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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.
| 33501. |
if 2 and 1 are the two zeros of quadratic polynomial ax square + bx + 2 find value of a and b |
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Answer» a=1, b=-3......... a= 1, b= -3 a= 1 b =-3 a= -1 b= -3 |
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| 33502. |
Find the smallest number which leves reminder 8and 17 when divided by 520 and 468? |
| Answer» We will subtract 8 from 520 and 17 from 468 . By subtracting these we get 451 and 512 . Now by long division method or either by euclid division algo.divide them you will get the ans.1 | |
| 33503. |
Find the smallest number which leves reminder 8and12 when division by 28 and 32 respectively? |
| Answer» First, we find L.C.M of 28 and 32 28= 2 . 2 . 732= 2 . 2 . 2 . 2 . 2 L. C. M (28,32)= 2 . 2 . 2 . 2 . 2 . 7=224The smallest number that leaves remainder 8 and 12 when divided by 28 and 32 = 224 -(8+12) = 204Thus, 204 is required smallest number. | |
| 33504. |
How form of comp math subject fill up |
| Answer» | |
| 33505. |
What sun |
| Answer» The sun is a huge glowing sphere of hot gas , most of this gas is hydrogen(about 70%) and helium(about 28%).Carbon , nitrogen and oxygen make up 1.5% and the other 0.5% is made up of small amounts of many other elements such as neon , iron , silicon , magnesium , sulphur. | |
| 33506. |
Use Euclid algorithm to find the hcf of1190 and 1445. Express as hcf in form 1190m+1445n |
| Answer» Here we have to find HCF of 1190 and 1445 and express the HCF in the form 1190m + 1445n.1445 = 1190 ×\xa01 + 2551190 = 255 ×\xa04 + 170255 = 170 ×\xa01 + 85170 = 85 ×\xa02 + 0So, now the remainder is 0, then HCF is 85Now,85 = 255 - 170= (1445 - 1190) - (1190 - 255 ×\xa04)= 1445 - 1190 - 1190 + 255 ×\xa04= 1445 - 1190 ×\xa02 + (1445 - 1190) ×\xa04= 1445 - 1190 ×\xa02 + 1445 × 4 - 1190 ×\xa04= 1445 ×\xa05 - 1190 ×\xa06= 1190 ×\xa0(- 6) + 1445 ×\xa05= 1190m + 1445n , where m = - 6 and n = 5 | |
| 33507. |
What is the HCF of 24&42 |
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Answer» 2×3×2×2 2×3×7 = 2 6 6 6 6 6 is the hcf of 24&42 |
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| 33508. |
if m and n are odd positive integers then m2+n2 is even but not divisible by 4 justify |
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Answer» and it\'s not m×2 its m square no M and n should be odd When m=2,n=3, 2m+2n=2×2+2×3=4+6= 10 Which is even |
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| 33509. |
a³-12(a-4）-64 |
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| 33510. |
What is unique solution |
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Answer» it doesn\'t means a single one but having all solutions be equal Single solution that satisfy both the equations |
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| 33511. |
find the HCF of 52 and 117 and Express in the form of 52 X + 117 y |
| Answer» By Euclid\'s lemma ,117=52×2+1352=13×4+0HCF is 1313=117-(52×2) =117×1-52×2 =117y+52x(where y=1,x=-2) | |
| 33512. |
How to calculate lcm |
| Answer» Lcm means lowest common multiple so,If we have to find Lcm of 12..then find llowest common multiple let\'s see,12=2×6 6=2×3Or 3=3×1Hence lcm of 12=2×2×3.Also if we want to find Lcm of Two or more than two no. First brought every no in form( as we have done above) and then find lowest common factors from them .. | |
| 33513. |
factories by middle term splitx²+2√5x-3 |
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| 33514. |
If sum of squares of 3 consecutive odd natural numbers is 155 then find their product |
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| 33515. |
Proof that x2n-1 -1 is divisible by x-1,xnot =1 |
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| 33516. |
Two lines are parallel the equation of one line is 4x+3y=4. Then find equation of the another line. |
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Answer» Many equation to solve this equation 16x+12y=16 12x+9y=12 8x+6y=8 |
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| 33517. |
Find smallest number which leaves reminders 8and 12 when divided by 28 and 32 respectively |
| Answer» The LCM of 28 and 3228 = 2× 2 × 7=22×732 = 2 × 2 × 2 × 2 × 2=25LCM = 25 × 7 = 224Smallest no: which leaves remainder 8 and 12 when divided by 28 and 32= LCM of 8 & 12 = 224 - 20 = 204Therefore, 204 is smallest number which when divided by 28 and 32 leaves remainder 8 and 12 respectively. | |
| 33518. |
Find greatest numbers that will divide 445,572 and 699 leaving remainders 4,5and 6 respectively |
| Answer» We have to find the greatest number that divides 445, 572 and 699 and leaves remainders of 4, 5 and 6 respectively. This means when the number divides 445, 572 and 699 leaves remainders 4, 5 and 6 is that445 - 4 = 441, 572 - 5 = 567 and 699 - 6 = 693 are completely divisible by the required number.For the highest number which divides the above numbers can be calculated by HCF .Therefore, the required number is the H.C.F. of 441, 567 and 693 Respectively.First, consider 441 and 567.By applying Euclid’s division lemma, we get567 = 441 {tex}\\times{/tex}\xa01 + 126441 = 126 {tex}\\times{/tex}\xa03 + 63126 = 63 {tex}\\times{/tex}\xa02 + 0.Therefore, H.C.F. of 441 and 567 = 63Now, consider 63 and 693again we have to apply Euclid’s division lemma, we get693 = 63 {tex}\\times{/tex}\xa011 + 0.Therefore, H.C.F. of 441, 567 and 693 is 63Hence, the required number is 63. 63 is the highest number which divides 445,572 and 699 will leave\xa04,5 and 6 as remainder respectively. | |
| 33519. |
Sin A value |
| Answer» Complete the question | |
| 33520. |
How to draw a graph for quadratic equation |
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Answer» And plot them to the graph....as graph is not a straight line you required more then 5 coordinate... Just , do = y to the equation ....and find the coodinates like the linear equation as putting values of X and Y |
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| 33521. |
What is value of sin30°? |
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Answer» 1/2 1/2 The value of sin30° 0.5 0.5 |
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| 33522. |
If 7 sin square+cos square=4, show that tan =1/√3 |
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Answer» If 7 sin square A+3cos squareA=4, show that tanA=1/√3. This question is wrong. Kindly check your question. It seems wrong.☺☺☺ |
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| 33523. |
Simplify 9-2✓3x-x*x |
| Answer» -x^2-2√3x+9-x^2-3√3x+√3x+9-x(x+3√3)+√3(x+3√3)(-x+√3) (x+3√3) | |
| 33524. |
When cbse release compartment form 2018 |
| Answer» Check CBSE updates here :\xa0https://mycbseguide.com/blog/ | |
| 33525. |
1÷a alpha +b + 1÷a beta + b |
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| 33526. |
4÷3×6+3_2 |
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Answer» 9 hi |
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| 33527. |
What is the relation between mean,mode and median |
| Answer» See ncert mathematics 10 ch statistics | |
| 33528. |
If roots are equal then find p, 5x2-px+1=0 |
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| 33529. |
Alpha minus beta Qawali square |
| Answer» Alpha^2+Beta^2+2×Alpha×Beta | |
| 33530. |
Write a rational number between √2 and √3 |
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Answer» Bro i can\'t find it plzz give space between and write once more √2 and √31/2*(√2+√3)(√2+√3)/2(√2+√3)/√2+√2 2=√2+√2√2 and √2 cancelled so√3/√2 answer Hi |
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| 33531. |
Express 150 as a product of its prime factors |
| Answer» 2*3*5² | |
| 33532. |
In an ap, a3 =15, s10 =125, find d and a10 |
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Answer» d. -10 and a10.8 How 125 125 |
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| 33533. |
If sec theta=5\\7,show that 2(cos theta-sin theta)/cot theta -tan theta =12/7 |
| Answer» It can\'t be possible bro because we know sec theta=h/b then,here \'b\' is greater than \'h\' which can\'t be possible. | |
| 33534. |
Explain why 11×13×15×17+47 is a composite numbera |
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| 33535. |
Solve for x:-12abx2 -(9a2 - 8b2)x - 6ab |
| Answer» 12 abx2 - (9a2 - 8b2)x - 6ab = 0Comparing with Ax2 + Bx + C = 0, we getA = 12 ab, B = -(9a2 - 8b2), C = -6 abUsing the quadratic formula, {tex}\\mathrm { x } = \\frac { - \\mathrm { B } \\pm \\sqrt { \\mathrm { B } ^ { 2 } - 4 \\mathrm { AC } } } { 2 \\mathrm { A } }{/tex}we get {tex}\\Rightarrow x = - \\left( - \\left( 9 a ^ { 2 } - 8 b ^ { 2 } \\right) \\right\\}{/tex}{tex}\\frac { \\pm \\sqrt { \\left( - \\left( 9 a ^ { 2 } - 8 b ^ { 2 } \\right) \\right) ^ { 2 } - 4 ( 12 a b ) ( - 6 a b ) } } { 2 ( 12 a b ) }{/tex}{tex}= \\frac { 9 a ^ { 2 } - 8 b ^ { 2 } \\pm \\sqrt { 81 a ^ { 4 } + 64 b ^ { 4 } - 144 a ^ { 2 } b ^ { 2 } + 288 a ^ { 2 } b ^ { 2 } } } { 24 a h }{/tex}{tex}= \\frac { 9 a ^ { 2 } - 8 b ^ { 2 } \\pm \\sqrt { \\left( 9 a ^ { 2 } + 8 b ^ { 2 } \\right) ^ { 2 } } } { 24 a b }{/tex}{tex}= \\frac { 9 a ^ { 2 } - 8 b ^ { 2 } \\pm \\left( 9 a ^ { 2 } + 8 b ^ { 2 } \\right) } { 24 a b }{/tex}{tex}= \\frac { 9 a ^ { 2 } - 8 b ^ { 2 } + 9 a ^ { 2 } + 8 b ^ { 2 } } { 24 a b }{/tex}{tex}= \\frac { 9 a ^ { 2 } - 8 b ^ { 2 } - 9 a ^ { 2 } - 8 b ^ { 2 } } { 24 a b }{/tex}{tex}= \\frac { 18 a ^ { 2 } } { 24 a b } , \\frac { - 16 b ^ { 2 } } { 24 a b } = \\frac { 3 a } { 4 b } , \\frac { - 2 b } { 3 a }{/tex}{tex}\\therefore{/tex} the solution of the given equation are\xa0{tex}\\frac { 3 a } { 4 b } \\text { and } \\frac { - 2 b } { 3 a }{/tex}. | |
| 33536. |
Find hcf of 81and237 and express it as linear combination of 81 and 237 |
| Answer» Since, 237 > 81On applying Euclid\'s division algorithm, we get237 = 81 {tex}\\times{/tex}\xa02 + 75 ..........(i)81 = 75 {tex}\\times{/tex}\xa01 + 6 ..........(ii)75 = 6 {tex}\\times{/tex}\xa012 + 3 .........(iii)6 = 3 {tex}\\times{/tex}\xa02 + 0 .............(iv)Hence, HCF (81,237) = 3.In order to write 3 in the form of 81x + 237y,Now,{tex}\\style{font-family:Arial}{\\begin{array}{l}3=75-6\\times12(\\;from(iii))\\\\=75-(81-75\\times1)\\times12\\\\=75-(81\\times12-75\\times12)\\\\=75-81\\times12+75\\times12\\\\=75(1+12)-81\\times12\\\\=75\\times13-81\\times12\\\\=13\\times(237-81\\times2)-81\\times12\\;(\\;Substuting\\;75\\;from(i))\\\\=13\\times237-13\\times81\\times2-81\\times12\\\\=237\\times13-81\\times(26+12)\\\\=81(-38)+237\\times13\\\\=81x+237y\\end{array}}{/tex}Hence, x = -38 and y = 13\xa0 | |
| 33537. |
What is the hcf of smallest composite number and the smallest prime number |
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Answer» Smallest composite number is 4 Smallest prime number is 2HCF of 2 & 4 Is =2 Smallest prime number=2Smallest composite number=8Their hcf will be 8 For prime :-1 and for composite :- 2 8 |
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| 33538. |
Chapter 7 |
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| 33539. |
Prove herons formula |
| Answer» Hjj | |
| 33540. |
Prove heron sformula |
| Answer» Are you really in 10th standard | |
| 33541. |
If remainder of [(5m+1)(5m+3)(5m+4)]/5 is a natural number then find it |
| Answer» | |
| 33542. |
Find a quadratic polynomial for which sum and product of its zeroes are 0 and -3/5 |
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Answer» Given:. x+y=0 and x × y=-3/5. The quadratic equations formula is. X^2-(sum of zero)x +(product of zero)X^2 -0x + (-3/5). X^2-0x-3/55X^2/5 -0x/5 -3/5=0. Therefore p(x)=5X^2 -0X -3. Or, 5X^2-3 5X² - 3 |
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| 33543. |
Cos^2 60 + 4sec^2 60 - tan^2 45 evaluate |
| Answer» Cos60°=1/2. ,sec 60°=2. ,tan 45°=1. (1/2)^2 + 4(2)^2 - (1)^2=1/4 + 16 - 1=1/4 +64/4 - 4/4= 61/4 . | |
| 33544. |
A^2+B^2 = |
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Answer» (a+b)^2-2ab C^2 |
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| 33545. |
Use Euclid division algorithm to find the HCF of 867 and 225 |
| Answer» 867=225*3+192225=192*1+33192=33*5+2733=27*1+627=6*4+36=3*2+0Therefore hcf of 867 and 225 is 3 | |
| 33546. |
I want to re-check my wards copy of 10th class maths subject.what is the correct process..? |
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| 33547. |
How to slove theorem bpt? |
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| 33548. |
( a+ b) ^2 x^2 + 8(a^2 -b^2 ) x + 16 (a-b )^2 = 0 solve by using completing the squares method. |
| Answer» {tex}(a + b)^2x^2\xa0+ 8(a^2 - b^2)x + 16(a - b)^2 = 0{/tex}{tex}A = (a + b)^2,\\ B = 8(a^2 - b^2),\\ C = 16(a - b)^2\xa0{/tex}D = B2 - 4AC= [8(a2 - b2)]2 - 4 {tex}\\times{/tex}\xa0(a + b)216(a - b)2\xa0= 64(a2 - b2)2 - 64(a2 - b2)2 = 0{tex}\\therefore{/tex}x =\xa0{tex}\\frac{-B}{2A}{/tex}\xa0=\xa0{tex}\\frac { - 8 \\left( a ^ { 2 } - b ^ { 2 } \\right) } { 2 ( a + b ) ^ { 2 } }{/tex}\xa0=\xa0{tex}\\frac { - 4 ( a - b ) } { a + b }{/tex} | |
| 33549. |
How to prove that root p is a irrational number |
| Answer» Let us assume, to the contrary, that √p is rational.So, we can find co-prime integers a and b(b ≠ 0){tex}\\begin{array}{l}\\sqrt p=\\frac ab\\\\\\end{array}{/tex}{tex}a = b _ { \\sqrt { P } }{/tex}on squaring both sides we geta2 = pb2 ...... (1)so a2 is divisible by phence a is divisible by p ....... (2)So, we can write a = pc for some integer c.Squaring both the sides we geta2 = p2 c2 ....⇒ pb2 = p2 c2 ....[From (1)]⇒ b2 = pc2⇒ b2 is divisible by p⇒ b is divisible by p ....... (3)From (2) and (3) we conclude that\xa0p divides both a and b.∴ a and b have at least p as a common factor.But this contradicts the fact that a and b are co-prime. (As per our assumption)This contradiction arises because we have assumed that √p is rational.∴ √p is irrational. | |
| 33550. |
How much marks ncert questions come in board |
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Answer» It depends upon the creators of board paper s Approx 90% About 60% |
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