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.
| 63351. |
\frac { \operatorname { sin } 30 ^ { \circ } + \operatorname { tan } 45 ^ { \circ } - \operatorname { cosec } 60 ^ { \circ } } { \operatorname { sec } 30 ^ { \circ } + \operatorname { cos } 60 ^ { \circ } + \operatorname { cot } 45 ^ { \circ } } |
| Answer» | |
| 63352. |
B. 4A. 2Ń.If 22x+4 = 1 o", then x3 =A. 2D. 16B. 4D. 16 |
|
Answer» 2^(2x+4)=16^x2^(2x+4)=(2^4)^x2^(2x+4)=2^(4x)same base; powers should be same2x+4=4x4=4x-2x4=2xx=2Therefore x^3=2^3=8 |
|
| 63353. |
( a + b ) ^ { 4 } - ( a - b ) ^ { 4 } \cdot \text { Hence, evaluate } ( \sqrt { 3 } + \sqrt { 2 } ) ^ { 4 } - ( \sqrt { 3 } - \sqrt { 2 } ) ^ { 4 } |
| Answer» | |
| 63354. |
43. If k 28 +16v3, then what will be the value of1/2-32A.C.2E. of these |
| Answer» | |
| 63355. |
13. Which of the following is true?sum of all the terms +1numberof termsMode is the element which has the lowestA. Mean is:B.frequencyC. Mode for 3, 4, 5, 6, 4, 5, 1, 5 is 4.D. Median for 3, 4, 5, 6, 7 is5. |
|
Answer» Option D is correct. Median = middle term no of terms = 5 Median = n+1/2 = 5+1/2 = 6/2 = 3rd term So the median is 5. Like my answer if you find it useful! |
|
| 63356. |
4*b^4 %2B a^4 - 5*a^2*b^2 |
| Answer» | |
| 63357. |
229. The difference between thegreatest and least five-digitnumbers formed by the digits2,5,0,6,8 is (repetition of digitsare not allowed) |
|
Answer» 86520-02568 = 83952. Greatest no. ➡86520Smallest no. ➡20568Their difference ➡86520-20568➡65952 Greatest five - digit number by the digits 2,5,0,6,8 is = 86520,Least five - digit number by the digits 2,5,0,6,8 is = 20568,Their difference is = 86520 - 20568, = 65952. |
|
| 63358. |
ALATO 28-80and find 25.2ट |
|
Answer» c =30 because three corner's sum 180 in triangle.. c= 30 total angle of triangle is 180 then a+b= 150 c=30 kyuki triangle ke teeno angles ka yog = 180 c=30 because three coroner's sum in 180 in triangle Angle A=70Angle B=80Angle C=?Now,As we know that Sum of all angles of a triangle is 180Therefore,A+B+C=18070+80+C=180C=180-70-80C=30 Therefore, Angle C=30 Angle A = 70Angle B = 80 Angle C = ?As we know that sum of all angles of a triangle is 180therefore A + B + C = 18070 + 80 + C = 180150 + C = 180C = 180 - 150C = 30 30 degree because angle sum property of triangle states that sum of all 3 angles of triangle is =180 degree c=30° become a+b+c= 180°a=70°B=80°70°+80°+C= 180°SO C= 30° A=70,B=80,c=?A+B+c=180,70+80=150,180_150=30 sequel 30 because three corners some 180 degree in triangle using angle sum property of triangleangle C is 30 degree X = + 2 - 2 because X + 2 equal to x power 4 angle C= 30 because sum of the triangle angle is 180 the third angle is 30 by applying angle sum property of triangle c=30 because triangle sum is 180 c=30 because three coroner`s sum in 180 in triangle 30° is the answer of this question a+b+c=180°70+80+c=180c=180-150c=30° teeno angles ka yog =180 to c=30 deegri total angles-180 angle70+angle80 -180-150_30 c is equal to 30 because all sum are 180 angle C= 30 degree because sum of angle of triangle is 180 If angle A=70 and B=80,we know that the sum of angles of a triangle is be 180 triangle c=30 hoga80+50+©=180=180-150=30 c=30 total angle of triangle since the sum of all angle of triangle is 180, Let third angle be Xthen 70 +80+X=180So,X=30 180-(80+70)=30 degree c=30 because a+b+c=180so, c=180-a-b =30 180= 70+80+c ? c=30 sum of all angles of triangle is 180°therefore<A+<B+<C=18070+80+<C=180<C=30° 30 degree is correct answer LA=70, LB=80 WE KNOW THAT LA+LB+LC=180 70+80+LC=180 150+LC=180 LC=180-150=30 SO, LC=30° c= 30° because there corner is 180° c=30 right answer hoga 30 is the right answer180-(sum of two angles) इस त्रिभुज में एंगल C का मान ३० डिग्री होगा Angle C is 30° C=180-(70+80)C=30° |
|
| 63359. |
5 x %2B 3 y = - 11 ; 2 x %2B 4 y = - 10 |
| Answer» | |
| 63360. |
\frac 5 \operatorname sin ^ 2 30 ^ \circ %2B \operatorname cos ^ 2 45 ^ 11 %2B 4 \operatorname tan ^ 2 60 ^ \circ 2 \operatorname sin 30 ^ \circ \operatorname cos 60 ^ 4 %2B \operatorname tan 45 ^ \circ |
|
Answer» 5sin^2 30° + cos^2 45° + 4tan^2 60° /2sin30°cos60°+ tan45° = 5(1/2)^2 + (1/√2)^2 + 4(√3)^2 /2(1/2)(1/2) + 1 = (5/4) + (1/2) + (12) /(1/2) + 1 = (5 + 2 + 48)/12 / (3/2) = (55/12)/(3/2) = 55/12*2/3= 55/18 |
|
| 63361. |
13*(11*(7*y)) %2B 2*(3*(4*(5*(6*(13*(text*(7*(a*(d*n))))))))) |
|
Answer» 7x11x13+13=1014/10=101.4; 7x6x5x4x3x2x1+5=5045/10=504.5 |
|
| 63362. |
Solve any one of the following sub-questions.03i Mr. Dinesh owns a rectangular agricultural farm at village Talvel. The length of thefarm is 10 metre more than twice the breadth. In order to harvest rain water, he dug asquare shaped pond inside the farm. The side of the pond is g of the breadth of thefarm. The area of the farm is 20 times the area of the pond. Find the length andbreadth of the farm and side of the pond. |
|
Answer» Let length and breadth of farm be L and B resp. According to ques, we are given L = 2B + 10 Area of farm = L * B or (2B + 10) * B Area of farm = B * (2B + 10) Side of square pond is given to be 1/3 of breadth of farm which is, Side = 1/3 B Area of square pond will be = Side² i.e. (1/3 B)² Area of pond = 1/9 B² Now it is given, area of farm = 20 x area of pond So, B * (2B + 10) = 20 x 1/9 B² 2B + 10 = 20/9 B 20/9 B - 2B = 10 20/9 B - 18/9 B = 10 2/9 B = 10 B = 45m So, L = 2(45) + 10 L = 100m and Side = 1/3 (45) Side = 15m Therefore, L and B are 45 and 100 of farm, and Side of pond is 15m |
|
| 63363. |
The base of a triangular farm is three times itsaltitude. If the cost of irrigation the farm is6662.60 per hectare 493.60. Find the heightand the base of the farm14. |
| Answer» | |
| 63364. |
In an isosceles Δ ABC, if AC-BC and AB2-2AC2 , then LC is euqal to :(A)450(B) 600(C) 30°(D)900 |
|
Answer» AB²=2AC²=AC²+AC²AB² =AC²+BC² { given AC=BC} Hence ABC is a right angled triangle with angle C = 90° because AB is the longest side and the angle opposite to longest side is largest. Option (D) is correct. |
|
| 63365. |
is always at the centre(A) Mean(B) Median (C Mode (D) Range |
|
Answer» Mean is always at the centre |
|
| 63366. |
The radius of the circle, in which an equilateral triangle of side 16 cm isinscribed, is_16v3cmcm |
|
Answer» thnx |
|
| 63367. |
VE-516,The value of 81a2 +9b2+54ab, if a - 2, b4 isVEL-51) 9002) 8003) 7004) 600 |
|
Answer» Answer: 1)900Explanation:Given a=-2 ;b=-481a^2+9b^2+54ab= 81(4)+9(16)+54(8)=324+144+432=900 thank you and answer another questions |
|
| 63368. |
17) The difference between largest and smallest 7-digit numbers formed bythe digits 2,8,6,0,5 is_ (using all the digits).(3)1) 8285592 2) 6885592 37 68859524) 6995852 |
|
Answer» 3......................... . largest no. 86520smallest no.02568 their difference is 6885952 largest 7 digit number formed by digits 2, 8, 6, 0, 5 = 8886520 smallest 7 digit number formed by the digits 2, 8, 6, 0, 5 = 2000568 ER. RAVI KUMAR ROY. difference between both = 8886520 - 2000568 = 6885952 (ANS) options c is the correct answer. |
|
| 63369. |
5. In Fig. 9.33p l qwith transversal mand Ll = 75°, find the p-measure of all the an-gles formed by paral-qlel lines p and q andthe transversal m3 64 15Fig. 9.33 |
|
Answer» angle 1 = angle 7 = angle 3 = angle 5 = 75° ( by vertically opposite and alternative angle)angle 1 + angle 8 = 180angle 8 = 180° - 75° = 105°angle 8 = angle 2 = angle 6 = angle 4 = 105° ( by vertically opposite and alternative angle) |
|
| 63370. |
Evaluate (2|11)^4 × (11|3)^2 × (3|2)^3 |
| Answer» | |
| 63371. |
((-5)/11)^(K %2B 2)/((-5)/11)^(-4*K %2B 5)=((-5)/11)^(2*K) |
|
Answer» (-5/11)^k+2 ÷ (-5/11)^-4k+5= (-5/11)^2k (-5/11)^[k+2-(-4k+5)] = (-5/11)^2k (-5/11)^[k+2+4k-5] = (-5/11)^2k (-5/11)^[5k-3] = (-5/11)^2k 5k - 3 = 2k 5k - 2k = 3 3k = 3 k = 3/3 = 1 |
|
| 63372. |
11^1/2/11^1/4 |
| Answer» | |
| 63373. |
6. Show that the line segments joining the mid-points of the opposite sides of aquadrilateral bisect each other.h tha mid-noint M of hypotenuse AB |
| Answer» | |
| 63374. |
11) 30% of 3000 is_a) 300c) * 600b9008 |
|
Answer» so 30 / 100 * 3000= 900 900 is the correct answer |
|
| 63375. |
As shown in figure 1.57, two poles ofheight 8 m and 4 m are perpendiculorto the ground. If the length of shadowof smaller pole due to sunlight is6 m then how long will be the shadowof the bigger pole at the same time ?3.4QB4Fig. 1.57 |
| Answer» | |
| 63376. |
basecuo , Scourtes -1) en nc.cs ss (am and-c con, ă§ |
| Answer» | |
| 63377. |
Wark)Rational number between 3 and 4.vwhichquadrantdochniaollowingpointslies(i)(-65) (i)(-33. Find zero of the polynomial, p(x)24. Write two solution of the equationx 4ySECTION-B(Each question carry 2 Marks)Find the value of K, i#x-1 is a factor of p (x): kx2in the form of where p and q are integers and qro.if the Point (3,4) lies on the graph of the equation 3y - ax +7. Find the value of a.If a point C lies between two points A and B such that AC BC, then prove that AC-SECTION- C(Each question carry 3 Marks)9 Evaluate 95 X 96Simplify (3+ v3) (3- V3)Show how V5 can be represented on the number line.Factorize x3 +13x2+32x + 20 |
|
Answer» Find the value of 64 ^-1/2 |
|
| 63378. |
の..-number concepts6) P(B) prenumberintegers lie betwemanylie between -10 and -iyow(B) 20(C) 19(D) 18 |
|
Answer» -9,-8,-7,-6,-5,-4,-3,-2,-1,0,1,2,3,4,5,6,7,8,9 hence, option (c) |
|
| 63379. |
and verify the relationship betwe(vi) 3x2-x-4bers as the sum and product of i(ii) 0, v5(vi) 4,1 |
| Answer» | |
| 63380. |
Sum of two numbers is 95IF one exceeds the other by3 find the numbers |
|
Answer» x+x+3 = 95 2x = 92 x = 46 |
|
| 63381. |
Sum of two numbers is 95. If one exceeds the other by 3, find the numbers. |
| Answer» | |
| 63382. |
1.Sum of two numbers is 48. If one exceeds the other by 4, find the numbers. |
|
Answer» let the two numbers be x and y, y<x.given, x+y=48.......(1) and, x=y+4.........(2) Now, Putting value of x from (2) in (1)We get, y+4+y=48 or 2y+4=48 or 2y=48-4 or 2y=44 or y=44/2 or y=22.ans.. Continuing....... Putting value of y in (2)We get, x=y+4 x=22+4 x=26 |
|
| 63383. |
8 SUm of two number is equal to 95if one exceeds the other by 15 findthe numbers |
|
Answer» suppose numbers are x and x+15so 2x+15=95so 2x=80 so x=40so numbers are 40 and 55 |
|
| 63384. |
en a Binomial distribution is said to be positively and negatively skewed? |
|
Answer» Inpositively skewed distributions, themeanis usually greater than themedian, which is always greater than themode. Innegatively skewed distributions, themeanis usually less than themedian, which is always less than themode |
|
| 63385. |
Question numbers 5 to 9 carry two marks each.Q.5) Expand (3a-2b) |
|
Answer» (3a-2b)²9a² + 4b² -12ab |
|
| 63386. |
The sun is never overhead in London and there is sunlight till8 o'clock in the evening, whereas the sun is overhead in Singapore,Why? |
|
Answer» because it is in the middle of the Earth and the Greenwich meridian line pass through it |
|
| 63387. |
petrol!17. The diameter of a garden roller is 1.4 m and it is 2 m long. How much area it willLiul uul in 2079 m3 and the diameter of its base is 21rmcover in 5 revolutions?t. |
| Answer» | |
| 63388. |
5. The sum of two numbers is 8 and the sum of theirreciprocals is . Find the numbers.15 |
| Answer» | |
| 63389. |
The sum of two numbers is 67. Their difference is 15.Find the numbers.6. |
| Answer» | |
| 63390. |
the jug has to be filled with WALI15. The product of two numbers is 15. If one of the numbers is 6. find the other. |
| Answer» | |
| 63391. |
Sum of two numbers is 95. If one exceeds the otherby 15. Find the numbers. |
|
Answer» Let the two numbers be x,ygiven x=y+15x+y=95y+15+y=952y+15=952y=80y=40x=55 thanks |
|
| 63392. |
15. The product of two numbers is 15If one of the numbers is 6. find the other. |
| Answer» | |
| 63393. |
The sum of two numbers is 15 and the sum of their reciprocals is . Find the numbers.10 |
| Answer» | |
| 63394. |
The sum of two numbers is 15 the sum of their reciprocals is 3/10, find thenumbers. |
|
Answer» Let the two numbers be A and B=>A+B=15 .............(1)and 1/A +1/B=3/10⇒(A+B)/AB = 3/10from (1)(A+B)/AB = 15/AB⇒15/AB = 3/10⇒AB = 50 ............(2)solving (1) and (2)A = 10 and B = 5orA = 5 and B = 10 |
|
| 63395. |
4. Malik Gas Agency (Chandigarh Union Teritory) purchased some gas cylinderfor industrial use foră24,500, and sold them to the local customers forFind the GST to be paid at the rate of 5% and hence the CGST and UTGST to bepaid for this transaction. (for Union Territories there is UTGST instead of SGST.2 |
|
Answer» (i) For Malik as Agency :Input Tax (GST on Purchase) = 5% of ₹24500 =24500∗510024500∗5100 = ₹1225 (ii) Output tax (GST on Sale) = 5% of ₹26500 =26500∗510026500∗5100 = ₹1325(iii) GST payable = Output tax – Input tax = 1325 – 1225∴ GST payable =₹100(iv) We know that, CGST and UTGST are similar for the article. The net GST amount isdivided into two equal parts for this. ∴ CGST paid =10021002 =₹50 and ∴ UTGST paid =10021002=₹50 |
|
| 63396. |
20 gram and 745 milligram is equal to(A) 0.20745 kg(B) 0:2745 kg(C) 0.020745 kg |
|
Answer» 745 milligram=0.745 gmso total=20+0.745=20.745gm1 gm=1/1000 kg20.745 gm=0.020745 kg Tq...... |
|
| 63397. |
The ratio of the prices of two fans was 16 23. Two years later, when the price of thefirst fan had risen by 10% and that of the second by 477, the ratio of their prices became11: 20. Find the original prices of two fans. |
|
Answer» Given:Ratio of two fans = 16:23multiply by common factor xthe prices will be 16x & 23x AFTER TWO YEARSfirst fan increaed by 10 % = 0.1*16x price of the fan =16x+1.6x= 17.6x increase in other fan = Rs. 477price = 23x+477 352x=11(23x+477) 352x=253x+5247352x-253x=5247 99x=5247 x=53Therefore, Price of one fan = 53*16 =Rs. 848 price of second fan = Rs. 1219Like if you find it useful |
|
| 63398. |
The ratio of the prices of two fans was 16: 23. Two years later, when the price of thefirst fan had risen by 10% and that of the second by 477, the ratio of their prices became11 20. Find the original prices of two fans.6. |
| Answer» | |
| 63399. |
20. The ratio of the present prices of two bycycles is 2:3. Two years later, when the price ofthe first is increased by 15% and that of the second by Rs.475, the ratio of their pricesbecomes 3:5. The present price ofthe first bycycle in rupees is |
|
Answer» Let present price of first bicycle be 2xLet present price of second bicycle be 3x After 2 yearsFirst bicycle price = 2x(1 + 15/100) = 2x*(23/20) = 23x/10Second bicycle price = 3x + 475 Then(23x/10):(3x + 475) = 3:523x/2 = 3(3x + 475)23x = 6(3x + 475)23x - 18x = 28505x = 2850x = 570 Therefore, present price of first bicycle = 2x = 2*570 = 1140 |
|
| 63400. |
ts. If each paSter7. The product of two numbers 15 IDJnumbers.A welfare association collected Rs 202500 as donation from the residents. Ifmany rupees as there were residents, find the number of residents.- collected as many paise as there |
| Answer» | |