If P (n, 4) = 12 . P (n, 2), find n.
Question: IfP(n, 4) = 12 .P(n, 2), findn. Solution: P(n, 4) = 12 .P(n, 2) $\Rightarrow \frac{n !}{(n-4) !}=12 \times \frac{n !}{(n-2) !}$ $\Rightarrow \frac{(n-2) !}{(n-4) !}=12 \times \frac{n !}{n !}$ $\Rightarrow \frac{(n-2)(n-3)(n-4) !}{(n-4) !}=12$ $\Rightarrow(n-2)(n-3)=12$ $\Rightarrow(n-2)(n-3)=4 \times 3$ On comparing the LHS and the RHS, we get : $n-2=4$ $\Rightarrow n=6$...
Read More →Find the value
Question: If $\frac{x}{y}+\frac{y}{x}=-1$, where $x \neq 0$ and $y \neq 0$, then the value of $\left(x^{3}-y^{3}\right)$ is (a) 1 (b) $-1$ (c) 0 (d) $\frac{1}{2}$ Solution: (c) 0 $\frac{x}{y}+\frac{y}{x}=-1$ $\Rightarrow \frac{x^{2}+y^{2}}{x y}=-1$ $\Rightarrow x^{2}+y^{2}=-x y$ $\Rightarrow x^{2}+y^{2}+x y=0$ Thus, we have: $\left(x^{3}-y^{3}\right)=(x-y)\left(x^{2}+y^{2}+x y\right)$ $=(x-y) \times 0$ $=0$...
Read More →A cottage industry produces a certain number of pottery articles in a day.
Question: A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production on each article (in rupees) was 3 more than twice the number of articles produced on that day. If the total cost of production on that day was Rs. 90, find the number of articles produced and the cost of each article. Solution: Let the number of article produced by the cottage industry be $x$. Then the cost of production of each article $=$ Rs. $(2 ...
Read More →If P(11, r) = P (12, r − 1) find r.
Question: IfP(11,r) =P(12,r 1) findr. Solution: $P(11, r)=P(12, r-1)$ $\Rightarrow \frac{11 !}{(11-r) !}=\frac{12 !}{(13-r) !}$ $\Rightarrow \frac{(13-r)}{(11-r) !}=\frac{12 !}{11 !}$ $\Rightarrow \frac{(13-r)(12-r)(11-r) !}{(11-r) !}=\frac{12 \times 11 !}{11 !}$ $\Rightarrow(13-r)(12-r)=12$ $\Rightarrow(13-r)(12-r)=4 \times 3$ On comparing the two sides, we get: $13-r=4$ $\Rightarrow r=9$...
Read More →If P(11, r) = P (12, r − 1) find r.
Question: IfP(11,r) =P(12,r 1) findr. Solution: $P(11, r)=P(12, r-1)$ $\Rightarrow \frac{11 !}{(11-r) !}=\frac{12 !}{(13-r) !}$ $\Rightarrow \frac{(13-r)}{(11-r) !}=\frac{12 !}{11 !}$ $\Rightarrow \frac{(13-r)(12-r)(11-r) !}{(11-r) !}=\frac{12 \times 11 !}{11 !}$ $\Rightarrow(13-r)(12-r)=12$ $\Rightarrow(13-r)(12-r)=4 \times 3$ On comparing the two sides, we get: $13-r=4$ $\Rightarrow r=9$...
Read More →In a class test, the sum of Shefali's marks in Mathematics and English is 30.
Question: In a class test, the sum of Shefali's marks in Mathematics and English is 30. Had she got 2 marks more in mathematics and 3 marks less in English, the product of her marks would have been 210. Find her marks in two subjects. Solution: Let marks obtained by Shefali in mathematics be $x$, then in english $=(30-x)$ It is given that, $(x+2) \times(30-x-3)=210$ $(x+2) \times(27-x)=210$ $27 x-x^{2}+54-2 x=210$ $-x^{2}+25 x+54-210=0$ $-\left(x^{2}-25 x+156\right)=0$ $x^{2}-25 x+156=0$ $x^{2}-...
Read More →The value of
Question: The value of $(249)^{2}-(248)^{2}$ is (a) $1^{2}$ (b) 477 (c) 487 (d) 497 Solution: $(249)^{2}-(248)^{2}$ We know $a^{2}-b^{2}=(a+b)(a-b)$ So, $(249)^{2}-(248)^{2}=(249-248)(249+248)=497$ Hence, the correct answer is option (d)....
Read More →If P (9, r) = 3024, find r.
Question: IfP(9,r) = 3024, findr. Solution: P(9,r) = 3024 $\Rightarrow \frac{9 !}{(9-r) !}=3024$ $\Rightarrow \frac{9 !}{(9-r) !}=9 \times 8 \times 7 \times 6$ $\Rightarrow \frac{9 !}{(9-r) !}=\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 !}$ $\Rightarrow \frac{9 !}{(9-r) !}=\frac{9 !}{5 !}$ $\Rightarrow(9-r) !=5 !$ $\Rightarrow 9-r=5$ $\Rightarrow r=4$...
Read More →If P (9, r) = 3024, find r.
Question: IfP(9,r) = 3024, findr. Solution: P(9,r) = 3024 $\Rightarrow \frac{9 !}{(9-r) !}=3024$ $\Rightarrow \frac{9 !}{(9-r) !}=9 \times 8 \times 7 \times 6$ $\Rightarrow \frac{9 !}{(9-r) !}=\frac{9 \times 8 \times 7 \times 6 \times 5 !}{5 !}$ $\Rightarrow \frac{9 !}{(9-r) !}=\frac{9 !}{5 !}$ $\Rightarrow(9-r) !=5 !$ $\Rightarrow 9-r=5$ $\Rightarrow r=4$...
Read More →Solve the following
Question: If ${ }^{n} P_{4}=360$, find the value of $n .$ Solution: ${ }^{n} P_{4}=360$ $\Rightarrow \frac{n !}{(n-4) !}=360$ $\Rightarrow \frac{n(n-1)(n-2)(n-3)(n-4) !}{(n-4) !}=360$ $\Rightarrow n(n-1)(n-2)(n-3)=360$ $\Rightarrow n(n-1)(n-2)(n-3)=6 \times 5 \times 4 \times 3$ On comparing the two sides, we get: $n=6$...
Read More →If (x + 1) is factor of the polynomial
Question: If $(x+1)$ is factor of the polynomial $\left(2 x^{2}+k x\right)$ then the value of $k$ is (a) 2(b) 3(c) 2(d) 3 Solution: (c) 2 $(x+1)$ is a factor of $2 x^{2}+k x$ So, $-1$ is a zero of $2 x^{2}+k x$. Thus, we have : $2 \times(-1)^{2}+k \times(-1)=0$ $\Rightarrow 2-k=0$ $\Rightarrow k=2$...
Read More →In a class test, the sum of the marks obtained by P in Mathematics and science is 28.
Question: In a class test, the sum of the marks obtained by P in Mathematics and science is 28. Had he got 3 marks more in mathematics and 4 marks less in Science. The product of his marks would have been 180. Find his marks in two subjects. Solution: Let marks obtained by $P$ in mathematics be $x$, then in science $=(28-x)$ It is given that, $(x+3) \times(28-x-4)=180$ $(x+3) \times(24-x)=180$ $24 x-x^{2}+72-3 x=180$ $-x^{2}+21 x+72-180=0$ $-\left(x^{2}-21 x+108\right)=0$ $x^{2}-21 x+108=0$ $x^{...
Read More →If P (n, 5) = 20. P(n, 3), find n
Question: IfP(n, 5) = 20.P(n, 3), findn Solution: P(n, 5) = 20.P(n, 3) $\Rightarrow \frac{n !}{(n-5) !}=20 \times \frac{n !}{(n-3) !}$ $\Rightarrow \frac{n !}{n !}=20 \times \frac{(n-5) !}{(n-3) !}$ $\Rightarrow 1=20 \times \frac{(n-5) !}{(n-3)(n-4)(n-5) !}$ $\Rightarrow(n-3)(n-4)=20$ $\Rightarrow(n-3)(n-4)=5 \times 4$ On comparing the two sides, we get: $\Rightarrow n-3=5$ $\Rightarrow n=8$...
Read More →If P (n, 5) = 20. P(n, 3), find n
Question: IfP(n, 5) = 20.P(n, 3), findn Solution: P(n, 5) = 20.P(n, 3) $\Rightarrow \frac{n !}{(n-5) !}=20 \times \frac{n !}{(n-3) !}$ $\Rightarrow \frac{n !}{n !}=20 \times \frac{(n-5) !}{(n-3) !}$ $\Rightarrow 1=20 \times \frac{(n-5) !}{(n-3)(n-4)(n-5) !}$ $\Rightarrow(n-3)(n-4)=20$ $\Rightarrow(n-3)(n-4)=5 \times 4$ On comparing the two sides, we get: $\Rightarrow n-3=5$ $\Rightarrow n=8$...
Read More →If a + b + c = 9 and
Question: If $a+b+c=9$ and $a^{2}+b^{2}+c^{2}=35$, find the value of $\left(a^{3}+b^{3}+c^{3}-3 a b c\right)$. Solution: $a+b+c=9$ $\Rightarrow(a+b+c)^{2}=9^{2}=81$ $\Rightarrow a^{2}+b^{2}+c^{2}+2(a b+b c+c a)=81$ $\Rightarrow 35+2(a b+b c+c a)=81$ $\Rightarrow(a b+b c+c a)=23$ We know, $\left(a^{3}+b^{3}+c^{3}-3 a b c\right)=(a+b+c)\left(a^{2}+b^{2}+c^{2}-a b-b c-c a\right)$ $=(9)(35-23)$ $=108$...
Read More →A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in
Question: A pole has to be erected at a point on the boundary of a circular park of diameter 13 meters in such a way that the difference of its distance from two diametrically opposite fixed gates A and B on the boundary is 7 meters. Is it the possible to do so? If yes, at what distances from the two gates should the pole be erected? Solution: Let $P$ be the required location on the boundary of a circular park such that its distance from gate $B$ is $=x$ metres that is $B P=x$ metres Then, $A P=...
Read More →If 5 P(4, n) = 6. P (5, n − 1), find n.
Question: If 5P(4,n) = 6.P(5,n 1), findn. Solution: 5P(4,n) = 6.P(5,n 1) 54Pn= 65Pn-1 $\Rightarrow 5 \times \frac{4 !}{(4-n) !}=6 \times \frac{5 !}{(5-n+1) !}$ $\Rightarrow 5 \times \frac{(6-n) !}{(4-n) !}=6 \times \frac{5 !}{4 !}$ $\Rightarrow 5 \times \frac{(6-n)(6-n-1)(6-n-2) !}{(4-n)}=6 \times \frac{5 \times 4 !}{4 !}$ $\Rightarrow 5 \times \frac{(6-n)(5-n)(4-n) !}{(4-n)}=6 \times 5$ $\Rightarrow(6-n)(5-n)=6$ $\Rightarrow(6-n)(5-n)=3 \times 2$ On comparing the LHS and the RHS, we get: $\Righ...
Read More →If a, b, c are all nonzero and a + b + c = 0, prove that
Question: If $a, b, c$ are all nonzero and $a+b+c=0$, prove that $\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}=3$. Solution: $a+b+c=0 \Rightarrow a^{3}+b^{3}+c^{3}=3 a b c$ Thus, we have: $\left(\frac{a^{2}}{b c}+\frac{b^{2}}{c a}+\frac{c^{2}}{a b}\right)=\frac{a^{3}+b^{3}+c^{3}}{a b c}$ $=\frac{3 a b c}{a b c}$ $=3$...
Read More →Some students planned a picnic.
Question: Some students planned a picnic. The budget for food was Rs. 500. But, 5 of them failed to go and thus the cost of food for each member increased by Rs. 5. How many students attended the picnic? Solution: Letxstudents planned a picnic. Then, the share of each student $=\frac{500}{x}$ According to question, 5 students fail to go picnic, then remaining students $=(x-5)$. Therefore, new share of each student $=\frac{500}{x-5}$ It is given that $\frac{500}{x-5}-\frac{500}{x}=5$ $\frac{500 x...
Read More →If P (5, r) = P (6, r − 1),
Question: IfP(5,r) =P(6,r 1), findr. Solution: $P(5, r)=P(6, r-1)$ or ${ }^{5} P_{r}={ }^{6} P_{r-1}$ $\frac{5 !}{(5-r) !}=\frac{6 !}{(6-r+1) !}$ $\Rightarrow \frac{(6-r+1) !}{(5-r) !}=\frac{6 !}{5 !}$ $\Rightarrow \frac{(7-r) !}{(5-r) !}=\frac{6(5 !)}{5 !}$ $\Rightarrow \frac{(7-r)(6-r)(5-r) !}{(5-r) !}=6$ $\Rightarrow(7-r)(6-r)=6$ $\Rightarrow(7-r)(6-r)=3 \times 2$ On comparing the above two equations, we get: $7-r=3$ $\Rightarrow r=4$...
Read More →Prove that
Question: Prove that $(a+b+c)^{3}-a^{3}-b^{3}-c^{3}=3(a+b)(b+c)(c+a)$ Solution: $(a+b+c)^{3}=[(a+b)+c]^{3}=(a+b)^{3}+c^{3}+3(a+b) c(a+b+c)$ $\Rightarrow(a+b+c)^{3}=a^{3}+b^{3}+3 a b(a+b)+c^{3}+3(a+b) c(a+b+c)$ $\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3 a b(a+b)+3(a+b) c(a+b+c)$ $\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3(a+b)\left[a b+c a+c b+c^{2}\right]$ $\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3(a+b)[a(b+c)+c(b+c)]$ $\Rightarrow(a+b+c)^{3}-a^{3}+b^{3}-c^{3}=3(a+b)(b+c)(a+c)$...
Read More →Rs. 9000 were divided equally among a certain number of persons.
Question: Rs. 9000 were divided equally among a certain number of persons. Had there been 20 more persons, each would have got Rs. 160 less. Find the original number of persons. Solution: Let the original number of persons bex.Then, by the given information, $\frac{9000}{x}-160=\frac{9000}{x+20}$ $\frac{9000-160 x}{x}=\frac{9000}{x+20}$ $(x+20)(9000-160 x)=9000 x$ $9000 x-160 x^{2}+180000-3200 x=9000 x$ $160 x^{2}-180000+3200 x=0$ $x^{2}-1125+20 x=0$ $x^{2}-1125+20 x+100=100$ $(x+10)^{2}=1225$ $...
Read More →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 →If the list price of a toy is reduced by Rs. 2,
Question: If the list price of a toy is reduced by Rs. 2, a person can buy 2 toys more for Rs. 360. Find the original price of the toy. Solution: Let the original list price of the toy be Rs.x. Then, the number of toys brought for Rs. $360=\frac{360}{x}$ According to question, reduced list price of the toys $=$ Rs. $(x-2)$. Therefore, the number of toys brought for Rs. $360=\frac{360}{x-2}$ It is given that $\frac{360}{x-2}-\frac{360}{x}=2$ $\frac{360 x-360(x-2)}{(x-2) x}=2$ $\frac{720}{(x-2) x}...
Read More →Evaluate
Question: Evaluate (i) $(-12)^{3}+7^{3}+5^{3}$ (ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$ Solution: (i) $(-12)^{3}+7^{3}+5^{3}$ 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, $x=(-12), y=7, z=5$ $(-12)^{3}+7^{3}+5^{3}$ $=(-12+7+5)\left[(-12)^{2}+7^{2}+5^{2}-7(-12)-35+60\right]+3(-12) \times 35$ $=0-1260=-1260$ (ii) $(28)^{3}+(-15)^{3}+(-13)^{3}$ We know $x^{3}+y^{3}+z^{3}-3 x y z=(...
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