Evaluate each of the following:
Question: Evaluate each of the following: (i) ${ }^{8} P_{3}$ (ii) ${ }^{10} P_{4}$ (iii) ${ }^{6} P_{6}$ (iv) $P(6,4)$ Solution: (i) ${ }^{8} P_{3}$ ${ }^{n} P_{r}=\frac{n !}{(n-r) !}$ $\therefore{ }^{8} P_{3}=\frac{8 !}{(8-3) !}$ $=\frac{8 !}{5 !}$ $=\frac{8(7)(6)(5 !)}{5 !}$ $=8 \times 7 \times 6$ $=336$ (ii) ${ }^{10} P_{4}=\frac{10 !}{(10-4) !}$ $=\frac{10 !}{6 !}$ $=\frac{10(9)(8)(7)(6 !)}{6 !}$ $=10 \times 9 \times 8 \times 7$ $=5040$ (iii) ${ }^{6} P_{6}=\frac{6 !}{(6-6) !}$ $=\frac{6 !}...
Read More →Out of a group of swans, 7/2 times the square root of the
Question: Out of a group of swans, 7/2 times the square root of the total number are playing on the share of a pond. The two remaining ones are swinging in water. Find the total number of swans. Solution: Let the total number of swans bex. Then, total numbers of swans are playing on the share of a pond $=\frac{7}{2} \sqrt{x}$ It is given that $\frac{7}{2} \sqrt{x}+2=x$ Let $x=y^{2}$, then $\frac{7}{2} y+2=y^{2}$ $\frac{7 y+4}{2}=y^{2}$ $2 y^{2}=7 y+4$ $2 y^{2}-7 y-4=0$ $2 y^{2}+8 y-y-4=0$ $2 y(y...
Read More →There are 10 lamps in a hall. Each one of them can be switched on independently.
Question: There are 10 lamps in a hall. Each one of them can be switched on independently. Find the number of ways in which the hall can be illuminated. Solution: Total number of lamps = 10 A lamp can be either switched on or switched off. Since, at least one lamp is to be kept switched on, $\therefore$ The total number of ways are $2^{10}-1=1023$. Thus, the number of ways in which the hall can be illuminated are 1023....
Read More →In how many ways can 4 prizes be distributed among 5 students,
Question: In how many ways can 4 prizes be distributed among 5 students, when (i) no student gets more than one prize? (ii) a student may get any number of prizes? (iii) no student gets all the prizes? Solution: (i) Since no student gets more than one prize; the first prize can be given to any one of the five students. The second prize can be given to anyone of the remaining 4 students. Similarly, the third prize can be given to any one of the remaining 3 students. The last prize can be given to...
Read More →Factorize:
Question: Factorize: $a^{3}(b-c)^{3}+b^{3}(c-a)^{3}+c^{3}(a-b)^{3}$ Solution: We have: $a^{3}(b-c)^{3}+b^{3}(c-a)^{3}+c^{3}(a-b)^{3}=[a(b-c)]^{3}+[b(c-a)]^{3}+[c(a-b)]^{3}$ Put $a(b-c)=x$ $b(c-a)=y$ $c(a-b)=z$ Here, $x+y+z=a(b-c)+b(c-a)+c(a-b)$ $=a b-a c+b c-a b+a c-b c$ $=0$ Thus, we have : $a^{3}(b-c)^{3}+b^{3}(c-a)^{3}+c^{3}(a-b)^{3}=x^{3}+y^{3}+z^{3}$ $=3 x y z \quad\left[\right.$ When $\left.x+y+z=0, x^{3}+y^{3}+z^{3}=3 x y z .\right]$ $=3 a(b-c) b(c-a) c(a-b)$ $=3 a b c(a-b)(b-c)(c-a)$...
Read More →A dealer sells an article for Rs. 24 and gains as much
Question: A dealer sells an article for Rs. 24 and gains as much percent as the cost price of the article. Find the cost price of the article. Solution: Let the cost price of article be Rs.x. Then, gain percent =x Therefore, the selling price of article $=\left(x+\frac{x}{100} \times x\right)$ $=\frac{x^{2}+100 x}{100}$ It is given that $\frac{x^{2}+100 x}{100}=24$ $x^{2}+100 x=2400$ $x^{2}+100 x-2400=0$ $x^{2}+120 x-20 x-2400=0$ $x(x+120)-20(x+120)=0$ $(x+120)(x-20)=0$ Becausexcannot be negativ...
Read More →In how many ways can 7 letters be posted in 4 letter boxes?
Question: In how many ways can 7 letters be posted in 4 letter boxes? Solution: Each letter can be posted in any one of the 4 letter boxes. Number of ways of posting one letter = 4 $\therefore$ Required number of ways of posting the 7 letters $=4 \times 4 \times 4 \times 4 \times 4 \times 4 \times 4=4^{7}$...
Read More →In how many ways can 5 different balls be distributed among three boxes?
Question: In how many ways can 5 different balls be distributed among three boxes? Solution: Each ball can be distributed in 3 ways. $\therefore$ Required number of ways in order to distribute 5 balls $=3 \times 3 \times 3 \times 3 \times 3=243$...
Read More →Some students planned a picnic.
Question: Some students planned a picnic. The budget for food was Rs. 480. But eight of these failed to go and thus the cost of food for each member increased by Rs. 10. How many students attended the picnic? Solution: Letxstudents planned a picnic. Then, the share of each student $=\frac{480}{x}$ According to question, 8 students fail to go picnic, then remaining students $=(x-8)$. Therefore, new share of each student $=\frac{480}{x-8}$ It is given that $\frac{480}{x-8}-\frac{480}{x}=10$ $\frac...
Read More →Factorize:
Question: Factorize: $(5 a-7 b)^{3}+(9 c-5 a)^{3}+(7 b-9 c)^{3}$ Solution: Put $(5 a-7 b)=x,(9 c-5 a)=z$ and $(7 b-9 c)=y$. Here, $x+y+z=5 a-7 b+9 c-5 a+7 b-9 c=0$ Thus, we have: $(5 a-7 b)^{3}+(9 c-5 a)^{3}+(7 b-9 c)^{3}=x^{3}+z^{3}+y^{3}$ $=3 x z y \quad\left[\right.$ When $\left.x+y+z=0, x^{3}+y^{3}+z^{3}=3 x y z .\right]$ $=3(5 a-7 b)(9 c-5 a)(7 b-9 c)$...
Read More →Find the total number of ways in which 20 balls can be put into 5 boxes so that first box contains just one ball.
Question: Find the total number of ways in which 20 balls can be put into 5 boxes so that first box contains just one ball. Solution: Any one of the twenty balls can be put in the first box. Thus, there are twenty different ways for this. Now, remaining 19 balls are to be put into the remaining 4 boxes. This can be done in $4^{19}$ ways because there are four choices for each ball. $\therefore$ Required number of ways $=20 \times 4^{19}$...
Read More →Factorize:
Question: Factorize: $(3 a-2 b)^{3}+(2 b-5 c)^{3}+(5 c-3 a)^{3}$ Solution: Put $(3 a-2 b)=x,(2 b-5 c)=y$ and $(5 c-3 a)=z$. We have : $x+y+z=3 a-2 b+2 b-5 c+5 c-3 a=0$ Now, $(3 a-2 b)^{3}+(2 b-5 c)^{3}+(5 c-3 a)^{3}=x^{3}+y^{3}+z^{3}$ $=3 x y z \quad\left[\right.$ Here,$x+y+z=0 .$ So, $\left.x^{3}+y^{3}+z^{3}=3 x y z\right]$ $=3(3 a-2 b)(2 b-5 c)(5 c-3 a)$...
Read More →Three dice are rolled. Find the number of possible outcomes in which at least one die shows 5.
Question: Three dice are rolled. Find the number of possible outcomes in which at least one die shows 5. Solution: Required number of possible outcomes $=($ Total number of outcomes $-$ Number of possible outcomes in which 5 does not appear on any of the dice.) Total number of outcomes when a single dice is rolled $=6$ $\therefore$ Total number of outcomes when two dice are rolled $=6 \times 6$ Similarly, total number of outcomes when three dice are rolled $=6 \times 6 \times 6=216$ Number of po...
Read More →Factorise:
Question: Factorise: $(a-3 b)^{3}+(3 b-c)^{3}+(c-a)^{3}$ Solution: We know $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ $x^{3}+y^{3}+z^{3}=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)+3 x y z$ Here, $x=(a-3 b), y=(3 b-c), z=(c-a)$ $(a-3 b)^{3}+(3 b-c)^{3}+(c-a)^{3}$ $=(a-3 b+3 b-c+c$ $-a)\left[(a-3 b)^{2}+(3 b-c)^{2}+(c-a)^{2}-(a-3 b)(3 b-c)-(3 b-c)(c-a)-(c-a)(a-3 b)\right]$ $+3(a-3 b)(3 b-c)(c-a)$ $=0+3(a-3 b)(3 b-c)(c-a)$ $=3(a-3 b)(3 b-c)(c-a)$...
Read More →Find the number of ways in which one can post 5 letters in 7 letter boxes.
Question: Find the number of ways in which one can post 5 letters in 7 letter boxes. Solution: Each of the 5 letters can be posted in any one of the 7 letter boxes. $\therefore$ Required number of ways of posting the letters $=7 \times 7 \times 7 \times 7 \times 7=7^{5}$...
Read More →A piece of cloth costs Rs. 35.
Question: A piece of cloth costs Rs. 35. If the piece were 4 m longer and each meter costs Rs. one less, the cost would remain unchanged. How long is the piece? Solution: Let the length of the piece bexmetres. Then, rate per metre $=\frac{35}{x}$ According to question, new length $=(x+4)$ meters. Since the cost remain same. Therefore, new rate per metre $=\frac{35}{x+4}$ It is given that $\frac{35}{x}-\frac{35}{x+4}=1$ $\frac{35(x+4)-35 x}{x(x+4)}=1$ $\frac{140}{x(x+4)}=1$ $x^{2}+4 x=140$ $x^{2}...
Read More →Find the number of ways in which 8 distinct toys can be distributed among 5 childrens.
Question: Find the number of ways in which 8 distinct toys can be distributed among 5 childrens. Solution: Each of the toy can be distributed in 5 ways. $\therefore$ Total number of ways of distributing the toys $=5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5 \times 5=5^{8}$...
Read More →How many five digit telephone numbers can be constructed using the digits 0 to 9.
Question: How many five digit telephone numbers can be constructed using the digits 0 to 9. If each number starts with 67 and no digit appears more than once? Solution: The first two digits are fixed as 67. As repetition of digits in not allowed, the number of available digits to fill the remaining places is 8. The third place can be filled in 8 ways. The fourth place can be filled in 7 ways. The fifth place can be filled in 6 ways. Total number of such telephone numbers $=8 \times 7 \times 6=33...
Read More →Factorize:
Question: Factorize: $(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$ Solution: $(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$ Putting $(a-b)=x,(b-c)=y$ and $(c-a)=z$, we get: $(a-b)^{3}+(b-c)^{3}+(c-a)^{3}$ $=x^{3}+y^{3}+z^{3} \quad[$ Where $(x+y+z)=(a-b)+(b-c)+(c-a)=0]$ $=3 x y z \quad\left[(x+y+z)=0 \Rightarrow x^{3}+y^{3}+z^{3}=3 x y z\right]$ $=3(a-b)(b-c)(c-a)$...
Read More →Two pipes running together can fill a tank in
Question: Two pipes running together can fill a tank in $11 \frac{1}{9}$ minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately. Solution: Let the first pipe takes $x$ minutes to fill the tank. Then the second pipe will takes $=(x+5)$ minutes to fill the tank. Since, the first pipe takes $x$ minutes to fill the tank. Therefore, portion of the tank filled by the first pipe in one minutes $=\frac{1}{x}$ ...
Read More →Factorise:
Question: Factorise: $3 \sqrt{3} a^{3}-b^{3}-5 \sqrt{5} c^{3}-3 \sqrt{15} a b c$ Solution: $3 \sqrt{3} a^{3}-b^{3}-5 \sqrt{5} c^{3}-3 \sqrt{15} a b c$ $=(\sqrt{3} a)^{3}+(-b)^{3}+(-\sqrt{5} c)^{3}-3(\sqrt{3} a)(-b)(-\sqrt{5} c)$ We know $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ Here, $x=(\sqrt{3} a), y=(-b), z=(-\sqrt{5} c)$ $3 \sqrt{3} a^{3}-b^{3}-5 \sqrt{5} c^{3}-3 \sqrt{15} a b c$ $=(\sqrt{3} a)^{3}+(-b)^{3}+(-\sqrt{5} c)^{3}-3(\sqrt{3} a)(-b)(-\sqrt{5} c)$...
Read More →How many natural numbers less than 1000 can be formed from the digits 0, 1, 2, 3, 4,
Question: How many natural numbers less than 1000 can be formed from the digits 0, 1, 2, 3, 4, 5 when a digit may be repeated any number of times? Solution: Since the number is less than 1000, it means that it is a three-digit number, a two-digit number or a single-digit number. Three-digit numbers: The hundred's place can be filled by 5 digits neglecting zero as it can't be zero. The ten's place and the unit's place can be filled by 6 digits. So, total number of three digit numbers $=5 \times 6...
Read More →Two water taps together can fill a tank in
Question: Two water taps together can fill a tank in $9 \frac{3}{8}$ hours. The tap of larger diameter takes 10 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Solution: Let the first water tape takes $x$ hours to fill the tank. Then the second water tape will takes $=(x+10)$ hours to fill the tank. Since, the faster water tape takes $x$ hours to fill the tank. Therefore, portion of the tank filled by the faster water tap...
Read More →Factorise:
Question: Factorise: $2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c$ Solution: $2 \sqrt{2} a^{3}+3 \sqrt{3} b^{3}+c^{3}-3 \sqrt{6} a b c$ $=(\sqrt{2} a)^{3}+(\sqrt{3} b)^{3}+c^{3}-3(\sqrt{2} a)(\sqrt{3} b) c$ We know $x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\right)$ $x=\sqrt{2} a, y=\sqrt{3} b, z=c$ $(\sqrt{2} a)^{3}+(\sqrt{3} b)^{3}+c^{3}-3(\sqrt{2} a)(\sqrt{3} b) c$ $=(\sqrt{2} a+\sqrt{3} b+c)\left(2 a^{2}+3 b^{2}+c^{2}-\sqrt{6 a b}-\sqrt{3 b c}-\sqrt{2} ...
Read More →How many three digit numbers can be formed by using the digits 0, 1, 3, 5, 7
Question: How many three digit numbers can be formed by using the digits 0, 1, 3, 5, 7 while each digit may be repeated any number of times? Solution: Since the hundred's place cannot be zero, it can be filled by any of the 4 digits (1, 3, 5 and 7). Number of ways of filling the hundred's place = 4 Since the digits can be repeated in the number, the ten's place and the unit's place can each be filled in 5 ways. $\therefore$ Total numbers $=4 \times 5 \times 5=100$...
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