n an AP, if a = 1, an = 20 and
Question: n an AP, if a = 1, an = 20 and Sn = 399, then n is equal to (a) 19 (b) 21 (c) 38 (d) 42 Solution: (c) $\because$ $S_{n}=\frac{n}{2}[2 a+(n-1) d]$ $399=\frac{n}{2}[2 \times 1+(n-1) d]$ $798=2 n+n(n-1) d$ ...(i) and $\quad a_{n}=20$ $\Rightarrow \quad a+(n-1) d=20 \quad\left[\because a_{n}=a+(n-1) d\right]$ $\Rightarrow \quad 1+(n-1) d=20 \Rightarrow(n-1) d=19$ ...(ii) Using Eq. (ii) in Eq. (j), we get $798=2 n+19 n$ $\Rightarrow \quad 798=21 n$ $\therefore$ $n=\frac{798}{21}=38$...
Read More →A bag contains 3 white, 4 red and 5 black balls.
Question: A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white? (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (C) $\frac{1}{3}$ (d) $\frac{3}{4}$ Solution: (c) $\frac{1}{3}$ Explanation:Total possible outcomes = Total number of balls $=(3+4+5)=12$ Total number of balls that are non-black and non-white = 4 $\therefore P$ (getting a ball that is neither black nor white) $=\frac{4}{12}=\frac{1}{3}$...
Read More →A bag contains 3 white, 4 red and 5 black balls.
Question: A bag contains 3 white, 4 red and 5 black balls. One ball is drawn at random. What is the probability that the ball drawn is neither black nor white? (a) $\frac{1}{4}$ (b) $\frac{1}{2}$ (C) $\frac{1}{3}$ (d) $\frac{3}{4}$ Solution: (c) $\frac{1}{3}$ Explanation:Total possible outcomes = Total number of balls $=(3+4+5)=12$ Total number of balls that are non-black and non-white = 4 $\therefore P$ (getting a ball that is neither black nor white) $=\frac{4}{12}=\frac{1}{3}$...
Read More →The system of linear equations:
Question: The system of linear equations:x+y+z= 22x+yz= 33x+ 2y+kz= 4 has a unique solution if(a)k 0(b) 1 k 1(c) 2 k 2(d)k= 0 Solution: (a) $k \neq 0$ For a unique solution, $|A| \neq 0$. The given system of equations can be written in matrix form as follows: $\left[\begin{array}{ccc}1 1 1 \\ 2 1 -1 \\ 3 2 k\end{array}\right] \neq 0$ $\Rightarrow 1(k+2)-1(2 k+3)+1(4-3) \neq 0$ $\Rightarrow k+2-2 k-3+1 \neq 0$ $\Rightarrow k \neq 0$ So, the given system of equations has a unique solution if $k$ i...
Read More →A bag contains 8 red, 2 black and 5 white balls.
Question: A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black? (a) $\frac{18}{15}$ (b) $\frac{2}{15}$ (c) $\frac{13}{15}$ (d) $\frac{1}{3}$ Solution: (c) $\frac{13}{15}$ Explanation:Total possible outcomes = Total number of balls = ( 8 + 2 + 5) = 15 Total number of non-black balls = (8 + 5) = 13 $\therefore P$ (getting a ball that is not black) $=\frac{13}{15}$...
Read More →Find the surface area of a cuboid whose
Question: Find the surface area of a cuboid whose (i) length = 10 cm, breadth = 12 cm, height = 14 cm (ii) length = 6 dm, breadth = 8 dm, height = 10 dm (iii) length = 2 m, breadth = 4 m, height = 5 m (iv) length = 3.2 m, breadth = 30 dm, height = 250 cm. Solution: (i) Dimension of the cuboid : Length $=10 \mathrm{~cm}$ Breadth $=12 \mathrm{~cm}$ Height $=14 \mathrm{~cm}$ Surface area of the cuboid $=2 \times($ length $\times$ breadth $+$ breadth $\times$ height $+$ length $\times$ height $)$ $=...
Read More →A bag contains 8 red, 2 black and 5 white balls.
Question: A bag contains 8 red, 2 black and 5 white balls. One ball is drawn at random. What is the probability that the ball drawn is not black? (a) $\frac{18}{15}$ (b) $\frac{2}{15}$ (c) $\frac{13}{15}$ (d) $\frac{1}{3}$ Solution: (c) $\frac{13}{15}$ Explanation:Total possible outcomes = Total number of balls = ( 8 + 2 + 5) = 15 Total number of non-black balls = (8 + 5) = 13 $\therefore P$ (getting a ball that is not black) $=\frac{13}{15}$...
Read More →A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?
Question: A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball? (a) $\frac{2}{5}$ (b) $\frac{3}{5}$ (C) $\frac{1}{10}$ (d) none of these Solution: (b) $\frac{3}{5}$ Explanation:Total possible outcomes = Total number of balls = (4 + 6) = 10 Number of black balls = 6 $\therefore P($ getting a black ball $)=\frac{6}{10}=\frac{3}{5}$...
Read More →A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball?
Question: A bag contains 4 red and 6 black balls. A ball is taken out of the bag at random. What is the probability of getting a black ball? (a) $\frac{2}{5}$ (b) $\frac{3}{5}$ (C) $\frac{1}{10}$ (d) none of these Solution: (b) $\frac{3}{5}$ Explanation:Total possible outcomes = Total number of balls = (4 + 6) = 10 Number of black balls = 6 $\therefore P($ getting a black ball $)=\frac{6}{10}=\frac{3}{5}$...
Read More →The sum of first 16 terms of
Question: The sum of first 16 terms of the AP 10, 6, 2, is (a)-320 (b) 320 (c)-352 (d)-400 Solution: (a)Given, AP is 10, 6, 2, $\therefore$$S_{16}=\frac{16}{2}[2 a+(16-1) d]$$\left[\because S_{n}=\frac{n}{2}\{2 a+(n-1) d\}\right]$ $=8[2 \times 10+15(-4)]$ $=8(20-60)=8(-40)=-320$...
Read More →If the first term of an AP is – 5
Question: If the first term of an AP is 5 and the common difference is 2, then the sum of the first 6 terms is (a) 0 (b) 5 (c) 6 (d) 15 Solution: $(a)$ Given $\quad a=-5$ and $d=2$ $\therefore$ $S_{B}=\frac{6}{2}[2 a+(6-1) d] \quad\left[\because S_{n}=\frac{n}{2}\{2 a+(n-1) d\}\right]$ $=3[2(-5)+5(2)]$ $=3(-10+10)=0$...
Read More →Fill in the blanks in each of the following so as to make the statement true:
Question: Fill in the blanks in each of the following so as to make the statement true: (i) 1 m3= .........cm3 (ii) 1 litre = ....... cubic decimetre (iii) 1 kl = ....... m3 (iv) The volume of a cube of side 8 cm is ........ (v) The volume of a wooden cuboid of length 10 cm and breadth 8 cm is 4000 cm3. The height of the cuboid is ........ cm. (vi) 1 cu.dm = ........ cu. mm (vii) 1 cu. km = ........ cu. m (viii) 1 litre = ........ cu. cm (ix) 1 ml = ........ cu. cm (x) 1 kl = ........ cu. dm = ....
Read More →A box contains 3 blue, 2 white and 4 red marbles. If a marbles.
Question: A box contains 3 blue, 2 white and 4 red marbles. If a marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble? (a) $\frac{1}{3}$ (b) $\frac{4}{9}$ (C) $\frac{7}{9}$ (d) $\frac{2}{9}$ Solution: (C) $\frac{7}{9}$ Explanation:Total possible outcomes = Total number of marbles = ( 3 + 2 + 4) = 9 Let E be the event of not getting a white marble.It means the marble can be either blue or red but not white.Number of favourable outcomes =...
Read More →A box contains 3 blue, 2 white and 4 red marbles. If a marbles.
Question: A box contains 3 blue, 2 white and 4 red marbles. If a marbles. If a marble is drawn at random from the box, what is the probability that it will not be a white marble? (a) $\frac{1}{3}$ (b) $\frac{4}{9}$ (C) $\frac{7}{9}$ (d) $\frac{2}{9}$ Solution: (C) $\frac{7}{9}$ Explanation:Total possible outcomes = Total number of marbles = ( 3 + 2 + 4) = 9 Let E be the event of not getting a white marble.It means the marble can be either blue or red but not white.Number of favourable outcomes =...
Read More →The number of solutions of the system of equations:
Question: The number of solutions of the system of equations:2x+yz= 7x 3y+ 2z= 1x+ 4y 3z= 5(a) 3(b) 2(c) 1(d) 0 Solution: $(\mathrm{d}) 0$ The given system of equations can be written in matrix form as follows: $\left[\begin{array}{ccc}2 1 -1 \\ 1 -3 2 \\ 1 4 -3\end{array}\right]\left[\begin{array}{l}x \\ y \\ z\end{array}\right]=\left[\begin{array}{l}7 \\ 1 \\ 5\end{array}\right]$ $A X=B$ Here, $A=\left[\begin{array}{ccc}2 1 -1 \\ 1 -3 2 \\ 1 4 -3\end{array}\right], X=\left[\begin{array}{l}x \\...
Read More →The famous mathematician associated
Question: The famous mathematician associated with finding the sum of the first 100 naturalnumbers is (a)Pythagoras (b) Newton (c) Gauss (d) Euclid Solution: (c)Gauss is the famous mathematician associated with finding the sum of the first 100 natural numbers i,e., 1,2, 3..100....
Read More →The 4th term from the end
Question: The 4th term from the end of an AP 11, -8,-5,, 49 is (a) 37 (b) 40 (c)43 (d) 58 Solution: (b)We know that, the n th term of an AP from the end is $a_{n}=l-(n-1) d \quad \ldots$ (i) Here, $l=$ Last term and $l=49$ [given] Common difference, $\quad d=-8-(-11)$ $=-8+11=3$ From Eq. (i), $a_{4}=49-(4-1) 3=49-9=40$...
Read More →In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize?
Question: In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize? (a) $\frac{3}{4}$ (b) $\frac{3}{5}$ (c) $\frac{4}{5}$ (d) none of these Solution: (c) $\frac{4}{5}$ Explanation:Number of prizes = 6Number of blanks = 24Total number of tickets = 6 + 24= 30 $\therefore P($ not getting a prize $)=\frac{24}{30}=\frac{4}{5}$...
Read More →In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize?
Question: In a lottery, there are 6 prizes and 24 blanks. What is the probability of not getting a prize? (a) $\frac{3}{4}$ (b) $\frac{3}{5}$ (c) $\frac{4}{5}$ (d) none of these Solution: (c) $\frac{4}{5}$ Explanation:Number of prizes = 6Number of blanks = 24Total number of tickets = 6 + 24= 30 $\therefore P($ not getting a prize $)=\frac{24}{30}=\frac{4}{5}$...
Read More →In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?
Question: In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize? (a) $\frac{1}{2}$ (b) $\frac{1}{3}$ (c) $\frac{2}{3}$ (d) none of these Solution: (b) $\frac{1}{3}$ Explanation:Number of prizes = 8Number of blanks = 16 Total number of tickets $=8+16=24$ $\therefore P($ getting a prize $)=\frac{8}{24}=\frac{1}{3}$...
Read More →In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize?
Question: In a lottery, there are 8 prizes and 16 blanks. What is the probability of getting a prize? (a) $\frac{1}{2}$ (b) $\frac{1}{3}$ (c) $\frac{2}{3}$ (d) none of these Solution: (b) $\frac{1}{3}$ Explanation:Number of prizes = 8Number of blanks = 16 Total number of tickets $=8+16=24$ $\therefore P($ getting a prize $)=\frac{8}{24}=\frac{1}{3}$...
Read More →If 7 times the 7th term of an AP
Question: If 7 times the 7th term of an AP is equal to 11 times its 11th term, then its 18th term will be (a) 7 (b) 11 (c) 18 (d) 0 Solution: (d) According to the question, $7 a_{7}=11 a_{11}$ $\Rightarrow \quad 7[a+(7-1) d]=11[a+(11-1) d] \quad\left[\because a_{n}=a+(n-1) d\right]$ $\Rightarrow \quad 7(a+6 d)=11(a+10 d)$ $\Rightarrow \quad 7 a+42 d=11 a+110 d$ $\Rightarrow \quad 4 a+68 d=0$ $\Rightarrow \quad 2(2 a+34 d)=0$ $\Rightarrow \quad 2 a+34 d=0$$[\because 2 \neq 0]$ $\Rightarrow \quad ...
Read More →Solve this
Question: Let $X=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right], A=\left[\begin{array}{rrr}1 -1 2 \\ 2 0 1 \\ 3 2 1\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $A X=B$, then $X$ is equal to (a)$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$ (b) $\left[\begin{array}{r}-1 \\ -2 \\ -3\end{array}\right]$ (c) $\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$ (d) $\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]$ (e) $\left[\begi...
Read More →Solve this
Question: Let $X=\left[\begin{array}{l}x_{1} \\ x_{2} \\ x_{3}\end{array}\right], A=\left[\begin{array}{rrr}1 -1 2 \\ 2 0 1 \\ 3 2 1\end{array}\right]$ and $B=\left[\begin{array}{l}3 \\ 1 \\ 4\end{array}\right]$. If $A X=B$, then $X$ is equal to (a)$\left[\begin{array}{l}1 \\ 2 \\ 3\end{array}\right]$ (b) $\left[\begin{array}{r}-1 \\ -2 \\ -3\end{array}\right]$ (c) $\left[\begin{array}{l}-1 \\ -2 \\ -3\end{array}\right]$ (d) $\left[\begin{array}{r}-1 \\ 2 \\ 3\end{array}\right]$ (e) $\left[\begi...
Read More →A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm.
Question: A solid rectangular piece of iron measures 6 m by 6 cm by 2 cm. Find the weight of this piece, if 1 cm3of iron weighs 8 gm. Solution: The dimensions of the an iron piece is $6 \mathrm{~m} \times 6 \mathrm{~cm} \times 2 \mathrm{~cm}$, i.e., $600 \mathrm{~cm} \times 6 \mathrm{~cm} \times 2 \mathrm{~cm}(\because 1 \mathrm{~m}=100 \mathrm{~cm}) .$ Its volume $=600 \times 6 \times 2=7200 \mathrm{~cm}^{3}$ Now, $1 \mathrm{~cm}^{3}=8 \mathrm{gm}$ i.e., $7200 \mathrm{~cm}^{3}=7200 \times 8 \ma...
Read More →