Free Modern Maths 03 Practice Test - CAT

Question 1

For a drama, 3 students need to be selected out of 6 students A,B,C,D,E and F. If A is already selected, what is the probability of selecting C also?

A. 0.5

B. 0.2

C. 0.3

D. 0.4

SOLUTION

Solution : D

If A is selected, then two more students need to be selected. This can be done in $^{5}$ C $_{2}$ ways= 10

If C is also selected, only one more out of the remaining 4 needs to be selected.
That can be done in 4 ways.

Thus, Probability = $\frac{4}{10}$ = 0.4

Question 2

If 30% of the visitors to an ice cream shop have chocolate ice-cream, then what is the probability that 2 out of 3 people who entered the store on Saturday at 5 pm will have chocolate ice-cream?

A. 0.16

B. 0.15

C. 0.19

D. 0.28

SOLUTION

Solution : C

Probability of a success= 0.3

Probability of a failure= 0.7

Required Probability = 0.3 $\times$ 0.3 $\times$ 0.7 $\times$ $^{3}$ C $_{1}$ = 0.189

Question 3

A Number is selected at random from first thirty natural numbers. What is the chance that it is a multiple of either 3 or 13?

\frac{17}{30}

\frac{2}{5}

\frac{11}{30}

\frac{4}{15}

SOLUTION

Solution : B

Total sample space is 30

The number of multiples of 3 are 10 and the number of multiples of 13 are 2, out of which none of them are common.
Hence, answer is $\frac{2}{5}$

Question 4

What is the probability of getting at least one six in a single throw of three unbiased dice?

\frac{1}{6}

\frac{125}{216}

\frac{1}{36}

\frac{91}{216}

SOLUTION

Solution : D

The probability of getting no six is ${(\frac{5}{6})}^{3}$ = $\frac{125}{216}$

Hence the probability of getting at least one six is 1 - $\frac{125}{216}$ = $\frac{91}{216}$

Question 5

What is the probability that a two digit number selected at random will be a multiple of '3' and not a multiple of '5'?

\frac{2}{15}

\frac{4}{15}

\frac{1}{15}

\frac{3}{10}

SOLUTION

Solution : B

Sample space is 90 numbers, of which , $\frac{90}{3}$ = 30 are multiples of 3. $\frac{90}{5}$ = 18 are multiples of 5, of which every 1 out of 3 are multiples of both 3 and 5.

Probability that a two-digit number is a multiple of 3 and not a multiple of 5 = 30-6 = $\frac{24}{90}$ = $\frac{4}{15}$

Question 6

When two dice are thrown simultaneously, what is the probability that the sum of the two numbers that turn up is less or equal to 10?

\frac{5}{6}

\frac{11}{12}

\frac{1}{6}

\frac{1}{12}

SOLUTION

Solution : B

First find the probability of sum being greater than 10. Combination whose sum of 12 is (6, 6)

Combinations whose sum of 11 is (5, 6), (6, 5).

Therefore, there are totally 3 occurrences out of 36 occurrences that go against the given condition.

Probability whose sum of two numbers is greater than or equal to 11 = $\frac{3}{36}$ = $\frac{1}{12}$ .

Hence probability whose sum of two numbers is lesser than 11 = 1 – $\frac{1}{12}$ = $\frac{11}{12}$ .

Question 7

4 out of 15 apples are rotten. They are taken out one by one at random and examined. The ones which are examined are not replaced. What is the probability that the 9 $^{t h}$ one examined is the last rotten one?

\frac{^{4} C_{3} \times^{11} C_{5}}{^{15} C_{3}} \times \frac{1}{^{7} C_{1}}

\frac{^{11} C_{5} \times^{4} C_{4}}{^{15} C_{9}}

\frac{^{4} C_{3} \times^{11} C_{5}}{^{15} C_{9}}

D. None of these

SOLUTION

Solution : A

Lets consider upto the 9 $^{t h}$ apple. The first 8 apples should have 3 rotten ones and remaining 5 good ones. This can be chosen in $^{4}$ C $_{3}$ × $^{11}$ C $_{5}$ ways.

Total number of selections of 8 apples out of 15 apples is $^{15}$ C $_{8}$ .
The last apple is the only rotten one left, which can be selected in 1 way.
Total number of ways of selecting that 1 apple from remaining 15 – 8 = 7 apples = $^{7}$ C $_{1}$ ways

Total probability = $\frac{^{4} C_{3} \times^{11} C_{5}}{^{15} C_{3}} \times \frac{1}{^{7} C_{1}}$

Question 8

A furniture shop has six identical steel cabinets of brand A and four identical steel cabinets of brand B. Three customers buy one cabinet each. Then the probability that two or more cabinets of brand A have been sold

\frac{1}{3}

\frac{2}{3}

\frac{1}{2}

\frac{3}{4}

SOLUTION

Solution : C

The total possibilities are 2 $\times$ 2 $\times$ 2 = 8 (each customer has two possibilities).
These are AAA, AAB, ABA, BAA, BBA, BAB, ABB, BBB.
The favourable outcomes are only 4 (AAA, AAB, ABA, BAA).
Thus the probability = $\frac{4}{8}$ = $\frac{1}{2}$ .

Question 9

Arpit and Rita will play one game of Rock, Paper, Scissors. In this game, each will select and show a hand sign for one of the three items. Rock beats Scissors, Scissors beat Paper, and Paper beats Rock. Assuming that both Arpit and Rita have an equal chance of choosing any one of the hand signs, what is the probability that Arpit will win?

\frac{5}{6}

\frac{2}{3}

\frac{1}{2}

\frac{1}{3}

SOLUTION

Solution : D

No matter what sign Arpit throws, there is one sign Rita could throw that would beat it, one that would tie, and one that would lose. Rita is equally likely to throw any one of the three signs. Therefore, the Probability that Arpit will win is $\frac{1}{3}$

Probability that Rita will win = $\frac{1}{3}$

Probability of a tie = $\frac{1}{3}$ .

Probability that Arpit will win = $\frac{1}{3}$

The correct answer is (d).

Question 10

Two dice are thrown simultaneously. The probability of getting an even number on both the dice is?

\frac{1}{2}

\frac{3}{4}

\frac{1}{4}

\frac{1}{5}

SOLUTION

Solution : C

Probability of getting an even number of the first dice and the second dice is= $\frac{3}{6}$ $\times$ $\frac{3}{6}$ = $\frac{1}{4}$ .

Question 11

Parul and Vijay throw 3 dice in a single throw. It is known that Parul throws a total of 16. Find Vijay’s probability of getting a higher value.

\frac{1}{4}

\frac{1}{3}

\frac{1}{52}

\frac{1}{36}

SOLUTION

Solution : C

The maximum total possible = 18

There are two options for Vijay

1) A+B+C=17 Possibilities = 6,6,5

Number of combinations = $\frac{3!}{2!}$ =3

2) A+B+C=18Possibilities = 6,6,6

Number of combinations = 1

Required Probability= $\frac{4}{216}$ = $\frac{1}{52}$ .

Option (c)

Question 12

Arun draws 3 balls at random from a basket which contains 4 red and 5 blue balls. What are the odds in favour of these being all blue?

A. 5 : 42

B. 5 : 37

C. 4 : 43

D. 4 : 50

SOLUTION

Solution : B

Probability of all three balls being blue = $\frac{^{5} C_{3}}{^{9} C_{3}}$ = $\frac{5}{52}$

Odds in favour of being blue = $\frac{5}{(42 - 5)}$ = 5 : 37

Question 13

In a single throw of 2 dice, the odds against drawing 7 are?

A. 1 : 5

B. 2 : 5

C. 1 : 6

D. 5 : 1

SOLUTION

Solution : D

7 can be drawn in the following ways $\to$ 1 + 6, 2 + 5, 3 + 4, 6 + 1, 5 + 2 and 4 + 3.

Total number of ways = 6

Total number of possibilities = 6 $\times$ 6

Probability of not drawing a seven = $\frac{(36 - 6)}{36}$ = $\frac{30}{36}$

Odds against drawing a 7 is = 30 : 6 or 5 : 1.

Option (d)

Question 14

A wooden cask contains some tools i.e. exactly 12 nuts and 24 bolts. ${(\frac{1}{3})}^{r d}$ of each of these are defective. If 2 tools are picked up one after the other without replacement, then what is the probability that both are not defective?

\frac{49}{106}

\frac{23}{135}

\frac{49}{105}

\frac{89}{105}

SOLUTION

Solution : C

Out of the 12 nuts, 4 are defective and out of the 24 bolts, 8 are defective the probability of picking up both non-defective tools = $\frac{24}{36}$ $\times$ $\frac{23}{135}$ = $\frac{49}{105}$

Question 15

Two squares are chosen at random from the small squares drawn on a chessboard. What is the chance that the two squares chosen have exactly one corner in common?

A. 0.052

B. 0.042

C. 0.048

D. 0.075

SOLUTION

Solution : C

Option (c)

If the first square chosen is one of the 4 corner squares, the second square can be chosen in 1 way = 4 $\times$ 1. If the first square chosen is one of the squares on the sides (other than corners) = 24, the second square can be chosen in 2 ways = 24 $\times$ 2

If the first square is any of the middle squares = 36, the second square can be chosen in 4 ways = 36 $\times$ 4. Total number of ways = 4 + 48 + 144 = 196

Number of ways in which 2 random squares can be selected in a chess board = 64 $\times$ 63

Required probability = $\frac{196}{(64 \times 63)}$ = 0.048

Question 16

If the XAT centre of 4 students can be any one of the 7 cities, then calculate the probability that all the 4 students get any one of exactly 2 centres.

\frac{3}{49}

\frac{6}{49}

\frac{7}{48}

\frac{12}{39}

SOLUTION

Solution : B

Total number of ways in which centres can be allocated = 7 $^{4}$ . Two centres can be chosen in $^{7}$ C $_{2}$ ways. The number of ways they can accommodate 4 students = 2 $^{4}$

But this includes the two cases where one centre is having all the four students. So, no. of ways = 2 $^{4}$ – 2 = 14. So, total number of favourable ways = $^{7}$ C $_{2}$ $\times$ 14

So, probability = $^{7}$ C $_{2}$ $\times$ $\frac{14}{7^{4}}$ = $\frac{6}{49}$ .

Hence option (b)

Question 17

6 boys and 3 girls are randomly placed in a row. Determine the probability of no two girls being placed adjacently.

\frac{7}{23}

\frac{7}{24}

\frac{8}{23}

\frac{9}{24}

SOLUTION

Solution : B

The number of ways that no two girls are placed together is the number of ways in which 3 places marked with G are selected out of the SEVEN places.

___B___B___B___B____B____B____

This can be done in $^{7}$ C $_{3}$ ways.

Total no. of ways in which balls can be arranged = $\frac{10!}{(7! \times 3!)}$ = $^{10}$ C $_{3}$ . So, required probability = $\frac{^{7} C_{3}}{^{10} C_{3}}$ = $\frac{7}{24}$ .

Hence option(b)

Question 18

What is the probability that product of two integers chosen at random has the same unit’s digit as the integers themselves?

\frac{3}{10}

\frac{1}{25}

\frac{4}{15}

\frac{7}{15}

SOLUTION

Solution : B

An integer can end with any of the ten’s digits (0, 1, 2 ... 9) out of which if it ends with one of the four (0, 1, 5, 6), the required condition will be satisfied. The probability of an integer ending with 0 or 1 or 5 or 6 is $\frac{4}{10}$ = $\frac{2}{5}$ Now the probability of second integer also ending with the digit that has come in the unit’s place of the first integer is $\frac{1}{10}$

Therefore, the required probability = $(\frac{2}{3})$ $\times$ $(\frac{1}{10})$ = $\frac{1}{25}$

Question 19

A man speaks truth 3 out of 4 times. He throws a die and reports that it is a six. Find the probability that it is actually a six.

\frac{2}{3}

\frac{3}{4}

\frac{1}{8}

\frac{2}{7}

SOLUTION

Solution : C

option (c)

For it to be an actual six two things are required to happen together.

1. He is speaking the truth. Let this be an event A.

2. The die shows 6. Let this be the event B.

We have: P(A) = $\frac{3}{4}$ P(B) = $\frac{1}{6}$

The probability that both events happen together = P(A) $\times$ P(B) = $\frac{3}{4}$ $\times$ $\frac{1}{6}$ = $\frac{1}{8}$ .

Question 20

An experiment succeeds twice as often as it fails. What is the probability that in the next 5 trials there will be four successes?

\frac{80}{232}

\frac{17}{53}

\frac{80}{213}

\frac{80}{243}

SOLUTION

Solution : D

An experiment succeeds twice as often as it fails. So, the probability of success is $\frac{2}{3}$ and the probability of failure is $\frac{1}{3}$ .

In the next 5 trials, the experiment needs to succeed in 4 out of the 5 trials. 4 out of 5 trials can be selected in $^{5}$ C $_{4}$ = 5 ways.

So, required probability = 5 × ${(\frac{2}{3})}^{4}$ × $(\frac{1}{3})$ = $\frac{80}{243}$

Hence option (d)

Free Modern Maths 03 Practice Test - CAT

Question 1

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Question 2

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Question 3

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Question 4

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Question 5

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Question 6

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Question 7

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Question 8

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Question 9

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Question 10

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Question 11

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Question 12

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Question 13

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Question 14

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Question 15

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Question 16

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Question 17

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Question 18

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Question 19

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Question 20

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