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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.
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trytopluvesIClatiollLet ts nowTheorem 9.1are equal in area.Theorem 9.1 : Parallelograns on the same base and between the same parallelA R |
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Practice set 5.41. In CIJKL, side IJ Il side KL △ = 108°K = 53° then find the measures ofJ and LL |
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| 3. |
thoct¡ balls on (2 hec totes st land |
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Answer» 1 hectare=10000sq.m2 hectares=20000sq.mVolume=Area*height=20000*5/100=1000 cubic metres |
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| 4. |
4 n^{2}-8 n+3 |
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*--इनमें से कोई =*3. विलुप्त संख्या की पूर्ति करें।| 18।| 1217 182413. 1419 20(1) 15 (2) 12 (3) 9 (418. राजेन्द्र प्रथम किसका पुत्र था?(1) बिन्दुसार प्रथम2) देवभूति प्रथम(3) स्कन्दगुप्त प्रथम (4) राजराज प्रथम |
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Answer» 18 is the right answer.. 18 is the right answer |
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\Gamma ( n + 1 ) = \underline { n } , n = 1,2,3 , \dots , ( n \in N ) |
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Answer» The given is gamma function propertieswhich is extension of factorial function . |
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1+3+3^{2}+\dots \dots+3^{n-1}=\frac{\left(3^{n}-1\right)}{2} |
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| 8. |
15.The perimeter of the given figure is(A) 10 cm(H)cm |
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Answer» perimeter = triangle perimeter + rectangle perimeter - length= 5/2+13/5+11/4+2(11/4+7/6)-11/4= 50+52+55/50+2(33+14/12)-11/4=157/50+47/6-11/4=157×6+47×50-11×75/300please solve you will get answer= |
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15. If the perimeter of a square is 64 cm, then find its area. |
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Answer» let the side of square 'a'perimeter of square is =4a 4a=64 a=64/4 a=16 area of square is =side×side =16×16 =256 |
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\left. \begin{array} { l } { 1.2 .3 + 2.3 .4 + 3.4 .5 + \dots } \\ { n ( n + 1 ) ( n + 2 ) ( n + 3 ) } \end{array} \right. |
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| 11. |
1 + 2 + 3 + \dots + n < \frac { 1 } { 8 } ( 2 n + 1 ) ^ { 2 } |
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Answer» thanks siri think you are the best |
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| 12. |
^{2 n} P_{n}=2^{n}\{1,3,5 \dots \ldots(2 n-1)\} |
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| 13. |
1 + 2 + 3 + \dots + n = 120 |
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| 14. |
The height of a cone is 16cm and its base radius is 12cm.Find the total surface area of the cone. [usell 314]Prove that the line segment joining the mid-points of two sidesof a triangle is parallel to the third side.9. Diagonals AC and BD of a quadrilateral ABCD interest eachother at P. Show that ar (AAPB) x ar (ACPD) = ar (AAPD) x ar(ABPC) |
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Answer» computer ka Khalistan diapride kaise hota hai |
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| 15. |
4 TheThe length and breadth of a rectangular field are in the ratio 3: 2. If the perimeter of the field is 250 mfind the cost of reaping the field at ?15 per 50 m2. |
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Answer» But it is wrong .here is asked the cost of reaping the field it is wrong all |
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Use a suitable identity to get(x+3) (x+ 3) |
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| 17. |
Practice Set 9.11700, selling price = * 1540 then find the discount.990 and percentage of discount is 10, then find the sellIl marked price =If marked price =price. |
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Answer» If marked =₹990 and percentage of discount is 10,then find the selling price |
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2Practice Set 3.11. Which of the following sequences are A.P.? If they are A.P. find the commondifference(3) -10, -6,-2,2,2,3' |
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Answer» 1.4-2=26-4=28-6=2 yes it is an AP with common difference 2 |
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Vu<bxc then(2) if axd=bx c thenII(3) if axd>bxc then>Practice Set 1.21. Compare the following numbers.(1) -7,-2 (2) 07 (3) 20(4)(10) |
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Answer» what is your question 1) -7 < -22) 0>-9/53) 8/7 > 0 |
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| 20. |
Practice set 2.Ientify, with reason, which of the following are Pythagorean triplet(ii) (4, 9, 12) i)(5, 12, 13)(v)(10, 24, 27) (vi)(11, 60, 61)(iv) (24, 70, 74) |
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Answer» for Pythagorean triplets , largest no. should be square root of the sum of the other 2 smaller no. this condition is following in 1) as 5²=3²+4²3) as 13²=5²+12²4) 74²=70²+24² 6) as 61²=60²+11² |
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(v) (x2-1)2 +8x (x2+1)+ 19x |
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| 22. |
umber lines and locate the points on themDraw n2 38' 8' 8' 8 |
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| 23. |
(i)1+ u1-u |
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Answer» what's the main question? What is the main question?? |
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| 24. |
.Ttiuitsareaat15GiThe perimeter of a square field is 8 km. Find its area in hectares.4oo ha |
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Answer» Perimeter of square field = 8 kmLet side length of square field is s Then,4s = 8s = 8/4 = 2 km Area of field = s*s = 2*2 = 4 sq km 1 sq km = 100 hectare Therefore, area of square field in hectares = 400 hectares |
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il) diea2.25 m(l(i) The perimeter of a square in 20 cm. Find its area.in)The perimeter of a square field in 8 km. Find its area in hectares. |
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| 26. |
4 (i) The perimeter of a square in 20 cm. Find its area.(i) The perimeter of a square field in 8 km. Find its area in hectares.t un into small squares o |
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Answer» i) Perimeter of a square= 4a 20=4aa=5 cmArea = a^2=5×5 =25 cm^2 ii) Perimeter=4a8 km =4aa=2 k mArea=a^2=4 km^2.1 km^2 =100 hectaresSo 4 km^2 =400 hectares. So it's area is 400 hectares i) p of sq. = 4×side20=4×side20÷4=side5cm=side Area =side×side =5×5 =25cm^2 ii)p=4×side 8km= 4×side 8km÷4 = side 2km = side area = side × side area = 2×2 area = 4km^2 1km^2=100 hectares therefore, 4km^2 = 400hectares since, area = 400hectares |
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The perimeter of a rectangular field is 0.6 km and its length is twice its breadth. Determine iarea of the field in hectares. |
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Answer» Please hit the like button if this helped you |
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| 28. |
9.A man owns a square plot measuring 400 m x 400 m. He wants to keep an area of 9 hectares with him andsell the remaining plot. How much money will he get if he sells the remaining plot at the rate of? 900 pesquare metre? |
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| 29. |
s. One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its costat the rate of ? 250 per m2. |
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lnud the le9. One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its cosat the rate of250 per m210. Find the |
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| 31. |
One side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its costat the rate of t 250 per mdun mensures (il 18 cm, (i) 20 cm. |
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| 32. |
เคนเฅฉ48x-11 |
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Answer» (1-x)/4 + (3x-1)/6 = x/2 - 1 1/61/4 - x/4 + x/2 - 1/6 = x/2 - 7/6x/4 - x/2 + x/2 = 7/6 + 1/4 -1/6x/4 = 6/6 + 1/4x/4 = 1 + 1/4x/4 = 5/4x = 5 Value of x = 5 |
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its legsOne side of a right-angled triangular scarf is 80 cm and its longest side is 1 m. Find its costat the rate of9.250 per m2le each of whose sides measures (i) 18 cm, (ii) 20 cm |
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9. Ibelow?f x = 9, what is the value of y in the equationy = 8x-( 1 + 4x) |
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Answer» when x=9 y= 8×9-(1+4×9)=72-37= 35 |
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8x-5x=3x+22 |
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Answer» 8x - 5x = 3x + 223x = 3x + 220 = 22, an absurd result therefore no solution for the given Equation. |
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6 \cdot 5 \div 0 \cdot 13 \times 1 \cdot 1 |
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Answer» by BODMAS rule6.5/0.13 ×1.1=50×1.1=55 |
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| 37. |
\tan 5^{\circ} \cdot \tan 25^{\circ} \cdot \tan 45^{\circ} \cdot \tan 65^{\circ} \cdot \tan 85^{\circ}=A. 5B. 0C. 2D. 1 |
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\tan 5^{\circ} \cdot \tan 25^{\circ} \cdot \tan 45^{\circ} \cdot \tan 65^{\circ} \cdot \tan 85^{\circ}=\ldots \ldots \ldotsA. 5B. 0C. 2D. 1 |
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| 39. |
How much paper is required to prepare 20 cone shaped caps of radius 14 cm and height 48 cm? |
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| 40. |
s A right circolar cylender has base radius 14 cm, and height 21 cm. Then find total surfacearea of the cy er(2 Marks) |
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| 41. |
( Rs 2068)13. A right circular cone is 3.6cm high and radius of its base is 1.6cm. It is melted and recast into a ricircular cone with radius of its base as 1.2cm. Find its height.(6.4 cm) |
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Answer» For cone, h = 3.6 cm, r = 1.6 cm Volume of the given cone = 1/3 x Π2h = (1/3 x 22/7 x 1.6 x .16 x 3.6) cm2 Radius of the second cone = 1.2 cm. Let its height be h1</sup> then, Volume of the second cone = 1/3 x Π r2h = 1/3 x 22/7 x 1.2 x h1 it is melted and recasted so ,the volume of the cones would be equal, therefore 1/3 x 22/7 x 1.2 x h1= 1/3 x 22/7 x 1.6 x .16 x 3.6 h1= (1.6 x 1.6 x 1.6) / (1.2 x 1.2) = 6.4 Hence the height of the second cone is 6.4 cm |
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| 42. |
The circumference of the circular base of a cylinder is 44 cm and its height is 15 cm. Thevolume of the cylinder is7. |
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Q. 20. The radius of the inetrele oftriangle is 4 cm and the segments intowhich one side ls divided by the point ofcontact 0.1e 6 cm ฼.nd 8 cm. Determinethe other twe sides of the triangleSol. |
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Answer» Area Δ ABC = 1/2 . 4 . (AB + BC + AC) = root( s(s – a)(s – b)(s – c))i.e., 4 s = root( s(s – a)(s – b)(s – c) )16 s = (s – a) (s – b) (s – c)i.e., 16 (14 + x) = x X 6 X 8, i.e., x = 7Therefore, AB = 15 cm and AC = 13 cm. |
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- 1+2.3 ing eQ. 20. The redius of the ineirele oftrianglo is cm ดเธ4 the segments intowhich one side is divided by the polnt ofcentaet aro 6 cm 8cm. Determinethe other twe sides of the triangle.Solgene-tan7s |
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Answer» Area Δ ABC = 1/2 . 4 . (AB + BC + AC) = root( s(s – a)(s – b)(s – c))i.e., 4 s = root( s(s – a)(s – b)(s – c) )16 s = (s – a) (s – b) (s – c)i.e., 16 (14 + x) = x X 6 X 8, i.e., x = 7Therefore, AB = 15 cm and AC = 13 cm. |
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Practice set 6.1Ifsin.:- , find the values of cosθ and tand25' |
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| 46. |
5. Find the perimeter of the following figure -0.5 cm4 cm IN2.5 cm1 cm4 cm2.5 cm0.5 cm |
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Answer» Perimeter of the figure =sum of all the sides. i.e. 1+4+0.5+2.5+2.5+0.5+4= 15cm is the best answer. 15 cm is the correct answer of the given question perimeter of the figure is the sum of its two sides so =1cm+4cm+0.5cm+2.5cm+2.5+0.5cm+4cm=15cm perimeter of the figure is the sum of its two sides so =1cm+4cm+0.5cm+2.5cm+2.5+0.5cm+4cm=15cm 1+4+4+0.5+0.5+2.5+2.5 = 15 cm |
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What is the circumference of a circular disc of radius 14 cm? |
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| 48. |
The length and the breadth of a rectangular park are in the ratio 8 : S. A path 1.5 m wide, running allaround the outside of the park has an area of 594 m. Find the dimensions of the park8x5x1.5 m |
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Answer» please give me the direct answer of this question |
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| 49. |
9. The length and the breadth of a rectangular park are in the ratio 8 :5. A path 1.5 m wide, running allaround the outside of the park has an area of 594 m2. Find the dimensions of the park8x1.5 m5x |
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Answer» Let Reactangle length = 8x andbreadth = 5xOuter area length = 8x + 3Outer area breadth = 5x + 3 Area of path = (8x+3)(5x+3) - 8x*5x594 = 40x^2 + 24x + 15x +9 - 40x^239x = 594-9 = 585x = 585/39 = 15 Length of park = 8*15 = 120 mBreadth of park = 5*15 = 75 m |
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Practice Set 42Complete the table below.Sr. No. Radius (r) Diameter (d) Circumference (c)(i) 7 cm28 cm(111)616 cm(IV) |
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Answer» i diameter-14cm circumference-44cm |
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