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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. |
उक्त140’829, A) tan? g B) cot? g C) sin? 6 D) cos? 6 |
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Answer» 1 + tan²--------------1 + cot² (1 + tan ²) tan²--------------------------- 1 + tan² tan² A) is correct |
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| 2. |
Exercise 3.21. Find the squares of the following numbers using the columnusing the usual method:howing numbers using the column method. Verify the result by finding the(i) 252. Find the squares of:(i) 54(iii) 71(iv) 92 |
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Answer» 1(i)=25^2=625, (ii)54^2=2916; (iii)71^2=5041; ( iv)92^2=8464 Above answers are absolutely correct... 👌 i) 25×25=625ii)54×54=2,916iii)71×71=5,041iv)92×92=8,464 25 square is 62554 square is 2916 71 square is 504192 square is 8464 25 is the answer yes 1(¡) =25^2=625,(¡¡)54^2=2916,(¡¡¡)71^2=5041,(¡v) 92^2=8464 |
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| 3. |
passenger train takes 2 hours less for a journey of 300 kms, if its speed is increased by 5usual speed. Find its usual speed.28. Akms/hr from its |
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| 4. |
5. A man with- of his usual speed reaches thedestination 2 h late. Find his usual time to reach thedestination. |
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Answer» When his speed becomes 3/4th, his time would increase by 1/3rd. Thus, the normal time = 7.5 hrs. (Since increased time = 2.5 hrs). |
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| 5. |
23. A train takes 2 hours less for a journey of 300km if its speed is increased by 5 km/h from itsusual speed. Find the usual speed of the train |
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Answer» DaaaaaasasssfgaufsdhjxAhb |
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| 6. |
Add 4 to eight times a number, you get 60 |
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| 7. |
e outer diameter of a spherical shell is 12 cm and its inner diameter is8 cm. Find the volume of metal contained in the shell. Also, find itsouter surface area.6. Th |
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| 8. |
2n2 ((1- tan A1 +cot A1-cot A |
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Answer» L.H.S1+tan^2A/1+cot^2A=(1+sin^2A/cos2A)/(1+cos^2A/sin^2A)=((cos^2A+sin^2A)/cos^2A)/(sin^2A+cos^2A)/sin^2A=(1/cos^2A)/(1/sin^2A)=1/cos^2A*sin^2A/1=sin^2A/cos^2A=tan^2A M.H.S(1-tanA/1-cotA)^2=(1+tan^2A-2tanA)/(1+cot^2-2cotA)=(sec^2A-2tanA)/(cosec^2A-2cosA/sinA)=(sec^2A-2*sinA/cos A)/(cosec^2A-2cosA/sinA)=(1/cos^2A-2sinA/cosA)/(1/sin^2A-2cosA/sinA)=[(1-2sinAcosA)/cos^2A]/[(1-2cosAsinA)sin^2A]=(1-2sinAcosA)/cos^2A*sin^2A/(1-2sinAcosA)=sin^2A/cos^2A=tan^2A R.H.S=tan^2A Hence L.H.S=M.H.S=R.H.S hit like if you find it useful |
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| 9. |
Mohan borrowed a sum of Rs 75,000 from a bank at a rate of 19% per annuminterest. How much sum he will return after 5%2 years?(1) Rs 1,13,375 (2) Rs 1,53,375 (3) Rs 1,83,375 (4) Rs 1,93,375 |
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Answer» Given,Principal amount P = 75000, Rate of Interest R = 19%, Time period T = 5 1/2 years SI = P*R*T/100 = 75000*19*11/200 = 78375 Amount he will return after 51/2 years = 75000 + 78375= 15375 (2) is correct option |
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| 10. |
(ii) Parallelogram HEARHE 5 cmEA 6 cmR 85 |
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| 11. |
(a) Add 4 to eight times a number, you get 60. |
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Answer» 4+4+4+4+4+4+4+4=3260-32=28If we add 32 in 28 then we will get 60. So your answer is 28. |
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| 12. |
get 12 cm2. Find the area of sector whose radius is 7 em. and angle be 60be 60°. |
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| 13. |
a) Add 4 to eight times a number, you get 60. |
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Answer» 4+4+4+4+4+4+4+4=3260-32=28If we add 32 in 28 then we will get 60. So your answer is 28. let the no.=x4+8x=808x=60-4x=56/8 =7 x=7 |
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| 14. |
5158.An open tank with a square base and vertical sides is to be constructed from a metal sheet soas to hold a given quantity of water. Show that the total surface area is least when depth of thetank is half of its width.(CBSE 2010)orAn open tank with a square base and vertical sides is to be constructed from a mctal shtet soas to hold a given quantity of water. Show that the cost of the material will be least when thedepth of the tank is half of its width.(CBSE 2003, 2007; P.B. 2007) |
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| 15. |
d the quotient when f(x) = 3x4 +573-7x2 + 2x + 2 is divided by (x + 3x + 1). |
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| 16. |
3) Find the sum of 7x2-4x+5, -3x2+2x-1 and 5x2-x+9. |
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Answer» 7x^2-4x+5+(-3x^2+2x-1)+5x^2-x+99x^2-3x+13 |
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| 17. |
3000 votes were polled. one candidate get 60 percent of votes. by how many votes he win |
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Answer» 60% votes of 3000hence 60/100*3000=1800votes |
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| 18. |
22. A journey odjourney of 192 km from Mumbai to Pune takes 2 hours less by the fast train than by a slowain. If the average speed of the slow train is 16 km/hr less than that of the fast train, find theaverage speed of each train. |
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Answer» Given, Total distance = 192 km Let the average speed of the faster train = x km/hr So, average speed of the slower train = (x-16) km/hr Now, Time taken to cover 192 km by faster train (t1)= 192/x And, Time taken to cover 192 km by slower train (t2) = 192/(x-16) According to question, t2 - t1 = 2192/(x-16) - 192/x=2192x-192(x-16)/x(x-16)=23072=2x^2-32xx^2-16x-1536=0x²-48x+32x-1536=0x(x-48)+32(x-48)=0(x-48)(x+32)=0x= 48 , -32 Speed can't be negative. So, x = 48 km/hr Therefore, speed of faster train = 48 km/hr And, speed of slower train = (48-16) = 32 km/hr |
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| 19. |
eorem 9.2 The lengths of tangents drawn fromexternal point to a circle are equal.of: We are given a circle with centre O, a pointing outside the cirequired to prove that PQ PR.rcle and two tangents PQ, PR on the circle from |
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Answer» Let two tangent PT and QT are drawn to circle of centre O as shown in figure. Both the given tangents PT and QT touch to the circle at P and Q respectively. We have to proof : length of PT = length of QT Construction :- draw a line segment ,from centre O to external point T { touching point of two tangents } . Now ∆POT and ∆QOT We know, tangent makes right angle with radius of circle. Here, PO and QO are radii . So, ∠OPT = ∠OQT = 90° Now, it is clear that both the triangles ∆POT and QOT are right angled triangle.nd a common hypotenuse OT of these [ as shown in figure ] Now, come to the concept , ∆POT and ∆QOT ∠OPT = OQT = 90° Common hypotenuse OT And OP = OQ [ OP and OQ are radii]So, R - H - S rule of similarity ∆POT ~ ∆QOT Hence, OP/OQ = PT/QT = OT/OT PT/QT = 1 PT = QT [ hence proved] |
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| 20. |
Example 2: Two tangents TP and TQ are ulalto a circle with centre O from an external point T.Prove that L PTQ 2 OPQolution : We are given a circle with centre Oan external point T and two tangents TP and TQto the circle, where P, Q are the points of contactcee Fig 10.9), We need to prove that |
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Answer» We know that, the lengths of tangents drawn from an external point to a circle are equal.∴ TP = TQIn ΔTPQ,TP = TQ⇒ ∠TQP = ∠TPQ ...(1) (In a triangle, equal sides have equal angles opposite to them)∠TQP + ∠TPQ + ∠PTQ = 180º (Angle sum property)∴ 2 ∠TPQ + ∠PTQ = 180º (Using(1))⇒ ∠PTQ = 180º – 2 ∠TPQ ...(1)We know that, a tangent to a circle is perpendicular to the radius through the point of contact.OP ⊥ PT,∴ ∠OPT = 90º⇒ ∠OPQ + ∠TPQ = 90º⇒ ∠OPQ = 90º – ∠TPQ⇒ 2∠OPQ = 2(90º – ∠TPQ) = 180º – 2 ∠TPQ ...(2)From (1) and (2), we get∠PTQ = 2∠OPQ |
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| 21. |
bohween the two tangents drawn from an external point to a circle is supplementary |
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| 22. |
9.Find the zeros of the polynomial 6c3-7x2-11x + 12, ifx-l is a factor of the polynomial. |
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Answer» Like my answer if you find it useful! |
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| 23. |
After spending Rs 85 1/2, a man has Rs. 34 1/2 left with him. How much money did he have in the beginning? |
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| 24. |
Exercise - 7.1Mohan saves Rs. 3,950 after spending 75% of his salary, what is his monsalary?hair number is 18, so what is the ratio of numbers of |
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| 25. |
Divide polynomial 2x4-2x3-7x2+3x+6 by x2-3 |
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| 26. |
Find all the zeroes of the polynomial p (x)=x4-7x3+9x2+13x-4,if two of its zeroes are 2+underroot 3 and 2-underroot 3 |
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Answer» x^4-7x^3+9x^2+13x-4 from (2)=(2)^4-7(2)^3+8(2)^2+13(2)-4=16-7(8)+8(4)+13(2)-4=16-56+48+26-4=16+48+26-56-4=42+48-60=90-60=30; (3)^4-7(3)^3+8(3)^2+13(3)-4=81-189+72+39-4=-1 |
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| 27. |
20. Find all other zeroes of the polynomialx"-2x'-7x2 + 8x + 12, if two of its zeroesare-1 and 2,CRSE 015 |
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Answer» hit like if you find it useful Thanks |
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| 28. |
12ttt...t.vuotras TRBOUR16. 1.4 4.7 7.101.11++-3.(2n +1) (2n+3)3(2n+3)GRE |
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Answer» 1/1.4+ 2/4.7+1/7.10+,+1/(3n-2)(3n+1)=n/(3n+1); n=1.4; 1/(3(1.4)-2)(3(-2)+1)=1/((3(1.4)-2)(3(1.4)+1)=1/(3.12)(-6+3)=1/(3.12)((-3)=1/(-9.36) |
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| 29. |
i Sum of squares of first 'n' odd natural numbers isgiven by1) nin-1(2n+1)3) n(2n-1(2n+1)2 nn-1i2n+1n(2n - 1)(2n +1)3 |
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| 30. |
Number of ways of forming a team consisting of atleast 2 and atmost (nstudents is(A) 2n- 2n -2(C) 2n-2n2) students from n(B) 2n - 2n-4(D) 2 -2n 3 |
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| 31. |
Q.4.1 minute. Find itsA train increases its speed from 60 kmh to 90 km/h inacceleration.i bob(2) |
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Answer» initial velocity is 60km/h final velocity is 90km/h initial velocity is 60 kilometre m hour final velocity is equals to 90 km per hour the formula to calculate acceleration is v-u/t 90-60/60=30/60=3/6 acceleration =3/6 m |
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| 32. |
SETInan AP, Sn=n2, what is tn ?(A) 2n(C) 2n+1(B) 2n-1(D)2n + 3 |
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Answer» Given,Sn = n^2 Then,Tn = Sn - Sn-1 = n^2 - (n - 1)^2 = n^2 - n^2 - 1 + 2n = 2n - 1 (B) is correct option |
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| 33. |
Danjeuo| 7.Two number are in the ratio 5 : 8. Ifthe sum of the numbers is 195, find thenumbers.(A) 15 and 190 (B) 50 and 145KC 75 and 120 (D) 150 and 45 |
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Answer» C)75 and 120 is a right answer C)75 and 120 is a right answer let the no. be 5x and 8x . 5x+8x=195 then x =13 .put the value of x and your answer come 75 and 120 I hope you help this answer plz follow me and mark a best answer |
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| 34. |
ells his house for ?8,00,000. If he gets 25%profit, find the cost price of the house. |
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Answer» Selling price = (100+25)/100 * Cost Price=> 800000 = 5/4 * Cost=> Cost = 800000*4 / 5=> Cost = Rs. 640000 |
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| 35. |
text \ rank of the matrix\left[ \begin{array} { c c c } { 1 } & { 4 } & { 5 } \\ { 2 } & { 6 } & { 8 } \\ { 3 } & { 7 } & { 2 } \end{array} \right] |
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| 36. |
Fig. 3.4EXERCISE 3.11. Aftab tells his daughter, "Seven years ago, I was seven times as old as you were theAlso, three years from now, I shall be three times as old as you will be." Isn't tinteresting?) Represent this situation algebraically and graphically. |
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Answer» Thnxxx u very much for answering the questions |
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| 37. |
The vertices of a \triangle A B C are A(-5 , 7), B(-4,-5) and C(4 , 5). Find the slopesof the altitudes of the triangle. |
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Answer» slope will be -1/ slope of side BCslope of BC = 5-(-5)/4-(4)10/8slope of attitude = 8/10 |
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| 38. |
2. Banti had some amount of money after spending Rsof his initial amount was left with him what amount30,of money was left with him? |
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| 39. |
10-1-227_DC vopombo out do stool autputa |
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Answer» x²+2x-1=0x=-b+/-√(b²-4ac) -------------------- 2ax=-2+/-√(4-(4)(1)(-1) ------------------------ 2x=-2+/-√8 ------------ 2= 2(-1+/-√2) ----------- 2x=-1+/-√2 |
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| 40. |
Le dan |
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| 41. |
ie ateas of two similar triangles ABC and PQR are 64 cm2 and 121 cm2, respectively. If QRthen the value of BC will be:15.4 cmof two si |
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| 42. |
(-5, 7, B-4, -5), C-1, -6) and D(4, 5) are the vertices of a quadrilateralarea of the quadrilateral ABCD5 |
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| 43. |
If A(-5, 7). B(-4, 5). C-1.-6) and D(4., 5) are the vertices of a quadrilateral, find thearea of the quadrilateral ABCD |
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| 44. |
I. Construet a 2 matrix 4 -la,J whose elements are given by a,-(ngchao cilu)o ,4 ndern 3 x 2 ๑aus1.etny cnidoil.ocada, |
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| 45. |
Construct a 3 x 4 matrix, whose elements are given by:-3i +(ii)a,-21 |
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| 46. |
25. Construct a 3 x 4 matrix, whose elements are given by(i)aijar3i + j |(i)a,-21-J2- |
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| 47. |
3. If the matrix 4 7-0-2 x |
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Answer» Determinant = 4 × x - 7 × (-2) = 0 4x + 14 = 0 4x = -14 x = -14/4 x = -7/2 |
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| 48. |
A = | x 0 | is an involutary matrix then,(2) 2(3) -1(4) 2 |
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| 49. |
31. The value of a car isăŚ5,00,000. If its value depreciates at 10% pa, what willbe its value after 3 years? |
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Answer» Value of the machine after 3 years = Rs.[500000*(1- 10/100)^3] = Rs.500000*9*9*9/10*10*10 = Rs. 500*729 = Rs. 364500 |
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| 50. |
o Sr money putaamount to in 3 years at the same rate?12 122 in 2 years . What will it17. A sum ofinvestemoney invested at 8% per annum amounts toamount to in 2 years 8 months at 9% per annum?in 31 years? |
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