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| 67001. |
the following exppessions.6p 12q2.De- 42U 7a 14a10 a- 15b +20 |
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Answer» Find the area of a triangle whose two sides are 18 cm and 10 cm and the circumference is 42 cm |
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| 67002. |
Write the following cubes in the expanded form:(i) (3a + 4b)3 (ii) (5p - 3q) |
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| 67003. |
7 x + 9 x + x - 16 x = 0 |
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Answer» 7x+9x+x-16x=016x+x-16x=0so x=0 |
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| 67004. |
7. Write the following cubes in the expanded form:(i) (3a + 46)(6) (5p - 3q)3 |
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| 67005. |
14?463351 |
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| 67006. |
: 4 x ^ { 2 } + 9 y ^ { 2 } + 16 z ^ { 2 } + 12 x y - 24 y z - 16 x z |
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| 67007. |
4 x ^ { 2 } + 9 y ^ { 2 } + 16 z ^ { 2 } + 12 x y - 24 y z - 16 x z |
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| 67008. |
x+8=24 \quad(x=3, x=15, x=16, x=0) |
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Answer» x + 8 = 24 x = 24 - 8 = 16 |
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| 67009. |
।।।If the quadratic equation (a? - b)x2 + (b2 - cx + c2-a2 = 0 has equal roots, then which of thefollowing is true :दि । सात समीकरण (a-b2) + (b’ - ४ + C -a = 0 के मूल बराबर है, तब निम्न में से कौनसाकिया रात्य होगा।(2) b(1) b* + c* = a+ ८ = 2a2(4) a = b + 2c(3) b -2 = 2a2 |
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| 67010. |
() (a') x (2a2) x (4a26) |
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| 67011. |
11. Simplify:(a) a(a+4) +3a (2a2-1)+ 4a2 +4 at a = -2(b) 1502 (2-3a)+5a(5+4b), at a =0, b=-1.12. Simplify :(1) 2xy(x - y)–3x2 (y-y2)-5y2 (2x2 – x)+2xy(y – x)(ii) a(a+5)+3a(2a2 –3)+5a2 +613. Perform the indicated multiplication :(095a(3–2a)6 ( 34) +a+o. |
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Answer» 13(i) 5a(3-2a) 5a×3-2a×5a = 15a-10a² bhenchod bhasdiwale thii ke lavde |
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| 67012. |
-3x(x +2 xy + 3y2)(taking 3 Ă x as common fDo Thisactorise (i) 9a-6atorisation by grouping the termsxpression ar +hr+ay+ by: You will find that there is no single common factorto(i) 15 ab-35ab(ii) 7Im 21mnBut the first two terms have the common factor 'x' and the last two terms have theIetus see how we can factorise such an expression. |
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Answer» solve it in elimination method |
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| 67013. |
(4) Find four consecutive terms in an A.P. whose sum is 88 and the sum of the 1st and3rd terms is 40 |
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Answer» Let four consecutive terms are a - 3d , a - d , a + d , a + 3d , where a and d is real numbers. A/C to question, sum of all four terms = 88(a - 3d) + (a - d) + (a + d) + (a + 3d) = 884a = 88 a = 22 ......(1) again, sum of 1st and 3rd term is 40e.g., (a - 3d) + (a + d) = 402a - 2d = 40a - d = 20 from equation (1), 22 - d = 20d = 2 hence, a = 22 and d = 2 then, a - 3d = 22 - 3 × 2 = 22 - 6 = 16 a - d = 22 - 2 = 20a + d = 22 + 2 = 24 a + 3d = 22 + 3× 2 = 22 + 6 = 28 therefore, four consecutive term in an A.P are 16, 20, 24, 28 |
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| 67014. |
3. Find the values of x and y so that -3, x, y, 12 may be fourconsecutive terms of an A.P |
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Answer» a=-3 x = -3 + d y = -3 + 2d 12 = -3 + 3d so 3d = 12 + 3 = 15 so d = 5 so x = -3+5 = 2 so y = -3+10 = 7 |
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| 67015. |
Express the following decimals as percent.i) 2.5ii) 39.6oc Dnd sells 8 fo |
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Answer» Thanks Did🐱 |
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| 67016. |
write the common factor of 14a square b and 35ab square |
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Answer» 14a²b = 2 × 7 × a × a × b35ab² = 5 × 7 × a × b × b common factors are1 ,7, a, b, 7a, 7b, 7ab thanks |
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| 67017. |
multiply the negative of 2/3by the inverse of 9/4 |
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Answer» inverse of 9/4 is 4/9negative of 2/3 is -2/34/9*-2/3=-8/27 -8/27 is the correct answer of the given question . -8/27 is the right answer -2/3×4/9=-8/27 is the right answer. -8/27 is the answer of the question |
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| 67018. |
Write the following cubes in expanded(i)(2r+1(u) (2a -3by |
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| 67019. |
(i) 2a2 + 5a+2 |
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Answer» 2a^2 + 5a + 2 = 0=> 2a^2 + 4a + a + 2 = 0=> 2a(a+2) + 1(a+2)= 0=> (2a + 1)(a + 2) = 0=> a = -2, a = -0.5 PLEASE HIT THE LIKE BUTTON |
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| 67020. |
(i) (a-b2) - (2a2-2b2) |
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| 67021. |
0) (a') x(2a2) x(4a26) |
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Answer» (a^2)*(2a^22)*(4a^26) = 8a^[2+22+26]= 8a^50 ans |
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| 67022. |
(1) If x +y 14 and 2x-y- 16, then x? |
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Answer» x + y = 14 .....(1)2x - y = 16 .....(2) Adding eq (1) and eq(2)3x = 30x = 10 |
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| 67023. |
8. Solve for x:(i) 1+6++ 16+..-+x=148 |
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Answer» Given 1 + 6+ 11+ 16 + ............+x = 148take the AP :1,6,11,16,..........,xIn this APa = 1d = 6-1 = 5Given Sn = 148we know that Sn = n/2[2a+ (n-1)d]⇒n/2 [ 2a +(n-1)d] = 148⇒n[2(1) + (n-1)5 ] = 148×2⇒n[ 2 +5n - 5 ] = 296⇒n [ -3 + 5n ] = 296⇒-3n + 5n² = 296⇒5n² - 3n -296 = 0⇒ 5n² - 40n + 37n - 296 = 0⇒5n( n - 8) + 37( n - 8) = 0⇒(n - 8) (5n + 37) = 0⇒n-8 = 0 or 5n +37 = 0⇒n = 8 or n= -37/5. As n is the no.of terms in the AP can not be fractional and negative∴n=8Theirfore x is the 8th termTn = T8 = a + (n-1)d = 1 + (8-1)5 = 1+7(5) = 1+35 = 36∴8th term = x = 36. eighth term is thirty six (36) |
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| 67024. |
2a b(a - a+ 1)- ab(2a4 -2a2 +a)-ab(a + a -1) |
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Answer» 2a^5b-2a^3b+2a^2b-2a^5b+2a^3b-a^2b-a^4b-a^2b+ab=ab-a^4b=ab(1-a^3)now X3-Y3 =(X-Y) • (X2 +XY +Y2).hence(1^3-a^3)=(1-a)(1+a+a^2)hence=ab(1-a)(1+a+a^2) |
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| 67025. |
6. Factorisei) 2a2 -5a -7ii) x-164i) x2 -7x + 12 |
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Answer» x²-7x+12=0x²-3x-4x+12=0x(x-3)-4(x-3)=0(x-4)(x-3)=0 x=3,4 |
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| 67026. |
(4) Find four consecutive terms in an A.P. whose sum is 88 and the sum of the 1st and the3rd terms is 40. |
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| 67027. |
\frac{\partial^{3} u}{\partial x \partial y \partial z}=\left(1+3 x y z+x^{2} y^{2} z^{2}\right) e^{x y z} |
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Answer» thankyou sir |
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| 67028. |
12-1 L = tanL.HS = tan+ tan ,--tan- 2 ll.tan202 11. (cost)·12 in the simplest fople 5 Express tanˊ--< x < _ in the simplest foution We write2COS1-sin x- tan2cos Sin2Sin coS |
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| 67029. |
\frac { a ^ { 3 } + 3 a b ^ { 2 } } { 3 a ^ { 2 } b + b ^ { 3 } } = \frac { x ^ { 3 } + 3 x y ^ { 2 } } { 3 x ^ { 2 } y + y ^ { 3 } } , \text { prove that } \frac { x } { y } = \frac { a } { b } |
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| 67030. |
If the coefficients of 2nd, 3rd and 4th terms inthe expansion of (1value of n is(A) 25.x) are in A.P., then the(B) 7(D) 14(C) 11 |
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| 67031. |
\begin{array} { l } { \text { If } \frac { a } { b } = \frac { 2 } { 3 } \text { then find the values of the following expressions. } } \\ { \text { (i) } \frac { 4 a + 3 b } { 3 b } } & { \text { (ii) } \frac { 5 a ^ { 2 } + 2 b ^ { 2 } } { 5 a ^ { 2 } - 2 b ^ { 2 } } } \\ { \text { (iii) } \frac { a ^ { 3 } + b ^ { 3 } } { b ^ { 3 } } } & { \text { (iv) } \frac { 7 b - 4 a } { 7 b + 4 a } } \end{array} |
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Answer» Here a=2 and b=3.Substitute the values in the ques.1)17/92)38/2=193)35/274)7/23 Ans |
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| 67032. |
( a %2B b ) x %2B ( a - b ) y = 2 a b , ( a %2B b ) x - ( a - b ) y = a b |
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Answer» Like if you find it useful |
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| 67033. |
0*a^3 %2B 27*b^6 |
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Answer» a^3+b^3=(a+b)(a^2-ab+b^2)hence1000a^3=(10a)^3and27b^6=(3b^2)^3hence1000a^3+27b^6=(10a+3b^2)(100a^2-30ab^2+9b^4) |
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| 67034. |
\left(\frac{3}{4} a^{2}+3 b^{2}\right) \text { and } 4\left(a^{2}-\frac{2}{3} b^{2}\right) |
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Answer» If addition is not the question post the complete question |
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| 67035. |
1. Which of the following numbers is not a perfect square(a) 7056(b) 3969(c) 5478 |
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Answer» (c) 5478 propertoes of square a number which ending 2,3,7,8 is not perfect square |
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| 67036. |
(2*a^2 %2B 3*b^2)/(2*a^2 - 3*b^2) |
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| 67037. |
15. Use the given figure to express:) a in terms of b and f;(ii) e in terms of f and g;(ii) d in terms of a and e;(iv) c in terms of e and g. |
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| 67038. |
lation tables.(b)7859x16944*17372(g) 463x 19(h)X 18(1)(m187x 14293x 16 |
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Answer» 78*17=1,326372*18=6,696463*19=874187*14=2,618293*16=4,688 1 answer 9442 answer13263 answer 66964 answer87965 answer 26186 answer 4688 |
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| 67039. |
917. Name the verify the property를 + (+) (+) |
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| 67040. |
a/b=7/3 then\frac{5 a+3 b}{5 a-3 b} \quad\left(\text { ii) } \frac{2 a^{2}+3 b^{2}}{2 a^{2}-3 b^{2}}\right. |
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| 67041. |
\left( \frac { 2 } { 3 } a ^ { 2 } b - \frac { 4 } { 5 } a b ^ { 2 } + \frac { 2 } { 7 } a b + 3 \right) \text { by } 35 a b |
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| 67042. |
The sum of 3 terms of an Ap is121 And the prodect of the firstand the 3rd terms x is the exsistthe Second terms by Lind 3 terms |
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Answer» question answer is 9 |
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| 67043. |
Uhe the given figure fo express:a in terms of b and fe in terms of fand gd in terms of a and e;e) e in terms of e and g. |
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| 67044. |
18)36 = 2 x 18 = 2x2x9=2x2x3x3363636 =........................What we observe from the above is that, the factorsthough the order of the factors is different. Usually, ththe greatest as 2x2x3x3. |
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Answer» 2×2×3×3= 36 right answer |
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| 67045. |
\frac { \operatorname { cos } 36 - \operatorname { sin } 36 } { \operatorname { cos } 36 + \operatorname { sin } 36 } |
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Answer» we have (cos x - sin x)/( cos x + sin x)= (1- sin x/ cos x)/(1+ sin x/ cos x)= (1- tan x)/ ( 1+ tan x)= tan (45-x) as tan (45-x) = (tan 45 - tan x)/((1+ tan 45 tan x) = (1- tan x)/(1+ tan x) putting x = 36 we get (cos36-sin36)/(cos36+sin36)= tan (45-36)= tan 9 = tan (180+9) = tan 189 how uh did d 2nd step |
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| 67046. |
\left. \begin{array} { l } { \text { Given } \frac { x ^ { 3 } + 12 x } { 6 x ^ { 2 } + 8 } = \frac { y ^ { 3 } + 27 y } { 9 y ^ { 2 } + 27 } . \text { Using componendo } } \\ { \text { and dividendo find } x : y . } \end{array} \right. |
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| 67047. |
Use the given figure to express:(i) a in terms of b and f;(i) e in terms of fand g;fii) d in terms of a and e(iv) c in terms of e and g.15. |
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Answer» (i) a = f - b(applying exterior angle theorem, f =a+b)(ii) e = g-f(iii) angle(AGF) = e (vertically opposite angles) By exterior angle theorem, d = a + angle(AGF)d = a + e(iv) c = 180 - f (linear pair) f = 180 -cby exterior angle theorem, g= e +f g = e + 180 - cans. c = 180 + e - g |
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| 67048. |
18 Given 6.12x = 9324227 . U18 Given z? + 12x-y3 + 27yUsing componendo and dividendo, find x :y (2015) |
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| 67049. |
vldendo, fhnd a:b.x3 +12x y" + 27 y6x2+8 9y: + 27a. GiverUsing componendo and dividendo, find r y. |
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Answer» 2 by 3 minus 2 by 3 into 3 by 5 + 5 by 2 minus 3 by 5 into 1 by 6 |
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| 67050. |
618+852×68 |
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Answer» =618+852×68=618+57936=58554 |
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