Explore topic-wise InterviewSolutions in Current Affairs.

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.	IQ of a person is given by the formula `I Q =(M A)/(C A)xx100` where MA is mental age and CA is chronological age. If `80lt=I Qlt=140` for a group of 12 years old children, find the range of their mental age.
Answer» `80 <= IQ <= 140` `80 <= (MA)/(CA)100 <= 140` `80 <= (MA)/12100 <= 140` `(8012)/100 <= MA <= 14012/100` `48/5 <= MA <= 84/5` `9.6 <= MA <= 16.8` `:. MA in [9.6, 16.8]` answer

2.	The water acidity in a pool is considered normal when the average pH reading of three daily measurements is between ` 8.2 and 8.5` If the first two pH readings are `8.48 and 8.35`, then find the range of pH value for the third reading that will result in the acidity level being normal.
Answer» Given, first pH value = `8.48` and second pH value = `8.35` Let third pH value be x. Since, it is given that average pH value lies between ` 8.2 and 8.5` `:. 8.2 lt (8.48+8.35+x)/3 lt 8.5` `rArr 8.2 lt (16.83+x)/3 lt 8.5` `rArr 8 xx 8.2 lt 16.83 +x lt 8.5 xx 3` `rArr 24.6 lt 16.83 + x lt 25.5` `rArr 24.6 - 16.83 lt x lt 25.5 - 16. 83` `rArr 7.77 lt x lt 8.67` Thus, third pH value lies between ` 7.77 and 8.67` .

3.	Find the linear inequalities for which the shaded region in the given figure is the solution set.
Answer» Consider the line ` 3x+ 2y=48`, we observe that the shaded and the origin are on the same side of the line ` 3x+2y=48 and (0,0)` satisfy the linear constraint ` 3x+2y le 48`. So, we must have one inequation as ` 3x+2 y le 48`. Now, condider the line x+ y = 20 . We find that the shaded region and the origin are on the same side of the line x + y = 20 and (0, 0) satisfy the constraints ` x+y le 20`. So, the second inequation is ` x + y le 20`. We also notice that the shaded region is above X- axis and is on the right side of y-axis, so we must have ` x ge 0, y ge 0`. Thus, the linear inequations corresponding to the given solution set are `3x + 2 y le 48, x + y le 20 and x ge 0, y ge 0`.

4.	`4/(x-1)le3le6/(x+1)(xgt0)`
Answer» Consider first two inequallities, `4/(x+1)le3` `rArr 4le3(x+1)` `rArr 4le 3x+3` `rArr 4-3le3X` [subtracting 3 on both sides] `rArr 1 le 3x` `:. Xge1/3` …(i) and consider last two inequalities, `3le6/(x+1)` `rArr 3(x+1)le6` `rArr 3x+3le 6` `rArr 3xle6-3` [subtracting 3 to both sides] `rArr 3xle3` [dividing by 3] `:. xle1` ....(ii) From Eqs. (i) and (ii), `x in[1/3,1]` `1/3lexle1`

5.	`-5 le (2-3x)/4 le 9`
Answer» We have, ` -5 le (2-3x)/4` `rArr -20 le 2-3x` [multiplying by 4 on both sides] `rArr 3x le 2+20` `rArr 3x le 22` `rArr x le (22)/3` and `(2-3x)/4 le 9` `rArr 2-3x le 36` `rArr 2- 3x le 36-2` `rArr -3x le 34` `rArr 3x ge - 34` `rArr x ge - (34)/3` `rArr -(34)/3 le x le (22)/3` `rArr x ni [(-34)/3,(22/3]`

6.	A company manufactures cassettes. Its cost and revenue functions are C(x)=26000+30x and R(x)= 43x, respectively, where x is the number of cassettes produced and sold in a week. How many cassettes must be sold by the company to realise some profit ?
Answer» Cost function, ` C(x)=26000+30x` and revenue function, ` R(x)=43x` For profit, ` R(x) lt C(x)` `rArr 26000+30x lt 43x` `rArr 30x-43x lt - 26000` `rArr -13x mlt - 26000` `rArr 13x gt 26000` ` rArr x gt (26000)/13` `:. x gt 2000` Hence, more than 2000 cassettes must be produced to get profit.

7.	The inequality representing the following graph is A. `x in (-infty,7/2)`B. `x in (-infty,7/2]`C. `x in [7/2, -infty)`D. `x in (7/2, infty)`
Answer» Correct Answer - A The given graph represents all the values less than `7/2` on the real line. `rArr x in (-infty, 7/2)`

8.	The inequality representing the following graph is A. `\|x\| lt 5`B. `\|x\| le 5`C. ` \|x\| gt 5`D. `\|x\| ge 5`
Answer» Correct Answer - A The given graph represent ` x gt -5 and x lt 5`. On combining these two result, we get `\|x\| lt 5`. Solution of a linear inequality in variable x is represented on number line in following questions.

9.	A solution is to be kept between `68oF`and `77oF`. What is the range in temperature in degree Celsius (C) if the Celsius / Fahrenheit (F) conversion formula is given by`F=9/5C+32 ?`
Answer» solution : `68^@ F - 77^@F` `68^@F < T < 77^@F` `68^@F < 9/5C+32 < 77^@F ` `68-32 < 9/5C <77-32` `36<9/5C <45` `= 20 < C < 25` Solution is in between `20^@C & 25^@ C` answer

10.	Let x and b are real numbers. If ` b gt 0 and \|x\| gtb`, thenA. `x in (-b, infty)`B. ` x in [-infty, b)`C. `x in (-b, b)`D. `x in (- infty, -b) cup(b, infty)`
Answer» Correct Answer - D Given, `\|x\|gt b and b gt 0` `rArr x lt - b or x gt b` ` rArr x in (-infty,-b) cup (b, infty)`

11.	`4x+3 ge 2x+ 17, 3x-5 lt -2`
Answer» We have, `4x+3 ge 2 x+17` `rArr 4x-2x ge 17-3 rArr 2x ge 14` `rArr x ge (14)/2` `rArr x ge 7 ` …(i) Also, we have ` 3x-5 lt - 2` `rArr 3x lt -2 +5 rArr 3x lt 3` `rArr x lt 1 ` …(ii) On combining Eqs. (i) and (ii), we see thet solution is not possible because nothing is common between these two solutions. (i.e.,`x lt 1,x ge 7`).

12.	`sin25^@cos115^@=1/2(sin40^@-1)`
Answer» LHS `sin25*cos115` `sin25(-sin25)` `-sin^2 25` `-(1-cos50)/2` `1/2(cos50-1)` `1/2(sin40-1)` RHS.

13.	The centres of a set of circles, each of radius 3, lie on the circle `x^2+y^2+25`. The locus of any point in the set is:
Answer» `OA<=OP<=OB` `S-3<=sqrt(x^2+y^2)<=S+3` `2<=sqrt(x^2+y^2)<=8` `4<=x^2+y^2<=8` option A is correct.

14.	The inequality representing the following graph is A. `x in (9/2,infty)`B. `x in [9/2,infty)`C. `x in -[infty,9/2)`D. `x in (-infty,9/2]`
Answer» Correct Answer - B The given graph represents all the values greater than `9/2 " including"9/2` as the real line. ` x in [ 9/2, infty)`

15.	Ravi obtained 70 and 75 marks in first two unit tests. Find the number of minimum marks he should get in the third test to have an average of at least 60 marks.
Answer» let the minimum marks he got be x `(x+70+75)/3 >= 60` `x+145 >= 180` `x >= 180-145` `x>=35` answer

16.	A solution of ` 9%` acid is to be diluted by adding `3%` acid solution to it. The resulting mixture is to be more then ` 5%` but less than ` 7%` acid. If there is 460 L of the ` 9%` solution, how many litres of ` 3%` solution will have to be added?
Answer» Let x L of ` 3%` solution be added to 460 L of ` 9%` solution of acid. Then, total quantity of mixture = (460+x)L Total acid content in the (460 + x) L of mixture ` (460 xx 9/100+x xx 3/100)` It is given that acid content in the resulting mixture must be more than ` 5%` but less than `7%` acid. Therefore, ` 5% "of "(460+x) lt 460 xx 9/100+(3x)/100 lt 7%" of "(460+x)` `rArr 5/100 xx(460+x) lt 460 xx 9/100 x lt 7/100 xx (460+x)` `rArr 5 xx (460+x) lt 460 xx 9+3x lt 7 xx (460+x)` [ multiplying by 100] ` rArr 2300+5x lt 4140 +3x lt 3220+7x` Taking first two inequalities, `2300+ 5x lt 4140+3x` `rArr 5x-3x lt 4140-2300` `rArr 2x lt 1840` `rArr x lt (1840)/2` `rArr x lt 920` ....(i) Taking last two inequalities, ` 4140+3x lt 3220+7xx` `rArr 3x-7x lt 3220-4140` `rArr -4x lt - 920` `rArr 4x gt 920` `rArr x gt (0=920)/4` `rArr x gt 230` ...(ii) Hence, the number of litres of the ` 3%` solution of acid must be more than 230 L and less than 920 L.

17.	A man wants to cut three lengths from a single piece of board of length 91 cm The second length is to be 3 cm longer than the shortest and the third length is to be twice as long as the shortest. What are the possible lengths of the shortest board?
Answer» minimum 8 cm , maximum 22 cm .

18.	Solve the inequalities graphically in two-dimensional plane: `y < 2`
Answer» Here, line representing the inequality will be, `y = -2` This line will be a line parallel to x-axis. Please refer to video in the graph. So, the region below this line is the feasible region and will be the solution.

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