JEE Main Practice — Mathematics Set 4 with Answer Key

Question 1

Mathematics · 4 marks · −1

The number of ways in which 6 persons can sit around a round table so that two particular persons do NOT sit next to each other is

A.72
B.48
C.120
D.96

Correct answer: A

Explanation: Total circular arrangements $=(6-1)!=120$ . With the two together (as a block, 2 internal orders): $2\times4!=48$ . Required $=120-48=72$ .

Question 2

Mathematics · 4 marks · −1

The number of non-negative integer solutions of $x+y+z=10$ is

A.55
B.36
C.66
D.78

Correct answer: C

Explanation: Stars and bars: $\binom{10+3-1}{3-1}=\binom{12}{2}=66$ .

Question 3

Mathematics · 4 marks · −1

The number of five-digit numbers divisible by 4 that can be formed using the digits 0, 1, 2, 3, 4, 5 without repetition is

A.120
B.168
C.192
D.144

Correct answer: D

Explanation: Divisible by 4 $\iff$ last two digits form a multiple of 4. Possible endings (distinct digits): 04, 20, 40 (contain 0) and 12, 24, 32, 52 (no 0). Endings with 0: the first three places are filled from the 4 remaining non-zero digits: $4\cdot3\cdot2=24$ each, $3\times24=72$ . Endings without 0: 4 digits remain including 0; first place 3 choices, then $3\cdot2$ : 18 each, $4\times18=72$ . Total $=144$ .

Question 4

Mathematics · 4 marks · −1

The number of diagonals of a decagon (10-sided polygon) is ______.

Correct answer: 35

Explanation: $\binom{10}{2}-10=45-10=35$ .

Question 5

Mathematics · 4 marks · −1

The number of arrangements of the letters of the word BANANA in which the two N's are not adjacent is ______.

Correct answer: 40

Explanation: Total $=\frac{6!}{3!\,2!}=60$ . N's together (treat NN as one unit): $\frac{5!}{3!}=20$ . Required $=60-20=40$ .

Question 6

Mathematics · 4 marks · −1

Two fair dice are thrown. The probability that the sum of the numbers shown is 8 is

A. $\frac16$
B. $\frac{5}{36}$
C. $\frac19$
D. $\frac{7}{36}$

Correct answer: B

Explanation: Favourable pairs: (2,6), (3,5), (4,4), (5,3), (6,2) — 5 out of 36.

Question 7

Mathematics · 4 marks · −1

A fair coin is tossed three times. The probability of getting at least one head is

A. $\frac78$
B. $\frac18$
C. $\frac34$
D. $\frac12$

Correct answer: A

Explanation: $P(\text{no head})=\left(\frac12\right)^3=\frac18$ , so $P(\text{at least one head})=1-\frac18=\frac78$ .

Question 8

Mathematics · 4 marks · −1

For events $A$ and $B$ , $P(A)=0.4$ , $P(B)=0.5$ and $P(A\cap B)=0.2$ . Then $P(A'\cap B')$ equals

A.0.7
B.0.1
C.0.3
D.0.2

Correct answer: C

Explanation: $P(A\cup B)=0.4+0.5-0.2=0.7$ . $P(A'\cap B')=1-P(A\cup B)=0.3$ . (Here $A$ , $B$ happen to be independent, and indeed $0.6\times0.5=0.3$ .)

Question 9

Mathematics · 4 marks · −1

Bag I contains 3 red and 2 black balls; Bag II contains 2 red and 3 black balls. A bag is chosen at random and a ball drawn from it is red. The probability that it came from Bag I is

A. $\frac12$
B. $\frac25$
C. $\frac{3}{10}$
D. $\frac35$

Correct answer: D

Explanation: Bayes: $P(\mathrm I|R)=\frac{\frac12\cdot\frac35}{\frac12\cdot\frac35+\frac12\cdot\frac25}=\frac{3}{5}$ .

Question 10

Mathematics · 4 marks · −1

A fair die is thrown 4 times. The probability of getting exactly two sixes is

A. $\frac{25}{216}$
B. $\frac{1}{36}$
C. $\frac{25}{1296}$
D. $\frac{5}{72}$

Correct answer: A

Explanation: Binomial: $\binom42\left(\frac16\right)^2\left(\frac56\right)^2=6\times\frac{25}{1296}=\frac{150}{1296}=\frac{25}{216}$ .

Question 11

Mathematics · 4 marks · −1

A and B throw a fair die alternately, A starting first. The first to throw a six wins. The probability that A wins is

A. $\frac12$
B. $\frac{6}{11}$
C. $\frac{5}{11}$
D. $\frac{1}{6}$

Correct answer: B

Explanation: A wins on throw 1, 3, 5, …: $P=\frac16+\left(\frac56\right)^2\frac16+\left(\frac56\right)^4\frac16+\cdots=\frac{1/6}{1-25/36}=\frac{1/6}{11/36}=\frac{6}{11}$ .

Question 12

Mathematics · 4 marks · −1

Let $X$ be the number shown when a fair six-faced die is thrown. If $\sigma^2$ is the variance of $X$ , then $12\sigma^2$ equals ______.

Correct answer: 35

Explanation: $E(X)=\frac72$ , $E(X^2)=\frac{91}{6}$ . $\sigma^2=\frac{91}{6}-\frac{49}{4}=\frac{182-147}{12}=\frac{35}{12}$ . So $12\sigma^2=35$ .

Question 13

Mathematics · 4 marks · −1

A leap year is selected at random. If $p$ is the probability that it has 53 Sundays, then $7p$ equals ______.

Correct answer: 2

Explanation: A leap year has 366 days = 52 weeks + 2 days. The two extra days are one of 7 equally likely consecutive pairs (Sun–Mon, Mon–Tue, …, Sat–Sun); 2 of them contain a Sunday. $p=\frac27$ , $7p=2$ .

Question 14

Mathematics · 4 marks · −1

The 10th term of the arithmetic progression $3, 7, 11, \ldots$ is

A.43
B.40
C.39
D.35

Correct answer: C

Explanation: $a_{10}=a+9d=3+9\times4=39$ .

Question 15

Mathematics · 4 marks · −1

The sum to infinity of the geometric series $6+2+\frac23+\cdots$ is

A.9
B.12
C.8
D.18

Correct answer: A

Explanation: $a=6$ , $r=\frac13$ : $S_\infty=\frac{6}{1-1/3}=9$ .

Question 16

Mathematics · 4 marks · −1

The value of $1\cdot2+2\cdot3+3\cdot4+\cdots+10\cdot11$ is

A.385
B.440
C.495
D.330

Correct answer: B

Explanation: $\sum_{k=1}^{10}k(k+1)=\sum k^2+\sum k=385+55=440$ (or $\frac{n(n+1)(n+2)}{3}=\frac{10\cdot11\cdot12}{3}=440$ ).

Question 17

Mathematics · 4 marks · −1

The third term of a geometric progression is 12 and its sixth term is 96. Its first term is

A.2
B.6
C.4
D.3

Correct answer: D

Explanation: $\frac{ar^5}{ar^2}=r^3=8\Rightarrow r=2$ . $ar^2=12\Rightarrow a=3$ .

Question 18

Mathematics · 4 marks · −1

The arithmetic mean of two positive numbers is 10 and their geometric mean is 8. The numbers are

A.16 and 4
B.12 and 8
C.18 and 2
D.15 and 5

Correct answer: A

Explanation: $a+b=20$ , $ab=64$ : roots of $t^2-20t+64=0$ , i.e. $t=16,4$ .

Question 19

Mathematics · 4 marks · −1

The sum $\displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$ equals

A. $\frac12$
B. $\frac16$
C. $\frac14$
D. $\frac13$

Correct answer: C

Explanation: $\frac{1}{n(n+1)(n+2)}=\frac12\left[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right]$ . The sum telescopes to $\frac12\cdot\frac{1}{1\cdot2}=\frac14$ .

Question 20

Mathematics · 4 marks · −1

The sum of all two-digit positive integers that are divisible by 7 is ______.

Correct answer: 728

Explanation: Terms 14, 21, …, 98: an AP with 13 terms. Sum $=\frac{13}{2}(14+98)=13\times56=728$ .

Question 21

Mathematics · 4 marks · −1

The value of $1+2\cdot2+3\cdot2^2+4\cdot2^3+\cdots+10\cdot2^9$ is ______.

Correct answer: 9217

Explanation: Let $S=\sum_{k=1}^{10}k2^{k-1}$ . Then $2S-S=-(1+2+\cdots+2^9)+10\cdot2^{10}=-(2^{10}-1)+10240$ . So $S=10240-1023=9217$ (formula $(n-1)2^n+1$ ).

Question 22

Mathematics · 4 marks · −1

If $n(A)=25$ , $n(B)=18$ and $n(A\cap B)=7$ , then $n(A\cup B)$ is

A.43
B.50
C.36
D.32

Correct answer: C

Explanation: $n(A\cup B)=n(A)+n(B)-n(A\cap B)=25+18-7=36$ .

Question 23

Mathematics · 4 marks · −1

The domain of the real function $f(x)=\sqrt{4-x^2}$ is

A. $(-2,2)$
B. $[-2,2]$
C. $[0,2]$
D. $\mathbb R$

Correct answer: B

Explanation: Need $4-x^2\ge0\iff|x|\le2$ , i.e. $x\in[-2,2]$ (endpoints allowed since $\sqrt0=0$ ).

Question 24

Mathematics · 4 marks · −1

The number of reflexive relations on a set with 3 elements is

A.512
B.8
C.27
D.64

Correct answer: D

Explanation: $A\times A$ has 9 ordered pairs; the 3 diagonal pairs must be included, and each of the other 6 may or may not be: $2^6=64$ .

Question 25

Mathematics · 4 marks · −1

If $f(x)=2x+3$ and $g(x)=x^2$ , then $(g\circ f)(2)$ equals

A.49
B.11
C.19
D.25

Correct answer: A

Explanation: $(g\circ f)(2)=g(f(2))=g(7)=49$ . ( $(f\circ g)(2)=f(4)=11$ .)

JEE Main Practice — Mathematics Set 4

Question 1

Question 2

Question 3

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22

Question 23

Question 24

Question 25

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