InterviewSolution
Saved Bookmarks
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.
| 1001. |
0-15153030 4S 45-60 607561064By changing the following frequeney distribution 'to less than type distribution, drawits ogiveClasses oFrequency0-1515-30 30-45 45-60 60-7561064 |
| Answer» | |
| 1002. |
By changing the following frequenc y distribution 'to less than type disirnbution, drawits ogiveClasses0Frequcecy 60-15 15-30 30-45 45-60 60- 7510 |
| Answer» | |
| 1003. |
(37) Himani ate half of a cake on Monday. She ate halfof the cake left on Tuesday and so on. She followedthis pattern till Thursday. What fraction of the cakehas she eaten till Thursday? |
| Answer» | |
| 1004. |
. In the figure, line AB line DE. Find the measuof LDRE and LARE using given measures of soangles.40°70° |
| Answer» | |
| 1005. |
(d) Writeas the sum of 4 unit fracoms.23- 2 1-371and corect to two decorrect to two deUsing this, find the aproximate values of 3+places. (31.732 21.414 |
| Answer» | |
| 1006. |
Q1. In the given figure,dABC-AADE. If AE-2 cm, EC-3 cm and DEthen find BC.1.6 cm,Ans. BC 4 (using Thales' theorem)] |
| Answer» | |
| 1007. |
10n figure 3.10, line AB || line DE. Findthe measures of ZDRE and LAREusing given measures of some angles.4070°Fig. 3.10 |
|
Answer» Thankz for answer |
|
| 1008. |
TION Ations number 1 to 6 carry 1 mark each.In the given figure,AABC AADE. If AE 2 cm, EC 3 cm and DE 1.6 cm,then find BCAns. BC 4 (Using Thales' theorem)liervations ar |
|
Answer» hit like if you find it useful |
|
| 1009. |
S Doraw a perpendiculat bisector of aling8cm6) construct 90' without using PorotactosIC DE=? (2) anionintructAARB-Golc40 EAIS |
|
Answer» angle dE=90°90° is the right ans |
|
| 1010. |
-f(x^2*y^2)/2 - f(x^2/y^2)/2 %2B f(x^2)*f(y^2) |
|
Answer» I think ,zero is right answer |
|
| 1011. |
4. Prove that v 4siny-o1S an increasing2-cos θfunction of θ in | 0,NCERT)2 |
| Answer» | |
| 1012. |
13 Prove that v-4sin θθis an increasing function of θIs an(2+cos θ)in | 0,- |
|
Answer» so dy/do is always positive as shown so function is increasing in following interval. |
|
| 1013. |
The radius of a circle is 28 cm and the area of the sector is filled with rain water is 205.4cm2. Find the central angle.8. |
|
Answer» 30° central angle/360 = 205.4/πr² central angle = 205.4*360*7/22*28*28 = 30° |
|
| 1014. |
(27*(hat*k) %2B 2*(hat*i) %2B 6*(hat*j))*(mu*(hat*k) %2B hat*i %2B lambda*(hat*j))=0*vec |
| Answer» | |
| 1015. |
ही. के ने सनक — =+ (2i-27+4K)-(3i +4 j-5K |
|
Answer» The given is dot product (2i - 2j + 4k) . ( 3i + 4j - 5k) (2×3) + (-2) × (4) + (4) × (-5) 6 - 8 - 20 6 - 28 -22 |
|
| 1016. |
(-7+9) +(7-6) |
|
Answer» (-7+9)+(7-6)= 2+1= 3 |
|
| 1017. |
27.The distance of the point 2i+j-k from the plane r.(i - 2j+4k) = 9 is |
| Answer» | |
| 1018. |
1-iample 14. |
| Answer» | |
| 1019. |
what is the difference between 7 times 9 and 9 times 10 |
|
Answer» -27 is the answer of the following 7 times 9=639 times 10=90there is a difference of 27 you can write it in positive or negative as your wish 7 times 9=639 times 10=90there is a difference of 27 |
|
| 1020. |
(5/7)—² * (5/7)⁴ ÷ (5/7)³ |
|
Answer» (5/7)^-2+4-3=(5/7)^-1=7/5 |
|
| 1021. |
1. किसी राशि पर 2 वर्ष के लिए 4% वार्षिक दर परसाधारण और चक्रवृद्धि ब्याज के बीच का अंतर 1रूपए है। वह राशि है -A. 625B. 630C. 640D. 650 |
|
Answer» the answer will be 650 |
|
| 1022. |
(7*(7/9))*((7/5)*(-5*5/7)) |
|
Answer» 560/9 is the correct answer of the given question 7/5(-40/7)×70/9=--8×70/9=-560/9 |
|
| 1023. |
0.20 ^ { \circ } \cdot \operatorname { cos } 40 ^ { \circ } \cdot \operatorname { cos } 60 ^ { \circ } \cdot \operatorname { cos } 80 ^ { \circ } = \frac { 1 } { 16 } |
|
Answer» Cos20°cos40°cos60°cos80°=(1/2)(2cos20°cos40°)(1/2)cos80° [∵,cos60°=1/2]=(1/4)[cos(20°+40°)+cos(20°-40°)]cos80°=(1/4)(cos60°+cos20°)cos80°=(1/4)(cos60°cos80°+cos20°cos80°)=(1/4)(1/2)cos80°+(1/4)cos20°cos80°=(1/8)cos80°+(1/4)(1/2)(2cos20°cos80°)=(1/8)cos80°+(1/8)[cos(20°+80°)+cos(20°-80°)]=(1/8)cos80°+(1/8)(cos100°+cos60°)=(1/8)cos80°+(1/8)cos100°+(1/8)cos60°=(1/8)(cos80°+cos100°)+(1/8)×(1/2)=(1/8)[{2cos(80°+100°)/2}{cos(80°-100°)/2}]+(1/16)=(1/8)(2cos90°cos10°)+(1/16)=0+(1/16) [cos90°=0]=1/16 (proved) what abt cos20° |
|
| 1024. |
-(-sqrt(5) %2B 7)/(sqrt(5) %2B 7) %2B (sqrt(5) %2B 7)/(-sqrt(5) %2B 7)=a %2B 7*(sqrt(5)*b) |
| Answer» | |
| 1025. |
Prove that the circle drawn with any side of a rhombus as a diameter, passesAMPLE 22the point of intersection of its diagonals |
| Answer» | |
| 1026. |
(&)\JmC| |
| Answer» | |
| 1027. |
Category I. निम्नलिखित रूपों के समाकलनिम्नलिखित को ज्ञात करें (Evaluate the1. JM NCERT]1/1+42 |
| Answer» | |
| 1028. |
( 59 ×59×59-9×9×9)/(59×59+59×9+9×9) = 50 |
| Answer» | |
| 1029. |
ax. Marks: 80SUsing Euclid's division algorithm, find the HCF of 56, 96 and 404. |
|
Answer» 96=56×1+4056=40×1+1640=16×2+816=8×2+0HCF=8 404=8×50+48=4×2+0 HCF(56,96,404)=4 |
|
| 1030. |
\frac { 59 \times 59 \times 59 - 9 \times 9 \times 9 } { 59 \times 59 + 59 \times 9 + 9 \times 9 } = 50 |
| Answer» | |
| 1031. |
12.Arrange the following fractions in2/9 , 8/9 , 6/9 , 7/9 1 4/9 |
| Answer» | |
| 1032. |
Velocity time graph of a vehicle is shown inthe figure. Find the time for which the brakeswere applied to it-000v (m/s) →ON A510 15 20 25T(Sec.) → |
|
Answer» brake means negative acceleration ie. negative slope which is between 10-15s. so brakes are applied between 10-15s between 10 to 15 sec |
|
| 1033. |
9 \times 9 \times 9 \times 9 |
|
Answer» answer= 9×9×9×9= 6561 9 × 9 × 9 × 9 81 × 9 × 9729 × 9 6561 😆😆😆😆😆 |
|
| 1034. |
\begin{array} { l } { \text { Which of the following fractions is NOT } } \\ { \text { the same as } \frac { 7 } { 8 } \text { ? } } \\ { \text { A) } \frac { 35 } { 40 } \text { (B) } \frac { 14 } { 16 } \text { (c) } \frac { 21 } { 40 } \text { (D) } \frac { 63 } { 72 } } \end{array} |
|
Answer» c is the right answer. |
|
| 1035. |
\frac { 25 } { 9 } \times \frac { 25 } { 9 } - \frac { p } { 9 } + \frac { 16 } { 9 } \times \frac { 16 } { 9 } = 1 \text { then } p = |
|
Answer» by BODMAS rule625/81-p/9+256/81=1so p/9=(625+256-81)/81=800/81so p=9*800/81=800/9 |
|
| 1036. |
-80*f*o*text %2B 4*(640/80) %2B 639 - 8 |
|
Answer» (b) |
|
| 1037. |
\left. \begin{array} { | c | c | c | c | c | } \hline - 20 & { 20 - 40 } & { 40 - 60 } & { 60 - 80 } & { 80 - 100 } & { 100 - 120 } \\ \hline 5 & { F _ { 1 } } & { 10 } & { F _ { 2 } } & { 7 } & { 8 } \\ \hline \end{array} \right. |
| Answer» | |
| 1038. |
7 \cdot 50 : \text { Rs. } 22 \cdot 80 |
|
Answer» Rs 7.50 = 750 paise Rs 22.80 = 2280 paise ratio = 750/2280 = 75/228 = 25/76 |
|
| 1039. |
(x-+Ifa+Âť-Tprove that a2 + b-(fIfa + ib, prove that a2+F= (222xt +1 |
|
Answer» thank you soo much😊 |
|
| 1040. |
3d + 50°2d +20°ーm(A) 22 (B) 30 (C) 45° (D) 80° |
|
Answer» line l and line m are parallel3d+50°+2d+20°=180° ------------(by interior angle test) 3d+2d+50+20=1805d+70=1805d=180-705d=110d=110/5d=22 |
|
| 1041. |
AMPLE 22 Ecaluate JM b0-00-3162oaluates, is being given that 5 2.236 and1510+V20+ V40 -V5 8080, is being given that./5 = 2.236 and51 |
| Answer» | |
| 1042. |
a man riding his bicycle covers 150m in 25s. what is his speed on km/h? |
| Answer» | |
| 1043. |
1. Describe the refrigeration and air conditioning. |
|
Answer» Refrigeration may be defined as the process of achieving and maintaining a temperature below that of the surroundings, the aim being to cool some product or space to the required temperature. Air conditioning is the process of altering the properties of air to more comfortable conditions, typically with the aim of distributing the conditioned air to an occupied space, such as a building or a vehicle, to improve the thermal comfort and indoor air quality |
|
| 1044. |
7. Bart drove 140 miles in 2 hours..What was Bart's average speed in milesper hour?.A. 66'miles per hourB. 65 miles per hourC. 56 miles per hourD. 55 miles per hour. |
|
Answer» Answer C. 56 miles per hour. C. 56 miles per hour. Distance covered by Bart in 2½ hours = 140 miles,distance covered by Bart in 1 hour = 140 * ( 2 / 5 ), = 56 miles per hour. distance=140milestime=5/2hrspeed=distance/time=140÷5/2=140×2/5=28×2=42miles/hr |
|
| 1045. |
A mechanic earns $5 more per hour thanhis helper. On a six-hour job the two menearn a total of $114. How much does eachearn per hour? |
|
Answer» let the cost /per hour of helper be x and that of the mechanic is x+5 now for 6 hr job total earning is 6(x+x+5) = 114 => 2x+5 = 19 so, 2x = 14 or x = 7 the earning of helper is = 7$ /hr and earning of mechanic is = 12$/hr |
|
| 1046. |
a) Complete what is missing from division:- 80 .............-40. |
|
Answer» 80÷x=4080/x=4040x=80x=80/40x=2so the missing term is 2 |
|
| 1047. |
I. Copy and complete the table ( -22)Radius of the circle (cm) 7Height (cm)Volume (cm)21401416 |
| Answer» | |
| 1048. |
8. A fast train covers 360 km in 3 hours 40 minHow much time will it take cover a distance of900 km? |
|
Answer» 360 km in 3h 40m so 720 km in 6h 80m=7h 20mso 900 km in 7h20m+1.5h20m=8.5h40m 3hrs 40min=(3×60)+40=180+40220min cover 360 kmso in 1 min cover 220/360=11/1811/18×900=550minin hour=550/60=9 hrs 10 min |
|
| 1049. |
MULTIPLICATION IN MEASURES OF TIMEMultiply 25 hours 20 min 15 sec by 7 |
| Answer» | |
| 1050. |
unice ll 3 hours.6. A person covered a distance of 15 km in 1 hr 30 min. How muchadditional time will be required to cover a further distance of 5 kmat the same speed ? |
| Answer» | |