Saved Bookmarks
This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.
| 1. |
Verify that3xyz =-(x +x-y) (y - z)+(z-x) |
|
Answer» First We take R.H.S & use the Formula [( a-b)²= a²+b²-2ab] & simplify it then R.H.S becomes equal to L.H.S R.H.S ⇒ 1/2×(x + y + z) (x²+ y²-2xy +y²+ z²-2yz+x²+z²-2xz) [( a-b)²= a²+b²-2ab] ⇒ 1/2×(x + y + z) (2x²+ 2y²+2z²-2xy -2yz-2xz) ⇒ 1/2×(x + y + z) 2(x² + y²+ z² – xy – yz – xz) =(x + y + z) (x² + y²+ z² – xy – yz – xz) = x³+y³+z³-3xyz= L.H.S We know that, [x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)] L.H.S = R.H.S [x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)] Please let me know if you have any doubt and like my solution as well |
|
| 2. |
. If.x + y + z =0, show that,x^3 + y^3 + z^3 =3xyz. |
| Answer» | |
| 3. |
What is the circumference of circle Pif the radius Is Bem?ВО64лО 10О 16o 87 |
|
Answer» Circumference= 2πr= 2πx8= 16π |
|
| 4. |
┬о (рей5): ( 6) 2 |
| Answer» | |
| 5. |
4m+3=6m+80 |
|
Answer» 6m - 4m = 3-80 2m = -77 m = -77/2 + +=? |
|
| 6. |
or (6m+), here f I5 SShow that any positive odd integer is of the form-(4m + 1) or (4m+3),where m is some integer. |
| Answer» | |
| 7. |
.8.lf.x + y + z-o then write the value of х, уз + z3-3xy2 |
| Answer» | |
| 8. |
= 3xyz |
| Answer» | |
| 9. |
z = 0 then which one of the following expression is correctय ि३ = 0 है, तब निम्न में से कौनसा व्यंजक सही होगा।11१३(A) x+ y + z= 0(C) x + y + 2 = 3xyz(B) x + y +1=3x3 y३(D) x2 + y +1=3xyz ।there |
|
Answer» x⅓ + y⅓ + z⅓ = 0 Let x⅓ = a y⅓ = b z⅓ = c then , a + b + c = 0 we know, when, a + b + c = 0then, a³ + b³ + c³ = 3abc hence, (x⅓)³ + (y⅓)³ + (z⅓)³ = 3(x⅓)(y⅓)(z⅓) x + y + z = 3(xyz)⅓ Therefore, x + y + z = 3 x⅓ y⅓ z⅓ (B) is correct option |
|
| 10. |
3xyz(x + y + |
|
Answer» First We take R.H.S & use the Formula [( a-b)²= a²+b²-2ab] & simplify it then R.H.S becomes equal to L.H.S R.H.S ⇒ 1/2×(x + y + z) (x²+ y²-2xy +y²+ z²-2yz+x²+z²-2xz) [( a-b)²= a²+b²-2ab] ⇒ 1/2×(x + y + z) (2x²+ 2y²+2z²-2xy -2yz-2xz) ⇒ 1/2×(x + y + z) 2(x² + y²+ z² – xy – yz – xz) =(x + y + z) (x² + y²+ z² – xy – yz – xz) = x³+y³+z³-3xyz= L.H.S We know that, [x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)] L.H.S = R.H.S [x³+ y³ + z³– 3xyz = (x + y + z)(x²+ y² + z² – xy – yz – xz)] Like my answer if you find it useful! |
|
| 11. |
Solve thefollowing problems.kg.orsugar costsäš400. What will be the cost of 25 kg of sugar?IMThat will be the cost of 24 tickets |
|
Answer» 1000 thousand in hindi ek hazar |
|
| 12. |
a) If xiyt-0 show that xty+23 3xyz |
| Answer» | |
| 13. |
present each of the following numbers on the number line(11) -1(1) asiorico(vii) -3(VI(viii) -2701 |
| Answer» | |
| 14. |
71132-3m째: 1, 3, 5, ,22, 19, 24qqq收甭pm: 1, 2, 3,, m-阿HERTRS, +S, +S, ++s_ =min(mn +1)E7% g, b, c氛オ角 |
| Answer» | |
| 15. |
Reciprocal of 1 isО 2О 1О10 |
|
Answer» reciprocal of 1 is 1 reciprocal of 1 is always 1 if we turn 1 again and we get 1 only reciprocal of one is one Reciprocal of 1 is option B that is 1 |
|
| 16. |
1. Represent each of the following numbers on the number line:(iv) 2011(ii) 1(vii) 4(1137(vi) 5.(viii) 8o(V) |
| Answer» | |
| 17. |
O69. ITX V3 tan 6 = 3sin6, 7 sin® 0 _ cos? 0 बराबर हैAT79, |
|
Answer» If you like the solution, Please give it a 👍 |
|
| 18. |
itx +-= 7, then find the value of x2 |
|
Answer» (x+1/x)^2= x^2+1/x^2+249=x^2+1/x^2+2x^2+1/x^2= 49-2= 47Answer please post again step by step |
|
| 19. |
13 If x+y +z-0, show that x3 +ys +z33xyz. |
|
Answer» Given, x^3+ y^3+ z^3= 3xyz Therefore, x^3+ y^3+ z^3- 3xyz =0 This means,x^3+ y^3+ z^3- 3xyz = (x + y + z) (x^2+ y^2+ z^2- xy - yz - zx) Now if x + y + z = 0, then...... x^3+ y^3+ z^3- 3xyz = ( 0 ) (x^2+ y^2+ z^2- xy - yz - zx) . . . . . . .... ..... ... .. . . .. . ..... [ subsituting the value of x + y + z ] x^3+ y^3+ z^3- 3xyz = 0 x^3+ y^3+ z^3= 3xyz . |
|
| 20. |
If the side of chess board is smaller than its perimeter by 21 cm. then find theside of the chess board.(CRA Module 2019) |
|
Answer» Let the side of chess board be x Then Perimeter of chess board = 4x As per given condition4x - 21 = x4x - x = 213x = 21x = 21/3 = 7 Therefore, Length of side = 7 cm |
|
| 21. |
Hint. There are 64 squares in the chess boardjA person has 2 parents, 4 grandparents, 8 great-grandparents and so on. Find thenumber of his ancestors during ten generations preceding his own. |
| Answer» | |
| 22. |
CLASS-VIII1.There were 4200people at a concert and 1400 of these were female. What percentage were males?ind the nercentage of |
|
Answer» Number of female= 1400 % of female = (1400/4200)×100 = 100/4 = 25% So percentage of male = 100- % of female = 100-25= 75% |
|
| 23. |
1610. Concert tickets usually costper person. For students they are priced atthe normal cost.How much will 6 tickets cost for students ? |
|
Answer» Given:Usual cost =120(4/5)=604/5cost for students=1/4(604/5)nowcost for 6 students=(604/5)(1/4)(6) =3624/20=Rs 181.2 |
|
| 24. |
How many squares are there in the given figure? |
|
Answer» total 10 squares are there |
|
| 25. |
A cistern 6 m long, 4 m wide, contains water to adepth of 1 m 25 cm, find the area of the wet surface. |
| Answer» | |
| 26. |
3.(c) RS. 1920एक घन के विकर्ण की लम्बाई ४-६, हैं तो इसका क्षेत्रफल होगा।(15122(b)384 cm |
|
Answer» 384cm^2 is the right answer |
|
| 27. |
1O.A card is drawn from a pack of cards. Find theprobability that it is a club. |
|
Answer» Total cards=52Number of cards with club=1*13=13 Probability=13/52=1/4 |
|
| 28. |
15 0010.00Quadratic Surdo - if x = 3 find itx+vixVith-vi-217.00 |
| Answer» | |
| 29. |
7.ABCD is a trapezium in which AB CD and P and Q are the mid-points ofADand BC. IfAB 4 cm, CD 7 cm then find PQHints in |
| Answer» | |
| 30. |
One card is drawn from a pack of 52 cards.Find the probability of drawing a face card. |
| Answer» | |
| 31. |
A chess board contains 64 equal squares and the area of each square is 6.25 cm2A border around the board is 2cm wide. Find the length of the side of the chess board. |
|
Answer» Let the length of the side of a chessboard be x cm. Area of 1 square in chessboard= 6.25 cm² area of 64 squares= 64 × 6.25 Area of 64 square = (x - 4)² 64 × 6.25 = (x - 4)²400 = (x - 4)²400 = x² + 4² -2×x×4[(a-b)²= a²+b² -2ab] 400 = x² +16 -8x400 -16 = x²+16384 = x²+16x² +16 -384 = 0x² -24x +16x -384= 0x(x -24) + 16 ( x - 24) = 0(x -24) (x +16) = 0 (x -24) = 0 or (x +16) = 0 x = 24 or x= -16 Length can't be negative, so x= 24. Hence the Length of the side of the chess board= 24 cm. |
|
| 32. |
18. AB Find the area of the shaded region in fig.if ABCD is a square of side 14 cm &APD & BPC are semicircles |
| Answer» | |
| 33. |
24) Find the area of the shaded region in figure, if ABCD is asquare of side 14 cm APD and BPC are semicircles.Color Length vi vius visquarc - ii. |
| Answer» | |
| 34. |
12. The first row of a concert hall has 25 seats, and each i a-first has one more seat than the row before it. There are 32 roses ofseats. 35 students from a class want to sit in the same row in whichrow would the class sit? |
| Answer» | |
| 35. |
Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm andAPD and BPC are semicircles.. |
| Answer» | |
| 36. |
Fig. 12.21 Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm and APD and BPC are semicircles |
| Answer» | |
| 37. |
Fig. 12.21Find the area of the shaded region in Fig. 12.21, if ABCD is a square of side 14 cm andAPD and BPC are semicircles.3. |
| Answer» | |
| 38. |
how many squares are there in chess board |
|
Answer» 64 There are many more different-sized squares on the chessboard. Therefore, there are actually64+ 49 + 36 + 25 +16+ 9 + 4 + 1 squares on a chessboard! (in total 204). |
|
| 39. |
Find the area of the shaded region in Fig 2, where arcs drawn with centres A, B.cand D intersect in pairs atith centresmid-points P, Q, R and S of the sides AB, BC, CD and DArespectively of a square ABCD ofside 12 cm/Use Ď= 3.14] |
| Answer» | |
| 40. |
If (3, 2) and (-3, 2) are two vertices of anequilateral triangle which contains the origin,find the third vertex.[Board Term-2, 2012 Set (12)] |
|
Answer» Let an equilateral triangleΔ ABC : AB = BC = AC.vertices are A(3,2) and B(-3,2) and C(x,y).The mid point of the side AB is M (0,2).(AB)²= (3+3)² +(2-2)² = 6² + 0² =36AB = 6 cmAB = BC = AC = 6 unit.AM = 3 unit.As two vertices are A(-3,2) and B(3,0) so that Third vertex will be atY-axis(x=0).so thatThird vertex C(0,y) and it is located below the origin.Hence it contains the origin.Now, y² = (AC)² -(AM)²⇒ y² = 6² - 3² = 36 - 9 =27 y =√27 = 3√3 unitHence Third vertices of the triangleare A(-3,2) , B(3,2) and C(0,3√3). |
|
| 41. |
=*74 2 73/200ionDriver oआपको स्No. (in Wber of linस्टेशन से ।1. प्राप्त किया गया प्रा |
|
Answer» rs 125 is the correct answer |
|
| 42. |
Bhalwani Tal.ParnerIExam 384 unittStd vithSub math.Q I Solve the following4 mks1. Rewrite the following using a letter.a. Four times a number is 24b. The sum of a certain nunmber and 3 |
|
Answer» 1.a: 4x = 241.b: x+3 |
|
| 43. |
The surface area of a cube is 384 cm). Find its volume |
| Answer» | |
| 44. |
\int \operatorname { sin } ^ { - 1 } \sqrt { \frac { x } { a + x } } d x \quad ( \text { Hints put } : x = a \operatorname { tan } ^ { 2 } t ) |
| Answer» | |
| 45. |
A-994UNDERSTAIDIIs. Two digits are selected at random from the digits 1 to 9. If the sum is even, thenprobability that both digits are oddellahuffled pack of 52 cards. If E is the event 'the ca |
| Answer» | |
| 46. |
1. Find the possible number of digits in thesquares of the following numbers.b) 98 c) 117d) 287a) 5e) 15 |
|
Answer» 1 only 5 2 only 23 only 13 a .5 only 2 only 2 3 only 13 |
|
| 47. |
racton becomes ind i18. The sum of a number of two digits and of the number formedby reversing the digits is 110, and the difference of the digits is 6: find.i13 nd the differenรงethe numbers. |
|
Answer» Let once place digit=a tens=bNumber =10b+a On reversing=10a+bdiff of digits=a-b=6. or a=6+bGiven, 10b+a+10a+b=11011(a+b)=110a+b=106+b+b=102b=4b=2a=6+2=8 Therefore the number is 28 and if b=6+1=82Therefore the number is =82,28Like if you find it useful super |
|
| 48. |
possiblee. A stadium has a capacity of 64070 persons How many persone shouldbone row if there are 86 rows? |
|
Answer» 64070 ---------- 86=745 ans |
|
| 49. |
2How many 3 diÄŸil no, are divisible by 4 |
|
Answer» Three digits numbers starts from 100 and end with 999 .The first three digit number which is divisible by 4 = 100 .The last three digit number which is divisible by 4 = 996 100 = 25 ( 4 )996 = 249 ( 4 ) Now, Number of numbers completely divisible by 4 = 249 - 24 = 225 . Therefore, 225 Three digit numbers are completely divided by 4 . |
|
| 50. |
24) Find the area of the shaded region in figure, if ABCD is asquare of side 14 cm APD and BPC are semicircles.Colwie-- -> (1)Area of square = 14 x 14 = 196 cm |
|
Answer» Shaded area= area of square- area of two semicircle= 14*14-(π*49) 156-154= 2 cm^2 |
|