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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.
| 1551. |
2 \operatorname { tan } ^ { - 1 } \frac { 1 } { 2 } + \operatorname { tan } ^ { - 1 } \frac { 1 } { 7 } = \operatorname { tan } ^ { - 1 } \frac { 31 } { 17 } |
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Answer» thanks Sir |
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| 1552. |
।।५।।।। -मन्तर-----(1) 0(2) 29997 (3) 30003 (4) 689025, 16, 48 और 54 में से किसके अपवर्तकों की संख्या अधिक है ?(1) 25 (2) 16 (3) 48 (4) 54 |
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Answer» 54 is the best answer please like my answer then I like your answer to all of you 54 is the correct answer the crrect answer is 48 |
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| 1553. |
48÷4(3÷3+3) |
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Answer» 12x(4)= 48thanksaccording to the bodmas i did it 48 is the most correct answer 48 is the correct answer. 48 is the right answer 48÷4(3÷3+3)=48 of the following 48÷4(3÷3+3)=12×(4)=48 answer 48the answer is right 48÷4(3÷3+3)48÷4(1+3)48÷4(4)48÷163. ans. 3 is the right answer. 48÷4(3÷3+3)=48÷4(1+3)=48÷4(4)=48÷16=3 3 is the right answer 3 is the right answer 48 is a correct answer the correct answer is 3. 48 is the correct answer of the given question 3 is the right answer 48 is the most correct answer 48 is the correct answer |
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| 1554. |
2 \operatorname { tan } ^ { - 1 } \frac { 1 } { 3 } + \operatorname { tan } ^ { - 1 } \frac { 1 } { 7 } = |
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| 1555. |
4. Let us show that, 3/48-4/75+/192 0 |
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| 1556. |
Practice Time1) 28 ÷ 2 =3) 48 ÷ 4 =5) 96 ÷ 8=2) 56÷ 7 =4) 66÷ 6 =6) 110÷10 = |
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| 1557. |
((5/4)*((y/((4/5)))*(75/48 %2B 125/64)))*(64/125 %2B 48/75) |
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| 1558. |
Do THISFind which of the following are not GP1. 6, 12, 24, 48,....4-4,-20,-100,-500, .. |
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Answer» 2.1, 4 , 9 , 16 .....is not in GP Reason :4/1 = 49/4 = 9/4 There is no common ratio |
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| 1559. |
Let us show that, 3/48 -4/75192 0 |
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| 1560. |
Q. 2.A coin is tossed for a certain number of times. Tthe probability of getting a head is 0.4 and headappears for 24 times, find the number of times,the coin was tossed. Hence, find probability of1.getting a tail and verify that P(1) + PCT) |
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Answer» P(H) = 0.4 number of times head came/total number of outcomes = 0.4 Total number of outcomes = 24/0.4 = 60 Number of time tail came = 60-24 = 36 P(T) = 36/60 = 12/20 = 0.6_________________________P(H) + P(T)= 0.4 + 0.6= 1 |
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| 1561. |
tossed thrice. Find the probability of getting an odd number at least on |
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| 1562. |
If two coins are tossed at a time, find the probability of a head. |
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Answer» I didn't get answer bro if two coins are tossed case 1 : then you can head on first coin and tail on second coin case 2 : Or you can get tail on first coin and head on second coin. |
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| 1563. |
A coin tossed twice. find the probability of getting at least one head. |
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Answer» Outcomes = HH,HT,TH,TT Probability of getting at least one head = 3/4 If you find this answer helpful then like it. |
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| 1564. |
. If three coins are tossed simultaneously, find the probability of getting atleast two heads |
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| 1565. |
three coins are tossed simultaneously. find the probability of getting at least two head |
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| 1566. |
Three different coins are tossed simultaneously. Find the probability ofgetting exactly one head |
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Answer» When three coins are tossed simultaneously, the total possible outcomes are {(H,H,H) , (H,H,T) , (H,T,H) , (H,T,T), (T,H,H), (T,H,T), (T,T,H), (T,T,T)}.The total no. of outcomes = 8at least one head means O or 1 so total favourable out comes = 4so probability = 4/8= 1/2 . |
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| 1567. |
Square root of 17 |
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| 1568. |
A coin is tossed twice. find the probability of getting at least one head. |
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Answer» Sample space S = {HH, HT, TH, TT} i.e. Favourable cases for atleast one head are{HH, HT, TH} i.e. = 4 Required probability = 3/ 4 |
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| 1569. |
The area of square A is 25 cm2. The periof square B is 12 cm. What is the area andperimeter of square C? |
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| 1570. |
9. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrieal taeher field, whichis 10 m in diameter and 2 m deep. If water flows through the pipeats |
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Answer» R = radius of cylinder = 6m H = height of cylinder = 2.5m r = radius of pipe = 25/2cm =1/8m Rate of flow of water = 3.6km/hr In 1 hr water upto a length of 3.6km = 3600m will come out of pipe. let the tank be filled in 'x' hrs. volume of water coming out of pipe in x hrs = volume of cylindrical tank. (22/7)(1/8)(1/8)(3600)(x) = (22/7)(6)(6)(2.5) (1/8)(1/8)(3600)( x) = 6 x 6 x 2.5 3600(x) = 36 x 2.5 x 8 x 8 100(x) = 8 x 8 x 2.5 1000(x) = 8 x 8 x 25 x = 16/10 = 1.6 hrs x = 1 hr.36 min. Now, cost of water = volume of cylindrical tank x 0.07 = 22/7 x 6 x 6 x 2.5 x 0.07 = Rs.19.80. The tank will be filled in 1hr.36 min and the cost of water will be Rs.19.80. |
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| 1571. |
Find the capacity of a cylindrical container with internal diameter 28 cm and height20 cm. |
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Answer» D=28cmh=20cmr=d/2 = 28/2 = 14cmv =pi *r^2 h= 22/7*14*14*20=12320cm^31cm^3=1/1000ltherefore 12320cm^3 = 12320/1000=12.32 l |
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| 1572. |
: cof‘(coss——h) का मान बऱबर है |
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Answer» You can write7π/6as(π+π/6)Thus we can clearly see the angle falls in the third quadrant. And the cosine value in third quadrant is always negative. Hence,cos(π+π/6)=−cos(π/6) coming back to the questioncos−1[cos(7π/6)]=cos−1[−cos(π/6)]=π−cos−1[cos(π/6)]=π−π6=5π6 right answer |
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| 1573. |
A farmer connects a pipe of internal diameter 20 cm from a canal into acylindrical tank which is 10 m in diameter and 2 m deep. If the waterflows through the pipe at the rate of 4 km/hr inthe tank be filled completely?how much time will |
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| 1574. |
22. A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank inher field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at therate of 3 km/h, in how much time will the tank be filled? |
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Answer» R=10 cm R=500cmh=200cmvolume of water flows in 1 hr =π×10×10×300000(cmcube)volume of cylindrical tank =π×500×500×200time taken to be fill the tank=volume of cylindrical tank÷volume of water flow in 1 hr =(π×500×500×200)÷π×10×10×300000=(5÷3)hr=(5÷3)×60=100min |
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| 1575. |
A farmer connects a pipe of internal diameter 20 cm from a canal into a cylindrical tank inher field, which is 10 m in diameter and 2 m deep. If water flows through the pipe at therate of 3 km/h, in how much time will the tank be filled? D9. |
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| 1576. |
. Prove thatsSin x +sin 3x > 81:22:605 X + COSs 3x |
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| 1577. |
27. Iof a property coss? 252,000; find the4cost ofof it. |
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| 1578. |
area will it irrigate in 30 minutes, nA farmer connects a pipe of internal diameter 20 cm from a canal into a cylindricalher field, which is 10 m in diameter and 2 m deep. If water flows through the git tanke9.rate of 3 km/h, in how much time will the tank be filled?pipe at the |
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| 1579. |
51. A farmer runs a pipe of internal diameter 20 cm from the canal into a cvindrical telhis field which is 10 m in diameter and 2 m deep. If water flows through therate of 3 km/h, in how much time will the tank be filled?tark inCBSE 2014, NCECTcircdar cylinthe volume oradius of the |
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| 1580. |
wherer is flowing at the rate of 3 km/hr through a circular pipe of 20 cm internal diametera circular cistern of diameter 10 m and depth 2 m. In how much time will the cistern beWaterfilled?nf Snlids (Gylinder Cone and Sphere)273 |
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| 1581. |
nalsofrhombusare 20 cm and 21acm respectively, then find the side of rhomhys and its perimeter |
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Answer» As the diagonals of rhombus bisects each other by 90° Then By Pythagoras theorem, (20/2)²+(21/2)²=side² 10²+(21/2)²=side² 100+441/4 =side² (400+441)/4=side² 841/4 =side² √{841/4} =side 29/2 =side Then Perimeter =4*side Perimeter =4*29/2 Perimeter =2*29 Perimeter =58cm |
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| 1582. |
9. OABCD is a parallelogram, AC, BD are the dianals & intersect at point O. X and Y are the cof A4DC and A4BC respectively. If BY 6 cm, tOX-7(a) 2 cm(c) 4 cm(b) 3 cm(d) 6 cm |
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Answer» Since centroid divides the median in 2 : 1 ,BO : OY = 2 : 1OY = 6/2 = 3.OY = OX = 3. |
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| 1583. |
\frac { 3 } { 5 } = \frac { 3 \times \square } { 5 \times \square } = \frac { \square } { 10 } = \square \quad ( 2 ) \frac { 25 } { 8 } = \frac { 25 \times \square } { 8 \times 125 } = \frac { \square } { 1000 } = 3.125 |
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Answer» 1)3/5=(3*2)/(5*2)=6/10=0.62)25/8=(25*125)/(8*125)=3125/1000=3.125 |
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| 1584. |
calculate the value of 83 83 square + 17 square upon 83 square - 83 into 17 + 17 square |
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| 1585. |
Find le sinlaiioothe sntaethe square root of the perfect square so obtained.Find the smallest number by which 7350 must be divided to make it a perfect square. Also findthe square root of the perfect square so obtainedhnt ir a nerfect square. |
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Answer» Thanks |
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| 1586. |
a square+b square+c square=? |
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Answer» a²+b²+c²=(a+b+c)²-2(ab+bc+ca) |
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| 1587. |
breadth.3 The sum of three consecutive numbers is 36. Find the numbers.(Hint : Consider three consecutive numbers as x, x + 1 and x + 2) |
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| 1588. |
Surn of a three consecutive numbers is 90.Find the numbers |
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| 1589. |
numbers.The sum of three consecutive numbers is 120. Find the |
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| 1590. |
3x2-30x~-72 |
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| 1591. |
30x=60, then calculate the value of x? |
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Answer» X=2 is the right answer |
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| 1592. |
if f: R-> R is defined by f(X)= X/ X square +1, then fig(2) is given by____? |
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| 1593. |
by 5% every year, how many stuudiA wh ch coss? 350000 depreciates by 10% every year. What will be theworth of the car after three years? |
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Answer» Every year depreciation is 10/100*3,50,000=35000rupeeshence for 3 years 3*35000=105000rupeeshence cost will be 3,50,000-1,05,000=rupees 2,45,000 |
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| 1594. |
35.Find three consecutivenumbers whose sum is 10 |
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| 1595. |
\left. \begin{array} { l } { \text { Evaluate } \frac { 15 } { \sqrt { 10 } + \sqrt { 20 } + \sqrt { 40 } - \sqrt { 50 } } , \text { it being given that } } \\ { \sqrt { 10 } = 3.162 } \end{array} \right. |
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| 1596. |
\sqrt { 21 - 4 \sqrt { 5 } + 8 \sqrt { 3 } - 4 \sqrt { 15 } } = |
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| 1597. |
The ascending order of three observations isand whose median is \frac{x}{7}, \frac{x}{5} \text { and } \frac{x}{3}3.14. Find 'x'.9,,一and3 |
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| 1598. |
Evaluate \sqrt{5+2 \sqrt{6}}+\sqrt{8-2 \sqrt{15}} |
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Answer» √(5+2√6) + √(8-2√15) = √(√2+√3)(√2+√3) + √(√5-√3)(√5-√3) = √2 +√3 +√5 -√3 = √2 +√5 |
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| 1599. |
1. The sum of three consecutive odd numbers is 63. Find the numbers. |
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Answer» Let the first odd number be x.Then, the next consecutive odd numbers will bex+2 andx+4According to the question:x +x+2+x+4 = 21or, 3x + 6 = 21or, 3x + 6-6= 21-6 Subtracting 6 from both the sidesor, 3x = 15or,3x/3=15/3 Dividing both the sides by 3or, x = 5∴ Required numbers are:x =5x + 2 =5 + 2 =7x + 4 = 5 + 4 =9 the three consecutive numbers are 19,21 and 23. the three consecutive number are 19,21,23 so the new numbers are7 and 9 |
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| 1600. |
The sum of three consecutive odd numbers is 63. Find the numbers.TE |
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