Free Quant Practice Test - CAT

Question 1

The remainder when $4^{45} + 4^{44} + 4^{43} + 4^{42} + 4^{41} + . . . . . . . . . . . + 4^{0}$ is divided by 17 is:-

None of these

SOLUTION

Solution : A

$S_{n} = 4^{0} + 4^{1} + 4^{2} + 4^{3} + . . . . . . . . . . . . . . . . . . . + 4^{43} + 4^{44} + 4^{45}$

$S_{n} = (1 + 4 + 16 + 64) + 4^{4} (1 + 4 + 16 + 64) + . . . . . . . . . . . . . . + 4^{40} (1 + 4 + 16 + 64) + 4^{44} + 4^{45}$

All the terms except $4^{44} + 4^{45}$ are divisible by 17 (as 1 + 4 + 16 + 64 = 85 which is a multiple of 17).

Now, $\frac{4^{45} + 4^{44}}{17} = \frac{4^{44} (4 + 1)}{17} = \frac{{(4^{4})}^{11} \times 5}{17} = \frac{{(255 + 1)}^{11} \times 5}{17} = \frac{1^{11} \times 5}{17} = \frac{5}{17} .$

Hence the required remainder is 5.

Question 2

If Farida can travel South and West along a line segment in the following diagram, in how many ways can she go from A to B?

none of these

SOLUTION

Solution : A

(Number of ways to reach at point B from point A, when p to q path is given - (Number of ways to reach point P from A × Number of ways to reach at point B from Q))
( $^{8}$ C $_{3}$ - $^{4}$ C $_{2}$ $\times$ $^{3}$ C $_{1}$ ) = 38. Option (a)

Question 3

ABCD is a square with length 15cm. Pattern of shading one quarter of the square is shown in the diagram. If this pattern is continued indefinitely, what is the area of shaded region?

56.25

SOLUTION

Solution : C

The largest shaded square has area 1/4 of whole square. The next one is 1/4 its size. Therefore quarter its area, so it is $\frac{1}{16}$ of the whole square. Thus the total shaded area , as a fraction of whole square, is
$\frac{1}{4} + \frac{1}{16} + \frac{1}{64} + . . . . . . . . . . . . . . . . . . . . . . . . . .$
$= \frac{\frac{1}{4}}{(1 - \frac{1}{4})}$
$= \frac{\frac{1}{4}}{\frac{3}{4}}$
$= \frac{1}{3}; \frac{1}{3} \times 225 = 75$

Question 4

A GP consists of 500 terms. Sum of the terms occupying the odd places is P1 and the sum of the terms occupying the even places is P2. Find the common ratio?

$\frac{p_{2}}{p_{1}}$

$\frac{p_{1}}{p_{2}}$

$p_{2} + \frac{p_{2}}{p_{1}}$

$p_{2} + \frac{p_{1}}{p_{2}}$

SOLUTION

Solution : A

Consider a simple GP = 2, 4 with two terms
Sum of terms occupying the odd places =2 and sum of terms occupying the even places =4

Thus, common ratio = 2 = $\frac{p_{2}}{p_{1}}$ .

Question 5

Set N consists of all natural numbers from 1 to 10. Three numbers are selected at random from this set. Find the probability that the product of two of the number is equal to the third

$\frac{3}{4}$

$\frac{1}{8}$

$\frac{2}{8}$

$\frac{1}{40}$

SOLUTION

Solution : D

The only set of numbers possible = (2,3,6) (2,4,8) and (2,5,10)
thus required probability = $\frac{3}{^{10} C_{3}}$ = $\frac{1}{40}$

Question 6

A number 777....7 containing 77 sevens is divisible by?

any multiple of 7

odd multiple of 7

by 11

None of these

SOLUTION

Solution : D

777....7 (77 times) is not necessarily divisible by any multiple of 7.
Option (a) is rejected as the given number cannot be divisible by even multiples of 7, since the number is odd.
Option (b) is rejected as the number is not divisible by 21 as the sum of the digits is not divisible by 3.
Option (c) is also rejected as the number is divisible by 11 only if there are even number of 7s.
Thus, it is evident that option (d) is the correct answer choice.

Question 7

A rectangle with perimeter 88 m is cut into 5 congruent rectangles as shown in the diagram. The perimeter of each of the congruent rectangle is

120

none of these

SOLUTION

Solution : B

Here AD = BC, 3a = 2b, a=2y, b=3y
The Perimeter of rectangle =4b +5a =4 x 3y + 5 x 2y = 22y; 22y =88; y =4; a =8 and b =12;
The perimeter of each of the congruent rectangle = 2(a+b) =40; Option (b).

Question 8

The integers 34041 and 32506, when divided by a three-digit integer N, leave the same remainder. What is the value of N?

289

367

453

307

SOLUTION

Solution : D

If 34041 and 32506 are the only two terms in an AP, they will have common difference of 32506 - 34041 = 1535. Thus, all numbers which are factors of this common difference will divide the integers such that they will leave a constant remainder.
Now go from answer options. We are looking for factors of 1535. Among the answer options, only 307 is a factor of 1535. Answer is option (d).

Question 9

A square field has length of 4 meter. On the one corner of the square, a cow is tied with the rope length of 5 meter. The maximum area, cow can graze is ___ $m^{2}$ (Approximately)

SOLUTION

Solution :

Required area = Area of triangle ABF + Area of triangle ADE + Area of sector AEF
This is approximately equal to area of square - area of triangle CFE = 16 - 0.5 = 15.5 meter $^{2}$ .

Question 10

$S = 100^{2} - 99^{2} + 98^{2} - 97^{2} + . . . . . . . . . . . . . . . . . . . . . . . . + 2^{2} - 1^{2};$ what is the value of S?

2020

5050

5024

5000

SOLUTION

Solution : B

$(100^{2} - 99^{2}) + (98^{2} - 97^{2}) + . . . . . . . . . . . . . . . . . . + (2^{2} - 1^{2})$

= 199 + 195 + 191 + 187 + ......................... + 3
$= (199 + 3) \times \frac{50}{2} = 5050.$
Even simpler way of calculation would be:
(100+99)+(98+97)+...+(2+1)
= Sum of first 100 natural numbers $= \frac{100 \times 101}{2} = 5050$

Question 11

The function f(x) satisfies the equation (x- 1)f(x) + f $(\frac{1}{x})$ = 1, for all x not equal to zero. The value of f(99) is

$\frac{9703}{197}$

$\frac{197}{9703}$

SOLUTION

Solution : C

Substituting x = 99 gives 98f(99) + f $(\frac{1}{99})$ =1 and substituting x = $\frac{1}{99}$ gives - $\frac{98}{99}$ f $(\frac{1}{99})$ + f(99) = 1.98
times of the first equation plus 99 times of the second equation simplify to $9703 \times f (99) = 197;$
f(99) = $\frac{197}{9703}$ . Option (c).
Substituting x = 99 gives 98

Question 12

In the given diagram ABCD , What is the value of x $^{\circ}$ ?

30 $^{\circ}$

75 $^{\circ}$

60 $^{\circ}$

None of these

SOLUTION

Solution : A

Given figure, we can divide into 4 kites. Where (s + r) = x $^{\circ}$ and (p + q) = 150 $^{\circ}$ and
2(p+ q) =300 $^{\circ}$ .
From the given figure 2(p +q) + 2(s + r) = 360 $^{\circ}$ ;
2(s+ r) = 60 $^{\circ}$ , so, (s + r) =30 $^{\circ}$ . Option(a)

Question 13

There are 170 people each one of them likes at least one of the fruits, 60 like grapes, 40 like papaya, 50 like watermelon and 70 like pomegranate. If there are exactly 10 people who like all four fruits. Then, find the maximum number of people who like only watermelon: -

SOLUTION

Solution : A

Maximum possible value is 40.
Option(a). The distribution in this case is as follows.

Question 14

For what values of x will the following inequality hold true $3 | a - 1 | + a^{2} - 7 > 0$ ?

$(- \infty, - 5) \cup (2, \infty)$

$(- \infty, - 1) \cup (2, \infty)$

$(- \infty, - 1) \cup (4, \infty)$

$(- \infty, - 5) \cup (4, \infty)$

SOLUTION

Solution : B

Let a = 3. the inequality holds good as 3(2)+9-7 $>$ 0
Options c $&$ d are eliminated.
Let a= -2, the inequality holds good as 3(3)+4-7 $>$ 0
Option a is eliminated
Answer = b

Question 15

Arnab needs to pay Rs.210 for his dinner. He has an unlimited supply of Rs 2, Rs 5 and Rs 10 notes; in how many ways can he pay?

210

253

231

None of these

SOLUTION

Solution : A

Equation become like 2x + 5y + 10z = 210, where
X should be a multiple of 5 and Y should be a multiple of 2. So, we can take X and Y as 5P and 2Q respectively.
Now equation becomes:-
10P + 10Q + 10Z = 210;
P + Q + Z = 21
where P, Q and Z are non-negative integers. Number o f solutions of the given equation is $^{23} C_{2}$ = 253. Option (b).

Question 16

A hare completes a straight 300-meter downhill run in t seconds and at an average speed of (x + 10) meters per second. The hare then cycles up the mountain (the same distance) at an average speed of (x - 8) meters per second. If the ride up the mountain took 135 seconds longer than its run down the mountain, what was the hare's average speed, in meters per second, during its downhill run?

B. 15

SOLUTION

Solution : C

Approach : Reverse gear
Since the difference in time is given, and

$T i m e = \frac{D i s t a n c e}{S p e e d}$

$\frac{300}{x - 8} - \frac{300}{x \pm 10} = 135$
Now going from answer options, we can easily substitute and see which value satisfies this equation. Option (c) is the correct answer choice.

Question 17

Vikrant drives to office from home every day at a steady rate of x kilometers per hour. On a particular day, Vikrant car tire punctured exactly half way to office. He immediately started walking to office at a steady pace of y kilometers per hour. He arrived at office exactly t hours after leaving his home. How many kilometers is it from the office to Vikrant's home?

$\frac{x + y}{t}$

$\frac{2 (x + t)}{x y}$

$\frac{2 x y t}{x + y}$

$\frac{2 (x + y + t)}{x y}$

SOLUTION

Solution : C

Note that it is a variable $\Rightarrow$ variable question. It can be solved in no time at all using assumption.
Let the total distance =12 km. Let x = 6 kmph and y = 6 kmph
Then t = 2

Look where you are getting 2 in the answer options. Answer is option (c)

Question 18

If m times the m^th term of an AP is equal to n times its n^th term, find the (m+n)^th term of that AP?

t_m + t_n

t_m - t_n

t_m * t_n

SOLUTION

Solution : D

option (d) ,0. This is a property of an AP
eg 1, 1/2 0 is one such AP when m=1 and n=2

Question 19

Given that p>1. What is the minimum value of ?

SOLUTION

Solution : D

Question 20

Disha wants to paint her room a unique brown. For that she uses a certain violet paint which contains $30 %$ blue pigment and $70 %$ red pigment by weight. She mixes that violet paint with a certain green paint which contains $50 %$ blue pigment and $50 %$ yellow pigment. When these paints are mixed to produce the brown paint which she uses, it contains $40 %$ blue pigment. If the brown paint weighs 10 grams, then the red pigment contributes how many grams of that weight?

2.8 gms

3.5 gms

4 gms

5.3 gms

SOLUTION

Solution : B

Use allegation

Considering only the blue pigment in all the paints

Thus, 5gms of each paint is used .Weight of red paint = $\frac{7}{10} \times 5 = 3.5$ gms.Option (b)

Question 21

During the festival season, A shopkeeper marks up the rate of his silver products by $40 %$ . If he increases the discount from $5 %$ to $10 %$ , then the profit will decrease by Rs 14. How much profit will he get if he gives a discount of $20 %$ on the marked price?

None of these

SOLUTION

Solution : C

Answer = Option c
Assume, CP =100
MP = 100 $\times$ 1.4=140
Initial discount = 5 $%$ of 140 = Rs 7
New discount = 10 $%$ of 140 = Rs 14
Profit decreases by Rs 7
When profit decrease by Rs 7, CP = Rs 100
When profit decreases by Rs 14, CP = Rs 200
Actual, MP = 1.4 $\times$ 200 = 280
Discount of 20 $%$ on MP implies SP = 0.8 $\times$ 280 =224
Profit = 224 - 200 = Rs.24

Question 22

Answer these questions based on the following data.
In a standard 5 class, the students are arranged according to the increasing order of their marks in an examination. The marks obtained by every student, starting from the third student, are the sum of the marks obtained by the earlier two students.

Marks scored by students are in A.P. If the $12^{t h}$ student got 809 marks and the $2^{n d}$ student got 6 marks, then what is the sum of the marks obtained by the first ten students?

815

803

800

Cannot be determined

SOLUTION

Solution : B

By observation, we get
$T_{12} = S_{(n - 2)} + t_{2}$
It is given that
$T_{12}$ = 809 and $t_{2}$ =6
$T_{12} = S_{(12 - 2)} + t_{2}$
$\Rightarrow t_{12} = S_{10} + t_{2}$
$\Rightarrow S_{10} = 809 - 6 = 803$

Question 23

Answer these questions based on the following data.
In a standard 5 class, the students are arranged according to the increasing order of their marks in an examination. The marks obtained by every student, starting from the third student, are the sum of the marks obtained by the earlier two students.

If the $3^{r d}$ student got 10 marks and $5^{t h}$ student got 26 marks, then how many marks does the $12^{t h}$ student get?

754

466

1220

Cannot be determined

SOLUTION

Solution : A

If the $3^{r d}$ student got 10 marks and the $5^{t h}$ student got 26 marks, then the 4” student got
(26-10) = 16 marks
The $2^{n d}$ student got (16-10) =6
The marks obtained by the students are
4, 6, 10, 16, 26, 42, 68, 110, 178, 288, 466, 754
Hence , the $12^{t h}$ student got 754 marks.

Question 24

You take a wrong turn on the way to Port Blair and end up in Forks, where 99% of the inhabitants are vampires and the rest are regular humans. For obvious reasons, you want to be able to figure out who's who. On an average, nine-tenths of the vampires are correctly identified as vampires and nine-tenths of humans are correct identified as humans. What is the probability that someone identified as a human is actually a human?

$\frac{9}{1000}$

$\frac{1}{12}$

$\frac{1}{26}$

$\frac{1}{9}$

SOLUTION

Solution : B

The probability is taken as : (the number of humans who were correctly identified as humans)/(the number of vampires incorrectly identified as humans)+(the number of humans who were correctly identified as humans)
$= \frac{0.01 \times 0.9}{0.99 \times 0.1 + 0.01 \times 0.9}$
$= \frac{1}{12}$

Question 25

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

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 = 9 $π^{2}$ - 4 .(approximately = 24.26). Hence the correct answer is option (b).

Question 26

How many pairs of positive integers (a, b) are there such that a and b have no common factor greater than 1 and $\frac{a}{b} + \frac{14 b}{9 a}$ is an integer?

SOLUTION

Solution : D

Let $u = \frac{a}{b} .$ Then the problem is equivalent to finding all possible rational numbers 'u' such that
$u + \frac{14}{9 u} = K$ , for some integer K.
This equation is equivalent to $9 u^{2} - 9 u k + 14 = 0$
$9 u^{2} - 9 u k + 14 = 0$ whose solutions are,
$u = \frac{9 k \pm \sqrt{81 k^{2} - 504}}{18}$
$= \frac{K}{2} \pm \frac{1}{6} \sqrt{9 k^{2} - 56}$
Hence u is rational if and onlly if
$\sqrt{9 k^{2} - 56}$ is rational, Which is true if and only if $9 k^{2} - 56$ is a perfect square. Suppose that $9 k^{2} - 56 = S^{2}$ 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 $u = \frac{1}{3}$ or $u = \frac{14}{3}$
If k=3 then $u = \frac{2}{3}$ of $u = \frac{7}{3} .$
Therefore there are four pairs (a,b) that satisfy the given conditions, namely (1,3), (2,3), (7,3) and (14,3)

Question 27

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?

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)
$\Rightarrow$ (6 - a)(6 - b) (6 - c) (6 - d) (6 - e) =(1) (-1) (3) (-3) 5
The 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 28

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.

-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 29

There is an Arithmetic Progression with 10 terms. The Sum of numbers at even places is 15 and Sum of numbers at odd places is $\frac{25}{2}$ . Find the common difference.

$\frac{1}{2}$

$\frac{1}{4}$

$\frac{2}{3}$

SOLUTION

Solution : B

Let a1 be the first term.. then the second term can be written as $a_{1} + d$ . Similarly we write all even in terms of odd.
$a_{1} + a_{3} + a_{5} + a_{7} + a_{9} = \frac{25}{2} a n d a_{2} + a_{4} + a_{6} + a_{8} + a_{10} = 15$

$5 a + 20 d = \frac{25}{2}$
$5 a + 25 d = 15$
$d = 0.5$

Question 30

What is the probability of forming the word "ADDS” by picking 4 cards at random and kept in the same order out of 50 cards given with 10 cards having "A”, 10 having "S” and the remaining 30 having "D” on it?

$\frac{2^{2}}{5^{4}}$

$\frac{15}{(48 - 1) (48 + 1) \times 4}$

$\frac{1}{50^{2}}$

None of these

SOLUTION

Solution : D

1) Probability of Getting an $A = \frac{10}{20} = \frac{1}{5}$
2) Probability of getting a $D = \frac{30}{49}$
3) Probability of getting a $D = \frac{29}{48}$
4) Probability of getting a $S = \frac{10}{47}$
Total probability= $\frac{1}{5} \times \frac{30}{49} \times \frac{29}{48} \times \frac{10}{47}$

Question 31

All the 6 sides of a cube are colored. Now this cube is broken into smaller cubes, with each side ${\frac{1}{4}}^{t h}$ the size of the side of the larger cube. How many of the smaller cubes will have all three sides colored?

Cannot be determined

SOLUTION

Solution : B

The number of smaller cubes with 3 sides coloured will be 8 (the 8 corner cubes).

Question 32

P, Q, R, S, and T run a 100 metre race. P beats Q by 20 m, R by 25 m, S by 40 m, and T by 50 m. Q, R, S, and T now run another 100 metre race. Assume that they run at exactly the same speeds as before. What is the approximate total of the distances by which Q beats R, S, and T? (Write your answer to the nearest possible natural number) Metre
___

SOLUTION

Solution : From the first race, we know that, when P has run 100 m, Q, R, S, and T have run 80 m, 75 m, 60 m, and 50 m respectively.

In the second race, Q completes 100 m. Hence R will complete $75 \times (\frac{100}{80})$ = 93.75 m, and will be beaten by 6.25 m.

Similarly, S will run $60 \times (\frac{100}{80})$ = 75 m, and will be beaten by 25 m.

T will run $50 \times (\frac{100}{80})$ = 62.5 m, and will be beaten by 37.5 m.

The total of the three distances is (6.25 + 25 + 37.5) = 68.75 m, or approximately 69 m.
Alternatively,
In the first race, when Q has run 80 m, R, S and T have run 75 m, 60 m and 50 m respectively. So, the sum of the distances by which Q is ahead of R, S, and T is (5 + 20 + 30) = 55 m
In the second race, Q runs 100 m. So the sum of distances by which Q beats R, S, and T is $55 \times (\frac{100}{80})$ = 68.75 m, or approximately 69 m.

Question 33

If Y is a prime number greater than 10000 and it leaves a remainder of 2 when divided by 3, then what remainder does Y leave on division by 6?
___

SOLUTION

Solution : The number is of the form 3y+ 2 and as it is a prime number, it is odd.
$∴$ y is odd and is of the form 2k+1.
$∴$ 3y+ 2 = 3(2k+ 1) + 2 = 6k+5

$∴$ The number leaves a remainder of 5 when divided by 6.

Question 34

Ajay traveled along a path which was straight up to 10 m (from point P to point T) and then went along an arc of a circle (from point T to point B) to reach his destination (B). Path PT is tangent to the circle at point T. If Ajay had travelled along the straight line joining his starting point P and his destination point B, he would have traveled 20 m and would have passed through the centre of the circle to which the arc belongs. The radius of the circle is(to which arc TB belongs) ___ m.

SOLUTION

Solution :
Let r be the radius of the circle of which the arc is a part.

$∵$ PT $^{2}$ = PA × PB

$∴$ 100 = (20 - 2r) × 20

$∴$ 40r = 400 − 100

$∴$ r = 7.5 m

Free Quant Practice Test - CAT

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Question 33

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