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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. |
Intelligence and CriticalReasoningArti' and 'Saurabh' are the children of Mr.and Mrs. Shah. "Ritu' and Shakti are thetionchildren of Mr. 'and Mrs. Mehra, 'Saurabh'ansand 'Ritu' are married to each other and twodaughters 'Mukti' and 'Shruti' are born tothem. Shakti' is married to 'Rina' and twochildren 'Shubash' and 'Reshma' are born tothem. How is 'Arti' related to 'Shruti'?(B) Mother-in-Law(C) Sister(A) Mother(DY Aunt |
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Answer» Ans is. a. unt |
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| 2. |
6. The age of father is twice the sum of ages of his two children Ten years hence, the age offather will be three- quarter of the sum of the ages of his children then. Find the present age of lather |
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| 3. |
NoteThe midpoint C'of the hypotenuše1.Find the midpoint of the line segn(i) (1, -1) and (-5, 3)ntroid of the triangle v |
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Answer» the mid point is given by (x1+x2)/2 , (y1+y2)/2 = 1-5/2 , (-1+3)/2 = -4/2, 2/2 = -2,1. thank you |
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| 4. |
divide 84 chocolates between two children in the ratio 5:7. |
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Answer» suppose both will get 5x and 7xso 5x+7x=12x=84so x=7so both will get 35 and 49 chocolates. thoda long mai bataye I can not understand |
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| 5. |
Mental Ability19. In the adjoining figure find the shaded area (in sq. units)D SUnits G(a) 77(b) 74(c) 72(d) 84.57 Units12 UnitsE 7 Units B |
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| 6. |
14. How many persons can be accommodated in a dining hall ofdimensions (20 m × 16 m × 4.5 m), assuming that each person requires5 cubic metres of air? |
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| 7. |
The dimensions of a cinema hall are 120 m ×70 m × 18 m. How many persons can sit inthe hall, if each person requires 216 m2 ofair. |
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| 8. |
e area of a circle inscribed in an equilateral triangle is 154 cmFind the perimeter of the triangle.Take 3-173] |
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| 9. |
4. 75 persons can sleep in a room 25 m by9-6 m. If each person requires 16 m3 of air;1find the height of the room. |
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Answer» Dimensions of roomLength = 25 m, Breadth = 9.6 m, Height =? Number of person in room = 75and each person require 16 m^3 air Then volume of room = 75*16 = 1200 m^3 We know, Volume of room = Length*Breadth*Height Therefore, Length*Breadth*Height = 120025*9.5*Height = 1200Height = 1200/(25*9.5) = 48/9.5 = 5.05 m Height of room = 5.05 m |
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| 10. |
The dimensions of a cinema hall are 120 m x70 m × 18 m. How many persons can sit inthe hall, if each person requires 216 m2 ofair. |
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| 11. |
water11. A steamer going downstream a river, covers the distance between two tos i2hCingback upstream, it covers this distance in 25 hours. The speed of water ish Find the detanbetween the two towns |
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| 12. |
The dimensions of a cinema hall are 120 mx70 m x 18 m. How many persons can sit inthe hall, if each person requires 216 m ofair. |
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Answer» 120×70×18=151,200m²if each person need 216m²then the person can saet is 151200/216=700 120×70×18=151200m151200÷216=700 |
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| 13. |
that they are congruent.5.; D, E and F are respectively and WABC., and F are respectively the mid-points of sides AB, BC and CA of A ABC. Find the11. |
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| 14. |
14. How many persons can be accommodated in a dining hall ofdimensions (20 m X 16 m X4.5 m), assuming that each person requires5 cubic metres of air? |
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Answer» thnx |
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| 15. |
Find the ratio in which the point (-3, p) divides the line segn(-2 3). Hence, find the value of p.19. The angles of depression of the top and bottom of a 50 m high building from the top of a tower are 45and 60 respectively. Find the height of the tower and the horizontal distance between the tower andthe building. (Use 3 -173) |
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Answer» Here CD is the tower and AB is the pole. In △CDB, CD/BC = tan 60° => 50m/BC = √3 => BC = 50/√3m AE = BC AE = 50/√3m In △ADE, DE/AE = tan 45° => DE/50/√3m = 1 => DE = 50/√3m so the height of the pole is 50/√3m |
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| 16. |
17. Find the area of both the segments of a circle of radiuscentral angle 120. (Given, sin 120 and 3-173.]42 cm with |
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Answer» thanks |
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| 17. |
Is 173 a prime? why? |
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Answer» For173, the answer is: yes,173is aprime numberbecause it has only two distinct divisors: 1 and itself (173). |
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| 18. |
(рек) 13 1/517 173 |
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| 19. |
7. In the given figure, I and m are←>A<--D→1two parallel tangents to a circlewith centre O, touching thecircle at A and B respectively.Another tangent at C intersectsLXthe line I at D and m at E. Provethat <DOE = 90°.> m[Delhi 2013] (4) |
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Answer» The exact solution is provided. Only replace the letters. |
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| 20. |
7. In the given figure/ l and m are ←_D->1|two parallel tangents to a circlewith centre O, touching thecircle at A and B respectively.Another tangent at C intersectsthe line I at D and m at E. Provethat <DOE = 90°.B E[Delhi 2013] (4) |
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| 21. |
How many questions did she answer correctly?ORA-man can row a boat at the rate of 4 km/hour in still water. Hegoing 30 km upstream as in going 30 km downstream. Find thercos θ + sin θ-V2 cos θ , prove that cos θ-sin θ V2 sin eOR |
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| 22. |
14. How many persons can be accommodated in a dining hall ofperson requiresdimensions (20 m 16 m x 4.5 m), assuming that each5 cubic metres of air? |
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| 23. |
3x + 2x2 +3x +1Ma(DZ)__3(x2 +1)उत्तर[उ. प्र. 2007 (Etane[उ. प्र. 2014 (Gउत्तर |
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| 24. |
After going 4 of my journey, I find that I have covered 18 km. How much journey isstill left ?3 |
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| 25. |
the steamer going down steam in a river cover the distance between in 25 hrs. the speed of water is 4 km/h find the distance between 2 towns? |
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Answer» soln:speed=distance/timedistance=speed × time=4× 25distance=100 km.Ans thanks |
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| 26. |
0.8A person can row a boat at the rate of 5 km/hr in still water. He takes thrice as much time in going 40 kmupstream as in going 40km downstream. Find the speed of the stream.(Ans. 2.5 km/hr] |
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Answer» 400km/hr was the write answer |
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| 27. |
(Hint: The parlour should be equidistant from A, B and C)Complete the hexagonal and star shaped Rangolies [see Fig. 7.53 (i) and (ii)l by fillingthem with as many equilateral triangles of side 1 cm as you can. Count the number oftriangles in each case. Which has more triangles?4. |
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| 28. |
The sides AB, BC & CA of a triangle ABC have 3, 4 & 5 interior points respectively on them. If thenumber of triangles that can be constructed using these interior points as vertices is k, theequal to5 |
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| 29. |
&, B are the roots of the quadratic polynomial 7 _ (k—6) x + (2k + 1). Find the|value of k, if o + 8 = of = =0 ke ४4. |
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Answer» Like if you find it useful |
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| 30. |
™ o) 2o i र(छ)[4v1dinax3 Lu30N] DYl 12५ 3 ॥- हि:Lo Mate Bl phE ‘DY % kit 3 1w = दहन के & + ;%6Indblnk PIRS] LIS ikl & kikb ke Job 7 kak |
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| 31. |
Type II.10 (0) How many numbers of four digits can be formed with the di1, 2, 4, 5, 7 no digit being repeated ?senn ke formed with the digite |
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Answer» So the total number is9*9*8*7=4536, beginning at1023, 1024, 1025, 1026, etc., and going up to9876. For each of these there is a negative counterpart (-1023, -1024, etc.), for a total of 2*4536=9072possibilities. There are4537 fourdigit-numbers with no digit repeated. |
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| 32. |
हि 7. oA oaR अलग «आल . - जज.| एक शेयर का बाजार मूल्य 200 रूपये है । वह शेयर खरीदते समय 0.3% दलाली दी, तो शेयर काक्रय मूल्य ज्ञात करो ।लि e अर - L ke b SR |
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Answer» Explanation : बाजार मूल्य = 200 0.3% दलाली क्रय मूल्य = 200+200×3/1000 = 200+0.6 = 200.6 Solution : 200.6 If you find this answer helpful then like it. |
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| 33. |
In a circular table cover of radius 42 cm, a design is formed, leavingan equilateral triangle ABC in the middle, as shown in the figure.Find the area of the design. [Use 3 173.]CBSE 2013C]O. |
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| 34. |
On a circular table cover of radius 42 cm, ais formed by a girl leaving an equilateralle ABC in the middle, as shown in thesignfigure. Find the covered area of the design.[Use/3 1.73and π 22 ]CBSE 2013C]7 |
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| 35. |
4. Venu reached Delhi on 6 March and left on 15 April. How lonestay in Delhi? |
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| 36. |
If|28 5H6-2( then write the value8 7 3of x. Delhi 20143x 7 |8 7-2 4 6 4If, then find the value |
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| 37. |
The size of a match box is 4 cm x 2.5 cm X 1.5 cnmwhat is the volume of a packet containing 144match boxes? How many such packets can beplaced in a carton of size 1.5 m × 84 cm× 60 cm? |
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| 38. |
(1) 6079 80(2) 70(4) 10066. Observe the following pattern oftriangles made of match-sticks :ΔΝ ΔΔΔΝ3 5 7 9Number of triangles 1 2 3 4Number of matchsticks 35719How many match-sticks do youthink are required to make 10triangles ?(1) 15(2) 19© 21(4) 25(67. The sum of the digits of a number issubtracted from the number. Theresulting number is |
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Answer» the answer of this Q. is 21 the answers 21 me name Justin the answer is the option is 3) 21 the answer of this question is 21 the answer of the following problem is 21 |
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| 39. |
A match box measures 4cm x 2.5cm x 1.5cm. What will be the volume of packet containing12 such boxes? |
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| 40. |
The measure of a match box is 3 cm Ă 2 cm x 1 cm. Find the volume of a packet of such 12match boxes.I. |
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Answer» Volume of one packet=3*2*1=6cm^Hence 12 boxes =12*6=72cm^3 |
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| 41. |
Number of triangles 1 2 3 4Number of matchsticks 3 5 7 9:12How many match-sticks do youthink are required to make 10triangles? |
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Answer» (9) is a right answers 16 match sticks are required |
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| 42. |
What is the shape of the following.a)Match boxb)brickc)your class roomd)chalk box |
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Answer» cuboid is the shape.The last one might be a cube as well. d)chalk box is the shape of the following |
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| 43. |
8 packets of Matchbox e s contain 640 matchsticks how many match box will contain 800 match stick? |
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Answer» 8 packets have 640 matchsticks So, 1 packet have 640/8 = 80 matchsticks So , 800 packets have 800 × 80 = 6400 matchsticks |
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| 44. |
\left| \begin{array}{cc}{x^{2}-x+1} & {x-1} \\ {x+1} & {x+1}\end{array}\right| |
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| 45. |
x, y \text { and } 2 \text { from } \left[ \begin{array}{cc}{4} & {3} \\ {x} & {5}\end{array}\right]=\left[ \begin{array}{cc}{y} & {z} \\ {1} & {5}\end{array}\right] |
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| 46. |
(a) 51°(c) 390(b)61°(d) 1 19。(x +40°)79。 |
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Answer» 79+x+40=180 linear pair x=180-(79+40) x=180-119 x=61 |
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| 47. |
\begin { equation } \begin{array}{l}{\frac{(-112)}{120}+\frac{105}{130}=?} \\ {\text { (1) } \frac{49}{390} \text { (2) } \frac{-49}{390} \text { (3) } \frac{94}{390} \text { (4) } \frac{-94}{390}}\end{array} \end { equation } |
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Answer» option (2) is correct. |
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| 48. |
झाइ्न (580 - 390 + 5b%) =डा (9) |
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| 49. |
20 \% \text { more than } 390 |
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Answer» 20% of ₹90(20/100) × 90 = ₹18so20% more will be ₹90+₹18 = ₹108 |
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| 50. |
5. 39 % WA ----- s(A) 0.39 (B) 0.039(0) 39 (0) 390 |
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Answer» 3.9 % = 3.9/100 = 0.039 Like my answer if you find it useful! |
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