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This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
`Let `veca=4veci+3vecj and vecb=3veci+4vecj`. a.Find the magnitudes of a. veca, b. `vecb, c. veca+vecb and d. veca-vecb.` |
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Answer» Correct Answer - (a) 5 (b) 5 (c) `7sqrt(2)` (d)2 `veca=4veci+vecj, vecb=3veci+4vecj` (a). `|vecalpha|=sqrt(9+16)=5` (b).`|vecb|=|3veci+4vech|=sqrt(9+16)=5` (c).`|veca-vecb|=|7veci+7vecj|=sqrt(49+49)` `sqrt(98)=7sqrt(2)` (d). `veca-vecb=(4veci+3vecj)-(3veci+4vecj)`=`veci-vecj, rarr |veca-vecb|=sqrt((1)^2+(-1)^2)=2` |
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| 2. |
Consider the quadratic equation `az^(2)+bz+c=0` where `a,b,c` are non-zero complex numbers. Now answer the following. The condition that the equation has both roots purely imaginary isA. `(bar(a))/a=(bar(b))/b=(bar(c))/c`B. `(bar(a))/a=-(bar(b))/b=(bar(c))/c`C. `(bar(a))/a=(bar(b))/b=-(bar(c))/c`D. `-(bar(a))/a=(bar(b))/b=(bar(c))/c` |
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Answer» Correct Answer - B Let `alpha` be one root of equation then `bar(alpha)=-alpha` Given equation is `az^(2)+bz+c=0` `:.aalpha^(2)+balpha+c=0` Taking conjugate we get `:.bar(a).bar(alpha)^(2)+bar(b).bar(alpha)+bar(c)=0` `bar(a).alpha^(2)-bar(b).alpha+bar(c)=0` so `alpha.beta` are roots of equation `az^(2)+bz+c=0` and `bar(a)z^(2)-bar(b)z+bar(c)=0` So `(bar(a))/a=-(bar(b))/b=(bar(c))/c` if `|alpha|=1` then `bar(alpha)=1/(alpha).` As `aalpha^(2)+balpha+c=0`..........1 `impliesbar(a)bar(alpha)^(2)+bar(b)bar(alpha)+bar(c)=0` `implies(bar(a))/(alpha^(2))+(bar(b))/(alpha)+bar(c)=0` So `bar(c).alpha^(2)+bar(b).alpha+bar(a)=0`.........2 Now applying condition for one common root for 1 and 2 we get `(a bar(b)-b bar(c))(b bar(a)-c bar(b))=(a bar(a)-c bar(c))^(2)` |
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| 3. |
if `E_(1)` and `E_(2)` are two events such that `P(E_(1))=(1)/(4),P((E_(2))/(E_(1)))=(1)/(2)` and `P=((E_(1))/(E_(2)))=(1)/(4)`A. then `E_(1)` ad `E_(2)` are independentB. `E_(1)` and `E_(2)` are exhaustiveC. `E_(2)` is twice as likely to occur as `E_(1)`D. probabilities of the events `E_(1)capE_(2),E_(1)` and `E_(2)` are in G.P. |
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Answer» Correct Answer - A::C::D `P(E_(1))=(1)/(4)` `P_((E_(2))/(E_(1)))=(P(E_(1)capE_(2)))/(P(E_(1)))=(1)/(2)` `P(E_(1)capE_(2))=(1)/(8)` `P((E_(1))/(E_(2)))=(P(E_(1)capE_(2)))/(P_(E_(2)))=(1)/(4)` `P(E_(2))=(1)/(2)` `P_(E_(1)).P_(E_(2))=P(E_(1)capE_(2))` |
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| 4. |
If the vertex = (2,0) and the extremities of the latus rectum are (3, 2) and (3, -2), then the equation of the parabola is: |
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Answer» we know equation of parabola is `y^2=4ax` `(y-0)^2=4a(x-2)` `y^2=4a(x-2)` from diagram we can see that the value of a is 1 unit `y^2=4(x-2)` `y^2=4x-8`(eq of parabola) |
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| 5. |
IF `veca=2veci+3vecj+4veck and vecb =4veci+3vecj+2veck` find the angle between `veca and vecb`. |
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Answer» We have `veca.vecb=ab costheta` or `costheta= (veca.vecb)/(ab)` where `theta` is the angle between `veca and vecb` Now `veca.veb=a_xb_x+a_yb_y+a_zb^z` `=2xx4+3xx3+4xx2=25` Also, `veca=sqrt(a_x^2+a_y^2+a_z^2) `=sqrt(4+9+16)= sqrt29` and ` b= sqrt(b_x^2+b_u^2+b_z^2= sqrt(16+9+4)= sqrt(16+9+4)=sqrt29` Thrus, `costheta = 25/29` `or, `theta= cos^-1 (25/29)`. |
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| 6. |
Let `veca=2veci+3vecj+4veck and vecb=3veci+4vecj+5veck`. Find the angle between them. |
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Answer» Correct Answer - `cos^-1((38)/sqrt(1450))` Here `veca=2veci+3veci+4veck` and`vecb=3veci+4vecj+5veck` `veca.vecb=abcostheta` `theta= cos^-1((veca.vecb)/(ab))` =`cos^-1 (2xx3+3xx4+4xx5)/(sqrt(2^2+3^2+4^2)sqrt(3^2+4^2+^2)) =cos^-1 ((38)/(sqrt(29)sqrt(50))) = cos^-1 ((38)/sqrt(1450))` |
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| 7. |
Consider the quadratic equation `az^(2)+bz+c=0` where a,b,c and non-zero complex numbers. Now answer the following: Q. The condition that the equation has one complex root `alpha` such that `|alpha|=1`, isA. `(overline(b)c-boverline(a))/(aoverline(a)-coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b+aoverline(b))`B. `(overline(b)(c)+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b+aoverline(b))`C. `(overline(b)c-boverline(a))(overline(c)b-aoverline(b))=(aoverline(a)-coverline(c))^(2)`D. `(overline(b)c+boverline(a))/(aoverline(a)+coverline(c))=(aoverline(a)+coverline(c))/(overline(c)b-aoverline(b))` |
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Answer» Correct Answer - C Let `alpha` be one root of equation then `overline(alpha)=-alpha` Given equation is `az^(2)+bz+c=0` `becauseaalpha^(2)+balpha+c=0` Taking conjugate we get `becauseoverline(alpha)overline(alpha)^(2)+overline(b).overline(c)=0` `overline(a).alpha^(2)-overline(b).apha+overline(c)=0` and `overline(a)z^(2)-overline(b)z+overline(c)=0` So `(overline(a))/(a)=(overline(b))/(b)=(overline(c))/(c)` if `|alpha|=1` then `overline(alpha)=(1)/(alpha)` as `aalpha^(2)+balpha+c=0` ..(i) `impliesoverline(a)overline(alpha)^(2)+overline(b)overlinealpha)+overline(c)=0` `implies(overline(a))/(alpha^(2))+(overline(b))/(alpha)+overline(c)=0` So `overline(c).alpha^(2)+overline(b).alpha+overline(a)=0` .(ii) Now appying condition for one common root for (1) and (@) we get `(aoverline(b)-boverline(c))(boverline(a)-coverline(b))=(aoverline(a)-coverline(c))^(2)` |
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| 8. |
Lines whose equation are `(x-3)/2=(y-2)/3=(z-1)/(lamda)` and `(x-2)/3=(y-3)/2=(z-2)/3` lie in same plane, then. Angle between the plane containing both lines and the plane `4x+y+2z=0` isA. `(pi)/3`B. `(pi)/2`C. `(pi)/6`D. `cos^(-1)(1/(sqrt(186)))` |
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Answer» Correct Answer - B Normal vector to plane containing lines is `vec(n)=|(hati, hatj, hatk),(2, 3, -1),(3, 2, 3)|=1hat(i)+6hatj-5hat(k)` So angle between planes `costheta=(4+6-10)/(sqrt(21)sqrt(62))=0impliestheta=(pi)/2` |
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| 9. |
Lines whose equation are `(x-3)/2=(y-2)/3=(z-1)/(lamda)` and `(x-2)/3=(y-3)/2=(z-2)/3` lie in same plane, then. The value of `sin^(-1)sinlamda` is equal toA. `3`B. `pi-3`C. `4`D. `pi-4` |
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Answer» Correct Answer - D Line are in same plane to `|(1, -1, -1),(2, 3, lamda),(3, 2, 3)|=0implies1(9-2lamda)+1(-3lamda)-1(-5)=0` `20-5lamda=0implieslamda=4` So `sin^(-1)sin4=sin^(-1)sin(pi-4)=pi-4` |
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| 10. |
A die is rolled and showing any number is directly proportional to that number. If prime number appears then a ball is chosen from urn A containing 2 white and 3 block balls otherwise a ball is chosen from urn B containing 3 white and 2 block balls then. Q. The probability of drawing a block ball isA. `(53)/(105)`B. `(52)/(105)`C. `(49)/(105)`D. `(51)/(105)` |
| Answer» Correct Answer - B | |
| 11. |
Lines whose equation are `(x-3)/2=(y-2)/3=(z-1)/(lamda)` and `(x-2)/3=(y-3)/2=(z-2)/3` lie in same plane, then. Angle between the plane containing both lines and the plane `4x+y+2z=0` isA. `(pI)/(3)`B. `(pi)/(2)`C. `(pi)/(6)`D. `cos^(-1)((1)/(sqrt(186)))` |
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Answer» Correct Answer - B normal vector to plane containing lines is `vecn=|{:(hati,hatj,hatk),(2,3,-1),(3,2,3):}|=hati+6hatj+5hatk` So angle between planes `costheta=(4+6-10)/(sqrt(21)sqrt(62))=0impliestheta=(pi)/(2)` |
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| 12. |
A die is rolled and showing any number is directly proportional to that number. If prime number appears then a ball is chosen from urn A containing 2 white and 3 block balls otherwise a ball is chosen from urn B containing 3 white and 2 block balls then. Q. If white ball is draw then the probability that it is from urn BA. `(52)/(53)`B. `(1)/(53)`C. `(20)/(53)`D. `(33)/(53)` |
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Answer» Correct Answer - D `P(i)alphai` `P(i)=Ki` `P(1)=K,P(2)=2K,…P(6)=6K` `K+2K+….+6K=1` `K(1+2+…+6)=1` `K=(1)/(21)` `P(1)=(1)/(21).P(2)=(2)/(21),….P(6)=(6)/(21)` P (prime number) `=(2)/(21)+(3)/(21)+(5)/(21)=(10)/(21)` P (not prime) `=(11)/(21)` ltbr `P(B)=(10)/(21)+(3)/(4)+(11)/(21)xx(2)/(5)=(52)/(105)` `P(w)=(10)/(21)xx(2)/(5)+(11)/(21)xx(3)/(5)=(53)/(105)` `P((urnB)/(w))=((33)/(105))/((53)/(105))=(33)/(53)` |
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| 13. |
If `(1+x+x^(2))^(3n+1)=a_(0)+a_(1)x+a_(2)x^(2)+…a_(6n+2)x^(6n+2)`, then find the value of `sum_(r=0)^(2n)(a_(3r)-(a_(3r+1)+a_(3r+2))/2)` is |
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Answer» `(1+x+x^(2))^(3n+1)=a_(0)+a_(1)x+a_(2)x^(2)` `+a_(6n)x^(6n)+a_(6n+1)x^(6n+1)+a_(6n+2)x^(6n+2)` putting `x=omega` and `omega^(2)` we get `(a+omega+omega^(2))^(3n+1)=0=a_(0)+a_(1)omega+a_(2)omega^(2)+a_(3)+…+a_(6n)+a_(6n+1)omega+a_(6n+2)omega^(2)` .(i) `(1+omega^(2)+omega^(4))^(3n+1)=0=a_(0)+a_(1)omega^(2)+a_(2)omega+a_(3)+....+a_(6n+1)omega^(2)+a_(6n+2)omega` ..(ii) adding (i) and (ii) we get `sum_(r=0)^(2n)(2a_(3r)-(a_(3r+1)+a_(3r+2)))=0` |
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| 14. |
Given `abclt0` and `a+b+cgt0`. If `(|a|)/a+(|b|)/b+(|c|)/c=x`,then find the value of `x^(3)+16x-8`. |
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Answer» Correct Answer - 9 Out of `a,b,c` any one is `-ve` and are `+ve`. |
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| 15. |
Given `abclt0` and `a+b+cgt0`. If `(|a|)/(a)+(|b|)/(b)(|c|)/(c)=x`, then find the value of `x^(3)+16x-8` |
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Answer» Correct Answer - 9 Out of a,b,c any one is `-ve` and two are `+ve` |
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| 16. |
The possible number of ordered triplets `(m,n,p)` where `m,n,p epsilon N` is `(6250k)` such that `1le-mle100,1lenle50,1leple25` and `2^(m)+2^(n)+2^(p)` is divisible by `3` then `k` is |
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Answer» Correct Answer - 5 Here `2^(m)+2^(n)+2^(p)=(3-1)^(m)+(3-1)^(n)+(3-1)^(p)=3k+(-1)^(m)+(-1)^(n)+(-1)^(p)` So that `2^(m)+2^(n)+2^(p)` is divisible by `3` if `m,n,p` all are odd or all are even `implies` Number of possible ordered triplets `=50xx25xx12+50xx25xx13=31250=6250xx5` |
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| 17. |
let `|{:(1+x,x,x^(2)),(x,1+x,x^(2)),(x^(2),x,1+x):}|=(1)/(6)(x-alpha_(1))(x-alpha_(2))(x-alpha_(3))(x-alpha_(4))` be an identity in x, where `alpha_(1),alpha_(2),alpha_(3),alpha_(4)` are independent of x. Then find the value of `alpha_(1)alpha_(2)alpha_(3)alpha_(4)` |
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Answer» Correct Answer - 6 Since it is an identity the value of `L.H.S` and R.H.S are equal for all values of x put `x=0` `impliesalpha_(1)alpha_(2)alpha_(3)alpha_(4)=6` |
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| 18. |
The possible number of ordered triplets (m, n, p) where `m,np in N` is (6250K) such that `1ltmlt100,1ltnlt50,1ltplt25` and `2^(m)+2n^(n)+2^(p)` is divisible by 3 then k is |
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Answer» Correct Answer - 5 Here `2^(m)+2^(n)+2^(p)=(3-1)^(m)+(3-1)^(n)+(3-1)^(p)=3k+(-1)^(m)+(-1)^(n)+(-1)^(p)` (`k in 1`) So that `2^(m)+2^(n)+2^(p)` is divisible by 3 if m, n , p all are odd or all are even `implies` number of possible ordered triples `=50xx25xx12+50xx25xx13=31250=6250xx5` |
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| 19. |
1 mole of a real gas changes it state from state-A(2bar, 3L, 100 K) to state -B (2bar, 5L, 200 K) at constant pressure and finally to state-C (3bar, 10 L, 300 K). If `DeltaU_(BC) = 110 J` and `C_(Pm)` of gas `=3R = 3 xx 8.3 JK^(-1)mol^(-1)` then thoose the correct option(s) :A. `W_(AB) = 830 J`B. `DeltaH_(AC) = 4600 J`C. `Delta U_(AC) = 2200 J`D. `DeltaU_(AC) = 1770 J` |
| Answer» Correct Answer - B::C | |
| 20. |
The order and degree of the differential equation \([1+(\frac{dx}{dy})^3]^\frac73 =7(\frac{d^2y}{dx^2})\) are respectively.(A) 2, 3 (B) 3, 2 (C) 2, 2 (D) 3, 3 |
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Answer» (A) 2,3 \([1+(\frac{dx}{dy})^3]^\frac73 =7(\frac{d^2y}{dx^2})\) Cubing on both sides, we get \([1+(\frac{dx}{dy})^3]^7 =7^3(\frac{d^2y}{dx^2})^3\) By definition of degree and order, Degree: 3 ; Order: 2 |
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| 21. |
Given three vectors `veca,vecbandvecc` each two of which are non-collinear. Further if `(veca+vecb)` is collinear with `vecc,(vecb+vecc)` is collinear with `vecaand|veca|=|vecb|=|vecc|=sqrt(2)`. Then the value of `veca.vecb=vecb.vecc+vecc.veca=`A. 3B. -3C. 0D. cannot be evaluated |
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Answer» Correct Answer - B `veca+vecb=lamdavecc` and `vecb+vecc=muveca` `therefore(lamdavecc-veca)+vecc=muveca("putting "vecb=lamdavecc-veca)` `rArr(lamda+1)vecc=(mu+1)veca` `rArrlamda=mu=-1` ltBrgt `rArrveca+vecb+vecc=0` ltBrgt `rArr|veca|^(2)+|vecb|^(2)+|vecc|^(2)+2(veca.vecb+vecb.vecc+vecc.veca)=0` `rArrveca.vecb+vecb.vecc+vecc.veca=-3` |
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| 22. |
The equation `2sin^3theta+(2lambda-3)sin^2theta-(3lambda+2)sintheta-2lambda=0`has exactly three roots in `(0,2pi)`, then `lambda`can be equal to0 (b)2 (c) 1(d) `-1`A. -1B. 0C. `(1)/(2)`D. 1 |
| Answer» Correct Answer - ad | |
| 23. |
If the equation`2^((2pi)/cos^(-1)x)-(a+1/2)2^((pi)/cos^(-1)x-a^2=0` has exactly one real solution the range of `a` is equal toA. `(-3, 1)`B. `(-oo,-3]`C. `[1,oo)`D. `[-3,oo)` |
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Answer» Correct Answer - B::C Let `2^((pi)/(cos^(-1)x))=timpliestge2` `:.` equation becomes `t^(2)-(a+1/2)t-a^(2)=0` has one root 2 or greater than 2 & other root less than `2impliesf(2)le0` `implies4-(a+1/2):2-a^(2)le0` `a^(2)+2a-3ge0` `(a+3)(a-1)ge0` `impliesale-3` or `age1` |
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| 24. |
Consider the integral `"l"_(1)=int_(1)^(e)(1+x)(x+Inx)^(100)dx. "l"_(2)=int_(sin^(-1)(1//e))^(pi//2) (1 + esinx + Insinx)^(101)cos x dx` If `l_(1)+e/101 l_(2)=(e(1+e)^(101)-k)/101` then `kge` |
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Answer» Correct Answer - A::B::D `I_(1)=int_(1)^(e)x ((1+x)/x)(x+In x)^(100) dx` `=(e(e+1)^(101)-1)/101-1/101 int_(1)^(e)(x+In)^(101)dx` `I_(2)=1/e int_(1)^(e)(t+Int)dt` on taking `e sin x=t` `implies I_(1)+(eI_(2))/101=(e(e+1)^(101)-1)/101impliesK=1` |
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| 25. |
The equation of the tangent to the curve `y=sqrt(9-2x^(2))` at the point where the ordinate & the abscissa are equal isA. `2x+y-sqrt3=0`B. `2x+y-3=0`C. `2x-y-3sqrt(3)=0`D. `2x+y-3sqrt(3)=0` |
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Answer» Correct Answer - D Let the point `(x,y)` on tangent to the curve Given that `x_(1)=y_(1)` `:.x_(1)=sqrt(9-2x_(1)^(2))` `impliesx_(1)^(2)=9-2x_(1)^(2)impliesx_(1)= +- sqrt(3)` Since `ygt0`, therefore the point is `(sqrt(3),sqrt(3))` Also `y=sqrt(9-2x^(2))impliesy^(2)=9-2x^(2)` differentiate it `2y.(dy)/(dx)=-4ximplies(dy)/(dx)=(-2x)/y` `:.((dy)/(dx))_(((sqrt(3),sqrt(3)))=-2` So, the equation of tangents is `(y-sqrt(3))=-2(x-sqrt(3))` `implies2x+y-3sqrt(3)=0` |
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| 26. |
`lim_(xrarroo) (e^(1/x^(2))-1)/(2tan^(1)(x^(2))-pi)=`A. `-1/4`B. `-1/2`C. `1/2`D. Does not exists |
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Answer» Correct Answer - B `lim_(xrarroo) (e^(1/(x^(2)))-1)/(2tan^(-1)(x^(2))-pi)=lim_(xrarroo) (e^(1/(x^(2)))-1)/(2((pi)/2-cot^(-1)x^(2))-pi)` `lim_(xraroo)=(e^(1/(x^(2)))-1)/(-2tan^(-1)1/(x^(2)))=-1/2` |
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| 27. |
Let `D_(r) = |(a,2^(r),2^(16) -1),(b,3(4^(r)),2(14^(16) -1)),(c,7(8^(r)),4(8^(16) -1))|`, then the value of `underset(k =1)overset(16)Sigma D_(k)`, is |
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Answer» Correct Answer - A `sum_(r=1)^(16)S_(r)=|(a, sum_(r=1)^(16)2^(r),2^(16)-1),(b, 3(sum_(r=1)^(16)4^(r)), 2(4^(16)-1)),(c, 7(sum_(r=1)^(16)8^(r)),4(8^(16)-1))|` `=|(a, (2(2^(16)-1))/(2-1),2^(16)-1),(b, 3xx4((4^(16)-1))/(4-1), 2(4^(16)-1)),(c, 7xx8((8^(16)-1))/(8-1), 4(8^(16)-1))|` `=|(a, 2(2^(16)-1),2^(16)-1),(b, 4(4^(16)-1),2(4^(16)-1)),(c,8(8^(16)-1),4(8^(16)-1))|=0` |
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| 28. |
Let `hatu, hatv, hatw` be three unit vectors such that `hatu+hatv+hatw=hata, hata.hatu=3/2, hata.hatv=7/4` & `|hata|=2`, thenA. `hatu hatv=3/4`B. `hatu.hatw=0`C. `hatu. Hatv=1/4`D. `hatu.hatw=-1/4` |
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Answer» Correct Answer - A::D `hatu.hatv+hatu.hatw=1/2`…(ii) (taking dot product with `hatu,hatv` & `hata`) `hatu.hatv+hatv.hatw=3/4`………(iii) `hata.hatw=3/4`…………..(iv) Again taking dot product with `hatw` we get `hatu.hatw+hatv.hatw+|hatw|^(2)=hata.hatw` `hatu.hatw+hatu.hatw+1=3/4` `hatu.hatw+hatu.hatw=(-1)/4`............(v) on adding (ii),(iii),(v) `hatu.hatv+hatv.hatw+hatu.hatw=1/2` on substrating equation (ii) from (iv) `hatv.hatw=0` |
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| 29. |
Let `veca` and `vecb` be unit vectors such that `|veca+vecb|=sqrt(3)`, then the value of `(2veca+5vecb)`. `(3veca+vecb+vecaxxvecb)=`A. `11/2`B. `13/2`C. `39/2`D. `23/2` |
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Answer» Correct Answer - C `|veca+vecb|=sqrt(3)` `implies1+1+2veca.vecb=3impliesveca.vecb=1/2` Now, `(2veca+5vecb).(3veca+vecb+vecaxxvecb)` `=6+5+17 veca.vecb=39/2` |
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| 30. |
Let `veca` and `vecb` be unit vectors such that `|veca+vecb|=sqrt(3)`, then the value of `(2veca+5vecb)*(3veca+vecb+vecaxxvecb)=`A. `11/2`B. `13/2`C. `39/2`D. `23/2` |
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Answer» Correct Answer - C `|veca+vecb|=sqrt(3)` `implies1+1+2veca.vecb=3impliesveca.vecb=1/2` Now, `(2veca+5vecb).(3veca+vecb+vecaxxvecb)` `=6+5+17 veca.vecb=39/2` |
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| 31. |
What is the value of b such that the scalar product of the vector î + ĵ + k̂ with the unit vector parallel to the sum of the vectors 2î + 4ĵ - 5k̂ and bî + 2ĵ + 3k̂ is unity?1. -22. -13. 04. 1 |
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Answer» Correct Answer - Option 4 : 1 Calculations: Parallel vector = the sum of the vectors 2î + 4ĵ - 5k̂ and bî + 2ĵ + 3k̂ ⇒ Parallel vector = (2î + 4ĵ - 5k̂) + (bî + 2ĵ + 3k̂) ⇒ Parallel vector = (2 + b)î + 6ĵ - 2k̂ Now, unit vector parallel to the sum of the vectors 2î + 4ĵ - 5k̂ and bî + 2ĵ + 3k̂ = \(\rm \dfrac {(2 + b) \hat i+ 6 \hat j - 2\hat k}{\sqrt {(2 + b)^2 + 6^2 + 2^2}}\) Given, scalar product of the vector î + ĵ + k̂ with the unit vector is unity So, \(\rm \dfrac {(2 + b) \hat i+ 6 \hat j - 2\hat k}{\sqrt {(2 + b)^2 + 6^2 + 2^2}}\) . (î + ĵ + k̂) = 1 ⇒ \(\rm \dfrac {(2 + b) + 6 - 2}{\sqrt {(2 + b)^2 + 6^2 + 2^2}}\) = 1 ⇒\(\rm {(2 + b) + 6 - 2} ={\sqrt {(2 + b)^2 + 6^2 + 2^2}}\) Squaring both sides, we get ⇒ (b + 6)2 = \(\rm(2 + b)^2 + 6^2+ 2^2\) ⇒ b2 + 12b + 36 = b2 + 4b + 44 ⇒ 8b = 8 ⇒ b = 1 |
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| 32. |
एक बर्तन में 60 लीटर दूध है उसमे से 12 लीटर दूध निकालकर पानी भर दिया जाता है फिर से उस मिश्रण में से 12 लीटर दूध निकालकर पानी भर दिया जाता है परिणामी मिश्रण में दूध और पानी का अनुपात क्या होगा ?A. `16:9`B. `15:10`C. `16:10`D. `9:5` |
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Answer» Correct Answer - A total milk = 60 litres drawn off = 12 litres `("Final Quantity")/("Initial Quantity")=(1-x/c)^(t)` x= Replaced Quantity C= Capacity T= number of process `("Final Quantity")/("Initial Quantity")=(1-12/60)^(2)` `=(4/5)^(2)=16/25` Ratio of milk and water in the resultant mixture (परिणामी मिश्रण में दूध और पानी का अनुपात ) `=16:9` |
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| 33. |
निकिता ने 30 किलो गेहू `9.50` रुपये प्रति किग्रा और 40 किलो गेहू `8.50` रुपये प्रति किग्रा से ख़रीदे दोनों को मिलाने के बाद वह मिश्रण को `8.90` रुपये प्रति किग्रा से बेचती है तो बताओ पूरी प्रक्रिया में कितना लाभ या कितनी हानि हुई !A. Rs 2 lossB. Rs. 2 profitC. Rs. 7 lossD. Rs. 7 profit |
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Answer» Correct Answer - A According to the question CP of the mixture (मिश्रण का क्रय मुल्य) `=30xx9.5+40xx8.5` `=285+340=Rs 625` S.P of mixture (मिश्रण का विक्रय मुल्य) `= 8.90xx70` =Rs 623 loss (हानि)=C.P-S.P loss `=625-623=Rs 2` |
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| 34. |
5 ली. के अल्कोहल-पानी मिश्रण में 40% अल्कोहल है। इसमें 1` लि. पानी मिलाया जाता है, तो नये मिश्रण में अल्कोहोल का प्रतिशत क्या होगा?A. `30%`B. `33%`C. `33(2)/(3)%`D. `33(1)/(3)%` |
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Answer» Correct Answer - d (d) `40%=2/5{:(to"Water"),(to"Mixture"):}` `{:("Water",:,"Alcohol"),(3,:,2):}` Required pecentage `=2/((5+1))xx100` `=2/6xx100` `=1/3xx100=33(1)/3%` |
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| 35. |
चार बर्तनो में दूध और पानी का अनुपात क्रमश `5:3, 2:1, 3:2` और `7:4` है कौन से बर्तन में दूध की मात्रा पानी के सम्बन्ध में कम है ?A. FirstB. SecondC. ThirdD. Fourth |
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Answer» Correct Answer - C According to the question Container:- `{:(," I","II","III","IV"),(,"M W","M W","M W","M W"),(,"5 ":3,"2 ":1,"3 ":2,"7 ":4):}` Container (I):- `("Milk")/("Water")=5/3=1.67` Container (II):- `("Milk")/("Water")=2/1=2` Container (III):- `("Milk")/("Water")=3/2=1.5` Container (IV):- `("Milk")/("Water")=7/4 = 1.75` `:.` The quantity of milk relative to water minimum in container III (तीसरे पात्र में दूध की मात्रा पानी की तुलना में न्यूनतम है ) |
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| 36. |
एक बर्तन में 81 लीटर शुद्ध दूध है एक - तिहाई दूध को पानी से विस्थापित किया जाता है दोबारा एक तिहाई मिश्रण निकाला जाता है और इसमें इतना ही पानी मिलाया जाता है तो नए मिश्रण में दूध और पानी का अनुपात क्या होगा ?A. `1:2`B. `1:1`C. `2:1`D. `4:5` |
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Answer» Correct Answer - D According to the question (प्रश्नानुसार) `1/3xx"Milk" =1/3xx81=27` Final Quantity of milk (दूध की अंतिम मात्रा ) = Initial quantity `(1-x/e)^(n)` x= Quantity taken out at a time (एक समय बाहर निकाली गयी मात्रा) c= Capacity of vessel (पात्र की क्षमता ) n= no. of process `=81(1-27/81)^(2)=81(1-1/3)^(2)` `=81xx2/3xx2/3=36` Quantity of water (पानी की क्षमता ) `81-36=45` Ratio of milk and water in final mixture (अंतिम मिश्रण में दूध तथा पानी का अनुपात ) `36/45=4/5` `4:5` |
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| 37. |
रू 8 कीमत वाली 10 किलोग्राम गेहू को रू 10 वाली 15 किलो गेहू के साथ मिलाया जाता है सम्पूर्ण मिश्रण की औसत कीमत कितनी होगी?A. रू `9.5 kg`B. रू `9.2 kg`C. रू `7.5 kg`D. रू `8.5 kg` |
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Answer» Correct Answer - B `Qty xxPrice = Total cost` `10xx8=80` `15xx10=150` `"Average price"=("Total Cost")/("Total qty")` `=(80+150)/(10+15)=230/25=रू 9.2//kg` |
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| 38. |
15 लीटर के मिश्रण में अल्कोहल तथा पानी का अनुपात 1:4 है। यदि 3 लीटर पानी इसमें मिला दिया जाए तो नये मिश्रण में अल्कोहल है प्रतिशत क्या होगा ?A. `15%`B. `16(2)/(3)%`C. 0.17D. `18(1)/(2)%` |
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Answer» Correct Answer - b (b) `{:("Alcohol",:,"Water"),(to,:,4):}` According to the question, (1 +4) units = 15 litres 5 units = 15 litres 1 units = 3 litres Quantity of alcohol = 1 `xx` 3 = 3 liters Quantity of water = 4 `xx` 3 = 12 litres New quantity of water = (12+3) = 15 litres Required `%=3/((15+3))xx100` `=3/18xx100` `=16(2)/3%` |
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| 39. |
12 संतरों का लागत मूल्य 10 संतरों कं वि. मू. के बराबर है तो प्रतिशत लाभ ज्ञात करें ।A. `16(2)/(3)%`B. `20%`C. `18%`D. `25%` |
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Answer» Correct Answer - B According to question, 12CP=10SP `(CP)/(SP)=(10)/(12)=(5)/(6) gt 1 `profit Profit%`=(1)/(5)xx100=20%`. |
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| 40. |
दूध से भरे हुए तीन बर्तनो की धारिता `3:2:1` है तीनो को पानी से मिला दिया जाता है बर्तनो में दूध और पानी का अनुपात `5:2, 4:1` और `4:1` हो जाता है पहले से `1//3` दूसरे से `1//2` और तीसरे से `1//7` मिश्रण निकालकर एक बर्तन में रखा जाता है मिश्रण में पानी का प्रतिशत बताओ ?A. 32B. 28C. 30D. 24 |
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Answer» Correct Answer - D Capacitaies of Vessels (पात्रो की क्षमता का अनुपात ) `=3:2:1` `{:(,"M",:,"W",,"T Mix",),(V-1rarr,"("5,:,2,=,7")"_(xx5),),(V-2rarr,"("4,:,1,=,5")"_(xx7),),(V-3rarr,"("4,:,1,=,5")"_(xx7),):}` Equate the mixture (मिश्रण को समान करे ) `{:(,"M",:,"W",,"T Mix",),(V-1rarr,"("25,:,10,=,35,),(V-2rarr,"("28,:,7,=,35,),(V-3rarr,"("28,:,7,=,35,):}` `{:("Capacities","M",:,"W",,"T Mix",),("("V-1")"xx"3"rarr,75,:,30,=,105,),("("V-2")"xx"2"rarr,56,:,14,=,70,),("("V-3")"xx"1"rarr,28,:,7,=,35,):}` Water taken out (निकाला गया पानी) `rArr 1/3` of water in `(V-1)+1/2` of water in `(V-2)+1/7` of water in `(V-3)` `rArr 1/3xx30+1/2xx14+1/7xx7` `rArr 10+7+1=18` Similarly mixture will be (इसी प्रकार मिश्रण होगा ) `=1/3xx105+1/2xx70+1/7xx35=75` % of water `=18/75xx100 rArr 24%` |
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| 41. |
यदि 3 खिलौनों को 4 खिलौनों के लागत मूल्य पर बेचा जाता है , तो प्रतिशत लाभ ज्ञात करें।A. `25%`B. `33(1)/(3) %`C. `66 (2)/(3) %`D. `50%` |
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Answer» Correct Answer - B According to question, SP of 3 toys =CP of 4 toys `(SP)/(CP) =(4)/(3)gt1` gain gain % `=("Gain")/(CP)xx100` `=(1)/(3)xx100=33(1)/(3)%` `=(1)/(3)xx100=33(1)/(3)%` |
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| 42. |
24 सेबों का क्र.मू. 18 सेबों के वि. मू. के बराबर है, तो प्रतिशत लाभ ज्ञात करें ।A. `12(1)/(2)`B. `14(2)/(3)`C. `16(2)/(3)`D. `33(1)/(3)` |
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Answer» Correct Answer - D According to question 24CP=18SP `(CP)/(SP)=(18)/(24)(3)/(4)gt1` units profit profit%`=(1)/(3)xx100=33(1)/(3)%` |
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| 43. |
25 लीटर वाले मिश्रण में अम्ल और पानी का अनुपात `4:1` है इसमें 3 लीटर पानी और मिलाया जाता है तो नए मिश्रण में अम्ल और पानी का अनुपात बताये !A. `5:2`B. `2:5`C. `3:5`D. `5:3` |
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Answer» Correct Answer - A According to the question Mixture = 25 litres Ratio of `("Acid")/("Water")=4/1` `:. 5 "units" rarr 25 "litres"` 1 unit `rarr` 5 litres `:. ("Acid")/("Water")=(4xx5)/(1xx5)=20/5` Acid water Initial `20:5` Final ratio `20:8` |
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| 44. |
एक नारियल व्यापारी को यह ज्ञात है कि 2750 नानियलों को क्र.मू. 2500 नारियलों के वि. मू. के बराबर है, तो उसका प्रतिशत लाभ या हानि क्या होगा ?A. `5%`B. `10%` gainC. `15%` lossD. `20%` gain |
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Answer» Correct Answer - B According to question 2750 CP=2500SP `(CP)/(SP)=(2500)/(2750)=(10)/(11) gt 1` units profit profit%`=(1)/(10)xx100=10%` gain, |
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| 45. |
एक कपड़ा व्यापारी 33 मी कपड़ा बेचकर 11 मी कपडे के वि. मू के बराबर लाभ कमाना है, तो प्रतिशत लाभ ज्ञात करें।A. 0.4B. 0.22C. 0.5D. 0.11 |
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Answer» Correct Answer - C ( c) Let the SP of 1 metre cloth ( माना की 1 मीटर कपड़े का विक्रय मूल्य )= Rs. 1 The SP of 33 metre cloth ( 33 मीटर कपड़े का विक्रय मूल्य )= Rs. 33 CP of 1 meter cloth ( 1 मीटर कपड़े का क्रय मूल्य )= Rs. x CP of 33 metres cloth= `x xx 33` = Rs. 33x According to question, Profit= SP-CP 11=33-33x 33x=22 `x=(22)/(33)=(2)/(3)` CP of 1 meter cloth= Rs. `(2)/(3)` CP of 33 metre cloth `=(2)/(3)xx33`= Rs. 22 SP of 33 metres cloth = Rs. 33 Profit =SP-CP =33-22=11 Profit% `(11)/(22)xx100=50%` |
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| 46. |
यदि 15 मेजों की लागत मूल्य 20 मेजों के वि. मू. के बराबर है, तो प्रतिशत हानि ज्ञात करें।A. 0.2B. 0.3C. 0.25D. 0.375 |
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Answer» Correct Answer - C According to question CP of 15 table =SP of 20 tables `(CP)/(SP)=(20)/(15) gt 15` units loss `therefore` loss% `=(5)/(20)xx100=25%` |
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| 47. |
12 किग्रा आलू 63 के बेचने पर किसी दुकानदार को 5 % लाभ होता है | यदि वह 50 किग्रा आलू 247.50 में बेचे तो उसका लाभ/हानि प्रतिशत बताइए |A. 1 % लाभB. 1 % हानिC. न ही लाभ न ही हानिD. 2.5 % हानि |
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Answer» Correct Answer - B SP of 1 potatoes = Rs. `(63)/(12)` = Rs. `(21)/(4)` `CPxx((100+P%))/(100)=SP` `CPxx(105)/(100)=21/4` CP = Rs. 5 Gain or loss percent by silling 50 kg of the same potatoes for Rs. 247.50 (50कि.ग्रा आलू को 247.50 रुपये में बेचने पर लाभ या हानि प्रतिशत ) CP of 50 kg potatoes = `50xx5` = Rs. 250 Loss = 250-247.50=Rs.2.50 `Loss %=(2.50)/(250)xx100=1%` |
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| 48. |
12 कलमों का लागत मूल्य 8 कलमों के वि. मू. के बराबर है, तो प्रतिशत लाभ ज्ञात करें।A. `33(1)/(3)%`B. `66(2)/(3)%`C. `25%`D. `50%` |
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Answer» Correct Answer - D According to question to question, 12CP=8SP `(CP)/(SP)=(8)/(12)=(2)/(3) gt 1` profit profit% `=(1)/(2)xx100=50%`. |
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| 49. |
15 लीटर वाले मिश्रण में एल्कोहल और पानी का अनुपात `1:4` है यदि इसमें 3 लीटर पानी मिला दिया जाता है तो नए मिश्रण में एल्कोहल का प्रतिशत बताओ !A. 15B. `16 2/3`C. 17D. `18 1/2` |
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Answer» Correct Answer - B According to the question Mixture `=15` lites Ratio of `("Alcohol")/("Water")=1/4=5` units `:.` 5 units `rarr15` litres 1 unit `rarr 3` litres `:. ("Alcohol")/("water")=(1xx3)/(4xx3)=3/12` Alcohol water Initially `rarr3 12` Final `rarr 3 15` `:.` Percentage of alcohol in new mixture (नए मिश्रण में अल्कोहल का प्रतिशत ) `=3/18xx100=16 2/3%` |
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| 50. |
100 पेसिल को बेचने पर एक दुकानदार को 20 पेसिल कै वि . मू . कै बराव्रर लाभ होता है, तो प्रतिशत लाभ ज्ञात करेंA. 25B. 20C. 15D. 12 |
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Answer» Correct Answer - A (a) Let SP of 1 Pencil is Rs. 1 SP of 100 pencil is Rs. 100 CP of 1 pencil is Rs. x CP of 100 pencil is Rs. 100x According to question, Gain =SP -CP 20=100-100x 100x=80 `x=(80)/(100)=(4)/(5),x=(4)/(5)` CP of 1 pencil = Rs. `(4)/(5)` CP of 100 penci = Rs. `(4)/(5)xx100`=Rs. 80 SP of 100 pencils = Rs. 100 `:.` gain % `=(20)/(80)xx100=25%` |
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