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. |
A student multiplied 8,315 by 189 instead of multiplying by 198. How much was hisanswer less than the correct answer?Property3 |
| Answer» | |
| 2. |
8 By which number, 192 is multiplied fo make it 19200? |
|
Answer» let 192 be multiplied by x to get 19200then 192 × x=19200x=19200÷192=100 |
|
| 3. |
4, How many 5 digit numbers are there in all?5, Fill in the blanksa) 1 million =hundred thousandb) 1 crore =million6, write in descending order45908, 451098, 405691, 4019657. Find the difference of the place value and face value of the digit 6 in 456309.8. Round off 25164 to the nearest hundreds, thousands and ten thousands.9. A factory produces 1075 metres of pipe in a day. How many metres of pipe will beproduced in the month of March. 3332510. Estimate 4170+ 19785= 23 95511. A student multiplied 6245 by 62 instead of 52. How much was the answer greater thecorrect answer12. Medicine is packed in boxes each weighing 3 kg 500 g. How many such boxes can beloaded in a van which can not carry bevond 800 kg13. Write the greatest and the smallest 4-digit numbers using four different digits when de8 is always in hundreds place.14 Write the number names of following numbers in both Indian and the International30024001b) 7451246311th |
|
Answer» 5(a) 105(b)10(6) 451098,405691,401965,45908(7) face value-6 place value 6000 |
|
| 4. |
227. A boy was asked to multiply a number by12. But he multiplied by 21 and got his answer63 more than the correct answer. The numberto be multiplied wasb) 8d) 7a) 12c) 92 |
|
Answer» Answer-7 Let the no. be Xthen According to the question:12x = 21x-6363 = 21x-12x63= 9x63÷9 =x7=x12x= 8421x+63= 210 thanks |
|
| 5. |
11. By what number should (-5)- be multiplied so that the product is (8) "? |
| Answer» | |
| 6. |
Fig, 10.37In Fig. 10.38, < ABC = 69,2 ACB=31°, findL BDC.69°31 |
| Answer» | |
| 7. |
If Z = 2+、31 , Find the value of ZZ |
|
Answer» wrong answer nd wrong solution... it is question of complex no.. |
|
| 8. |
7. In the given figure,AABC is equilateralFind (i) 4BDC, (ii) ZBEC. |
| Answer» | |
| 9. |
3.Thesidesofarectangularpark are in the ratio 4: 3. If its area is 1728 m2. find the cost ofencing it at30 per metre. |
| Answer» | |
| 10. |
0.22) In the figure, LABC 69 and LACB 31. Find 4BDCA D6931 |
| Answer» | |
| 11. |
(iv) cosec31°—sec 59° |
|
Answer» cosec(31°) - sec(59°) cosec(90° - 59°) - sec(59°) sec(59°) - sec(59°) 0 |
|
| 12. |
In Fig 10.38,ABC =69°, ACB =31°,find BDC. |
|
Answer» 6)∠ABC = 69°, ∠ACB = 31° [Given] InΔABC∠ABC + ∠ACB+∠BAC = 180°⇒ ∠BAC = 180° - ∠ABC - ∠ACB⇒ ∠BAC = 180° -69° - 31° = 80° ∠BDC = ∠BAC ( Angle drawn from same chord) ∴ ∠BDC= 80° |
|
| 13. |
(59 Evaluate: /6sin 23°# sec 79° + 3 tan 48° |
|
Answer» 6 sin23 + sec79 + 3 tan48 ]/ [cosec 11 + 3 cot 42 + 6 cos 67 ] sin 23 = cos(90-23) = cos 67Sec79 = cosec(90- 79) = cosec 11tan 48 = cot (90 - 48) = cot 42 so answer is = 1 as the numerator and denominator are equal. |
|
| 14. |
In Fig. 10.38, ABC= 69, ACB= 31, find BDC. |
| Answer» | |
| 15. |
\frac{5 \sin 17^{\circ}}{\cos 73^{\circ}}+\frac{2 \cos 31^{\circ}}{\sin 59^{\circ}}-\frac{6 \sin 80^{\circ}}{\cos 10^{\circ}} \overline{\mathrm{d}} |
| Answer» | |
| 16. |
\frac { \operatorname { sin } 80 ^ { \circ } } { \operatorname { cos } 10 ^ { \circ } } + \operatorname { sin } 59 ^ { \circ } \operatorname { sec } 31 ^ { \circ } |
|
Answer» i) tan63° tan27°tan27° = cot(90- 63) = cot63°so tan63° cot63° = 1 ii) cos 13° cosec77° cosec77° = sec( 90° - 13°) = sec13°cos 13° sec13° = 1 |
|
| 17. |
श्रेणी 35, 31, 27 के कितने पदों का योग 170 होगा |
| Answer» | |
| 18. |
4. In Fig. 10.38, LABC 69, LACB 31. findZBDC |
|
Answer» In ∆ ABC ∠ABC + ∠BAC + ∠ACB = 180°∠BAC = 180° - 31° - 69° = 80°BC is same base to ∆ ABC and ∆BDCSo, ∠BAC = ∠BDCTherefore ∠BAC = 80° |
|
| 19. |
27311318 (C) |
|
Answer» 5/7 = 0.7142857113/18 = 0.7222222227/31 = 0.870967742/7 = 0.28571428 option c |
|
| 20. |
12 If the mean of 11,x, 27, 31, 47 is 27.8 find the value of x. |
|
Answer» answer galat hai 23 aaraha hai solve ve galat hai Mean= 11+27+31+47+x/5= 27.8116+x= 139x= 23thanks |
|
| 21. |
5. A cylindrical pillar is 1 m in diameter and 4.2 m in height. Find the cost of white washing the curvedsurface of the pillar at the rate of 15 per m26. Find the height of the solid circular cylinder of totol auafi |
|
Answer» Curved surface area of cylinder= 2*pi*r*h r = 1/2 m, h = 4.2 m = 2*(22/7)*(1/2)**4.2= 22*(.6)= 13.2 m^2 Cost of painting = 15*13.2 = Rs 198 |
|
| 22. |
The atio of tho humben ofs and ails in a class isEind the numb of Atds4if the totol number of students in theClass is 42.oris |
|
Answer» ratio 4:3 total students=42 4x+3x=427x=42x=42/7x=6 no. of girls=3x=3×6=18 iska ans 21 q bata raha hi book me ji 21 to nhi aa skta 21 tb ayega agar total students 49 h to.....check baar question check krlo aap kahi galat to copy nhi kiya ya fr ans. galat ho skta h....becoz total students 42 h to 18 girls and 24 boys ayege okk di thank you you are always welcome😊😊 |
|
| 23. |
, निम्नलिखित का मान निकालिए: oSS Sind2® () cosec 31° - see 59" |
|
Answer» i) sin 18°-----------cos 72° sin ( 90 - 72)------------------- cos 72 cos (72)------------ cos 72 1 |
|
| 24. |
Find the value of sec 59-cosec 31 |
|
Answer» sec (59°) - cosec (31°) sec(59°) - cosec(90 - 59°) sec(59°) - sec(59°) 0 |
|
| 25. |
\operatorname { cosec } 31 ^ { \circ } - \operatorname { sec } 59 ^ { \circ } |
| Answer» | |
| 26. |
1.Find the value of sec 59-cosec 31.ip 11 |
|
Answer» sec (59°) - cosec (31°) sec(59°) - cosec(90 - 59°) sec(59°) - sec(59°) 0 |
|
| 27. |
lii. ITy em of standing water is desired?is The dimensions of a room are 8 m x 6 m × h. It has two doors each of size 2 m × 1 m and onealmirah of size 3 m x 2 m. The cost of covering the walls by wall paper which is 40 em wideat t 1.25 per metre is 362.50. Find h (height).An open box at the ton hog itB |
|
Answer» Area of four walls 28hm^2Area of both the doors and almirah is 10m^2Area to be painted=28h-10Painted area= 362.50/1.25=290Now290*0.40=116According to the question116=28h-10126=28hh=126/28h=4.5hence the height is 4.5m |
|
| 28. |
2*cos(59^circ*(31^circ*cosec)) %2B 3*cos(80^circ*(10^circ*cosec)) |
|
Answer» 3 cos80 cosec10+2 cos59 cosec31=3 cos80/ sin10+2 cos59/ sim31= 3 cos80_/sin(90-10) + 2 cos59/ sin(90-31)= 3 cos80/ cos80+2 cos59/ cos59=3+2=5 |
|
| 29. |
partial fractions covering the types(x+ a)(x+b)(x+c) |
| Answer» | |
| 30. |
Infigure 6-59, if AB II CD, ifFig. 6.59/1 (5x-45)° and 25 (3x +15)°, find 21 and 25.3: 46 7:5, find the measures of 23 and 46.i)Ans. i)21-45 105 |
|
Answer» Hahahzhsgdvdhsj6sysuwiw8w |
|
| 31. |
59.यदि ।== 40 तब 0 का मूल्य3+31 1434*)२ |
|
Answer» 1/10 |
|
| 32. |
ह वर्षी में(4) 602.७2 >८* तथा &2 >८८ हो, तब हैजलकर28. यदि दर |
|
Answer» Given ax= by= czand b2= acLet ax= by= cz= kConsider, ax= k ⇒ a = k1/xइसी प्रकार हमें प्राप्त होगाwe get b = k1/yand c = k1/z हमें ज्ञात है, b2= ac⇒ (k1/y)2= (k1/x)(k1/z)⇒ k2/y= k1/x +1/z दोनों जगह की तुलना करने परComparing both the sides, we get(2/y) = (1/x) + (1/z)⇒ (2/y) = (x + z)/xz∴ y = (2xz)/(x + z) |
|
| 33. |
evulate cosec 31- sec 59 |
|
Answer» To find:cosec31°- sec59° Formula used: cosecA = sec(90°-A) So, cosec31° as sec(90-31°) = sec59° => sec59° -sec59° => 0 Like if you find it useful |
|
| 34. |
\cosec 31 ^ { \circ } - \sec 59 ^ { \circ }} |
|
Answer» cosec(31°) - sec(59°) = cosec (31°) - sec(90-31)cosec(31) - cosec(31) = 0 |
|
| 35. |
दर न 2 A ह |
|
Answer» 8 iska answer Hai Ab Mujhe like karo |
|
| 36. |
cos(59^circ)/sin(31^circ) |
|
Answer» 1 is the correct answer of the given question 1 is correct answer |
|
| 37. |
6. एक त्रिभुज में c=2,6-2, sin B = ह.तब कोण A का मान है : |
| Answer» | |
| 38. |
A race-boat covers a distance of 60 km downstream in one and a half hour. Itcovers this distance upstream in 2 hours. The speed of the race-boat in stillwater is 35 km/hr. Find the speed of the stream. |
| Answer» | |
| 39. |
I am a whole number. Between me and-7, there are 9 integers, then how many whole numbersare there smaller than me? |
| Answer» | |
| 40. |
vesseli) radius of the base,i) capacity of the vessel.acity of a closed cylindrical vessel of height 1 m is 15.4s of metal sheet would be needed to make it?square metlitres. How many |
| Answer» | |
| 41. |
LCM of 2 × 3 × 5 and 3x5 × 7 is equal to |
| Answer» | |
| 42. |
1.नीचे दिए गए कथन सही है।भी लिखो।गए कथन सही हैं या गलत। उनके सामने लिखो और साथ में ग1.23 = 324x 2 = 42ax 2 = axaab.c=axbxbxcxcxca = axaxa32 = 2x2x2१ |
|
Answer» 1)wrong2) wrong3) wrong4) wrong5) right6)wrong only 5 is correct and other are false wrong wrong wrongwrong right wrong same same above ...... 1 w2 w3 w4 w5 r6 w only 5 is true and all the other false 1. False2. False3. False4. False5. True6. False 1: wrong2: wrong3: wrong4: wrong5: right6: wrong only 5th question is right 1,wrong2,wrong3,wrong4,wrong5,right6,wrong |
|
| 43. |
प्रश्नावली-8.1निम्नलिखित लो घातलीय रूप में व्यक्त कीजिए-(i) 5x5x5X5 (ii) cXcXc (iii) 2X2X3X3X3(iv) 6x6xbxb (v) aXa><b><b<b><bXbxd |
|
Answer» ans |
|
| 44. |
axa |
| Answer» | |
| 45. |
1) consider A {-11)a) write all elements in AXA |
|
Answer» -1,1 all elementin A×A |
|
| 46. |
After covering a distance of 30 km with a uniform speed there is somedefect in a train engine and therefore its speed gets reduced to 4/5of its original speed. Consequently, the train reaches its destinationlate by 45 minutes. If, the defect in engine would have occurred aftercovering 18 kilometres more distance, the train would have reached9 minutes earlier. The original speed of the train is: |
| Answer» | |
| 47. |
है g — ky T 2 k2 |
|
Answer» y²-ky-2k²=0=>y²-2ky+ky-2k²=0=>y(y-2k)+k(y-2k)=0=>(y+k)(y-2k)=0=>y=-k or y=2k |
|
| 48. |
3x5 |
|
Answer» 5 times 3 is 15it's primary math |
|
| 49. |
29.2) 3S | kY]54* अपूर्णाकाचा ल. सा. वि. किती ? |
|
Answer» Lcm of (2,3,4)= 24hcf of (2,3,4)= 1lcm= 24/1 |
|
| 50. |
(a) 3x5(a) 3x 5- |
| Answer» | |