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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.
| 72701. |
Example1. Verify Rolle's Theorem for the function f(x) rn for the function(4 x 3 in the interval |
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| 72702. |
In an acute angled triangle ABC, if tan(A+B-c) = I and sec(B + C-A)=2, find the values ofA, B and C. |
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Answer» Tan ( A + b - c) = 1 tan ( a+ b - c) = tan 45°a+b+-c = 45°. {1}By angle sum property a +b+ c = 180°. { 2}also, sec( b +c - a) = 2b + c - a = 60°. { 3}from eqn 1 we get a + b = 45 + Cput this value in eqn 245+c + c = 1802c = 180 - 45 c = 135/2 = 67.5 °put this value in eqn 1 a +b = 45+67.5a+ b = 112.5 (4)also in eqn 3 B + 67.5 - A = 60-A + B = - 7.5A - B = 7.5 (5)eliminating (4) from (5)A = 105/2 ° = 52.5° B = 45° |
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| 72703. |
58. In an acute-angled triangle ABC, if tan(A+ B-C) = 1and sec(B + C-A)= 2 , find the value of A, B and C |
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Answer» From question,tan(theta¹)=1.So, theta¹=45°sec(theta²)=2So,theta²=60° A+B-C=45° .......eq(1)B+C-A=60° .......eq(2)ALSO, A+B+C=180°.....eq(3) NOW, ADDING eq(1) & eq(2) 2(B)=105B=52.5° ADDING eq(1) & eq(3) 2(A+B)=225°A=112.5°-52.5°A=60° now adding eq(2) & eq(3) 2(B+C)=240C=120°-52.5°C=67.5° |
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| 72704. |
[a] Find graphically in R the solution set of the equation : X 2-4 Ď + 3 =0in the interval [-1,5] |
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Answer» the given is graph of function x²-4x + 3= 0so, the number of times the graph cuts the x axis give us the numbers of zero. Therefore , we can see there are two cuts between 0 to 5 so there are two zeroes between 0 to 5. |
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| 72705. |
23. ABC in another at right angles.4BC in an acute angled triangle. The perpendicular fromangle. The perpendicular from B on AC is equal to the perpendicularfrom C to AB. Prove that AB = AC.Tu that the hypotenuse is- Inlad trinnl |
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Answer» There will be the identity of sides opposite to equal angles a're equal.because both the angle which are opposite to side ab and ac will be 90 |
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| 72706. |
2. Find the HCF of the following using long division method.a. 92,64 b. 90, 110 c. 216, 124 d. |
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| 72707. |
tand Six Rational Nmber betespen |
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Answer» a) rational number between 3 and 4 are 3.1 , 3.2 , 3.3, 3.4 , 3.5, 3.6. b) 3/4 = 0.75 and 4/5= 0.8so rational no. between them are,0.76, 0.77, 0. 78, 0.79 , 0.791, 0.792 |
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| 72708. |
comperc th nmberand 53 |
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Answer» -7/3*(2/2)= -14/6now(-5/2)*(3/3)= -10/6-10/6>-14/6 |
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| 72709. |
4. Which of the following can be the sides of a righttriangle?(6) 2.5 cm,6.5 cm, 6 cm.(ii) 2cm, 2cm, 5cm.(ii) 1.5 cm, 2cm, 2.5 cm.In the case of right-angled triangles, identify theright angles. |
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| 72710. |
Construct the ΔΡits centroid.4.QR such that PQ = 5cm, PR-6cm and LQPI-60° and loc |
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Answer» 1 2 |
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| 72711. |
let2AL 12 -IC |
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| 72712. |
Find the whole quantity if(a) 5% of it is 600(b)1 2% of it is1080. |
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Answer» 5% (x)=6005/100 (x)=600x=(600×100)/5x=12000 12%(x)=108012/100(x)=1080x=(1080×100)/12x=9000 |
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| 72713. |
nčľ.miCum 22 |
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| 72714. |
010 notebooks were purchased at 300 per dozen and sold at a profit of236 per dozen. What is the profit?72 (2) 360*720 (4) * 480db |
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Answer» 240 books=20 dozen Profit of one dozen=36 Profit of 20 dozens=36x20=720 72 per dozen ha bro very easy question thnx jn you are so cute 720 rs is the right answer profit of 1 dozen =36rsprofit of 20dozen=20*36=720rs3)720rs is the best answer 720₹ is answer write 720 is the correct answer. |
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| 72715. |
ove that sin (180° + θ) cos (90° + θ) tan (270-0) cot (360°-8)sin ( 3600-0) cos ( 360° + 0) cosec (-0) sin (270° + θ) = 1. |
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| 72716. |
Prove that the sum of the exterior angles of triangle ABC is 360 degrees (angle 1 + angle 2 + angle 3 = 360 degrees) |
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| 72717. |
Q. 26. In an acute angled triangle ABC, ifsin (A + B-O)- 2 and cos (B C-A)-rfind LA, B and LC.are |
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| 72718. |
3. NON UI31952.933.964.97A man invests half of his capital at the rate of 10%p.a., one-third ataverage rate of interest p.a., which he gets is:me-third at 9%p.a. and the rest at 12% p.a. The5.15%,י והווה |
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Answer» Let the total capital = Rs. 600 According to the question, Total interest = (30+18+12)= Rs.60 Average Rate of Interest= (60/600) ×100= 10% |
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| 72719. |
5.Find the average rate of change for the function f(x) = 3r on the interval [-3,3]. |
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Answer» For avg. rate of change of function f(x) = 3x², in the interval [-3,3] is , = f(3)-f(-3)/(3-(-3)) = 3*(3)² -3(-3)²/6 = 0/6 = 0. |
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| 72720. |
Prove that V2 is an irrational number using long division method? |
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| 72721. |
x3-7x2+6x+4 divided by x-6 by long division method. |
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| 72722. |
Radha made a picture of an aeroplane with coloredpaper as shown in fig. 12.15. Find the total area of the paper used. |
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| 72723. |
Radha made a picture of an aeroplane with coloured paper as shown in the fig. Find the total area of the paper used. |
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| 72724. |
e, AC | BD and AE | BF. Find the values of x, y and z.given figure,h the gven thgure,A115°85° |
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Answer» x+115=180(sum of adjacent angles). x=180-115 x=65x+y+85=180(straight angles)65+y+85=180 y+150=180 y=180-150 y=30 y+z=180 30+z=180 (sum of adjacent angles) z=180-30 z=150 |
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| 72725. |
Ls,28, Gven that α and β are the roots of the equation x2-7x + 4,α() show that α3-53a + 28(ii) find the value of+ |
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Answer» :aandßare the roots of the equation x2= 7x+ 4 i.e. x2– 7x– 4 = 0. now a should satisfy the equation.. as it is a root of the equation..so ⇒a2– 7a+ 4 = 0 ⇒ (a– 7) (a2– 7a–4) = 0 ⇒a3– 7a2– 4a+7a2–49a– 28 = 0 ⇒a3– 53a– 28 = 0 ⇒a3= 53a + 28 here sum of roots = a+ß = 7 and aß = -4 so, a/ß +ß/a = (a²+ß²)/(aß) = (a+ß)²-2aß/(aß) = ((7)²-2*4)/(4) = (49-8)/4 = 41/4 |
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| 72726. |
e. In the given figure some triangles have been gven. State for each triange whetherscalene, isosceles or equilateral.t isE 2.4E 2.4 cmF2.6 cm(in)60°70%60°60°55°(vi) |
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| 72727. |
22. Find the least nmber icn iceach case23. Find the least number of five digits that is exactly divisible by 16. 18, 24 and 30her of Gve diaits evactly divisible by 9. 12, 15, 18 and 24. |
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| 72728. |
colouredpaperasshowninFig8.15.Fin. Radha made a picture of an aeroplane withthe total area of the paper used. A5cm6cmIVl 5cm6.ScmlcmlcmFig, 8.15 |
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| 72729. |
PUIf the sum of the1080%, find the sunof the interior angles of a polygon isind the sum of its exterior angles. 360 |
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Answer» Sum of interior angles of polygon = (n - 2)*180 Where n is number of sides of polygon 1080 = (n - 2)*180 But sum of exterior angles of polygon is always 360° |
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| 72730. |
This triangle is divided into nine parallel strips of equal height. The total area of the shaded stripes is145 sq unit What is the total area (in square unit) of the unshaded strips ?2.(A) 116(B) 120(C) 135(D) 145 |
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| 72731. |
cos(270+x)cos(360+x)(cot(270-x)+cot(360+x))=1 |
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| 72732. |
Sum of angles of triangle isO180°O 100°O 360° |
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Answer» sum of triangle is 180 sum of tha angles in a triangle is 180 degrees 𝐬𝐮𝐦 𝐨𝐟.... 𝐢𝐬 𝟏𝟖𝟎 sum of all angles of a triangle is 180 degree Sum of the angles in a triangle is 180° the question of answers is=180° answer is 180 degree sum of angle of triangle is 180 180 is the answer of this question 180 degrees is the correct answer sum of triangle 180 degree 180 degree is the right answer 180°degree is the sum of angle of triangle Sum of angles of Triangles is 180° The sum of angle of a triangle is180° 180° is the correct answer sum of angle of a triangle is 180 degree 180 degree is the correct answer 180 degree is the correct answer for this question will be the right answer 180 sum of the angle in a triangle is 180 degrees the sum of triangle is 180° 180 sum of angle of triangle 180 is Sum of angles of triangle Its answer should be 180degree sum of angles of triangle is 180/💘💘💘 180 is right answer 💯👈🙏🙏👍 the sum of angles of triangle is 180 degree 180 degree is a right answer . sum of the angle of triangle is 180°...... this is correct answer sum of 180° sum of is triangle 180 180° is the correct answer. sum of angle of traingle is 180 degree 180 is the sum of triangt some of triangle is 180 |
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| 72733. |
x+60= 360 |
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Answer» X+60=360X=360-60X=300 thanks |
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| 72734. |
W 75, TUU, 140 (g) 150, 330, 360 (h)me HCF of the following numbers by the long division method.(a) 60, 84(b) 32, 64(c) 14, 98(e) 207, 230(1 65, 104(g) 320, 4700 45, 75, 105 ( 594. 792. 1848 (k) 270,945, 2175 (1)Show that foll(h) |
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Answer» Hcf 1) 60/2=30/2=15/3=5, 84/2=42/2=21/3=7; 2x2x3=12; 2) 32/2=16/2=8/2=4/2=2; 64/2=32/2=16/2=8/2=4/2=2; hcf = 16. 3) 14, 98; 14/2=7; 98/2=49/7=7, hcf 2x7 |
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| 72735. |
(1.1.1. February 2015, Turner)12. Which one is a greater fraction?01A Blorcolo colaTurnar |
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Answer» 5/12=0.41664/9=0.44444/6=0.66660.375=3/8hence the greatest fraction is 4/6 the answer of this question is 4/6 Ess qustion kaa sahi answer hoga 4/6 Very easy correct answer is 4/6 Very easy correct answer is 4/6 c) 4/6 is the correct answer 4/6 is a greater fraction 4/6 is a greater fraction. a correct answer is (C) 4/6 is the greatest fraction |
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| 72736. |
Radha made a picture of an aeroplane with coloured paperthe total area of the paper used.3. |
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| 72737. |
Radha made a picture of an aeroplane with coloured paper as shown in Fig. Find the total area of the paper used. |
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| 72738. |
made a picture of an aeroplane with coloured paper as shownind the total area of the paper tised.s cm1.5 cm |
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| 72739. |
Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Find the total area of the paper used. |
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Answer» For finding the area of the paper used determine the area of each part separately and then find the sum of the areas to get the area of used paper. For region I (Triangle) Length of the sides of the triangle section I = a=5cm, b=1cm and c=5cm Semi Perimeter of the triangle, s =( a+b+c)/2s=(5 + 5 + 1)/2= 11/2cm Semi perimeter = 11/2 cm = 5.5cm Using heron’s formula,Area of section I = √s (s-a) (s-b) (s-c) = √5.5(5.5 – 5) (5.5 – 5) (5.5 – 1) cm2 = √5.5 × 0.5 × 0.5 × 4.5 cm2 = √5.5 × 0.5 × 0.5 × 4.5 cm2 = 0.75√11 cm²= 0.75 ×3.32 cm²= 2.49 cm² (approx) Section II( rectangle) Length of the sides of the rectangle of section II = 6.5cm and 1cm Area of section II = l ×b= 6.5 × 1= 6.5cm² Section III is an isosceles trapezium Figure is in the attachment: In ∆ AMDAD = 1cm (given)AM + NB = AB – MN = 1cmTherefore, AM = 0.5cmNow,AD² =AM² +MD²MD²= 1² – 0.5²MD²= 1- 0.25= 0.75MD = √0.75= √75/100=√3/4cm Now, area of trapezium = ½(sum of parallel sides)×height =1/2×(AB+DC)×MD =1/2×(2+1)×√3/4 = ½(3)×√(3/4)= ½×3×√3×2=(3/4)√3 = (3/4)×1.73= 1.30cm²(approx)[√3=1.73....] Hence, area of trapezium = 1.30cm² Section IV and V are 2 congruent right angled triangles with base 6cm and height 1.5cm Area of region IV and V = 2 × 1/2 × 6 × 1.5cm² = 9cm² Total area of the paper used = (2.49+ 6.5 + 1.30 + 9) = 19.3 cm² (approx). |
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| 72740. |
Radha made a picture of an aeroplane with coloured paper as shown in Fig 12.15. Findthe total area of the paper used. |
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| 72741. |
4. Find: ( of a litre4. Find:12of a litre25coloof a kilogramof a kilogram (1of an houran hour |
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Answer» (i) 480(ii) 625(iii) 36 1) 12/25*1000 = 12*4 = 48 Since, 1litre = 1000 2) 5/8*1000 = 5*125 = 625Since, 1Kg = 1000 3) 3/5*60 = 3*12=36Since, 1hr = 60 minutes Its 480 sorry for typing error 480ml,625g,36min is the answer (1) (12/25)1000 ml = 480 ml ER. RAVI KUMAR ROY (2) (5/8)1000g = 625 g (3) (3/5)60min = 36 min (i) 480(ii) 625(iii) 36 (1) 480 ml(2) 625 g(3) 36 min 480, 625, 36 is the correct answer of the given question (I) 480(ii) 625(iii). 36 1) 4802) 625 3) 36 480, 625,36 is the correct one |
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| 72742. |
domom Hght = coloNCERT4.SA, one और SeCA को डालक पदोंidoEnd?AndoCOSA = V1- Sin3Atant - SinaCOIA11- SinaSee A E LACOSA. |
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Answer» cosA=1tonA=13ans=tanAsecA=tanAsecA=tanAsecA=tanAtonA=tonA |
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| 72743. |
1. A person purchased a chair forて700, spent? 170 on its repairs and30 on the cartachair for1080, what is his gain per cent? |
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Answer» total expenditure= 700+170+30=900selling price=1080 % profit = (1080-900)/900 x 100= 20% |
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| 72744. |
A fruit-seller purchased 48 dozens oranges at the rate ofof that 1 dozen oranges were spoiled. He sold the remaining oranges at therate of 6 per orange. Find the profit that he earned in this transaction.2.48 per dozen, out(1) R 1128(2)1220(3)ă1080(4) 1440 |
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Answer» Profit = (48-1)*12*6 - 48*48= 47*72 - 48*48= 3384 - 2304 = Rs. 1080 |
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| 72745. |
A dealer sold a camera for Rs 1080 gaining 1/8 of its cost price.camera, and (ii) the gain per cent earned by the dealer |
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| 72746. |
on the sanle suIITTOIrate of interest per annum.6.)On what sum will the difference between the simple and compound interest for 3for 3 years is 1080 and the compound intereat 10% pa, is232.50?ain cuum |
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Answer» Let the sum be x. SI = x * 3 * 10 / 100 = 30x/100=0.3x CI = x[1+ 10/100]^3 – x → 0.331x Therefore, (0.331-0.3)x = 232.5 Solving which gives x,sum = 7500 |
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| 72747. |
A person lent out, a certain sum on simple interest and thesame sum on compound interest at a certain interest peranum. He noticed that the ratio between the difference ofC.I and S.I ab 3 parts and that of 2 years is 25:8. Therate of interest is â ??. |
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| 72748. |
-x 360 =10802701080* 360 =255x 360 =1080105 360100 X 360 |
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Answer» 270/1080×360=90; 255/1080×360=83.33 105/1080×360 = 35 |
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| 72749. |
Example 7 (Allocation, problem) A cooperative society of farmers has 50 hectareof land to grow two crops X and Y. The profit from crops X and Y per hectare areestimated as Rs 10,500 and Rs 9,000 respectively. To control weeds, a liquid herbicidehas to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further,no more than 800 litres of herbicide should be used in order to protect fish and wild lifeusing a pond which collects drainage from this land. How much land should be allocatedto each crop so as to maximise the totaI profit of the society? |
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| 72750. |
Tortoi2+1WIN |
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