1. The additive inverse of rational number [latex]\frac{3}{4}[/latex]?
A. [latex]\frac{4}{3}[/latex]
B. [latex]\frac{-4}{3}[/latex]
C. [latex]\frac{-3}{4}[/latex]
D. 0
Answer: Option C
2. If a number divisible by 8 then it also divisible by
A. 2 and 4
B. 2 and 6
C. 2 and 3
D. 4 and 6
Answer: Option A
3. How many circles can pass through the non- col linear points?
A. 0
B. 1
C. 2
D. Infinity
Answer: Option B
4. When no unit of measurement is specified for an angle, then unit associated with the angles is
A. degree
B. radiant
C. cms
D. geometric measure
Answer: Option B
5. A dishonest dealer defrauds to the extent of x% in buying as well as selling his goods by using faulty weight. What will be the gain per cent on his outlay?
A. 2x%
B. ([latex]\frac{10}{x+x²}[/latex])%
C. ([[latex]\frac{2x + x²}{100}[/latex])%
D. ([latex]\frac{x + x²}{100}[/latex])%
Answer: Option C
Explanation:
Total profit = x + x + [latex]\frac{x²}{100}[/latex]
=([latex]\frac{2x + x²}{100}[/latex])%
6. The ratio of cost price and selling price of an article is 20: 21. The gain per cent on it is
Answer: Option B
Explanation:
[latex]\frac{CP}{SP[}[/latex] = [latex]\frac{20}{21}[/latex]
Profit = 21 – 20 = 1
Profit % = [latex]\frac{1}{20}[/latex] × 100 = 5%
7. The ratio of cost price and selling price 25: 26. The per cent of the profit will be
A. 26%
B. 25%
C. 1%
D. 4%
Answer: Option D
Explanation:
[latex]\frac{CP}{SP}[/latex] = [latex]\frac{25}{26}[/latex]
Profit = 26 – 25 = 1
Profit % = 1/25 × 100 = 4%
8. A shopkeeper buys a product of Rs. 150 per kg. 15% of the product was damaged. At what price (per kg) should he sell the remaining so as to earn a profit of 20%?
A. Rs. 218 ([latex]\frac{13}{187}[/latex])
B. Rs. 207 ([latex]\frac{13}{17}[/latex])
C. Rs. 225 ([latex]\frac{13}{1 7}[/latex])
D. Rs. 211 ([latex]\frac{13}{17}[/latex])
Answer: Option D
Explanation:
15% ? [latex]\frac{3}{20}[/latex]
Let total quantity = 20 kg
C.P = 150 × 20 = 3000 Rs.
Money he will get at 20%
Profit = 3000 × [latex]\frac{120}{100}[/latex]
= 3600 Rs.
S.P of = [latex]\frac{3600}{(20 – 3)}[/latex]
= [latex]\frac{3600}{17}[/latex]
= 211[latex]\frac{13}{17}[/latex] Rs.
9. Mr Kapur purchased two toy cycles for Rs. 750 each. He sold these cycles, gaining 6% on one and losing 4% on the other. The gain or loss per cent in the whole transaction is
A. 1% loss
B. 1% gain
C. 1.5% loss
D. 1.5% gain
Answer: Option B
Explanation:
Total S.P.
= [latex]\frac{(750 ×106)}{100}[/latex] + [latex]\frac{(750 × 96)}{100}[/latex]
= 795 + 720
= 1515
Profit = 1515 – 1500 = 15 Rs.
Overall Profit = [latex]\frac{15}{500}[/latex] × 100 = 1%
10. The third proportional of 12 and 18 is
Answer: Option C
Explanation:
18 × [latex]\frac{18}{12}[/latex]
11. Ram got twice as many marks in English as in Science. His total marks in English, Science and Maths are 180. If the ratio of his marks in English and Maths is 2 : 3, what are his marks in Science?
Answer: Option A
Explanation:
marks in English = 2x
marks Maths = 3x
marks Science = x
x + 2x + 3x = 180
6x = 180 x= 30
12. Three numbers are in the ratio 2 : 3: 4. If the sum of their squares is 1856, then the numbers are
A. 8, 12 and 16
B. 16, 24 and 32
C. 12, 18 and 24
D. None of the above
Answer: Option B
Explanation:
(x )[latex]^{2}[/latex]+ (2x)[latex]^{2}[/latex] + (3x )[latex]^{2}[/latex] = 1856
29x[latex]^{2}[/latex] = 1856
x = 8
13. If runs are scored by A, y runs by B and z runs by C, then x : y = y : z = 3 : 2. If total number of runs scored by A, B and C is 342, the runs scored by each would be respectively
A. 144 , 96,64
B. 162 , 108,72
C. 180 , 120 , 80
D. 189,126,84
Answer: Option B
Explanation:
x : y = 3 : 2
y : z = 3 : 2
x : y : z = 9 : 6 : 4
9a + 6a + 4a = 342
a = 18
A = 162 B = 108 C = 72
14. If the average of 6 consecutive even numbers is 25, the difference between the largest and the smallest number is
Answer: Option B
Explanation:
Nubers = x, x + 2, ------ , x + 10
Req diff = x + 10 - x = 10
15. A train goes from Ballygunge to Sealdah at an average speed of 20 km/hour and comes back at an average speed of 30 km/hour. The average speed of the train for the whole journey is
A. 27 km/hr
B. 26 km/hr
C. 25 km/hr
D. 24 km/hr
Answer: Option D
Explanation:
Req avg speed = 2 × 30 × [latex]\frac{20}{30}[/latex] + 20 =24kmph
16. The arithmetic mean of 100 observations is 24.6 is added to each of the observations and, then each of them is multiplied by 2.5. Find the new arithmetic mean.
Answer: Option B
Explanation:
on adding arithmetic mean = 24 + 6 = 30
on multiplying by 2.5 arithmetic mean = 30 × 2.5 = 75
17. There are five bells which start ringing together at intervals of 3, 6, 9, 12 and 15 seconds respectively. In 36 minutes, how many times will the bells ring simultaneously?
Answer: Option A
Explanation:
lcm of 9, 12, 15 = 180 sec
req = 36 × [latex]\frac{60}{180}[/latex] + 1 = 13
18. n is a whole number which when divided by 4 gives the remainder 3. The remainder when 2n is divided by 4 is
Answer: Option B
Explanation:
take n = 7 because 4 × 1 + 3 = 7
so, 2n = 14 the remainder will be 2
19. Two buses travel to a place at 45 km/hr and 60 km/hr respectively. If the second bus takes 5½ hours less than the first for the journey, the length of the journey is:
A. 900 km
B. 945 km
C. 990 km
D. 1350 km
Answer: Option C
Explanation:
Speed ratio = 45 : 60 = 3 : 4
Time ratio = 4 : 3 [S = 1/time when distance is same]
(4 – 3) ratio = [latex]\frac{11}{2}[/latex] hours
1 ratio = [latex]\frac{11}{2}[/latex] hours
time taken by bus travelling at 45 km/hr = 4 × [latex]\frac{11}{2}[/latex] = 22 hours
Distance = 45 × 22= 990 km
20. At an average of 80 km/hr, Shatabdi Express reaches Ranchi from Kolkata in 7 hrs. Then the distance between Kolkata and Ranchi is
A. 560 km
B. 506 km
C. 560 m
D. 650 m
Answer: Option C
Explanation:
Distance = 80 × 7
= 560 m
21. Two donkeys are standing 400 metres apart. The first donkey can run at a speed of 3 m/sec and the second can run at 2 m/sec. If two donkeys run towards each other after how much time (in a sec) will they bump into each other?
Answer: Option B
Explanation:
time = [latex]\frac{400}{5}[/latex]
= 80 seconds
22.A and B undertake to do a piece of work for Rs 600. An alone can do it in 6 days while B alone can do it in 8 days. With the help of C, they can finish it in 3 days, Find the share of C?
A. Rs 50
B. Rs 75
C. Rs.100
D. Rs.150
Answer: Option B
Explanation:
C’s one day’s work=([latex]\frac{1}{3}[/latex])-([latex]\frac{1}{6}[/latex] + [latex]\frac{1}{8}[/latex]) = [latex]\frac{1}{24}[/latex]
Therefore, A:B:C= Ratio of their one day’s work = [latex]\frac{1}{6}[/latex] : [latex]\frac{1}{8}[/latex]) : [latex]\frac{1}{24}[/latex] = 4 : 3 : 1
A’s share=Rs (600 × [latex]\frac{4}{8}[/latex]) =3 00
B’s share= Rs (600 × [latex]\frac{3}{8}[/latex]) = 225
C’s share=Rs [600 - (300 + 225)] = Rs 75
23.P, Q and R are three typists who working simultaneously can type 216 pages in 4 hours In one hour, R can type as many pages more than Q as Q can type more than P. During a period of five hours, R can type as many pages as P can during seven hours. How many pages does each of them type per hour?
A. x=15,y=18, and z=21
B. x=15,y=21, and z=21
C. x=18,y=18, and z=21
D. x=15,y=18, and z=15
Answer: Option B=A
Explanation:
Let the number of pages typed in one hour by P, Q and R be x,y and z
respectively
Then x + y + z = [latex]\frac{216}{4}[/latex] = 54 —————1
z - y = y - x => 2y = x + z ———–2
5z = 7x => x = [latex]\frac{5x}{7}[/latex] —————3
Solving 1,2 and 3 we get x = 15,y = 18, and z = 21
24.Ronald and Elan are working on an assignment. Ronald takes 6 hours to type 32 pages on a computer, while Elan takes 5 hours to type 40 pages. How much time will they take, working together on two different computers to type an assignment of 110 pages?
A. 6 hours
B. 8 hours
C. 10 hours
D. 12 hours
Answer: Option B
Explanation:
Number of pages typed by Ronald in one hour = [latex]\frac{32}{6}[/latex] = [latex]\frac{16}{3}[/latex]
Number of pages typed by Elan in one hour = [latex]\frac{40}{5}[/latex] = 8
Number of pages typed by both in one hour = (([latex]\frac{16}{3}[/latex]) + 8) = [latex]\frac{40}{3}[/latex]
Time taken by both to type 110 pages = 110 × [latex]\frac{3}{40}[/latex] = 8 hours.
25.A can do certain work at the same time in which B and C together can do it. If A and B together could do it in 10 days and C alone in 50 days, then B alone could do it in how many days?
A. 80 days
B. 50 days
C. 25 days
D. 12 days
Answer: Option C
Explanation:
(A+B)’s one day’s work = [latex]\frac{1}{10}[/latex];
C’s one day’s work = [latex]\frac{1}{50}[/latex]
(A+B+C)’s one day’s work = ([latex]\frac{1}{10}[/latex] + [latex]\frac{1}{50}[/latex]) = [latex]\frac{6}{50}[/latex] = [latex]\frac{3}{25}[/latex]
Also, A’s one day’s work = (B+C)’s one day’s work
From i and ii ,we get :2 × (A’s one day’s work) = [latex]\frac{3}{25}[/latex]
=> A’s one day’s work =3/50
B’s one day’s work=([latex]\frac{1}{10}[/latex] - [latex]\frac{3}{50}[/latex])
= [latex]\frac{2}{50}[/latex]
= [latex]\frac{1}{25}[/latex]
B alone could complete the work in 25 days.
26. The profit percentage on the three articles A, B and C is 10%, 20% and 25% and the ratio of the cost prices is 1 : 2: 4. Also the ratio of the number of articles sold of A, B and C is 2: 5: 2 then the overall profit percentage is-
A. 21%
B. 43%
C. 25%
D. 12%
Answer: Option A
Explanation:
Ratio of CP = 1 : 2 : 4
Ratio of No.
of articles
sold = 2 : 5 : 2
2 : 10 : 8
Ratio of % profit = 10% : 20% : 25%
SP = 1 × 1.1 : 5 × 1.2 : 4 × 1.25
Total SP = 1.1 : 6 : 5 = 12.1
So, Net % profit = [latex]\frac{(12.1- 10)}{10}[/latex] × 100
= 21%
27. Three friends A, B and C started a business by investing the amount in the ratio of 5: 7: 6 respectively. After a period of six months, C withdrew half of the amount invested by him. If the amount invested by A is Rs. 40,000 and the total profit earned at the end of one year is Rs. 33,000, what is C’s share in the profit?
A. 8000
B. 7200
C. 9000
D. 8800
Answer: Option C
Explanation:
Given that:-
Invested ratio of A : B : C = 5 : 7 : 6
After 6 months:-
Invested ratio of A : B : C = 60 : 84 : 54
Now, ratio = 40000 × 12 : 56000 × 12 : (48000 × 6 + 24000 × 6)
Profit ratio = 10: 14 : 9
Share of profit of C = [latex]\frac{9}{33}[/latex] × 33000
= Rs. 9000
28. Sohan starts a business by investing Rs. 25,000. 6 months later Aditya joins him by investing Rs. 15,000. After another 6 months, Aditya invests an additional amount of Rs. 15,000. At the end of 3 years they earn a profit of Rs. 2,47,000. What is Aditya’s share in the profit?
A. 2, 27, 500
B. 1, 12, 000
C. 1, 17, 000
D. 1, 12, 250
Answer: Option C
Explanation:
According to question:-
Sohan = 25000 × (36 months)
= Rs. 900000
Aditya = [15000 × 30 + 15000 × 24]
= Rs. 810000
Profit share of Aditya
= [latex]\frac{Sohan}{(Sohan+Mohan)}[/latex] × 247000
= [latex]\frac{9}{19}[/latex] × 247000
= 1,17,000
29. A tap can fill a tank in 6 hours. After half the tank is filled, three more similar taps are opened. What is the total time taken to fill the tank completely?
A. 3 hours 15 min
B. 3 hours 45 min
C. 3 hours 40 min
D. 3 hours 50 min
Answer: Option B
Explanation:
A tap can fill a tank in 6 hours.
After half the tank is filled i.e. after 3 hrs.
Three more similar taps are opened i.e.
no. of taps to fill remaining half tank = 4 taps
when 1 tap is used = (3 hrs/1 tap)
when 4 taps are used = (3 hrs/4 taps) = 45min
total= 3hrs + 45min
30. An electric pump can fill a tank in 3 hours. Because of a leak in , the tank it took 3([latex]\frac{1}{2}[/latex]) hours to fill the tank. If the tank is full, how much time will the leak take to empty it?
A. 28 hours
B. 25 hours
C. 21 hours
D. 20 hours
Answer: Option D
Explanation:
work done by the leak in 1 hour=([latex]\frac{1}{3}[/latex])-([latex]\frac{2}{7}[/latex]) = ([latex]\frac{1}{21}[/latex]).
The leak will empty .the tank in 21 hours.
31. A T.V was sold at a profit of 5%. If it had been sold at a profit of 10% the profit would have been Rs. 1000 more. What is its cost price?
A. Rs. 20000
B. Rs. 5000
C. Rs. 10000
D. Rs. 15000
Answer: Option A
Explanation:
+10% – 5% → 1000 Rs.
5% → 1000 Rs.
1% → 200 Rs.
100% → 20,000 Rs.
35. If an article is sold at 5% gain instead of 5% loss, the man gains Rs. 5 more. Find the cost price of that article
A. Rs. 100
B. Rs. 105
C. Rs. 50
D. Rs. 110
Answer: Option C
Explanation:
+5% – (–5%) ⇒ Rs. 5
10% = Rs. 5
1% = [latex]\frac{1}{2}[/latex]
100% = Rs. 50
33. By selling a table for Rs. 350 instead of Rs. 400, loss percent increases by 5%. The cost price of table is:
A. Rs. 1,050
B. Rs. 417.50
C. Rs. 435
D. Rs. 1,000
Answer: Option D
Explanation:
5% ⇒ Rs. 50
1% ⇒ Rs. 10
100% ⇒ Rs. 1000
34. The marked price of an article is Rs. 500. It is sold at successive discounts of 20% and 10%. The selling price of the article (Rs.) is:
A. Rs. 350
C. Rs. 375
C. Rs. 360
D. Rs. 400
Answer: Option C
Explanation:
Single discount = –20 – 10 + 2
= –28%
S.P = 500 × [latex]\frac{72}{100}[/latex]
= 360 Rs.
35. The marked price of an article is 10% higher than the cost price. A discount of 10% is given on the marked price. In this kind of sale, the seller bears
A. No Loss / No Gain
B.5% Loss
C. 1% Gain
D. 1% Loss
Answer: Option D
Explanation:
P = 10 – 10 – 1
= –1%
⇒ loss → 1%
36. From a solid cylinder of height 10 cm and radius of the base 6 cm, a cone of the same height and the same base is removed. The volume of the remaining solid is:
A. 240π cu.cm
B. 5280 cu.cm
C. 620π cu.cm
D. 360π cu.cm
Answer: Option A
Explanation:
Volume of cylinder = πr²h
= (6)² × 10 × π
= 360 π cm³
Volume of cone = [latex]\frac{1}{3}[/latex]πr²h
= [latex]\frac{1}{3}[/latex]× π × 36 × 10
= 120 π cm³
Volume of Remaining Solid
= 360 π – 120 π
= 240 π cm³
37. What part of a ditch, 48 metres long 16.5 metres broad and 4 metres deep can be filled by the sand got by digging a cylindrical tunnel of diameter 4 metres and length 56 metres?
A. [latex]\frac{1}{9}[/latex]
B. [latex]\frac{2}{9}[/latex]
C. [latex]\frac{7}{9}[/latex]
D. [latex]\frac{8}{9}[/latex]
Answer: Option B
Explanation:
Volume of ditch = lbh
= 48 × 16.5 × 4
= 48 × 66
Volume of sand = πr²h
= [latex]\frac{22}{7}[/latex] × 4 × 56
= 22 × 4 × 8
Part of Ditch filled
= [latex]\frac{(22 × 4 ×8)}{(48 × 66)}[/latex]
= [latex]\frac{2}{9}[/latex]
38. The size of a rectangular piece of paper is 100 cm × 44 cm. A cylinder is formed by rolling the paper along its length. The volume of the cylinder is
A. 4400 cm³
B. 15400 cm³
C. 35000 cm³
D. 144 cm³
Answer: Option B
Explanation:
2πr = 44
2 × [latex]\frac{22}{7}[/latex] × r = 44
r = 7 cm
h = 100 cm
Volume of cylinder = πr²h
= [latex]\frac{22}{7}[/latex] × 49 × 100
= 15400 cm³
39. A circus tent is cylindrical up to a height of 3 m and conical above it. If its diameter is 105 m and the slant height of the conical part is 63 m, then the total area of the canvas required to make the tent is
A. 11385 m²
B. 10395 m²
C. 9900 m²
D. 990 m²
Answer: Option A
Explanation:
Total area of canvas required
= Surface area of cylindrical part + Surface area of conical part
= 2πr₁h₁ + πr₂l₂
= π × 105 × 3 + π × [latex]\frac{105}{2}[/latex]× 63
= 11385 m²
40. If a cubic cm of cast iron weights 21 gms, then the weight of a cast-iron pipe of length 1 m with a bore of 3 cm and in which the thickness of the metal is 1 cm, is:
A. 46.2 kg
B. 24.2 kg
C. 26.4 kg
D. 18.6 kg
Answer: Option A
Explanation:
Bore = Inner radius = 3 cm
outer radius = 3 + 1 = 4 cm
volume of iron pipe = π(4[latex]^{2}[/latex] - 3[latex]^{2}[/latex]) × 100
= 2200
weight of iron = 2200 × 21 gm = 46200 gm = 46.2 kg
41. The value of (256) [latex]^{\frac{5}{4}}[/latex] is :
A. 512
B. 984
C. 1024
D. 1032
Answer: Option C
Explanation:
From the given equation :
(256) [latex]^{\frac{5}{4}}[/latex]
= (4[latex]^{4}[/latex])[latex]^{\frac{5}{4}}[/latex]
= 4[latex]^4 ×{\frac{5}{4}}[/latex]
= 4[latex]^{5}[/latex]
= 1024.
42. (2.4 × 10[latex]^{3}[/latex]) ÷ (8 × 10[latex]^{-2}[/latex]) = ?
A. 3 × 10[latex]^{-5}[/latex]
B. 3 × 10[latex]^{4}[/latex]
C. 3 × 10[latex]^{5}[/latex]
D. 30
Answer: Option B
Explanation:
Given equation
= (2.4 × 103) ÷ (8 × 10[latex]^{-2}[/latex])
then, 2.4 × 10[latex]^{3}[/latex]
8 × 10[latex]^{-2}[/latex]
= 24 × 10[latex]^{2}[/latex]
8 × 10[latex]\frac{}{}[/latex]– 2
= (3 × 10[latex]^{4}[/latex])
43. (1000)[latex]^{7}[/latex] ÷ 10[latex]^{18}[/latex] = ?
A. 10
B. 100
C. 1000
D. 10000
Correct Option: C
Given equation = (1000)[latex]^{7}[/latex] ÷ 1018
⇒ [latex]\frac{{1000}^{7}}{{10}^{18}}[/latex] ⇒ [latex]\frac{{10}^{(3 × 7)}}{{10}^{18}}[/latex]
⇒ [latex]\frac{{10}^{21}}{{10}^{18}}[/latex] = 10[latex]^{(21 - 18)}[/latex] ⇒ 10[latex]^{3}[/latex] = 1000.
44. 49 × 49 × 49 × 49 = 7[latex]^{?}[/latex]
Answer: Option C
Explanation:
From the given equation :
49 × 49 × 49 × 49
⇒ (7[latex]^{2}[/latex]× 7[latex]^{2}[/latex] × 7[latex]^{2}[/latex] × [latex]^{2}[/latex])
⇒ 7[latex]^{(2 + 2+ 2+ 2)}[/latex]
⇒ 7[latex]^{8}[/latex]
45. The value of (8 [latex]^{-25}[/latex] – 8 [latex]^{-26}[/latex]) is :
A. 7 × 8 8[latex]^{-25}[/latex]
B. 7 × 8 8[latex]^{-26}[/latex]
C. 8 × 8 8[latex]^{-26}[/latex]
D. 8 × 8 8[latex]^{-25}[/latex]
Answer: Option B
Explanation:
= ([latex]\frac{1}{8}^{25}[/latex] − [latex]\frac{1}{8}^{26}[/latex])
= [latex]\frac{(8 − 1)}{8}^{26}[/latex]
= 7 × 8[latex]^{-26}[/latex]
46. If cot ([latex]\frac{Π}{2}[/latex] – θ ) = √3, then the value of cos θ is
A. 0
B. [latex]\frac{1}{√2}[/latex]
C. [latex]\frac{1}{2}[/latex]
D. 1
Answer: Option C
Explanation:
cot ([latex]\frac{Π}{2}[/latex] – θ) = √3
⇒ tan θ = √3
[ ∵ cot ([latex]\frac{Π}{2}[/latex] – θ) = tan θ]
∴ θ = 60°
Hence, cos 60° = [latex]\frac{1}{2}[/latex]
47. If tan4θ + tan2θ = 1, then the value of cos4θ + cos2θ is
Answer: Option C
Explanation:
tan4θ + tan2θ = 1
tan2θ (tan2θ + 1)
⇒ tan2θ. sec2θ = 1 {∵ sec2θ = 1 + tan2θ}
⇒ [latex]\frac{sin2θ}{cos4θ}[/latex] = 1
⇒ 1– cos2θ = cos4 θ {∵ sin2 θ = 1 – cos2θ}
⇒ cos4 θ + cos2 θ = 1
48. The value of sin (45° + θ) – cos (45° – θ) is
A. 1
B. 0
C. 2 cos θ
D. 2 sin θ
Answer: Option B
Explanation:
sin(45° + θ) – cos(45° – θ)
= sin{90° – (45° – θ)} – cos (45° – θ)
= cos (45° – θ) – cos (45° – θ)
{∵sin(90 – A) = cos A }
= 0
49. If cot A + cosec A = and A is an acute angle, then the value of cos A is
A. [latex]\frac{4}{5}[/latex]
B. 1
C. [latex]\frac{1}{2}[/latex]
D. [latex]\frac{1}{√3}[/latex]
Answer: Option A
Explanation:
Here, cot A + cosec A = 3
⇒ [latex]\frac{cosA}{sinA}[/latex] + [latex]\frac{cosA}{sinA}[/latex] = 3
⇒ cos A + 1 = 3sin A
Squaring both sides.
⇒ (cos A + 1)[latex]^{2}[/latex] = 9sin[latex]^{2}[/latex] A
⇒ cos[latex]^{2}[/latex] A + 2cos A + 1 = 9 (1 – cos[latex]^{2}[/latex] A)
⇒ 10 cos2A + 2 cos A – 8 = 0
⇒ 5 cos2 A + cosA – 4 = 0
⇒ (5 cos A – 4)(cos A + 1) = 0
(∵ A is an acute angle)
∴ cos A = [latex]\frac{4}{5}[/latex]
50. If tan (x + y) tan (x – y) = 1, then the value of tan x is
A. √3
B. 1
C. [latex]\frac{1}{2}[/latex]
D. [latex]\frac{1}{√3}[/latex]
Answer: Option B
Explanation:
Here, tan (x + y) tan (x – y) = 1
⇒tan (x + y) tan (x – y) = 1
⇒ tan (x + y)
= cot (x – y) = tan ([latex]\frac{Π}{2}[/latex] – x + y )
{∵ tan (90° – θ) = cotθ}
∴ x + y = [latex]\frac{Π}{2}[/latex] – x +y ⇒ x = [latex]\frac{}{}[/latex]π/4
∴ tan x = tan [latex]\frac{Π}{4}[/latex] = 1
Directions(51-55): Study the following table chart carefully and answer the questions given beside.
The following table gives information about the populations of five different cities in the year 2017.
City |
Total populations |
Percentage of men |
Percentage of women |
Percentage of children |
A |
4860 |
45% |
35% |
20% |
B |
4540 |
35% |
55% |
10% |
C |
1500 |
44% |
38% |
18% |
D |
4850 |
38% |
48% |
14% |
E |
3650 |
56% |
36% |
8% |
51. What is the ratio of the total number of men in the city C to the total number of women in the city E?
A. 11 : 9
B. 110 : 221
C. 110 : 217
D. 110 : 219
Answer: Option D
Explanation:
The required ratio = 44% of 1500 : 36% of 3650
= 44 × 1500 : 36 × 3650 = 11 × 30: 9 × 73 = 110 : 219
52. In the year 2018, the population of children in the city A increased by 25% over the previous year and the population of children in the city B increased by 50% over the previous year, then what is the average of the population of children in the city A and B together in the year 2018?
A. 984
B. 948
C. 931
D. 924
Answer: Option B
Explanation:
In the year 2017, the population of children in the city A = 20% of 4860
= [latex]\frac{20 }{100}[/latex] × 4860 = 972
The new population of children in the year 2018 = 125% of 972
= [latex]\frac{125}{100}[/latex] × 972 = [latex]\frac{5}{4}[/latex] × 972 = 1215
In the year 2017, the population of children in the city B
= 10% of 4540 = [latex]\frac{10}{100}[/latex] × 4540 = 1 × 454 = 454
The new population of children in the year 2018 = 150% of 454
= [latex]\frac{150 }{100}[/latex] × 454 = [latex]\frac{3}{2}[/latex] × 454 = 681
The reqd. average = [latex]\frac{1215 + 681}{2}[/latex] = [latex]\frac{1896}{2}[/latex] = 948
53. In the city C, among the children the ratio of the boys to the girls was 2 : 3 and in the city D among the children the ratio of the boys to the girls was 4: 3 , then what was the total number of girls (among children) in the city C and city D together?
A. 453
B. 395
C. 473
D. 415
Answer: Option A
Explanation:
The population of children in the city C = 18% of 1500 = 270
The number of girls in the city C = [latex]\frac{3}{5}[/latex] × 270 = 162
the population of children in the city D = 14% OF 4850 = 679
The number of girls in the city D = [latex]\frac{3}{7}[/latex] × 679 = 291
the total number of girls children in the city C and city D together = 162 + 291 = 453
Hence, option A is correct.
54. The number of women in city D forms approximately what percentage of the number of men in city E? (approximately)
A. 106%
B. 102%
C. 110%
D. 114%
Answer: Option D
Explanation:
The number of women in the city D = 48% of 4850 = 2328
The number of men in the city E = 56% of 3650 = 2044
The reqd. % = [latex]\frac{2328}{2044}[/latex]2328 × 100 = 113.89% = 114%(approx)
55. If the population of all the cities are combined together then what is the sum of the total number of men and women in all the five cities together?
A. 15873
B. 19458
C. 15993
D. 16733
Answer: Option D
Explanation:
If we subtract the total number of children from all the five cities together with the total population of all the five cities together then we can get the total number of men and women in all the five cities together
The total population of all the five cities together = 4860 + 4540 + 1500 + 4850 + 3650 = 19400
The total population of children from all the five cities together = 20% of 4860 + 10% of 4540 + 18% of 1500 + 14% of 4850 + 8% of 3650 = 972 + 454 + 270 + 679 + 292 = 2667
The required answer = 19400 – 2667 = 16733
56. Common energy source in Indian villages is:
A. Electricity
B. Coal
C. Sun
D. Wood and animal dung
Answer: Option D
57. The one thing that is common to all fossil fuels is that they:
A. Were originally formed in marine environment
B. Contain carbon
C. Have undergone the same set of geological processes during their formation
D. Represent the remains of one living organisms
Answer: Option B
58. The process that converts solid coal into liquid hydrocarbon fuel is called:
A. Liquefaction
B. Carbonation
C. Catalytic conversion
D. Cracking
Answer: Option A
59. Lignite, bituminous and anthracite are different ranks of:
A. Nuclear fuel
B. Coal
C. Natural gas
D. Bio gas
Answer: Option B
60. Cruid oil is:
A. Colour less
B. Odorless
C. Smelly yellow to black liquid
D. Odorless yellow to black liquid
Answer: Option C
61. BTU is measurement of:
A. Volume
B. Area
C. Heat content
D. Temperature
Answer: Option C
62. The first controlled fission of an atom was carried out in Germany in:
A. 1920
B. 1928
C. 1925
D. 1938
Answer: Option D
63. Boiling water reactor and pressurized water reactors are:
A. Nuclear reactor
B. Solar reactor
C. OTEC
D. Bio-gas reactor
Answer: Option A
64. In Human beings the process of digestion of food begins in:
A. Stomach
B. Food Pipe
C. Mouth
D. Small Intestine
Answer: Option C
65. Which of the following organism have parasitic mode of nutrition?
A. Penicillium
B. Plasmodium
C. Paramecium
D. Parrot
Answer: Option B
66. Name the first enzyme that mix with food in the digestive tract?
A. Pepsin
B. Trypsin
C. Amylase
D. None of the above
Answer: Option C
67. Which of the following in biology is the energy currency of cells?
A. PDP
B. DTP
C. ATP
D. ADP
Answer: Option C
68. In the stem of a plant respiration and breathing takes place through:
A. Lenticels
B. Stomata
C. Root hair
D. Air tubes
Answer: Option A
69. Which animal has three-chambered heart?
A. Pigeon
B. Lizard
C. Fish
D. Lion
Answer: Option B
70. Uricotelism is found in ____________
A. Birds reptiles and insects
B. Frogs and toads
C. Mammals and birds
D. Fishes and fresh water protozoans
Answer: Option c
71. A terrestrial animal must be able to ________
A. Excrete large amount of water
B. Actively pump salts through skin
C. Excrete large amount of salts in urine
D. Conserve water
Answer: Option D
72. Animals which excrete urea are known as ____________
A. Aminotelism
B. Ureotelism
C. Uricotelism
D. Ammonotelism
Answer: Option C
73. Which of the following nephridia are not found in earthworm?
A. Septal nephridia
B. Macronephric nephridia
C. Pharyngeal nephridia
D. Integumentary nephridia
Answer: Option B
74. Excretory waste of birds and reptiles are _________
A. Urea
B. Uric acid and urea
C. Uric acid
D. Ammonia and uric acid
Answer: Option D
75. Exchange of genetic material takes place in
A. Vegetative reproduction
B. Asexual reproduction
C. Sexual reproduction
D. Budding
Answer: Option C
76. Two pink coloured flowers on crossing resulted in 1 red, 2 pink and 1 white flower progeny. The nature of the cross will be
A. Double fertilization
B. Self pollination
C. Cross fertilization
D. No fertilization
Answer: Option C
77. A cross between a tall plant (TT) and short pea plant (tt) resulted in progeny that were all tall plants because
A. Tallness is the dominant trait
B. Shortness is the dominant trait
C. Tallness is the recessive trait
D. Height of pea plant is not governed by gene 'T' or 't'
Answer: Option A
78. Which of the following statement is incorrect?
A. For every hormone there is a gene
B. For every protein there is a gene
C. For production of every enzyme there is a gene
D. For every molecule of fat there is a gene
Answer: Option D
79. If a round, green seeded pea plant (RR yy) is crossed with wrinkled, yellow seeded pea plant, (rr YY) the seeds production in F1 generation are
A. Round and yellow
B. Round and green
C. Wrinkled and green
D. Wrinkled and yellow
Answer: Option A
80. The isomeric pair is
A. ethane and propane
B. propane and butane
C. ethane and ethane
D. butane and 2-methyl propane
Answer: Option D
81. Which of the following is used to oxidise ethanol to ethanoic acid?
A. Alkaline KMnO4
B. Conc. H2SO4
C. Acidified K2Cr2O7
D. All of above
Answer: Option D
82. Which is denatured spirit?
A. ethanol only
B. ethanol and methanol (50%)
C. ethanol and methanol (5%)
D. methanol only
Answer: Option C
83. Tertiary butane gets oxidised with oxidising agents like alkaline KMNO4 to
A. Isobutane
B. Ter-butyl alcohol
C. Secondary-propyl alcohol
D. All of above
Answer: Option B
84. The substance not responsible for the hardness of water is
A. Sodium nitrate
B. calcium hydrogen carbonate
C. calcium carbonate
D. magnesium carbonate
Answer: Option A
85. The by product of soap is
A. isoprene
B. glycerol
C. butane
D. ethylene glycol
Answer: Option B
86. Biosphere is
A. The solid shell of inorganic materials on the surface of the Earth
B. The thin shell of organic matter on the surface of earth comprising of all the living things
C. The sphere which occupies the maximum volume of all the spheres
D. All of the above
Answer: Option B
87. Which of the following conceptual sphere of the environment is having the least storage capacity for the matter?
A. Atmosphere
B. Lithosphere
C. Hydrosphere
D. Biosphere
Answer: Option A
88. Which of the following is an example of the impact of development activities on the Hydrosphere?
A. Air pollution
B. Soil pollution
C. Soil erosion
D. Water pollution
Answer: Option D
89. Weather occurs in the Earths:
A. Troposphere
B. Mesosphere
C. Ionosphere
D. Thermosphere
Answer: Option A
90. Biosphere is a term used to represent the:
A. Entire atmosphere consisting of troposphere, stratosphere, mesosphere, and thermosphere
B. Entire hydrosphere-representing the entire collection of water over the Earth as well as inside the Earth
C. A small zone of Earth, where the lithosphere, hydrosphere, and atmosphere come in contact with one another
D. Entire lithosphere-representing the solid Earth and its interior
Answer: Option C
91. The period of one revolution of sun around the centre of galaxy is called:
A. Cosmic year
B. Astronomical year
C. Light year
D. Parsec
Answer: Option A
92. Which of the following is concerned with the description and mapping of the main features of the universe?
A. Cosmology
B. Cosmography
C. Astronomy
D. None of these
Answer: Option B
93. The same side of the moon always faces the earth because:
A. Moon and earth have gravitational force
B. Moon cannot change its position
C. The period of rotation of the moon on its axis and period of revolution around the earth is almost the same
D. None of these
Answer: Option C
94. The light coming from stars gives the idea of their:
A. Size
B. Rotational speed
C. Mass
D. Density
Answer: Option B
95 The constellation of stars appears at different positions in the sky at different times during night mainly:
A. Because earth rotates about its axis
B. Because earth revolves around the sun
C. Because of optical illusion
D. Because celestial bodies are changing their position
Answer: Option A
96. The problems for criticism about large dams are that they
A. Displace large number of peasants and trebles without proper rehabilitation
B. Swallow up huge amounts of public money without the generation of proportionate benefits
C. Contribute enormously to deforestation and the loss of biological diversity
D. all of the above.
Answer: Option D
97. The following are stakeholders of forests which one of these causes the maximum damage to forest?
A. People who live in or around the forest
B. The industrial
C. The wildlife and native enthusiasts
D. The forest department of the government.
Answer: Option B
98. The concept of ‘Biosphere Reserve’ was evolved by
A. Government of India.
B. Botanical Survey of India
C. UNESCO
D. UNDP.
Answer: Option C
99 Chipko Andolan is concerned with
A. Conservation of natural resources
B. Development of new breeds of forest plants
C. Zoological survey of India
D. Forest conservation.
Answer: Option A
100. Which energy of water is used to produce hydroelectricity?
A. Potential energy
B. Kinetic energy
C. Both A and B of these
D. None of these.
Answer: Option A