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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.
| 76501. |
For what values of k will the following pair of linear equations kx + 3y-(k-3) = 0,12x+ ky- k(A) k 4●0 have infintely many solution?(B) k 6(C) k 5(D) of these |
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| 76502. |
\left. \begin{array} { l } { \text { Find the value of } 4 k + 1 \text { if } k = 1 ; k = 2 } \\ { k = - 1 ; \text { and } k = \frac { 1 } { 2 } } \end{array} \right. |
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| 76503. |
If 3 women and 5 girls take 17 days to complete a piece of work, how long will 7 women and 11 girls working together take to complete the work? |
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Answer» Given 3 women or 5 girls cqn complete a piece of work in 17 days. 3women = 5 girls 1 women = 5/3 girls -------( 1 ) Now we have to convert both in either of women or girls. 7 women + 11 girls = 7 × (5/3)girls + 11 girls [ from ( 1 ) ] = ( 35 + 33 ) / 3 = 68/3 girls can complete the work in how many days . If 5 girls can finish the work in 17 days Then 68/3 girls can finish the same work in how many days ( 68 / 3 ) × 17 /5 = 1156 /15 days. |
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| 76504. |
7. If 3 women or 5 girls take 17 days to complete a piece of work, how long wil7 women and 11 girls working together take to complete the work? |
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Answer» Given 3 women or 5 girls cqn complete a piece of work in 17 days. 3women = 5 girls 1 women = 5/3 girls -------( 1 ) Now we have to convert both in either of women or girls. 7 women + 11 girls = 7 × (5/3)girls + 11 girls [ from ( 1 ) ] = ( 35 + 33 ) / 3 = 68/3 girls can complete the work in how many days . If 5 girls can finish the work in 17 days Then 68/3 girls can finish the same work in how many days ( 68 / 3 ) × 17 /5 = 1156 /15 days. |
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| 76505. |
61. 12 men or 15 women can finish a work in 24 days.In how many days the same work can be finished by8 men and 8 women ?(A) 16 days (B) 20 days(C) 24 days (D) 28 days |
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Answer» 16is the correct answer 20 is the correct answer |
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| 76506. |
IfA. B. C are three mutually exclusive and exhaustive events such that P (A)-2P(B)-Pc)P (A)2b)a) zc)d) |
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Answer» A ,B,C are three mutually exclusive and exhaustive events so P(A) + P(B) + P(C) = 1 P(A) = 2P(B) = 3P(C) P(B) = P(A)/2 P(C) = P(A)/3 P(A) + P(A)/2 + P(A)/3 = 1 (6P(A)+3P(A)+2P(A))/6 = 1 11P(A) = 6 P(A) = 6/11 P(B) = P(A)/2 = 1/2 * 6/11 = 3/11 P(C) = P(A)/3 = 1/3 * 6/11 = 2/11 If you find this answer helpful then like it. |
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| 76507. |
code10. A function / :1-5,9) -- R is defined as follows:16x +1 if-55x<2f(t) = 5x2-1 if 25x<63r - 4 if 65159boFind(ii) f(7)-f(1)(iv)(1)/(-3) + f(2)(iii) 2 f(4) + f(8)2-T6)f(4) + f(-2 |
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Answer» Social science 1st lesson ....................... 1) x^2+2x+3x+6=x( x+2)+3( x+2)=( x+2)( x+3); x= 2,3 |
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| 76508. |
\frac { 4 ^ { n + 4 } - 5 \times 4 ^ { n + 2 } } { 4 ^ { n + 1 } \times 11 } |
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| 76509. |
1 \times 2 %2B 2 \times 3 %2B 3 \times 4 %2B 4 \times 5 %2B \ldots |
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Answer» 40 is the answer of this question 40, is answer of this question |
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| 76510. |
03 A parachutist is descending vertically and makes angle of depression of 45 and 60 at twoobservation points 100m apart from each other on the left side of himself. find in meters theapproximate height from which he falls and also find in meters the approximate distance of thepoint where he fall on the ground from the first observation points. ANS (236.6 m) |
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| 76511. |
Find real numbers A and B. if A+ B-3-27 +4i |
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Answer» v gud |
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| 76512. |
b B e.‘A_’_[-S 4i]fi?4TB— हि हि2) का मान ज्ञात कोजिए 1... | |
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| 76513. |
1. Find a b wheni) a=i-21 + k and b = 31-4i-2k |
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Answer» answer |
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| 76514. |
16. Mihika constructed an angle of 120째 and trisectedit. Measure of two angles taken together will beA. 90째40째C. 80째D. of these17. Find the difference between the place value of 4 in3286.4023 and face value of 2 in 4568234.A. 1.6B. 2.4 |
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| 76515. |
48.Mean age of 6 boys is 12 years andmean of 9 girls is 10 years. What isthe mean age, in years, of all theseboys and girls taken together ?(A) 10.8 (B) 11(C) 10(D) 10.5 |
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Answer» Mean = sum of all observations / number of observations Given,Mean age of 6 boys is 12 yrs12 = sum of age of 6 boys/6sum of age of 6 boys = 12*6 = 72 Mean age of 9 girls is 10 yrs10 = sum of age of 9 girls/9sum of age of 9 girls = 10*9 = 90 Now, total sum of age of 6 boys and 9 girls = 72 + 90 = 162 Mean of 6 boys and 9 girls = 162/15 = 10.8 (A) is correct option |
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| 76516. |
30. The area of a rectangular field is 3584 m2 and its length is 64 m. A boy runs arod 1 m hrnad, It is to be paved with stones, each meafencing it atper metre.f dimensionsnd the cost offield at the rate of 6 km/h. How long ill he take to go 5 times around it? |
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Answer» Area of rectangular field=length*breadth64*breadth=3584 breadth=3584/64=56m Permeter of rectangular field=2(l+b)=2(64+56)=240 m Distance covered by boy when he completes 5 rounds of the field=5×240 m=1200 m=1.2km Time taken= Distance/speed=1.2/6 hours=0.2 hours |
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| 76517. |
8. There are three events A, B. C, one of which must, and one can happen. The odds are 7 to3against A and 6 to 4 against B, find the odds against C |
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| 76518. |
(2 x+5)^{2}-(2 x-5)^{2} |
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| 76519. |
Write the condition under which three numbers a, b, c may be in Aand G.P. both. |
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Answer» If we consider a set of three numbers a, b and c which are in AP, we know that a + c =2b. Now, if they are in GP, a*c =b^2. Now we have to see if it is possible to have three numbers satisfy both the equations. If we express a as 2b - c and replace it in a*c = b^2, it gives us (2b-c)*c = b^2 => 2bc – c^2 = b^2 => b^2 + c^2 – 2bc =0 => (b-c) ^2 =0 => b = c Also, as a = 2b-c = 2c-c = c we find that all the three number have to be equal. So a set of three numbers which are in AP as well as GP are a set of three equal numbers. |
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| 76520. |
1 \times 2 ^ { 2 } + 2 \times 3 ^ { 2 } + 3 \times 4 ^ { 2 } + \ldots \text { to } n |
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| 76521. |
5. The LCM of two polynomials p(x) and q(x) isx - 7x+6. If p(x) = (x² + 2x - 3) andg(x) = (x 2 + x - 6), then the HCF is(a) x + 3(b) x + 1(C) (x + 2)(x + 1)(d) of these |
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| 76522. |
ma z.If the ratio of two numbers is 5:6 and their HCF is 8, then find their LCM(A) 300(B) 120(C) 240(D) 160(E) of these |
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Answer» 120 is the correct answer me Arshad bhai k answer wrong h 120 is the anns for ur questions 120 is the correct answer let the x is included in the rationumbers =5x,6xHCF=8 Then LCM= 8×5×6=240 240 correct answer for this question 120 is a complete answer the correct answer is 240 rong it's answer are rong |
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| 76523. |
14. The LCM and HCF of two polynomials p(2) and (2)are 36x (x + a) (x3 -a) and (-a).respectively. If p(x) = 4.x(x2-a), then the valueof g(x) is(a) 9x2(x - 2)(b) 98°(x2-a)(c) 4x21x2-a)(d) of these |
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Answer» We know,p(x) * q(x) = HCF * LCM 4x(x^2 - a^2)*q(x) = 36x^3(x + a)(x^3 - a^3) * x^2(x - a) q(x) = 9x^3 (x^3 - a^3) * x q(x) = 9x^4 (x^3 - a^3) (d) is correct option |
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| 76524. |
T\ ol B Sumef 32 dams बूटी 2 अन्न कि 2et rom B =l £ पी |
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Answer» Let 'a' be the first term and 'd' be the common difference.3rd term = a + (n - 1)dHere n = 3= a + 2d3rd term = 1. ............... (given)Hence,a + 2d = 1 ................ (1)Now6th term = a +5d6th term = -11. ............. (given)Hence,a + 5d = -11. ............... (2)Subtracting equations (1) from (2)a + 5d = -11- a + 2d = 1——————3d = -12d = -12/3 = -4HencePutting d = -4 in equation (1)a + 2d = 1a + 2*(-4) = 1a - 8 = 1a = 9Sum formulaS(n) = n/2 [2a+ (n - 1)d]Sum of 32 terms =S(32) = 32/2 [ 2*9 + (32 - 1)*(-4)]= 16 [ 18 + 31*(-4)]= 16 ( 18 - 124)= 16 ( - 106)= - 1696 |
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| 76525. |
39Out of a number of Saras birds, one fourth the number aremoving about in lotus plants; th coupled (along) withth as well as 7 times the square root of the number move4on a hill; 56 birds remain in vakula trees. What is the totalnumber of birds? [HOTS] (Mahavira around 850 A.D.) |
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| 76526. |
Find the numbers.5. The sum of two numbers is 8 and the sum of theirireciprocals is. Find the numbers.15 |
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Answer» answer is a=3 and b=5 |
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| 76527. |
If pa 5-2 16 bind the value of per 1nd |
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| 76528. |
\cos ^{2} \frac{\pi}{16}+\cos ^{2} \frac{3 \pi}{16}+\cos ^{2} \frac{5 \pi}{16}+\cos ^{2} \frac{7 \pi}{16} |
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| 76529. |
\frac { 16 \times 2 ^ { n + 1 } - 4 \times 2 ^ { n } } { 16 \times 2 ^ { n + 2 } - 2 \times 2 ^ { n + 2 } } |
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Answer» 4(2^(n+3)-2^n)/4(2^(n+4)-2^n)=2^n(2^3-1)/2^n(2^4-1)=7/15 |
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| 76530. |
SECTION- Aé Ź:(A) HCF LCM(C) HCF=LCM(B)HCF < LCM |
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Answer» 12=2×2×320=2×2×5 HCF=4LCM=60 ऑप्शन B) सही है। |
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| 76531. |
4 x ^ { 2 } + 9 y ^ { 2 } + 16 z ^ { 2 } + 12 x y - 24 y z - 16 |
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| 76532. |
1. The relation between HCF and LCM of 12 and 20 will be:(d)(a) HCF > LCM (b) HCF < LCM(c) HCF = LCM of thesemial? |
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| 76533. |
Fill the missing numerals suchhat the sum of each side ofhe square is 18. Rememberhat you have to use numeralsrom2 to 9 without repeatingy of them. |
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| 76534. |
1¡lf'stan0-3sĂn e, find sino-cose |
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Answer» how did u get sin= 1 Its based on the formula, Sin^2 theta + cos ^2 theta =1 |
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| 76535. |
(e) Find the value of 16.016 +0.4. |
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Answer» 16.016÷0.440.04is correct answer 1 6 . 0 1 6 ÷ 0. 4 =40.04 1 6.0 1 6÷ 0.4=40.04 =16.016/0.4=40.04.is the division of it 16.016÷0.4=40.04is the answer 16.016÷0.4 = 40.04is the correct answer 40.04 is the right answer |
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| 76536. |
e. Find sik rauional numbers between and |
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Answer» 3/8 and 1*4/2*4=4/8hence rational numbers between 3/8 and 1/2=3.1/8,3.2/8,3.3/8,3.4/8,3.5/8,3.6/8 |
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| 76537. |
4A 63,65, 6-2, au eurd ) Also-bind 4h01 to m |
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| 76538. |
do clo 、epain work, uhatShouldbe he lecth the addeuhich he Should cue, uhen nehe poleasn30 |
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| 76539. |
Simplify : (64) 6 x(216) 3x(81%(512) 3 x(16)#xBP2. |
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Answer» ((2^6)^(-1/6)×(6^3)^(-1/3)×(3^4)^(1/4))/((8^3)^(-1/3)×(2^4)^(1/4)×(3^2)^(-1/2))=(2^(-1)×6^(-1)×(3))/(8^(-1)×2×3^(-1))=(1/2×1/6×3)/(1/8×2×1/3)=(1/4)/(1/12)=12/4=3 |
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| 76540. |
494-648+5497 |
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Answer» 494-648+54975991-648=5343 494-648+5497=5991-648=5343 is best answer. |
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| 76541. |
LCM of 648 |
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| 76542. |
9674*648-9674*548 |
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Answer» 9674*648=62687529674*548=5301352 62687552-5301352=967400 967400 |
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| 76543. |
EXERCISE 25Form 3-digit numbers using all the three given digitswithout repeating any digit and ring the largest number:I. 3, 5, 7: _II. 4, 6, 8:III. 9, 6, 3:IV. 5, 7, 0:V. 2, 3, 1: |
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Answer» 1. ( 3,5,7,9,11,13,15,17,19) 2.(4,6,8,10,12,14,16,18,20) |
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| 76544. |
Bind DistancaDictance |
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| 76545. |
(265) Find two part of 40 such that the sum of their squares is 850. |
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Answer» Let one part of 40 be x Another part is 40-x x²+(40-x)² = 850 x²+40²+x²-80x = 850 2x²-80x = 850-1600 2x²-80x+750 = 0 x = 80 +- √(-80)²+4*2*750 / 2*2 = 80 +- √6400 + 6000 /4 = 80 +- √12400/4 = 80+- √3100 |
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| 76546. |
प्रश्नावली 26 21. निम्नलिखित को 1 मी के प्रतिशत में बदलिए-(क) 40 सेमी (ख) 168 सेमी(ङ) 7823 सेमी (च) 80 सेमी |
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Answer» 80 is correct answer here your answer 80 is right answer 80 is the correct answer. 80 is the best answer 80 is the correct answer of the given question 80 is the correct answer 80 is the answer of the following a). 40cm= 0.40%b). 168cm= 1.68%c). 7823cm= 78.23%d). 80 cm= 0.80% answer 80 is the correct answer of the given question |
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| 76547. |
7823-128 /16 of 4 +3973 |
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Answer» using BODMAS7823-8*(-3969)7815x(-3969)=-31,017,735 |
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| 76548. |
2x+x=45 |
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Answer» 2x + x = 453x = 45x = 45/3x = 15 |
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| 76549. |
Simplify\left(\frac{8}{27}\right)^{-1} \times\left(\frac{25}{4}\right)^{2} \times\left(\frac{4}{9}\right)^{2}+\left(\frac{125}{64}\right)^{3} |
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| 76550. |
12. Three numbers are in A.P. If the sum of these numbers be 27 andthe product 648, find the numbers. |
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