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| 67351. |
* Measure the side of the red square on the dotthere as many rectangles as possible using 12 su* How many rectangles could you make?the dotted sheeto 12 such squareHere suEach rectangle is made out of 12 equalsquares, so all have the same area, but thelength of the boundary will be different.Length of theboundary is cokeperimeter.perimeter?* Which of these rectangles has the longest+ Which of these rectangles has the smallest perimeterdefinition of the term 'area', but develop a sensepportunities in the classroomChildren are not expected to learn the definition of the term 'areaconcept through suitable examples. Give them many opportunities interms of area and guess which is bigger. Things like stalfootprints, walls of the classroom etc. can be comparedgs like stamps, leal |
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| 67352. |
और ।3Bir tan (A - B) ==:0°<A+ B <90°: A> B तो Aaf tan (A + B) =कीजिए। |
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Answer» tan(A+B)=V3, A+B=60, (tan60=V3)________________(1) Tan(A-B)=1/V3 (tan30=1/V3)______________(2) solving 1&2 equations; A+B=60; A-B=30/2A=90. A=45, 45+B=60, B=60-45=15, A=45, B=15 |
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| 67353. |
(i) x+y+z=11, x2 + y2 + 2 = 45 si xyz = 40 |
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| 67354. |
44. If the product as the series of polgromet an-bn-bir4, find the value of a..Tap] |
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Answer» ax^2-6x-6 ; is 4, ; a(4)^2-6(4)-6=16a-24-6=0; 16a=24+6=30;; a=30/16=15/4 a=30/16 =15/4 answer |
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| 67355. |
Use Euclid's division algorithm to find the HCF of:0) 135 and 225Show that any positive oddi1.(ii)196 and 38220(iii) 867 and 2552.nt |
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Answer» thank you very much |
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| 67356. |
2)Find the height of a triangle having an area of 72 cm2 and base 16 cm |
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| 67357. |
one number is six times the other. If their sum is 147, find the numbers |
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Answer» The sum is 24 because if you Divide 147÷6=24 |
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| 67358. |
(b)If AB and MN bisect each other at O and AM and BN are perpendicularsto XY, prove that Δ。AM and Δ。BN are congruent and hence prove thatAM = BN. |
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Answer» AB and MN bisect each other therefore AO=BO,MO=NOin TriangleAMO and BNOangle AMO=angleBNOAO=BOMO=NOby RHS congruence rule triangleAMO and BNO are congruenttherefore triangleOAM andOBN are also congruent and hence AM = BN |
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| 67359. |
Rhombus BENDBN 5.6 cmDE 6.5 cm |
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| 67360. |
ConstructRhombus BENDBN= 5.6 cmDE= 6.5 cm |
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| 67361. |
Find the volume of a sphere whose surface area is 154 cmicnr. |
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| 67362. |
The diameter of a roller is 84 cm and its lengthǐsT20emTt takes 500revc itions to move once over to level a playground. Find the area of the playgroundin mcompleteナ |
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Answer» Thanks |
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| 67363. |
(ii) Rhombus BENDBN = 5.6 cmDE = 6.5 cm |
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| 67364. |
2olutions of the each of the folloii) y6x |
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Answer» send the full question . possible solution arex=1 then y = 6x= 0 then y = 0x = 2 then y= 12 |
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| 67365. |
2. Factorise the following expressions :(a) x2 + 4y2 + z² + 4xy - 2xz - 4yz |
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Answer» x^2 + 4y^2 + z^2 + 4xy - 2xz - 4yz Using identity(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = (x)^2 + (2y)^2 + (-z)^2 + 2*(x)*(2y) + 2*(x)*(-z) + 2*(2y)*(-z) = (x + 2y - z)^2 = (x + 2y - z)(x + 2y - z) |
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| 67366. |
How many times hands of a clock coincide in a day :-(A) 10(B) 22(C) 11 (D) 24 |
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| 67367. |
How many times hands of a clock coincide in a day -(A) 10(B) 22(C) 11(D) 24 |
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Answer» option B The hands of a clock coincide11 timesin every 12 hours (Since between 11 and 1, they coincide onlyonce, i.e., at 12 o'clock). The hands overlap about every 65 minutes, not every 60 minutes. The hands coincide22 timesin a day. (B) 22 times. At 12:00 am since between 11:00 and 1:00--> 1 time few min. after 1 am, 2 am,...upto 10am. --> 10 times Total 11 times similarly 11 times in on so, total 22 |
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| 67368. |
02) Howmany times hands of a clock coincide in a day :-(A) 10(8) 22(C) 11(D) 24 |
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Answer» Explanation: The hands of a clock coincide11 timesin every 12 hours (Since between 11 and 1, they coincide onlyonce, i.e., at 12 o'clock). The hands overlap about every 65 minutes, not every 60 minutes. The hands coincide22 timesin a day. |
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| 67369. |
If the base of rectangle is increased by 10 % and the area is unchanged then the corresponding altitude must be decreased by |
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| 67370. |
relationship between the zeroes and the coetbcFind a cubic polynomial with the sum of its zeroes,vo at a time and product of its zesum of the products of ILL I76. Froes being given beloso. 4, 1 and 617, 5 and-152.51. 1, 1 and 26and 1nt and the remainder on dividing p(x) by g(x) for |
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| 67371. |
(i) x+y+z= 11, x2 + y2 + 2 = 45 Bir xyz = 40 |
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| 67372. |
Tind the surface area of the sphere of radius 9 cm.(3.14) |
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Answer» Surface area is sphere is 4πr^2=4*3.14*9*9=1017.36cm^2 the radius of sphere(r)=9cm =4πr2 =4×3.14×9×9 =1017.36cmsqu. total surface area of sphare =1017.36cmsq. |
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| 67373. |
Find the volume of a sphere whose surface area is 154 cm2 |
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| 67374. |
L.ength of a rectangular ficld is six times its breadth. IF the length of the feldthe breadth and perimeter of the field.readth. If the length of the feldis 114 m, indn: |
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Answer» Length of a field = 114mBreadth = 114/6 m= 19 m Perimeter of the rectangle= 2( Length + Breadth)= 2( 114 + 19)= 2 x 133= 266 m |
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| 67375. |
hemisphere of lead of radius 9 cm is cast into a right circular cone ofheight 72 cm. Find the radius of the base of the cone.13. A |
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| 67376. |
.Find the radius of a sphere whose surface area is 154 cm'. |
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| 67377. |
n+ia anan+bnind the value on21 |
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| 67378. |
C) bo011.11, 12, 13, 14 3永ㄨ矾13 |
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| 67379. |
(a) Find the value of a, if x - a is factor of x3 - ax+x+2.(b) If a + b + c 9 and ab + bc + ca-26, find value of аз + b3 + c3-3abc. |
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| 67380. |
How many times hands of a clock coincide in a day-(A) 10(B) 22 (C) 11 P)2422(c) 11 )24 |
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Answer» The hands of a clock coincide11 timesin every 12 hours (Since between 11 and 1, they coincide onlyonce, i.e., at 12 o'clock). The hands overlap about every 65 minutes, not every 60 minutes. The hands coincide22 timesin a day. |
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| 67381. |
17. If the base of a rectangle is increased by10% and the area is unchanged, then thecorresponding altitude must be decreasedby(a)10%(b) 944%(c) 11%(d) 11-% |
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| 67382. |
W) g °C, A°C, =11:5 तो / का मान ज्ञात कॉजिए। |
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Answer» =>15Cr / 11 = 15Cr-1 / 5=>15Cr / 15Cr-1 = 11/5=>15-r + 1 / r = 11/5(nCr / nCr-1 = n- r+1/ r )=>80 = 11r + 5r=>r= 5 hit like if you find it useful |
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| 67383. |
Find the product.1. (x+y-z)(x2+y2 + z2-xy + yz + zx)4. (Gr-5y+4)9+25y +15xy-20y+ 12r+ 16)Factorise:5. 125a +b3 +64c-60abc6. a3+8b3+64c3-24abcs. 216+27b3+c3-10Sbc10. Sa +125b-64c+120abo12. 125-8x3-27y3-90xy9. 27a-b+8c3+18abc27b3-343c3-126bc |
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Answer» you should always specify the question you wants to be Answered |
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| 67384. |
Alfa + b6and ab-8, find : a3 + b3 |
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| 67385. |
If a-b=4 , ab = 5, find = a3 - b3 |
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| 67386. |
d the missing measures:(11) Area of 11 gmPORS = 252 cm"PO 9 cm |
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Answer» Area of parallelogram=base*height252 cm²=PQ*RT 252 cm²=9*RTRT=252/9=28 cm |
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| 67387. |
If the circumference of a circular sheet is 154 m, find its radius. Also find the area ofthe sheet. (Take π ) |
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| 67388. |
rcumference of a circular sheet is 154 m,find its radius. Also find the atthe sheet. (Take π-) |
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| 67389. |
2. Draw a circle of radius 9 cm and write its longest chord after measuring,basd pO of the circle |
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Answer» Circle of radius 9cm will have longest cord of 18 centimeters i.e it's diameter. |
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| 67390. |
i in the tao caesUtass has inner diameter 105 m.em.Fid the cot eit on the inside at the rate of16sphere whose surface area is 154 emeter of the imoon is approu6- |
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Answer» 6.Given:-The sphere has a surface area of 154 cm².∴4πr²=154=>r²=154/(4π)=12.25=>r=√12.25=3.5 cm ∴Radius of the sphere=3.5 cm |
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| 67391. |
evaluate the following limits in exercise lim x +3x=3 |
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| 67392. |
4: In Fig. 12.15, two circular flower bedsbeen shown on two sides of a square lawnof side 56 m. If the centre of each circularer bed is the point of intersection O of theof the square lawn, find the sum of the Dof the lawn and the flower beds. |
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Answer» thannks mam😀 |
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| 67393. |
Q.9 F"pr-720 and nCr= 120ind n and r if |
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Answer» Like my answer if you find it useful |
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| 67394. |
í¸Example 4: In Fig. 12.15, two circular flower bedshave been shown on two sides of a square lawnABCD of side 56 m. If the centre of each circularflower bed is the point of intersect ion O of thediagonals of the square lawn, find the sum of theC.areas ofthe lawn and the flower beds.Fig. 12.15 |
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Answer» Given:Side of a square ABCD= 56 mAC = BD (diagonals of a square are equal in length)Diagonal of a square (AC) =√2×side of a square.Diagonal of a square (AC) =√2 × 56 = 56√2 m. OA= OB = 1/2AC = ½(56√2)= 28√2 m. [Diagonals of a square bisect each other] Let OA = OB = r m (ràdius of sector) Area of sector OAB = (90°/360°) πr² Area of sector OAB =(1/4)πr²= (1/4)(22/7)(28√2)² m²= (1/4)(22/7)(28×28 ×2) m²= 22 × 4 × 7 ×2= 22× 56= 1232 m² Area of sector OAB = 1232 m² Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m² Area of flower bed AB =area of sector OAB - area of ∆OAB= 1232 - 784 = 448 m² Area of flower bed AB = 448 m² Similarly, area of the other flower bed CD = 448 m² Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD=(56× 56) + 448 +448= 3136 + 896= 4032 m² Hence, the sum of the areas of the lawns and the flower beds are 4032 m². Like my answer if you find it useful! |
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| 67395. |
:In Fig. 12.15, two circular flower bedshave been shown on two sides of a square lawnABCD of side 56 m. If the centre of each circularlower bed is the point of intersect ion O of thediagonals of the square lawn, find the sum of theareas of the lawn and the flower beds. |
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Answer» total area = area of sector OAB+ area of sector ODC + area of triangle OAD + area of triangle OBC=(90/360 × 22/7 × 28 × 56 + 90/360 × 22/7 × 28 × 56 + 1/4 × 56 × 56 + 1/4 × 56 × 56) m^2=1/4 × 28 × 56( 22/7 + 22/7 + 2 + 2)m^2=7×56/7(22+22+14+14)m^2=56×72 m^2=4032m^2 |
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| 67396. |
If the circumference of a circular sheet is 154 m, find its radius. Also find the areathe sheet. (Take π =22/7 ) |
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Answer» circumference of circle=2πr154=2*22/7*r (given circumference=154m)r=(154*7)/(2*22)r= 7*7/2r=24.5m area of circle=πr^2 =22/7*(24.5)^2 =1885.7m |
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| 67397. |
enamplesExample 4: In Fig. 12.15, two circular flower bedshave been shown on two sides of a square lawnABCD of side 56 m. If the centre of each circularflower bed is the point of intersection O of thediagonals of the square lawn, find the sum of theareas of the lawn and the flower beds.56mFig. 12.15 |
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Answer» Given:Side of a square ABCD= 56 mAC = BD (diagonals of a square are equal in length)Diagonal of a square (AC) =√2×side of a square.Diagonal of a square (AC) =√2 × 56 = 56√2 m.OA= OB = 1/2AC = ½(56√2)= 28√2 m.[Diagonals of a square bisect each other]Let OA = OB = r m (ràdius of sector)Area of sector OAB = (90°/360°) πr² Area of sector OAB =(1/4)πr²= (1/4)(22/7)(28√2)² m²= (1/4)(22/7)(28×28 ×2) m²= 22 × 4 × 7 ×2= 22× 56= 1232 m²Area of sector OAB = 1232 m²Area of ΔOAB = ½ × base ×height= 1/2×OB × OA= ½(28√2)(28√2) = ½(28×28×2)= 28×28=784 m²Area of flower bed AB =area of sector OAB - area of ∆OAB= 1232 - 784 = 448 m²Area of flower bed AB = 448 m²Similarly, area of the other flower bed CD = 448 m²Therefore, total area = Area of square ABCD + area of flower bed AB + area of flower bed CD=(56× 56) + 448 +448= 3136 + 896= 4032 m²Hence, the sum of the areas of the lawns and the flower beds are 4032 m². Like if you find it useful! |
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| 67398. |
Example 4: In Fig. 12.15, two circular flower bedshave been shown on two sides of a square lawnABCD of side 56 m. If the centre of each circularflower bed is the point of intersection O of thediagonals of the square lawn, find the sum of theareas of the lawn and the flower beds.O 56mFig. 12.15 |
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| 67399. |
a a aक का मान जिसके fow agag 13 + 4x? - px + 8 Ui(#-2) से भाज्य है-- |
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Answer» Please hit the like button if this helped you |
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| 67400. |
O 77 A bus covered 150 kila disiance in luuss(a) /fow muell time , s re lilred toă0ăŤ1 250 km with the same lee(bj Find the dislanec covered in 4 hours with the sane speed |
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