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1.	Find the value of `\|[5,3],[-7,-4]\|`A. `-1`B. `-41`C. 41D. 1
Answer» Correct Answer - D `\|[5,3],[-7,-4]\|= 5 xx ( - 4) -3 xx ( - 7) = -20 + 21 = 1 `

2.	In right-angled triangle `PQR`, if hypotenuse `PR = 12`, `PQ = 6` then what is the measure of `/_ P` ?A. `30^(@)`B. `60^(@)`C. `90^(@)`D. `45^(@)`
Answer» Correct Answer - B

3.	A die is rolled. What is the probability that the number appearing on the upper surface is less than 4 ?A. `(1)/(6)`B. `(1)/(4)`C. `(1)/(3)`D. `(1)/(2)`
Answer» Correct Answer - D

4.	Find the 18th terms of the A.P. 7, 13,19, 25
Answer» Correct Answer - The 18th term of the given A.P. Is 109 Here, `a= 7,d= 13-7= 6 , t_(18) = ?` `t_(n) = a+ ( n -1) d ` …( Formula ) ` :. T_(18) = 7+ ( 18-1) xx 6` …( Substituting the values ) ` = 7 + 17 xx 6` `= 7 + 102 ` `:. t_(18) = 109`

5.	For a given A.P. A=3.5 , d=0, then `t_(n) = "_ _ _ _ _ _ _ _"`A. `0`B. 3.5C. 103.5D. 104.5
Answer» Correct Answer - B `t_(n) = a+ ( n-1) d = 3.5 + ( n-1) xx x0` `= 3.5 + 0 = 3.5 `

6.	If `Delta ABC ~ Delta PQR` and `4A (Delta ABC) = 25 A (Delta PQR)` then AB : PQ = ?A. `4 : 25`B. `2 : 5`C. `5 : 2`D. `25 : 4`
Answer» Correct Answer - C

7.	Find the side and perimeter of a square whose diagonal is `13 sqrt(2)` cm`.
Answer» Correct Answer - Side of square is 13 cm and perimeter of square is 52 cm. Diagonal of a Square `= 13 sqrt(2) cm` ….(Given) Side of square `= (1)/(sqrt(2)) xx` diagonal `= (1)/(sqrt(2)) xx 13 sqrt(2)` = 13 cm Perimeter of square = `4 xx` side `= 4 xx 13` = 52 cm

8.	A bag contains in all 180 balls. Some of them are white, some are blue and some are red. The number of white balls is 12 times the number of blum balls. The number of red balls is less than the number of white balls but more than the number of blub balls. IF one ball is selected at random from the bag, what is the probability that it is red ?
Answer» Correct Answer - The probability is `( 7)/( 25)` Let the number of blue balls of x. Then the number of white balls is 11x …...(From the given conditon ) Total number of balls is 50. …(Given ) `:. ` the number of red balls `= ( 50 - 12x ) ` As per the given condition , `11x gt ( 50 - 12x ) gt x ` (a) `11x gt 50 - 12x " " :. 11x + 12x gt 50" " 23x gt 50` `:. x gt ( 50)/( 23)` i.e. ` x gt 2 (4)/(23)` ...(1) (b) `50 - 12x gt x " " :. 50 gt x + 12 x " " 50 gt 13 x " "` i.e. `13x lt 50` `:. x lt ( 50)/( 13) ` `" ":. x lt 3 (11)/( 13)` ...(2) From (1) and (2) , ` x gt 2(4)/(23)` but less than `3 (11)/(13)` `:. x = 3 ` The number of red balls `= 50 - 12x = 50 -12 xx 3 ` `= 50 - 36 = 14 ` `:. n (R ) = 14 ` The probability of taking at randoma red ball `=P (R )` `P ( R ) = ( n (R ))/( n (S)) = (14)/( 50)` `:. P (R ) = ( 7)/( 25)`

9.	Obtain the sum of the first 56 terms of an A.P. Whose 18th and 39th terms are 52 and 148 respectively. (1) Usint `t_(18)` and `t_(39)` find two simultaneous equations in variables a and d. (2) Using these equations, find `S_(56)`
Answer» Correct Answer - The sum of the first 56 terms is 5600 Let the first term of A.P. be a and the common difference d. `t_(n) = a+ ( n -1) d ` …(Formula) `:. T_(18) = a + ( 18-1) d ` `:. 52 = a+ 17d` ....(1) and `t_(39) a + ( 39-1)d` `:. 148 = a + 38d` ...(2) Adding equations (1) and (2) `52= a + 17d ` ...(1) `148 = a + 38 d ` ...(2) `bar( 200 = 2a + 55d)` ....(3) We have to find `S_(56 )` `S_(n) = ( n )/( 2) [2a + (n -1) d]` `:. S_(56) = ( 56)/( 2) [2a + ( 56-1) d ]` `= 28 (2a+ 55d]` `= 28( 200) ` ...[From (30] ` = 5600`

10.	What is the value of the discriminant for the quadratic equation `6x^(2) - 7x + 2 = 0` ?A. 1B. -1C. 3D. -3
Answer» Correct Answer - A

11.	Find the mode from the following information `:` `L= 10 , h = 2,f_(0)=58 , f_(1) = 70 , f_(2) = 42`
Answer» Correct Answer - Mode = 10.6 Mode `= L + [ ( f_(1)- f_(0))/(2f_(1) - f_(0)-f_(2))] xx h ` `= 1 0 + [ ( 70 - 58)/( 2(70) - 58 - 42 ) ] xx2 ` ` = 10 + (( 12)/( 140 - 100)) xx 2` `= 10 + ( (12)/( 40)) xx 2 ` ltbr. ` = 10 + 0.3 xx 2 ` `= 10 + 0.6 = 10.6 `

12.	Fine `bar( u)` , if `Sigma f_(i) u_(i ) 252` and `Sigmaf_(i) = 420`
Answer» `bar( u ) = ( Sigma f_(i) u _(i))/( Sigmaf_(i))` ` = ( 252 )/( 420 ) = 0.6 ` `:. bar( u ) = 0.6 `

13.	A two-digit number is to be formed from the digits 0,1,2,3. Repetition of the digits is not allowed. Complete the following activity to find the probability that th enumber so formed is (1) a prime number (2) a multiple of 4. The sample space `S = { 10,12,13,20,21,23,30,31,32}` `:. n (S) = 9` ( 1) Let A be the event that the number so formed is a prime number. Then `A = { " "}, n (A) = square ` `P(A) = ( n(A))/( n (S)) =" ______"= "______"` (2) Let B be the event that the number so formed is a multiple of 4. Then B `={" "}, n (B) = square ` `P (B) = ( n (B))/( n (S))="______"=" ________"`
Answer» Activity `:` The sample space `S = { 10,12,13,30,21,23,30,31,32 }` `:. n(S) = 9 ` (1) `A = { 13,23,31}, n(A) = 3 ` `P(A) = ( n(A)) /(n(S))= ( 3)/( 4) = (1)/(3)` (2) ` B = {12,20,32}, n (B) = 3 ` ` P(B) = ( n (B))/( n (S)) = ( 3)/( 9) = (1)/(3)`

14.	If `x= - 5 ` is a root of the quadratic equation `kx^(2) + 13x - 10 = 0 `, then complete the following activity to find the value of k .
Answer» Activity `:` One root of the quadratic equation `kx^(2) + 3x - 10 = 0 ` is -5. Substitute `x= - 5 ` in the equation . `:. K ( - 5)^(2) + 13( - 5) - 10 = 0 ` `:. 25k - 65 - 10 = 0 ` `:. 25 k = 75 ` `:. K = ( 75)/( 25) = 3 ` The value of k is 3.

15.	A trader purchased cosmetics worth Rupee 8000 and sold it to a customer for Rupee 10,000. The rate of GST is 18%. Complete the following activity to calculate CGST and SGST paid by the trader.
Answer» Activity `:` Input tax by the trader `= 18 % ` of Rupee 8000 `= ( 18)/( 100) xx 8000 ` `= Rupee 1440` `:. ` ITC for the trader `= Rupee1440 ` Output tax ( Tax collected on sale ) `= 18%` of Rupee 10000 `= ( 18)/( 100) xx 10000` ` = Rupee 1800` `:. ` GST to be paid by the trader `= `Output tax - ITC `= Rupee 360` CGST = SGST `= ( 1)/(2) xx`GST `= Rupee 180`