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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 2851. |
7^(2*m)/7^(-4)=7^(-6) |
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Answer» 4 is the correct answer 7²m/7-⁴=7^-67^(2m+4)=7^-62m+4=-62m=-6-42m=-10m=-5 |
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| 2852. |
99 \frac{1}{7}+99 \frac{2}{7}+99 \frac{3}{7}+99 \frac{4}{7}+99 \frac{5}{7}+99 \frac{6}{7}=? |
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| 2853. |
(4 x ^ { 4 } - 5 x ^ { 3 } - 7 x + 1 ) \div ( 4 x - 1 ) |
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| 2854. |
4*(x*(-4*x - 1)) - 5*(4*x - 1) |
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| 2855. |
over.tuse t=22/716. How many spherical lead shots each 4.2cm in diameter can be obtained from a rectanlead with dimensions 66cm, 42cm, 21cm.(use Tt-22/7)gular(1500)solid of |
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Answer» For sphereR = 4.2/2 = 2.1 cmLet the no. of spheres be x For cuboidl = 66 cmb = 42 cmh = 21 cm A/Q Volume of cuboid = volume of spheres x nlxbxh = 4/3πR³ x n66x42x21 = 4x22x2.1x2.1x2.1x n / 3x7n = 66x42x21x3x7 / 4x22x2.1x2.1x2.1n = 1500 spheres Ans. |
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| 2856. |
Unless stated otherwise, use Tt 22.1. Find the area of the shaded region in Fig. 12.19, ifpO 24 cm, PR 7 cm and O is the centre of thecircle |
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| 2857. |
sin A -2sin3 AtanA2cos3 A - cos A21. Prove that:ch-22.cubical block of side 10 cm is surmounted by a hemisphere. What is the largest diameter that thehemisphere can have ? Find the cost of painting the totalsurface area of the solid so formed, at the raternr5 per 100 sq. crn. I Use Tt-3.141 |
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| 2858. |
7]19.In the given figure, three circles each of radius 3.5 cm aredrawn in such a way that each of them touches the othertwo. Find the area of the shaded region.Use TT 22 and 3 1.732]7 |
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Answer» Area which is enclosed in between these three circles = area of equilateral triangle - area of sectors= Ar(ABC) - 3(area of sector) = √3/4 r² - 3(∅/360)(πr²) = (√3/4) - 3(60/360)(22/7)(3.5)²= (√3/4) - (1/2)(22)(7/4)= (√3/4) - (77/4)= (√3 -77)/4 cm² |
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| 2859. |
6.7.Find the circumference of a circle with diameter 35 cm. (UseFind the area of a circle whose circumference is 22 cmTt- |
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Answer» The solution for 1st question - Diameter = 35 cm Circumference = 2πr or πd = 22/7 × 35 = 22 × 5 (Since, 7 × 5 = 35) = 110 cm
The answer for 2nd question - circumference = 22 diameter = C ÷ π = 22 ÷ 22/7 = 22 × 7/22 = 7 (Since 22 ÷ 22 = 1)radius. = 7/2 = 3.5 Area of the circle = πr² = 22/7 × 3.5 × 3.5 = 22/7 × 7 = 22 cm² (As we cut 7 and 7) Thanks |
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| 2860. |
(1) Find the surface area and volume of a beach ball of diameter(Tt-22)42 cm. |
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Answer» Surface area of sphere = 4πr²= 4*(22/7)*(42/2)²= 4*22*21*3= 5544 cm² Volume = (4/3)πr³= (4/3)*(22/7)*(21)³= 38808 cm³ Please hit the like button if this helped you. |
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| 2861. |
a square of side 20 crn, verify that the areaofthecircumcircle is twice that of the incircle. [Use Tt 3.14] |
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| 2862. |
in Fig. 12.31, a square OABC is inscribed in a quadrant OPBQ. IrOA-20 em, find thearea of the shaded region. (Use Tt-3.14) |
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| 2863. |
Find the mode of above distribution.22 Water is flowing at the rate of 2.52 km/h through a cylindrical pipe into a cylindrical tank, theradius of whose base is 40 cm. If the increase in the level of waterin the tank, in half an hour is 3.15 m,find the internal diameter of the pipe. |
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| 2864. |
कटी IR 1 9.9x191 D1ज्ञात कीजिए ।एक वृत्त के परिगत एकचतुर्भुज 28009 खींचा गयाहै सिद्ध कीजिए किAB +CD=AD+ BCi |
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Answer» AP=AS ---1(Tangents drawn from an PB=BQ ---2 external point to a circle CR=CQ ---3 are equal)DR=DS ---4 By adding equation 1,2,3 and 4AP+PB+CR+DR=AS+BQ+CQ+DS (AP+PB)+(CR+DR)=(AS+DS)+(BQ+CQ) AB+CD=AD+BC |
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| 2865. |
DILM(neD(D) इनमें से कोई नहीं(प्र.8यदि cosx%D1-sin2x.O<x < , तो x का एकमान है -(A) tan-12 (B)(C)(D) इनमें से कोई नहीं |
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Answer» Finally tan inverse 2 is the answer (A) is the right answer finally tan inverse 2 is the correct answer |
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| 2866. |
Hiwicantwo numbers have 18 as theirHer and 380 en their Limyes reason no reason |
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Answer» No. Justification: HCF of two numbers is always a factor of their LCM. But here 18 is not a factor of 380. So, Two numbers cannot have 18 as their HCF and 380 as their LCM |
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| 2867. |
(x + 8) (x- 10) |
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| 2868. |
(x+8)(x-10) |
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Answer» hello |
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| 2869. |
a cone of height 15 cm and base diameter 30 cm is carried out of a wooden sphere of radius 15 m the percentage of wasted wood is |
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Answer» Volume of cone = 1/3 × π(15)^2×15=1/3π(15)^3 Volume of sphere = 4/3 π(15)^3 Required percentage = volumeofcone/volumeofsphere×100 = volumeofcone/volumeofsphere×100 = 1/3 × π × (15)^3/4/3 ×π(15)^3 × 100 = 1/4 × 100 = 25% |
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| 2870. |
A powder tin has square base with side 8 cm and height 13 cm. Another is cylindrical withbase radius 7 cm and height 15 cm. Which of the two contains more powder ? How muchis the difference in their capacities? |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 2871. |
The base of a powder packet is a square whose each side is 8 cm and its height is 13 cm. another packet is cylindrical whose base radius is 7 cm and height 15 cm. which packet will hold more quantity of powder ? |
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Answer» compare volumes of both and whichever is greater will have more powder |
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| 2872. |
The height of a cone is 15 cm. If its volume is 1570 cm^3, find the radiusheightcm, slant height ln)12he(3.(Use pi =3.14) |
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| 2873. |
lose it 22/ 7]20. In a group of 3 girls, one girl forgot to bring her lunch, so other two girls decided to share theirlunch with her lunch box. 1st girl's lunch box is in the shape of a cuboidal box measures 6 cm x 8 em xal shaped having radius 7 cm and height 15 cm. Which box15 em and of 2nd girl's lunch box is cylindrical shaped having radius 7 em and height 15 cm. Which boxhas moresurface area and which box has more volume ? Which value is depicted by girls ? |
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Answer» First girl's lunch box Volume v1 = length*breadth*height= 6*8*15 = 720 cm^3Surface area A1= 2(l*b + b*h + l*h)= 2(6*8 + 8*15 + 6*15)= 2(48 + 120 + 90)= 2(258) = 516 cm^2 Second girl's lunch box Volume v2= pi*r*r*h= 22/7*7*7*15= 22*105 = 2310 cm^2 Surface area A2= 2*pi*r(r + h) = 2*22/7*7(7 + 15)= 44(22)= 968 cm^2 Therefore, second girl's lunch box has more surface area and more volume. |
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| 2874. |
3. How many hours will it take to cover a distance of (x + 120) km with the speed40 km/h?(1) x +3 (2)x (3) +40 (4)+33340 |
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Answer» Speed = distance /timeTime= distance/speed= x+120/40x/40+3 km/h |
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| 2875. |
5. A submarine travelled 10 m below from the sea level and went 32m further down. Find the total distancetravelled by the submarine. What sign will you put in your final answer? |
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| 2876. |
10 Find the perimeter of a triangle whose sides are 3x +y, 5x 2y, and 3x -y.11 Sonu travelled (p2+3p+ 5) kilometres by bus and (2p2 -Sp-7) kilometres by train. What is the totaldistance travelled by Sonu? |
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| 2877. |
t6. A train travels for 12 hours, the first half of thedistance at kinph and otlHi hali D1 the distanceat 40 kmph. Find the total distance travelled |
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Answer» We knowDistance = speed*time For speed = 80 km/hrDistance D1 = 12*80 = 960 km For speed = 40 km/hrDistance D2 = 12*40 = 480 km Total distance travelled= D1 + D2= 960 + 480 = 1440 km |
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| 2878. |
1Shone that:a+b?-4abc |
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Answer» multiply R1 by c R2 by a R3 by b Then solve it further |
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| 2879. |
12. A man swimming in a river finds that it takes him four times as long to swim 1 km upstream as itdoes to swim back. The river flows at 1.5 km/h. Assuring that the man can swim x km per hourll water, form an equation and find x |
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| 2880. |
(e) wri5 Find the time interval between 6.IJFind the time interval from 9:45 a.m. to 2:30 p.m.for morning walk at 5:50 a.m. She came back at 7:0in the morning. It takeskilkreach Kilk |
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Answer» 9:45am to 1:45pm=4 hoursHencefrom 1:45pm-2:30=45minuteshence total time 4 hours 45minutes |
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| 2881. |
EXPRESS THE FOLLOWING IN THE FORM Of a+ibi^9+i^29 |
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| 2882. |
7 In winters, a school opens at 10 a.m. and closes at 3.30 p.m. If the lunch interval is of30 minutes, find the ratio of lunch interval to total time of the class periods. |
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Answer» class time = 5 hours ratio = 30:5*60 minutes = 30:300 = 1:10 |
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| 2883. |
77 In winters, a school opens at 10 am. and closes at 3.30 p.m. If the lunch interval is of30 minutes, find the ratio of lunch interval to total time of the class periods. |
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Answer» no of school hours =5h30min=330minlunch time is 30minremaining class periods = 300minratio=30/300=1/10 is the answer. |
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| 2884. |
Write the expanded form of 854.HTC8 5 450800Short form = 854Expanded form = 800 + 50 + 4Maths-210 |
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Answer» expanded Form -800+50+4 |
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| 2885. |
I. Simplify("[一0xl_I 지__|×(기in exponential formform |
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| 2886. |
\begin{array} { l } { \text { In } \triangle A B C \text { , if } a \cos A = b \cos B , \text { then prove that } } \\ { \text { the triangle is either a right-angled or an } } \\ { \text { isosceles triangle. } } \end{array} |
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| 2887. |
1271) Write127/200 decimai form.1n decimai form. |
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| 2888. |
Change the following to general formto-29 general form9 |
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| 2889. |
X't y = 11of value is ac oryI toc=7 |
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Answer» x=4y=9 are the answers. x=4 is the write answer x=4y=9it is the correct answer 4259908566888588996556 x=4 andy=9 is the correct answer x =4 and y= 9 are the correct answer the solving answer is x=4, y=9 this is the right answer that,s qustion x = 4 and y = 9 is right answer |
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| 2890. |
sqrt(cosec^2*A(-cos(A)^2 %2B 1))=1 |
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Answer» To prove: √(cosec²A(1-cos ²A)=1 Solution√(cosec²A(1-cos ²A)=√(cosec²A sin²A)=√(1/sin²A)(sin²A)=1RHS Hence proved |
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| 2891. |
((tan(A)^2 %2B 1)*cot(A))/((A*cosec^2)) |
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| 2892. |
1. Which of the following quadratic equations are in standard form? Those., which arenot in standard form, rewrite them in standard form:(i)3y,-2 y+1(ii) 5-3x -2x2-0(iv) 5-x = 3x2 |
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| 2893. |
EXERCISE 2.41. Determine which of the following polynomials has a+ 1)a factor:(i) x+x+x +1(iii) x4 + 3x1+ 3x' +x + 1 |
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Answer» thanks |
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| 2894. |
3x²-2√6x+2=0 |
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Answer» 3x²-2√(6)x+2=0 3x²-√(6)x-√(6)x+2=0 √(3)x(√(3)x-√(2))-√(2)(√(3)x-√(2))=0 (√(3)x-√(2))(√(3)x-√(2))=0x=√(3/2),√(3/2) |
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| 2895. |
ich ofthe following can be the sides of a right btriangle?(i) 2.5 cm,65cm, 6cm-(ii) 2 cm, 2 cm, 5 cm. g(ii) 1.5 cm, 2cm, 2.5 cm. 304. Wh |
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| 2896. |
10. Pradeep has 10,080, in whe the number ofich 500-rupee notes ar100-rupee notes, the 50-rupee notes are half the numbernotes andthe number of 100-rupee notes that Pradeep has.f 100-rupee1-rupee notes are twice the number of 100-rupee notes. Find(2017)(1) 20(2) 30(3) 40(4) 50 |
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| 2897. |
(ii) 5x2 -6x-2 0 |
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Answer» 5x^2 -6x -2 = 0 x^2 -(6/5)x -(2/5) = 0 (x)^2 - [2.(3/5).x] + (3/5)^2 - (3/5)^2 -(2/5) = 0 [x-(3/5)]^2 = (2/5)+(9/25) [x-(3/5)]^2 = (19/25) x-(3/5) = √19/5 x = (3/5) ± √19/5 Roots of the equation are x = [3+(√19)]/5& x = [3-(√19)]/5 Given: 5x^2-6x-2=0 Now we follow the algorithm Step-1: x^2-6÷5x-2÷5=0 (dividing both sides by 5 ) Step-2: x^2-6÷5=2÷5 Step-3: x^2-6÷5+[3÷5]^2=2÷5+[3÷5]^2 (adding [3÷5]^2 to both sides ) Step-4: [x-3÷5]^2=2÷5+9÷25 Step-5: [x-3÷5]^2=19÷25 Step-6: x-3÷5=+-√19÷25 Step-7: x=3÷5+√19÷5 or x=3÷5-√19÷5 Therefore... x=3+√19÷5 or x=3-√19÷5 |
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| 2898. |
Solve 3x2-6x+2=0 by completing the square. |
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| 2899. |
neYU- In the given Sigor ich ory= utz, theforove that dob ich a cline e |
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Answer» Given=x+y=w+z=>we can say( w+z= x+y)=>(x+y)+(x+y)=360. (complete angle)=>2(x+y) =360=>x+y =180. Hence sum of angle of straight line is 180=>AOB is a straight line. |
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| 2900. |
1/(1/(1/(1 %2B 1) %2B 1) %2B 1) %2B 1 |
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Answer» 8/5 please correct answer answer the question 8/5 is the right answer 8/5 is right answer..... |
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