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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.
| 1501. |
Find a number such that one-sixth of thenumber is 3 more than 7. |
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Answer» The number is sixty (60 ) |
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| 1502. |
Find three rational numbers between:and(); and(inandFind three prestíonal numbers between |
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| 1503. |
Answer:The florist sold 252 tulips and tied them into bundles with the same number of tulips in each bundle.the florist tied together 14 full bundles, how many tulips did she tie in each bundle?swer: |
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Answer» Total tulips = 252Total number of bundles of tulip = 14 Then, in each bundle number of tulips= 252/14= 18 |
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| 1504. |
valed a*A bumbmumberebpectively if they change together ac9 a.m., when will they change together again3. Find the least 5-digit number which on dividing by 4, 12,20 and 24HOFleaves a remainder 3 in each caseHCF is4 Retermito |
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Answer» Least 5 digit number = 10000. LCM(4,12,20,24) = 120 Dividing 10000 by 120 leaves a remainder 40 , Subtract 40 from 10000 = 9960 So this means 9960 is divisible by 120. 9960 + 120 = 10080. To find the number which leaves remainder 3, we have to add 3 to 10080. 10080 + 3 = 10083. |
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| 1505. |
bicadtl.tiepenuits length and breadth.13. How much would it cost to lay a wall-to-wall carpet5 cmin a room 8 m long and 6 m wide, with a carpet thatcosts125 per m2? |
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Answer» Area=8*6=48m^21m^2=125Rs48m^2=125*48=6000Rs |
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| 1506. |
Find the largest three digit number having three distinct digits. |
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Answer» 987 is the correct answer. 987 is the right answer 987 is the right answer 987 is the right answer 987 is the correct answer |
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| 1507. |
Write three irrational number between 13 and V5.Find three different irrational numbers between |
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| 1508. |
IN A EXERCISE 2.5Solve the following linear equations.\frac{x}{2}-\frac{1}{5}=\frac{x}{3}+\frac{1}{4} \quad \text { 2. } \frac{n}{2}-\frac{3 n}{4}+\frac{5 n}{6}=21 \quad \text { 3. } x+7-\frac{8 x}{3}= |
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| 1509. |
\frac{6^{1 / 4}}{6^{1 / 5}} \quad\left(\text { ii) } \frac{8^{1 / 2}}{8^{2 / 3}} \quad \text { (iii) } \frac{5^{6 / 7}}{5^{2 / 3}}\right. |
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Answer» -7mn+5,12mn+2,9mn-8,-2mn-3 |
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| 1510. |
\begin{array}{l}{\text { Factorise the following expressions. }} \\ {\text { (i) } p^{2}+6 p+8 \quad \text { (ii) } q^{2}-10 q+21}\end{array} |
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| 1511. |
\begin{array}{l}{\text { 4. Factorise. }} \\ {\text { (i) } a^{4}-b^{4}} \\ {\text { (iv) } x^{4}-(x-z)^{4}} \\ {\text { 5. Factorise the following expressions. }} \\ {\text { (i) } p^{2}+6 p+8 \quad \text { (ii) } q^{2}-10 q+21}\end{array} |
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| 1512. |
.Show that the given fractions are equivalent.\frac{6}{8}, \frac{3}{4} \quad \text { (ii) } \frac{2}{6}, \frac{1}{3} |
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Answer» i) both given fraction are equivalent 6/8 = 3/43/4 = 3/4 ii) both given fraction are equivalent 2/6 = 1/31/3 = 1/3 |
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| 1513. |
\left. \begin{array} { l } { \text { The solution of the pair of equations } x + \frac { 6 } { y } = 11,3 x + \frac { 8 } { y } = 28 \text { is } } \\ { ( a ) x = 5 , y = 1 \quad ( b ) x = 8 , y = 2 \quad ( c ) x = 4 , y = \frac { 1 } { 2 } ( d ) } \end{array} \right. |
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| 1514. |
(33/8)*(quad*text) |
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Answer» We know that for collinear points, the area of the triangle formed by them is 0. Therefore, Area of triangle = 1/2(x1(y2- y3) + x2(y3-y1) + x3(y1-y2) = 0 Here, x1 = - 2, x2 = k, x3 = 5y1 = - 3, y2 = 4, y3 = 5 Hence,1/2(-2(4-5) + k(5-(-3)) + 5(-3-4)) = 0 -2(-1) + k(8) + 5(-7) = 0 2 + 8k - 35 = 0 8k - 33 = 0 k = 33/8 Therefore the value of k = 33/8 (b) is correct option. |
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| 1515. |
x + 4 y = 9 \text { and } 6 x + 8 y = 15 \text { is } - a ) \frac { 3 } { 2 } \quad , b ) \frac { 3 } { 10 } c ) - \frac { 3 } { 2 } , d ) |
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Answer» 6x+8y=153x+4y=9 so 6x+8y=18so distance=(18-15)/root(36+64) =3/root(100)=3/10 |
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| 1516. |
\left. \begin array l \text value of \operatorname sec ^ - 1 ( \operatorname sec \frac 8 \pi 5 ) \text is : \\ \frac 8 \pi 5 \quad \phi ( b ) \frac 3 \pi 5 \quad \text (c) \frac 2 \pi 5 \text (d) \end array \right. |
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Answer» Option c is correct. |
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| 1517. |
7. Simplify4+/5 4-54-5 4+/53 |
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Answer» (4+√5)^2+(4-√5)^2/16-5=16+5+8√5+16+5-8√5/11=21+21/11=42/11 |
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| 1518. |
- \operatorname { cot } A \quad 1 |
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Answer» Tan a/(1-cot a) +cot a/(1-tan a) =(sin a/,cos a) /(1-cos a/sin a) + (cos a/sin a) /(1-sin a/cos a) =sin ^2 a/cosa(sina - cosa) +cos^2 a/sina (cosa-sina) =sin^2a/cosa(sina-cosa) - cos^2a/sina (sina-cosa) =(sin^3a-cos^3a)/sina.cosa(sina-cosa)=(sina-cosa)(sin^2a+cos^2a+sinacosa)/sina.cosa(sina-cosa)=(1+sinacosa)/sina.cosa=(1/sinacosa)+1=1+seca.coseca |
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| 1519. |
\frac { 4 + \sqrt { 5 } } { 4 - \sqrt { 5 } } + \frac { 4 - \sqrt { 5 } } { 4 + \sqrt { 5 } } |
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| 1520. |
What is the name of horizontal and vertical lines drawn to determine the posi-tion of any point in the Cartesian plane ?3. |
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Answer» Horizontal line is called X-axis. Vertical line is called Y-axis. |
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| 1521. |
I=Integral(sec(-4*x %2B 7)^2, x) |
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Answer» I=⌡sec²(7-4x) dx Let u=7-4x dx = (-1/4)*(du) I = (-1/4)*⌡sec²(u) du Apply :⌡sec²(x) dx = tan(x)+C Therefore :I = (-1/4)*(tan(u)) +C I = (-1/4)*{tan(7-4x)} +C |
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| 1522. |
\int ( \sec ^ { 2 } x - \frac { 4 } { x } + \frac { 1 } { x \sqrt { x } } - 7 ) d x |
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| 1523. |
(1) What is theWhat is the name of the horizontal and vertical lines drawn to determine thenosition of any point in the Cartesian plane?+bonomo of the |
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Answer» The name of horizontal and vertical lines drawn to determine the position of any point in the Cartesian plane is x-axis and y-axis respectively. The name of each part of the plane formed by these two lines x-axis and the y-axis is quadrants. The point where these two lines intersect is called the origin. |
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| 1524. |
the1. Write the answer of each of the following questions:What is the name of horizontal and the vertical lines drawn to determine theposition of any point in the Cartesian plane?he(ii) What is the name of each part of the plane formed by these two lines?(iii) Write the name of the point where these two lines intersect |
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Answer» who is selfish according to the poet |
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| 1525. |
7. From the figure given alongside identifty and namea. the horizontal lines.b. the vertical lines.c- the oblique lines.d. the point of intersection of lines AC and BD.e. the point of intersection of lines AB and BC.A8 |
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Answer» a) horizontal lines are which are parallel to along x-axisAB and CD are horizontal linesb) vertical lines are parallel to or along the y-axis AD and BC are vertical linesc) oblique lines are which make angle with x and y axisAC and BD are oblique linesd)point of intersection is the point where two lines meetpoint of intersection of AC and BD is Oe) point of intersection of AB and BC is B |
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| 1526. |
7. From the figure given alongside identify and namea. the horizontal lines.b. the vertical lines.c. the oblique lines.d. the point of intersection of lines AC and BD.e. the point of intersection of lines AB and BC.BMEASURING THE UNE r |
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Answer» DC and AB are the horizontal lines.DA and BC are the vertical lines.AC and BD are the oblique lines.O is the point of interaction of AC and BDB is the point of interaction for AB and BC |
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| 1527. |
Name the quadrilateral with both horizontaland vertical lines,symmetry |
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Answer» A square or a rectangle. thankss😊😊 |
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| 1528. |
Logical ReasoningAnswer the questions:1) Joseph roils a cube in which the letters Y, S.O. E. L and G appear on faces 1.2.3.4.5and 6 respectivelyWhat is the fraction of symmetric letters (either horizontal or vertical) on the cube?2) What fraction of the following letters can not be drawn using only straight lines?ALBI DUE3) 9999 = 9000 - + 9. What is the missing number?4 What is the weight of big banana? |
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Answer» (1)Y, O, E is symmetricso, 1/2 is ans (1)Y,O,E are symmetric so ans is 1/2 |
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| 1529. |
15. The total mass of 8 packets, each of same size is 10 kg 600 g. What is the mass of each suchpacket? |
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Answer» 1kg=1000g10kg 600g = 10×1000+600=10,600no. of packets=8the mass of each such packets=10,600 ÷ 8 = 1,325 g1,325 g = 1,325÷1000=1kg 325gso the mass of one packet is 1 kg 325 g. |
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| 1530. |
\frac{1}{1,3}+\frac{1}{3.5}+\frac{1}{5.7}+\frac{1}{(2 n-1)(2 n+1)}=\frac{n}{2 n+1} |
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Answer» 2 thanks |
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| 1531. |
1 ^ { 2 } + 3 ^ { 2 } + 5 ^ { 2 } + \ldots + ( 2 n - 1 ) ^ { 2 } = \frac { n } { 3 } ( 2 n - 1 ) ( 2 n + 1 ) |
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| 1532. |
1 ^ { 2 } + 3 ^ { 2 } + 5 ^ { 2 } + \ldots + ( 2 n - 1 ) ^ { 2 } = \frac { n ( 2 n - 1 ) ( 2 n + 1 ) } { 3 } |
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Answer» plz send correct answer prove with mathematical induction |
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| 1533. |
Example 1 For all n > = 1, prove that1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6} |
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Answer» 1 2 |
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| 1534. |
\frac { ( 2 n + 1 ) ! } { n ! } = 2 ^ { n } [ 1.3 .5 \dots \ldots ( 2 n - 1 ) ( 2 n + 1 ) ! |
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| 1535. |
xample 1 For all n 2 1, prove that1^{2}+2^{2}+3^{2}+4^{2}+\ldots+n^{2}=\frac{n(n+1)(2 n+1)}{6} |
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| 1536. |
(7) sec"X-tan"X = 1 + 3secxx tan"X |
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| 1537. |
2 tan x -7 sec x |
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Answer» Change into sin and cos term may be it helps u |
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| 1538. |
1.3 + 3.5 + 5.7 + \ldots + ( 2 n - 1 ) ( 2 n + 1 ) = \frac { n ( 4 n ^ { 2 } + 6 n - 1 ) } { 3 } |
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| 1539. |
6 On reducing the value of a wall clock by 8%, it becomes 115 The original value of the wall clock is(a) 125(b) 120(c)110Percentage and lts Applications 147 |
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Answer» thank you |
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| 1540. |
उदाहरण 7. tan'2r+tan-13 =को सरल कीजिए। |
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| 1541. |
\frac { 1 + \sin 2 \theta + \cos 2 \theta } { 1 + \sin 2 \theta - \cos 2 \theta } |
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| 1542. |
प्रति75/ नीचे दी गई आकृतियों में प्रश्नवाचक चिह्न (?) के स्थान पर केसंख्या आएगी?(144)192)(A) 152176(B) 162(D) 104 |
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Answer» it is question of math let there be the X. answer dijiye sir iska solution ke sath |
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| 1543. |
6.Find the value of x, if:(0) 2 + 2 + 2 = 192 |
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Answer» धन्यवाद 2^x + 2^x + 2^x = 1923 x 2^ x= 1922^x = 192/32^x= 64x=6 2^x+2^x+2^x=192 3*2^x=1922^x=192/32^x=64x=6 |
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| 1544. |
\tan ^ { - 1 } \frac { 1 } { 5 } + \tan ^ { - 1 } \frac { 1 } { 7 } + \tan ^ { - 1 } \frac { 1 } { 3 } + \tan ^ { - 1 } \frac { 1 } { 8 } = \frac { \pi } { 4 } |
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| 1545. |
prove that tan-1 (2/7)+tan-1(1/4)=tan-1(15/26) |
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Answer» tan^-1A+tan^-1B=tan^-1(a+b/1-ab)hence put in given equationtan^-1(2/7+1/4/1-2/7*1/4)=tan^-1(8+7/28/1-2/28)tan^-1(15/28/26/28)tan^-1(15/26) |
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| 1546. |
2 \tan ^{-1} \frac{1}{2}+\tan ^{-1} \frac{1}{7}=\tan ^{-1} \frac{31}{17} |
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| 1547. |
2 \tan ^ - 1 \frac 1 2 %2B \tan ^ - 1 \frac 1 7 = \tan ^ - 1 \frac 31 17 |
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| 1548. |
saTECUT 7. tan-1 2 sin 2 cos 1 |
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Answer» Thanks |
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| 1549. |
\tan ^ - 1 ( \tan \frac 7 \pi 6 ) %2B \cot ^ - 1 ( \cot \frac 7 \pi 6 ) |
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Answer» tan^-1(tan 7pi/6) + cot^-1(cot 7pi/6) = tan^-1(tan(pi + pi/6)) + cot^-1(cot(pi + pi/6)) = tan^-1(tan pi/6) + cot^-1(cot(pi/6)) = pi/6 + pi/6 = 2pi/6 = pi/3 |
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| 1550. |
\operatorname { tan } ^ { - 1 } \frac { 1 } { 3 } + \operatorname { tan } ^ { - 1 } \frac { 1 } { 5 } + \operatorname { tan } ^ { - 1 } \frac { 1 } { 7 } + \operatorname { tan } ^ { - 1 } \frac { 1 } { 8 } = \frac { \pi } { 4 } |
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Answer» 1/3 aur 1/5 wale ko ek sath Nahi let sakte h you can try doing that also but waise Karne per answer Nahi aa Raha hai Aapne first ke dono ko Nahi lekar bich Ka kyun liya ? iska koi rule h kya? answer will come from both. Check this yes .. mistake Kar rahe the.. thanks |
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