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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. |
(b) Prove \begin{array}{l}{\vec{(a} \times \vec{b} ) \cdot(\vec{c} \times \vec{d})} \\ {\vec{a} \cdot \vec{c} )(\vec{b} \cdot \vec{d})-(\vec{a} \cdot \vec{d})(\vec{b} \cdot \vec{c})}\end{array} |
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Answer» We need to prove that (A x B).(C x D) = (A.C)(B.D) - (A.D)(B.C) Lets consider (A x B) = Pnow RHS = P.(C x D) = (P x C).D ={(A x B)x C}.D ={B(A.C) - A(B.C)}.D [vector triple product expansion] =(B.D)(A.C) - (A.D)(B.C) = LHS |
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| 2. |
If $\vec{a}, \vec{b}, \vec{c}$ are vertices of triangle then prove thatits area $\frac{1}{2}|\vec{a} \times \vec{b}+\vec{b} \times \vec{c}+\vec{c} \times \vec{a}|$ |
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| 3. |
(a*vec)*((b*vec)*(c*vec))=-a*b*c*vec^3 %2B (b*vec)*((a*vec)*(c*vec)) |
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Answer» bhi kosis ha like da dana koi baat ha to massage kar dana reply |
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| 4. |
2.Simplify |
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| 5. |
**(/5)$=(2/5)ZSECTION -C(EachSimplify ab(a2-b?)+F = 5/57**(4/3) find the value of p/allSimplify 2011 (2/3)*x3*x(1/6)Test, mare lines of summetry of line segments Y1-2 |
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Answer» 2.ans= p/q = (5/6)^-2 × (4/3)^2 = (6/5)^2 × (4/3)^2= = 36/25 × 16/9 = (4₩ |
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| 6. |
. Simplify:224 |
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Answer» the power y into 2 into 2 into 1 it will be 4 so minus 1 by 4 in bracket with 4 power so it will be my it will be multiplied multiplied with both numerator and denominator so answer will be minus 1 by 256 we will multiply power 2 into 2 into 1 there will be 4 and then we will multiply in both numerator and denominator minus 1 into 1 into 14 x and minus 4 into 4 into 4 into 4 4 x 2 is the answer will be -1 by 256 |
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| 7. |
2.What will be the marked price, if sellingprice Rs 792 and trade discount10%? |
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Answer» SP=792DISCOUNT=10%so MP=SP/0.9=792/0.9=880 |
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| 8. |
A bought a car for rs100000 and spent rs10000 on its repairs. He sold this car to B at a gain of 10%, who later on sold it to C at a gain of 5% What did C pay for the car? |
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Answer» Price of th car = Rs 100000Cp of the car = Rs 10000 +100000 = 110000Gain = 10%Sp of the car = Cp + Gain%110000 + (10/100 x 110000)110000 + 11000 = Rs 121000 Cp of the car purchased by B =Rs121000Gain = 5%Sp of the car =Cp + Gain%121000 + (5/100 x 121000)121000 + 6050 = Rs 127050Therefore C has to pay Rs 127050 for the car |
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| 9. |
6. The price of nee is increased by 10%. If the new price is50 per kg, find the old price. |
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Answer» Like if you find it useful |
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| 10. |
DA shopkeeper increased the price of a cycle by 15%. The increasednew price was 1610. What was the original price? |
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| 11. |
Price of a commodity is increased by 15% the price before the increase was Rs 220 Calculate itsnew price. [ Ans Rs 253]37 |
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| 12. |
car is 530000 rupees now. The manufacturerecides to reduce the price by 2% next month ,what wouldTheprice of abe the reduction in price? What would be the new price? |
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Answer» Reduction in price=2*530000/100=10600New price=530000-10600=519400 |
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| 13. |
Q15. If a , B and y are three zeroes of the cubic polynomial p(x) = 3x3 - 6x2 + 5x - 3, then, find their sum andproduct. |
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Answer» Sum=-(-6) /3=2& product=-(-3) /3=1 Sum=(-6)/3=2& product =(-3)/3 |
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| 14. |
1. Sum of the zeros of the cubic polynomialP(x) = 3x - 5x2 - 11x-3 is [HSLC'15](a) 5(b) 11 (c) 3(d)whh |
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Answer» f(x)=3(3)^3-5(3)^2-11(3)-3=3(27)-5(9)-33-3=81-45-33-3=81-81=0 (d) option is correct -b/a P(x)= 3x^3-5x-11x-3sum of zeros= -b/a= -(-5)/3 = 5/3option D is right option D is the correct answer Sum of zero=-b/a=-(-5)/3=5/3 |
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| 15. |
11. Using that 5,-2 and 3 are the zeroes of thecubic polynomial P(x) = 3x3-10x2-27x + 10then verify the relationship between thezeroes and its coefficients. |
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| 16. |
Tick () the correct opt9. The catalogue price of an article is? 16, but the retailer is allowed a trade discount of 25%,must the retailer sell it to gain 25% of what he pays for it?(b)き14(a)20(c)15(d)18olTo |
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Answer» Catalog price = rs. 16;Price for seller =rs. (16-( 25%of 16))=12;To make 25%gain on what he pays he has to sell it at 12+(25%of 12);=12+3=₹15 |
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| 17. |
6) The catalogue price of an article is? 16, but the retailer is allowed a trade discount of 25%. At whatprice must the retailer sell it to gain 25% of what he pays for it ? |
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Answer» Marked price = 16Cost price = 16(1-25/100) = 16(1-1/4)= 16(3/4) = 12 (SP - CP)/CP = 25/100SP - 16 = 1/4*16SP = 16 + 4SP = 20Rs.20 ans no this anwer is not correct which you gave , right answer is 15.. |
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| 18. |
A sold an article to B at a profit of 2.5%. B sold the same article to C at a loss of 10%C paid R 9.90 for it, how much did it cost A? |
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Answer» Let the article be x. With the profit of 2.5%=x+2.5x/100 With loss of 10% = = (x+2.5x/ 100)*10/100 = 102.5x/100 Now new cost = 102.5x/ 100-102.5x/1000 = 92250x/100000 = 9.9 x=10.73 So cost=10.73 |
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| 19. |
A sold an aricle to B at a profit of 2.5%, B soldthe same article to C at a loss of 10%. If C paid9.90 for it, how much did it cost to A? |
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Answer» Let article =x With profit of 2.5%=x+2.5x/100 With loss of 10% = (x+2.5x/100)*10/100 = 102.5x/100 Now new cost = 102.5x/100-102.5x/1000= 92250x/100000 = 9.9 X= 10.73 So cost = 10. 73 Let the article be x. With the profit of 2.5%=x+2.5x/100 With loss of 10% = = (x+2.5x/ 100)*10/100 = 102.5x/100 Now new cost = 102.5x/ 100-102.5x/1000 = 92250x/100000 = 9.9 x=10.73 So cost=10.73 |
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| 20. |
18, A sold an aricle to B at a profit of 2.5%, B soldthe same article to C at a loss of 10%. If C paid9.90 for it, how much did it cost to A? |
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Answer» Please like the solution 👍 Let the article be x. With the profit of 2.5%=x+2.5x/100 With loss of 10% = = (x+2.5x/ 100)*10/100 = 102.5x/100 Now new cost = 102.5x/ 100-102.5x/1000 = 92250x/100000 = 9.9 x=10.73 So cost=10.73 |
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| 21. |
Verify that 1,-1 and-3 are the zeroes of the cubic polynomial ㎡ + 3x2-x-3 andcheck the relationship between zeroes and the coefficients. |
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| 22. |
Raghav sold his chair at a loss of 20%. Had he sold it forprofit of 5%Find the cost price.15.800 more, he would have received a |
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| 23. |
4. Find a cubic polynomial whose zeros are 3,anaWhen f(x) = 4.83 - 872 +8x+1 is divided by a polynomial g(x), we get(2x-1) as quotient and (x + 3) as remainder. Find g(x).11 |
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| 24. |
Consider cubic polynomial x + 2x + 3x + 4. What is sum of all roots. |
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Answer» For a cubic polynomial, sum of roots = -(coefficent of x²/coefficient of x³) = -2/1 = -2 =-2/1=-2 |
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| 25. |
14, गुणनखण्ड विधि से 1001-20x10 के।Find the roots of 100x-- 20x + 1 - by factors |
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| 26. |
Find the value of x in each of the following figure if m100°t2x100X80° |
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| 27. |
III. Answer the following questions.2 X 4 = 8M7. a) If the zeroes of the cubic polynomial p(x) = x- 3x2 + x + 1 are a-d, a, a + d findthe zeroes. |
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| 28. |
100x=200 |
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Answer» x = 2 answer. 100*2=200 so the answer is 2 100x=200then x=200/100i.e., x=100 100x=200×=200/100×=2 100x=200x=200/100X=2 |
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| 29. |
+1)2-20x -3) |
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Answer» LHS = (x+1)² = x² + 2x + 2 RHS = 2(x - 3) = 2x - 6 Dono saman nai hai |
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| 30. |
4. Find the zeroes of the following quadratic polynomials and verify therelationship between the zeroes and the coefficients.(i) x2 - 12x +32 (ii) x2 + 15x + 54 (iii) x2 – 20x + 96 |
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Answer» continued 1st one.................... |
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| 31. |
85=2x+1=25 mzmıx2-3x-10=0(ii) 21+x-6=0(iv) 2x2-x+-=0)100x2-20x + 1 =0 |
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Answer» please give question in English. |
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| 32. |
EXERCISE 2.2と'n Divide the polynomial p(x) by the polynomial g(x) and find the quotient andmainder in each of the following:(i)p(x) =x3-3x2 + 5x-3, g(x) = x2-2 |
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| 33. |
14 Find the zeroes of the following polynomial:55x2 +30x +85 |
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Answer» 5sqrt(5)x^2 + 30x + 8sqrt(5) = 05sqrt(5)x^2 + 20x + 10x + 8sqrt(5) = 0sqrt(5)x(5x + 4sqrt(5)) + 2(5x + 4sqrt(5)) = 0(sqrt(5)x + 2)(5x + 4sqrt(5)) = 0x = - 2/sqrt(5), - 4/sqrt(5) Zeros of given equation are - 2/sqrt(5) and - 4/sqrt(5) |
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| 34. |
and B sold it to C at a profit of 20%. If A had sold theA sold an article to B at a loss of 20%article for the pricc C paid, what would have been his loss or gain per c |
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| 35. |
18. A sold an article to B at a loss of 20%, and B sold it to C at a profit of 20%. If A had soldarticle for the price C paid, what would have been his loss or gain per cent? |
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| 36. |
A loss of 20% is incurred when 6 articles are sold for a rupee. To gain 20should be sold for a rupeehow many articles |
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Answer» Let cost price be x He incurrs a loss of 20% .80x=1 X = 5/4 or 1.25 rupees To gain 20% , he has to sell 1.25 + .2 × 1.25 = 1.5 rupees |
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| 37. |
3. A sold an article to B at a loss of 20%, and B sold it to C at a profit of 20%. If A had sold tarticle for the price C paid, what would have been his loss or gain per cent? |
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| 38. |
8, A sold an article to B at a loss of 20%, and B sold it to C at a profit of 20%. If A had sold thearticle for the price C paid, what would have been his loss or gain per cent? |
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Answer» ruppes 4600 |
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| 39. |
\frac { 1 ^ { 2 } + 1 } { 5 ^ { - 1 } + 2 \times 3 ^ { - 1 } } - \left( \frac { 13 } { 2 } \right) ^ { - 1 } |
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| 40. |
18. A sold an article to B at a loss of 20%, and B sold it to C at a profit of 20%. If A had sold thearticle for the price C paid, what would have been his loss or gain per cent? |
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| 41. |
two equal circles in the figure is 308 cm. Find the perimeter oThearcathe rectangle. |
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| 42. |
Evaluate : 20xx2 +1 |
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| 43. |
31. 55x2 +20x +3/5 |
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Answer» If you find this solution helpful, Please give it a 👍 |
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| 44. |
(v)100x2-20x +1=0 |
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Answer» 100x² - 20x + 1 = 0(10x)² - 2× 10x × 1 + 1² = 0(10x - 1)² = 0x = 1/10 |
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| 45. |
(ii)25x2-20x+4-0 |
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Answer» 25x^2-20x+4=25x^2-10x-10x+4=5x(5x-2)-2(5x+2)=(5x-2) (5x+2) |
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| 46. |
() +12=20x-3) рел. W |
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Answer» x²+1+2x=2x-6x²+7=0x=√(-7) x=7i |
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| 47. |
A circus artist is climbing a 20 m lontightly stretctched and tied from the top of a verticaltidhity the ground. Find the height of the pole, fhe angle made by the rope with the ground level is30° (see Fig. 9.11). |
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Answer» part 1 part 2 |
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| 48. |
us artist is climbing a 20 mlong rope, which istched and tied from the top of a venicalle to the ground. Find the height of the pole, ifangle made by the rope with the ground levelis30° (see Fig. 9.11). |
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| 49. |
EXSolve the following egク |
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Answer» x-2=7 x=7+2x=9 7+2=x9=xkydditditd |
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| 50. |
Solve the following equat1.3x =2x +18 |
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