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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.
| 26501. |
Determine the AP whose third term is 16 and 7th term exceeds the 5th term by 12. |
| Answer» Let the first term and the common difference of the AP be a and d respectively.Then,Third term = 16{tex} \\Rightarrow {/tex}\xa0a + (3 - 1)d = 16\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}{tex} \\Rightarrow {/tex}\xa0a + 2d = 16 ...... (1)and, 7th term = 5th term + 12{tex} \\Rightarrow {/tex}\xa0a + (7 - 1)d = a + (5 - 1)d + 12\xa0{tex}\\because {a_n} = a + (n - 1)d{/tex}{tex} \\Rightarrow {/tex}\xa0a + 6d = a + 4d + 12{tex} \\Rightarrow {/tex}\xa06d - 4d = 12{tex} \\Rightarrow {/tex}\xa02d = 12{tex} \\Rightarrow d = \\frac{{12}}{2} = 6{/tex}Put d = 6 in (1), we geta + 2(6) = 16{tex} \\Rightarrow {/tex}\xa0a + 12 = 16{tex} \\Rightarrow {/tex}\xa0a = 16 - 12{tex} \\Rightarrow {/tex}\xa0a = 4Hence, the required APs also4, 4 + 6, 4 + 6 + 6, 4 + 6 + 6 + 6, .......i.e. 4, 10, 16, 22, ....... | |
| 26502. |
Cos 45 |
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Answer» Sorry its mistake ?? 1/root2 1/root 2 Root3/2 |
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| 26503. |
What is the sum of all natural numbers ??? |
| Answer» | |
| 26504. |
Find the distance between the points P(a+b,a-b) and Q(a-b,a+b) |
| Answer» 8b sqare | |
| 26505. |
Prove that the points (a,0),(0,b,(1,1)are collinear if 1/a+1/b=1 |
| Answer» Since (a, 0), (0, b) and (1, 1) are collinearArea = 0{tex}\\frac{1}{2}\\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \\right] = 0{/tex}{tex} \\Rightarrow \\frac{1}{2}\\left[ {a(b - 1) + 0(1 - 0) +1 (0 - b)} \\right] = 0{/tex}{tex} \\Rightarrow {/tex}\xa0ab - a - b = 0{tex} \\Rightarrow {/tex}\xa0ab = a + bDividing by ab,{tex}\\frac{{ab}}{{ab}} = \\frac{a}{{ab}} + \\frac{b}{{ab}}{/tex}{tex}\\frac{1}{a} + \\frac{1}{b} = 1{/tex} | |
| 26506. |
If tanA=ntanB nd sinA= msinB, then prove that cos^2=m^2-1/n^2-1. |
| Answer» Given,\xa0tan A = n tan B{tex} \\Rightarrow{/tex} tanB = {tex}\\frac{1}{n}{/tex}tan A{tex}\\Rightarrow{/tex}\xa0cotB =\xa0{tex}\\frac { n } { \\tan A }{/tex}..........(1)Also given,\xa0sin A = m sin B{tex}\\Rightarrow{/tex}\xa0sin B =\xa0{tex}\\frac{1}{m}{/tex}sin A{tex}\\Rightarrow{/tex}\xa0cosec B =\xa0{tex}\\frac { m } { \\sin A }{/tex}.....(2)We know that, cosec2B - cot2B = 1, hence from (1) & (2) :-{tex} \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } } { \\tan ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } } { \\sin ^ { 2 } A } - \\frac { n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - n ^ { 2 } \\cos ^ { 2 } A } { \\sin ^ { 2 } A } = 1{/tex}{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = sin2A{tex}\\Rightarrow{/tex}\xa0m2 - n2cos2A = 1 - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = n2cos2A - cos2A{tex}\\Rightarrow{/tex}\xa0m2 - 1 = (n2 - 1) cos2A{tex}\\Rightarrow \\quad \\frac { m ^ { 2 } - 1 } { n ^ { 2 } - 1 } ={/tex}\xa0cos2A | |
| 26507. |
The exam of board will come from ncert almost is it true? |
| Answer» | |
| 26508. |
Is the triangle with side 12cm, 18cm a right triangle? Give reason |
| Answer» We know, sum of two sides is greater than third sideSo, given Sum of two sides = 6 + 12 cm = 18cmTherefore, measure of third side should be less than 18cm. | |
| 26509. |
Defination of mean median mode |
| Answer» | |
| 26510. |
Aaj time table release Hoga Ki Nahi |
| Answer» Bhai 10 tak | |
| 26511. |
The area of circle is 301.84 cm2 find its circumference |
| Answer» We have,Area of circle = 301.84 cm2{tex}\\Rightarrow \\pi {r^2} = 301.84{/tex}{tex}\\Rightarrow \\frac{{22}}{7} \\times {r^2} = 301.84{/tex}{tex} \\Rightarrow {r^2} = \\frac{{301.84 \\times 7}}{{22}}{/tex}{tex} \\Rightarrow {r^2} = \\frac{{301.84 \\times 7}}{{22}}{/tex}{tex}\\Rightarrow {r^2} = \\frac{{1372 \\times 7}}{{100}}{/tex}{tex} \\Rightarrow r = \\sqrt {\\frac{{9604}}{{100}}} = \\frac{{98}}{{10}} = 9.8cm{/tex}{tex}\\therefore {/tex}\xa0circumference of circle =\xa0{tex} 2\\pi r{/tex}{tex} = 2 \\times \\frac{{22}}{7} \\times 9.8{/tex}= 61.6 cm | |
| 26512. |
weight of 72 bags is 8kg how manysuch bags weight 20 kg |
| Answer» 180 | |
| 26513. |
Properties of a tangents |
| Answer» In a circle a line intersects at only one point. | |
| 26514. |
A triangle ABC at b 90 and c tan =^3 |
| Answer» Let ABC be a triangle right angled at C such that tan {tex}A = \\frac{1}{{\\sqrt 3 }}{/tex} and tan B ={tex}\\sqrt 3 {/tex}tan {tex}A = \\frac{1}{{\\sqrt 3 }}{/tex}{tex}\\Rightarrow {/tex}{tex}\\frac{{BC}}{{AC}} = \\frac{1}{{\\sqrt 3 }}{/tex}{tex}\\Rightarrow {/tex}{tex}\\frac{{BC}}{1} = \\frac{{AC}}{{\\sqrt 3 }} = k{/tex}(say)where k is a positive number{tex} \\Rightarrow \\left. \\begin{gathered} BC = k \\hfill \\\\ AC = \\sqrt 3 k \\hfill \\\\ \\end{gathered} \\right\\}{/tex}....(1){tex} \\Rightarrow \\frac{{AC}}{{BC}} = \\sqrt 3 {/tex}{tex} \\Rightarrow \\frac{{AC}}{{\\sqrt 3 }} = \\frac{{BC}}{1} = k(say){/tex}where k is a positive number{tex}\\left. \\begin{gathered} AC = \\sqrt 3 k \\hfill \\\\ BC = k \\hfill \\\\ \\end{gathered} \\right\\}{/tex} ...(2)From (1) and (2),AC ={tex}\\sqrt 3 k{/tex}BC = KIn {tex}\\vartriangle {/tex} ABC,{tex}\\because {/tex}{tex}\\angle{/tex} C={tex}{90^ \\circ }{/tex}{tex}\\therefore {/tex} AB2\xa0= AC2\xa0+ BC2 ...By Pythagoras theorem{tex}\\Rightarrow {/tex} AB2 = ({tex}\\sqrt 3 k{/tex})2\xa0+ (k)2{tex}\\Rightarrow {/tex} AB2 = 3k2\xa0+ k2{tex}\\Rightarrow {/tex} AB2\xa0= 4k2{tex}\\Rightarrow {/tex} AB =\xa0{tex}\\sqrt {4{k^2}} {/tex}{tex}\\Rightarrow {/tex} AB = 2kTherefore,sin A =\xa0{tex}\\frac{{BC}}{{AB}} = \\frac{k}{{2k}} = \\frac{1}{2}{/tex}cos B =\xa0{tex}\\frac{{BC}}{{AB}} = \\frac{k}{{2k}} = \\frac{1}{2}{/tex}cos A =\xa0{tex}\\frac{{AC}}{{AB}} = \\frac{{\\sqrt 3k }}{{2k}} = \\frac{{\\sqrt 3 }}{2}{/tex}sin B ={tex}\\frac{{AC}}{{AB}} = \\frac{{\\sqrt 3 k}}{{2k}} = \\frac{{\\sqrt 3 }}{2}{/tex}Now, sin A cos B + cos A sin B{tex}\\frac{1}{2}.\\frac{1}{2} + \\frac{{\\sqrt 3 }}{2}.\\frac{{\\sqrt 3 }}{2} = \\frac{1}{4} + \\frac{3}{4} = 1{/tex} | |
| 26515. |
If x =9-80 then find X+1/x |
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| 26516. |
Find the four number in AP whose sum is 50 and in which the greatest number is 4 times the least. |
| Answer» Take no as (a-3d), (a-d), (a+d), (a+3d) nd sve | |
| 26517. |
The nth term of an A.P is 3n+5 find its common difference? |
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Answer» 3 D=3 |
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| 26518. |
I have completed NCERT by hard and I want 70 marks out of 80 in class 10 is it enough |
| Answer» Ok I will follow this can u send me all sample papers of math my number is 8218867053 | |
| 26519. |
If 2 is divisible by 5 then find remainder |
| Answer» Quotient :- 0.4 nd remainders :- 0 | |
| 26520. |
Defferent between mandelevs periodic table and modern periodic table? |
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Answer» Devil ...plz check !! U r wrong Mandeleev use the concept of reactivity if elements with oxygen Sorry meendleve used atomic mass Mostbasic diff is meelev table uses atomic no whereas mordern table uses atom8c no Esy bt long .... |
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| 26521. |
(x+1)÷x whole sq.X=2, 1÷x=4 |
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| 26522. |
kisi pe maths ka board ka blue print aaya kya |
| Answer» | |
| 26523. |
√sec€-1/sec€+1 + √sec€+1/sec€-1 =2cosec€ |
| Answer» LHS ={tex}\\sqrt { \\frac { \\sec \\theta - 1 } { \\sec \\theta + 1 } } + \\sqrt { \\frac { \\sec \\theta + 1 } { \\sec \\theta - 1 } }{/tex}\xa0Rationalise the denominator and we get,\xa0{tex}= \\sqrt{\\frac{(sec\\theta-1)^2}{sec^2\\theta-1}}+ \\sqrt{\\frac{(sec\\theta+1)^2}{sec^2\\theta-1}}{/tex}=\xa0{tex}\\frac { ( \\sec \\theta - 1 ) + ( \\sec \\theta + 1 ) } { \\sqrt { ( \\sec \\theta + 1 ) ( \\sec \\theta - 1 ) } }{/tex}=\xa0{tex}\\frac { 2 \\sec \\theta } { \\sqrt { \\sec ^ { 2 } \\theta -1 } } = \\frac { 2 \\sec \\theta } { \\sqrt { \\tan ^ { 2 } \\theta } } = \\frac { 2 \\sec \\theta } { \\tan \\theta }{/tex}=\xa0{tex}2 \\times \\frac { 1 } { \\cos \\theta } \\times \\frac { \\cos \\theta } { \\sin \\theta }{/tex}=\xa0{tex}2 \\times \\frac { 1 } { \\sin \\theta }{/tex}= 2 cosec{tex}\\theta{/tex}= RHSHence Proved | |
| 26524. |
The values of trigonometric ratios for angles0, 30,45,60,and90 |
| Answer» Given in ncert | |
| 26525. |
i have an doubt in completing square method |
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| 26526. |
Write cos theta in terms of cos square_6447 |
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| 26527. |
What is the value of tanA+tanB+tanC |
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| 26528. |
How to find median |
| Answer» The median is also the number that is halfway into the set. | |
| 26529. |
Upper limit lower limit statistics |
| Answer» Which has the highest frequency is the modal class........ | |
| 26530. |
All formula of chapter circle |
| Answer» Check formulae in revision notes :\xa0https://mycbseguide.com/cbse-revision-notes.html | |
| 26531. |
What is bit coin |
| Answer» Cryptocurrency and world wide payment | |
| 26532. |
Construct tangents to a circle of radius 4cm inclined at an angle of45 |
| Answer» Steps of Construction:\tDraw a circle with centre O and radius = 4 cm.\tDraw any radius OA.\tDraw another radius OB such that {tex}\\angle{/tex}AOB = 180° - 45° = 135°\tAt point A draw AP {tex}\\bot{/tex}\xa0OA.\tAt point B draw BR {tex}\\bot{/tex} OB, intersecting AP at C. AC and BC are required tangents. | |
| 26533. |
WhT is division algorithm |
| Answer» Euclid\'s division algorithm is a technique to compute the (HCF) of two given positive integers | |
| 26534. |
12-61 |
| Answer» _49 | |
| 26535. |
Datesheet of all paper of 10 class when ? |
| Answer» Check datesheet here :\xa0https://mycbseguide.com/cbse-datesheet.html | |
| 26536. |
Any body can tell me when an equation called an \'identity\'? (very important) |
| Answer» when there r a number of roots more than the degree of the polynomial . | |
| 26537. |
Date sheet 10th |
| Answer» Check datesheet here :\xa0https://mycbseguide.com/cbse-datesheet.html | |
| 26538. |
Prove that :-SinA +(1+tan A)+cot A (1+cot A)=sec A +cosec A |
| Answer» | |
| 26539. |
About chemotropism |
| Answer» | |
| 26540. |
When will CBSE release class 10th datesheet |
| Answer» | |
| 26541. |
If p is the HCF of 45 and 27 , find a and b such that p= 27a+45b |
| Answer» 45 = 27 {tex} \\times {/tex}\xa01 + 1827 =18 {tex} \\times {/tex}\xa01 + 918 = 9 {tex} \\times {/tex}\xa02 + 0So H.C.F. =9Now 9 = 27 – 18 {tex} \\times {/tex}\xa01{tex}\\style{font-family:Arial}{\\begin{array}{l}9=27-18\\\\=27-(45-27\\times1)\\times1\\\\=27-45+27\\\\=2\\times27-(1)\\times45=27x+45y\\end{array}}{/tex}⇒ a\xa0= 2, b\xa0= – 1. | |
| 26542. |
16x-10/x=27 find natural roots |
| Answer» X=2;-5/16 | |
| 26543. |
In fig ST parallel to QR where PS =3 cm SR=4cm find ratio of area of triangle PST to triangle PRQ |
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Answer» Correct ayush Bhai yaha figure wala question nahi ban pauega ....sorry ??? |
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| 26544. |
In an equilateral triangle ABC, D is a point on side BC such that BD=1/3BC. Proove that 9AD²=7AB². |
| Answer» Bina diagram nhi hoga | |
| 26545. |
If sinθ= cosθ so let find theθ |
| Answer» | |
| 26546. |
When an equation is called \'identity\'? |
| Answer» Identity equations are equations that are true no matter what value is plugged in for the variable. If you simplify an identity equation, you\'ll ALWAYS get a true statement. | |
| 26547. |
How we can prove the converse of BPT |
| Answer» | |
| 26548. |
Cosec6-cot6 = 1+ 3 cot2+3cot4 |
| Answer» | |
| 26549. |
If 13 cos A-5=0, evaluate 5sin A - 2cos A/ tan A |
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| 26550. |
If the radiys |
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