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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Decide whether the number `0.120120012000012….` Is rational or not. Give reason to support your answer. |
| Answer» Clearly the given number 0.120120012000012… is a nonterminating and nonrepeating decimal. So, it is not rational. | |
| 2. |
Show that each of the following numbers is rational. What can you say about the prime factors of their denominators ? (i) 23.123456789 (ii) 32. `overline(123456789)` |
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Answer» (i) Clearly , the given number 23.123456789 is a terminationg decimal. So, it is rational and the prime factors of its denominatior are 2 or 5 or both. (ii) Clearly, the given number 32. `overline(123456789)` is a nonterminating repeating decimal. So, it is rational and the prime factors of its denominator are other than 2 or 5 also. |
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| 3. |
In a school there are two sections, namely A and B, of class X. there are `30` students in section A and `28` students in section B. find the minimum number of books required for their class library so that they can be distributed equally among students of section A ore section B. |
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Answer» Clearly, the required number of books are to be distributed equally among the students of section A or B. So , the number of these books must be a multiple of 30 as well as that of 28. Consequently, the required number is LCM( 30,28) Now ,` 30 = 2xx3xx5` and ` 28 = 2^(2)xx7` LCM ( 30,28) = product of prime factors with highest power . ` ( 2^(2) xx 3 xx 5xx7) = ( 4xx 3xx 5xx7) = 420` Hence , the required number of books = 420. |
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| 4. |
The mode of the following frequency distribution is 46 and the total frequency is 150. Find the missing frequencies x and y. |
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Answer» Here, we are given total frequency is `150`. `6+8+17+x+42+30+y+8 = 150` `=> x+y+111 = 150` `=>x+y = 39` It means, both `x` and `y` does not exceed `42`. So, our modal class wil be `40-50`. Now, Mode can be given by formula, `Mode = l+(f_1-f_0)/(2f_1-f_0-f_2)**h` Here, `l =` Lower limit of modal class `= 40` `f_1 = ` Frequency of Modal class `= 42` `f_2 = ` Frequency of Pre Modal class `= x` `f_3 = ` Frequency of Succeeding Modal class `= 30` `h =` Class interval `= 10` Putting these values in Mode formula, `Mode = 40+(42-x)/(84-30-x)**10 = 40+(42-x)/(54-x)**10` We are given, `Mode = 46` `=>40+(42-x)/(54-x)**10 = 46` `=> (42-x)/(54-x)**10 = 6` `=> (42-x)/(54-x) = 6/10` `=> (42-x)/(54-x) = 3/5` `=>210-5x = 162-3x` `=>2x = 48` `=> x = 24` `=>y = 39-24 => y = 15` |
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| 5. |
Prove that `sqrt(11)` is irrational. |
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Answer» Let, if possible, `sqrt(11)` be rational and its simplest form be`(a)/(b).` Then a and b are integers and having no common factor other than 1 and `b ne 0`. Now, `sqrt(11) = (a)/(b)` `rArr` `11 = (a^(2))/(b^(2))` `rArr` `a^(2) = 11b^(2)" "`...(1) `therefore` As `11b^(2)` is divisible by 11. `rArr` `a^(2)` is divisible by 11. `rArr` a is divisible by 11. Let a = 11c, for some integer c. From equation (1) `(11c)^(2) = 11b^(2)` `rArr` `b^(2) = 11c^(2)` But `11c^(2)` is divisible by 11. `therefore b^(2)` is divisible by 11. `rArr` b is divisible by 11. Let b = 11d, for some integer d. Thus, 11 is a common factor of a and b both. But it contradicts the fact that a and b have no common factor other than 1. So, our supposition is wrong. Hence, `sqrt(11)` is irrational. `" "`Hence Proved. |
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| 6. |
Prove that `sqrt(5)` is irrational. |
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Answer» `sqrt5=a/b` (wnere a and b are Co-prime Numbers) `a=sqrt5b` squaring both side `a^2=5b^2` `a^2` is divisible by 5 a is also divisible by 5 similarly, b is divisible by 5 which means 5 is common factor of a and b so, a/b is irrerational which makes `sqrt5` irrerational |
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| 7. |
In the given figure, ABCD is a cyclic quadrilateral; O is the centre of the circle.If `/_BOD=160^@`, then find `/_BPD`. |
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Answer» BD,`theta and theta/2` `/_BAD=1/2/_BOD=1/2*160=80` ABPD is also cyclic quadrilateral. `/_BAD+/_BPD=180^0` `/_BPD=180-80=100` `/_BPD=100^0`. |
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| 8. |
The product of two consecutive integers is divisible by 2. Is this statement true or false. Give Reason? |
| Answer» Yes, two consecutive integers can be `n(n+1)`. So, one number of these two must be divisible by 2. Hence, product of number is divisiblw by 2. | |
| 9. |
A girl is twice as old as her sister. four years hence , the product of their ages (in years) will be 160 . Find their present ages? |
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Answer» Let age of the girl in years ` = 2x` Then, age of her sister in years ` = x` We are given, `(x+4)**(2x+4) = 160` `=>2x^2+12x+16 = 160` `=>2x^2+12x-144 = 0` `=>x^2+6x-72 = 0` `=>x^2+12x-6x-72 = 0` `=>(x+12)(x-6) = 0` `x = -12 and x = 6` But, age can not be negative. `:. x = 6` `:.` Age of girl ` = 2**6 = 12` years Age of her sister `= 6 ` years |
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| 10. |
Prove that one of every three consecutive positive integers isdivisible by 3. |
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Answer» Any three consecutive positive integers must be of the form n, (n+1) and (n+2), where n is any natural number i.e. n=1,2,3… Let, a=n,b=n+1 and c=n+2 Order triplet is (a,b,c) (n,n+1,n+2), where n=1,2,3… (i) `"At n=1 (a,b,c)=(1,1+1,1+2)=(1,2,3)"` `"At n=2, (a,b,c)=(1,2+1,2+2)=(2,3,4)"` `"At n=3, (a,b,c)=(3,3+1,3+2)=(3,4,5)"` `"At n=4, (a,b,c)=(4,4+1,4+2)=(4,5,6)"` `"At n=5, (a,b,c)=(5,5+1,5+2)=(5,6,7)"` `"At n=6, (a,b,c)=(6,6+1,6+2)=(6,7,8)"` `"At n=7, (a,b,c)=(7,7+1,7+2)=(7,8,9)"` `"At n=8, (a,b,c)=(8,8+1,8+2)=(8,9,10)"` We observe that each triplet consist of one and only one number which is multiple of 3 i.e. divisible by 3 Hence one of any three consecutive positive integers must be divisible by 3. |
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| 11. |
If two positive integers `m`and `n` are expressible in the form `m=p q^3`and `n=p^3q^2`, where `p , q`are prime numbers, thenHCF `(m , n)=`A. `pq`B. `pq^(2)`C. `p^(3)q^(3)`D. `p^(2)q^(2)` |
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Answer» Correct Answer - B Given that `a-x^(3)y^(2)=x xx x xx x xx y xx y` and `b=xy^(3)= x xx y xx y xx y` `therefore HCF` of a and b `=HCF(x^(2)y^(2),xy^(3))=x xx y xx y=xy^(2)` [Since, HCF is the product of the smallest power of each common prime facter involed is the number] |
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| 12. |
If (m+1)th term of an A.P is twice the (n+1)th term, prove that (3m+1)th term is twice the (m+n+1)th term. |
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Answer» `n^(th)` term in an AP is given by, `T_n = a+(n-1)d` `:. T_(m+1) = a+md` `T_(n+1) = a+nd` We are given, `a+md = 2(a+nd)` `=> a+md = 2a+2nd` `=>a = (m-2n)d->(1)` Now, `T_(3m+1) = a+3md` From (1), `T_(3m+1) = (m-2n)d+3md` `= (m-2n+3m)d` `=(4m-2n)d` `T_(3m+1)=2(2m-n)d->(2)` Now, `T_(m+n+1) = a+(m+n)d` `=(m-2n)d+(m+n)d` `=(m-2n+m+n)d` `T_(m+n+1)=(2m-n)d->(3)` From (2) and (3), `T_(3m+1)= 2(T_(m+n+1))` |
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| 13. |
Show that one and only one out of `n`, `n +1` and `n +2` is divisible by `3`, where `n` is any positive integer. |
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Answer» On dividing n by 3, let q be the quotient and r be the remainder . Then, n = 3q +r , where `0 ge r gt 3` ` Rightarrow n = 3q + r , " where r " = 0 ,1 or 2 ` m ` Rightarrow n = 3q +r or n ( 3q +1) or n = ( 3q +2)` Case I If n = 3q then n is clearly divisible by 3. Case II If n= ( 3q +1) then ( n+2) = (3q +3) = 3(q +1) , which is clearly divisible by 3. In this case, ( n+2) is divisble 3. Case III If n = ( 3n +2) then (n+1) = ( 3q +3) = 3(q +1) , whihc is clearly divisible by 3. In this case, (n+2) is divsible by3 . In this case , (n +1) is divisible by3. Hence, one and only one out of n,( n +1) and (n+2) is divisible by 3. |
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| 14. |
A positive integer is the form of 3q+1 q, being a natural number. Can you write its square in any form other than 3m+1 i.e. 3m or 3m+2 for some integer? Justify your answer. |
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Answer» No, by Euclid Lemma, b=aq+rm `0 le r lt a` Here, b is any positive integer `a=3, b=3q+r for `0 le r lt 3` So, this must be in the form `3q,3q+1 or 3q+2` Now, `(3q)^(2) =9q^(2)=3m` [here, `m=3q^(2)`] and `(3q+1)^(2)=9q^(2)+6q+1` `=3(3q^(2)+2q)+1=3m+1` [where, `m=3q^(2)+2q`] Also, `(3q+2)^(2)=9q^(2)+12q+4` `=9q^(2)+12q+3+1` `3=(3q^(2)+4q+1)` `3m=1` [here, `m=3q^(2)+4q+1]` Hence, square of a positive integer is of the form 3q+1 is always in the form 3m+1 for some integer m. |
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| 15. |
Show that square of any positive integer is of the form of `5m,5m+1,5m-1` for some integer m. |
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Answer» `(5n-2)^2=25n^2+4-20n=5(n^2-4n+1)-1=5m-1` `(5n-1)^2=25n^2+1-10n=5(5n^2-22n)+1=5m+1` `(5n)^2=25n^2=5(5n^2)=5m`. |
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| 16. |
In the figure `AB=AC` angle `BAD=30^@ and AE=AD,` then the value of `x` is |
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Answer» In the given figure, `AB = AC` `:. /_ABC = ACB ` Let `/_ABC = ACB = theta` Also, `AE = AD` `:. /_ADE = /_AED` Let, `/_ADE = /_AED = phi` `:. /_CED = 180-phi` Now, in `Delta CED`, `x+theta+(180-phi) = 180` `=> phi-theta =x ->(1)` Now, in `Delta ABC`, `/_ABC+/_ACB+/_BAC = 180` `theta+theta+/_BAD+/_DAC = 180` `2 theta+30 +/_DAC = 180` `/_DAC = 150-2theta->(2)` In `Delta ADE` `2 phi+/_DAC = 180` `/_DAC = 180-2phi->(3)` From (2) and (3), `150-2theta = 180-2phi` `=>phi-theta = 15->(4)` From (1) and (4),`x = 15^@` |
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| 17. |
A can do as much work in 3 days as B can do in 5 days and B can do as much work in 2 days as C can do in 3 days. All of them together can do a certain piece of work in 20 days. The no. of days A alone take to do the work will be : |
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Answer» Ratio of ability of `A` and `B` to do a piece of work `= 5:3` Ratio of ability of `B` and `C` to do a piece of work `= 3:2` `:.` Ratio of ability of `A` and `C` to do a piece of work `= 5:2` `:.` Ratio of ability of `A`,`B` and `C` to do a piece of work = `5:3:2` Together `A,B and C` complete a certain work in `20` days. So, time taken by `A` to complete that work alone `= (5+3+2)/5**20 = 10/5**20 = 40` days |
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| 18. |
In the figure, CD, AE and BF are one-third of their respective sides. It follows that `AN_2 : N_2 N_1 : N_1 D= 3 : 3 : 1` and similarly for lines BE and CF. Then the area of triangle `N_1 N_2 N_3`. in term of`Delta ABC` is |
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Answer» `AE=1/3AC``area(/_ABE)=1/3area(/_ABC)` `area(/_ABN)=6/7area(/_ABC)` `=6/7*1/3area(/_ABC)` `=2/7 area(/_ABC)` `area(/_BFC)=1/3(/_ABC)` `area(BNC)=6/7area(/_BFC)` similarly, `area(/_BNC)=2/7area(/_ABC)` `area(ANC)=2/7area(/_ABC)` `area(N_1N_2N_3)=area(ABC)-3(2/7area(ABC))` =`area(ABC)/7`. |
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| 19. |
In the given figure, ABCD is a quadrilateral in which `/_D = 90^@`. A circle C(O,r) touches the side AB, BC, CD and DA at P,Q,R and S, respectively. If BC= 38 cm, CD = 25 cm and BP = 27 cm, then find the value of r. |
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Answer» We can create a diagram with the given details. Pleae refer to video for the diagram. We know, length of tangents from any common point outside a circle are always equal. `:. OS = RD = r` `RC = CD-RD = 25-r` Also, `CQ = RC = 25-r` `BQ = BC - CQ = 38-25-r = 13+r` Now, `BP = BQ = 13+r` It is given that `BP =27cm` `:. 13+r = 27 => r = 14cm` |
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| 20. |
Find the greatest possible length which can be used to measure exactly the lengths `7 m`, `3 m` `85 cm` and `12 m` `95 cm`. |
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Answer» Correct Answer - 35 cm Required length = HCF ( 700cm, 385cm, 1295 cm) = 35 cm |
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| 21. |
The sum of a number and its positive square root is 6/25.Find the number. |
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Answer» let number=n `n+sqrtn=6/25` `sqrtn=6/25-n` `n=36/625+n^2-12/25n` `n^2-12/25n-n+36/625=0` `625n^2-925n+36=0` `n=(925pmsqrt((925)^2-4*36*625))/(2*625)` `n=1.44,0.077` `n=1/25`. |
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| 22. |
Points X and Y are on sides BC and CD of square ABCD, as shown in the figure. The lengths of XY, AX and AY are 3, 4 and 5 units respectively. Then the side length of square ABCD is: |
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Answer» In triangle AXY, sides are `3,4, and 5`. These are Pythagoras triplets which means, triangle AXY is aright angle triangle. Let `/_XAB = theta` Then, `/_CXB` will also be `theta`. As sides of a square are equal.so, `AB =BC=>AB = BX+CX` `=>4costheta = 4sintheta+3costheta` `=>costheta = 4sintheta=>sintheta/costheta = 1/4 ` `tantheta = 1/4=> costheta = 4/sqrt(4^2+1^2) = 4/sqrt17` So, `AB = 4costheta = 16/sqrt17` so, side of the square will be `16/sqrt17`. |
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| 23. |
Points X and Y are on sides BC and CD of square ABCD,as shown in the figure. The lengths of XY, AX and AY are 3,4 and 5 respectively.Then the side length of the square ABCD is :a. `15/sqrt7`b. `16/sqrt7`c. `13/sqrt7`d. none of these |
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Answer» AD=CD, AD=DY+CY `5Costheta=5sintheta+3cos(90+theta-phi)` `5costheta=5sintheta+3cos(theta+37)` `5costheta=5sintheta+3(4/5costheta=3/5sintheta)` `5costheta-12/5costheta=5sintheta=9/5sintheta` `13/5costheta=16/5sintheta` `tantheta=13/16` `costheta=16/20.6` `a=5costheta=(16*5)/20.6=3.88`. |
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| 24. |
In the given figure, ABCD is a rectangle of 20 cm x 10 cm. A semicircle is drawn with centre O, radius `10sqrt2` cm and it passes through A and B. Find the area of the shaded region. [Use `pi=22/7`] |
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Answer» Area of APB=Area of sector OAPB-Area of `/_OAB` `=90/360*pi(10sqrt2)^2-1/2*20*10` Area of Area APB=400/7 Area of shaded region= Area of rectangle ABCD- Area of triangle APB `20*10-400/7` `(1400-400)/7` `1000/7 cm^2`. |
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| 25. |
Which term of an AP 20,19[1/4],18[1/2], 17[3/4],... is the first negative term? |
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Answer» `20,77/4,37/2,71/4,.....` `d=77/4-20=3/4` `d=(77-80)/4=-3/4` Let,` a_r<0` `a+(r-1)d<0` `20+(r-1)(-3/4)<0` `(r-1)(3/4)>20` `(r-1)>80/3` `r>83/3` `r=28` 28th term. |
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| 26. |
If a,b,c and d are four positive real numbers such that abcd=1 , what is the minimum value of `(1+a)(1+b)(1+c)(1+d)`. |
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Answer» Here, we are given `abcd = 1`. Now, `(1+a)(1+b)(1+c)(1+d) = 1+a+b+c+d+(ab+bc+cd+ad+bd+ac)+(abc+bcd+cda+dab)+abcd` `=1+a+b+c+d+(ab+bc+ac+1/(ab)+1/(bc)+1/(cd))+(1/a+1/b+1/c+1/d)+1`...(As abcd = 1) `=1+1+(a+1/a)+(b+1/b)+(c+1/c)+(d+1/d)+(ab+1/(ab))+(bc+1/(bc))+(ac+1/(ac))` Now, we know minimum value of `x+1/x` is `2`. `:. (1+a)(1+b)(1+c)(1+d) ge 1+1+2+2+2+2+2+2+2` `=> (1+a)(1+b)(1+c)(1+d) ge 16` `:.` Minimum value of given expression is `16`. |
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| 27. |
The sum of three numbers in an AP is 27 and their product is 405.Find the numbers. |
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Answer» `a-d,d,a+d` `a-d+d+a+d=27` `a=9` `(9-d)*9*(9+d)=45` `9^2-d^2=45` `d=6` `a-d=3` `a=9` `a+d=15`. |
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| 28. |
Show that the expression `(px^2 + 3x - 4)/( p + 3x - 4x^2)` will be capable of all values when x is real,provided that p has any value between 1 and 7. |
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Answer» `(px^2+3x-4)/(p+3x-4x^2)=y` `px^2+3x-4=py+3xy-4yx^2` `(4y+p)x^2+(3-3y)x-4-py=0` `D>=0` `9(1-y)^2+4(4+py)(4y+p)>=0` `8y^2-18y+9+64y+16p+16py^2+4p^2y>=0` `(9+16P)y^2+(46+4p^2)y+9+16p>=0` `D<=0` `(46+4p^2)^2-4(9+16p)^2<=0` `4[(23+2p^2)^2-(9+16p)^2]<=0` `(p^2+8p+16)(p^2-8p+7)<=0` `D<0` `p^2-8p+7<=0` `(p-1)(p-7)<=0` `P in [1,7]`. |
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| 29. |
For what value of k the following system of simultaneous equation has a unique solution-2kx + 3y + 3= 04x + y + 4= 0 |
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Answer» `a_1x+b_1y+c_1=0` `a_2x+b_2y+c_2=0` `a_1/a_1!=b_1/b_2` unique solution `2k+3y+3=0` `4x+y+4=0` `(2k)/4!=3/1` `2k!=12` `k!=6` k can have value energy=6. |
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| 30. |
A shopkeeper sells a table at 8% profit and a chair at 10% discount, thereby getting Rs 1008. If he had sold the table at 10% profit and chair at 8% discount, he would have got Rs. 20 more. Find the cost price of the table and the list price of the chair. |
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Answer» CP of chair -x CP of table=y Case I SP of table=`y+(8y)/100=108/100y` CP of chair=`x-10/100x=9/10x` According to question `108/100y+9/10x=1008` `108y+90x=100800` `6y+5x=5600-(1)` Case II SP of chair=`y+10/100y=11/10y` SP of table=`x-8/100x=92/100x` According to question `11/10y+92/100x=1028` `110y+92x=102800-(2)` solving equation 1 and2 x=400 y=600 Cost of chair=400Rs Cost of table=600Rs. |
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| 31. |
An army of pilot is flying an aeroplane at an altitude of 1800 m observes suspicious activity of two ships which are sailing towards it in the same directions and immediately report it to navy chief.The angles of depression of the ships as observed from the aeroplane are `60^@` and `30^@`,respectively.1. Find the distance between two ships.2. What values of the pilot are shown. |
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Answer» `In/_PAQ` `tan60^o=(QA)/(PQ)` `QA=1800sqrt3m` `In/_PQB` `tan30^o=(OB)/(PQ)` `QB=1800/sqrt3=600sqrt3m` Distance between ships=`QA-QB=1800sqrt3-600sqrt3` `=1200sqrt3m`. |
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| 32. |
The length and breadth of a rectangle are in the ratio 4:3 If the diagonal measures 25 cm then perimeter of the rectangle is? |
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Answer» let breadth = `3x` length = `4x` by pythagoras theorem `(4x)^2 + (3x)^2 = (25)^2 ` `16x^2 + 9x^2 = 25 xx 25` `25x^2 = 25 xx 25` `x= 5` L=`4x = 4 xx5 = 20` B= `3x = 3 xx 5 = 15` Perimeter= `2(L+B)` `= 2[20 + 15]` `=70 ` cm L= `4x = 4 xx 5 = 20cm` B= `3x = 3 xx 5 = 15 cm` Answer |
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| 33. |
If 36 is exactly divisible by a, then the GCD of 36 and a is _____ |
| Answer» Correct Answer - a | |
| 34. |
Find the greatest number of four digits which is exactly divisible by 15, 24 and 36. |
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Answer» Correct Answer - 9720 Greatest number of four digits = 9999 LCM ( 15, 24,36) = 360 on dividing 9999 by 360, reaminder = 279. required number = ( 9999 - 279)= 9720 |
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| 35. |
How many composite numbers are in between 50 and 100 (inclusive of 50 and 100)?A. 39B. 40C. 41D. 42 |
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Answer» Correct Answer - C (a) Find the numebers of primes between 50 and 100. (b) Subtract that number from 51. |
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| 36. |
Write the denominator of the rational number `257/5000` in the form `2^m xx 5^n`, where m, n and non-negative integers. Hence, write its decimal expansion without actual division. |
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Answer» Denominator of the raitonal number `(257)/(5000) "is" 5000`. Now, factors of 5000 `=2 xx 2xx 2xx 5xx 5xx 5xx 5=(2)^(3)`, which is of the type `2^(m)xx6(n)`, where m=3 and n=4 are non-negative integers `therefore` Rational number `=(257)/(5000)=(257)/(2^(3)xx5^(4)xx(2)/(2)` [Since, multiplying numerator and denominator by 2] `=(514)/(2^(4)xx5^(4))=0.00514` Hence, which is the required decimal expansion of the rational `(257)/(5000)` and it is also terminating decimal number. |
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| 37. |
Find the least number which should be added to 2497 so that the sum is exactly divisible by 5,6,4and 3 |
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Answer» Correct Answer - 23 LCM of 5,6,4 and 3 = 60 On dividing 2497 by 60 remainder = 37 |
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| 38. |
Find the least number that is divisible by allthe numbers between 1 and 10 (both inclusive).A. 100B. 1260C. 2520D. 5040 |
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Answer» Correct Answer - C Required number = LCM { 1,2,3,4,5,6,7,8,9,10} , = LCM : 1,2,3,`2^(2) , 5,2 xx 3, 7,2^(2) , 3^(2) , 2xx5` `= ( 1xx 2^(3) xx 3^(2) xx 5 xx 7) = ( 8 xx 9 xx 5xx 7) = 2520` |
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| 39. |
What is the least number that is divisible by all the numbers `1` to `10`A. `10`B. `100`C. `504`D. `2520` |
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Answer» Correct Answer - D Factors of 1 to 10 1=1 `2=1xx2` `3=1xx3` `4=1 xx 2 xx 2` `5=1 xx 5` `6=1 xx 2 xx 3` `7=1 xx 7` `8=1 xx 2 xx 2 xx 2` `9=1 xx 3 xx 3` `10=1 xx 2xx 5` `therefore` LCM of number 1 to 10 =LCM (1,2,3,4,5,6,7,8,9,10) `=1 xx 2 xx 2 xx 2xx 3xx 3xx 5xx 7=2550` |
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| 40. |
The least number that is divisible by all the numbers from 1 to 5 is.A. 70B. 60C. 50D. 40 |
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Answer» Correct Answer - B |
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| 41. |
Find the greatest four - digits number which when divided by 4,7 and 13 leaves a remainder 3 in each case. |
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Answer» Correct Answer - 9831 Required number = ( greatest number of 4-digits divible by 4,7 and 13) + 3. |
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| 42. |
The multiplication of two irrational numbers is:A. always irrationalB. always rationalC. can be irrational or rationalD. none of these |
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Answer» Correct Answer - C |
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| 43. |
Find the LCM and HCF of the following pairs of integers and verify that `L C MxxH C F=`product of the two numbers. `(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54` |
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Answer» 1)510=2*3*5*17 92=2*2*23 HCF=2 LCM=2*2*3*5*17*23 HXF*LCM=510*92 Proved 2)336=2*2*2*2*3*7 54=2*3*3*3 HCF=2*3=6 LCM=2*3*9*8*7 LCM*HCF=2*2*2*2*2*3*3*3*37=5*36 Proved |
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| 44. |
Sum of a number and its reciprocal is `5/2`. Find the number. |
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Answer» Let the number is `x`. Then, its reciprocal will be `1/x`. So, `x+1/x = 5/2` `=>(x^2+1)/x = 5/2``=>2(x^2+1) = 5x` `=>2x^2-5x+2 = 0` `=>2x^2-4x-x+2 = 0` `=>2x(x-2)-1(x-2) = 0` `=>(2x-1)(x-2) = 0` `:. x = 2 and x = 1/2` So, `2` and `1/2` are the required numbers. |
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| 45. |
Range of the function `f(x)= "^(7-x)P_(x-3) `is given by(1) (13, 4, 5)(3) (1, 2, 3) |
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Answer» `nP_m` stands form `P_m=(n!)/((n-m)!)` `m<=n` n,m shall be there n,m=0,1,2,3... `x-3<=7-x` `2x<=10` `x<=5` 0,1,2,3,4,5. `7-x>=0,x-3>=0` `x<=T` at `x>=3` `f(x)=((7-x)!)/((10-2x)!)` `f(3)=(4!)/(4!)=1` `f(4)=(3!)/(2!)=3` `f(5)=(2!)/(0!)=2` Range{1,2,3} option 3 is correct. |
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| 46. |
A rational number in its decimal expansion is 327.7081. What would be the prime factors of q when the number is expressed in the p/q form?A. 2 and 3 B. 3 and 5C. 2, 3 and 5 D. 2 and 5 |
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Answer» Correct Answer - D |
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| 47. |
Find the LCM and HCF of the following integers by applying the prime factorisation method.` (i) 12,15 and 21 (ii) 17, 23 and 29 (iii) 8,9 and 25` |
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Answer» 1)12,15,21 factors 12=2*2*3 15=3*5 21=3*7 HCF=3 LCM=3*2*2*5*7=420 2)17,23,29 factors 17=1*17 23=1*23 29=1*29 HCF=1 LCM=17*23*29=11339 3)8,9,25 factors 8=2*2*2 9=3*3 25=5*5 HCF=1 LCM=8*9*25=1800 |
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| 48. |
Give prime factrisation of 4620. |
| Answer» Correct Answer - 72 | |
| 49. |
Find the HCF of 1008 and 1080 by prime factorisation method. |
| Answer» Correct Answer - HCF =` 2/81 , LCM = `80/9` | |
| 50. |
Using prime factorisation, find the HCF and LCM of : , (i) 36, 84 (ii) 23,31 (iii) 96,404 (iv) 144,198 (v) 396, 1080 (vi) 1152, 1664 In each case, verify that : HCF ` xx` LCM = product of given number . |
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Answer» Correct Answer - (i) HCF= 12, LCM= 252, (ii) HCF= 1, LCM= 713 (iii) HCF = 4, LCM= 9696, (iv) HCF = 18, LCM= 1548 (v) HCF= 36, LCM= 11880, (iv) HCF = 18, LCM= 14976 |
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