Explore topic-wise InterviewSolutions in .

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.

1.

One tank can be filled up by two taps in 6 hours. The smaller tap alone takes 5 hours more than the bigger tap alone. Find the time required by each tap to fill the tank separately.

Answer» Let the bigger tap alone take `x` hours to fill the tank.
Then the smaller tap alone takes `(x+5)` hours to fill the tank.
The bigger tap fills `1/x` part of the tank in 1 hour and the smalller tap fills `1/(x+5)` part of the tank in 1 hour.
`:.` both the taps together fill `(1/x+1/(x+5))` part of the tank in 1 hour. Both the taps together fill the tank in 6 hours. (Given)
`:.` Both the taps together fill `1/6` part of the tank in 1 hour.
`:.1/x+1/(x+5)=1/6`
`:.(x+5+x)/(x(x+5))=1/6`
`:.(2x+5)/(x^(2)+5x)=1/6`
`:.6(2x+5)=x^(2)+5x`
`:.12x+30=x^(2)+5x`
`:.x^(2)+5x-12x-30=0`
`:.x^(2)-7x-30=0`
`:.x^(2)-10x+3x-30=0`
`:.x(x-10)+3(x-10)=0`
`:.(x-10)(x+3)=0`
`:.x-10=0` or `x+3=0`
`:.x=10` or `x=-3`
But the time cannot be negative.
`:.x=-3` is unacceptable.
`:.x=10` and `x+5=10+5=15`.
Ans. The bigger tap alone fills the tank in 10 hours and the smaller tap alone in 15 hours.
Discussion
2.

Solve : `(x^(2002)+10x^(2001))/(10x^(2000))=957.9`

Answer» `(x^(2002)+10x^(2001))/(10x^(2000))=957.9`
Multiplying both the sides by 10
`(x^(2002)+10x^(2001))/(x^(2000))=9579`
`:.(x^(2002))/(x^(2000))+(10x^(2001))/(x^(2000))=9579`
`:.x^(2002-2000)+10x^(2001-2000)=9579` ……… `((a^(m))/(a^(n))=a^(m-n))`
`:.x^(2)+10x-9579=0`
`:.x^(2)+10x+25-25-9579=0` .[Add 25 to make `x^(2)+10x` a perfect square]
`:.lx^(2)+10x+25-9604=0`
`:.(x+5)^(2)-(98)^(2)=0`
`:.(x+5+98)`
`(x+5-98)=0`
`:.(x+103)(x-93)=0`
`:.x+103=0` or `x-93=0`
`:.x=-103` or `x=93`
`93, -103` are the roots of the give quadratic equation.
Discussion
3.

The sum of x - coordinate of the vertices of the Delta is 18 and that of y - coorcinates is 24 then the coorinates of its centroid are `…….`A. (6,8)B. (8,6)C. (9,12)D. (12,9)

Answer» Correct Answer - A
Discussion
4.

Find the values of `a` and `b` for which the simultaneous equations `x+2y=1` and `(a-b)x+(a+b)y=a+b-2` have infinitely many solutions.

Answer» The condition for simultaneous equations having infinitely many solution is `(a_(1))/(a_(2))=(b_(1))/(b_(2))=(c_(1))/(c_(2))` ……………..1
For `x+2y=1,a_(1)=1, b_(1)=2, c_(1)=1`
For `(a-b)x+(a+b)y=a+b=2`
`a_(2)=a-b,b_(2)a+b,c_(2)=a+b-2`
Substituting these values in (1)
`1/(a-b)=2/(a+b)=1/(a+b-2)`
Now `1/(a-b)=1/(a+b-2)`
`:.a+b-2=a-b :. b-2=-b :. b+b=2`
`:.2b-2 :. b=2/2 :.b=1`
Substituting `b=1`
`1/(a-b)=2/(a+b)`
`:.1/(a-1)=2/(a+1) :. a+1=2(a-1)`
`:. a+1=2a-2 :. a-2a=-2-1`
`-a=-3 :.a=(-3)/(-1):.a=3`
Ans The values of a and b are 3 and 1 respectively.
Discussion
5.

Seg AB is a diameter of a circle with cente P , seg AC is a chord . A through P and parallel to seg AC intersects the tangent drawn at C is D . Prove that line DB is a tangent to the circle .

Answer» In `Delta PAC` ,
seg `PA cong seg PC " "` …(Radii of the same circle )
` :. Angle PAC cong angle PCA " "` …(Isosceles Delta theorem ) …(1)
seg AC `||` seg PD nd PA is the transversal ,
` angle PAC cong angle BPD " " ` ...(Corresponding angles ) ... (2)
seg AC `||` seg PD and PC is the transversal ,
` angle PCA cong angle CPD " "` ...(ALternate angles ) ...(3)
` :. angle BPD cong angle CPD " "` ... [ From (1) , (2) and (3) ] ...(4)
In `Delta BPD and Delta CPD,`
seg `BP cong seg CP" "` ...(Radii of the same circle )
` angle BPD cong angle CPD " " ` ...[ From (4)]
seg `PD cong seg PD " "` ...(Common side )
` :. Delta BPD cong Delta CPD " "` (SAS test)
` :. angle PBD cong angle PCD " " ` ... (c.a.c.t.)
` angle PCD = 90^(@) " "` ... (Tangent is perpendicular to radius )
` :. angle PBD = 90^(@)`
` :. ` line DB is tangent to the circle at point B
... ( Converse of tangent theorem )
Discussion
6.

In ` Delta ABC, angle ACB = 90^(@) ," seg "CD bot ` side AB and seg CE is angle bisector of `angle ACB` Prove : ` (AD)/(BD) = (AE^(2))/(BE^(2))`

Answer» In ` Delta ACB`,
ray CE bisects ` angle ACB`
` :. ` by theorem of angle bisector of a Delta
` (AC)/(CB)= (AE)/(EB)`
Squaring both the sides , we get ,
`(AC^(2))/(CB^(2)) = (AE^(2))/(EB^(2))`
In ` Delta ACB, angle ACB = 90^(@)`
seg CD `bot` hypotenuse AB
` Delta ACB ~ Delta ADC ~ Delta CDB`
... (Similarity of right angled Delta ) ... (2)
` Delta ACB ~ Delta ADC " " ` ... [ From (2)]
` (AC)/(AD) = (AB)/(AC) " "` (Corresponding sides of similar triangfle )
` :. AC^(2) = AB xx AD " "` ...(3)
also, ` Delta ACB ~ Delta CDB " "` ... [ From (2)]
` :. (AB)/(BC) = (BC)/(BD)" "` ( Correesponding sides of similar Delta )
` :. BC^(2) = AB xx BD " "` ... (4)
Substituting the values of (3) and (4) in (1) , we get
` (ABxx AD)/(ABxxBD) = (AE^(2))/(EB^(2))`
` :. (AD)/(BD) = (AE^(2))/(EB^(2))`
Discussion
7.

Height of a cylinfrical barrel is 50 cm and radius of its base is 20 cm . Anurag started to fill the barrel with water, when it was empty by cylindrical mug . The diameter and height of the mug are 10 cm and 15 cm respectively. Find the no. of mugs required for the barrel to overflow ?

Answer» `("For barrel")/("height (h)")` = 50 cm , radius (r ) = 20 cm
Volume of the barrel ` = pi r^(2)h`
` = pi xx (20)^(2) xx 50`
` = 20000 pi " cm"^(3)`
`("For mug")/("diameter") = 10 cm `
` :. " its radius " (r_(1)) = 5 cm ," height "(h_(1)) = 15 cm`
Volume of mug ` = pi e_(1)^(2) h_(1)`
` = 375 pi " cm"^(3)`
`("Volume of barrel")/("Volume of mug") = (2000 pi)/(375 pi) = 160/3 = 53 /3`
` :. ` When 54 th mug is poured in the barrel it will overflow.
Discussion
8.

How many terms of the A.P. 16,14,12……….. are needed to given the sum `60` ? Explain why do we get two answers.

Answer» Here `S_(n)=60, a=16, d=-2, n=?`
`S_(hn)=n/2[(2a+(n-1)d]`………..(Formula)
`:.60=n/2[2xx16+(n-1)xx(-2)]` ….[Substitutig the values]
`:.60=n/2(32-2n+2)`
`:.60=n/2(34-2n)`
`:.60=n/2xx2(17-n)`
`:.60=n(17-n)`
`:.60=17n-n^(2)`
`:.n^(2)-17n+60=0`
`:.n^(2)-5n-12n+60=0`
`:.n(n-5)-12(n-5)=0`
`:.(n-5)(n-12)=0`
`:.n-5=0` or `n-12=0`
`:.n=5` or `n=12`
The required terms are 5 or 12.
Explanation:
THe common difference d of the A.P. is `-2`
`:.` The terms of the A.P. are in descending order.
Taking `n=5` the first 5 terms are 16,14,12,10,8. The sum is 60.
Taking `n=12`, the last 7 terms `(12-5)` are `6,4,2,0,-2,-4,-5`
The sum of these seven terms is 0.
`:.` The sum of first 12 terms is also 60.
The sum of the first terms `=` the sum of the first twelve terms.
`:.` we get two answers.
Ans. 5 terms or 12 terms.
Discussion
9.

(A) Choose the correct alternative : O is a centre of a circle, Tangents TP and TQ of the circles itersect at point T in the exterior of the circle. Points P and Q lie on the circle . If ` anglePOQ = 120^(@) ` then ` angle PTQ = ?`A. `120^(@)`B. `30^(@)`C. `60^(@)`D. `90^(@)`

Answer» Correct Answer - C
Discussion
10.

In a right Delta, the sum of the squares of sides containing right angle is 225 , then what is the length of its hypotenuseA. 14B. 13C. 12D. 15

Answer» Correct Answer - D
Discussion
11.

Find the common difference of an AP, whose first term is 100 and the sum of whose first six terms is five times the sum of the next six terms.

Answer» Here `a=100`. Let the common diference be `d`.
The sum of the first six terms `t_(1)+t_(2)+…………+t_(6)`
`=a+(a+d)+...+(a+5d)`.
The sum of the six terms `=t_(7)+t_(8)+….+t_(12)`
`=(a+6d)+(a+8d)+….+(a+11d)`
From the given condition,
`(t_(1)+t_(2)+t_(3)+.........+t_(6))=5(t_(7)+t_(8)+.....+t_(12))`
`[(a+(a+d)+.............+a+5d)]=5[(a+6d)+(a+7d)+...........+(a+11d)]`
`:.[6a+(1+2+....+5)d]=[(6a+6+7+.............+11)d]`.............1
Now `1+2+.......+5=n/2(t_(1)+t_(n))=5/2(1+5)=5/2(1+5)=5/2xx6=15`.........2
and `6+7+............+11=n/2(t_(1)+t_(n))=6/2(6+11)=3xx17=51`.............3
From 1, 2 and 3
`6a+15d=(6a+51d)`
`:.6a+15d=30a+255d`
`:.30a-6a=-255d+15d`
`:.24a=-240d`
`:.a=-10d`........(Dividing both the sides by 24)
`:.100=-10d` ........(Substituting `a=100`) .........(Given)
`:.d=-10`
Now `a=-t_(1)=100,d=-10`
`:.t_(2)+t_(1)+d=100-10=90,t_(3)=t_(2)+d=90-10=80`
Ans. The required A.P. is 100, 90, 70............
Discussion
12.

Point P divides the lne segment joning the points A (2,1) and `B(5,-8)` such that `(AP)/(AB)=1/3` . If P lies on the line ` 2x - y + k = 0 ,` find the value of k

Answer» Correct Answer - Values of k is -8
`(AP)/(AB) = 1/3`
` :. 3 AP = AB`
` 3AP - AP = PB " " …(A-P-B)`
` :. 2AP = PB`
` (AP)/(PB) = 1/2`
` :. ` P divides the segment AB in the ratio AP : PB
i.e 1:2
A (2,1) and B `(5,-8)`
Let `A(x_(1),y_(1))and B (x_(2),y_(2)) and P (x,y)`
` :. x_(1) = 2 , y_(1) = 1 , x_(2) = 5 and y_(2) = -8`
` m : n = 1 :2`
By section formula ,
` x = (mx_(2)+nx_(1))/ (m+n) " " and y = (my_(2)+ny_(1))/(m+n)`
` :. x = (1(5)+2(2))/(1+2) " " :. y = (1(-8)+2(1))/(1+2)`
` :. x = 9/3 " " :. y = (-6)/3 `
` :. x = 3 " " :. y = -2`
` :. ` the coordinates of point P will be `(3,-2)`
`P (3,-2) " lies on the line " 2x - y + k = 0`
` :. ` its coordinates satisfy the equation of the line substituting x = 3
and ` y = -2 "in " 2x - y +k = 0,` we get
` 2(3) -(-2) +k =0`
` :. 6+ 2+ k =0`
` :. 8+k = 0`
` :. k = -8`
Discussion