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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.
| 75551. |
EXERCISE 5.1Which of the following statements are true and which are false? Give reasons for answerOnly one line can pass through a single point. |
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Answer» (a)Infinite number of lines can pass through a single point. (b)Only one line can pass through two given points. |
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| 75552. |
A solid sphere hung at the lower end of a wireis suspended from a fixed point so as to give anelongation of 0.4mm. When the first solidsphere is replaced by another one made of samematerial but twice the radius, the newelongation is |
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| 75553. |
t. InaDABC, D, E,and F are, respectively, the mid- points of BC, CAand AB. Ifthelemgtand 9 cm, respectively, find the perimeter of Δ DEF.and CA are 7an, 8 am |
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| 75554. |
\frac { \operatorname { sin } \theta - \operatorname { sin } 5 \theta + \operatorname { sin } 9 \theta - \operatorname { sin } 13 \theta } { \operatorname { cos } \theta - \operatorname { cos } 5 \theta - \operatorname { cos } 9 \theta + \operatorname { cos } 13 \theta } |
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Answer» Thanks bhai |
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| 75555. |
\left. \begin{array} { | c | c | c | c | } \hline 5 & { 9 } & { 8 } & { 7 } \\ \hline 8 & { 6 } & { 9 } & { 10 } \\ \hline 7 & { 13 } & { ? } & { 19 } \\ \hline 5 & { 7 } & { 8 } & { 9 } \\ \hline ( a ) & { 10 } & { ( b ) } & { 12 } & { ( c ) } & { 9 } & { ( d ) } & { 15 } \\ \hline \end{array} \right. |
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Answer» The sum of columns is 25,35, 45. The third column sum can be 40 so option d is correct. |
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| 75556. |
Find the smallest number by which 2560 must be multiplied so that the prperfect cube.That is the smallest number by which 1600 must be divided so that the q |
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Answer» 2650 should be multiplied by 106 as 25 is the biggest square to divide 2650 RO MAKE THE NUMBER PERFECT CUBE IS MILTIPLY BY THE NUMBER 25 |
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| 75557. |
Solv.HONAMatt2+1 |
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| 75558. |
Two pipes running together can fill a tank in11 minutes. If one pipe takes 5 minutes morethan the other to fill the tank separately, findthe time in which each pipe would fill the tankseparately[O.D. Set III, 2016] |
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| 75559. |
Two pipes running together can fill a tank in 11 1/9 minutespipetakes 5 minutes more than the other to fill the tank separately, find thetime in which each pipe would fill the tank separately. CBSE 2010 |
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| 75560. |
OA sua very was condected by a group of students asa part of their environment awarness programmein which they collected the following data regardingthe number of plant in 20 hause in to cality find themeen number of Plant per hause!0-2 2-4 4-6 6-8 8-10 10-2 12-14NOOBPlantsnoophouse1215623 |
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| 75561. |
dia sas f irinODthetiou ian |
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Answer» Please keep drawing the figure as you follow below.Let ABC be the triangle with sides BC, AC, AB opposite angles A, B, Cbeing in lengthequal to a, b, c respectively. Draw perpendiculars AD, BE and CF from A, B, C to opposite sides meeting sides BC, AC and AB at D, E, F respectively. Now perpendicular AD² = AC² - CD² ==> AD² < AC² or AD < AC or AD < b-----(1) Also, BE²= AB² - AE² ==> BE² < AB² or BE < AB or BE < c ------(2) Likewise , CF² = BC² - BF² ==> CF² < BC² or CF < BC or CF < a ----(3)Adding inequalities (1), (2), (3), AD + BE + CF < a + b + c |
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| 75562. |
Md s and 13599 |
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Answer» 5/9 + (-13)/9(5-13)/9-8/9 |
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| 75563. |
Q3) What smallest number should be added to 4456 so that the sum is completely divisible by 6?Anamn time the quotient and 5 times the remainder If the mai |
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Answer» When we divide 4456 by 6 we get 4 as a remainder.So to make it a complete no divisible by 6 wuthout leaving any remainder, we have to add 2 to 4456.So the smallest no is 2 . smallest no is 02 is the correct answer |
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| 75564. |
fasal Chakra ke Siddhant Hindi mai |
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Answer» फसलों की अच्छी पैदावार लेने के साथ-साथ मृदा की उर्वरा शक्ति स्थिर बनाये रखने के लियेे कृषक को सस्य चक्रों के सिद्धांतों का ज्ञान होना आवश्यक है। विभिन्न फसलों से फसल चक्र बनाते समय निम्नलिखित सिद्धांत ध्यान में रखने चाहिए (i) गहरी जड़ों वाली फसलों के बाद कम गहरी जड़ों वाली फसलें उगाई जायें जिससे कि फसलें विभिन्न मृदा सतहों से अपने पोषक तत्व व नमी ग्रहण कर सकें। (ii) अदलहनी फसलों के बाद दलहनी फसलें उगानीचाहिएँ। अदलहनी फसलेे मृदा को नत्रजन की मात्रा अधिक लेकर कमजोर करती हैं। दलहनी फसलों की जड़ों में पाया जाने वाला राइजोबियम जीवाणु मृदा में “नत्रजन” का स्थिरीकरण वातावरण से करता है। (iii) अधिक खाद चाहने वाली फसलों के बाद कम खाद चाहने वाली फसलें उगानी चाहिएँ। जिससे कि मृदा उर्वरता भी सुरक्षित रहती है व किसान, खाद आदि का प्रबंध भी सुगमतापूर्वक करता रहता है। (iv) अधिक जल चाहने वाली फसलों के बाद कम जल चाहने वाली फसलें उगानी चाहिएँ। लगातार अधिक जल चाहने वाली फसलों के उगाने से मृदा में वायु का संचार रुक जाता है। इससे जड़ो की वृद्धि व मृदा जीवाणुओं की क्रियाओं पर हानिकारक प्रभाव पड़ता है।सभी फसलों के लिए कृषि कार्य आसानी से हो जाते हैं क्योंकि आपसी प्रतियोगिता नहीं होती एवं निरीक्षण कार्य भी सुगमता से हो जाता है। (v) कृषि के विभिन्न साधनों का वर्षभर क्षमतापूर्ण ढंग से उपयोग होना चाहिये। फसल चक्र बनाते समय उसमें फसलों का समावेश ऐसा होना चाहिये कि श्रम, सिंचाई, उर्वरक, बीज व धन आदि जो भी कृषक के पास उपलब्ध हों, उनका पूर्ण उपयोग भी हो तथा कृृषक को आवश्यकता की सभी वस्तुएँ; जैसे- अनाज, सब्जी, दाल, पशुओं के लिए चारा, कपड़े के लिये रेशे वाली फसलें तथा नकद रुपया भी आवश्यकतानुसार प्राप्त होता रहे। (vi) फसल चक्र ऐसा बनाना चाहिए कि दो ऐसी फसलें जिनमें हानिकारक कीट-पतंगे व बीमारियां एक हों, कभी भी लगातार नहीं उगानी चाहिएँ। (vii) फसलों की छाँट फसल चक्र बनाते समय मृदा तथा जलवायु को ध्यान में रखकर करनी चाहिए। Kripya like kar ke samarthan kare ! |
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| 75565. |
ek latri mai 15 inam or 25empty hai inam pane ki probability gyaat karo. |
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Answer» Total possibilities = 15 + 25 = 40Possiblity of winning = 15Probablity = 15/40 = 3/8 thank you. |
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| 75566. |
A tank can be filled by one pipe in x minutes and emptied by another pipe in (x+5)minutes. Both the pipes when opened together can fill the empty tank in 16.8minutes. Find x.(EXEMPLAR PROBLEM) |
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Answer» In one minute 1/x part of tank is filled by the first pipe only. When both pipes are opened, in one minute part of tank filled is1/x-1/x+5 = 5/x(x+5) The problem tells us that the tank gets filled in 16.8 minutes. In one minute 1/16.8 part is filled. So, 5/x(x+5) = 1/16.8x(x+5)/5 = 16.8x(x+5) = 84 x^2+5x-84 = 0 (x+12)(x-7) = 0 x = -12 or x = 7x = -12 is rejected. The only solution is x = 7 |
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| 75567. |
Two pipes running together can fill a tank in 11 minutes. If one pipetakes 5 minutes more than the other to fill the tank separately, find thetime in which each pipe would fill the tank separately. e |
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Answer» Let the time taken by 1 pipe a be xthen pipe taken by 2nd pipe b =x+5time taken by both pipe together 11 1/9 or 100/91/a+1/b=1/t1/x+1/(x+5)=9/100(x+5+x)/x(x+5)=100/9100(2x+5)=9(x2+5x)200x+500=9x2+45x9x2-155x-500=09x2-180x+25x-500=09x(x-20)+25(x-20)=0(x-20)(9x+25)=0since time cannot be negativex=20minx+5=25min |
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| 75568. |
25-Two pipes running together can fill a tank in 11: minutes. If one pipe takes 5 minutes more thanthe other to fill the tank separately. Find the time in which each pipe would fill the tankseparately. |
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| 75569. |
. Sua polyneNow f oym |
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Answer» Polynomial- 2x²-3x+12x²-3x+1=02x²-2x-x+1=02x(x-1)-1(x-1)=0(x-1)(2x-1)=0Thus, x= 1(ALPHA) OR x=1/2(BETA)3ALPHA=33BETA=3/2 Sum of Zeroes= 9/2Product of Zeroes= 9/2 Thus, Polynomial= x²-(Sum of Zeroes)x +(Product of Zeroes)Thus, Polynomial= x²-9/2x+9/2 |
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| 75570. |
A wooden bookshelf has external dimensions asfollows: Height 110 cm, Depth 25 cm,Breadth 85 cm (see Fig. 13.31). The thickness ofthe plankare to be polished and the inner faces are to bepainted. If the rate of polishing is 20 paise percm2 and the rate of painting is 10 paise per emfind the total expenses required for polishing andis 5 cm everywhere. The external faces2inting the surface of the bookshelf. |
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Answer» External faces to be polished = Area of six faces of cuboidal bookshelf – 3 (Area of open portion ABCD) = 2 (110 x 25 + 25 x 85 + 85 x 110) – 3 (75 x 30) [AB = 85 – 5 – 5 = 75 cm and AD =x 110 – 5 – 5 – 5 – 5 = 30 cm] = 2 (2750 + 2125 + 9350) – 3 x 2250 = 2 x 14225 – 6750 = 28450 – 6750 = 21700 cm2 Now, cost of painting outer faces of wodden bookshelf at the rate of 20 paise. = Rs. 0.20 per cm2= Rs. 0.20 x 21700 = Rs. 4340 Here, three equal five sides inner faces. Therefore total surface area = 3 [ 2 (30 + 75) 20 + 30 x 75] [Therefore, Depth = 25 – 5 = 20 cm] = 3 [ 2 x 105 x 20 + 2250] = 3 [ 4200 + 2250] = 3 x 6450 = 19350 cm2 Now, cost of painting inner faces at the rate of 10 paise i.e. Rs. 0.10 per cm2. = Rs. 0.10 x 19350 = Rs. 1935 Total expenses required = Rs. 4340 + Rs. 1935 = Rs. 6275 |
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| 75571. |
lnly hat RĂĽmesh will lose the gameWater is flowing at 7 m/s through a circular pipe of internal diameter of 4 cm intothe radius of whose base is 40 cm. Find the increase in water level in 30 minutes.lindrical tank,a cy |
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| 75572. |
Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hourHow much area will it irrigate in 30 min. if 9 cm of standing water is desired |
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| 75573. |
)Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocity of 20 km per hourHow much area will it irrigate in 30 min. if 9 cm of standing water is desired ? |
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| 75574. |
7. 6 pipes can fill a tank in 24 minutes. One pipe ca(d)(a) 4 minutes (b) 30 minutes(c) 72 minutesbuild it in |
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| 75575. |
7. If the points A(6, 1). B(8, 2), C(9, 4) and Dip, 3) are thelogram, taken inorder, find the value ofp. |
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| 75576. |
010. Using quadratic formula, solve the following,(1) pogo2(932 - q?) x -- que = 0(0) -- 2ax + (622-02) = 09x2 - 3(a + b) x + a²b2 = 09x2-3(a+b)xtab |
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Answer» p^2x^2+(p^2-q^2)x-q^2=0p^2x^2+p^2x-q^2x-q^2=0p^2x(x+1)-q^2(x+1)=0(x+1)(p^2x-q^2)=0either x+1=0 , x= -1or p^2x-q^2=0p^2x=q^2x= q^2/p^2 2) x^2-2ax+(a-b)(a+b)=0x^2-x{(a+b)+(a-b)+(a-b)(a+b)=0x^2-x(a+b)-x(a-b)+(a+b)(a-b)=0x{x-(a+b)}-(a-b){x-(a+b)}=0{x-(a+b)}{x-(a-b)}=0either x--(a+b)=0x= a+bor x-(a-b)=)x= a-b |
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| 75577. |
sua para elogram form a rectangle.07. Show that a median of a triangle divides it into two triangles of equal areas01 Find the area of a triangle to see |
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| 75578. |
(A) 359 H(C) 2874From a solid cube of edge 14 cm, a sphere of largest volume is cut. Approximate volume ofIs -(B)1437亩sphereA) 359 cm3(B) 1437 cm3C) 2874 cm3(D) of thesePage 2 of 16 |
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Answer» Answer:B) 1437 cm3 |
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| 75579. |
VOLUME OF SPHERE AND HEMISPHEREA hemispherical bowl is made of steel 0.5 cm thick. The inside radius of the bowl is 4 cm. Thernthe volume of the steel used in making the bowl?1)55.85 cm 2) 56.83 cm33) 57. 85 cm34) 58.83 cm3 |
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| 75580. |
Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocityof 20 km per hour. How much area will it irrigate, if 9 cm of standingwater is desired? |
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Answer» Like if you find it useful |
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| 75581. |
1. A collection ofl numbers gathered to get some |to get some 3. Theinformation is called(a) Range (b) Data (c) Frequency (d) none.in |
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Answer» A collection of numbers gathered to get some information is called data. Option b is correct. |
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| 75582. |
27. Water in a canal, 30 dm wide and 12 dm deep, is flowing with a velocityof 20 km per hour. How much area will it irrigate, if 9 cm of standingwater is desired? |
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Answer» Depth = 12dm = 1.2mwidth = 30dm = 3mflow velocity = 20km/h = 20,000 m/htime = 30minute = 0.5 hourstanding water required = 9cm = 0.09mLet the area = A m² Volume of water will be same1.2m×3m×20,000 m/h× 0.5 h = A× 0.09⇒ A = 36000/0.09⇒ A = 4,00,000 m² = 40 hectare 40 hectare plot can be irrigated. |
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| 75583. |
ouds22. Water is flowingat the rate of 7m per sec nd through a circular pipe whose intnternal diameter is2cm in to a cylindrical tank the radiulevel in hour.rease in the water |
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| 75584. |
Water in a canal,area will it irrigate in 30 minutes, if 8 cm of standing water is needed?8.6 m wide and 1.5 m deep, is flowing with a speed of 10 km/h. How much |
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| 75585. |
ar(EFUIlPand Q are any two points lying on the sides DC and AD respectively of a parallelogramABCD Show that at (APB) - ar (BQC). |
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| 75586. |
i Oare any two points lying on the sides DC and AD respectively of a paralelogramPanriri, show that ar (APB) = ar (IOC). |
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| 75587. |
A 15 cm E8 cmB 15 cm C 6 cm DFig. (i) |
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Answer» Area of rectangle= 15*8= 120= lengthx breadtharea of triangle= 1/2*6*8= 24= 1/2* basex height |
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| 75588. |
Evaluate (16 / 81)^{1 / 4} |
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| 75589. |
622+622-2+1 by |
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| 75590. |
Two cubes, each of volume 512 cm3 are joined endto end. Find the surface area of the resulting cuboid. |
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Answer» Since its a cube , V = s^3root 512 = 8side is 8 cmwhen you join two cubes , you will get a cuboid whose length is 16 cm , h= 8 , b=8 Total s.area of cuboid = 2(lh+bh+lb) = 2( 128+128+64)= 640 cm sq Lateral surface area = 2h(l+b)2*8 (16+8)= 324cm sq, |
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| 75591. |
A man is known to speak the truth 3 out of 5 times. He throws a dice and reports that it is 1. Find theprobability that it is actually 1. |
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| 75592. |
Pand Q are any two points lying on the sides DC and AD respectively of a parallelogramABCD. Show that ar (APB)-ar (BQC).3. |
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| 75593. |
5.Pand Q are any two points lying on the sDC and AD respectively of a paralleABCD show that ar(AAPB)- ar A(B |
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| 75594. |
7\left(a^{2}+b^{2}\right)^{2}-15\left(a^{4}-b^{4}\right)+8\left(a^{2}-b^{2}\right)^{2} |
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Answer» 7a*a*a*a+14a*a*b*b+7b*b*b*b-15a*a*a*a+15b*b*b*b+8a*a*a*a-16a*a*b*b+8b*b*b*b=30*b*b*b*b-2a*a*b*b=2*b*b(15b*b-a*a) |
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| 75595. |
(3 a-4 b)^{2}+24 a b=9 a^{2}+16 b^{2} |
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| 75596. |
81^(-1/4)/81^(1/4) |
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Answer» (81)^-1/4 ÷ (81)^1/4 = 1/(3)^4*1/4 ÷ (3)^4*1/4 = 1/3 ÷ 3 = 1/3 * 1/3 = 1/9 thanks aunty |
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| 75597. |
( a - 2 b ) \left( a ^ { 2 } + 2 a b + 4 b ^ { 2 } \right) \left( a ^ { 3 } + 8 b ^ { 3 } \right) |
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Answer» Simplifying (a + 2b)(a2 + -2ab + 4b2) = 0 Reorder the terms: (a + 2b)(-2ab + a2 + 4b2) = 0 Multiply (a + 2b) * (-2ab + a2 + 4b2) (a(-2ab + a2 + 4b2) + 2b * (-2ab + a2 + 4b2)) = 0 ((-2ab * a + a2 * a + 4b2 * a) + 2b * (-2ab + a2 + 4b2)) = 0 Reorder the terms: ((4ab2 + -2a2b + a3) + 2b * (-2ab + a2 + 4b2)) = 0 ((4ab2 + -2a2b + a3) + 2b * (-2ab + a2 + 4b2)) = 0 (4ab2 + -2a2b + a3 + (-2ab * 2b + a2 * 2b + 4b2 * 2b)) = 0 (4ab2 + -2a2b + a3 + (-4ab2 + 2a2b + 8b3)) = 0 Reorder the terms: (4ab2 + -4ab2 + -2a2b + 2a2b + a3 + 8b3) = 0 Combine like terms: 4ab2 + -4ab2 = 0 (0 + -2a2b + 2a2b + a3 + 8b3) = 0 (-2a2b + 2a2b + a3 + 8b3) = 0 Combine like terms: -2a2b + 2a2b = 0 (0 + a3 + 8b3) = 0 (a3 + 8b3) = 0 Solving a3 + 8b3 = 0 Solving for variable 'a'. Move all terms containing a to the left, all other terms to the right. Add '-8b3' to each side of the equation. a3 + 8b3 + -8b3 = 0 + -8b3 Combine like terms: 8b3 + -8b3 = 0 a3 + 0 = 0 + -8b3 a3 = 0 + -8b3 Remove the zero: a3 = -8b3 Simplifying a3 = -8b3 Combine like terms: -8b3 + 8b3 = 0 a3 + 8b3 = 0 The solution to this equation could not be determined. |
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| 75598. |
1. Three equal cubes are placed adjacently in a row. The ratio of the total surface area of turesulting cuboid to that of the sum of the surface areas of three cubes is:fa) 7:9(b) 49:81(c) 9:7(d) 27 : 23 |
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| 75599. |
38 m (c) 36 m2 (BT 40 m2A cube of edge 5 cm is cut into cubes, each of edge 1 cm. The ratio ofthe total surface area of one of small cubes to that of the large cube is :8.(C) 1:625 (D) 1:259, Find the cost of |
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Answer» Surface area of cube of 5 cm=6*5²=150cm²surface area of 1 cube of 1 cm=6*1²=6cm²ratio=6:150=1:25 |
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| 75600. |
se and height of a parallelogram are 40 cm and 15.S em, then the area of the parallelogtam ishe base and height of a pequal toa) 622 cm100 en(d) none of these(b) 620 cm2(c) 628 cm2.(d) nonr of these |
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Answer» Area= base*height= 40*15.5= 620 sq cm |
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