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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.
| 76101. |
EXERCISE 1.11. Using appropriate properties find.1 2 3 5 3 1KtX3 5 2 5 6Write the additive inverse of each of the |
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Answer» the answer of these questions is -1/10 |
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| 76102. |
4n²+5n-313 |
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| 76103. |
(2)Underline the verb in the following sentence and say if it is Transitive or Intransitive.A rainbow colours the entire sky. |
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Answer» In this Verb is colours" |
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| 76104. |
which is large r2 and 313 |
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Answer» 3√3 is greater than √2.Because 3√3 has a factor 3 but √2 has only 1. 3√3 is large because 3×3 is 9 and 9×2 =18 and √2 =2 3√3 is greater than √2 because 3√3 = 18 and √2 = 2 so, 18 is greater than 2.. 3√3 > √2because 3√3 has a factor 3 but √2 has only 1 |
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| 76105. |
16. Simplify: 313-2V2 |
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| 76106. |
Rationalize the Denominator313- da |
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| 76107. |
18 (A) İF4 SinO-1-0 and o is less than 900, find the value of θ and thevalue of Cos o+Tan'o(OR) |
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| 76108. |
10. In the given figure line AB and CD intersect at 0. ZBOC-36. FindZy and836°2xFig. Q. 11acting each other at point O'. If4Fig. Q. 10 |
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| 76109. |
8. Find the value of x if(ICSE 1992) |
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Answer» sqrt(p/q) = (q/p)^1-2x (p/q)^1/2 = (p/q)^2x-1 As base are same then power should be also equal 2x - 1 = 1/24x - 2 = 14x = 3x = 3/4 Therefore, value of x = 3/4 |
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| 76110. |
: Insert two numbers between 3 and 81such that the resulting sequence is a G.P |
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| 76111. |
nside a rectangular garden 24 m long and 18 m wide, a 2 m marginfor planting tulip flowers. If4 tulip stems can be planted in1 m2 and each tulip stem costs 30, find the total expenditureincurred. |
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Answer» Area for planting flowers=24*2+18*2-4(2*2)=48+36-16=48+20=68m²Flowers in 1m²=4Flowers in 68m²=4*68=272 Cost of 1 flower=30RsCost of 272 flower=30*272=8160Rs |
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| 76112. |
EXERCISE 1Using appropriate properties find. |
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| 76113. |
\left| \begin{array} { c c c } { 3 - x } & { 5 } & { 4 } \\ { 5 } & { 4 - x } & { 3 } \\ { 4 } & { 3 } & { 5 - x } \end{array} \right| = 0 |
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| 76114. |
-4*a*c - 4*b*c %2B 2*(a*b) %2B 4*c^2 %2B a^2 %2B b^2 |
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Answer» We know,(a+b+c) ²=a²+b²+c²+2ab +2bc +2ac. Then,a^2 + b^2 + (-2c)^2 + 2ab - 4bc - 4ac= (a + b - 2c)^2= (a + b - 2c)(a + b - 2c) |
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| 76115. |
EXERCISE 1.1I.Using appropriate properties find. |
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| 76116. |
the adjoining figure, pll q. Find the unknown angles125° |
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| 76117. |
9. In the adjoining figure, ABCD is a quadrilateral Dand AC is one of its diagonals. Prove that(i) AB+ BC+ CD+ DA 2AC(ii) AB+ BC+CD> DA(ii) AB+BC+CD+ DA> AC+ BD. |
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| 76118. |
9. In the adjoining figure, ABCD is a quadrilateral Dand AC is one of its diagonals. Prove that(i) AB+BC+ CD +DA> 2AC(ii) AB+ BC+ CD > DA(ii) AB+ BC+CD + DA> AC+ BD |
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| 76119. |
Quadrilaterals3139. In the adjoining figure, ABCD is a quadrilateral Dand AC is one of its diagonals. Prove that(G) AB + BC+ CD+ DA 2AC(i) AB+ BC+CD> DA(ii) AB + BC+ CD+ DA > AC+ BD.drilateral is 360 |
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Answer» thanks |
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| 76120. |
2. Ashwin was born on 15 July 1992. His friend Gautam wasborn on 24 Feb 1992. Who is older? By how many days? |
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Answer» gautam was elder than ashwin |
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| 76121. |
AGRASAIN BOYS' SCHOOLWhated to the Council for the Indian School Certificate Examin21/A, AGRASAIN STREET, LILUAH, HOWRAH -PERIODICAL EXAMINATION: 2018-19Class :IXDate: 26-04-2018Subject - Mathematics"ime: 11/2 hr.SECTION: A (Answer all the questionsinsert two irrational numbers betweenand |
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Answer» 3/7and 6/7using the formula n+1 (1/7) = 0.14285714 (nearly)(2/7) = 0.28571428 (nearly)Irrrational numbers between these two rational numbers are:0.1505005000500005....0.20200200020000.... 3/7and6/7 using the formula n+1 |
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| 76122. |
Name the younger son of Shivaji whoalso the third Chhatrapati.036was |
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Answer» sambhaji maharaj is son of shivaji maharaj |
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| 76123. |
For the class 21 to 30, write lower and upper class limit. |
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Answer» upper 30.5lower 20.5 To find the class limit, in inclusive method, subtract 0.5 from lower score to get lower class and add 0.5 to the upper score to get upper class limit |
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| 76124. |
If4 x 5* 500, then the value ofx is |
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Answer» 4 x 5ˣ=5004 x 5ˣ=5 x 1005ˣ =5x1255ˣ=6255ˣ=5⁴x=4xˣ=4⁴=256 |
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| 76125. |
(-16/15 x 20/ 8) - ( 15/5 x 35/5) |
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Answer» (-16/15 × 20/8) - (15/5×35/5) = (-2×5×4/(5×3)) - (3×7) = -8/3 - 21 = -71/3 |
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| 76126. |
8*x - 35=3*x |
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Answer» 8x-35=3x 5x=35 x=7 8x-35=3x-35=3x-8x-35=-5x35/5=7 x=7 is right answer for this question x=7 is the right answer X=7 is the correct answer |
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| 76127. |
L. The ratio between the present ages of Vijay and Sanjayis 6:7. Three years hence the ratio will become 7:8,respectively. What is the present age of Vijay? |
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Answer» Let present age of Vijay be x and Sanjay be y x:y=6:77x=6yx=6y/7 After 3 years,x+3:y+3=7:88(x+3)=7(y+3)8x+24=7y+218x-7y=-3 Put x=6y/7 8(6y/7)-7y=-348y-49y=-3*7-y=-21y=21 x=6(21)/7=18 So, present age of Vijay is 18 and Sanjay is 21 |
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| 76128. |
, In the adjoining figure, ABCD is a quadrilateraland AC is one of its diagonals. Prove that6) AB+ BC+ CD+ DA 2AC6i) AB+ BC+ CD > DAGii) AB+ BC+CD+ DA> AC+BD |
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| 76129. |
,. In the adjoining figure, ABCD is a quadrilateral Dand AC is one of its diagonals. Prove that( AB+BC+ CD +DA> 2AC(i) AB+ BC+ CD> DAii) AB+ BC+CD+DA> AC + BD. |
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| 76130. |
Find the value of lambda, if the mode of the following data is 2015, 20, 25, 18, 13, 15, 25, 15, 18, 17, 20, 25, 20, Îť, 18.12.: |
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Answer» Mode is the data which has the highest frequency. Excluding the unknown value, there are equal numbers of 20 and 25. So, for 20 to be the mode, the unknown value must be 20. |
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| 76131. |
ofthefollowingdistribution is 18. Find the frequency fof the class 19-21Class11-1313-1515-1717-1919-21 21-23 -213 |
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| 76132. |
(oFindthevalueofx, if the mode of the following data is 25:15, 20, 25, 18, 14, 15,25, 15, 18, 16, 20, 25, 20, x, 18.(ii) If both 20 and 18 are changed to 15 in the above data, find the new mode. |
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Answer» If 20 and 18 are changed to 15, the new mode is 15. good |
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| 76133. |
Find the missing euencypfor thedistrbution. Find the missing frequency p for the ollowing frequency distributionwhose mean is 28.252530354015201415 |
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| 76134. |
e four walls of the rodlill2. A dice is rolled 100 times and its outcomes are as follows56215OutcomesFrequency 12Find the probability of getting201 4a. A multiple of 3. |
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Answer» not getting the question Multiples of 3 are 3 and 6hence total frequency 20+18=38hence probability will be 38/100=0.38 |
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| 76135. |
Calculate the mode of the following distribution:Class:10-15 15-20 20-25 25-30 30-35Frequency4 72058 |
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| 76136. |
9. A GP. consists of an even number of terms. If the sum of all the terms i5 times the sum of terms occupying odd places, then find its common raia |
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| 76137. |
ii) 1 head. (iii) 0 head?A dice is thrown 100 times and the outcomes are noted as given below:ChitcoiE342114189Frequency1523When a dice is thrown at random, what is the probability of getting a (1) 3. (i) 6. ()4. (v) 1? |
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| 76138. |
11. A die is rolled 200 times & its outcomes are released as below:Outcomes2842253540Frequency |
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Answer» (A) The only even prime number Is 2.Which occurs 35 times. So it's probability = 35/200= 7/40. (B)there are only two multiples of 3 I.e 3&6.So both of these occur = 40.+ 30 = 70 times So, it's probability = 70/200 = 7/20. |
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| 76139. |
An unbiased dice is thrown 100 times and the data is recorded as beOutcomeFrequency 10415 2025151S(i) What is the probability of getting an odd number.(i) What is the probability of getting a number less than 4. |
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Answer» 1) probability of odd number= 40/100= 0.42) probability of getting a number less than 4= 50/100= 1/2= 0.5 |
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| 76140. |
12. A dice is rolled 100 times and its outcomes are as follows24OutcomesFrequencyFind the probability of gettinga A multiple of 3. |
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Answer» Probability of getting a multiple of 3 = Probability of getting a 3 + Probability of getting a 6= 18/100 + 20/100= 38/100= 19/50. |
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| 76141. |
27. An unbiased dice is thrown 100 times and the data is recorded as below4Outcome 1Frequency 10(Gi) What is the probability of getting an odd number.(ii) What is the probability of getting a number less than 4.25 15 201515 |
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Answer» 1) probability of odd number= 40/100= 0.42) probability of getting a number less than 4= 50/100= 1/2= 0.5 |
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| 76142. |
5. A coin is thrown 35 times and it showed an even number 15 times. Findthe probabiliy of getting a od uncer |
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| 76143. |
In the adjoining figure line p | line q.40°Line t and line s are transversals.Find measure of x and Lyusing the measures of angles givenin the figure. |
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| 76144. |
- In the adjoining figureline p || line q. Line tand line s are transver-sals. Find measure ofZx and Z y using the xmeasures of anglesgiven in the figure. |
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Answer» b = 40° ( corresponding angles)c = 40° ( vertically opposite angles) a + 40° = 180° a = 180° - 40°a = 140° So, x = 140° ( corresponding angle) y = 180° - 70° (co interior angles)y = 110° |
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| 76145. |
Evaluate 25 power 3|2 × 243 power 3|5 divided by 16 power 5|4 × 8 power 4|3 |
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Answer» (25)^3/2 * (243)^3/5 /(16)^5/4 * 8^4/3 = (5)^3 * (3)^3 / (2)^5 * (2)^4 = 125 * 27/ (2)^9 = 3375/512 |
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| 76146. |
0.44. Describe in brief operationsthat can beperformed on a horizontal boring machine. |
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Answer» Ans :- Ahorizontal boring machineorhorizontal boring millis amachine toolwhichboresholes in ahorizontaldirection. There are three main types— table, planer and floor.The table type is the most common and, as it is the most versatile, it is also known as the universal type. A horizontal boring machine has its work spindle parallel to the ground and work table. Typically there are three linear axes in which the tool head and part move. Convention dictates that the main axis that drives the part towards the work spindle is the Z axis, with a cross-traversing X axis and a vertically traversing Y axis. The work spindle is referred to as the C axis and, if a rotary table is incorporated, its centre line is the B axis. |
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| 76147. |
(a) 1 million =lakh |
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Answer» 10 lakh |
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| 76148. |
222° +5in”68°)222° + c0s°68°)+in263° + cos 63° sin ?.7\” =(cy 248 |
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Answer» [sin²22° sin²68°/ cos²22° cos²68°] + sin²63° + cos63° sin27° = [sin²22° sin²68°/ cos²(90°-68°) cos²(90°-22°)] + sin²63° + cos63° sin(90°-63°) = [sin²22° sin²68°/ sin²68° sin²22°] + (sin²63° + cos²63°) = 1 + 1 = 2 Like my answer if you find it useful! |
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| 76149. |
If the die is rolled 300 times, how many times would you predict a roll of a lor a 6? |
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Answer» Probability = possible outcomes/ Total outcomesProbability of getting 1 or 6 in any given roll of dice= 2/6 = 1/3 For 300 roll probability of getting 1 or 6= 1/3 * 300= 100 predicted rolls |
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| 76150. |
26. A die is thrown 100 times.If the probability of getting an even number isHow manytimes an odd number is obtained? |
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