If the nth term of the A.P. 9, 7, 5, ... is

Question: If thenth term of the A.P. 9, 7, 5, ... is same as thenth term of the A.P. 15, 12, 9, ... findn. Solution: Here, we are given two A.P. sequences whosenthterms are equal. We need to findn. So let us first find thenthterm for both the A.P. First A.P. is 9, 7, 5 Here, First term (a) = 9 Common difference of the A.P. $(d)=7-9$ $=-2$ Now, as we know, $a_{n}=a+(n-1) d$ So, for $n^{\text {th }}$ term, $a_{n}=9+(n-1)(-2)$ $=9-2 n+2$ $=11-2 n$......(1) Second A.P. is 15, 12, 9 Here, First term ...

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Boundaries of surfaces are

Question: Boundaries of surfaces are(a) lines(b) curves(c) polygons(d) none of these Solution: (b) curves...

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If g(x)

Question: If $g(x)=x^{2}+x-2$ and $\frac{1}{2} g o f(x)=2 x^{2}-5 x+2$, then $f(x)$ is equal to (a) $2 x-3$ (b) $2 x+3$ (c) $2 x^{2}+3 x+1$ (d) $2 x^{2}-3 x-1$ Solution: We will solve this problem by the trial-and-error method.Let us check option (a) first. If $f(x)=2 x-3$ $\frac{1}{2}(g o f)(x)=g(f(x))$ $=\frac{1}{2} g(2 x-3)$ $=\frac{1}{2}\left[(2 x-3)^{2}+(2 x-3)-2\right]$ $=\frac{1}{2}\left[4 x^{2}+9-12 x+2 x-3-2\right]$ $=\frac{1}{2}\left[4 x^{2}-10 x+4\right]$ $=2 x^{2}-5 x+2$ The given co...

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Boundaries of solids are

Question: Boundaries of solids are(a) lines(b) curves(c) surfaces(d) none of these Solution: (c) surfaces...

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How many dimensions does a point have

Question: How many dimensions does a point have(a) 0(b) 1(c) 2(d) 3 Solution: A point is a fine dot which represents an exact position. It hasnolength,nobreadth andnoheight. Thus, a point hasnodimension or a point haszerodimension.Hence, the correct answer is option (a)....

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The number of dimensions of a surface are

Question: The number of dimensions of a surface are(a) 1(b) 2(c) 3(d) 0 Solution: A plane surface has length and breadth, but it hasnoheight. Thus, a plane surface hastwodimensions. Hence, the correct answer is option (b)....

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Three persons enter a railway compartment. If there are 5 seats vacant,

Question: Three persons enter a railway compartment. If there are 5 seats vacant, in how many ways can they take these seats? (a) 60 (b) 20 (c) 15 (d) 125 Solution: (a) 60Three persons can take 5 seats in5C3 ways. Moreover, 3 persons can sit in 3! ways. $\therefore$ Required number of ways $={ }^{5} C_{3} \times 3 !=10 \times 6=60$...

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If (m + 1)th term of an A.P is twice the (n + 1)th term,

Question: If $(m+1)^{\text {th }}$ term of an A.P is twice the $(n+1)^{\text {th }}$ term, prove that $(3 m+1)^{\text {th }}$ term is twice the $(m+n+1)^{\text {th }}$ term. Solution: Here, we are given that $(m+1)^{\text {th }}$ term is twice the $(n+1)^{\text {th }}$ term, for a certain A.P. Here, let us take the first term of the A.P. as a and the common difference as d We need to prove that $a_{3 m+1}=2 a_{m+n+1}$ So, let us first find the two terms. As we know, $a_{n^{\prime}}=a+\left(n^{\p...

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The number of dimensions of a solid are

Question: The number of dimensions of a solid are(a) 1(b) 2(c) 3(d) 5 Solution: A solid shape has length, breadth and height. Thus, a solid hasthreedimensions.Hence, the correct answer is option (c)...

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The side faces of a pyramid are

Question: The side faces of a pyramid are(a) triangles(b) squares(c) trapeziums(d) polygons Solution: (a) triangles...

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Let [x] denote the greatest integer less than or equal to x.

Question: Let $[x]$ denote the greatest integer less than or equal to $x$. If $f(x)=\sin ^{-1} x, g(x)=\left[x^{2}\right]$ and $h(x)=2 x, \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}$, then (a) $f \circ g \circ h(x)=\frac{\pi}{2}$ (b) $\operatorname{fogoh}(x)=\pi$ (c) hofog $=$ hogof (d) $h o f o g \neq h o g o f$ Solution: (c) hofog $=$ hogof We have, $g(x)=\left[x^{2}\right]$ $=0 \quad\left(A s \frac{1}{2} \leq x \leq \frac{1}{\sqrt{2}}\right.$ $\left.\therefore \frac{1}{4} \leq x^{2} \leq \frac...

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A pyramid is a solid figure, whose base is

Question: A pyramid is a solid figure, whose base is(a) only a triangle(b) only a square(c) only a rectangle(d) any polygon Solution: (d) any polygon...

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There are 12 points in a plane. The number of the straight lines joining any two of them when 3 of them are collinear,

Question: There are 12 points in a plane. The number of the straight lines joining any two of them when 3 of them are collinear, is (a) 62 (b) 63 (c) 64 (d) 65 Solution: (c) 64 Number of straight lines joining 12 points if we take 2 points at a time $={ }^{12} C_{2}=\frac{12 !}{2 ! 10 !}=66$ Number of straight lines joining 3 points if we take 2 points at a time =3C2= 3 But, 3 collinear points, when joined in pairs, give only one line. $\therefore$ Required number of straight lines $=66-3+1=64$...

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Euclid stated that 'all right angles are equal to each other', in the form of

Question: Euclid stated that 'all right angles are equal to each other', in the form of(a) a definition(b) an axiom(c) a postulate(d) a proof Solution: (b) an axiom...

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The statement that 'the lines are parallel if they do not intersect' is in the form of

Question: The statement that 'the lines are parallel if they do not intersect' is in the form of(a) a definition(b) an axiom(c) a postulate(d) a theorem Solution: (a) a definition...

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Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to

Question: Total number of words formed by 2 vowels and 3 consonants taken from 4 vowels and 5 consonants is equal to (a) 60 (b) 120 (c) 7200 (d) none of these Solution: (c) 7200 2 out of 4 vowels can be chosen in4C2ways and 3 out of 5 consonants can be chosen in5C3ways. Thus, there are $\left(C_{2} \times{ }^{5} C^{4}{ }_{3}\right)$ groups, each containing 2 vowels and 3 consonants. Each group contains 5 letters that can be arranged in $5 !$ ways. $\therefore$ Required number of words $=\left({ ...

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In a certain A.P. the 24th term is twice the 10th term.

Question: In a certain A.P. the 24thterm is twice the 10thterm. Prove that the 72nd term is twice the 34th term. Solution: Here, we are given that 24thterm is twice the 10thterm, for a certain A.P. Here, let us take the first term of the A.P. asaand the common difference asd We have to prove that So, let us first find the two terms. As we know, $a_{n}=a+(n-1) d$ For $10^{\text {th }}$ term $(n=10)$, $a_{10}=a+(10-1) d$ $=a+9 d$ For $24^{\text {th }}$ term $(n=24)$, $a_{24}=a+(24-1) d$ $=a+23 d$ ...

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Which of the following needs a proof?

Question: Which of the following needs a proof?(a) axiom(b) postulate(c) definition(d) theorem Solution: (d) theorem...

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Pythagoras was a student of

Question: Pythagoras was a student of(i) Euclid(ii) Thales(iii) Archimedes(iv) Bhaskara Solution: (ii) Thales...

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Solve the following

Question: 5C1+5C2+5C3+5C4+5C5is equal to (a) 30 (b) 31 (c) 32 (d) 33 Solution: (b) 31 ${ }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}$ = ${ }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{2}+{ }^{5} C_{1}+{ }^{5} C_{5}$ $=2 \times{ }^{5} C_{1}+2 \times{ }^{5} C_{2}+{ }^{5} C_{5}$ $=2 \times 5+2 \times \frac{5 !}{2 ! 3 !}+1$ $=10+20+1$ $=31$...

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Solve the following

Question: 5C1+5C2+5C3+5C4+5C5is equal to (a) 30 (b) 31 (c) 32 (d) 33 Solution: (b) 31 ${ }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{3}+{ }^{5} C_{4}+{ }^{5} C_{5}$ = ${ }^{5} C_{1}+{ }^{5} C_{2}+{ }^{5} C_{2}+{ }^{5} C_{1}+{ }^{5} C_{5}$ $=2 \times{ }^{5} C_{1}+2 \times{ }^{5} C_{2}+{ }^{5} C_{5}$ $=2 \times 5+2 \times \frac{5 !}{2 ! 3 !}+1$ $=10+20+1$ $=31$...

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Thales belongs to the country

Question: Thales belongs to the country(a) India(b) Egypt(c) Greece(d) Babylonia Solution: (c) Greece...

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Solve the following

Question: If ${ }^{\left(a^{2}-a\right)} C_{2}={ }^{\left(a^{2}-a\right)} C_{4}$, then $a=$ (a) 2 (b) 3 (c) 4 (d) none of these Solution: (b) 3 $a^{2}-a=2+4 \quad\left[\because{ }^{n} C_{x}={ }^{n} C_{y} \Rightarrow n=x+y\right.$ or $\left.x=y\right]$ $\Rightarrow a^{2}-a-6=0$ $\Rightarrow a^{2}-3 a+2 a-6=0$ $\Rightarrow a(a-3)+2(a-3)=0$ $\Rightarrow(a+2)(a-3)=0$ $\Rightarrow a=-2$ or $a=3$ But, $a=-2$ is not possible. $\therefore a=3$...

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Euclid belongs to the country

Question: Euclid belongs to the country(a) India(b) Greece(c) Japan(d) Egypt Solution: (b) Greece...

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The 10th and 18th terms of an A.P.

Question: The 10thand 18thterms of an A.P. are 41 and 73 respectively. Find 26thterm. Solution: In the given problem, we are given 10thand 18thterm of an A.P. We need to find the 26thterm Here, $a_{10}=41$ $a_{18}=73$ Now, we will find $a_{10}$ and $a_{18}$ using the formula $a_{n}=a+(n-1) d$ So, $a_{10}=a+(10-1) d$ $41=a+9 d$........(1) Also, $a_{18}=a+(18-1) d$ $73=a+17 d$.......(2) So, to solve foraandd On subtracting (1) from (2), we get $8 d=32$ $d=\frac{32}{8}$ $d=4$ Substitutingd=4 in (1)...

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