Saved Bookmarks
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. |
e sides of a triangle are 42 cm, 34 cm and 20 cm. Calculate its area andheight on the longest side. |
| Answer» | |
| 2. |
4.Solve the following:14 x+1= 5x |
|
Answer» solve(4)4*x-3*(x+1)=54x-3x-3 =5x-3=5x=5+3x=8 ans answers is 1/11 hai iss ka |
|
| 3. |
अ ... लि.pxample 17 I nC, ZWAC,., find "Cy7. |
|
Answer» please like my answer if you find it useful |
|
| 4. |
Determine n if(i) 2nC3 : nC,-12 : 1 |
|
Answer» part 1 of solution part 2 of solution |
|
| 5. |
(iii) Difference of squares of two nis 56.(iv) A number divided by 5 givessthe number. Solve the following equation for18 33a 2a 74. Solve the equation :-+-=-124. Sol |
|
Answer» 12y-4+3y+5 = 18/3 15y = 6-1 15y = 5 y = 1/3 not good answer |
|
| 6. |
Make 3 magic squares of 3x3,4x4,5x5 |
|
Answer» 3x3=9; 4x4=16; 5x5=25 3x3=9=81=6561; 4x4=16,256,55536, 5x5=25,625, 411625 |
|
| 7. |
4)solveT DE |
| Answer» | |
| 8. |
3. evono "thaナthe area d) adaomgee with vertices (Leba) |
|
Answer» Given three points are (t, t-2) , (t+2, t+2) and (t+3, t) use formula of traingle form of co-ordinate geometry . area of traingle =1/2 {x1 (y2-y3)+x2 (y3-y1)+x3(y1-y2)} now,ar of triangle=1/2 {t (t+2-t)+(t+2)(t-t+2)+(t+3 (t-2-t-2)}=1/2 {2t +2t +4 -4t -12}=1/2 {-8}=-4 but here mention only magnitude of area of triangle so , area of traingle =4 sq unit So area is not dependent to t |
|
| 9. |
2. The circumference of a circle is 44 cm. Findthaarea of the circle. |
| Answer» | |
| 10. |
te area o tha shade075 |
| Answer» | |
| 11. |
If there are two points A(-3, -2) and B(5, 9), thenline 2x+ y = 2 divides the line joining the pointsA, B in the ratio |
| Answer» | |
| 12. |
s Find the equation to the straight line which bse t ll),ld, b) and also bisects the distance between the points (-a, b) and (a',-b)divides the join of points (3, -1) and (8,9) in the ratio |
| Answer» | |
| 13. |
Find a point which is equidistant from the points A=(-5,4)and B=(-1,6).How many such points are here? |
|
Answer» let the point be P(x,y)m1= m2= 1A(-5,4) and B(-1,6)x =m1x2+m2x1/m1+m2 = 1(-1) + 1(-5) / 2 = -6/2 =-3y =m1y2+m2y1/m1+m2 = 1(6) + 1(4) /2 = 10 /2 =5the point P(-3,5) is equidistant from A(-5,4) and B(-1,6). |
|
| 14. |
The area of a parallelogram is 780 sq. cm.If the length of one side is 26 cm, find thelength of the other side. 3 |
|
Answer» Area of the parallelogram=length×base=>780=26×length of the other side=>length of the other side=780/26=30 cm sir, this is perpendicular distance between sides. sir, this is height(perpendicular) on side( base) 26 any idea I am also not gat answer |
|
| 15. |
area of a rectangle is 300 sq. cm it's length is 20 cm calculate it's breadth |
|
Answer» L = 20 cm B = ? Area = 300 cm² Area of recatngle = L × B 300 = 20 × B B = 300/20 = 15 cm |
|
| 16. |
Find the value of Laudadethat 20+ P x3 +02002- setaes divisible by 22-12 |
| Answer» | |
| 17. |
Given $a=3^{p}$ and $b=3^{q}$ find $\log _{9} a+\log _{27} b$ in terms of $p$ and $q$ |
| Answer» | |
| 18. |
(iii) 6x2-3-7xNC |
| Answer» | |
| 19. |
2 \a=2+2^{2 / 3}+2^{1 / 3}Prove that a^{3}-6 a^{2}+6 a-2=0 |
|
Answer» answer will be -56 |
|
| 20. |
(iii)6x2-6x-12-0; |
|
Answer» 6x²-6x-12=0 Dividing both sides by 6, equation becomes,x²-x-2=0x²-2x+x-2=0x(x-2)+1(x-2)=0(x+1)(x-2)=0 x=-1,2 |
|
| 21. |
5. Divide:(i) 6m2 - 16m3 + 10 m4 by -2m(i) 9x3 - 6x2 by 3x(iii) 15x3уг + 25x2ys_ 36xy4 by 5x2y2(iv) 36a3x5 - 24ax+18a5x3 by -6a3x3 |
|
Answer» thanks |
|
| 22. |
Q.4Solve any THREE of the following. (3 Marks each)) Prove that the sum of the squares of diagonals of a parallelogram is equal to the sum09of squares of its sides. |
|
Answer» Figure In parallelogram ABCD, AB = CD, BC = ADDraw perpendiculars from C and D on AB as shown. In right angled ΔAEC, AC²= AE²+ CE²[By Pythagoras theorem] ⇒ AC²= (AB + BE)2+ CE²⇒ AC²= AB²+ BE²+ 2 AB × BE + CE² → (1)From the figure CD = EF (Since CDFE is a rectangle) But CD= AB⇒ AB = CD = EFAlso CE = DF (Distance between two parallel lines)ΔAFD ≅ ΔBEC (RHS congruence rule)⇒ AF = BE Consider right angled ΔDFB BD²= BF²+ DF²[By Pythagoras theorem] = (EF – BE)²+ CE²[Since DF = CE] = (AB – BE)²+ CE² [Since EF = AB]⇒ BD²= AB²+ BE²– 2 AB × BE + CE²→ (2) Add (1) and (2), we getAC²+ BD²= (AB²+ BE²+ 2 AB × BE + CE2) + (AB²+ BE²– 2 AB × BE + CE²) = 2AB²+ 2BE²+ 2CE² AC²+ BD²= 2AB²+ 2(BE²+ CE²) → (3)From right angled ΔBEC, BC²= BE²+ CE²[By Pythagoras theorem] Hence equation (3) becomes, AC²+ BD²= 2AB²+ 2BC² = AB²+ AB²+ BC²+ BC2² = AB²+ CD²+ BC²+ AD²∴ AC²+ BD²= AB²+ BC²+ CD²+ AD²Thus the sum of the squares of the diagonals of a parallelogram is equal to the sum of the squares of its sides. |
|
| 23. |
Find the squares of the following:(i) 1(ii) 46 |
|
Answer» (I) 1^2=1(ii) 46^2=2116 1^2=146^2=2116 1×146×46 1×1=1 46×46=2116. |
|
| 24. |
find the lcm of the following 14 and 56 |
|
Answer» 14=2×756=2×2×2×7is the right answer lcm of 14 and 56 is 14=2*756=2*2*2*7lcm=2*2*2*7=56 is answer 14 = 2 x 7; 56/2=28 /2=14/2=7 GCF(14,56) = 14LCM(14,56) = ( 14 × 56) / 14LCM(14,56) = 784 / 14LCM(14,56) = 56 14 = 2 × 56 = 2 × 2 × 2 × 7 LCM = 2 × 2 × 2 × 7LCM =56 right answer lcm of 14and 56id14=2*756=2*2*27LCM=2*2*2*7=56 56 is the correct answer 56 is the correct answer 56 is the right answer 56= is the correct answer ( 56 )= is right answer |
|
| 25. |
V6. Show that the following points form a equilateraltriangle A(a,0), B(-a,0), C(0,a (3) |
|
Answer» AB=V(-a-a)^2+(0-0)^2=V(-2a)^2+0=2a_____________(1), Bc=(-a,0)(0,aV3)=V( 0-(-a)^2+( aV3)^2-0=a+3a^2=2a____(2), CA=(0, aV3)(a, 0)=Va^2+a^2(3)=2a______(3); 2a+2a+2a=equational triangles |
|
| 26. |
Show thart the points A42 B(75) and C(9,7) do not form a triangle |
| Answer» | |
| 27. |
Show that the points (1,7) ,(4,2),(-1,-1) and (-4,4) form a square Find the area |
| Answer» | |
| 28. |
\begin{array}{c}{\text { Evaluate: }} \\ {\text { (i) } \frac{\sin 18^{\circ}}{\cos 72^{\circ}}}\end{array} |
| Answer» | |
| 29. |
if y=x+1/x,show that x square dy/dx- xy+2=0 |
| Answer» | |
| 30. |
\begin{array}{l}{\text { Evaluate }} \\ {\text { (i) } 2^{4} \times 2^{5} \text { (ii) } 2^{14}+2^{12} \text { (iii) } 12^{3} \times 3^{3}}\end{array} |
| Answer» | |
| 31. |
\begin{array} { l } { \text { Evaluate: } } \\ { \text { (i) } \frac { \log 8 \times \log 9 } { \log 27 } } \end{array} |
| Answer» | |
| 32. |
What is the value of ( \operatorname { cos } ^ { 2 } 28 ^ { 0 } - \operatorname { sin } ^ { 2 } 62 ^ { 0 } ) |
| Answer» | |
| 33. |
Find the breadth of the following rectangle:(a) Area = 520 sq cm Length = 40 cm. |
|
Answer» Breadth=Area/length520/4013cm l*b=A520=40*xx=520/40=13cm is the best answer area = L×BB = area/ L = 520/ 40 = 13therefore B= 13 13 cm is a right answer 13 cm is a right answer 13 is the correct answer B=A/L =520/40 =13cm answer. 13cm is the correct answer breadth= area/length=520/40=13 |
|
| 34. |
3.(VII) (XYZ-4) (xyz-2)Find the following squares by using the identities.(ii)(xy +32P(iii)(6x2-5y)223 2 |
| Answer» | |
| 35. |
(iii)(6x2-5p)2 |
| Answer» | |
| 36. |
\left. \begin{array} { l } { 30 \div ( 3 + 2 ) - 3 \text { of } 3 + 2 \times ( 6 + 3 ) } \\ { 200 \div 2 ( 2 \text { of } 5 ) - ( 10 \times 10 - 5 ) } \end{array} \right. |
|
Answer» a) 30÷(3+2) -3of3 +2×(6+3)= 30/5 -9 +2*(9)= 6-9+18 = 15 b) 200÷2*(2of5) -(10×10-5)= 200÷(2*10) -(100-5)= 10-(100-5)= 10+5-100= 15-100= -85 Your Ans:a is correct but And:b is wrong |
|
| 37. |
(3*(5*sqrt(5)))*(((3*(2*sqrt(2)))/sqrt(10))/((10/((2*sqrt(11)))))) |
|
Answer» 2/V10 x 7/10x5/1==7/10 |
|
| 38. |
Find the following squares by using the identities.(i) (b-7)^2(ii) (xy +3z)^2(iii)(6x2-5y)^2 |
|
Answer» i) (b-7)² = b² +49 - 14b ii) (xy+3z)² = x²y² +9z² +6xyz iii) (6x²-5y)² = 36x² +25y² -60x²y very very thanks to you |
|
| 39. |
Find the positive integer n so that lim\lim _{x \rightarrow 3} \frac{x^{n}-3^{n}}{x-3}=108 |
| Answer» | |
| 40. |
15.The length and breadth of a rectangle are in the ratio 3 :2. If the area of the rectangleis 726 m, find its perimeter.16.Find the following sun56/HIS/2 |
| Answer» | |
| 41. |
Prove thatsin 80° cos 20° + cos 80° sin 20° = √3/2 |
| Answer» | |
| 42. |
CeSaPardiShow that the following sets of points form a rectangle. |
|
Answer» suppose. A={-5,-2}B= {3,-1}now AB = √(3+5)^2+(-1+2)^2 √64+1= √65 unitsame wise CD√(1+3)^2+(-5-2)^2√16+49√65 as AB=CD it is a rectangle |
|
| 43. |
(1) Write the following statements in conditional formEvery rectangle is a parallelogram. |
|
Answer» If parallelogram have all interior angle 90° then it is rectangle. |
|
| 44. |
6*(x %2B 2*y) %2B 8*z(x %2B 2*y) |
|
Answer» the answer is=6x+12y+8zx+16yz Your correct answer is6x+12y+8zx+16yz 6x+12y+8zx+16yz is the correct answer of the given question |
|
| 45. |
8 I %2B x z = x \varepsilon ( x ) |
|
Answer» 3x = 2x+18 3x-2x = 18 x = 18 Like my answer if you find it useful! |
|
| 46. |
\left. \begin array l \text WAMPLE 22 Evaluate \frac 15 \sqrt 10 %2B \sqrt 20 %2B \sqrt 40 - \sqrt 80 , \text is being given that \sqrt 5 = \\ \sqrt 10 = 3.162 \end array \right. |
| Answer» | |
| 47. |
what is the fourth proportional of (x^2-y^2),(x^2y-xy^2) and (x+y) |
| Answer» | |
| 48. |
2.Take any point O in the interior of a triangle POR. Is.O(ii) OQ + OR > QR?(iii) OR+ OP> Rp? |
| Answer» | |
| 49. |
2802 xy2 |
| Answer» | |
| 50. |
In Δ POR, right-angled at Q. PRsin P, cos P and tan PR-25 cm and PQ-5cm. Determine the values |
| Answer» | |