Modern Maths(Exams > Cat > Quantitaitve Aptitude ) Questions and answers for exam Preparation

Exams > Cat > Quantitaitve Aptitude

MODERN MATHS MCQs

Total Questions : 95 | Page 4 of 10 pages

Question 31. 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?

49106
23135
49105
89105

Discuss Question

Answer: Option C. -> 49105
:
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 32. 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

Discuss Question

Answer: Option C. -> 12
:
C
The total possibilities are 2

\times

\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 33. In a single throw of 2 dice, the odds against drawing 7 are?

1 : 5
2 : 5
1 : 6
5 : 1

Discuss Question

Answer: Option D. -> 5 : 1
:
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 34. What is the probability that product of two integers chosen at random has the same unit’s digit as the integers themselves?

Discuss Question

Answer: Option B. -> 125
:
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 35. A bag has 3 red balls, 2 yellow balls and 3 black balls. They are drawn one by one and placed in a row. Find the number of ways they can be arranged.

Discuss Question

Answer: Option C. -> 560
:
C
Soln:
There are a total of 8 balls with 3 of one kind, 2 of other kind and rest 3 of the same kind. These can be arranged in

\frac{8!}{(3! \times 2! \times 3!)} = 560

Question 36. At a board meeting, each person shook hands with everyone present. Totally there were 45 handshakes. Mid- way through the meeting, the 3 foreign delegates left as they had a flight to catch. The number of women is now 1 more than the number of men. When the meeting concluded, each person shook hands with only the person of the same gender. What were the total number of handshakes in the meeting?

Discuss Question

Answer: Option B. -> 54
:
B
Answer = b
Initially

^{n} C_{2}

= 45 => n=10
3 people leave, implies that 7 people are left.
number of women = one more than number of men
no. of women = 4. men =3
Handshakes at the end =

^{4} C_{2} +^{3} C_{2} = 9

Total handshakes = 45+9 =54.

Question 37. Consider a Master Set S= {1,2,3,4….12} How many subsets can be formed which will contain one or more elements of S (including all S) such that the elements of the sets are integral multiples of the smallest subset of the set.

2246
2824
3452
2102
1857

Discuss Question

Answer: Option D. -> 2102
:
D
If 1 is the smallest element of the set, all or none of the other elements can be selected in

2^{11}

ways, as for each number from 2 to 12, there are two options, of getting selected or not getting selected. There are 11 such numbers 2-12 including both, therefore 2.2.2…..11 times =

2^{11}

Similarly, If 2 is the smallest element in the set, 4,6,8,10 and 12 can be selected in

2^{5}

ways
If 3 is the smallest element in the set 6,9,12

2^{3}

different sets
Required Solution =

2^{11}

2^{5}

2^{3}

2^{2}

2^{1}

2^{1}

+ 6 = 2102

Question 38. Find the value of x for which the value of

^{16} C_{x}

is maximum for any natural number x.

Discuss Question

Answer: Option D. -> 8
:
D
Soln: Maximum value of

^{16} C_{x}

occurs when x =

\frac{16}{2}

= 8.
Hence option (d)

Question 39. Find the value of x such that:

^{7} C_{x - 1} +^{7} C_{x} =^{8} C_{x + 2}

Discuss Question

Answer: Option B. -> 3
:
B
Soln:We have

^{n} C_{r - 1} + n C_{r} =^{n + 1} C_{r}

^{7} C_{x - 1} +^{7} C_{x} =^{8} C_{x}

^{8} C_{x} =^{8} C_{x + 2} =^{8} C_{(8 - x - 2)}

Either x = x + 2 or x + 2 = (8 - x - 2)
But

x \neq x + 2

2x + 2 = 8
x = 3
Hence option (b)
Alternatively, go from answer options.

Question 40. How many different three digit numbers can be formed with the digits 1, 2, 3, 4, 5 and 6 so that none of the digits are repeated?

Discuss Question

Answer: Option A. -> 120
:
A
Suppose we start with the unit’s place. Unit place can be filled by any of the 6 digits given, so it can be filled in 6 ways. Now coming the tens place, it can be filled by any of the six digits except the one which has been used at unit’s place. So, it can be filled in 5 ways. Similarly the hundred’s place can be filled by any of the six digits except those two which have been used at unit’s and tens’ place. So, it can be filled in 4 ways.
4 ways 5 ways 6 ways
So, total numbers of numbers possible =