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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.
| 852. |
In the figure, pointP is the centre of the circle and line AB is the tangent to thecircle at the pointT. The radius of the circle is 6 cm . Find PB, if /_ TPB= 60^(@) |
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Answer» `5sqrt(5)cm ` |
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| 853. |
Prove that sqrt(5) is irrational. |
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| 854. |
Solve the given pair of equations using subtitution method. 2x-y=5 3x+2y=11 |
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| 857. |
If sin(A-B)=(1)/(2),cos(A+B)=(1)/(2),0^(@)ltA+Ble90^(@), A gt B,then find A and B. |
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| 858. |
Angle C of DeltaACB in right angle, AB = 29 units, BC = 2 units and angleABC=theta. The value of cos theta is : |
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Answer» `(21)/(29)` |
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| 859. |
A self help group wants to manufacture joker's caps (conical caps) of 3 cm radius and 4 cm height. If the available colour paper sheet is 1000 cm^(2), then how many caps can be manufactured from that paper sheet ? |
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| 860. |
Prove that : (cot A - cos A)/(cot A + cos A) = ("cosec" A - 1)/("cosec" A + 1). |
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| 861. |
Jawaharlal Nehru Stadium is a multi-purpose sports stadium and a very popular sports stadium of Delhi. It has a capacity to seat 60,000 people. It is the third largest multi-purpose stadium in India and owned by the Indian Olympic Association. In 2010, the Jawaharlal Nehru Stadium was the main stadium for XIX Commonwealth Games, a major sporting Jawaharlal Nehru Stadium is conducting the annual sports competition soon. The curator of the stadium is tasked to figuring out the dimensions for carving out some areas allotted for a hockey court and a shooting range, as shown in the figure below. The shapes of the hockey court and the shooting range are square and triangle respectively. Both of the courts have a common edge that touches the centre of stadium. The construction of the shooting range is such that the angle to centre is 90^(@). The radius of the stadium is 180 metres. On the basis of the above information, answer any four of the following questions: What is the area allotted to shooting range ? |
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Answer» `12,600" m"^(2)` NowAO = OB = radius of CIRCLE = 200 m Thus area of `DeltaAOB`, `=(1)/(2)xxOAxxOB` `=(1)/(2)xx200xx200=20,000" m"^(2)` Thus ( c ) is correct OPTION. |
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| 862. |
A pack of 52 cards is divided into two sets, one with Reds and the other with blacks. The face cards are removed from both the sets. One card in picked from each set. What is probability that the numbers on the pair of cards give a product of 4 ? |
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| 863. |
Rupesh invests Rs. 45,000 partly in Rs. 150 shares at Rs. 180 for 8% and partly in Rs. 75 shares at Rs. 135 for 12%. If the annual income from the investments are in the ratio 2 : 3 respectively, then find the investments made by Rupesh in the two types of sphares. |
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| 864. |
Find the 12th term from the end of AP 8, 10, 12, ..... , 130 |
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| 865. |
From a point 50 m above the ground, the angle of elevation of a cloud is 30^(@) and the angle of depression of its reflection is 60^(@). Find the height of the cloud above the ground. |
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| 866. |
Draw a right triangle in which the sides (other than hypotenuse) are of lengths 4 cm and 3 cm. Then construct another triangle whose sides are 5/3 times the corresponding sides of the given triangle. |
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Answer» Solution :Steps of Construction : 1. DRAW a line segment BC = 4 cm. 2. From B. draw a line AB = 3 cm which making right angle at B. 3. Join `AC. DeltaABC` is the given right triangle. 4. From B, draw an acute `angleCBY` downwards. 5. On BY, take five POINTS `B_(1),B_(2),B_(3),B_(4)andB_(5)` such that`BB_(1)=B_(1)B_(2)=B_(2)B_(3)=B_(3)B_(4)=B_(4)B_(5).` 6. Join `B_(3)C.` 7. From point `B_(5),` draw `B_(5)C'"||"B_(3)C` producijng BC to C'. 8. Form point C', draw `C'A"||"CA` PRODUCING BA to A' Hence, `DeltaA'BC'` is the required triangle. Justification : By construction, `B_(5)C'"||"B_(3)C` `therefore""(BC)/(C C')-3/2` `Now,""(BC')/(BC)=(BC+C C')/(BC)=1+(C C')/(BC)=1+2/3=5/3` `Also,""C'A'"||"CA` `therefore""DeltaABC~DeltaA'BC'` `and""(A'B)/(AB)=(BC')/(BC)=(A'C')/(CA)=5/3.` 
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| 868. |
Draw a circle of radius 3 cm. Take a point P on it. Without using the centre of the circle, draw a tangent to the circle at point P. |
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Answer» Solution :Steps of Construction : 1. Draw any chord PQ through the GIVEN point P on the CIRCLE. 2. Take a point R on the circle and join P and Q to a point R. 3. CONSTRUCT `angleQPY=anglePRQ` and on the opposite side of the chrod PQ. 4. Produce YP to X to get YPX as the required tangent . 
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| 869. |
A(-2, 4) and B(-4, 2) are reflected in the y-axis. If A and B' are images of A and B respectively. Assigna special name to quadrilateral AA' B'B |
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| 870. |
On which day of the week was birthday of Sahil? I. Sahil celebrated his birthday the very next day on which Arun celebrated his birth day. II. The sister of Sahil was born on the third day of the week and two days after Sahil was born. |
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Answer» if the data in statement I alone are sufficient to answer the question, while the data in statement II alone are not sufficient to answer the question, |
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| 871. |
Jawaharlal Nehru Stadium is a multi-purpose sports stadium and a very popular sports stadium of Delhi. It has a capacity to seat 60,000 people. It is the third largest multi-purpose stadium in India and owned by the Indian Olympic Association. In 2010, the Jawaharlal Nehru Stadium was the main stadium for XIX Commonwealth Games, a major sporting Jawaharlal Nehru Stadium is conducting the annual sports competition soon. The curator of the stadium is tasked to figuring out the dimensions for carving out some areas allotted for a hockey court and a shooting range, as shown in the figure below. The shapes of the hockey court and the shooting range are square and triangle respectively. Both of the courts have a common edge that touches the centre of stadium. The construction of the shooting range is such that the angle to centre is 90^(@). The radius of the stadium is 180 metres. On the basis of the above information, answer any four of the following questions: What is the area allotted to hockey court ? |
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Answer» `12, 600" m"^(2)` Now `""a^(2)+a^(2)=(200)^(2)` `2a^(2)=200xx200` `a=sqrt(100xx100xx2)=100sqrt(2)` cm Area of square `""a^(2)=(100sqrt(2))^(2)=20,000" m"^(2)` Area of hockey court is equal to area of shooting court. THUS ( c ) is correct option. |
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| 872. |
Calculate the angles x,y and z if (x)/(3) = (y)/(4) =(z)/(5)(##SEL_RKB_ICSE_MAT_X_C17_E02_042_Q01.png" width="80%"> |
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| 873. |
Find the missing values in the following frequency distribution. |
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| 874. |
Check whether the given sequences form a G.P. or not : (1)/(8),(1)/(24),(1)/(72),(1)/(216), . . . . . . .. . |
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| 875. |
If a polygon of ‘n’ sides has (1)/(2) n (n-3) diagonals. How many sides are there in a polygon with 65 diagonals? Is there a polygon with 50 diagonals? |
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| 876. |
Find the values of m and n so that x-1 and x + 2 both are factors of x^3+(3m+1)x^2+nx-18. |
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| 877. |
A women selfhelp group (DWACRA) is supplied a rectangular solid (cuboid shape) of wax block with dimensions 66cm , 42 cm, 21 cm , to prepare cylindrical candles each 4.2 cm in diameter and 2.8 cm of height . Find the number of candles prepared using this so solid . |
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| 878. |
Find the median of the following frequency distibution : |
Answer» Solution :First of all we will PREPARE an EXCLUSIVE series from the GIVEN (inclusive) series. To make it we subtract and add the same number, i.e.,`(70-68)/(2)=1` in each class. So, new table is :  Here, `(N)/(2)=(98)/(2)=49` So, cumulative frequency just greater than or equal to 49 is 70 and its corresponding class is the medianl class. `:. """Median"=l_(1)+((N)/(2)-C)/(f)xxi=89+(49-40)/(30)xx10+3=92` |
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| 879. |
Rita went to a shop to purchase an article A with MRP = rupes 850 and rate of GST = 12% How muchwill Rita pay for this article? If instead of article A , Rita purchases some other article B withMRP = rupes 1,200 and rate of GST = 18%find now much extrea money will she pay to the shopkeper? |
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| 880. |
If A is a 2times3 matrix and B is 3times4 matrix, how many columns does AB have |
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Answer» `3` |
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| 881. |
A solid is in the form of a right circular with a hemisphere at one end and a cone at the other end. The radius of the common base is 8 cmand the heights of the cylindrical and conical portions are 10 cm and 6 cm respectivly. Find the total surface area of the solid [use pi=3.14] |
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| 882. |
Which one of the following is always found in 'Bravery'? |
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Answer» Courage |
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| 883. |
A metallic sphere of radius 4.2 cm is melted and recast into the shape of a cylinder of radius 6cm. Find the height of the cylinder. |
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| 884. |
Find the area of the shaded region. In the given figure common between the two quadrants of circles of radius 8cm each. OR In fig. ABC is a quadrant of a circle of radius 14cm and a semicircle is drawn with BC as diameter. Find the area of the shaded region. |
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Answer» OR 98 sq.cms. |
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| 885. |
The total area of a solid metallic sphere is 1256 cm^(2). It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate : the number of cones recast. |
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| 886. |
The total area of a solid metallic sphere is 1256 cm^(2). It is melted and recast into solid right circular cones of radius 2.5 cm and height 8 cm. Calculate : the radius of the solid sphere, and the no. of cones cast (π=3.14). |
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| 887. |
10 students of class-X took part in a mathematics quiz. If the number of girls is 4 more than the number of noys then, find the number of boys and the number of girls who took part in the quiz. |
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| 888. |
Determine the A.P whose 3rd term is 16 and the 7th term exceeds the 5th term by 12. |
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| 889. |
Construct an angle ABC = 45°. Mark a point P on BC such that BP = 4-8 cm. Construct a circle to touch AB at B and also to pass through P. |
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| 890. |
If x gt 0, then the minimum value of 11/3+5(x-7/2)^2 is……… |
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Answer» `3/11` |
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| 891. |
The whole surface area of a hollow hemisphere will be |
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Answer» 308 sq-cm |
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| 892. |
Solve following pair of equations by using the substitution method x - y = 8 3x + 3y = 16 |
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| 893. |
In a certain code, 15789 is written as XTZAL and 2346 is written as NPSU. How is 23549 written in that code? |
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Answer» NPTUL |
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| 894. |
Draw a parallelogram ABCD in which BC=5cm and angleABC=60^(@), divide it into triangles BCD and ABD by the diagonal BD. Construct the triangle BD'C' similar to DeltaBDCwith scale factor (4)/(3). Draw the line segment D'A' parallel of DA, where A' lies on extended side BA. Is A'BC'D' a parallelogram ? |
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Answer» Solution :Steps of construction 1. Draw a line segment AB= 3 cm 2. Now,draw a ray BY making an acute `angleABY=60^(@)`. 3.With B as centre and radius equal to 5 cm draw an arc cut the POINT C on BY. 4. Againdraw a ray AZ making an acute `angleZAX'=60^(@)`,[`:. BY ||AZ,:.angleYBX'=ZAX'=60^(@)`] 5. With A as centre and radius equal to 5 cm draw an arc cu the point D on AZ .  6. Now, join CD and finally make a parallelogram ABCD. 7. Join BD, which is a diagonal of parallelogram ABCD. 8. From B draw any ray BX downwards making an acute `angleCBX`. 9. Locate 4 points `B_(1),B_(2),B_(3),B_(4)` on BX, such that `BB_(1)=B_(1)B_(2)=B_(2)B_(3)=B_(3)B_(4)`. 10.Join `B_(4)C` and from `B_(3)C` draw a line `B_(4)C'||B_(3)C` intersecting the extended line segment BC at C'. 11. From C' draw C'D'||CD intersecting the extended line segment BD at D'. Then, `DeltaD'BC'` is the required triangle whose sides are `(4)/(3)` of the corresponding sides of `DeltaDBC`. 12. Now draw a line segment D'A' parallel to DA, where A' lies on extended SIDE BA i.e., a ray BX'. 13. Finally, we observer that A'BC'D' is a parallelogram in which A'D'=6.5 cm A'B'=4 cm and `angleA'BD'=60^(@)` divide it into triangle BC'D' and A'BD' by the diagonal BD'. |
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| 895. |
Solve: costheta cos2theta cos3theta =(1)/(4) |
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| 896. |
Find the ratio in which point T(-1, 6) divides the line segment joining the points P(-3, 10) and Q(6, -8) |
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| 897. |
Find the probabilty of drawing an ace or a jack from a pack of 52 cards. |
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| 898. |
What should be subtracted from 3x^(3)-8x^(2)+4x - 3, so that the resulting expression has (x+2) as a factor |
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| 899. |
Use the Remainder Theorem to factorise the following expression 2x^(3)+x^(2)-13x+6. |
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