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| 64951. |
ही okl g Al nemlpnlnank %Mj&‘ >एज Hag%‘;&_LQJ:L_QM ¢ Cpuskien a ele L)1 901.६ न 20. 5 t5= |
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| 64952. |
(a) 36Round each of the following numbers to the nearest ten:(b) 173(c) 3869 |
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Answer» 40 40,170,3870are the rounded off those no |
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| 64953. |
1. Round off the following numbers to the nearest ten.a) 267b) 4,232c) 94 |
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Answer» 270423290 .............. a)= 270b)= 4230c)=90 is round off correctly to nearest ten ¶¶ 270, 4230,90 is the correct answer of the given question A,267=270B,4,232=4,230C,94=90 2704,23090 is the correct answer A,267=270B,4,232=4,230C=94=90 A = 270 B = 4230C = 90 |
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| 64954. |
ant I (On dividing -3+2 by a polynomial gx) thequotient and remainder were x-2 and-2x+4. respectively.Find g(x).) |
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| 64955. |
Section-A' / 7gu-' 37'1. What is the nature of roots of quadratic equation 5x2-7x2 0? |
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Answer» 5x²-7x+2 = 0 D = b²-4ac = (-7)² - (4*5*2) = 49-40 = 9 = +ve so, roots are real and unequal.. and rational also because D is a perfect square of 3 |
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| 64956. |
checsethelargestnumber |
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Answer» As 20/2= 10 and 56/8= 7 now 10>7 10 is the largest number |
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| 64957. |
\operatorname { lim } p \operatorname { log } \frac { \sqrt { 2 } 5 } { \sqrt [ 3 ] { 64 } } + ( \frac { 256 } { 625 } ) ^ { - \frac { 1 } { 4 } } + \frac { 1 } { ( \frac { 64 } { 125 } ) ^ { \frac { 2 } { 3 } } } |
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Answer» (√25/³√64) + (256/625)^(-1/4) + 1/(64/125)^⅔= 5/4 + (625/256)^(1/4) + 1/(4/5)²= 5/4 + 5/4 + 25/16= 10/4 +25/16= 40/16+25/16= 65/16 |
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| 64958. |
as a rational number with denominator1. Express(ii) -30(iii) 35(i) 20(iv) -40 |
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| 64959. |
If 7 times 7th term of a AP is equal to 11 times the 11th term, thenterm.findthe18h |
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Answer» 7th term = (a+6d) 11th term = (a+10d)(a+6d)7 since 7th term of the ap is 7 times and the same for 11th term(a+6d)7=(A+10d)11 (given)7a + 42d =11a +110d (on solving)7a - 11a = 110d - 42d-4a = -68da= -17d we know 18th term = (a+17d)................................. (i)now we substitute the value of a in (i)which emplies, -17d + 17d = 0hence prove.. |
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| 64960. |
SA1. Express(1) 20as a rational number with denominator(11) -30(1) 35(iv) --40 |
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Answer» Puri detail mein samjhaie |
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| 64961. |
8) Write an equivalent fraction ofin which25the:(i) Numerator is 110(ii) Numerator is 176(iii) Denominator is 125(iv) Denominator is 275 |
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Answer» 11/25=(11×10)/(25×10)=110/250so numerator is 110 thanks sir u r great |
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| 64962. |
-3as a rational number with denominator1. Express(i) 20(ii) -30(iii) 35(iv) -40 |
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| 64963. |
1. What are rational numbers? Give ten examples of rational numbers. |
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| 64964. |
Rounding numbers nearest ten - 36 |
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Answer» The number is 36and the units place is 6hence greater then 5 so it can be rounded off to 40 |
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| 64965. |
Expressi) 20as a rational number with denominator(ii) -30(iii) 35-42innol number with denominator 7. |
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| 64966. |
If the roots of the equation 5x2 + 13 x + K 0 are reciprocal of each other. Find thevalue of K. |
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Answer» Let Alpha be the one zero of the given polynomial. Then the other will be 1/Alpha P(X) = 5X²+13X+K Here, A = 5 , B = 13 and C= K Product of zeros = C/A Alpha × 1/Alpha = K/5 1 = K/5 K = 5. |
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| 64967. |
(iv) 5x2 -3xy+ 4y- 9, 7y" + 5xy 2x13au_ant-73p -p 12 |
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Answer» Thank you |
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| 64968. |
19. Find the largest number that divides 2053 and 967 and leaves a remainder of 5 and 7respectively |
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Answer» It is given that on dividing 2053 by the required number there is a remainder of 5 .this mean that 2053 - 5 = 2048 is exactly divisible by required number . similarly,967- 7 = 960the required number is the largest number satisfying the above property .therefore ,it isthe HCF of 2048 & 960HCF of 2048 & 960 is 64hence required number is 64 |
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| 64969. |
\lim _{x \rightarrow 0} \frac{e^{x}-e^{-x}}{x} |
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| 64970. |
Sum of polynomials322- Tx t5 and 623+54-7 |
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Answer» 3x²-7x+5 + (6x³+5x-7) 6x³+3x²-2x-2 |
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| 64971. |
ta The dimensions of a closed oil tin are 30 cm x 30 cm x 45 cm. Find the area of the metal sheet requiredmaking 10 such tins. |
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Answer» Dimension of cuboidal tin = 30cm×30cm×45cmlet length = 30cmbreath = 30cmheight = 45cm Total surface area = 2(lb+bh+lh) = 2(30×30 + 30×45 + 45×30) = 2(900 + 1350 + 1350) = 2(3600) = 7200 cm²TSA of one tin box = 7200 cm²TSA of 20 tin boxes = 7200×10 = 72000 cm² |
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| 64972. |
\operatorname { lim } _ { x \rightarrow 0 } \frac { e ^ { x } - e ^ { - x } - 2 x } { x - \operatorname { sin } x } |
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| 64973. |
Theorem 6.2: If a line divides any two sides of atriangle in the same ratio, then the line is parallelto the third side. |
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| 64974. |
12. A.P. 10,7,4,322.Find 11th term from the last term of A.P. 10,7,4....,.-62 |
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| 64975. |
2. In the given figure, 4BTO 30. Find ZATO, where O is centre of circle and TA and TB are tangents.30 |
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| 64976. |
12. The mean of 20 numbers is 18. If 3 is added to each of the frstnumbers, find the mean of the new set of 20 numbersten |
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Answer» please like the solution 👍 ✔️ |
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| 64977. |
Prove using mathematical induction that(x2n−y2n)(x2n−y2n)is divisible by(x+y)(x+y). |
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Answer» please post full question with proper information x2n is in product of something else. |
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| 64978. |
If tan2x- 3 0, then find the value ofvaluate : 2 sin2 30-2 cos2 45+ tan2 60ヲ孤ボ双2 30°-2cos2 45° +ta: 2 sirn |
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Answer» tan2x = √3 2x = tan^-1(√3) 2x = 60 x = 30° |
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| 64979. |
If one root of 5x2 13 x + k 0 is the reciprocal of the other root,then find the value of k |
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| 64980. |
bxe* Express10 202. ExpressLEHUISE TAas a rational number with denominatormiss30IV- 402. Expres421 |
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| 64981. |
In the given figure, angle BTO=30°. Find angleATO, where O is centre of circle and TA and TB are tangents. |
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Answer» In ∆ AOT and ∆BOTAO = BOangle A = angle B (90°)OT = OT ( common)So, by RHS rule ∆ AOT and ∆BOT are congurent.So, by CPCT angle BTO = angle OTAangle ATO = 30° |
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| 64982. |
7स्तविक स.कस प्रकार प्रदर्शित ।नव प्रसारों पर विचार व।खण्ड 11857।। ५५५ सारकार प्रदर्शित किया जाता है।पर विचार करते हैं। ध्यानउपर्युक्त तरबनकर आती हैं या नहीं।38 710 7 और 4 के दशमला1 के दशमलव प्रसार ज्ञात कीहोता है। अंक ५। (i) या तो शेतब हमें शेषफ-स्थिति पर अटINCERT SOLVED EXAMSस्थिति ।। वाले| ().8758 | 7.)61| 0.142810।हो जाता है।603028506.06||()1)208दशमलवसांत (ta60361013)की पु-दूसरे| 4तब । |
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| 64983. |
IS. The number x is 2 more than the number y ex Ifthe sum of the squares of x and y is 34the product of x and y |
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Answer» Given x = y + 2 . . . . . (1) x² + y² = 34From (1) x - y = 2 squaring both sides, (x - y)² = 2² x² + y² - 2 x y = 4 34 - 2 x y = 4 x y = 15 |
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| 64984. |
The number x is 2 more than the number y. Ifthe sum of the squares of x and y is 34; findthe product of x and y. |
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Answer» x=y+2 x*x+y*y=34 x=y+2 so x-y=2 by taking square both the sides x*x-2xy+y*y = 4 34-2xy = 4 so 2xy = 34-4 = 30 so xy = 15 |
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| 64985. |
Theorem 6.2 : Ifa line divides any two sides of atriangle in the same ratio, then the line is parato the third side.lutoking a line DE such |
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| 64986. |
MATHEMATICALThe number 144 can be represented by12 x 12 square grid. Which of the followingcan also be represented on a square grid?(A) 39(C) 70(B) 50(D) 81 |
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Answer» So we actually finding perfect square there answer will be (D) 81 which sqaure of 9 81 can be represented as 9 x 9 square grid |
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| 64987. |
If x4 +-=322, prove that x-= 4 or-4 ; x being a real number.x.7. |
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| 64988. |
6.The polynomial p(x) = x4-2x3 + 3x2-ax + 30-7 when divided by x + 1, leaves theremainder 19. Find the value of a. Also, find the remainder when p(x) is divided byx +2. |
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Answer» Given Polynomial⇒ P(x) = x⁴ - 2x³ + 3x² - ax + 3a - 7. Divisor = x + 1∴ x + 1 = 0∴ x = -1 Thus, P(-1) = (-1)⁴ - 2(-1)³ + 3(-1)² - a(-1) + 3a - 7.19 = 1 + 2 + 3 + a + 3a - 719 = 6 - 7 + 4a4a - 1 = 194a = 20⇒ a = 4 ∴ Value of a is 4. Now, the Polynomial will be ---→ P(x) = x⁴ - 2x³ + 3x² - (4)x + 3(4) - 7P(x) = x⁴ - 2x³ + 3x² - 4x + 12 - 7P(x) = x⁴ - 2x³ + 3x² - 4x + 5 Now, When this polynomial is divided by (x + 2), then, x + 2= 0x = - 2 ∴ P(-2) = (-2)⁴ - 2(-2)³ + 3(-2)² - 4(-2) + 5⇒ P(-2) = 16 + 16 + 12 + 8 + 5⇒ P(-2) = 57 Thus, Remainder will be 57. |
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| 64989. |
Principle of Mathematical Induction |
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Answer» According to theprinciple of mathematical induction, to prove a statement that is asserted about every natural number n, there are two things to prove. If the statement is true for n = k, then it will be true for its successor, k + 1. |
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| 64990. |
4. If one root of the equation x2 -30 x+p=0 is square of the other, then p is equal to- |
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| 64991. |
The angle of rotation of a figure is 36 What is its order of rotational symmetry? |
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Answer» For rotation symmetry, divide it 360 by the angle of rotation.so the order of symmetry is 360°/36°=10 |
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| 64992. |
27.The order of rotational symmetry for an equilateral triangleA) 2B) 1C) 3D) 10 |
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| 64993. |
13. SimplifyA.B. 19+22 +((693 × 64) ÷ 6883 × a4b82 × a6b56 x3xb91 x7D.C. |
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| 64994. |
___ + ___ = 8+ +___ - ___ = 6= =13. 8 |
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| 64995. |
6. If the mean of the following data is 18.75, find the value of p.| x | 10 | 15 | p | 25 | 30 ||TA 5 10 1718/21 |
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| 64996. |
30.Find the value of p, when/X-1 = 0 and x =-is oneroot of this equation.CBSE 2013 |
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| 64997. |
3) . एक बेलन का आयतनं 693 सेमी & | यदि बेलन की त्रिज्या 10.5 से मी. है, तो उसकी ऊँचाईज्ञात क़ीजिए। |
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| 64998. |
A number P is aivide byThe sum of a number x and twice the number y is 30. |
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Answer» the first number = x the second number = y The sum of two numbers is 30. x + y = 30 The first number is twice the second number. x = 2y substitute 2y in for x in the first equation x+ y = 30 2y+ y = 30 3y = 30 y = 30/3 y = 10 then put 10 in for y x = 2*(10) x =20 |
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| 64999. |
किसी वस्तु को ₹ 693 में बेच कर, स्मृति को 26%लाभ प्राप्त होता है। स्मृति ने कितने रूपए में उसवस्तु की खरीदी की थी? |
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Answer» cp. sp100. 126x(let). 693126x= 693×100x= 693×100/126 =550₹ cost pric (cp)= 550₹ COST PRICE=(693×100) /126=11×50=550 cost price=SP×100/(100+%P) =693×100/(100+26) =693×100/126 =11×50=550 |
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| 65000. |
Upper Quartile:Maximum:mphmphQ2: The ages of the children, in years, at a birthday party are 8, 12, 10,9,12,9,11, and 9. Use the mathematicalformula for the mean to find the mean age of the children at the birthday party.years old.The mean age is |
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Answer» Mean age=8+12+10+9+12+9+11+9/8=80/8=10 yrs old Men is 10 years old the mean age is 10 year old mean age =8+12+10+9+12+9+11+9/8=80/8=10 yers old |
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