Every year SSC conducts many exams like CGL, CHSL, MTS, etc. SSC CGL provides leading jobs in ministries and other government department posts. SSC CGL exam has four phases tier 1 , tier 2 , written exam and interview. Tier 2 exam covers quantitative aptitude and English section. To get the proper idea and pattern of the question papers a candidate should be focused on previous years question papers. Here we provide the SSC CGL tier 2 question paper for 2020 for the quantitative aptitude section.
Que 1. A person saves 33 ⅓% of his income. If the saving Increases by 22% and the expenditure Increases by 10%, then the percentage Increase In his income is:
1. 18%
2. 14%
3. 16%
4. 22%
Answer : 2
Exp :
Savings = 33⅓% = 1/3
Expenditure = (1 – 1/3) = 2/3
Ratio of expenditure & Savings = 2/3 : 1/3 = 2 : 1
Let income increase percentage = x
Applying alligation method,
Expenditure Savings
10 22
x
(22 – x) (x – 10)
Now, (22 – x) : (x – 10) = 2 : 1
=> 22 – x = 2x – 20
=> 3x = 42
=> x = 14%
∴ The percentage increase in his income is 14%.
Here the correct option is 2.
Que 2. P can finish a work in 18 days. When he had worked for 5 days, Q joined him. If both of them together completed the remaining work in 13/5 days, then in how many days can Q alone finish 66 ⅔% of the same work?
1. 5
2. 4
3. 2
4. 3
Answer : 4
Exp :
M1 × D1 = M2 × D2
=> P × 18 = (P × 5) + (P + Q) × 13/5
=> 90P = 25P + (P + Q) × 13 [ Multiplying both sides by 5]
=> 52P = 13Q
=> 4P = Q
Here efficiency of Q is 4 times of P.
66⅔% = 2/3
Total work = 1
Let Q complete the 66⅔% of the work in D2 days.
(P × 18)/work = (Q × D2)/work
=> (P × 18)/1 = (4P × D2)/(2/3)
=> 18P = 6P × D2
=> D2 = 3
∴ The 66⅔% of work was completed by Q in 3 days.
Here the correct option is 4.
Que 3. In ∆ABC, AB = 20 cm, BC = 7 cm and CA = 15 cm. Side BC is produced to D such that ∆DAB ~ ∆DCA. DC is equal to:
1. 9 cm
2. 8 cm
3. 10 cm
4. 7 cm
Answer : 1
Exp :
Let CD = p cm and AD = q cm
AB/AC = DB/AD = AD/DC ( Since ∆DAB ~ ∆DCA)
now taking AB/AC = DB/AD
20/15 = (7 + p)/q
=> 4/3 = (7 + p)/q
=> q = (21 + 3p)/4 ————– (1)
Now taking AB/AC = AD/DC
20/15 = q/p
=> q = 4p/3 ———————- (2)
Equating equations (1) and (2), we get
4p/3 = (21 + 3p)/4
=> 16p = 63 + 9p
=> 7p = 63
=> p = 9
∴ The length of DC = 9 cm.
Here the correct option is 1.
Que 4. The areas of three adjacent faces of a cuboidal solid block of wax are 216 cm², 96 cm² and 144 cm². It is melted and 8 cubes of the same size are formed from it. What is the lateral surface area (in cm²) of 3 such cubes?
1. 648
2. 432
3. 576
4. 288
Answer : 2
Exp :
Let the side of the cube = a
We know,
Square of the volume of a cuboid = Product of three adjacent faces of the cuboid
So,
(8 × a³)² = (216 × 96 × 144)
=> (8 × a³) = √( 6 × 6 × 6 × 6 × 4 × 4 × 12 × 12)
=> (8 × a³) = (6 × 6 × 4 × 12)
=> a³ = 6 × 6 × 6
=> a = 6
Lateral surface area of a cube = 4 × side² = 4 × 6² = 144 cm²
∴ Lateral surface area of 3 cubes = (3 × 144) = 432 cm²
Here the correct option is 2.
Que 5. The ratio of the Incomes of A and B in 2020 was 5 : 4. The ratios of their Individual Incomes in 2020 and 2021 were 4 : 5 and 2 : 3, respectively. If the total income A and B in 2021 was Rs.7,05,600, then what was the Income (in Rs.) of B in 2021?
1. 3,45,600
2. 2,79,700
3. 3,60,000
4. 4,25,900
Answer : 1
Exp :
A : B in 2020 = 5 : 4 —-(×4) = 20 : 16
A in 2020 : 2021 = 4 : 5 —-(×5) = 20 : 25
B in 2020 : 2021 = 2 : 3 —-(×8) = 16 : 24
Income of A in 2021 = 25 units
Income of B in 2021 = 24 units
Now according to the question,
(25 + 24) units = 705600
=> 1 unit = 705600/49
=> 24 units = 14400 × 24
=> 24 units = 345600
∴ The income of B in 2021 is Rs.3,45,600.
Here the correct option is 1.
Que 6. In a medical transaction, 17 times the cost price Is equal to 8 times the sum of the cost price and the selling price. What is the gain or loss percentage?
1. Loss 15%
2. Gain 17.5%
3. Gain 12.5%
4. Loss 30%
Answer : 3
Exp :
17 × CP = 8 × ( CP + SP)
=> 9 CP = 8 SP
= CP/SP = 8/9
Profit = (9 – 8) = 1
Profit percentage = 1/8 × 100 % = 12.5%
∴ The gain percentage is 12.5%.
Here the correct option is 3.
Que 7. In an examination, B obtained 20% more marks than those obtained by A, and A obtained 10% less marks than those obtained by C. D obtained 20% more marks than those obtained by C. By what percentage are the marks obtained by D more than those obtained by A?
1. 33 ⅓%
2 13 ⅓%
3. 43 ⅓%
4. 23 ⅓%
Answer : 1
Exp :
Let C get 100 marks.
D = 100 × 120/100 = 120
A = 100 × 90/100 = 90
B = 90 × 120/100 = 108
Required percentage = (120 – 90)/90 × 100% = 1/3 × 100 = 33⅓%
∴ 33⅓% marks more marks obtained by D than by A.
Here the correct option is 1.
Que 8. The value of [(2.53)³ + (2.47)³]/(25.3 × 25.3 – 624.91 + 24.7 × 24.7) is (5 × 10K) where the value of k is :
1. – 2
2. – 1
3. 1
4. 2
Answer : 1
Exp :
[(2.53)³ + (2.47)³]/(25.3 × 25.3 — 624.91 + 24.7 × 24.7)
= [(2.53)³ + (2.47)³]/[(25.3)² — 25.3 × 24.7 + (24.7)²]
= [(2.53)³ + (2.47)³]/[100{(2.53)² — 2.53 × 2.47 + (2.47)²}]
= [(2.53 + 2.47)(2.53² — 2.53 × 2.47 + 2.47²]/[(2.53² — 2.53 × 2.47 + 2.47²) × 100]
= 5/100
= 5 × 10–2
According to the question,
5 × 10k = 5 × 10–2
=> K = — 2
∴ The required value of K = — 2
Here the correct option is 1.
Que 9. The driver of a car, which travels at a speed of 75 km/h, locates a bus 80 m ahead of him, travelling in the same direction. After 15 seconds, he finds that the bus is 40 m behind the car. What is the speed of the bus (in km/h)?
1. 44.2
2. 42.5
3. 47.5
4. 46.2
Answer : 4
Exp :
Let the speed of the bus = x km/hr
Since both the car and bus are travelling in the same direction so their relative speed = (75 — x) km/hr
They travel in 15 sec = (80 + 40) = 120 m
Speed = 120/15 m/s = 8 × 18/5 km/hr
(75 — x) = 8 × 18/5
=> 75 — x = 144/5
=> x = 75 — 144/5
=> x = 46.2 km/hr
∴ The speed of the bus = 46.2 km/hr
Here the correct option is 4.
Que 10. The value of 17 ½% of 3 ¼% of 33 ⅓% of 7200 is :
1. 7.65
2. 11.68
3. 13.65
4. 9.65
Answer : 3
Exp :
17½% of 3¼% of 33⅓% of 7200
= (35/200) × (13/400) × (100/300) × 7200
= 273/20
= 13.65
∴ The required value is 13.65
Here the correct option is 3
Que 11. A1 and A2 are two regular polygons. The sum of all the interior angles of A1 is 1080°. Each interior angle of A2 exceeds its exterior angle by 132°. The sum of the number of sides A1 and A2 is :
1. 21
2. 22
3. 23
4. 24
Answer : 3
Exp :
We know,
Exterior angle(E) + Interior angle(I) = 180° —(1)
the sum of all interior angle = (n — 2) × 180°
According to the question,
(n — 2) × 180° = 1080°
=> n — 2 = 6
=> n = 8
Side of A1 = 8
According to the question,
Interior angle (I) — Exterior angle (E) = 132° –(2)
Subtracting equation (2) from equation (1),
2E = 48°
=> E = 24°
Number of sides of polygon A2 = 360/24 = 15
The sum of the number of sides of A1 and A2 is = (8 + 15) = 23.
Here the correct option is 3.
Directions (12 – 14): Study the given graph and answer the question that follows.
Que 12. In 2020, the production of cement by company C increased by the same percentage as in 2019, over its previous year. The production (in million tonnes) of cement by company C in 2020 (correct to one decimal place) was:
1. 454.6
2. 455.8
3. 457.1
4. 452.4
Answer: 3
Exp :
Increased Percentage by company C from year 2018 to 2019 = (400 – 350)/350 × 100 = 100/7%
According to the question, the same percentage increased in 2020 over the previous year.
So, Production in 2020 = 400 × (100+100/7)/100 = 400 × 800/700 = 3200/7 = 457.1
∴ The production of cement by company C in 2020 is 457.1
Here the correct option is 3.
Que 13. The ratio of the total production of cement by company A in 2016 and company C in 2018 to the total production of cement by company B in 2017 and 2019 is:
1. 9 : 8
2. 7 : 6
3. 8 : 7
4. 10 : 9
Answer : 1
Exp :
Total production by A in 2016 and by C in 2018 = (280 + 350) = 630
Total production by B in 2017 and 2019 = ( 200 + 360) = 560
Required ratio = 630 : 560 = 9 : 8
∴ The required ratio = 9 : 8.
Here the correct option is 1.
Que 14. The average production of cement by company B in 2015, 2016 and 2018 is what percentage less than the average production of cement by company C in 2015 and 2017?
1. 7 ⅐%
2. 7 ⅔%
3. 5 ⅓%
4. 6 ⅔%
Answer : 4
Exp :
Average production by B in 2015, 2016 and 2018 = ( 150 + 180 + 300)/3 = 630/3 = 210
Average production of cement by C in 2015 and 2017 = (200 + 250)/2 = 225
Required less percentage = (225 – 210)/225 × 100 = 6⅔%
∴ The required less percentage = 6⅔%
Here the correct option is 4.
Que 15. The value of [3(cosec²26° — tan²64°) + (cot²42° — sec²48°)]/[cot(22° — θ) — cosec²(62° + θ) — tan(θ + 68°) + tan²(28° — θ)] is :
1. 3
2. 4
3. -1
4. -2
Answer : 4
Exp :
3(cosec²26° — tan²64°) + (cot²42° — sec²48°)] / [cot(22° — θ) — cosec²(62° + θ) — tan(θ + 68°) + tan²(28° — θ)]
= [3{cosec²(90° — 64°) — tan²64°} + {cot²(90° — 48°) — sec²48°}] / [cot22° — cosec²62° — tan 68°) + tan²28°]
Applying value putting method, θ = 0°
= [3(sec²64° — tan²64°) + (tan²48° — sec²48°)] / (tan68° — cosec²62° — tan68° + cot²62°)
= (3 ×1 — 1) / (—1)
= —2
∴ The required value is —2.
Here the correct option is 4.
Note :
sec²θ — tan²θ = 1
cosec²θ — cot²θ = 1
Que 16. The value of sin²9° + sec² 89° + cos²9° – tan²89° + 8 is equal to –
1. 7
2. 10
3. 16
4. 2
Answer : 2
Exp :
sin²9° + sec² 89° + cos²9° – tan²89° + 8
= sin²9° + cos²9° + sec²89° – tan²89° + 8
= 1 + 1 + 8
= 10
Here the correct option is 2.
Note : we know , sin² θ + cos² θ = 1
sec² θ – tan² θ = 1
Que 17. P and Q start a shop with a capital of Rs. 1,50,000 and Rs. 4,50,000, respectively. After a year, out of the profit of Rs. 2,00,000, P gets his share of the profit plus some money as his salary which is not a part of the profit. If P gets a total of Rs. 90,000, what is the amount of salary (in Rs.) that he received?
1. 20,000
2. 25,000
3. 50,000
4. 40,000
Answer : 4
Exp :
Investment (/profit) ratio of P and Q = 150000 : 450000 = 1 : 3
Profit share of P = 200000 × 1/4 = 50,000
P gets a total amount of Rs.90,000.
P’s salary = (90,000 – 50,000) = 40,000
∴ The amount of salary of P = Rs.40,000
Here the correct option is 4.
Que 18. If 91% of A is 39% of B, and B is x% of A, then the value of x is :
1. 200/3
2. 700/3
3. 400/3
4. 500/3
Answer : 2
Exp :
91% of A = 39% of B
=> 91 × A = 39 × x/100 × A
=> 91 = 39 × x/100
=> x = 9100/39
=> x = 700/3
∴ The value of x = 700/3
Here the correct option is 2.
Que 19. The average of n numbers is 45. If 60% of the numbers are increased by 5 each and the remaining numbers are decreased by 10 each, then what is the average of the numbers so obtained?
1. 42
2. 43
3. 46
4. 44
Answer : 4
Exp :
Let the number(n) be 100.
60% of 100 (= 60) increased by 5 each.
So total increase = 60 × 5 = 300
The remaining 40% (= 40) decreased by 10 each.
Total decreased = 40 × 10 = 400
Net decreased = (400 — 300) = 100
Net average decreased = 100/100 = 1
Thus the required average = (45 — 1) = 44
Here the correct option is 4.
Que 20. ABCD is a cyclic quadrilateral. Sides AB and DC, when produced, meet at E and sides AD and BC when produced, meet at F. If ∠ADC = 76° and ∠AED = 55°, then ∠AFB is equal to:
1. 34°
2. 26°
3. 29°
4. 27°
Answer : 4
Exp :
From the ∆ADE,
∠DAE = 180° — (∠ADE + ∠DEA)
=> ∠DAE = 180° — (76° + 55°)
=> ∠DAE = 49°
We know, the sum of the opposite angles of a cyclic quadrilateral is 180°.
So ∠ABF = 180° — ∠ADE = 180° — 76°
∠ABF = 104°
Now from ∆ABF,
∠ABF + ∠BAF + ∠AFB = 180°
=> 104° + 49° + ∠AFB = 180°
=> ∠AFB = 180° — 153°
=> ∠AFB = 27°
∴ The required value is 27°
Here the correct option is 4.
Que 21. If cosθ = 12/13, then the value of sin θ(1 – tan θ)/tan θ(1 + cosec θ) is :
1. 25/78
2. 35/234
3. 35/108
4. 25/156
Answer : 2
Exp :
cosθ = 12/13 = Base/hypotenuse
Perpendicular = √(13² – 12²) = √25 = 5
Now,
sin θ(1 – tan θ)/tan θ(1 + cosec θ)
= {5/13 × (1 – 5/12)}/{5/12 × (1 + 13/5)}
= (5/13 × 7/12)/(5/12 × 18/5)
= 5/13 × 7/12 × 12/5 × 5/18
= 35/234
The required value is 35/234.
Here the correct option is 2.
Que 22. If x² — 3x + 1 = 0, then the value of (x⁴ + 1/x²)/(x² + 5x + 1) is :
1. 9/4
2. 27/8
3. 5/2
4. 2
Answer : 1
Exp :
x² — 3x + 1 = 0
=> x — 3 + 1/x = 0 (Dividing both sides by x)
=> (x + 1/x) = 3
=> (x + 1/x)³ = 3³ (Cubing on both sides)
=> x³ + 1/x³ + 9 = 27
=> x³ + 1/x³ = 18
Now,
(x⁴ + 1/x²)/(x² + 5x + 1)
= (x³ + 1/x³) / (x + 5 + 1/x) (Dividing by x)
= 18/(3+5)
= 9/4
∴ The required value is 9/4.
Here the correct option is 1.
Que 23. A shopkeeper marks an article at such a price that after giving a discount of 12 ½% on the marked price, he still earns a profit of 15%. If the cost price of the article is Rs. 385, then the sum of the marked price and the selling price (in Rs.) of the article is:
1. 948.75
2. 849.50
3. 984.75
4. 954.75
Answer : 1
Exp :
Discount = 12½ % = 1/8
Market price : Selling price = 8 : 7 —(×23) = 184 : 161
Profit = 15 % = 3/20
Cost price : Selling price = 20 : 23 —–(×7) = 140 : 161
Now, MP : CP : SP = 184 : 161 : 140
140 unit = 385
=> 1 unit = 2.75
=> (184 + 161) unit = 345 × 2.75 = 948.75
∴ The required price = 948.75
Here the correct option is 1.
Que 24. A and B worked together and received a total of Rs. 18,000 for 15 days. A’s efficiency in the work was 5 times that of B’s. The daily wage of A (in Rs.) was:
1. 800
2. 600
3. 1,200
4. 1,000
Answer : 4
Exp :
(A and B)’s 1 day wage = 18000/15 = 1200
Efficiency ratio of A and B = 5 : 1
Wages share of A = 1200 × 5/6 = 1000
∴ The daily wage of A is Rs.1000
Here the correct option is 4.
Que 25. If x = 32.5, y = 34.6 and z = 30.9, then the value of x³ + y³ + z³ — 3xyz is 0.98k, where k is equal to :
1. 1033
2. 933
3. 1026
4. 921
Answer : 1
Exp :
We know,
x³ + y³ + z³ – 3xyz = 1/2(x + y + z)[(x — y)² + (y —z)² + (z — x)²
1/2(x + y + z)[(x — y)² + (y —z)² + (z — x)² = 0.98k
=> 1/2(32.5 + 34.6 + 30.9)[(32.5 — 34.6)² + (34.6 — 30.9)² + (30.9 — 32.5)²] = 0.98k
=> (1/2) × 98 [(—2.1)² + (3.7)² + (—1.6)²] = 0.98k
=> 49(4.41 + 13.69 + 2.56) = 0.98k
=> 49 × 20.66 = 0.98k
=> 1012.34 = 0.98k
=> k = 1012.34/0.98
=> k = 1033
∴ The required value of k is 1033.
Here the correct option is 1.
Que 26. If (secθ – tanθ)/(secθ + tanθ) = 1/7, θ lies in the first quadrant, then the value of (cosecθ + cot²θ)/(cosecθ – cot²θ) is :
1. 19/5
2. 22/3
3. 37/12
4. 37/19
Answer : 1
Exp :
(secθ – tanθ)/(secθ + tanθ) = 1/7
=> 7secθ — 7tanθ = secθ + tanθ
=> 6secθ = 8tanθ
=> 6/cosθ = 8sinθ/cosθ [secθ = 1/cosθ & tanθ = sinθ/cosθ]
=> 6 = 8sinθ
=> sinθ = 3/4
cosecθ = 4/3 = hypotenuse (h)/perpendicular(p)
Applying Pythagoras theorem,
Base² = h² — p²
=> Base = √(4² — 3²)
=> Base = √7
Now, cotθ = √7/3 (= b/p)
(cosecθ + cot²θ)/(cosecθ – cot²θ)
= [4/3 + (√7/3)²] / [4/3 — (√7/3)²]
= (4/3 + 7/9)/(4/3 — 7/9)
= (19/9)/(5/9)
= 19/9 × 9/5
= 19/5
∴ The required value is 19/5
Here the correct option is 1.
Que 27. A sum of money at simple interest amounts to Rs. 6,000 in 4 years and to Rs. 6,750 in 7 years at the same rate percent p.a. of interest. The sum (In Rs.):
1. 5,100
2. 4,800
3. 4,000
4. 5,000
Answer : 4
Exp :
7 years = 6750
4 years = 6000
3 years interest = (6750 — 6000) = 750
1 year interest = 750/3 = 250
Now, interest for 4 years = 250 × 4 = 1000
Principal = Amount — Simple interest
= 6000 — 1000
Principal = 5000
∴ The required sum is Rs.5000
Que 28. If the sides of a triangle are in the ratio 3 : 1¼ : 3¼, then the triangle is
1. Right angle triangle
2. Obtuse triangle
3. Equiangular triangle
4. Acute triangle
Answer : 1
Exp :
Ratio of sides = 3 : 1¼ : 3¼ = 3 : 5/4 : 13/4
= 12 : 5 : 13
Now , 5² + 12² = 169 = 13²
∴ It is a Right angle triangle.
Que 29. The expression (tanθ + cotθ)(secθ + tanθ)(1 – sinθ), 0° < θ < 90°, is equal to :
1. sec θ
2. cosec θ
3. cot θ
4. sin θ
Answer : 2
Exp :
Applying value putting method, let θ = 30°
(tan30° + cot30°)(sec30° + tan30°)(1 — sin30°)
= (1/√3 + √3)(2/√3 + 1/√3)(1 — 1/2)
= (4/√3) × (3/√3) × (1/2)
= 2
Now we do the option tests.
sec30° = 2/√3
cosec30° = 2
cosec30° satisfied the answer.
∴ The required value is cosecθ.
Que 30. A sum of Rs. 8,400 amounts to Rs. 11,046 at 8.75% p.a. simple interest in a certain time. What will be the simple Interest (In Rs.) on a sum of Rs. 10,800 at the same rate for the same time?
1. 3,402
2. 3,204
3. 3,024
4. 3,420
Answer : 1
Exp :
P1 = 8400 ; R = 8.75% ; S.I1 = (11046 – 8400) = 2646
S.I1 = (8400 × 8.75 × T)/100
=> 2646 × 100 = 8400 × 8.75 × T
=> T = 3.6 years
Now,
S.I2 = (P2 × R × T)/100 [ T and R are same ]
= (10800 × 8.75 × 3.6)/100
S.I2 = 3402
∴ The required simple interest is Rs.3402
Que 31. The circumference of the base of a cylindrical vessel is 264 cm and its height is 50 cm. The capacity (in litres) of the vessel is: (Take π = 22/7)
1. 277.2
2. 278.4
3. 280.6
4. 267.4
Answer : 1
Exp :
Circumference = 2πr = 264 ( r = radius)
2 × 22/7 × r = 264
=> r = 42 cm
Now, the volume of the cylindrical vessel = πr²h
πr²h = 22/7 × 42 × 42 × 50 cm³
= 277200 × (1/1000) litres
= 277.2 litre
∴ The capacity of the vessel = 277.2 litres.
Que 32. The value of 9 ÷ {1/2 + 1/3 + 1/4 + 1/6 ÷ (3/4 — 1/3) of 2/9} is :
1. 540/173
2. 340/173
3. 480/173
4. 2540/173
Answer : 1
Exp :
9 ÷ {1/2 + 1/3 + 1/4 + 1/6 ÷ (3/4 — 1/3) of 2/9}
= 9 ÷ {1/2 + 1/3 + 1/4 + 1/6 ÷ ( 5/12) of 2/9}
= 9 ÷ (1/2 + 1/3 + 1/4 + 1/6 ÷ 5/54)
= 9 ÷ (1/2 + 1/3 + 1/4 + 1/6 × 54/5)
= 9 ÷ (1/2 + 1/3 + 1/4 + 9/5)
= 9 ÷ 173/60
= 9 × 60/173
= 540/173
∴ The required value is 540/173.
Rule : BODMAS ( B FOR BRACKETS, O FOR OF, D FOR DIVISION, M FOR MULTIPLICATION, A FOR ADDITION AND S FOR SUBTRACTION)
Que 33. A well with an inner radius of 3 m, is dug 6 m deep. The soil taken out of it has been spread evenly all around it to a width of 2 m to form an embankment. The height (in m) of the embankment is:
1. 4 ½
2 4 ¼
3. 3 ¼
4. 3 ⅜
Answer : 4
Exp :
The volume of a cylinder (well ) = πr²h1
Volume of hollow cylinder (embankment) = π(R² — r²)h2
Here r = 3 m , width = 2 m , R = (3 + 2) = 5 m , h1 = 6 m , h2 = height of embankment
According to the question,
πr²h1 = π(R² — r²)h2
=> 3 × 3 × 6 = (5² — 3²) h2
=> h2 = 54/16 = 3⅜
∴ The height of the embankment is 3⅜ m.
Que 34. The expression (tan⁶θ – sec⁶θ + 3sec²θtan²θ)/(tan²θ + cot²θ + 2) , 0° < θ < 90° is equal to
1. sec²θcosec²θ
2. – sec²θcosec²θ
3. cos²θsin²θ
4. – cos²θsin²θ
Answer : 4
Exp :
(tan⁶θ – sec⁶θ + 3sec²θtan²θ)/(tan²θ + cot²θ + 2)
= [(tan²θ)³ — (sec²θ)³ — 3sec²θ tan²θ(tan²θ — sec²θ)] / (tan²θ + cot²θ + 2tanθcotθ)
= (tan²θ — sec²θ)³/(tanθ + cotθ)²
= (—1)³/(sinθ/cosθ + cosθ/sinθ)²
= — 1/{(sin²θ + cos²θ)²/sin²θ cos²θ}
= — sin²θ cos²θ
∴ This is the required answer(4).
Que 35. A trader bought 640 kg of rice. He sold a part of the rice at 20% profit and the rest at 5% loss. He earned a profit of 15% in the entire transaction. What is the ratio of the quantity of rice that he sold at a loss of 5% to that of the quantity that he sold at a profit of 20%?
1. 1:3
2. 4:1
3. 1:4
4. 3:1
Answer : 3
Exp :
Applying alligation method,
20 —5
15
15—(—5) = 20 20 — 15 = 5
20 : 5 = 4 : 1
∴ The required ratio = 1 : 4
Que 36. 5⁷¹ + 5⁷² + 5⁷³ + 5⁷⁴ + 5⁷⁵ is divisible by which of the following number?
1. 71
2. 69
3. 89
4. 73
Answer : 1
Exp :
5⁷¹ + 5⁷² + 5⁷³ + 5⁷⁴ + 5⁷⁵
= 5⁷¹(1 + 5¹ + 5² + 5³ + 5⁴)
= 5⁷¹(1 + 5 + 25 + 125 + 625)
= 5⁷¹ × 781
= 5⁷¹ × 11 × 71
Here the given number is divisible by 11, 71 and 5n.
∴ The given number is divisible by 71.
Que 37. Let x, y, z be fractions such that x < y < z. If z is divided by x, the result is 5/2, which exceeds y by 7/4. If x + y + z = 23/12, then the ratio of (z — x) : (y — x) is
1. 6:5
2. 9:5
3. 5:6
4. 5:9
Answer : 1
Exp :
x + y + z = 23/12 ———(1)
z/x = 5/2 ———(2)
z/x = y + 7/4 ———-(3)
From equation (3), we get
5/2 = y + 7/4 (putting the value of z/x = 5/2)
=> y = 3/4
putting the value of y in equation (1), we get
x + z + 3/4 = 23/12
=> x + z = 14/12
=> x + z = 7/6
=> x + 5x/2 = 7/6 ( z/x = 5/2)
=> 7x/2 = 7/6
=> x = 1/3
And z = 5x/2
= (5/2) × (1/3)
z = 5/6
Now,
(z — x) : (y — x)
= (5/6 — 1/3) : (3/4 — 1/3)
= 3/6 : 5/12
= 6 : 5
∴ The required answer is 6 : 5
Que 38. In ∆ABC, D and E are points on the sides BC and AB, respectively, such that ∠ACB = ∠DEB. If AB = 12 cm, BE = 5 cm and BD : CD = 1 : 2, then BC is equal to:
1. 8√3 cm
2 5√5 cm
3. 6√5 cm
4. 6√3 cm
Answer : 3
Exp :
BD : CD = 1 : 2
Let , BD = x and CD = 2x
BC = x + 2x = 3x
Now from ∆ABC and ∆BED,
∠ACB = ∠DEB
∠ABC = ∠DBE (common)
So ∆ABC ~ ∆BED
BC : BE = AB : DB
=> 3x/5 = 12/x
=> 3x² = 60
=> x = 2√5
BC = 3x = 6√5
∴ The value of BC = 6√5 cm.
Que 39. An athlete runs 8 times around a circular field of radius 7 m in 3 minutes 40 seconds. His speed (in km/h) is : (take π = 22/7)
1. 72/25
2. 118/25
3. 144/25
4. 108/25
Answer : 3
Exp :
3 minutes 40 seconds = 220 seconds
Time takes to cover 1 round = 220/8 = 55/2 sec
Distance covered in 1 round = 2πr = 2 × 22/7 × 7 = 44 m
Speed = Distance/Time = 44/(55/2) = 88/55m/s
88/55 m/s = 88/55 × 18/5 km/hr = 144/25 km/h
∴ The required speed = 144/25 km/hr
Que 40. Three numbers are in the ratio 1/2 : 2/3 : 3/4. If the difference between the greatest number and the smallest number is 33, then HCF of the three numbers is :
1. 9
2. 5
3. 13
4. 11
Answer : 4
Exp :
Ratio = 1/2 : 2/3 : 3/4
Difference between greatest and smallest number = (3/4 — 1/2) = 1/4
According to question,
1/4 unit = 33
1 unit = 132
The numbers are (132 × 1/2), (132 × 2/3) and (132 × 3/4) = 66, 88, 99
HCF = 11
∴ The HCF of the three numbers is 11
Que 41. The number of students in a class is 45, out of which 33 ⅓% are boys and the rest are girls. The average score of girls in Science is 66 ⅔% more than that of boys. If the average score of all the students is 78, then the average score of girls is:
1. 78
2. 54
3. 90
4. 65
Answer : 3
Exp :
33⅓ % = 1/3
Number of boys = 45 × (1/3) = 15
Number of girls = (45 — 15) = 30
Ratio of number of boys and girls = 15 : 30 = 2 : 1
66⅔ % = 2/3
Boys : girls = 3 : (3 + 2 =) 5
Let average score of boys and girls are 3x and 5x respectively.
Now applying alligation method,
3x 5x
78
(5x — 78) (78 — 3x)
Now,
(5x — 78) : (78 — 3x) = 1 : 2
=> (5x — 78)/(78 — 3x) = 1/2
=> 10x — 156 = 78 — 3x
=> 13x = 234
=> x = 234/13
x = 18
=> 5x = 18 × 5 = 90
∴ The average score of girls is 90.
Que 42. In a square ABCD, diagonals AC and BD Intersect at Q. The angle bisector of ∠CAB meets BD and BC at F and G, respectively. OF : CG is equal to:
1. 1 : 2
2. 1 : 3
3. 1 : √2
4. 1 : √8
Answer : 1
Exp :
Let each side of the square = x
Diagonal = AC = BD = √2x
OA = OB = OC = √2x/2 = x/√2
From ∆AOB,
AB : AO = BF : OF
=> x/(x/√2) = BF/OF
=> √2 : 1 = BF : OF
Let BF = √2y and OF = y
BF + OF = OB
=> √2y + y = x/√2
=> y = x/√2(√2+1)
OF = x/√2(√2+1)
From ∆ABC,
AB : AC = BG : CG
=> x/√2x = BG : CG
=> BG : CG = 1 : √2
Let BG = z and CG = √2z
BG + CG = BC
=> z + √2z = x
=> z = x/(√2+1)
CG = √2z = √2x/(√2+1)
Now,
OF : CG = x/√2(√2+1) : √2x/(√2+1)
= 1/√2 : √2
OF : CG = 1 : 2
∴ The required answer is 1 : 2
Que 43. A sum of Rs. 5,000 is divided into two parts such that the simple interest the first part for 4⅕ years at 6⅔% p.a. is double the simple interest on the second part for 2¾ years at 4% p.a. The ratio of the second part to the first part is:
1. 11 : 14
2. 11 : 13
3. 14 : 11
4. 13 : 11
Answer: 3
Exp :
According to question, S.I1 = 2 × S.I2
S.I1 (= 2 S.I2) = (P1 × 21/5 × 20/3)/100
=> S.I2 = (P1 × 21/5 × 20/3)/200 ——-(1)
And S.I2 = (P2 × 11/4 × 4)/100 ——-(2)
Equating equations (1) and (2),
(P1 × 21/5 × 20/3)/200 = (P2 × 11/4 × 4)/100
=> 14 × P1 = 11 × P2
=> P2 : P1 = 14 : 11
∴ The required ratio = 14 : 11
Que 44. The distance between two places A and B is 140 km. Two cars x and y start simultaneously from A and B, respectively. If they move in the same direction, they meet after 7 hours. If they move towards each other, they meet after one hour. What is the speed (In km/h) of car y if its speed is more than that of car x?
1. 60
2. 100
3. 80
4. 90
Answer: 3
Exp :
Let the speed of y car and x car be A km/hr and B km/hr respectively.
When two cars travel in the opposite direction, then the relative speed = A + B
When they travel in the same direction, then relative speed = A — B
Now, A + B = 140/1
A + B = 140 ——–(1)
And , (A — B) = 140/7
A — B = 20 ——-(2)
From equations (1) and (2), we get
2A = 160
=> A = 80
∴ The speed of car y is 80 km/hr.
Que 45. Study the given pie charts and answer the question that follows.
If the ratio of the number of boys to that of the girls who passed from institute A is 5 : 6, and 40% of the students who passed from institute D are boys, then the ratio of the number of boys who passed from institute A to that of boys who passed from institute D is:
1. 25 : 24
2. 4 : 3
3. 5 : 4
4. 3 : 2
Answer : 3
Exp :
For A institute :
Total Passed students = 1200 × (22/100) = 264
Total passed boys = 264 × (5/11) = 120
For D institute :
Total passed students = 1200 × (20/100) = 240
Total passed boys = 240 × (2/5) = 96
Required Ratio = 120 : 96 = 5 : 4
∴ The ratio of passed boys from institute A to the institute D is 5 : 4
Que 46.The value of 3/70 + 1/42 + 1/66 + 3/286 + 1/130 + 1/170 is :
1. 7/85
2. 11/85
3. 9/85
4. 3/85
Answer : 3
Exp :
3/70 + 1/42 + 1/66 + 3/286 + 1/130 + 1/170
= (18 + 10)/420 + (13 + 9)/858 + 1/130 + 1/170
= 1/15 + 1/39 + 1/130 + 1/170
= 1/15 + (10 + 3)/390 + 1/170
= 1/15 + 1/30 + 1/170
= 3/30 + 1/170
= 1/10 + 1/170
= (17 + 1)/170
= 18/170
= 9/85
∴ The required value is 9/85
Que 47. The monthly salary of a person was Rs. 1,60,000. He used to spend on three heads – Personal and family expenses (P), Taxes (T) and Education loan (E). The rest were his savings. P was 50% of the income. E was 20% of P, and T was 15% of E. When his salary got raised by 30%, he maintained the percentage level of P, but E became 30% of P and T became 20% of E. The sum of the two savings (in Rs.) Is:
1. 2,11,680
2. 1,28,160
3. 1,18,620
4. 1,62,810
Answer : 2
Exp :
P1 = 160000 × (50/100) = 80000
E1 = 80000 × (20/100) = 16000
T1 = 16000 × (15/100) = 2400
Initial Savings = 160000 — (80000 + 16000 + 2400) = 61600
New Salary = 160000 × (13/10) = 208000
Now,
P2 = 208000 × (50/100) = 104000
E2 = 104000 × (30/100) = 31200
T2 = 31200 × (20/100) = 6240
New Savings = 208000 — (104000 + 31200 + 6240) = 66560
Sum of two savings = (61600 + 66560) = 128160
∴ The required sum of two savings is Rs.128160
Que 48. The radius of a spherical balloon is inflated from 7 cm to 10.5 cm. The percentage increase in its surface area is:
7. 150%
2. 125%
3. 120%
4. 135%
Answer : 2
Exp :
R1 , R2 are the radius and A1 and A2 are the areas
R1 : R2 = 7 : 10.5 = 1 : 1.5
R1² : R2² = 1 : 2.25
We know, Surface area ∝ Radius²
So,
A1 : A2 = 1 : 2.25
Increase of area = (2.25 — 1) = 1.25
Percentage increase in area = (1.25/1) × 100% = 125%
∴ The required percentage increase in its surface area is 125%
Que 49. The volume of a cylinder is 4312 cm³. Its curved surface area is one-third of its total surface area. Its curved surface area (in cm²) Is: (Take It π = 22/7)
1. 572
2. 528
3. 660
4. 616
Answer : 4
Exp :
Curved surface area = 1/3 × total surface area
2πrh = 1/3 × 2πr(r + h)
=> 3h = r + h
=> 2h = r
According to the question,
πr²h = 4312
=> 22/7 × r² × r/2 = 4312 ( putting h = r/2)
=> r³ = 14³
=> r = 14 cm
h = 14/2 = 7 cm
Curved surface area = 2πrh
= 2 × 22/7 × 14 × 7 = 616 cm²
∴ The required curved surface area is 616 cm²
Que 50. The graphs of the equations 7x + 11y = 3 and 8x + y = 15 intersect at the point P, which also lies on the graph of the equation:
7. 2x + y = 2
2 2x – y = 1
3. 3x + 5y = 1
4. 3x + 2y = 3
Answer : 3
Exp :
7x + 11y = 3 ——(1)
8x + y = 15 ——(2)
Multiplying equation (1) by 1 and equation (2) by 11, we get
7x + 11y = 3
88x + 11y = 165
=> —81x = —162
=> x = 2
Putting x = 2 in equation (2), we get
16 + y = 15
=> y = —1
So the point intersection is (2, —1)
Now testing the option,
2x + y = —2 + 2 = 0 ( not matched)
2x — y = —2 — 2 = — 4 (not matched)
3x + 5y = 6 + 5(—1) = 1 (satisfied)
∴ The required answer is 3x + 5y = 1
Que 51. A dealer gains 20% by selling an article at 25% discount on its marked price. If the cost price of the article is decreased by 15%, how much discount percentage should he now give on the same marked price so as to earn the same percentage of profit as before?
7. 32.50%
2. 35%
3. 36.25%
4. 37.75%
Answer : 3
Exp :
CP : MP = (100 — 25) : (100 + 20) = 5 : 8
New CP = 5 × (85/100) = 4.25
New SP = 4.25 × (120/100) = 5.1
New discount percentage = (8 — 5.1)/8 × 100% = 36.25%
∴ Required discount percentage = 36.25%
Que 52. The total surface area of a solid hemisphere is 942 cm². Its volume (in cm³) is closest to: (Take n = 3.14)
1. 2089
2. 2093
3. 2037
4. 2097
Answer : 2
Exp :
The total surface area of a solid hemisphere = 3πr²
According to the question,
3πr² = 942
=> r² = 100
=> r = 10
Volume = 2/3 πr³ = 2/3 × 22/7 × 10³ = 2093 cm³
∴ The required volume is 2093 cm³.
Que 53. Two pipes A and B can fill a tank in 12 minutes and 24 minutes, respectively, while a third pipe C can empty the full tank in 32 minutes. All three pipes are opened simultaneously. However, pipe C is closed 2 minutes before the tank is filled. In how much time (in minutes) will the tank be full?
1. 9
2. 10
3. 12
4. 8
Answer : 2
Exp :
LCM of 12, 24 and 32 is 96.
A’s efficiency = 96/12 = 8
B’s efficiency = 96/24 = 4
C’s efficiency = 96/32 = (—3)
(A + B — C) together fill the tank in 1 min = (8+4—3) = 9 unit
C alone can empty the tank in 2 min = 3×2 = 6 unit
Total work done by all pipes = (96 — 6) = 90 unit
Required time = 90/9 = 10 mins
∴ Required time = 10 mins
Que 54. If x + 1/2x = 2, find the value of 8x³ + 1/8x³
1. 48
2.88
3. 40
4. 44
Answer : 1
Exp :
x + 1/2x = 2
=> 2x + 1/x = 4 (multiplying both sides by 2)
Cubing on both sides,
(2x + 1/x)³ = 4³
=> 8x³ + 1/x³ + 3 × 2x × 1/x (2x + 1/x) = 64
=> 8x³ + 1/x³ + 6 × 4 = 64
=> 8x³ + 1/x³ = 64 – 24
∴ 8x³ + 1/x³ = 40
∴ The required value is 40
Que 55. A solid metallic sphere of radius 4 cm is melted and recast into spheres of 2 cm each. What is the ratio of the surface area of the original sphere to the sum of the surface area of the spheres, so formed?
1. 2 : 1
2. 2 : 3
3. 1 : 2
4. 1 : 4
Answer : 3
Exp :
Number of small spheres =(1/3×4πR³)/(1/3×4πr³) = R³/r³ = (4/2)³ = 8
Surface area of original sphere = 4πr² = 64π
Sum of surface area of small spheres = 8 × 4π2² = 128π
Required ratio = 64π : 128π = 1 : 2
Que 56. Two circles of radius 13 cm and 15 cm intersect each other at points A and B. If the length of the common chord is 24 cm, then what is the distance between their centers?
1. 12 cm
2. 16 cm
3. 14 cm
4. 18 cm
Answer : 3
Exp :
Let x be the midpoint of AB.
From ∆APX and ∆BPX
AP = BP (radius of a circle)
PX = BX (common )
∴ ∆APX ~ ∆BPX
∠AXP = ∠BXP = 90°
AX = BX = AB/2 = 12 cm
Now from ∆APX,
PX² = AP² — AX²
=> PX² = 13² — 12² = 25
=> PX = 5
And from ∆AQX ,
QX² = AQ² — AX² = 15² — 12² = 81
=> QX = 9
PQ = PX + QX = 5 + 9 = 14 cm
∴ The distance between their centre is 14 cm.
Que 57. The value of 15 ÷ 8 — 5/4 of (8/3 × 9/16) + (9/8 × 3/4) — (5/32 ÷ 5/7) + 3/8 is :
1. 0
2. 1
3. 2
4. 3
Answer : 2
Exp :
15 ÷ 8 — 5/4 of (8/3 × 9/16) + (9/8 × 3/4) — (5/32 ÷ 5/7) + 3/8
= 15 ÷ 8 — 5/4 of 3/2 + 27/32 — (5/32 × 7/5) + 3/8
= 15 ÷ 8 — 5/4 of 3/2 + 27/32 — 7/32 + 3/8
= 15 /8 — 15/8 + 27/32 — 7/32 + 3/8
= (27 — 7 + 12)/32
= 32/32
= 1
∴ The required value is 1
Que 58. By selling an article for Rs. 2,200, a profit of 10% is earned. If the same article is sold for Rs. 2,600, then what will be the gain percentage?
1. 20%
2. 15%
3. 37%
4. 30%
Answer : 4
Exp :
Cost price = 2200 × 100/120 = 2000
Selling price = 2600
Profit = (2600 — 2000) = 600
Profit percentage = (600/2000) × 100% = 30%
∴ The required gain percentage = 30%
Que 59. If √(1 + √3/2) — √(1 — √3/2) = c, then the value of c is :
1. 1
2. 4
3. 3
4. 2
Answer : 1
Exp :
√(1 + √3/2) — √(1 — √3/2) = c
=> √{(2 + √3)/2} —√{(2 — √3)/2} = c
=> √{(4 + 2√3)/4 — √{(4 — 2√3)/4 = c
=> √{(√3 + 1)²/4} — √{(√3 — 1)²/4} = c
=> (√3 + 1)/2 — (√3 — 1)/2 = c
=> (√3 + 1 — √3 + 1)/2 = c
=> 2/2 = c
∴ c = 1
∴ The required value of c is 1
Que 60. A dealer offers a cash discount of 20% and still makes a profit of 20%. If he further sells 8 articles at a rate of 6 articles, then how much percentage above the cost price does he mark on each article?
1. 77.5%
2. 100%
3. 112.5%
4. 87.5%
Answer : 2
Exp :
Market price = 100
Selling price = 80
Selling price of 8 articles = 80 × 6 = 240
Cost price of 8 articles = 480 × 100/120 = 400
Cost price of 1 article = 400/8 = 50
Market price percentage = (100 — 50)/50 × 100% = 100%
∴ The required percentage above the cost price is 100%
Que 61. Let p, q, r and s be positive natural numbers having three exact factors including 1 and the number itself. If q > p and both are two-digit numbers, and r > s and both are one-digit numbers, then the value of the expression (p – q – 1)/(r – s) is :
1. – s – 1
2. s – 1
3. 1 – s
4. s + 1
Answer : 1
Exp :
q > p and both are two-digit numbers,
So their possible value = 25(5²) and 49(7²)
q = 49 or 7² [ factors 1, 7 and 49]
p = 25 or 5² [ factors 1, 5 and 25]
r > s and both are one-digit numbers,
So their possible value = 4(2²) and 9(3²)
r = 9 [ factors 1, 3 and 9]
s = 4 [ factors 1, 2 and 4]
Now,
(p – q – 1)/(r – s)
= (25 – 49 – 1)/(9 – 4)
= – 25/5
= – 5
∴ The required value is –5
Que 62. A sum of Rs. 50,250 is divided into two parts such that the simple Interest on the first part for 7½ years at 8⅓% p.a. is 5/2 times the simple interest on the second part for 5¼ years at 8% p.a. What is the difference (in Rs.) between the two parts?
1. 10,275
2. 12,750
3. 12,570
4. 15,270
Answer : 2
Exp :
Let, first part = P and 2nd part = Q
Now According to the question,
P × 15/2 × 25/3 = 5/2 × Q × 21/4 × 8
=> 25P = 42Q
=> P : Q = 42 : 25
Total = (42 + 25) = 67 unit
Difference = (42 – 25) = 17 unit
According to the question,
67 unit = 50250
=> 17 unit = (50250 × 17)/67 = 12750
∴ Thus the difference between the two parts is 12750
Que 63. If 27x³ – 64y³ = (Ax + By)(Cx² – Dy² + 12xy), then the value of 4A + B + 3C + 2D is :
1. 5
2. 3
3. 5
4. 8
Answer : 2
Exp :
27x³ – 64y³
= (3x)³ – (4y)³
= (3x – 4y)(9x² + 16y² + 12xy)
Now, A = 3, B = –4, C = 9 and D = –16
4A + B + 3C + 2D
= 4 × 3 + (–4) + 3 × 9 + 2 × (–16)
= 12 – 4 + 27 – 32
= 3
∴ The required value is 3
Que 64. The expression (cos⁴θ – sin⁴θ + 2sin²θ + 3)/(cosecθ + cotθ + 1)(cosecθ – cotθ + 1) – 2 , where 0° < θ < 90° , is equal to :
1. 1/2sinθ
2. 2sinθ
3. secθ
4. 2cosecθ
Answer : 2
Exp :
(cos⁴θ – sin⁴θ + 2sin²θ + 3)/[(cosecθ + cotθ + 1)(cosecθ – cotθ + 1) – 2]
Applying value putting method, θ = 30°
= (cos⁴30° – sin⁴30° + 2sin²30° + 3)/[(cosec 30° + cot 30° + 1)(cosec 30° – cot 30° + 1) – 2]
= [(√3/2)⁴ – (1/2)⁴ + 2 (1/2)² + 3]/[(2 + √3 + 1)(2 – √3 + 1) – 2]
= (9/16 – 1/16 + 1/2 + 3)/[(3 + √3)(3 – √3) – 2]
= (8/16 + 7/2)/(3² – √3² – 2)
= [(8 + 56)/16]/(9 – 3 – 2)
= 64/(16 × 4)
= 1
Now, we are applying the option test,
1/2 sin 30° = 1/2 × 1/2 = 1/4
2sin30° = 2 × 1/2 = 1
∴ Thus the value is 2sinθ
Que 65. The value of (0.3)[{(200 – 146)/(3 × 3 × 3)} – 3] is :
1. 10/3
2. 5/3
3. 7/3
4. 8/3
Answer : 1
Exp :
(0.3)[{(200 – 146)/(3 × 3 × 3)} – 3]
= (0.3)[(54/27) – 3]
= (0.3)(2 – 3)
= (0.3)–1
= 1/0.3
= 10/3
∴ The required value is 10/3
Que 66. The expression [(1 – sinθ + cosθ)²(1 – cosθ) sec³θ cosec²θ]/(secθ – tanθ)(tanθ + cotθ) where 0° < θ < 90°, is equal to :
1. sinθ
2. 2cosθ
3. cotθ
4. 2tanθ
Answer : 4
Exp :
Applying value putting method, θ = 45°
[(1 – sin45° + cos45°)²(1 – cos45°) sec³45° cosec²45°]/(sec45° – tan45°)(tan45° + cot45°)
= [(1 – 1/√2 + 1/√2)²(1 – 1/√2)(√2)³(√2)²]/[(√2 – 1)(1 + 1)
= [{(√2 – 1)/√2} × 2√2 × 2]/[2 × (√2 – 1)]
= 2(√2 – 1)/(√2 – 1)
= 2
Now from the options,
Sin 45° = 1/√2
2cos 45° = 2 × 1/√2 = 2/√2
cot 45° = 1
2tan 45° = 2 × 1 = 2
∴ Thus the correct value is 2tanθ
Que 67. The value of (cos 9° + sin 81°)(sec 9° + cosec 81°)/(cosec² 71° + cos² 15° – tan² 19° + cos² 75°) is :
1. 1
2. 4
3. 3
4. 2
Answer : 4
Exp :
(cos 9° + sin 81°)(sec 9° + cosec 81°)/(cosec² 71° + cos² 15° – tan² 19° + cos² 75°)
= (cos 9° + cos 9°) (sec 9° + sec 9°)/(sec² 19° – tan² 19° + cos² 15° + sin² 15°)
= (2cos 9° × 2sec 9°)/(1 + 1) [ cos 9° = 1/sec 9°]
= 4/2
= 2
∴ Thus the correct value is 2.
Note :
cosθ = sin (90° – θ)
secθ = cosec (90° – θ)
cos²θ + sin²θ = 1
sec²θ – tan²θ = 1
Que 68. In ∆ABC, A = 50°, BE and CF are perpendiculars on AC and AB at E and F, respectively. BE and CF intersect at H. The bisectors of ∠HBC and ∠HCB Intersect at P. ∠BPC is equal to:
1. 155°
2. 100°
3. 115°
4. 120°
Answer : 1
Exp :
Here AFHE is a quadrilateral ,
So ∠FHE + ∠E + ∠A + ∠F = 360°
=> ∠FHE + 90° + 50° + 90° = 360°
=> ∠FHE = 130°
∠FHE and ∠BHC are opposite angles, so
∠FHE = ∠BHC = 130°
Point P is incentre of ∆BHC,
∠BPC = ∠BHC/2 + 90°
=> ∠BPC = 130°/2 + 90°
=> ∠BPC = 155°
∴ The value of ∠BPC is 155°
Que 69. If √(26 – 7√3)/√(14 + 5√3) = (b + a√3)/11, b > 0, then what is the value of √(b – a)?
1. 5
2. 7
3. 9
4. 1
Answer : 1
Exp :
√(26 – 7√3)/√(14 + 5√3)
= [√(26 – 7√3) × √(14 – 5√3)]/[√(14 + 5√3)(14 – 5√3)]
= √(364 – 98√3 – 130√3 + 105)/√[14² – (5√3)²]
= √(361 – 228√3 + 108)/√(196 – 75)
= √[(19² – (6√3)²]/√121
= (19 – 6√3)/11
Now , b = 19 and a = – 6
√(b – a) = √(19 + 6) = 5
∴The required value is 5
Que 70. Study the given pie charts and the question that follows.
Total Number of Students passed = 1200. The number of students who passed institute C exceeds the number of students who appeared from institute E is x. The value of x lies between:
1. 18 and 22
2. 14 and 18
3. 10 and 14
4. 22 and 26
Answer : 2
Exp :
In institute C, passed students = 1200 × 18/100 = 216
In institute E, appeared students = 1800 × 40/360 = 200
Difference = (216 – 200) = 16
∴ The value of x lies between 14 and 18.
Que 71. If sin²θ/(cos²θ – 3cosθ + 2) = 1, θ lies in the first quadrant, then the value of (tan²θ/2 + sin²θ/2)/(tanθ + sinθ) is :
1. 2√3/27
2. 5√3/27
3. 2√3/9
4. 7√3/54
Answer : 4
Exp :
sin²θ/(cos²θ – 3cosθ + 2) = 1
=> cos²θ – 3cosθ + 2 = sin²θ
=> cos²θ – 3cosθ + 2 – 1 + cos²θ = 0
=> 2cos²θ – 2cosθ – cosθ + 1 = 0
=> (Cos θ – 1)(2cosθ – 1) = 0
cosθ – 1 = 0 2cosθ – 1 = 0
=> cosθ = 1 = cos0°. => cosθ = 1/2 = cos60°
=> θ = 0 => θ = 60°
θ lies in the first quadrant, so θ = 60° acceptable
[tan²(60°/2) + sin²(60°/2)]/(tan60° + sin60°)
= [(1/√3)² + (1/2)²]/(√3 + √3/2)
= (1/3 + 1/4)/(√3 + √3/2)
= 7/12 × 2/(2√3 + √3)
= 7√3/54
∴ Thus the required value is 7√3/54
Que 72. Study the given pie charts and answer the question that follows.
The number of students who appeared from institute B is what percentage more than the total numbers of students who passed from institutes A and C?
1. 16 ⅔%
2 15 ⅓%
3. 14 ⅐%
4. 7 ⅛%
Answer : 1
Exp :
In B , appeared students = 1800 × 112/360 =560
In A , passed students = 1200 × 22/100 = 264
In C, passed students = 1200 × 18/100 = 216
In A and C , total passed students = (264 + 216) = 480
Difference = (560 – 480) = 80
Required more percentage = (80/480) × 100 = 16⅔%
Que 73. Three sides of a triangle are √(a² + b²), √(4a² + b²), √(a² + 4b²) units. What is the area (in unit squares) of the triangle?
1. 5/2 ab
2. 3/2 ab
3. 4ab
4. 3ab
Answer : 2
Exp :
Suppose the triangle is an isosceles triangle.
So a = b
√(a² + a²) = √(2a²)
√(4a² + a²) = √(5a²)
√{(a² + (2a)²} = √(5a²)
Now area of an isosceles triangle
= 1/2 × [√{(√5a²)² – (√2a²)²/4} × √(2a²)
= 1/2 × √(5a² – a²/2) × √2 a
= 1/2 × √(9a²/2) × a√2
= 1/2 × 3a/√2 × a√2
= 3/2 ab ( a = b)
∴ The area of the triangle is 3/2 ab
Que 74. If √(38 – 5√3)/√(26 + 7√3) = (a + b√3)/23, b > 0, then the value of (b – a) is :
1. 7
2. 16
3. 29
4. 11
Answer : 3
Exp :
√(38 – 5√3)/√(26 + 7√3)
= √(38 – 5√3)√(26 – 7√3)/√(26 + 7√3)√(26 – 7√3)
= √(988 – 130√3 – 266√3 + 105)/√{26² –(7√3)²}
= √(121 + 972 – 396√3)/√529
= √(18√3 – 11)²/23
Here , a = –11 and b = 18
Now, (b – a) = 18 + 11 = 29
∴ The required value is 29
Que 75. In an examination, the number of students who passed and the number of students who failed were in the ratio 25 : 4. If one more student had appeared and passed and the number of failed students was 3 less than earlier, the ratio of passed students to failed students would have become 22 : 3. What is the difference between the number of students who, Initially, passed the examination and the number of students who failed the examination?
1. 132
2. 126
3. 174
4. 150
Answer : 2
Exp :
Let passed students’ and failed students’ numbers be 25x and 4x respectively.
Total appeared students = (25x + 4x) = 29x
According to the question,
(29x + 1)/(4x – 3) = 25/3
=> 87x + 3 = 100x – 75
=> 13x = 78
=> x = 6
Difference = (25x – 4x) = 21 × 6 = 126
∴ Thus the required difference is 126.
Que 76. What is the area (in unit squares) of the triangle enclosed by the graphs of 2x + 5y = 12, x + y = 3 and the x-axis?
1. 2.5
2. 3.5
3. 3
4. 4
Answer : 3
Exp :
2x + 5y = 12 —– (1)
x + y = 3 ——(2)
Multiplying (1) by 1 and equation (2) by 2 and then subtracting, we get
3y = 6
=> y = 2
y = 2 , putting in equation (2), we get
x = 1
Point of intersection (1,2)
Point of intersection in equation (1) and x – axis is (6, 0)
Point of intersection in equation (2) and x – axis is (3, 0)
Base(AB) = 6 – 3 = 3 unit
Height (CM) = 2 unit
Area = 1/2 × 3 × 2 = 3 sq unit
∴ The required area is 3 sq. unit
Que 77. In ∆PQR, the bisector of ∠R meets side PQ at S, PR = 10 cm, RQ = 14 cm and PQ = 12 cm. What is the length of SQ?
1. 5cm
2. 6 cm
3. 7 cm
4. 8 cm
Answer : 3
Exp :
RRS is the angle bisector of ∠R, So
RP/PS = RQ/SQ
Let, SQ = x cm and PS = (12 – x) cm
10/(12 – x) = 14/x
=> 10x = 168 – 14x
=> 24x = 168
=> x = 7 cm
∴ The length of SQ = 7 cm
Que 78. If x + 1/x = 3, x is not equal to zero, then the value of x⁷ + 1/x⁷ is :
1. 749
2. 843
3. 746
4. 849
Answer : 2
Exp :
x + 1/x = 3
(x + 1/x)³ = 3³ (Cubing both sides )
=> x³ + 1/x³ + 3 (x + 1/x) = 27
=> x³ + 1/x³ + 3 × 3 = 27
=> x³ + 1/x³ = 18
x + 1/x = 3
=> x² + 1/x² + 2x× 1/x = 9 (squaring both sides)
=> x² + 1/x² = 7
=> x⁴ + 1/x⁴ + 2 x² × 1/x² = 49. ( Squaring both sides)
=> x⁴ + 1/x⁴ = 47
Now,
x⁷ + 1/x⁷
= (x³ + 1/x³)(x⁴ + 1/x⁴) – (x + 1/x)
= 18 × 47 – 3
= 846 – 3
= 843
∴ The Required value = 843
Que 79. In equilateral ∆ABC, D and E are points on the sides AB and AC, respectively, such that AD = CE. BE and CD Intersect at F. The measure (in degrees) of ∠CFB is:
1. 120°
2. 135°
3. 125°
4. 105°
Answer : 1
Exp :
Let, ∠EBC = θ
Now , ∠BEC = 180° – ∠BCE – ∠EBC
= 180° – 60° – θ
∠BEC = 120° – θ
We know, the exterior angle is equal to the sum of opposite interior angles.
Exterior angle ∠CFB = (120° – θ) + θ = 120°
∴ The value of ∠CFB = 120°
Que 80. The volume of a right circular cone is 308 cm³ and the radius of its base is 7 cm. What is the curved surface area (in cm²) of the cone? (Take π = 22/7)
1. 22√21
2. 44√21
3. 22√85
4. 11√85
Answer : 3
Exp :
Volume of cone = (1/3)πr²h
According to the question,
(1/3)πr²h = 308
=> 1/3 × 22/7 × 7² × h = 308
=> h = 6 cm
We know,
l² = r² + h² = 7² + 6²
l = √85 cm
Curved surface area = πrl = 22/7 × 7 × √85 = 22√85 sq.cm
∴ The curved surface area of the cone is 22√85 cm²
Que 81. In ∆ABC, D is a point on BC such that ∠ADB = 2∠DAC, ∠BAC = 70° and ∠B = 56°. What is the measure of ∠ADC?
1. 72°
2. 54°
3. 74°
4. 81°
Answer : 1
Exp :
Let, ∠DAC = x
∠ABB = 2x
∠BAD = 70° – x
Now from ∆ABD,
56° + 2x + (70° – x) = 180°
=> x = 54°
=> 2x = 108°
∠ADB and ∠ADC lies in a straight line, so their sum is 180°
∠ADB + ∠ADC = 180°
=> 108° + ∠ADC = 180°
=> ∠ADC = 72°
Que 82. The ratio of the Investments of A and B in a business is 7 : 5, and the ratio of their profits at the end of a year is 2 : 5. If A Invested the money for 6 months, then for how much time (in months) has B invested his money?
1. 12
2. 21
3. 24
4. 18
Answer : 2
Exp :
Profit ratio = Investment amount × Time
Let, B invested his money for T months.
2/5 = (7 × 6) : (5 × T)
=> 10 T = 42 × 5
=> T = 21
∴ B invested his money for 21 months.
Que 83. A cylindrical tube, open at both ends, is made of a metal sheet that is 0.5 cm thick. Its outer radius is 4 cm and length is 2 m. How much metal (in cm³) has been used in making the tube?
1. 800 π
2. 450 π
3. 750 π
4. 550 π
Answer : 3
Exp :
Inner radius (r) = 4 – 0.5 = 3.5 cm
Volume = π(R² – r²)h
= π × (4² – 3.5²) × 200
= π × 7.35 × 0.5 × 200
= 750π cm³
∴ The required metal is 750π cm³.
Que 84. A vessel contained a solution of acid and water, in which water was 64%. Four litres of the solution was taken out of the vessel and the same quantity of water was added. If the resulting solution contains 30% acid, the quantity (in litres) of the water in the solution, at the beginning in the vessel, was:
1. 11.36
2. 15.36
3. 8.64
4. 12.64
Answer : 2
Exp :
Water = 64% ; Acid = (100% – 64%) = 36%
Ratio of water and acid = 64 : 36 = 16 : 9
Let, water = 16x and acid = 9x
Quantity Of Water after taken out and water added = 16x – (16/25) × 4 + 4 = (16x + 1.44)
Quantity of acid after 4 lit mixture taken out = 9x – (9/25) × 4 = (9x – 1.44)
According to the question,
(16 + 1.44)/(9x – 1.44) = 7/3
=> 48x + 4.32 = 63x –10.08
=> 15x = 14.40
=> x = 0.96
∴ 16X = 0.96 × 16 = 15.36
∴ Initially, the quantity of water is 15.36 lit.
Que 85. If x⁴ + y⁴ + x²y² = 17 (1/16) and x² – xy + y² = 5¼ then one of the values of (x – y) is :
1. 5/2
2. 3/4
3. 5/4
4. 3/2
Answer : 1
Exp :
x⁴ + y⁴ + x²y² = 273/16
=> (x² + y²)² – 2x²y² + x²y² = 273/16
=> (x² + y²)² – (xy)² = 273/16
=> (x² + y² –xy)(x² + y² + xy) = 273/16
=> 21/4 × (x² + y² + xy) = 273/16 (putting x² – xy + y² = 21/4)
=> x² + xy + y² = 13/4 ——–(1)
x² – xy + y² = 21/4. ——— (2)
Adding equations (1) and (2) , we get
2x² + 2y² = 34/4
=> x² + y² = 17/4
Subtracting equation (2) from equation (1),
2xy = –8/4 = – 2
Now,
(x – y)² = x² + y² – 2xy = 17/4 + 2 = 25/4
(x – y) = √(25/4) = 5/2
∴ The required value of (x – y) is 5/2
Que 86. The base of a right pyramid is a square of side 8√2 cm and each of its slant edges is of length 10 cm. What is the volume (in cm³) of the pyramid?
1. 256
2. 224
3. 426
4. 96√2
Answer : 1
Exp :
OABCD is a right pyramid.
Diagonal of square ABCD = √2 × 8√2 = 16 cm
DP = 1/2 of diagonal = 8 cm
OP² = OD² – DP² = 10² – 8² = 36
OP = 6 cm (height)
Volume of pyramid = 1/3 × area of base × height = 1/3 × (8√2)² × 6 = 256 cm³
∴ The volume of the pyramid is 256 cm³.
Que 87. The base of a right prism is a triangle whose sides are 8 cm, 15 cm and 17 cm, and its lateral surface area is 480 cm². What is the volume (in cm³) of the prism?
1. 540
2. 600
3. 720
4. 640
Answer : 3
Exp :
8, 15 , 17 is a triplets.
So the triangle is a right angle triangle.
Perimeter of the triangle = (8 + 15 + 17) = 40 cm
The lateral surface area of prism = perimeter of base × height
=> 480 = 40 × h
=> h = 12 cm
Area of base = 1/2 × 8 × 15 = 60 cm²
Volume = area of base × height = 60 × 12 = 720 cm³
∴ The volume of the prism is 720 cm³
Que 88. X and Y travel a distance of 90 km each such that the speed of Y is greater than that of X. The sum of their speeds is 100 km/h and the total time taken by both is 3 hours 45 minutes. The ratio of the speed of X to that of Y is:
1. 2 : 3
2. 1 : 3
3. 2 : 4
4. 1 : 4
Answer : 1
Exp :
3 hours 45 mins = 3¾ = 15/4 hours
Let, Speed of x is S km/hr and y be (100 – S) km/hr.
According to the question,
90/S + 90/(100 – S) = 15/4
=> (9000 – 90S + 90S)/(100S – S²) = 15/4
=> 1500S – 15S² = 36000
=> S² – 100S + 2400 = 0 (Dividing by 15)
=> S² – 60S – 40S + 2400 = 0
=> (S – 60)(S – 40) = 0
=> S = 60, 40
y = 60 and x = 40. ( y greater than x)
Ratio = 40 : 60 = 2 : 3
∴ The required ratio of x and y is 2 : 3
Que 89. Pipes A, B and C can fill a tank in 20, 30 and 60 hours, respectively. Pipes A, B and C are opened at 7 a.m., 8 a.m., and 9 a.m., respectively, on the same day. When will the tank be full?
1. 4:40 p.m.
2. 5:40 p.m.
3. 6:20 p.m.
4. 7:20 p.m.
Answer : 2
Exp :
LCM of 20, 30 and 60 is = 60(total work)
A’s efficiency = 60/20 = 3
B’s efficiency = 60/30 = 2
C’s efficiency = 60/60 = 1
A’s work for 2 hours (7 am to 9 am) = 3 × 2 = 6
B’s work for 1 hour (8 am to 9 am ) = 2
Remaining work = (60 – 6 – 2) = 52
Now they all work from 9 am
(A + B + C)’s efficiency = 3 + 2 + 1 = 6
(A + B + C)’s time taken = 52/6 = 8 hours 40 mins.
9 am + 8 hours 40 mins = 5 : 40 pm
∴ The tank will be full at 5:40 pm
Que 90. If a nine-digit number 789x6378y is divisible by 72, then the value of xy is –
1. 10
2. 12
3. 8
4. 15
Answer : 3
Exp :
789x6378y is divisible by 72.
72 = 8 × 9
A number is divisible by 72, which means it is divisible by 8 and 9 both.
A number is divisible by 8 when its last three digits are divisible by 8.
78y , here y should be 4 , so that it can divisible by 8.
A number is divisible by 9 when the sum of all digits is divisible by 9.
7 + 8 + 9 + x + 6 + 3 + 7 + 8 + 4 = (52 + x), here x should be 2 so that I can divisible by 9.
x = 2 and y = 4
xy = 2 × 4 = 8
∴ The required value is 8.
Que 91. The curved surface area of a right circular cone is 2310 cm² and its radius is 21 cm. If its radius is increased by 100% and height is reduced by 50%, then its capacity (in litres) will be (correct to one decimal place): (Taken π = 22/7)
1. 27.8
2. 28.2
3. 26.7
4. 25.9
Answer : 4
Exp :
The curved surface area of cone = πrl
=> 22/7 × 21 × l = 2310
=> l = 35 cm
We know,
l² – r² = h²
=> h² = 35² – 21² = 784
=> h = 28 cm
New Radius = 21 × (200/100) = 42 cm
New height = 28 × (50/100) = 14 cm
Volume = 1/3 πr²h = 1/3 × 22/7 × 42² × 14 = 25872 cm³
∴ The capacity is (25872/1000) = 25.90 lit (approx)
Que 92. Two pillars A and B of the same height are on opposite sides of a road which is 40 m wide. The angles of elevation of the tops of pillars A and B are 30° and 45°, respectively, at a point on the road between the pillars. What is the distance (in m) of the point from the foot of pillar A?
1. 40(√3-1)
2. 20(2-√3)
3. 20(3-√3)
4. 39√3
Answer : 3
Exp :
tan 30° = 1/√3 = PQ/QO
tan 45° = 1 = MN/ON
(√3 + 1) = 40
=> 1 unit = 40/(√3+1)
=> √3 unit = 40√3/(√3+1) = 40√3(√3 – 1)/(√3+1)(√3 – 1) = (120 – 40√3)/2
=> 3 unit = 20(3 – √3) m
∴ The required distance = 20(3 – √3) m.
Que 93. A lady sold an article for Rs. 960 at some profit. Had she sold it for Rs. 800, then there would have been a loss equal to 1/3 of the initial profit. What was the profit percentage of the article?
1. 150/7 %
2. 50/7 %
3. 10/7 %
4. 100/7 %
Answer : 4
Exp :
Let, initial profit = P and S.P1 = 960 ; CP = (960 – P)
Now loss = P/3 and S.P2 = 800 ; CP = (800 + P/3)
Now,
960 – P = 800 + P/3
=> P + P/3 = 960 – 800
=> 4P/3 = 160
=> P = 120
CP = (960 – 120) = 840
Profit % = (120/840) × 100% = 100/7%
Que 94. In ∆ABC, AB = 48 cm, BC = 55 cm and AC = 73 cm. If O is the centroid of the triangle, then the length (in cm) of BO (correct to one decimal place) is:
1. 25.6
2. 24.3
3. 20.4
4. 18.3
Answer : 2
Exp :
48,55 and 73 is a triple.
So the triangle is a right-angle triangle.
∠B = 90°
O is the centroid and the triangle is a right angle,
So, BP = (1/2) of AC = 73/2
And, OB : OP = 2 : 1
OB = 73/2 × 2/3 = 24.33 cm
∴ The length of BO = 24.33 cm
Que 95. Study the given histogram and answer the question that follows.
The total number of neon lamps having a lifetime of 800 or more hours is approximately what percentage more than the total number of neon lamps having a lifetime of 400 or more hours but less than 800 hours?
1. 22.7 %
2. 12.5 %
3. 32.2 %
4. 31.8 %
Answer : 2
Exp :
Number of Neon lamps having a lifetime of 800 or more hours = (350+300+275+325+100) = 1350
Number of Neon lamps having a lifetime of 400 or more hours but less than 800 hours = (225 + 260 + 340 + 375) = 1200
Difference = (1350 – 1200) = 150
Required percentage = (150/1200) × 100% = 12.5%
Que 96. In ∆PQR, PQ = PR and S is a point on QR such that ∠PSQ = 96° + ∠QPS and ∠QPR = 132°. What is the measure of ∠PSR?
1. 45°
2. 56°
3. 54°
4. 52°
Answer : 3
Exp :
Since PQ = PR , ∠PRQ = ∠PQR
∠PRQ = ∠PQR = (180° – 132°)/2 = 24°
Now, ∠PQS + ∠QPS + ∠PSQ = 180°
=> 24° + ∠PSQ – 96° + ∠PSQ = 180° (Given ∠PSQ = 96° + ∠QPS)
=> 2∠PSQ – 72° = 180°
=> ∠PSQ = 252°/2
=> ∠PSQ = 126°
∠PSR = (180° – 126°) = 54°
Here the correct option is 3.
Que 97. The sum of three fractions A, B, and C, A > B > C, is 121/60. When C is divided by B, the resulting fraction is 9/10, which exceeds A by 3/20. What is the difference between B and C?
1. 1/15
2. 1/10
3. 3/10
4. 7/15
Answer : 1
Exp :
A + B + C = 121/60 ——(1)
C/B = 9/10 ——(2)
C/B = A + 3/20. ——-(3)
From equations (2) and (3),
A + 3/20 = 9/10
=> A = 9/10 – 3/20 = (18 – 3)/20
=> A = 3/4
Putting the value of A in equation (1),
B + C = 121/60 – 3/4 = 19/15
=> B + 9B/10 = 19/15. ( C = 9B/10)
=> 19B/10 = 19/15
=> B = 2/3
Now,
2/3 + C = 19/15
=> C = 19/15 – 2/3
=> C = (19 – 10)/15 = 3/5
B – C = 2/3 – 3/5 = (10 – 9)/15 = 1/15
∴ The required difference = 1/15
Que 98. Mixture A contains chocolate and milk in the ratio 4 : 3 and mixture B contains chocolate and milk in the ratio 5 : 2. A and B are taken in the ratio 5 : 6 and mixed to form a new mixture. The percentage of chocolate in the new mixture is closest to:
1. 35%
2. 69%
3. 31%
4. 65%
Answer : 4
Exp :
Chocolate : Milk = 4 : 3 —(×5) = 20 : 15
Chocolate : Milk = 5 : 2 —(×6) = 30 : 12
Total ratio = 50 : 27
% of Chocolate in the mixture = (50/77) × 100% ≈ 65%
The percentage of chocolate in the new mixture is 65%.
Que 99. The expression [(1 – 2sin²θ cos²θ)(cotθ + 1)cosθ]/[(sin⁴θ + cos⁴θ)(1 + tanθ)cosecθ] — 1, 0° < θ < 90° , is equal to –
1. cos²θ
2. – sin²θ
3. cosecθ
4. – sec²θ
Answer : 2
Exp :
[(1 – 2sin²θ cos²θ)(cotθ + 1)cosθ]/[(sin⁴θ + cos⁴θ)(1 + tanθ)cosecθ] — 1
= [(1 — 2sin²θ cos²θ )(cosθ/sinθ + 1)cosθ]/[(sin²θ + cos²θ)² — 2sin²θ cos²θ(1 + sinθ/cos)cosecθ] — 1
= (cosθ/sinθ) × (cosθ/cosecθ) — 1
= (cosθ/sinθ) × cosθ sinθ — 1
= cos²θ — 1
= — sin²θ
The required value is — sin²θ.
Que 100. The average of 25 numbers is 64. The averages of the first 13 numbers and that of the last 13 numbers are 62.8 and 72.2, respectively. If the 12th number is 61, and if the 12th and 13th numbers are excluded, then what is the average of the remaining numbers (correct to one decimal place)?
1. 59.2
2. 62.5
3. 60.2
4. 61.5
Answer : 3
Exp :
Sum of 25 numbers = 25 × 64 = 1600
Sum of the first 13 numbers = 13 × 62.8 = 816.4
Sum of the last 13 numbers = 13 × 72.2 = 938.6
12th number is given = 61
13th number = (816.4 + 938.6) — 1600 = 155
Sum of 23 numbers = (1600 — 61 — 155) = 1384
Average of the 23 numbers = 1384/23 = 60.17 ≈ 60.2
The average of the remaining numbers is 60.2
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