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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.
| 1. |
adatic polynomial (6x-7x -3) and verify the relation between its zand coefficients.Prove that sin2 θ + cos2 θ = 1 |
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Answer» Let theta = x F(x) = sin^2(x) + cos^2(x) f '(x) = 2 sin(x) cos(x) + 2 cos(x) (-sin(x))= 2 sin(x) cos(x) - 2 cos(x) sin(x)= 0 Since the derivative is zero everywhere the function must be a constant. Take f(0) = sin^2(0) + cos^2(0) = 0 + 1 = 1 So, sin^2(x) + cos^2(x) = 1 everywhere. |
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| 2. |
11.Find the zeros and verify the relation between thezeros and the coefficients (i) x2 + 7x + 10 (i) x2 -3 |
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| 3. |
Find the mean of perimeters of two squares of sides x units and y units. |
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| 4. |
Show me an example of an equation, and show me an example that is not an equation. |
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Answer» An equation is a mathematical statement that two things are equal. It consists of two expressions, one on each side of an 'equals' sign. For example: 12 = 7 + 5 This equation states that 12 is equal to the sum of 7 and 5, which is obviously true. In an equation, the left side is always equal to the right side.Another example with variables is x - 2 = 3.An expression without equal sign is a non-equation.Eg: x + 2, y - 5 |
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| 5. |
8. Find the zeroes of the quadratic polynomial and verify the relationship between the zeroesand the coefficients 6x-3-7x.ietot if |
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Answer» this is not correct |
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| 6. |
- A sweet seller has 420 Kaju burfis and 130 Badam burfis. She wants to stacisuch a way that each stack has the same number, and they take up the leathe tray. What is the number of burfis that can be placed in each stack for thi |
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Answer» FindHCF (420, 130). Then, this number will give the maximum number of barfis in eachstack and the number of stacks will then be the least. The area of the tray that is usedup will be the least. Use Euclid’s algorithm to find their HCF. 420 = 130 × 3 + 30 130 = 30 × 4 + 10 30 = 10 × 3 + 0 So, the HCF of 420 and 130 is 10. Therefore, the sweet seller can make stacks of 10 for both kinds of barfi. Like my answer if you find it useful! |
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| 7. |
1kg = ____g |
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Answer» 1 kg = 10^3 gramsor 1000 gm the answer is 1000 gm iska answer hai 1000g 1killogram = 1000gram 1killogram = 1000 gram |
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| 8. |
65g:1kg |
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| 9. |
1kg=....gram |
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Answer» 1 kilogram = 1000 gram 1 Kilogram = 1000 grams. |
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| 10. |
x% of 1kg is 50g |
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Answer» 1kg = 1000g (x×1000)/100 = 50x × 10 = 50x = 50/10 = 5 |
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| 11. |
EXAMPLE 23 In a finite G.P. the product of the terms equidistant from the beginning and thealways same and equal to the product of first and last term |
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| 12. |
Show that the number 23 × 17 × 5 × 3 x 2,+ 23 is a composite |
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Answer» 23 × 17 × 5 × 3 × 2 + 23Take 23 common23 ( 17 × 5 × 3 × 2 + 1)So, the number have two factors other then 1 and itself so the number is composite |
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| 13. |
If the ratio of the areas of two squares is 9:1 ratio of their perimeters is: |
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| 14. |
44BC~ADEF and their perimeters are 32 cm and 24 cm respectively. If AB-10 cm, find DE |
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| 15. |
4What will be the ratio of perimeters of a square and a circle if theirareas are equal.2 |
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| 16. |
Prove that the ratio of the perimeters of two similar triangles is the sameas the ratio of their corresponding sides. |
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| 17. |
Q3.Give expressions of the following statements.(a) 7 added to twice of p(b) 12 is subtracted from m(c) Y is multiplied by -8 and added to 15 |
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Answer» a) 2p + 7b) m - 12c) -8Y + 15 |
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| 18. |
2r 3eroes of each of the followirthe relations4-4rand verifynd their coefficients7x 3s of the quadratic-1)x-5 are equal in mn then find the value n |
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Answer» 2x^2 + 7x/2 + 3/4 = 08x^2 + 14x + 3 = 08x^2 + 12x + 2x + 3 = 04x(2x + 3) + 1(2x + 3) = 0(4x + 1)(2x + 3) = 0x = - 1/4, - 3/2 Sum of zeroes should be - 7/2Verify = - 1/4 + - 3/2 = - 7/2 Product of zeroes should be 3/8Verify = - 1/4*-3/2 = 3/8 |
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| 19. |
Mow many Triangles are there in the followनीचे दिये गये चित्र में कितने त्रिभुण हैं?(A) 10(c) 14(B) 12(b) 16 |
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Answer» there are 14 triangle in the diagrae three multiplied by 4 so 12 there are 16 triangles in this figure.. there are 16 triangle in figure 16 is the right answer. option D ) 16 is the correct answer 10 is the correct answer 20 is the correct answer 14 is the correct answer 14 is the best answer |
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| 20. |
USB |
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Answer» USB stands for Universal service bus. |
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| 21. |
5. Subtract the sum of (8a -6a2 +9) and (-10a 8+ 8a2) from -3. |
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Answer» thanks |
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| 22. |
s. Subtract the sum of (8a -6a2 +9) and (-10a-8+8a2) from-3. |
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| 23. |
\ Q_ap_,eué/m 18 e 12 1kl ¢BB 1D 210० ॥स्व नी पर b |usb] bl |
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Answer» hit like if you find it useful |
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| 24. |
if the cost of 1kg badam is Rs1350 then what is the cost of 100g badam? |
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Answer» As 1 kg =1000gmHence price of 1 GM =1350/1000=1.350rsHence price of 100gm=1.350*100=135.0rs |
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| 25. |
y=m_{1} x+c_{1}, y=m_{2} x+c_{2} \text { and } x=0 \text { is } \frac{\left(c_{1}-c_{2}\right)^{2}}{2\left|m_{1}-m_{2}\right|} |
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| 26. |
112If middle term is Kxm in expansion ofthen m_ |
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| 27. |
I U T . eg urhich weak e MA.MJ}M_,_'L bके. «न. |
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Answer» Apartheid was particularly oppressive against the blacks. it denied third fundamental rights such as right to vote and it also it gave separate toilets and swimming pools and beaches etc. |
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| 28. |
If the areas of two similar triangles, tPQR and ΔΧΥΖ are 162sq .cm and. 200 sq.cmrespectively. Find the ratio of their perimeters. |
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Answer» Area of ABC/Area of PQR=162/200Sides Ratio =9/10perimeter ratio=9+9+9/10+10+10=27/30 |
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| 29. |
\beta\left(m_{1}, y_{2}\right)=2^{2 m-1} \beta(m \cdot m) |
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Answer» B(m,1/2)=2^(m/(1/2))-1=2^(2m-1)so B(m,n)=2^(m/n -1) |
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| 30. |
5. The perimeters of two similar triangles ABC and PQR are 32 cm and[CBSE 2001]24 cm respectively. If PQ12 cm, find AB |
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| 31. |
Give expressions in the follow(a) 11 added to 2m5. |
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Answer» we get expression11+2m |
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| 32. |
The number of permutations of the letters a, b, c, d suchthat b does not follow a, and c does not follow b and d doesnot follow c, is(a) 9(c) 13(d) 14 |
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| 33. |
1. Carry out the followir2. Carry out the followMultiply the follow |
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Answer» 1) i)5(7)+6(6)/78 = 35+ 36/78=71/78; 2) i) 7(7)-3(11)/77 = 49-33/77 =16/77; 3) i)3/11× 2/5 = 3(5)+2(11)/55 = 15+22/55=37/55 3(5) +2(11) /55=15+22/55=37/55 3(5)+2(11)/55=15+22/55=37/55 is the right answer |
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| 34. |
e various patterns with numbers in our previous classpattern for each of the following? If yes, completeTRY THESEWe have done various patterns with numCan you fird a pattern for each of the follow(a) 7, 3,-1,-5, -(b) -2,-4, -6, - 8, 10_ 12_,-14(c) 15, 10.5, 0, -(d) - 11, - 8,- 5,- 2,Make some more such patterns and ask your friends to com |
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Answer» A)7,3,-1,-5,-9,-13,-17)-2,-4,-6,-8,-10,-12,-14)15,10,5,0,-5,-10,-15)-11,-8,-5,-2,1,4,7 |
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| 35. |
Thus, they meet 3xercise 10AExpress each of the follow() 36 km/hExpress each of the follow |
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| 36. |
Q33. Prove that : tan10® . tan15° . tan75° . (पाप छे0 = 1. i e 3 g.b R |
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Answer» tan 10.tan 15.tan 75. tan 80 = cot(90 - 10).cot(90 - 15).tan 75.tan 80 = cot 80.cot 75.tan 75.tan 80 = (1/tan 80).(1/tan 15).tan 75.tan 80 = 1 Hence proved |
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| 37. |
44.What is the weight of the apple ?75g100g 100g(1) 250 g(2) 150 g(3) 75 g(4) 125 g |
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Answer» Since the weighing balance is balanced, which means total weight on left and right sides are equal. Therefore, weight of apple = 100+100+25-75weight of apple = 150 g Option (2) is correct. |
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| 38. |
sinsin20 +sin3esin6esinecos29 +sin30cos60Q33. Prove thattan50 |
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Answer» We will use the following formulae in this solution-- (1) Sin A . Sin B = (1/2) [ Cos (A-B) - Cos (A+B) ](2) Sin A Cos B = (1/2) [ Sin (A+B) + Sin (A-B) ] (3) Sin A + Sin B = 2 Sin (1/2) (A+B) . Cos (1/2)(A-B) (4) Sin A - Cos B = 2 Sin (1/2) (A-B) . Cos (1/2)(A+B) (5) Sin ( -x ) = - Sin x (6) Cos ( - x ) = Cos x Solution -- Numerator = (1/2) [ 2 SinA Sin 2A + 2 Sin 3A. Sin 6A ] --- Use formula (1) => (1/2) [ Cos (A-2A) - Cos (A+2A) + Cos (3A-6A) - Cos (3A+6A) ] => (1/2) [ Cos A - Cos 3A + Cos 3A - Cos 9A ] => (1/2) [ Cos A - Cos 9A ] .... Again use formula (1) => Sin 5A . Sin 4A Denominator = Sin A Cos 2A + Sin 3A.Cos 6A => (1/2) [ Sin (A+2A) + Sin (A-2A) + Sin ( 3A + 6A ) + Sin ( 3A - 6A ) ] => (1/2) [ Sin 3A - Sin A + Sin 9A - Sin 3A ] => (1/2) [ Sin 9A - Sin A ] => (1/2) [ 2 Cos 5A . Sin 4A ] => Cos 5A . Sin 4A Hence the Left Hand Side of the given identity = Numerator / Denominator => Sin 5A . Sin 4A / Cos 5A . Sin 4A => Tan 5A = RHS ………………. QED . |
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| 39. |
-Simplily(a) 13-15 + s(a) 13- + isSolution(b) 6 2 3 + 3 2 33 -EXERCISE 36 a1. Simplify following: |
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Answer» 3/5+2/53+2/55/5=1 is the right ans is 1 because 3/5+2/5=5/5=1 1 is the correct answer 1 is the correct5/5=1 1 is the correct answer of the given question 3/5+2/53+2/55/51 is the answer 3/5+2/55/51 is the best answer |
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| 40. |
4)AHsurvey was conducted to know the hobby of 220 students of class l130 students informed about their hobby as rock climbing anabout their hobby as sky watching. There are 110 students who follow both the hobbies.Then how many students do not have any of the two hobbies ? How many of themfollow the hobby of rock climbing only 2 How many students follow the hobby of skywatching only?FitIX. Out of which 2)(i)180 students informed(i |
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Answer» Thank yoy |
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| 41. |
3. Attempt the following:13 marks each/(1) Area of the base of a cone is 1386 cm2 and itsheight is 28 cm. Find its curved surface area.Solution: |
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Answer» Area of the base of a cone = 1386sq.cm. Height,h= 28cm. Surface are = ?? Therefore by using the formula , Area of base of cone = pi*r^2 1386 = 22/7*r^2 1386*7/22 = r^2 63*7 = r^2 441 = r^2 r = 21cm. We got the radius. Now we find the slant height, We know that, l^2 = r^2 + h^2 l^2= (21)^2 + (28)^2 l^2 = 441 + 784 l^2 = 1225 l = 35cm. So we got the value of radius and slant height now we can easily find the surface area of cone. By using the formula, Surface are of a cone= pi * r*l= 22/7*21*35=22*3*5= 330sq.cm. |
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| 42. |
7. Calculate the missing frequency from the following distribution, itbeing given that the median of the distribution is 24.Class0-1010-20 20-30 30-4040-502518Frequency |
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Answer» thank |
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| 43. |
No. ofboys10 18 227. Calculate the missing frequency from the following distribution, itbeing given that the median of the distribution is 24.Class0-1010-20 20-30 30-4040-50Frequency25187 |
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| 44. |
22.The median of followifrequency xFind the mCI 45ing frequency distribution is 24 years. Find theAge (In years)No. of personsFind the median of the following data0-10 10-20 20-30 30-40 40-5023.2518Find the7Marks |
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| 45. |
NO. 010boysCalculate the missing frequency from the following distribution, itbeing given that the median of the distribution is 24.Class10-20 20-30 30-40 40-500-10182Frequency251h. Find the missing frequencies a |
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| 46. |
Due to reduction of 9 1/3% in the price ofsugar a person is able to buy 125 g moresugar in Rs. 150. Find the reduced price of1 kg of sugar(A) 11.2(C) 1120(B) 112(D) 1.12 |
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Answer» 9.33 % of 150 = 14.So the previous quantity now can be bought at 136 rs. So at new price 14 rs buy us 125 g sugar. Therefore 1 kg sugar costs 112 rs. |
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| 47. |
3ab÷( )=a |
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Answer» the answer is 3b . 3ab÷3b=a is question ka answer hai 3a the answer is 3b 3ab/3b= a (3b/3b=1) this answer is 3b h ramkaran Beniwal the answer is = (3b) 3ab/3b=a by cancelling the right answer is 3b. 3ab÷3b=a Divided by 3b is Answer the answer is 3ab/3b=a answer these question are 3b 3b because of 3ab÷3b=a 3ab÷(3b)=a,,,,,,,,,,,,, 3b is the right answer 3b is correct answer 3b is the correct answer of these question.3ab/3b=a the answer is 3b, 3ab÷a=3b the answer is 3ab ÷(3b) = a answer is 3b 3ab ÷ 3b = a 3ab ÷ 3b = a answer is 3b the answer of this question is 3b the answer is 3b, 3ab÷3b=a 3ab divide b equal a 3ab divide 3b is answer answer of this question 3b because 3ab/3b=a answer is the 3b.3ab÷3b=a 3ab/3b=a correct answer the answer of this question is 3b=3ab÷3b=a 3b,because 3ab÷3b=a then answer is 3b 3ab÷( )=aAnswer ia 3b because 3b cancelled 3bso answer is = 3b this is a answer 3ab÷3b=a is question ka answer 3ab÷3b=a itna hoga 3b ÷de denge the answer is 3b. 3ab÷3b=a 3ab÷(3b)=a this is correct answer |
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| 48. |
8. 35 workers build a house in 160 days. How many days will 28 workers take to build the samehouse, working at the same rate? |
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Answer» 35 workers in 160 days~1 worker in (35×160) days~28 workers in(35×160÷28)days=200 days |
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| 49. |
Factorize a³–b³+1+3ab |
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| 50. |
If 120 is reduced to 96, what is the reduction per cent? |
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Answer» As per the question, Reduction=120-96.=24. Reduction%= 24/120 ×100.=1/5 ×100.=20%. |
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