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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.
| 61651. |
is. The price of sugar goes up by 30%. By what per cent should a houwe wife reduce her consumption sod a house wife reduce her consumpithat the expenditure does not increase? |
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Answer» Let price of 100kg sugar be=100rsPrice goes up by 30% , therefore new price= 130rs In 130rs quantity of sugar can be purchased =100kgIn 100rs quantity of sugar can be purchased =100×100/130=1000/13kgDifference in consumption =100/1-1000/13=1300-1000/13=300/13kg % of reduction= 300/13/100×100= 23×1/13%. |
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| 61652. |
linngistrueTriangleO Tris1. Classify the following triangles according to theirsides and (i) anglesScraScm8 cmde distribution by T.S. Government 2018-19TRIANGLE AND ITS PROPERTIES |
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Answer» ML= ✓98 the question 1. Isosceles triangle2. Right angled triangle3. right angled triangle isoscelesright angledright angled isosceles tringleright angledright angled is the correct answer |
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| 61653. |
Optimal solution in L.P.POptimal solution of L.P.PConstraints in L.P.P (M |
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Answer» 1. Anoptimal solutionis a feasiblesolutionwhere the objective function reaches its maximum (or minimum) value – for example, the most profit or the least cost. A globallyoptimal solutionis one where there are no other feasiblesolutionswith better objective function values.3.The variables are also subject to conditions, in the form of linearinequalities. These are called constraints.The variables must also satisfy the non-negativity condition: they can’tbe negative.The set of points, or values of the variables, which satisfy the con-straints and the non-negativity condition is called the feasible set. |
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| 61654. |
.. A cylindrical candle has base radius 3.5 cm. How high it must be made to holdw high it must be made to hold 308 cm of wax |
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Answer» Volume of cylinder with radius r and height h is given by V = πr²h It is given that,a cylindrical candle has base radius 3.5 The volume of cylindrical candle = 308 cm cube To find the height of candle Volume of candle is given by V = πr²h 308 = 3.14 x 3.5 x 3.5 x h h = 8cm Therefore the height of cylindrical candle= 8 cm |
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| 61655. |
EXERCISWhich term of the AP: 121, 117, 113its first negative term?[Hint : Find η fora.<0] |
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| 61656. |
Classify the triangle by its sides and by the measure of its angles.13514.S5Classify each triangle by its sides.15. |
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Answer» 13. Obtuse angled triangle14. Right angled triangle15. equilateral triangle |
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| 61657. |
Classify the quadrilateral |
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Answer» It is a rhombus because it's all four sides are equal and opposite sides are parallel. |
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| 61658. |
Whichterm of the AP: 121, r17, 113,...,isits first negative term?[Hint : Find n for an < 0] |
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| 61659. |
Find the first negative term of the sequence 999, 995, 991, 987,10. In an A.P. if mth term is |
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Answer» the term next to 0 will be the first negative termN=a+(n-1)dhere we take N=0given,a=999d=-4putting the values we get n=250.75it is clear that 0 is not the term of series but next term will be first negative term251th term will be next termN=999+(251-1) (-4)N=-1so -1 is the first negative term of the given series |
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| 61660. |
if the sum off two number is 33 and their difference is 15. find the smallest numbers |
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Answer» x+y=33x-y=152x=48x=24y=9so smallest no. is 9 Let the numbers be x&yx+y=33..........(1)x-y=15.............(2)(1)+(2)=>2x=48=>x=24now substituting x in (1) ,we get,y=9the smallest number is y which equals to9 x + y = 33x - y = 152× = 48× = 48/2x = 24So y = 24 - 15 = 9So smaller number is 9 x + y = 33, ; x - y = 15; x - y = 15/ 2y =18; y = 18/2 = 9; x = 15 +9=24; smallest number = 9; x+y=33x-y=152x=48x=24so y=24-15 =9so small er number is 9 let there be two variable x and yas per questionx+y =33x-y= 15adding above equation 2x= 48x=24therefore y = 9hence the smaller number is 9 |
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| 61661. |
J5+43J_ WAx+ —यदि % J का मान ज्ञात कीजिए। |
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Answer» x=√5+√3/√5-√3*(√5+√3)/(√5+√3)x=(√5+√3)^2/5-3=5+3+2√15/2=4+√15hence 1/X=1/4+√15*4-√15/4-√151/X=4-√15/16-15=4-√15Hence X+1/X=4+√15+4-√15=8 |
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| 61662. |
The angles of a triangle ABC are in AP. Ifof the triangle ABCthe smallest angle is 40° then find the other two angles |
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| 61663. |
X+5=5x-15 |
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Answer» x+5=5x-155x-x=5+154x=20x=20/4x=5 |
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| 61664. |
5(5x-4) = 30 |
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Answer» 5 ( 5x - 4) = 30 25x - 20 = 30 25x = 20 + 30 25x = 50 x = 50/25 = 2 x = 2 |
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| 61665. |
Q.4. Solve ANY TWO of the following:1.Find first negative term from following A.P. 122, 116, 110,(Note : find smallest n such that t 0) |
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Answer» First term of the given AP a = 122 common difference d = 116-122 = -6 therefore nth term of the AP Please like the solution 👍 ✔️ 👍👍👍 |
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| 61666. |
1 3x²+2x-1x² - 7x + 12x² – 5x + 4x² – 5x + 6Vx² - 2x - 8 |
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| 61667. |
if f(x) = (3x+4)/(5x-7) and g(x)=(7x+4)/(5x-3). show that fog(x)=x. |
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| 61668. |
(f) (5x-1)(x + 3) = (x-5)(5x + 1) + 40 |
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| 61669. |
Part - B: Knowing Our Numbers11. The sum of largest 3-digit number and the smallest two-digit number is |
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Answer» The sum of largest 3- digit number and the smallest two --digit number is - 1009 the sum of largest 3-digit number and the smallest two-digit number is -1009 1009 is the answer........... 1009 is the correct answer of the given question |
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| 61670. |
p ( x ) = x + 4 , \text { then find } p ( x ) + p ( - x ) |
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Answer» p(x) +p(-y)=(x+4)+(-x+4)=(x×-x)+(x×4)+(4×-x)+(4×4)=-x+4x+-4x+16=-x+16 |
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| 61671. |
47.If P(A)รทAB) = 3 and P(AnB) = 4 . Then find P(AUB). |
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| 61672. |
P(x)-5x+4x-3. Find P(0) and P(2). |
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| 61673. |
p ( x ) = x ^ { 3 } + 2 x ^ { 2 } + x - 4 . \text { Find } p ( 1 ) \text { and } p ( - 2 |
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| 61674. |
If 3,4 + p and 7-p are in A.P, then find the value of p. |
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Answer» 3,4,p+2,7-p are in A.P. so 4-3 = p+2-4 1 = p-2 so p=3 |
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| 61675. |
1+2/1-1/1+1/2 |
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Answer» 2+1/2=5/2 answer is 5/2 5/2 is a correct answer 2+1/2=5/2 answer is 5/2 5/2 is the correct answer 1+2/1-1/1+1/2=1+2-1+0.5=3.5-1=2.5 1 + 2/1 -1/1+1/2 = 2/1 + 1/2 = 4+1/2=5/2 correct answer is 2.5 2.5 is the right answer 2.2 is the correct answet 2.2 is the correct answer 5/2 is the correct answer of the given question 5/2 is the answer of this question |
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| 61676. |
1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+\ldots+n}=\frac{2 n}{n+1} |
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| 61677. |
. Ir O is a point within a quadrilateral ABCD, show thatOA +OB+OC+ OD> AC+BD |
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| 61678. |
If o is a point within a quadrilateral ABCD,show thatOA+OB+OC+OD>AC+BA |
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| 61679. |
4. AB and CD are respectively the smallest andlongest sides of a quadrilateral ABCD(see Fig. 7.50). Show that Z A> L C and |
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Answer» Thank you |
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| 61680. |
4. AB and CD are respectively the smallest andlongest sides of a quadrilateral ABCD(see Fig. 7.50). Show that A>L C and4BLD. |
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Answer» thank u |
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| 61681. |
vertices of a(3, -3) are collinear(5, -1) are collinearints (1. 1). (-2, 7) a21 (0 Show that the points (1. 1(i)Find p such that the points 1, 3), (2 |
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| 61682. |
vertices21. 0 Show that the points (1, 1). (-2, 7() Find p such that the points (-1, 3), (2and(3, -3) are collinearp)ang(5,-1) are collinear |
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| 61683. |
7. Passing through the points (-1, 1) and (2,-4). |
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Answer» points are (1,-1) and (2,-4)so equation is(y-y1)/(y2-y1)=(x-x1)/(x2-x1)(y-(-1))/(-4-(-1))=(x-1)/(2-1)(y+1)/(-3)=(x-1)/(1)y+1=-3x+33x+y=2 |
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| 61684. |
2Find the slope of line passing through the points (1, -3) and (-1,-1) |
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Answer» slope=y2-y1/x2-x1=-1+3/-1-1=-2please like❤👍 my answer -2 is the correct answer of the given question -2 is your required answer..... |
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| 61685. |
22. Show that the points (1, -1), (5, 2) and (9, 5) are collinear. |
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Answer» For collinear, area of triangle=0=(1/2)[1(2-5)+5(5+1)+9(-1-2)]=(1/2)[-3+30-27]=(1/2)[27-27]=0 Hence collinear points |
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| 61686. |
20how that the points (1, 7). (4, 2), (-1,-1) and (-4, 4) are the vertices of a square.Or |
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Answer» sabhi bindu ki duriya nikalo sabhi dooriyan barabar hongi to square hoga okay.. |
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| 61687. |
E4 Find the equation of a line passing through the points (-1, 1) and (2,-4). |
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Answer» wrong |
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| 61688. |
Find the equations of the plane passing through the points (1, 2, 3) (0, 1, 4) and (0, 0, 1)(07 Ma |
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| 61689. |
Vig7AAB and CD are respectively the smallest andlongest sides of a quadrilateral ABCD(see Fig. 7.50). Show that A > C andZB>2D. |
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| 61690. |
AB and CD are respectively the smallest andlongest sides of a quadrilateral ABCID(see Fig. 7.50). Show that Z A> Z C and |
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| 61691. |
. AB and CD are respectively the smallest andlongest sides of a quadrilateral ABCD(see Fig. 7.50). Show that Z A L C andFig. 7.50 |
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| 61692. |
4. AB and CD are respectively the smallest andlongest sides of a quadrilateral ABCD(see Fig. 7.50). Show that A > C andA.Fig. 7.50 |
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| 61693. |
1. If O is a point inside a quadrilateralABCD, show thatOA + OB+OC + OD > AC + BD. |
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| 61694. |
In right triangle ABC, right angled at C, M isthe mid-point of hypotenuse AB. C İs joinedto M and produced to a point D such thatDM = CM. Point D is joined to point B(see Fig. 7.23). Show that:(i) Δ AMC Δ BMD8.heTh(i)(iii) Δ DBCDBC is a right angle.Δ ACBFig. 7.23(iv) CM =-AB2me Properties of a Triangle |
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| 61695. |
8. If o is a point within a quadrilateral ABCD, showthatOA +OB + OC+OD>AC +BD |
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| 61696. |
8. If O is a point within a quadrilateral ABCD, show thatOA+ OB + OC + OD> AC +BD. |
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| 61697. |
8, If O is a point within a quadrilateral ABCD, show thatOA +OB+OC+OD> AC+ BD. |
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| 61698. |
In △ ABC if L A-45 determine the shortest and the longest sides of the triangle |
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| 61699. |
Find the eguation of plane which bisect theAlso find the angle of the line (21+3+9)+(21+3]+ 4k)line joining the points (-1,2.3) and (3-5.6 a right angleth |
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Answer» Maine pehle se hi solve kr rha tha bs answer check kr rha tha |
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| 61700. |
Tuia fiangle is losa than the sum of the other two, ahow thatthe iangle in at14 If one angle of a trongle ia amatolr than |
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Answer» Let∠A∠B and∠C be the interior angles ofΔABC Each angle is less than the sum of the other two angles (given) Consider∠A <∠B +∠C =>∠A < 180° -∠A [∠A +∠B +∠C = 180°] => 2∠A < 180° =>∠A < 90° So∠A is an acute angle ΔABC is an acute angled triangle. |
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