times^{5} C_{2}$ ). But A, B and C are not equal to 5. Number of cases which are not possible (5, 5, 0), (1, 4, 5) and (2, 3, 5). Total number of cases which are not possible = 3 + 6 + 6= 15.
Total number of different sequences = $^{13} C_{2} - 3 \times^{5} C_{2}$ - 15 = 21

Question 16

What is the ratio of shaded area to the area of hexagon?

$\frac{1}{10}$

$\frac{1}{14}$

$\frac{1}{18}$

$\frac{1}{21}$

SOLUTION

Solution : C

From the given figure, we can easily determine ratio = $\frac{3}{2 (7 + 9 + 11)} = \frac{1}{18}$

Question 17

Find the sum of the series $\frac{1}{4 \times 7} + \frac{1}{7 \times 10} + \frac{1}{10 \times 13} + . . . . + \frac{1}{25 \times 28} .$

$\frac{6}{28}$

$\frac{1}{14}$

$\frac{3}{28}$

$\frac{1}{28}$

SOLUTION

Solution : B

We can rewrite the given series in the form : $(\frac{1}{3}) [(\frac{1}{4} - \frac{1}{7}) + (\frac{1}{7} - \frac{1}{10}) + (\frac{1}{10} - \frac{1}{13}) + . . . . . . . . + (\frac{1}{25} - \frac{1}{28})]$

Therefore, sum of the series, $S = (\frac{1}{3}) (\frac{1}{4} - \frac{1}{28}) = (\frac{1}{3}) \times (\frac{6}{28}) = \frac{1}{14} .$

Question 18

A and B solved a quadratic equation. In solving it , A made a mistake in the constant term and obtained the roots as 5, -3 while B made a mistake in the coefficient of x and obtained the roots as 1, -3. The correct roots of the equation are

1,3

-1,3

-1,-3

1,-1

SOLUTION

Solution : B

Using the reverse gear approach
If the equation is $a x^{2} + b x + c = 0.$
A's mistake will be in "c”, hence sum of roots = $- \frac{b}{a} = 2..... (i)$ will be the correct value. Product of roots = $\frac{c}{a}$ will be wrong in this case
B's mistake will be in "b”, hence product of roots = $\frac{c}{a} = - 3...... (i i)$ will yield the correct value. In this case, sum of roots value = $- \frac{b}{a}$ will be wrong

Look in the answer options for the pair of roots which satisfy sum of roots = 2 and product of roots = -3. Answer is option (b).

Question 19

Given below are two overlapping unit circles. Find the area of the shaded region.

$\frac{2 π - 3 \sqrt{3}}{3}$

$\frac{2 π + 3 \sqrt{3}}{3}$

$\frac{2 π + \sqrt{3}}{3}$

$\frac{2 π - \sqrt{3}}{3}$

SOLUTION

Solution : A

Approach 1: With the use of given figure. We can find out Option(a) is right answer.

Shortcut: From the 4 - $π$ approach, each of the shaded region will be 10% of the square's area. Hence, the required area = 40% of 1= 0.4. Look in the answer options for this value. Only option(a) comes the closest. All answer options are far apart.

Question 20

If the diameter of the circle circumscribing the hexagon is 6 cm, then find the area of the shaded region

$\frac{6 \sqrt{3}}{2}$

$\frac{9 \sqrt{3}}{2}$

SOLUTION

Solution : B

The answer can be easily obtained usinggraphical division. The hexagon is made of 6 equal triangles, of which we are looking for the area of 2 triangles as shown below

The side of each triangle is 3 cm. Thus the area of 2 such triangles is 2 $\times$ $\frac{\sqrt{3}}{4}$ $\times$ a $^{2}$ = $\frac{9 \sqrt{3}}{2}$

Question 21

Person A wrote first N consecutive natural numbers on a board and found their sum. Another Person B came and deleted the smallest number and found the sum of remaining numbers. The process continued until just N remained on the board. The average of all the sums is found to out to be 105. Find N.

SOLUTION

Solution : A

The problem can be expressed as follows:
$1 + 2 + 3 + 4 + 5................ n 2 + 3 + 4 + 5................ n 3 + 4 + 5................ n$
This is nothing but $= \frac{\in n^{2}}{n}$ .
It is given that: $\frac{n (n + 1) (2 n + 1)}{6 n} = \frac{(n + 1) (2 n + 1)}{6} = 105$
Solving we get:
2 $n^{2}$ +3n-629=0 or (2n+37)(n-17)=0 or n=17.

Question 22

A table is set up which can seat 5 people from 4 adults and 6 children. Find the probability that the table seats atleast 2 children?

$\frac{31}{42}$

$\frac{124}{126}$

$\frac{41}{42}$

$\frac{3}{26}$

SOLUTION

Solution : C

Probability of seating at least 2 children= 1- (Probability that the table seats none or exactly one child)
5 people can be selected out of 10 in $^{10}$ C $_{5}$ ways= 252
Number of ways of seating a table of 5 with 0 children= 0 (only 4 adults)
Number of ways of seating a table with exactly one child= $^{6}$ C $_{1}$ =6 ways
Required probability = 1- $(\frac{6}{252})$ = $\frac{41}{42}$

Question 23

Find the least number which when divided by 9,10 and 11 give remainders of 2,3 and 5 respectively.

1037

897

1541

none of these

SOLUTION

Solution : D

Use Chinese remainder theorem
Let the number be N. N is of the form 9A+2 = 10B+3=11C+5
We will first find the number which is of the form 9A+2=10B+3.-----(1)
For that, reduce (1) to 9A=10B+1.
Find the first integer value of B which satisfies the equation such that A is also an integer.
Here for B=8, A=9, which are the first integer solutions.
Substituting this in equation (1) we get the value 83, the first number which when divided by 9 gives a remainder 2 and which when divided by 10 gives a remainder 3.
The next number which will give a remainder 2 when divided by 9 and a remainder 3 when divided by 10 will be 83+ (9*10) =90
The numbers will fall in an AP with the first digit as 83 and common difference 90 (9*10). The general form of the AP is 83+90k when k=0,1,2...
Now to include the $3^{r d}$ condition( i.e divisor 11 and remainder 5) we can write
83+90k=11C+5 -------(2)
reduce it to
90k+ 78 = 11C
to make it easier cast out 11's from LHS
i.e 2K + 1= 11D
To find the integer solutions
Find the first integer value of K which satisfies the equation such that D is also an integer.
Here for K = 5, D = 1, which are the first integer solutions. Substituting this in equation (2) we get the value 90 (5) + 83= 533, the first number which when divided by 9 gives a remainder 2, which when divided by 10 gives a remainder 3 and which when divided by 11 gives a remainder 5.

Question 24

Four goats are tethered at four corners of a square plot of side 2 metres (m) so that the adjacent goats can just reach one another. There is a small circular pond of area 0.5 m² at the centre. Find the ungrazed area.

0.2m $^{2}$

0.36m $^{2}$

0.5m $^{2}$

1m $^{2}$

SOLUTION

Solution : B

If the 4 goats are tethered at the corners of the square, the ungrazed area can be shown in the shaded region below.

The above diagram excluding the area of the pond can also be shown as, where the ratio between the area of the square & the circle will be 4: $π$ . The ungrazed area = 4 - $π$ - 0.5 = 0.36 m $^{2}$ . Answer is option (b)

Question 25

Bangalore's Metro urban planners expect the city's population to increase by $10 %$ per year over the next two years. If that projection were to come true, the population two years from now would be exactly double the population of one year ago. Which of the following is closest to the percent population increase in Bangalore over the last year?

SOLUTION

Solution : D

If we assume that today's population is 100, next year it would be 1.1 $\times$ 100 = 110, and the following year it would be 1.1 $\times$ 110 = 121. If this is double the population of one year ago, the population at that time must have been 0.5 $\times$ 121 = 60.5. Because the problem seeks the "closest” answer choice, we can round 60.5 to 60.
In this scenario, the population has increased from 60 to 100 over the last year, a net increase of 40 residents. To determine the percentage increase over the last year, divide the net increase by the initial population: , or roughly $67 % .$

Question 26

A beats B by 150 m in a race of 1000 m. B beats C by 100 m in a 1200m race. Approx by how many meters can A beat C in a race of 500 m?

110 m

150m

90 m

130 m

SOLUTION

Solution : A

We can solve this question using ratios as follows:

Question 27

If xxyy is a perfect square then x + y =?

SOLUTION

Solution : C

xxyy = xx00 + yy = 11(x00 + y) = 11(x0y)
xxyy is a perfect square, so if one divisor is 11, then there should be one more 11 to get $11^{2}$
As xxyy = 11(x0y) .x0y should be divisible by 11, so x + y has to 11.

Question 28

In a public school there is an increase in number of admissions every year. The number of students in 2000 was 780 and it increased to 3000 in the year 2005. Find the annual increase?

$25 %$

$31 %$

$39 %$

$20 %$

SOLUTION

Solution : B

We need to use Compound Annual Growth Rate (CAGR) for this question :
Conventionally, CAGR is computed using the formula $[(\frac{F i n a l}{I n i t i a l})^{\frac{1}{n}} - 1]$
Using the reverse gear approach ,we can obtain the answer much faster
$\frac{F i n a l}{I n i t i a l} = \frac{3000}{780} \approx 3.8$

We need to look for something close to 3.8 in the answer options
here n = 5.
Choose a middle answer option,assuming the answer is $30 %$

$(1.3)^{5} = (1.3)^{2} \times (1.3)^{3} = 17 \times 17 \times 1.3 = 3.75.$
Hence, the answer will be slightly higher than this = $31 %$
Option (b)

Question 29

If 2a² - 3a + 15 < 20, then the solution set of 'a' for the inequality is

SOLUTION

Solution : B

Reverse Gear Approach
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 which is there in option (b) and not in option (a). Put a =0
LHS<RHS. Hence, option (a) can also be eliminated. Answer is option (b)

Question 30

I usually take 72 minutes to reach college. On one fateful day, after traveling ${\frac{1}{3}}^{r d}$ of the distance, I realized that I left my ID card at home and returned to pick it up and then went back to college. To reach on time, I increased my speed by 30 kmph from the point I turned back home till the time I reached college. Assuming that no time was lost at home, what is my usual speed?

20 kmph

30 kmph

10 kmph

15 kmph

SOLUTION

Solution : B

Go from answer options .
Take a middle answer option.
Normal speed = 30kmph
Then distance between college and home = 30 $\times \frac{72}{60}$ =36 km
Also, since time =72 minutes, Speed = 30 kmph.
Distance travelled before leaving back $= \frac{1}{3}$ of 36 =12 km and 24 minutes.
Therefore, time wasted = 24 minutes. Time left to reach = 72 - 24 =48 minutes.
New Speed = 30+30=60 kmph
Distance ot be traveled = 48 km
Time taken $= \frac{48}{60} \times 60$ =48 minutes. Hence answer = option b.

Question 31

In the figure shown below the circles with centers O and R each have a radius of 2. If PQ=1, then the perimeter of rectangle KLMN is ___.

SOLUTION

Solution :
Since PQ=1 and the radius r of each circle is 2, then OP= QR= 1 and OR= 3.

Then KL=2+3+2=7. Sincer= 2,LM= $2 \times$ 2 = 4.
The perimeter of KLMN is P= $2 \times 7 + 2 \times 4$ = 22.

Question 32

Find the last digit of $7^{2009}$ .___

SOLUTION

Solution : This question is based on the power cycle of 7.
2009 is of the form 4k+1.
The power cycle of 7 is 7,9,3,1. Thus, 4k+1 is 7 itself. Answer is option (a).

Question 33

An integer N, $100 < N < 200$ , when expressed in base 5 notation has a final (i.e. rightmost) digit of 0. When N is expressed in base 8 notation and base 11 notation, the leading (i.e. leftmost) digit is 1 in both cases. Then N when divided by 7 gives a remainder of:
___

SOLUTION

Solution :
Expressing in base 5 has final digit 0. So, it has to be a multiple of 5.
Expressing in base 8 notation has 1 as leftmost digit. So, it has to be (1xy) in base 8.
So, the number varies from 100 to 127 (177 in base 8 = 127 in base 10).
Expressing in base 11 notation, 1 is the left most digit. Again, it has to be (1ab) in base 11.
So, the integer is greater than 121 (100 in base 11 = 121 in base 10).
Combining three conditions, we get the number as 125.
$∴ (\frac{125}{7})$ , remainder = 6.

Question 34

Find the second right most non-zero digit in the number $N = 2470^{1613}$ .___

SOLUTION

Solution :
Approach 1: Conventional
$2470^{1613} = 247^{1613} \times 10^{1613} .$
The second right most non-zero digit in the number would be the digit at the ten's place of the number $247^{1613}$ .
$\frac{247^{1613}}{100} = \frac{247^{1613}}{4 \times 25}$
The remainder when $247^{1613}$ is divided by 4 is 3.
So, $247^{1613} = 4 k + 3 .$
Now we need to find the remainder when $247^{1613}$ is divided by 25.
$\frac{247^{1613}}{25} = \frac{((250 - 3)^{1613})}{25}$
$R = \frac{(- 3^{1613})}{25} = \frac{((- 3^{3}) \times ((3^{10})^{161}))}{25} = (\frac{- 27 \times (59049^{161})}{25}) = (\frac{(- 2) \times (- 1)^{161}}{25})$
Therefore the remainder is 2. So, $247^{1613} = 25 n + 2$ .
4k+3 =25n+2.
For k to be an integer, possible values of n are 1, 5, 9, 13...
Corresponding values of k are 6, 31, 56, 81...
Therefore, values of k are of the form: 25m -19.
$247^{1613} = 4 (25 m - 19) + 3 = 100 m - 73.$
Therefore, the last two digits is 100-73=27.
So, the second right most non-zero digit in the number $N = 2470^{1613}$ is 2.

Approach 2: Using the techique of last two digits
We need to consider only the last two digits.
The number can be written as:
$(47)^{1613} = (47^{4})^{403} \times 47 = (47^{2} \times 47^{2})^{403} \times 47$
$= (09 \times 09)^{403} \times 47 = (81)^{403} \times 47 = 41 \times 47 = . . . .27$

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