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| 5901. |
verify a-(-b)= a+b if a=21, b=18. |
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Answer» a-(-b)= a+bif a= 21b = 1821-(-18)21+1839 |
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| 5902. |
4The areas of two similar triangles are 169 cm2 and 121 cm2 respectively.If the longest side of the larger triangle is 26 cm, find the longest side ofthe smaller triangle. |
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| 5903. |
_ डिक) (3लिड) +iV5 ) ( 3-45 ।(+21)-(B-ix2) |
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Answer» ((3 + i√5) ( 3 - i√5)---------------------------------(√3 + √2i) - (√3 -i√2) 3² - (i√5)²---------------- 2√2i 9 + 5----------- 2√2i 7----------- √2 i 7√2i-----------√2×√2i² -7√2i--------- 2 |
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| 5904. |
The longest side of a triangle is 3 times the shortest side and the third side is 2 cmshorter than the longest side. If the perimeter of the triangle is at least 61 cm, find theminimum length of the shortest side. |
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| 5905. |
The longest side of a triangle is 3 times the shortest side and the third side is 2 cmshorter than the longest side. If the perimeter of the triangle is at least 61 cm, findthe minimum length of the shortest side. |
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| 5906. |
Expressas a rational number with(a) numerator = -21(b) |
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Answer» To express - 3/7 with numerator = - 21 Multiply and divide given rational by 7 Therefore,-3/7 = - 3*7/7*7 = - 21/49 |
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| 5907. |
2-√5 ka har parimay karn karo |
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Answer» (2+√5)(2-√5)/(2+√5)4-5/(√2+√5)-1/(√2+√5) |
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| 5908. |
Fig. 13.88Fig. 13.8925. In Fig. 13.89, an equilateral triangle ABC of side 6 cm has been inscribed in a circle.Find the area of the shaded region. (Taket 3.14).ar Geld has a perimeter of 650 m. A square plot having its vertices on the |
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| 5909. |
UCULU(1) Every natural number is a whole number.Fier intencris a whole nur har |
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Answer» Natural numbersare the set of positive integers, that is, integers from 1 to ∞ excluding fractional n decimal part. They arewhole numbers excluding zero.Natural Numbersare also called countingnumbers. Zero is the onlywhole numberwhich is not anatural number. |
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| 5910. |
6. Find the difference between7. How many 6-digit numbers are there in all?har are there in all? |
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| 5911. |
(B)8.4cm(C) 4.2 cm(D) 0.525 cm4.In the adjoining figure, ABC and DBC are two triangles on the same base BC, AL i BC and DM.L BCarea (AABC)n, area (ADBc) is equal toAO2AO(A) OD(B) ODo ML.OD(C)(D) AOTAD |
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Answer» Construction:Draw AL ⟂ BC and DM ⟂BC . In ∆ ALO and ∆ DOM∠ALO = ∠DMO [each 90°]∠AOL = ∠DOM (vertically opposite angles)∆ALO ~ ∆ DMO (By AA Similarity criteria)AL/DM = AO/DO……………….(1) [corresponding sides of similar triangles are proportional]ar(∆ABC) =1/2 × BC × AL.. .. ... ...(2)ar(∆DBC) = 1/2×BC×DM………….(3)On dividing eq 2 and 3,ar(∆ABC)/ar(∆DBC) = 1/2×BC× AL/ ½ × BC ×DM ar(ABC) / ar (DBC) = AL/DMar(ABC) / ar (DBC) = AO/DO [from eq 1 ] |
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| 5912. |
xty 27(8) a+b 15b + c # 15, andthen what is the value of a+b c?a, 18c. 19b. 20d. 14 |
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Answer» Option c is correct. |
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| 5913. |
Day-Day)1 ± Man.(A) 6075(C) 5040(B) 4050瓦(D) 6000(A) 15(B) 18(C) 20. |
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| 5914. |
Find AUB, A1, AUL13.6.9. 12, 15, 18.21): B = (4.8, 12, 16, 20)14. 161: D (5, 10, 15, 20) then find (1) A B A C(vi) D-A (vii) B-C (viii) B-D (ix) C-B (x)D-B?C = (2,4,6,8,10.12.(iv) B-A(v)C-AAD |
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Answer» A-B={3, 6,15,18,21}A-C={3,9,15,18,21}A-D={3,6,9,12,18,21}B-A={4,8,16,20}C-A={2,4,8,10,14,16}D-A={5,10,20}B-C={20}B-D={4,8,12,16}C-B={2,6,10,14}D-B={5,10,20} |
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| 5915. |
3.The discriminant of the quadratic equation 4x2 - 6x + 3.0 is(A) 12(B) 84(c) 23D) - 12For the following frequency distribution:Class: 0-5 5-10 10-15 15 - 20 20 - 25Frequency: 8 10 19 25 8The upper limit of median class is(A) 15(B) 10(C) 20(D) 25 |
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Answer» kuch ni mode median ka hai formula lagega Bhai sahab question no. 3 ans:DD=b²-4ac. =(-6)²-4(4)(3). a=4,b=-6,=36-48. c=3=-12 question 4total(n)=60=n/2=60/2=30so,median class=20-25upper limit=20 median is a 20 and frequency is and his sum is upper limit is 15 -12 (3) D Are the answer to given questions (4) D Question no÷3Answer÷ D) -12 answer is c of this question 4rth question answer-12 upper limit of median class is 20 d=b^2-4ac36-4*4*3d=-12 20_25 it is a correct answer answer c is correct answer for this question iska answer 15 aayega |
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| 5916. |
If A = {3, 6, 9, 12, 15, 18, 21}; B = {4, 8, 12, 16, 20;C = {2, 4, 6, 8, 10, 12, 14, 16); D = {5, 10, 15, 20) find(1)A-B (i)A-C (iii)A-D (iv)B-A(vi) D-A (vii) B-C (viii) B-D (ix)C-B(v)C-A(x)D-B |
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Answer» (i)12. (ii)6,12. (iii)15. (iv)4,8. (v)15, (vi)(vii)(viii)(ix)(x) |
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| 5917. |
2. Sana and Fatima participated in an apple race. The race was conductein 6 parts. In the first part, Sana won by 10 seconds. In the secondpart she lost by 1 minute, then won by 20 seconds in the third parand lost by 25 seconds in the fourth part, she lost by 37 seconds inthe fifth part and won by 12 seconds in the last part. Who won therace finally? |
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Answer» sana was won the race because she had doing the all part of race better then fatima Let difference in time denoted by positive, when Sana wins the race and negative, when Sana loses the race Total difference in time taken by Sana in all the six parts = 10-16+20-25-37+12 = -80 secs Hence , Fatima won the race by 80 seconds Sana,because final is the fifth. sana won the race ,because fifth race is the final part of the race fatima won the race because she is leading the + timing |
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| 5918. |
2Il ll TI17]3 The eighth term of an A.P. is half of its second term and the eleventh term exceeds onethird of its fourth term by 1. Find the 15th term.luno third term is 16 and seventh term exceeds its |
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| 5919. |
Find a single discount equivalent to the discount series 20%, 10% and5% |
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Answer» suppose the value is x on 20% discount the price will be (100-20)%*x = 0.8x on 10% discount = (100-10)%*(0.8x) = 0.72x on 5% discount on it, the value will become 0.95*0.72x = 0.684x so the single discount will be (x-0.684x)*(100)/x = 0.316 = 31.6%> how do u get 0.8 |
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| 5920. |
HEIGHIS AND DISTANCES8. A chinney 20m height standing on the top of a building subtends an angle θ where tan θ= 1 / 6 at adistance 70m from the foot of the building. The height of the building isa) 30mc) 40md) 60m |
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| 5921. |
neangleofdepressionof two ships from the top ofare 45° and 30° towards east. If the ships are17200 metres apart, find the height of the light house. |
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| 5922. |
A 1.2tall girl spots a balloon movingwith the wind in a horizontal line at aheight of 88.2m from the ground. Theangle of elevation of the balloon fromthe eyes of the girl at any instant is5". After some time, the angle of 60deviation reduces to 30° (see Fig. 9.13).30Find the distance traveled by theballoon. |
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Answer» Hi aparna will u help me |
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| 5923. |
The longest side of a triangle is 3 times the shortest side and the third side is 2 cmshorter than the longest side. If the perimeter of the triangle is at least 61 cm, findthe minimum length of the shortest side.25. |
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Answer» ok thanx |
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| 5924. |
LINEAR INEQUALITIES 12The longest side of a triangle is 3 times the shortest side and the third side is 2 cmshorter than the longest side. If the perimeter of the triangle is at least 61 cm, findoths frmma single nicce of board of length 91cm25.the minimum length of the shortest side. |
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| 5925. |
s. A triangle has sides 35 cm, 54 cm and 61 cm long. Find its area. Also, find te |
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Answer» thank you very much |
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| 5926. |
S. If third term of an A.P. is four times of first term and 6th term is 17, find the series.UP Diploma 84] |
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| 5927. |
r. Find three such numbers which are in A.P. and sum of first and third term is 12, productof first and second terms is 24.UP Diploma 1990] |
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| 5928. |
13., How many terms of the series 1+2+4+8t.. should be taken so that their sum isIUP Diploma 2006]1023? |
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| 5929. |
0.4Atrianglehassides35cm,54cmand61cmlong.Finditsarea,alsofindthelengthofitssmallestaltitude |
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| 5930. |
A godown is in the form of cuboid of size 60m x 40m x 30m. How many cuboidal boxes can be stored in it if the volumeof one box is 10m3?1) |
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Answer» Volume of godown = 60×40×30 = 72000 m³ volume of a box = 10 m³so,number of boxes = 72000/10 = 720000/10 = 72000 boxes Like my answer if you find it useful! |
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| 5931. |
A) 30°Suppose you are shooting an arrow from the top of a building at ato a target on the ground at an angle of depressions of 60 , Then the distancebetween you and the objec.A) 23 mn height of 6 mB) 33mC) 4 3 mD) 5 3m |
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Answer» Refer to the attachment for the diagram. Let AB be the building you are standing. So AB = 6 m And let Angle of Depression =∠ C. =>∠ C = 60° To find The distance between you and the object. So According to the Diagram, the distance between you and the object is AC, which is the hypotenuse. So SinФ = Opposite / Hypotenuse Sin 60° = 6 m / Hypotenuse =>√ 3 / 2 = 6 / Hypotenuse => Hypotenuse = 6 * 2 /√ 3 => Hypotenuse = 12 /√ 3 => Hypotenuse = 12 √ 3 / 3 => Hypotenuse = 4√ 3 m Hence distance between the target and building is 4√ 3 m. |
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| 5932. |
Example 9: A godown is in the form of a cuboid of measures 60 m x 40 m x 30mHow many cuboidal boxescan be stored in it if the volume of one box is 0.8 m3? |
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Answer» Number of boxes= Volume of godown cuboid / Volume of one box= (60*40*30)/0.8= 90000 boxes. Please hit the like button if this helped you. thanks |
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| 5933. |
A2mall girl spotswind in ahorizontal live at a height of 88 2 m from he gound The angle of elevationof the hafloon from the of the girf at an instant is 60 Afer some time, the angle oflevation roduces to 30 Find the distance ravelled by the balloona balloon on the eve of Independence Day, moving with the |
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| 5934. |
(iii) 25m2 + 30m + 9 |
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Answer» 25m2+15m+15m+9=(25m2+15m)+(15m+9)=5m(5m+3)+3(5m+3)=(5m+3)(5m+3)correct answer (5m+3)(5m+3) is the correct answer of the given question (5m+3)(5m+3) is the correct answer 25m^2+15m+15m+9. =(25m^2+15m)+(15m+9). =5m(5m+3)+3(5m+3). =(5m+3)(5m+3). correct answer 25 m square +30x+925m square +15x+15x+95m(5m+3)+3m(5m+3)(5m+3) (5m+3) 25m²+30m+925m²+15m+15m+95m(5m+3)+3(5m+3)(5m+3)(5m+3) |
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| 5935. |
25 m² +30m +9factouse) |
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| 5936. |
The sum of two positive numbers is 5 and sumof their squares is 17, then their products is- |
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| 5937. |
18Product of three numbers in G.P. is 216. Sum of the products of two terms at a time is 156.JUP Diploma 19911Find the numbers. |
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| 5938. |
(iii) 25m2 + 30m +9 |
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Answer» 25m^2+30m+925m^2+(15+15)m+9=025m^2+15m+15m+9=05m(5m+3)+3(5m+3)=0(5m+3)(5m+3)=0either,. or,5m+3=0. 5m+3=05m=-3. 5m=-3m=-3/5. m=-3/5so that, m=(-3/5,-3/5). 25m²+30m+9 gives664m³ ,664m³is the correct answer (5m+3)^2 is the correct answer. (5 m+3)(5m+3) answer👈🇨🇮🇨🇮👍🌎👏🙏 |
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| 5939. |
\begin { equation } \cos \theta+\sin \theta=m \text { and } \sec \theta+\csc \theta=n \text { then } \operatorname{show} \text { that } n\left(m^{2}-1\right)=2 m \end { equation } |
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| 5940. |
Example 5: Find the area of the shaded region inFig. 12.16, where ABCD is a square of side 14 cnm. |
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| 5941. |
If two sides of a right angle triangle are 4and 5 cm, then the hypotenuse of the triangeis of length(a) V41 cnm(c) 9 cm(b) 3 cm(d) of these |
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Answer» Answer: a) root 41Explanation:Formula: hyp^2= a1^2+a2^2a1=4 : a2= 5hypo^2= 4^2+5^2=16+25=41Therefore hypotenuse = root 41 cm |
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| 5942. |
\begin { equation } a=5, d=3, a_{n}=50, \text { find } n \text { and } S_{n^{*}} \end { equation } |
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| 5943. |
Divide 207 in three parts, such that all parts are in A.P. and product of two smalleparts will be 4623. |
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Answer» Let the three parts of 207 in A.P be (a - d) , a , (a + d) where, a > d Now, clearly, (a + d) > a > (a - d) Now, A to Q, (a - d) + a + (a + d) = 207 => 3a = 207 => a = 69 --- (i) and, (a - d) x a = 4623 => 69 (69 - d) = 4623 => d = (4761 - 4623)/69 = 2 Hence, a = 69 and d = 2 so, (a - d) = 67, a = 69 and (a + d) = 71 |
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| 5944. |
Split 207 into three parts such that these are in an AP and the product of twosmaller parts is 4623. |
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Answer» Hit like if you find it useful Let ( a - d ) , a , ( a + d ) are three parts are in A.p according to the problem given , a - d + a + a + d = 207 3a = 207 a = 207 / 3 a = 69 product of two smaller parts = 4623 ( a - d ) × a = 4623 ( 69 - d ) × 69 = 4623 69 - d = 4623/69 69 - d = 1541/23 69 - d = 67 -d = 67 - 69 - d = - 2 d = 2 Therefore , required three parts are , a - d = 69 - 2 = 67 a = 69 a + d = 69 + 2 = 71 |
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| 5945. |
ine segment of length 24 cm is divided into three parts in the ratio 1:2:3, Find the length of eachpart7. Divide? 1400 into three parts such that the firstand the ratio between the |
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Answer» let length of each part of line be x,2x,3xx+2x+3x=246x=24x=4length of each part is x=4cm;2x=8cm;3x=12cm |
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| 5946. |
Divide 207 into three parts such that these are in A.P. and theproduct of the two smaller parts is 46232.2 |
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| 5947. |
10. Split 207 into three parts such that these are in A.P. and the product of the two smalINCERTEXEMPLARparts is 4623 |
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| 5948. |
3.Divide 30 into three parts such that the continued product of the first, thesquare of the second and the cube of the third be a maximum.Divide 120 into three parts so that the sum of the products taken two ata time shall be maximum.duet of the first square |
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Answer» square-1 cube-1 maximum -281+1+28=30 2)10×10^2×10^3=1000000 that's the maximum square 100 and cube 1000 40+40+40 =120 and some of the part 40 x 40 =1600 so correct answer is 40 2.)dividing 30 into three parts=10 ,10,10 10*100*1000= 1000000 2.) division 30.into three parts= 10' 10' 10.10*100*1000=1000000 1000000 is answer....... |
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| 5949. |
the value oe inn5-3 |
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| 5950. |
1.Ray OE bisectsangle AOB and OF is the ray opposite to OE. Show that angle FOB = angle FOA. |
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