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.
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4. From a window of a house in a street, h metres above the ground, the angles afelevation and depression of the top and the foot of another house on the opposite sideof the street are α and β respectively. Show that the height of the opposite hourh(1+ tan a.cot 8) metres. |
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Answer» you have to prove height h(1+tan A.cot B) m and what are u proving.....60(1+......) |
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| 2. |
1x Find the perimeter of a rectangle whose area is 600 cm and breadth is 25 cm |
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| 3. |
Give possible expressions for the length and breadth of the rectangle whose area is25a2 - 35a +12. |
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Answer» So here |
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| 4. |
|| का अनुपात 1:46 है तथा इसका गणनफल 1944 है. इनमें सबसे बड़ी संख्या क्या है।(b) 12(c) 18(d) इनमें से कोई।। |
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Answer» option C is the correct answer 3x:4x:6x=194472x^3=1944x^3=1944/72x^3=27X=36x=3x6=18 answer 18 is the right answer for this question C) 18 is the right answer 18 ( option C) is the correct answer A/B/C/D is the following optionsoption C is the correct answer of the following question 18 is the correct answer of the given question (3x)×(4x)×(6x)=194472x^3=1944x^3=1944/72x^3=27 => x^3=(3)^3x=3बङी संख्या=6x=6×3=18. ans. 3x : 4x : 6x =1944, (3×4×6)x^3=1944, 72 x^3=1944; x^3=1944/72= 27; x^3=3^3; x=3 6x = 6 x 3 = 1 8 3:4:6=19443x × 4x × 6x = 1944 72x^3 = 1944x^3 = 1944/72x^3 = 27x = 33x=3×3=94x=4×3=126x=6×3=18option (c) 18 is the correct answer. 18 is the best answer c) 18 is the right answer. |
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| 5. |
Identify the following in the given figure.KUA. Vertically opposite anglesB. Corresponding angles |
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Answer» vertically opposite= 1 and 3 , 2 and 4 , 5 and 7 , 6 and 8 corresponding angles= 2 and 3, 1 and 4, 6 and 7, 5 and 8 |
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| 6. |
If α β are the roots of the equation 8x2-3x+27 =0,32then the value ofis3 1tb72(d) 44 |
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| 7. |
10 = {B + (4% 208 |
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Answer» 10-{8+(4÷2)}=10-{8+2}=10-10 =0 |
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| 8. |
24From the given figure, Identify?a) Diagonalsb) Opposite sidesC) Adjacent angles |
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Answer» a) AC and BD b) AB and DC also AD and BCc) A and D, A and B , B and C, C and D |
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4 From the given figure, Identify?a) Diagonalsb) Opposite sidesc) Adjacent angles |
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Answer» a) AC and BD b) AB and DC also AD and BCc) A and D, A and B , B and C, C and D |
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| 10. |
8. In the given figure, name the pairs ofvertically opposite angles.s/e9. In the following figure, find the measure of other angles, if 41-50 |
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Answer» 8. Pairs of vertically opposite angles: (1 and 3), (2 and 4), (6and 8) and (5and7) 9. angle 2 = angle 1 =50 degrees (vertically opposite angles). angle 1 +angle 3 =180 (linear pair of angles)angle 3 =130 angle 4 = angle 3 =130 degrees (vertically opposite angles) |
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| 11. |
2 Angles opposite to equal sides of an isosceles triangle are equat.17.2orts result can be praved in many ways. One ofthe proofs is given here.We are given an isosceles triangle ABCin which AB = AC. We need to prove that |
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| 12. |
Let us consider Sohle exampies.Example 2 : Find the zeroes of the quadratic polynomialrelationship between the zeroes and the coefficients.+7x + 10, and verify |
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Answer» X²+7x+10=0 x²+2x+5x+10=0 x(x+2)+5(x+2)=0 (x+2)(x+5) = 0 x+2 = 0 ; x = -2 x+5 = 0 ; x = -5 Relationship between the zeroes and coefficients :- Sum of zeroes = -2+(-5) = -2-5 = -7/1 = -x coefficient /x² coefficient Product of zeroes = (-2)(-5) = 10/1 = constant/x² coefficient |
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| 13. |
The sum of two natural numbers is 117. If one number is twelve times the other ,findthe two numbers. |
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TherefcrossVerify this answer by checking if it asExample 9 Let us consider Example 2 in Section 3.3. i.e., the cost of 2 pencils and18. Find the cost of each3 erasers is? 9 and the cost of 4 pencils and 6 erasers ispencil and each eraser.Sointion The pair of linear equations formed were: |
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8. Two natural numbers differ by 3 and their product is 504. Find thenumbers. |
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Answer» thanks |
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| 16. |
2If2 , let us show thaty a+4 |
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| 17. |
The sum of two natural numbers is 20 while their difference is 4. Findthe numbers. |
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| 18. |
If the roots of the equation ax2 + bx + c = 0 are | 2α and β, then the quadratic equation whoseroots are -a and -p is2,(A) ax2-bx-c=0(C) ax2 + bx-c=0(B) ax2-bx + c 0(D) ax2-bx + 2c = 0 |
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| 19. |
The difference of two natural numbers is 5 and the difference of their reciprocats is. Find the numbers.410 |
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| 20. |
5. Sum of two natural numbers is 8 and thedifference of their reciprocals = 2Find thenumbers.[ICSE 2015] |
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| 21. |
The sum of two natural numbers is 15 and the sum of their reciprocais Find the numbers.CBSE 20810 |
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| 22. |
OrWhat happened on 22 february 1944 |
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Answer» Tuesday 22 February 1944. It was Tuesday, under the sign of Pisces (see planets position on February 22, 1944). The US president was Franklin D. Roosevelt (Democratic). Famous people born on this day include Robert Kardashian and Jonathan Demme |
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| 23. |
Solve the following quadratic equation for x :la a+bx+1=02 +(a+baOR |
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Answer» x. x+2a.a+xb.b+2abx+1=0 is the best answer x.x+2a.a+xb.b+2abx+1=0 is the best answer |
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| 24. |
If the discriminant of the quadratic equation 8x2 -bx +2 0 is 1, find the value ofb' |
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Answer» Discrimanat is(-b)² - 4 × 8 × 2 = 1b² - 64 = 1b² = 65 b = + or - √(65) |
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12. In the Quadratic equation ax2+bx+c=0 if b2-4ac>0 then write its Natureof Roots. |
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| 26. |
819. Ifsec O â tan 0 = 4, then prove that cos 0= â" . |
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Answer» secθ - tanθ = 4 -----> (1) Multiplying and dividing bysecθ +tanθ =secθ - tanθ×secθ +tanθ/secθ +tanθ = sec²θ - tan²θ/secθ + tanθ = 4 we know that --> sec²θ - tan²θ = 1 1/secθ + tanθ = 4 therefore , secθ + tanθ = 1/4----> (2) Adding equations (1) and (2) :- secθ - tanθ = 4 secθ + tanθ = 1/4 --------------------------2secθ = 4 + 1/42secθ = 17/4 secθ = 17/4*2 = 17/8 we know that cosθ = 1/secθ = 1/(17/8) = 8/17 |
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| 27. |
___limx â 0 |
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| 28. |
st' â] 0) |
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| 29. |
Find x â 0 |
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| 30. |
sin B âcos 0 +11 "=, using the identitâano " o Exsmp F % sin 0+ cos Osect =1+t 0, O |
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Answer» LHS = (sinA-cosA+1)/(sinA+cos-1) = (1 + sinA) (1 - sinA)/cos A(1 - sinA) = 1 - sin2A/cos A(1 - sinA) = cos2A/cos A(1 - sinA) = cosA/(1 - sinA) = 1/ (1/cosA - sinA/cosA) = 1/(secA - tanA) = RHS Hence Proved |
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which of the following equationshave the ones that are exact:e^{y} d x+\left(x e^{y}+2 y\right) d y=0 |
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| 32. |
o xl0 â |0 =8 |
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Answer» 100*10-10÷100100*10-1/101000-1/109999/10999.9 |
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| 33. |
Q. Ifa andſ are the zeroes of f(x) = 4x2- 5x-1, then evaluate:Ha? + B² (ija? +B3 Eini) at +84 (iv) a ?-B?C |
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Answer» usea+b=5/4ab=1/4and now solve each term |
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| 34. |
31/01रामकल्याण जीत्रा--2-__५-+___aludu+५dy५dz- |
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| 35. |
eV Sg———के... .e (SOd 7 - 500 7 Tz =909 >ही RRYS DY 1 e b के¥ SO b Bk % YO N dO fudey 13wk 20 Dy1B 3O Loy 13 B} 22 OO 5119 ydae ® |
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| 36. |
Dy e फिट है!) |
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| 37. |
The differential equation of the family of curvesy e (Acos.x + Bsin x), where A and B arearbitrary constants, is :69.dy dydx dxdy dy |
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| 38. |
The solution of D. Edy- secx -y tan x is |
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| 39. |
7. Find , when x-e cot, y- e sint.dy |
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Answer» please like my answer if you find it useful |
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| 40. |
e following equations and verify the solutions1. Solve the following equations a(a) 5(x + 3) = 20 |
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Answer» X= 1 is the correct answer of the given question 5( x + 3) = 20; 5x + 15 = 20;; 5x = 20 - 15; x = 5/5 =1 x=1 is right answer. |
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| 41. |
Find the value of x and y in the following equations the following equations:2xy/x+y = 3/2 ; xy/2x-y = -3/10Where x + y ≠ 0 and 2x - y ≠ 0 |
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| 42. |
following equations |
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Answer» 3x-7/5 = 1-x/-3= -3(3x-7) = 5(1-x) [cross multiply]= -9x+21 = 5-5x= -9x+5x = 5-21= -4x = -16= x = 16/4= x = 4 |
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है 1 P) (l+x—}<2—[l—x—}<4 |
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Answer» (x+1)*2/x = (x-1)*4/x2x+2/x = 4x-4/x 4x-4 = 2x+24x-2x = 4+22x = 6x = 6/2x = 3 |
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Solve the following equations. |
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Answer» = n / 5 - 5 / 7 = 2 / 3 = n / 5 = 2 / 3 + 5 / 7 = n / 5 = (2 * 7 + 5 * 3 ) / 21 = n / 5 = (14 + 15 ) / 21 = n = 29 / 21 * 5 = n = 145 / 21. |
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| 45. |
L kb p R'BIIRLOLE L +x= x 2y |
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| 46. |
\left. \begin array l l ( x %2B 2 ) ^ 3 = x ( x ^ 2 - 1 ) & ( B ) \\ ( x - 2 ) ( x %2B 2 ) = 5 & \text (D) x ^ 2 %2B \frac 1 x ^ 2 = 2 \end array \right. |
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Answer» Option Dx²+1/x² = 2As, x² comes in denominator. |
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| 47. |
7 kg \text { to } 140 g |
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Answer» thanks didi |
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| 48. |
Solve the following equations: |
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| 49. |
EXERCISolve the following equations. |
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| 50. |
1+/x+1 Gy‘_+.‘l‘-\/—-x+l & |
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Answer» sqrt(3x +1) + sqrt(x+1)/sqrt(3x+1) - sqrt(x+1) = 4 [sqrt(3x +1) + sqrt(x+1)] ^2/[(3x+1) - (x+1)] = 4 (3x+1 + x + 1) + 2sqrt[(3x+1)*(x+1)] = 2x*4 2sqrt[(3x+1)*(x+1)] = 4x - 2 sqrt[(3x+1)*(x+1)] = 2x - 1 Squaring both sides, we get(3x+ 1)(x +1) = 4x^2 + 1 - 4x3x^2 + 4x + 1 = 4x^2 - 4x + 1x^2 - 8x = 0x(x - 8) = 0x = 0, 8 |
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