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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.
| 14451. |
If pth term of ap=1÷q and qth term of ap =1÷p then prove that Spq=(1÷p_1÷q) |
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| 14452. |
If the nth term of an arthmetic progression a n =24-3n then its 2n term is |
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Answer» an = 24 - 3nWe need the second term and therefore :n = 2We substitute this value of n in the formula to get :a2 = 24 - 3 × 2= 24 - 6 = 18The second term is thus 18. 18 |
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| 14453. |
All formulas of 13 ch |
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| 14454. |
Exact answer of 7 unit example 7 plz tell me answer in one minute |
| Answer» 2:7 | |
| 14455. |
Find the area of the segment AYB ,if radius of the cricle is 21 cm and ∆AOB =120. (π= 22/7) |
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Answer» 461 Lgta hai aap hi merried mein aayge In ∆AOB, draw a perpendicular line from O which intersect AB at M.In ∆AOM, angle AMO = 90angle OAM = 30cos 30 = AM/AO√3/2 = AM/21AM = 21×√3/2AB = 2(AM)=2(21×√3/2)=21√3{tex}OM^2 = AO^2-AM^2{/tex}{tex}=21^2-(21√3/2)^2{/tex}=441-330.51=110.48OM =√110.48OM =10.51OM = 10.51cmArea of ∆AOM = 1/2 AB × OM=1/2 ×21√3 ×10.51=191.14cm^2Area of sector AOBY = 120πr^2/360=120×21×21×22/2520=462cm^2Area of segment AYB = Area of sector OAYB -Area of∆OAB=462-191.14=270.86Area of segment AYB is 270.86cm2.\xa0 |
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| 14456. |
For some a and b HCF of 55and 210 is 210a + 55b ,then find the value ofa and b |
| Answer» HCF of 210 and 55255= 55×3+45...(1)55=45×1+10...(2)45=10×4+5...(3)10=5×2+0. remainder =0HCF of 210 and 55 =5NOW,from (3) we get,45=10×4+545-10×4=545-(55-45)×4 =5 ....(from 2)45-55×4+45×4=5(210-55×3)5-55×4=5 ...(from 1)210×5-55×15-55×4=55=210×5-55×19=210A+55Btherefore,a=5 and b= -19\xa0 | |
| 14457. |
Find largest no which divide 398 ,436, and 542 leaving remainder 7 ,11 and 15 resp |
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Answer» On dividing 398 by the required number, there is a remainder of 7. This means that 398 7 = 391 is exactly divisible by the required number. Similarly, 436 -11 = 425 and 542 15 = 527 are exactly divisible by the required number.The HCF of two positive integers is the largest positive integer that divides both the integers.So, the required number will be the HCF of 391, 425 and 527. And that can be found by using Euclids division algorithm.425 = 391 x 1 + 34391 = 34 x 11 + 1734 = 17 x 2 + 0Thus, HCF = 17Hence, the required number is 17. 17 |
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| 14458. |
X US A POINT ON SIDE BC OF TRIANGLE ABC XM AND XN ARE DRAWN PARALLEL |
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| 14459. |
If lcm of two no is1200 whatis the hcf |
| Answer» LCM- “Lowest Common Multiple”\xa0Smallest integer that are the multiple of both numbers.HCF- “Highest Common Factor”\xa0Largest whole number that divides both numbers.LCM of two numbers is\xa01200.Factorizing 1200, we get,∴ Prime Factors Since we cannot find\xa0500 or multiples of 500 in the factors of 1200, hence 500 cannot be the HCF of the two numbers. | |
| 14460. |
If f(x) is cos2x +sec2x then what is f(x) |
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| 14461. |
In what ratio is the line joining the points A (4,4) and B(7,7) divided by P (-1,-1)? |
| Answer» Jjgg | |
| 14462. |
Chapter10 example 3 |
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| 14463. |
Find two numbers whose sum is 80 and difference is 40 |
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Answer» 60 and 20.. Let the first no. be X and second no. be Y A/Q X+Y=80....(1). X-Y=40...(2). Equate both eq. ||. 2x=120. ||. X=60 , Y=20. Let the first no. be x. Then second no. be (80-x) |
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| 14464. |
Show that one and only one out of N N + 2 and + 4 is divisible by 3 Where n is any positive integer |
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| 14465. |
Find the HCF of 336 and 68 by prime factorisation method |
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Answer» 336 = 2*2*2*2*3*7;68 = 2*2*17;H.c.f. = 2*2=4 Ans= 4. |
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| 14466. |
Radias and tangent of circle are incined at |
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Answer» 90 degree Any point on the circle |
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| 14467. |
A bus goes downward |
| Answer» K next | |
| 14468. |
Chapter 11 example 1 |
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| 14469. |
Find the lower limit of median class of the following data if median is 43 |
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Answer» Data kha hain Table knha hai |
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| 14470. |
If p(9a-2,-b) divides A(3a+1,-3) and B(8a,5) in 3:1 then the respective values of a and b |
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Answer» P(9a-2,b) = (x,y)A(3a+1.-3) = (x1,y1)B(8a,5) = (x2,y2)Ratio (m1,m2) = 3,1x = m1*x2 + m2*x1 --------------------- m1 + m29a-2 =3*8a +1* 3a+1 ----------------- 3 + 19a-2 = 24a + 3a + 1 ---------------- 436a - 8 = 27a + 136a - 27a= 1 + 89a = 9a = 9 ----- 9a = 1Now,y = m1*y2 + m2*y1 --------------------- m1 + m2-b = 3*5 + 1 * -3 ---------------- 4-b = 15 - 3 -------- 4-b = 12/4-b = 3b = -3Therefore a = 1, b = -3\xa0 9a= (24a+3a+1)/436a=27a+19a=1a=1/9-b= 15-3-b=12b=-12 |
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| 14471. |
What is the blueprint for standard maths 2020? |
| Answer» You can get it from mycbseguide app | |
| 14472. |
What is the blueprint for maths 2020? |
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| 14473. |
(a-b)x2+(b-c)x+(c-a) |
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Answer» Tell me what to find What i have to do with this |
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| 14474. |
How can i score above 80marks in science? |
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Answer» Ashu - outof 100 Paper is only for 80marks then hoe can u score above 80 ????????????? Mere paas aa jao...ek din Study ncert book and questions and clear your doughts from other books like s chand. Practice sample papers with the setting of 3 hrs. |
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| 14475. |
If area of quadrant of a circle is 38.5cm^2 then find its diameter |
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Answer» Solve this 14cm |
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| 14476. |
Find the mode of following data |
| Answer» It belongs to computer language i think so | |
| 14477. |
Reena purchased a dress ₹5400 including 12%GST . Find the price of dress before GST was added? |
| Answer» Let the original price be : Rs XAfter GST the price gone up = X + (12/100)XThe equation will beX + (12/100)X = 5400X + 3X / 25 = 5400(25X + 3X)/25 = 540028X = 5400 x 25X = 5400 x 25 /28X = 2700x25 /14X = 1350x25 / 7X = 33750 / 7X = 4821.42(approx)Therefore, originally it was Rs 4821.42(approx)\xa0 | |
| 14478. |
Reena purchased A dress ₹5400 including 12%gst . Find the price of dress |
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Answer» Approx 4821 or 4822 Let the original price be : Rs XAfter GST the price gone up = X + (12/100)XThe equation will beX + (12/100)X = 5400X + 3X / 25 = 5400(25X + 3X)/25 = 540028X = 5400 x 25X = 5400 x 25 /28X = 2700x25 /14X = 1350x25 / 7X = 33750 / 7X = 4821.42(approx)Therefore, originally it was Rs 4821.42(approx) |
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| 14479. |
What is alternating method |
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| 14480. |
CBSE EXAM TIMETABLE ANNOUNCED |
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Answer» 24feb to 18march 15feb to 20march 26 feb to 20 march.. ? 26 dec- 18 jan Yess Yes...I know.. |
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| 14481. |
43/2*4*5*3 will terminat e after how many decimal places |
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Answer» Since 3 is present in denominator,hence it is not terminating. yeah it will not terminate 43/2×4×5×3 = 43/120 = 43/2^3×5×3 .It will not terminate... |
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| 14482. |
Find the common difference ofthe A.p whose first termis 12 and fifth term is 0 |
| Answer» Given;In AP , a =12 and fifth term(a5)=0So;a5=a+4d0=a+4d0=12+4d-12=4dd=-12/4d = -3 | |
| 14483. |
The midpoint of the line segment A(2a,4) & B(-2, 3b)is C(1,2a+1) .find the values a & b. |
| Answer» Since the point C is midpoint of AB , Ratio=1:1So, applying the section formulax =x1+x2/21=2a-2/2. a=2 Same case with y y=y1+y2/22a+1 = 4-3b / 2 Put value of a solved above2(2)+1=4-3b/210-4 = -3bb= -2 | |
| 14484. |
Check whether x³-3x+1 is factor of x⅝-4x³+x²+3x+1 |
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| 14485. |
How to prove root five as irriational number |
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| 14486. |
Please show as all assessment paper maths |
| Answer» Yes | |
| 14487. |
Let sn be the sum of the first n terms of an AP.if S2n =3n +14n2.then what is the common difference |
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Answer» Yuuhhh Maybe -56 Kkkk |
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| 14488. |
In fig.10.37,angle pqr=100 degree , where p,q and r points on circle with centre o.find angle opr |
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Answer» 80° Answer |
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| 14489. |
Site 30 degree cos 30 degree + Cos 60 degree sin 30 degrees solve |
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Answer» Khud kro easy h The correct answer underroot3 +1/4 Hi Or √3/2.. (2√3)/4 |
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| 14490. |
Hots question with answer |
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| 14491. |
If Sn denotes the sum of first n terms of an AP then prove that S12= 3 (S8 - S4) |
| Answer» L.H.S ; S12= 12/2{2a + (12-1)d} = 6{2a + 11*d} = 12a + 66dR.H.S ; 3(S8 - S4) =3[n/2{2a+(n-1)d} - n/2{ 2a+(n-1)d}] = 3{8/2(2a+7d) - 4/2{2a+3d)} =3{4(2a+7d) - 2(2a+3d)} =3(8a+28d - 4a-6d) =3(4a+22d) =12a+66d Therefore ; L.H.S = R.H.S Proved......????????????????????????????????????????????????? | |
| 14492. |
√2xsquar-2x-√3 |
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| 14493. |
Find the sum of all natural number less than 100 which are divisible by 5 |
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Answer» Last term should be 95 not 100 bcos in question it is less than 100 so sum will be 950 Parishi ur answer is wrong AP formed will be-5,10,15.......100So, a=5d=5an=100n=? an= a+(n-1)d 100=5+(n-1)5 95= 5n-5 5n = 100 n=20 Putting these values in the sum formula we get, Sn=n/2 [ 2a+(n-1)d] S20= 20/2 [ 2(5) + (19)5] S20=10[10+95] S20=10[105] S20=1050 ANS.Hope it helps |
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| 14494. |
What is the identity of (a+b)³=? |
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Answer» a^3+b^3+3a^2b+3ab^2 a3+b3 3ab(a+b) a^3+b^3+3ab(a+b) |
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| 14495. |
HCF of 726 and 275 |
| Answer» 726 = 275*2 + 176275 = 176*1 + 99176 = 99*1 + 7799 = 77*2 + 2277 = 22*3 + 1122 = 11*2 +0Therefore ; H.C.F = 11.. ?????? | |
| 14496. |
7x_15y=2 ,x+2y=3 solved |
| Answer» 7x-15y=2.........17(X+2y)=3*7..........2 From eq.1 &eq.2 7x - 15y = 2 7x +14y = 21 ...↑-....↑-........↑-........... 0 -29y = -19 Therefore; y = -19/-29 Y = 19/29???????? Putting the value of "Y" in eq.2 X + 2y = 3 X + 2(19/29)=3 X +2*19/29 = 3 X = 3 - 38/29 X = 49/29???????Hence; X = 49/29 and Y = 19/29.... | |
| 14497. |
The remaider when the square of any prime no. greater than 3is divided by 6 is |
| Answer» 1 | |
| 14498. |
Important theorem |
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Answer» Theorem 6.1, 6.2, 6.6, 6.8, 6.9, 10.1, 10.2 Important theorem is 6. First five are in the chapter triangles and one is in the chapter construction Triangle me acha theorem hai |
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| 14499. |
If 2 and 3 are the zeros of polynomial 3x²-2kx+2m.find the value of k and m? |
| Answer» Let p(x) = 3x²-2kx+2mGiven, 2 & 3 are zeroes of p(x).Then , p (2) = 0 & p(3)= 0Now, p(2)= 3×2²-2k(2)+2m=03× 4-4kx+2m=012-4k+2m=0………..(1)p(3) = 3×3²-2k(3)+2m=03× 9-6kx+2m=027-6k+2m=0………..(2)On Subtracting Equation 1 from equation 2.27-6k+2m=012-4k+2m=0(-) (+) (-)------------------15-2k =015=2kk= 15/2On putting k= 15/2 in eq 112-4k+2m=012-4(15/2)+2m=012-30+2m=0-18+2m=02m=18m=18/2m=9Hence, k=15/2, m= 9 | |
| 14500. |
The points (7,2) &(-1,0) lie on a line |
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