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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.
| 77751. |
\frac { 3 \times ( 27 ) ^ { n + 1 } + 9 \times 3 ^ { ( 3 n - 1 ) } } { 8 x ^ { 3 } - 5 x ( 27 ) ^ { n } } |
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Answer» (27)^(n+1)=(3^3)^(n+1)=3^(3n+3)27^n=(3^3)^n=3^3ntake 3^3n common from numerator and denominator 3^3n{(3×3^3)+(9×1/3)}/3^3n {8-5}(81+3)/384/328 |
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| 77752. |
the viuen he receives anually.20. Aman wants to buy 62 shares available at * 132(par value being * 100).(1) How much he will have to invest.(ii) If the dividend is 7.5%, what will be his annualincome?(iii) If he wants to increase his annual income by* 150, how many extra shares should be buy?(ICSE 2002) |
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Answer» 1. He will have to invest= 62× Rs 132= Rs 8184 2. Dividend on 1 share= 7.5% of Rs 100=Rs 7.50 Therefore, his annual income = 62× Rs 7.50= Rs 465 |
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| 77753. |
A shopkeeper bought 30 kg of wheat at the rate of Rs. 45 per kg. He sold forty percent of thetotal quantity at the rate of Rs. 50 per kg. Approximately at what price per kg should he sellthe remaining quantity to make 25% overall profit?1. 702. 753. 504. 555. 60 |
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Answer» Like my answer if you find it useful! |
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| 77754. |
eld is in the shape of a trapezium whose parallelsides are 25 m and 10 m. The non-parallel sidesare 14 m and 13 m. Find the area of the field. |
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Answer» AB || CDAB =25 m CD= 10m BC=14mAD= 13m From point C draw CE || DA . Hence ADCE is a ||grm having CD || AE & AD||CE AE= CD= 10mCE= AD= 13mBE= AB- AE =25- 10=15 m BE= 15m In ∆BCEBC= 14m, CE= 13m, BE= 15m Semiperimeter (s)= (a+b+c)/2 Semiperimeter(s) =( 14+13+15)/2s= 42/2= 21ms= 21m Area of ∆BCE= √ s(s-a)(s-b)(s-c)Area of ∆BCE=√ 21(21-14)(21-13)(21-15) Area of ∆BCE= √ 21×7× 8×6 Area of ∆BCE= √ 7×3× 7× 4×2×2×3Area of ∆BCE=√7×7×3×3×2×2×4Area of ∆BCE= 7×3×2×2= 21× 4= 84m² Area of ∆BCE= 84m² Area of ∆BCE= 1/2 × base × altitudeArea of ∆BCE= 1/2 × BE ×CL 84= 1/2×15×CL 84×2= 15CL168= 15CLCL= 168/15CL= 56 /5m Height of trapezium= 56/ 5m Area of trapezium= 1/2( sum of || sides)( height) Area of trapezium=1/2(25+10)(56/5)Area of trapezium= 1/2(35)(56/5) Area of trapezium= 7×28= 196m²_____________________________Hence the area of field is 196m² |
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| 77755. |
10 3(m) 35. (a) Simplify 3x (4x-5) + 3 and find ts values for (i) x = 3 (ii)r-(b) Simplify ala++)+5 and find its value for (i) a - 0, Ginaaq-r) and r (r- p))x=3(ii) x=-2-(iii) a =-1. |
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Answer» thank you |
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| 77756. |
An airplane takes off at 4:00 p.m. and lands at an airport 8kilometers away at 6:00 p.m. What was the average speed of theairplane?A 360 km/hrB 410 km/hrC 720 km/hrD 1640 km/h |
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Answer» Total Distance = 820 kmTotal time = 2 hrs Average speed = Total Distance / Total time = 820/2 = 410 km/hr |
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| 77757. |
6. If a car travels 67.5 km in 4.5 litres of petrol, how many kilometreswillittravelin26.4 litres of petrol? |
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Answer» अच्छा saa lekho |
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| 77758. |
In each of the following assume that the base is 10. Prove that:(i) log (1/2) + log (2/3) + log (3/4) + log (4/5) - log (1/5) = 0 |
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| 77759. |
(iii)3x-5y-4-0 and 9x = 2y + 7 |
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Answer» 3x-5y=4and 9x-2y= 7 |
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| 77760. |
(iii) 3x-5y-4-0 and 9x=2y+7 |
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| 77761. |
2x-3 / 3x+2 = 2/3 |
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Answer» like if you find it useful please like if you find it useful |
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| 77762. |
Solve for x: 3(3x-1)-2(2x+3)-5x#1,-34.2x +3(3r-1 |
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| 77763. |
3 ( \operatorname { log } 5 - \operatorname { log } 3 ) - ( \operatorname { log } 5 - 2 \operatorname { log } 6 ) = 2 - \operatorname { log } a , f |
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Answer» thanks |
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| 77764. |
f two numbers is 27. The numbers can not be(B) 1, 27(A) 3, 27(E) of these(C) 3, 9(D) 9, 27 |
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Answer» option c) 3,9 |
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| 77765. |
thewritemoltiply Cor 5 (m²+2m-2.)the degree of product (3 m |
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| 77766. |
.Findthe value of 2 (cost 60°+sin 30)(tan2 60° +cor 45°)+3sec2 300 |
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Answer» 2[ cos⁴60 + sin⁴60] - [tan²60 + cot²45] + 3sec²30 = 2 [ (1/2)⁴ + (√3/2)⁴] - [(√3)² + (1)²] + 3(2/√3)² [Putting values of cos, tan sin 60; cot 45 sec 30] = 2[ 1/16 + 9/16] - [ 3 + 1] + 3 (4/3) [ Evaluating the values] = 2[10/16] - 4 + 4 = 10/8 = 5/4 Like my answer if you find it useful! |
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| 77767. |
(27^(-1/3)*27^(1/3))*(-9 %2B 27^(1/3)) |
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| 77768. |
A company with 4000 shares of nominal value of 110 declares annual dividendof 15%. Calculate:() the total amount of dividend paid by the company.(i) the annual income of Shah Rukh who holds 88 shares in the company.(iii) if he received only 10% on his investment, find the price Shah Rukh paid for each(2008)share. |
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Answer» (i) Total dividend = r/100*N.V*no. of shares = 15/100*110*4000 Rs.66,000 (ii) Annual Income = r/100 * N.V * No. of shares = 15/100 * 110 * 88 Rs.1,452 (iii) Return Percent = 10% Rate of dividend/100* N.V = Rate of Return/100 * M.V 15/100 * 110 = 10/100 * x x = Rs.165 |
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| 77769. |
11. Find the cost of watering a trapezoidal ficld whose parallel sides are 10 m and 25 m respectively, tieperpendicular distance between them is 15 cm and the rate of watering is R 4 per m'. |
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Answer» Considering the distance between parallel sides (h) = 15 MLet one parallelside of the trapezoid be `a` = 10mAnother parallel side be `b` = 25mThe height that is the distance between the parallel sides = 15mArea = [(a+b)/2]h = [(10+25)/2×15] m² = [35/2× 15] m² = [17.5× 15]m² = 262.5 m²Cost of watering one m² area = Rs 4Cost of watering 262.5 m²area = Rs 1050 thanks |
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| 77770. |
7/A train covers a distance of 51 km in 45 minutes. How long will it take to cover 221 km?R If22 5 metreo |
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| 77771. |
Travelling 900 km by rail costs 2520. What would be the fare for a journeywhen a person travels by the same class?of360km221 km? |
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| 77772. |
ntegrate the functions in Exercises 1 to 37:2xlog x1.2.3.·x + x log x" |
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| 77773. |
a car uses 30 litres of petrol to cover distance of 375 km how many litres will be needed to cover distance of 1200 km |
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Answer» It's a question of direct variationIf 375 km distance require petrol = 30 lit1 km distance require petrol = 30/375Then to cover 1200 km distance petrol required = (30/375)*1200 = (10/125)*1200 = 12000/125 = 96 litres |
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| 77774. |
If(5.k) is a solution of the equation 2x +y-7 0, find the value of k.Solve the following equations: |
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Answer» (5,k) is solution of 2x +y -7 = 0Therefore put x = 5 and y = k , in the given equation.So,2×5 + k - 7 = 010 + k -7 = 0k = -3 |
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| 77775. |
Solve the system of equation by elimination methodi) 2x - 5y 4 3x -2y16li) 3x +2y 5, 2x -3y 7 |
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Answer» like my answer if you find it useful 2 tq i another question sets |
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| 77776. |
\log 360=3 \log 2+2 \log 3+\log 5 |
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| 77777. |
Integrate the following(log x)222x3.1.2or |
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Answer» sir , integration of t to the power 2/3 is thank you |
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| 77778. |
log 3 + 3 log 5 - 5 log 2 as a single logarithm is2 log 3 + 3 log 5 - 5 log 2 esa as nossosort 4 |
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Answer» log (1125/32 ) is the correct answer of the given question |
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| 77779. |
\log _{2}\left[\log _{2}\left\{\log _{3}\left(\log _{3} 27^{3}\right)\right\}\right] |
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Answer» log2(log2(log3(log3 3^9)))=log2(log2(log3 9))=log2(log2(log3 3^2))=log2(log2 2)=log2 1=0 |
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| 77780. |
divide rupees 3600 into two parts such that if one part be lent at 9% per annum and the other at 10% per annum the total annual income is rupees 333 |
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| 77781. |
Find the cost of watering a trapezoidal field whose parallel sides are 10 m and 25 m respectively, the perpendicular distance between them is 15 cm and the rate of watering is 4 rupees per m |
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Answer» Considering the distance between parallel sides (h) = 15 mLet one parallelside of the trapezoid be `a` = 10mAnother parallel side be `b` = 25mThe height that is the distance between the parallel sides = 15mArea = [(a+b)/2]h = [(10+25)/2×15] m² = [35/2× 15] m² = [17.5× 15]m² = 262.5 m²Cost of watering one m² area = Rs 4Cost of watering 262.5 m²area = Rs 1050 right answer can you help me second answer The area of a Trapezium is 300 m square the perpendicular distance between the two parallel side is 15 M if the difference of The parallel sides is 16 m find the length of the parallel side |
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| 77782. |
What is this symbol Ď |
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Answer» it denotes magnetic flux in physics. |
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| 77783. |
length 80 m. Which plot has the gie13. A rectangular field is 60 m wide and its perimeter is 320 m. If laying of grass costs.50 per squaremetre, how much would it cost for grass to be laid on the field leaving a bare path 4 m wide all aroundthe field?What is the area of unshaded region? |
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Answer» B=60mp=320mnow,2(l+b)=320l+60=160l=160-60l = 100mnow,new length =100-(2×4)=92mnew bredth=60 -8=52marea=92×52=4784 m squrecost =4784×7.5=35,880 |
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| 77784. |
A train covers a distance of 51 km in 45 minutes. How long will it take to cover 221 km |
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Answer» 221*45/51=195 minutes |
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| 77785. |
What is this symbol |
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Answer» (^) this symbol is known as exponent or power ^ is the power or exponent. |
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| 77786. |
(ICSE 2155. Solve the equation 2x--= 7 . Write your ansvICSE 20correct to two decimal places. |
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| 77787. |
3. Solve the equation 3x2x-7 0 and give youranswer correct to two decimal places. ICSE 2004) |
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| 77788. |
12. If log, 2, log, (25) and log, 2"are in A.P., thenx is equal to2(d) of these2 |
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| 77789. |
If y=2^{1 / \log _{x} 4}, then x is equal to |
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| 77790. |
40/ The value of sin! To1S181818(d) 1/1641. The value of log tan 1° + log tan 2 +... +log tan 89, is(a) 0(b) -1(c) 1(d) oo42. If 1 + sin x + sin2x+sin3x +...+... o is equal to4 + 2,13, 0 < x < Ď, then x= |
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Answer» Crop only the question that you want a solution for. We will not be able to provide solutions to multiple questions. |
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| 77791. |
(D) x €COINQuestion LogarithmsLogarithmsbased on.22. - 1-+log/be abc log.co abc log jab abcthe value equal to(A)(B) 1(C) 2(D) 4. |
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| 77792. |
\log x^{2}-\log 2 x=3 \log 3-\log 6 \text { then } x= |
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| 77793. |
t is the degree of a zero polynomial? |
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| 77794. |
18.The taxi fare in a city is charged according to the rate stated as: for the first3 km distance, the fare is Rs 10 and for the subsequent distance is Rs 4 perkm. Taking the distance covered as x Km and the total fare as Rs y, write alinear equation for the above and draw its graph From the graph find thedistance travelled when the amount paid is Rs 60. |
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Answer» Taxi charges for the first km = Rs 20Taxi charges for the subsequent distance covered = Rs 12 per kmDistance covered = x kmgivengivenTotal fare = RsygivengivenFare charged for the distance covered leaving the first kilometre = Rs 12x−1x−1So, taxi charge = 20 + 12x−1x−1But total fare = Rs ygivengiven∴ y = 20 + 12x - 12 = 12x + 8Hence, the required relation between x and y is y = 12x + 8For,x=0x=0⇒y=8⇒y=8For,x=3x=3⇒y=44⇒y=44Plot these pointsA(0,8),B(3,44)A(0,8),B(3,44)on the graph paper.Join them and extend in both the directions.Then,x=16x=16givesy=200y=200.So, for 16 km taxi charge = Rs 200. |
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| 77795. |
The taxi fare in the city is as follows: For the first two kilometers, the fare is 25 rupees and theremaining distance is 10 rupees per kilometer. Taking the distance covered as x and the totalfare as 'y, write a linear equation for this situation. |
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| 77796. |
8. If D and E are points on sides AB and AC respectively of a Δ ABC stCE. Prove that Δ ABC is isosceles.DE 11 BC and BDCBSE 200T |
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| 77797. |
2 In ΔABC, D and E are points on the sides AB and ACrespectively such that DE 11 BC. If AD = 4, DB-x-4AE = 8 and EC = 3x-19, then find the value of x.[Ans. x =11] |
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Answer» if DE || BC ,. then AD/DB = AE/EC => 4/(X-4) = 8/(3X-19)=> 3X-19 = 2X-8=> X = 19-8 = 11 |
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| 77798. |
The taxi fare in the city is as follows: For the first two kilometers, the fare is 25 rupees and theremaining distance is 10 rupees per kilometer. Taking the distance covered as x and the totalfare as 'y', write a linear equation for this situation. |
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| 77799. |
11.A car covers a certain distance in 50 minutes if it runs at 45 km per hour. How much time will ittake to cover the same distance at a speed of 30 km per hour? |
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| 77800. |
Rohan covers a distance of 12 km in45 minutes by his bicycle. Find the speed ofthe bicycle per hour. |
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Answer» a cyclist cover 0.9 km in 3 min . find the speed in km per hr 1 hour 45 minutes past 11 A. M is |
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