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InterviewSolution
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.
| 1901. |
42. The sides of a right athe sides of a right angled triangle containing the right angle are 3(x + 1) cm and 2the area of this triangle is 30 sq. cm. Find the sides of the triangle,43. T.Dni |
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| 1902. |
The difference between the two adjoining sides containing right angle of a right-angled triangle is 14 cm. 1area of triangle is 120 cm Verify this area by using Heron's formulaC is rin ht angled triangle withD = |
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Answer» Let the perpendicular be a cmBase = a+14 cm According to question Area = 1201/2 b *p =120a(a+14) = 240a^2 +14a-240 = 0a^2 +24a - 10a - 240=0a(a+24) - 10(a+24)= 0(a+24)(a-10)=0 a= - 24 and a=10Neglecting negative value, we get a=10 Perpendicular = 10 cmBase = 24 cmHypotenuse = 25 cm Now, semi perimeter = 59/2(s-a) = 39/2(s-b) = 11/2(s-c) = 9/2 Area of triangle = root( 59*39*11*9/16)= 3/4 * root 59*39*11= 3/4 * 160(approx.)= 3*40=120 cm sq. |
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| 1903. |
Find the area of a right angled triangle, whose sides containing the right angle are of lenand 14.7 m. |
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| 1904. |
5० +8/ _ 5070 £/ 8 ; ¢. + = Amduns के |
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| 1905. |
L0Thediferencebetweenthetwoadjoiningsidescontainingtherightanigleofrightangled triangle is 14 cm. The area of triangle is 120 cm . Verify this area by usingherons formula.(4) |
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Answer» Let the perpendicular be a cmBase = a+14 cm According to questionArea = 1201/2 b *p =120a(a+14) = 240a^2 +14a-240 = 0a^2 +24a - 10a - 240=0a(a+24) - 10(a+24)= 0(a+24)(a-10)=0 a= - 24 and a=10Neglecting negative value, we geta=10 Perpendicular = 10 cmBase = 24 cmHypotenuse = 25 cm Now, semi perimeter = 59/2(s-a) = 39/2(s-b) = 11/2(s-c) = 9/2 Area of triangle = root( 59*39*11*9/16)= 3/4 * root 59*39*11= 3/4 * 160 (approx.)= 3*40=120 cm sq. when perpendicular is 10cm and base is 24cmthen how can hypotenuse be 25cmaccording to my calculation hypotenuse is 26cm. |
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| 1906. |
8 m9. Find the area of a right-angled triangle whose two sides containing the right angle measure8cmand 15 cm, respectively. |
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| 1907. |
The base of a right angled triangle is 15 cm and height is 24 cmFind the area of right angled triangle1· |
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Answer» Like if you find it useful |
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| 1908. |
The area of a right angled triangle is 62 cm.If one of the sides containing right angle is 12.4 cm, then find thesides. |
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Answer» Area=(1/2)base*height62*2=12.4*heightheight=124/12.4=10cm Other side ²+(10)²=(12.4)²other side²=(12.4+10)*(12.4-10)=22.4*2.4=53.76cm Sorry your Answer and in my book Answer is wrong |
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| 1909. |
(2) In a right angled triangle, one of the acute angles exceeds the other by 20°. Find themeasure of both the acute angles in the right angled triangle |
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Answer» Let the three angles are A,B and C it's an rt angle triangle let B=90° also given that one angle exceed other by 20° i.e.A=C+20° by angle sum property of triangle A+B+C=180° C+20+90+C=180 C=35° A=C+20=35+20=55° so 35° and 55° are required angles |
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| 1910. |
Q22. Define.Acute angled triangleObtuse angled triangleRight angled triangle |
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Answer» An acute triangle is a triangle with all three angles acute. An obtuse triangle is one with one obtuse angle and two acute angles. Since a triangle's angles must sum to 180°, no triangle can have more than one obtuse angle A right triangle or right-angled triangle is a triangle in which one angle is a right angle. The relation between the sides and angles of a right triangle is the basis for trigonometry. The side opposite the right angle is called the hypotenuse. The sides adjacent to the right angle are called legs. please like my answer if you find it useful |
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| 1911. |
8. Prove that the triangle formed by joining the midpoints of the sides of a right-angled triangle isalso right angled. |
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| 1912. |
sin-12 (2012, 14)1-x21 +15. |16. | cos |-1x tanx17. an13xdx218.J 1x-1sinx19.dxx2,/2 d(1+x2)3/222. [sec (tan-1x)de「e3x cos 2x dr(121。1 cosec"r dr-124、(a x3)3/2 |
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Answer» please jaldi sabhi Ko solve Karo ek ek question KR k post kro..so that other tutors can also help you to solve your problem |
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| 1913. |
EXAMPLE 15 In a right angled triangle, ohypotenuse is double the smallest sidea right angled triangle, one acute augle is double the other.ne acute angle is double theProve thatthe |
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| 1914. |
2. Draw the following triangles. Mark their medians and altitudesa) acute-angled triangle b) right-angled trianglefor ooh of the given triangles |
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Answer» option a is correct answer. |
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| 1915. |
6. Prove thatthe areas of two similar triangles are proportional to the squares of theircorresponding sides. (Triangles are acute angled triangle and right angled triangle |
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| 1916. |
उदाहरण 1. यदि (In r = a cos ti+ a sin tj+tk(Evaluate) dr dr der!dt de dal- के मान ज्ञात कीजिये। |
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Answer» the above answer is the correct answer |
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| 1917. |
(a) (1 + cos 0) dr-r sin θ do |
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Answer» wat to find out?,....... mtlb |
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| 1918. |
15. Prove that Ă 3r C, = 4".r=0te unDut |
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| 1919. |
le differential equation day + 4y + 5y = 0.- + 5y = 0.dr6dr2 |
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| 1920. |
tan xdrsin r cos x |
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| 1921. |
A sum of money becomes Rs 5070 in 18 months and Rs 5610 in 4 years. Find theprincipal. rate of interest, and the amount after 5 years. |
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| 1922. |
(c) 1, yearsof the sum. The rate per cent per anmum(c)lb) 8 months(d)1 years24The simple interest on a sum for 5 years is3510.(a) 8%(b)10%12%(d) 12 %2 |
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Answer» Thanks |
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| 1923. |
A man saves Rs. 200 in each of the first three months of his service. In each of the subsequentmonths his saving increases by Rs. 40 more than the saving of his immediatelmonth. Find the total time for his total saving from the start of the service is Ry previouss. 11040. |
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| 1924. |
10. Yasmeen saves ?32 during the first month, ? 36in the second month and ?40 in third month. Ifshe continues to save in this manner, in howmany months will she save 2,000? |
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Answer» AP formed is 32,36,40.......nSn=2000a=32, d=4Sn=n/2 [2a+(n-1)d]2000=n/2[64+4n-4]4000=4n2+60n4n2+60n-4000=0n2+15n-1000=0n2+40n-25n-1000=0by solving we get n = 25 and -40.As months cant be negative we take positive value. Therefore She will save Rs 2000 in 25 months. |
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| 1925. |
15 Yasmeen saves32 during the first month,36in the second month and 40 in the third month.If she continues to save in this manner, in howmany months will she save 2000?NCERT Exemplar |
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Answer» Here a=32 d=4 an=2000so using the formulaan=a+(n-1)d2000=32+(n-1)42000-32=(n-1)41968=(n-1)41968÷4= n-1492=n-1n= 492+1n=493 |
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| 1926. |
15 Yasmeen saves32 during the first month, R 36in the second month and 40 in the third monthIf she continues to save in this manner, in howmany months will she save 2000?NCERT Exemplar |
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| 1927. |
The average age o 40 students in a class is 14 years 10 month. If by the admissionof 5 more students average age decrease by 2 months Find average age or newstudents(a) 13 5 year (b) 13 25 year (Ci 14 year (d) of these |
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Answer» none of these because it is decreasing by 2 months and |
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| 1928. |
19. A trust fund has Rs. 30,000 that has to be invested in two different types of bonds. The first bondinpays 5 percent interest per year and the second bond pays 7 percent interest per year. Using matrixmultiplication, determine how to divide Rs. 30,000 among the two types of bonds if the fund mustobtain an annual interest of (a) Rs. 1800 (b) Rs. 2000 (c) Rs. 1600. |
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| 1929. |
11. A man borrows 10,000 at 10% compoundinterest compounded yearly. At the end ofeach year, he pays back 30% of the sumborrowed. How much money is left unpaidjust after the second year ? |
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Answer» For first year: P = 10000 R = 10% T = 1yr Interest = 1000 A = P + I = 10000 + 1000 = 11000 We know that he pays 20% of amount every year 20 / 100 x 11000 = 2200 So the total amount paid = 11000 - 2200 = 8800 For second year: P = 8800 R = 10% T = 1 yr Interest = 880 A = P+ I = 8800 + 800 = 9680 He pays 20% of amount again 20/100 x 9680 = 1936 Total amount paid = 9680 - 1936 = 7744 So the loan is 7744 |
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| 1930. |
11. A man borrows 10,000 at 10% compoundinterest compounded yearly. At the end ofeach year, he pays back 30% of the sumborrowed. How much money is left unpaidjust after the second year? |
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Answer» For first year: P = 10000 R = 10% T = 1yr Interest = 1000 A = P + I = 10000 + 1000 = 11000 We know that he pays 30% of amount every year 30 / 100 x 11000 = 3300 So the total amount paid = 11000 - 3300 = 7700For second year: P = 7700 R = 10% T = 1 yr Interest = 770 A = P+ I = 7700 + 770 = 8470 He pays 30% of amount again 30/100 x 8470 = 2541 Total amount paid = 8470-2541 So the loan is 6091 |
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| 1931. |
Find the minimum value of sin e+cos 9 |
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Answer» Sin^2x Also Cos^2x >= Cos^4x Add sin^2x both sides Cos^2x + sin^2x>= Cos^4x+sin^2x 1>= Cos^4x+sin^2x Therefore max of A is 1 Also both terms Cos^4x+sin^2x = (1 –sin^2x)^2 +sin^2x = 1 + sin^4x – 2sin^2x + sin^2x =Sin^4x – sinn^2x + 1 = (sin^2x – 1/2)^2 + 3/4 now Min =3/4 and Max = 1 |
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| 1932. |
In a right triangle ABC, B 90o, BC- 6 cm and AB 8cm. Inside this triangle,a circle with center o and radius x. Find the value of x. |
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| 1933. |
e. If a = 9-415, find the value of Ja-. |
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| 1934. |
right-angled ASTU, hypotenuseला and /(ST) = 4 cm. |
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Answer» SU = 5 CM, ST = 4 CM, TU=4 CM Correct Ans of this question =6.4 the correct answer is 6.4 of the following question |
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| 1935. |
. Suresh and Ramesh together invested144000 rupees in the ratio 4:5 andbought a plot of land. After some years 4they sold it at a profit of 20%. What isthe profit each of them got? |
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Answer» Suresh: Ramesh = 4:5.=9 units total9 units= 144000 , so 1 unit = 16000 .suresh= 4*16000=64000, Ramesh=5*16000=80000. profit of suresh = 64000*20%=12800 . profit of ramesh =80000*20%=16000 |
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| 1936. |
1. Suresh and Ramesh together invested144000 rupees in the ratio 4:5 andbought a plot of land. After some yearsthey sold it at a profit of 20%. What isthe profit each of them got? |
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| 1937. |
Example 26 In a survey of 400 students in a school, 100 were listed as taking applejuice, 150 as taking orange juice and 75 were listed as taking both apple as well asorange juice. Find how many students were taking neither apple juice nor orange juice. |
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| 1938. |
\begin{array} { l } { \text { The sum of the present ages of } A \text { and } B \text { is } } \\ { 110 \text { years. } 20 \text { years ago, ratio of their ages } } \\ { \text { was } 4 : 3 . \text { Find the age of } A . } \\ { \text { (1) } 60 } \\{\ text {2} } 55 }\\ { \text { (3) } 50 } \\{\ text {4} } 35} \\ { \text { (5) of these } } \end{array} |
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Answer» A+B=11020 years agoA+B=110-20-20=70now A:B=4:3so A=4B/34B/3+B=707B/3=70so B=30 so A=40so current age of A=40+20=60 years. |
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| 1939. |
xample 26 In a survey of 400 students in a school, 100 were listed as taking auice, 150 as taking orange juice and 75 were listed as taking both apple as welrange juice. Find how many students were taking neither apple juice nor orange j |
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| 1940. |
12. In a class of 33 studcils,play cricket and 5 students like to play both the games; how ihuily s13. In a class of 400 students, 100 were listed as taking apple juice, 1 50 as taking orange juice and75 were listed as taking both apple as well as orange juice. Find how many students wereraking neither apple juice nor orange juice?nofnonl, 40 like cricket, 10 like both cricket and tennis. How many like tennis |
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Answer» Total students = 400Students taking apple juice = 100Students taking orange juice = 150Students take both apple and orange = 75 Then, Students taking neither apple nor orange juice = 400 - [ (100 + 150) + 75] = 400 - [ 250 +75] = 400 - 325= 75 |
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| 1941. |
\begin{array} { l } { \text { A cistern is normally filled in } 8 \text { hours; but takes } } \\ { \text { two hours more to fill because of a leak in its bot- } } \\ { \text { tom. If the cistern is full, the leak will make it } } \\ { \text { empty in } } \\ { \text { a) } 16 \text { hours } } & { \text { b) } 40 \text { hours } } \\ { \text { c) } 20 \text { hours } } & { \text { d) } 34 \text { hours } } \end{array} |
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Answer» option b is the answer |
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| 1942. |
\begin{array} { l } { \text { 776. A reporter writes } 23 \text { articles a } } \\ { \text { month. How many articles will } } \\ { \text { he have written in } 1.5 \text { years? } } \\ { \text { (1) } 414 \text { (2) } 424 } \\ { \text { (3) } 418 \text { (4) } 404 } \\ { \text { (5) } 423 } \end{array} |
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| 1943. |
\begin{array} { l } { \text { The locus given by } x \text { cos } \alpha + y \sin \alpha = p \text { , } } \\ { \text { where } \alpha \neq 0 , \alpha \neq \frac { \pi } { 2 } \text { and p is a constant. } } \\ { \text { intersects the } X \text { and } Y \text { axes at the points } A \text { and } B } \\ { \text { respectively. Find the equation of locus of } } \\ { \text { midpoint of segment AB. } } \end{array} |
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| 1944. |
\begin{array} { l } { \text { In a division sum, the remainder is } 0 \text { . As } } \\ { \text { student. mistook the divisor by } 12 \text { instead of } } \\ { 21 \text { and obtained } 35 \text { as quotient. What is the } } \\ { \text { correct quotient? } } \\ { \text { (a) } 0 } \\ { \text { (c) } 13 \text { . } } \end{array} |
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Answer» division algorithma = bq + ra = 12×35 a = 21p so21p = 12×35p = (12×35)/21 = (12 ×5)/3 = 4×5p = 20d) is correct option |
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| 1945. |
\begin{array} { l } { \text { In a triangle } X Y Z , \text { if } m \angle X \text { is } 20 \text { more than } m \angle Y \text { and } 25 \text { more than } \angle Z \text { , then } } \\ { \text { find } m \angle X \text { . } } \end{array} |
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| 1946. |
1313The sides a, b, c of Δ ABC are in GP, and loga-log2b, log2b-log3c, log3c-loga are ithen Δ ABC is1) acute angled3) right angled2) obtuse angled4) triangle does not exist |
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| 1947. |
The corresponding sides of the similar ifriangle a |
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Answer» Two triangles are said to be similar if their corresponding angles are congruent and the corresponding sides are in proportion . In other words, similar triangles are the same shape, but not necessarily the same size. The triangles are congruent if, in addition to this, their corresponding sides are of equal length. Therefore, the corresponding sides of similar triangle are in propotion. |
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| 1948. |
Draw rough sketches of altitudes from A to BC for the following triangles :8Acute AngledRight AngledObtuse Angled |
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Answer» thanks,can you answer on more question on this topic |
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| 1949. |
s P, Q and R of the sides of the A ABC are (3, 3). (3,4) and (2. 4)If the mid-pointrespectively, then A ABC is(A) right angled(C) obtuse angled27.(B) acute angled(D) isosceles |
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| 1950. |
What is the whole sum of 7÷70 |
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Answer» the right answer is 10 |
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