Every year SSC conducts CGL examination for various posts in ministries and departmental post of group B and group C. SSC CGL provides India’s most prestigious jobs. So its exam quality and question pattern is quite standard. A candidate has to be focused on its previous years question papers. Solving previous years question papers, a candidate can get the clear knowledge of question pattern and along with can check his/her accuracy. Here we provide SSC CGL 2019 tier 2 question paper of quantitative aptitude section.
Que 1. In an examination, 92% of students passed and 480 students failed. If so, how many students appeared in the examination?
A) 6000
B) 5000
C) 5800
D) 6200
Answer : A
Exp :
% of failed students = (100 – 92) % = 8%
8% = 480
1% = 60
100% = 6000
∴ The total number of students who appeared in the examination is 6000.
Que 2. An article is listed at Rs. 7600 and the discount offered per unit is 10%. What additional discount must be given to bring the net selling price to Rs.5814?
A) 15%
B) 12%
C) 8%
D) 10%
Answer : A
Exp :
MP = 7600
Discount = 10%
Selling price = 7600 × 90/100 = 6840
Difference of selling price = (6840 – 5814) = 1026
Additional discount % = (1026/6840) × 100 = 15%
∴ The required additional discount = 15%
Que 3. A and B can do work together in 18 days. A is three times as efficient as B. In how many days can B alone complete the work?
A) 54 days
B) 62 days
C) 72 days
D) 84 days
Answer : C
Exp :
Efficiency of A = 3 × efficiency of B
A + B together do a work in 18 days.
Since both working together for 18 days,
B alone can do the work = (18 + 18 × 3) days
= 72 days
Que 4. The number of lead balls, each 3 cm in diameter, that can be made from a solid lead sphere of diameter 42 cm is :
A) 2744
B) 2740
C) 5300
D) 2570
Answer : A
Exp :
Volume of a sphere = (4/3) πr³
Required number of balls = (4/3 πR³)/(4/3πr³)
= (42 × 42 × 42)/(3 × 3 × 3)
= 14 × 14 × 14
= 2744
∴ The required number of lead balls = 2744
Que 5. The sum of the length, breadth, and height of a cuboid is 20 cm. If the length of the diagonal is 12 cm, then find out the total surface area of the cuboid.
A) 364 cm²
B) 264 cm²
C) 356 cm²
D) 256 cm²
Answer : D
Exp :
( L + B + H) = 20
Diagonal = √(L² + B² + H²)
=> 12 = √(L² + B² + H²)
=> (L² + B² + H²) = 144
We know,
(L + B + H)² = (L² + B² + H²) + 2(LB + BH + HL)
=> 20² = 144 + 2(LB + BH + HL)
=> 2(LB + BH + HL) = 400 – 144 = 256
Total surface area of a cuboid = 2(LB + BH + HL) = 256 cm²
∴ The required total surface area of the cuboid = 256 cm²
Que 6. Anil bought two articles A and B at a total cost of Rs.10000. He sold article A at 15% profit and article B at 10% loss. In the whole deal, he made no profit or no loss. Find the selling price of article A.
A) 4200
B) 4400
C) 4600
D) 4700
Answer : C
Exp :
Applying Alligation method,
15% – 10%
0
10 15
(0 + 10) : (15 – 0) = 2 : 3
(2 + 3 =) 5 units = 10000
=> 2 unit = (10000/5) × 2 = 4000
The cost price of article A = Rs.4000
Selling price of article A = 4000 × 115/100 =4600
∴ The selling price of article A = 4600
Que 7. The ratio of boys and girls in a school is 27 : 23. If the difference between the number of boys and girls is 200, then find the number of boys.
A) 1200
B) 1300
C) 1250
D) 1350
Answer : D
Exp :
Difference = (27 – 23) = 4 units
According to the question,
4 units = 200
=> 1 unit = 50
=> 27 units = (27 × 50) = 1350
∴ The number of boys is 1350.
Que 8. The interior angle of a regular polygon exceeds its exterior angle by 90°. The number of sides of the polygon is :
A) 6
B) 10
C) 8
D) 12
Answer : C
Exp :
We know, (Exterior angle + interior angle) =180°
Now, According to the question,
Exterior angle + exterior angle + 90° = 180°
=> 2 × exterior angle = 90°
=> Exterior angle = 45°
Number of sides = 360°/exterior angle = 360/45 = 8
∴ The required number of sides of the polygon = 8
Que 9. The radii of the two cylinders are in the ratio 3 : 4 and their heights are in the ratio 8 : 5. The ratio of their volume is –
A) 7 : 10
B) 8 : 11
C) 9 : 10
D) 11 : 13
Answer : C
Exp :
Let the radius of two cylinders is 3x and 4x respectively.
The height of the two cylinders is 8y and 5y respectively.
Ratio of volume of cylinders = πr1²h1 : πr2²h2 = (3x)² × 8y : (4x)² × 5y
= 9 × 8 : 16 × 5
= 9 : 10
∴ Required ratio of volume = 9 : 10
Que 10. Find the least number which when divided by 12, 18, 24 and 30 leaves 4 as remainder in each case, but when divided by 7 leaves no remainder.
A) 385
B) 364
C) 384
D) 374
Answer: B
Exp :
LCM of 12, 18, 24, and 30 = 360
Remainder = 4 in each case
So the number is (360 + 4) = 364 , it is divisible by 7.
∴ The required number is 7
Que 11. At what rate of interest will a sum of Rs.4500 amount to Rs.6525 at simple interest for 5 years?
A) 9%
B) 10%
C) 12%
D) 7%
Answer : A
Exp :
Simple interest = (6525 – 4500) = 2025
S.I = ( P × R × T)/100
=> (2025 × 100) = 4500 × R × 5
=> R = (2025 × 100)/(4500 × 5)
=> R = 9%
∴ Required rate of interest = 9%
Que 12. An athlete runs an 800 m race in 96 seconds. His speed ( in km/h) is –
A) 25 km/h
B) 27 km/h
C) 30 km/h
D) 32 km/h
Answer : C
Exp :
An athlete runs in 1 second = 800/96 m
1 m/s = 18/5 km/h
Speed = (800/96) × (18/5) km/h
= 30 km/h
∴ His speed is 30 km/h
Que 13. An umbrella is marked for Rs.150 and sold for Rs.138. The rate of discount is –
A) 9%
B) 8%
C) 12%
D) 15%
Answer : B
Exp :
MP = 150
Selling price = 138
Discount = (150 – 138) = 12
Discount % = (12/150) × 100 % = 8%
∴ The rate of discount is 8%
Que 14. The volume of a hemisphere is 2425½ cm³. Find its radius.
A) 8 cm
B) 8.5 cm
C) 9.5 cm
D) 10.5 cm
Answer : D
Exp :
2425½ = 4851/2
Volume of hemisphere = (2/3) πr³
According to the question,
(2/3) πr³ = 4851/2
=> r³ = (21/2)³
=> r = 21/2 = 10.5 cm
∴ The required radius is 10.5 cm.
Que 15. On selling 38 balls at Rs.2240, there is a loss equal to the cost price of 6 balls. The cost price of a ball is –
A) 60
B) 65
C) 70
D) 80
Answer: C
Exp :
Since selling 38 balls, the loss occurred cost price of 6 balls.
(38 – 6) = 32
Cost price of 32 balls = SP of 38 balls = 2240
CP of a ball = 2240/32 = 70
∴ The cost price of a ball is Rs.70
Que 16. The exterior angle obtained on producing the base of a triangle both ways are 121° and 104°. What is the measure of the largest angle of the triangle?
A) 65°
B) 70°
C) 72°
D) 76°
Answer : D
Exp :
We know, in a straight line, external angle + internal angle = 180°
∠ACB = 180° – 121° = 59°
∠ABC = 180° – 104° = 76°
∠BAC = 180° – 59° – 76° = 45°
∴ The largest angle is 76°
Que 17. In the given figure, ∠DBC = 65° , ∠BAC = 35° and AB = BC then the measure of ∠ECD is equal to –
A) 45°
B) 35°
C) 40°
D) 55°
Answer : A
Exp :
ABCD is a cyclic quadrilateral.
From ∆ABC,
Since AB = BC
∠BAC = ∠ACB = 35°
Angles are the same which are made by the same chord on the same side.
So , ∠DBC = ∠DAC = 65°
The Sum of opposite angles of a cyclic quadrilateral is 180°.
∠ DAB + ∠DCB = 180°
=> ∠DAC + ∠BAC + ∠DCA + ∠ACB = 180°
=> 65° + 35° + ∠DCA + 35° = 180°
=> ∠DCA = 180° – 135°
=> ∠DCA = 45°
Now, ∠DCA = ∠ECD
∠ECD = 45°
∴ The required value 45°
Que 18. Ramesh started a business investing a sum of Rs.40,000. Six months later, Kevin joined by investing Rs.20,000. If they make a profit of Rs.10,000 at the end of the year, how much is the share of Kevin?
A) Rs. 2700
B) Rs. 3200
C) Rs. 2000
D) Rs. 2300
Answer : C
Exp :
The investment ratio of Ramesh and Kavin
= (40,000 × 12) : (20,000 × 6)
= 4 : 1
Profit share of Kavin = (1/5) × 10,000 = 2000
∴ The share of Kavin is Rs.2000
Que 19. The train ticket fare from places A to B in 2nd class AC and 3rd class AC is Rs.2500 and Rs.2000 respectively. If the fares of 2nd class AC and 3rd class AC are increased by 20% and 10% respectively, then find the ratio of the new fares of 2nd class AC and 3rd class AC.
A) 13 : 12
B) 15 : 11
C) 13 : 11
D) 17 : 12
Answer : B
Exp :
Required ratio = 2500 × (120/100) : 2000 × (110/100)
= 3000 : 2200
= 15 : 11
∴ The required ratio = 15 : 11
Que 20. Evaluate : 1/15 + 1/35 + 1/63 + 1/99 + 1/143.
A) 11/39
B) 7/39
C) 5/39
D) 1/39
Answer : C
Exp :
1/15 + 1/35 + 1/63 + 1/99 + 1/143
= (21 + 9 + 5)/315 + (13 + 9)/1287
= 35/315 + 22/1287
= 1/9 + 2/117
= (13 + 2)/117
= 15/117
= 5/39
∴ The required value is 5/39
Que 21. In the figure, chords AB and CD of a circle intersect externally at P. If AB = 4 cm, CD = 11 cm and PD = 15 cm , then the length of PB is –
A) 6 cm
B) 10 cm
C) 13 cm
D) 18 cm
Answer : B
Exp :
Two chords AB and CD intersect at point P, so
PA × PB = PC × PD
Let , PA = X cm
PB = PA + AB = (X + 4 ) cm
PC = PD – CD = 15 – 11 = 4 cm
Now, from the formula
PA × PB = PC × PD
=> X × (X + 4) = 4 × 15
=> X² + 4X – 60 = 0
=> X² + 10X – 6X – 60 = 0
=> (X + 10) (X – 6) = 0
=> X = 6, – 10
Length can’t be negative, so the value of X = 6
PB = 6 + 4 = 10 cm
∴ The length of PB is 10 cm
Que 22 . The sum of the three numbers is 280. If the ratio between the first and 2nd numbers is 2 : 3 and the ratio between 2nd and 3rd numbers is 4 : 5, then find the 2nd number.
A) 96
B) 103
C) 80
D) 85
Answer : A
Exp :
First : Second = 2 : 3 —-(×4) = 8 : 12
Second : Third = 4 : 5 —-(×3) = 12 : 15
1st : 2nd : 3rd = 8 : 12 : 15
2nd number = (12/35) × 280 = 96
∴ The second number is 96
Que 23 . If (8 + 2√3)/(3√3 + 5) = a√3 – b, then the value of a + b is equal to –
A) 15
B) 16
C) 18
D) 24
Answer : C
Exp :
(8 + 2√3)/(3√3 + 5) = a√3 – b
=> [(8 + 2√3)(3√3 – 5)]/[(3√3 + 5)(3√3 – 5)] = a√3 – b
=> (24√3 – 40 + 18 – 10√3)/[(3√3)² – 5²] = a√3 – b
=> ( 14√3 – 22)/(27 – 25) = a√3 – b
=> 7√3 – 11 = a√3 – b
Here, a = 7 and b = 11
Now,
a + b
= 7 + 11
= 18
∴ The required value is 18.
Que 24 . If (secθ + tan θ)/(secθ – tan θ) = 2(51/79), then the value of sin θ is equal to –
A) 65/144
B) 55/144
C) 67/144
D) 7/144
Answer : A
Exp :
(secθ + tan θ)/(secθ – tan θ) = 209/79
=> 79 secθ + 79 tanθ = 209 secθ – 209 tanθ
=> 288 tanθ = 130 secθ
=> tanθ/secθ = 130/288
=> (sinθ/cosθ) × cosθ = 65/144
Sinθ = 65/144
∴ The required value is 65/144
Que 25 . The graphs of the linear equations 4x – 2y = 10 and 4x + ky = 2 intersect at a point (a, 4). Find the value of k.
A) – 4
B) 4
C) – 5
D) 5
Answer : A
Exp :
4x – 2y = 10 —–(1)
4x + ky = 2 ——(2)
Points (a, 4) satisfied the equations,
4a – 8 = 10
=> 4a = 18 —-(3)
4a + 4k = 2 —–(4)
Equating equations (3) and (4), we get
– 4k = 18 – 2
=> k = – 16/4
=> k = – 4
∴ The required value of k is – 4.
Que 26 . Find the sum of 6 + 8 + 10 + 12 + 14 + ……….. + 40
A) 310
B) 350
C) 396
D) 414
Answer : D
Exp :
6 + 8 + 10 + 12 + 14 + ………. + 40
= 2 + 4 – 2 – 4 + 6 + 8 + 10 + 12 + …………… + 40
= 2 + 4 + 6 + 8 + 10 + 12 + ……… + 40 – 6
= 20(20 + 1) – 6
= 420 – 6
= 414
∴ The required value is 414
Note :
- Sum of first n even numbers = n(n + 1)
Que 27 . In the given figure, ABCD is a rectangle and P is a point on DC such that BC = 24 cm, DP = 10 cm and CD = 15 cm . If AP produced intersects BC produced at Q, then find the length of AQ.
A) 31 cm
B) 39 cm
C) 45 cm
D) 50 cm
Answer : B
Exp :
From ∆APD,
AP ² = AD² + DP² = 24² + 10²
=> AP = 26 cm
Now from ∆APD and ∆CPQ,
∠ADP = ∠QCP ( right angle)
∠APD = ∠QPC ( vertically opposite angles)
AD || CQ
∴ ∆APD ~ ∆CPQ
∴ AD/QC = DP/CP = AP/PQ
Now, 10/5 = 26/PQ
PQ = 13 cm
AQ = AP + PQ = 26 + 13
AQ = 39 cm
∴ The required length of AQ = 39 cm
Que 28. A delivery boy started from his office at 10 am to deliver an article. He rode his scooter at a speed of 32 km/h. He delivered the article and waited for 15 minutes to get the payment. After the payment was made, he reached his office at 11.25 am travelling at a speed of 24 km/h. Find the total distance travelled by the boy.
A) 24 km
B) 27 km
C) 32 km
D) 40 km
Answer : C
Exp :
10 a.m to 11.25 am = 1.25 hr
15 minutes rest due to payment.
Travelling time = (1.25 h – 15 min) = 1.10 h = 70 minutes
Average speed = (2 × 32 × 24)/(32 + 24) = 192/7 km/h
Distance = (192/7) × 70/60 = 32 km
∴ The total distance travelled by the boy is 32 km.
Que 29. If 3sinx + 4cosx = 2, then find the value of 3cosx – 4 sinx.
A) √21
B) √23
C) 5
D) 23
Answer : A
Exp :
Let, 3cosx – 4sinx = k ———(1)
3sinx + 4cosx = 2 ———(2)
(1)² + (2)² , we get
(3cosx – 4sinx)² + (3sinx + 4cosx)² = k² + 2²
=> 9cos²x – 24sinxcosx + 16sin²x + 9sin²x + 24sinxcosx + 16cos²x = k² + 4
=> 9(cos²x + sin²x) + 16(sin²x + cos²x) = k² + 4
=> 9 + 16 – 4 = k²
=> k = √21
∴ the required value is √21
Directions (30 – 32): Study the following bar graph and answer the questions given below.
The total number of boys and girls in schools A, B, C, D, and E.
Difference between the number of boys and girls in schools A, B, C, D and E.
Solution ( 30 – 32) :
In School A
Boys + girls = 1800
Boys – girls = 350
Adding these, we get
2 × boys = 1800 + 350
=> Boys = 2150/2
Boys = 1075
Girls = (1800 – 1075) = 725
In School B
Boys + girls = 2600
Boys – girls = 520
2 × boys = 3120
Boys = 1560
Girls = (2600 – 1560) = 1040
In School C
Boys + girls = 2000
Boys – girls = 500
Boys = 2500/2 = 1250
Girls = (2000 – 1250) = 750
In School D
Boys + girls = 3200
Boys – girls = 850
Boys = 4050/2 = 2025
Girls = (3200 – 2025) = 1175
In School E
Boys + girls = 2800
Boys – girls = 700
Boys = 3500/2 = 1750
Girls = (2800 – 1750) = 1050
| SCHOOL |
TOTAL |
BOYS |
GIRLS |
| A |
1800 |
1075 |
725 |
| B |
2600 |
1560 |
1040 |
| C |
2000 |
1250 |
750 |
| D |
3200 |
2025 |
1175 |
| E |
2800 |
1750 |
1050 |
Que 30 . The number of boys in school B is what percentage of the total number of students in that school?
A) 40%
B) 50%
C) 60%
D) 70%
Answer : C
Exp :
Required percentage = (1560/2600) × 100% = 60%
Que 31 . What is the ratio of boys to the number of girls in school E?
A) 5 : 3
B) 3 : 2
C) 5 : 2
D) 4 : 3
Answer : A
Exp :
Required ratio = 1750 : 1050 = 5 : 3
Que 32 . What is the difference between the number of girls in school A and the number of girls in school C?
A) 15
B) 25
C) 30
D) 38
Answer : B
Exp :
Required difference = (750 – 725) = 25
Que 33. The average of five positive numbers is 56. If the first number is three – fourth of the sum of the last four numbers, then the average of the last four numbers is –
A) 26
B) 32
C) 40
D) 50
Answer : C
Exp :
Let the sum of the last four numbers = 8N
First number = 3/4 × 8N = 6N
Total sum of five numbers = 5 × 56 = 280
Now,
6N + 8N = 280
=> N = 280/14
=> N = 20
Sum of last four numbers = 8 × 20
Average of last four numbers = (8 × 20)/4 = 40
Que 34. If the radius of a cylinder is decreased by 20% and the height is increased by 20% to form a new cylinder, then the volume will be decreased by –
A) 21 %
B) 22.20 %
C) 23 %
Answer : D
Exp :
20% = 1/5
Original radius = 5
New radius = (5 – 1) = 4
Original height = 5
New height = (5 + 1) = 6
Original volume = π × 5² × 5 = 125π
New volume = π × 4² × 6 = 96π
Volume decrease = (125π – 96π) = 29π
Decrease% = (29π/125π) × 100% = 23.2%
Que 35 . The length of the shadow of a vertical tower on level ground increases by 10m when the altitude of the sun changes from 45° to 30°. The height of the tower is –
A) 2(√3 + 1) m
B) 3(√3 + 3) m
C) 5(√3 + 1) m
D) 6(√3 + 2) m
Answer : C
Exp :
tan45° = 1/1
tan30° = 1/√3
From fig,
(√3 – 1) units = 10 m
1 unit = 10/(√3 – 1)
1 unit = 5(√3 + 1)
∴ The height of the tower = 5(√3 + 1) m
Que 36 . In a triangle ABC, P and Q are points on AB and AC respectively, such that AP = 1 cm, PB = 3 cm, AQ = 1.5 cm and CQ = 4.5 cm. If the area of ∆APQ is 12 cm², then find the area of BPQC.
A) 190 cm²
B) 182 cm²
C) 180 cm²
D) 200 cm²
Answer : C
Exp :
AP : AB = 1 : (1 + 3) = 1 : 4
AQ : AC = 1.5 : (1.5 + 4.5) = 1 : 4
Here, the ratio of sides of the ∆APQ and ∆ABC are the same, so the triangles are similar.
∆APQ ~ ∆ABC
Now,
Area of ∆APQ/area of ∆ABC = (AP/AB)² = (PQ/BC)² = (AQ/AC)²
=> 12/∆ABC = (1/4)²
=> 12 × 16 = AREA OF ∆ABC
Area of ∆ABC = 192 cm²
Area of BPCQ = ∆ABC – ∆APQ = 192 – 12 = 180 cm²
∴ The area of BPCQ is 180 cm²
Que 37. Study the following histogram and answer the given question.
What is the ratio of the number of students who scored 30 or more marks, but below 40 marks, to the total number of students in the entrance examination?
A) 2 : 5
B) 3 : 5
C) 1 : 5
D) 3 : 2
Answer : C
Exp :
Number of Students who scored 30 – 40 = 20
Total students = (12 + 16 + 20 + 28 + 8 + 12 + 4) = 100
Required ratio = 20 : 100 = 1 : 5
∴ The required ratio is 1 : 5
Que 38. If the perimeter of an isosceles right triangle is 8(√2 + 1) cm, then find the length of the hypotenuse of the triangle.
A) 12 cm
B) 16 cm
C) 18 cm
D) 8 cm
Answer : D
Exp :
Let, equal sides of the isosceles triangle = x
Hypotenuse = √(x² + x²) = √2 x
Perimeter = x + x + √2x
According to the question,
2x + √2x = 8(√2 + 1)
=> √2x(√2 + 1) = 8(√2 + 1)
=> x = 8/√2 = 4√2 cm
∴ The length of the hypotenuse is 4√2 × √2 = 8 cm
Que 39. A container contains 20 L mixture in which there is 10% sulphuric acid. Find the quantity of sulphuric acid to be added in it to make the solution to contain 25% sulphuric acid.
A) 2 L
B) 4 L
C) 5 L
D) 8 L
Answer : B
Exp :
Sulphuric acid = 20 × (1/10) = 2 L
Remaining mixture = (20 – 2) = 18 L
Increasing sulphuric acid to 25% means the remaining mixture is 75%
75% = 18 L
100% = 24 L
Sulphuric acid = 24 × (25/100) = 6 L
Sulphuric acid added = (6 – 2) = 4 L
Que 40. If x² + 1/x² = 7, then find the value of x³ + 1/x³ where x > o.
A) 15
B) 12
C) 14
D) 18
Answer : D
Exp :
(x + 1/x)² = x² + 1/x² + 2
=> (x + 1/x)² = 7 + 2 = 9
=> x + 1/x = 3
Now,
x³ + 1/x³
= (x + 1/x)³ – 3 × x × 1/x(x + 1/x)
= 3³ – 3 × 3
= 18
∴ The required value is 18
Que 41. Rahul invested an equal sum of money at compound interest under two schemes A and B. Under scheme A, the interest rate was 10% per annum and under scheme B, the interest rate was 12% per annum. The compound interest after two years on the sum invested in scheme A was Rs.1050. How much is the interest earned under scheme B after two years, if the interest is compounded annually in both schemes?
A) 1136
B) 1186
C) 1272
D) 1395
Answer : C
Exp :
Scheme A
Rate for 2 years = 10 + 10 + (10 × 10)/100 = 21 %
Scheme B
Rate for 2 years = 12 + 12 + (12 × 12)/100 = 25.44 %
Here, 21 % = 1050
=> 25.44 % = (1050/21) × 25.44 = Rs 1272
∴ The compound interest in scheme B is rs.1272
Que 42 . A and B can do a piece of work in 18 days. B and C together can do it in 30 days. If A is twice as good a workman as C, then find in how many days B alone can do the work.
A) 90 days
B) 75 days
C) 60 days
D) 65 days
Answer : A
Exp :
LCM of 18 and 30 = 90
Efficiency of (A + B) = 90/18 = 5
Efficiency of (B + C) = 90/30 = 3
Efficiency of (A – C) = 5 – 3 = 2
Here, A is twice efficient as C. It means if A can do 4 units of work, C does 2 units of work.
Efficiency of B = ( 5 – 4) = 1
Time is taken by B = 90/1 = 90 days.
∴ B alone complete the work in 90 days
Que 43 . In a triangle ABC, AB = 6√3 cm, AC = 12 cm, and BC = 6 cm, Then measure the value of ∠B.
A) 60°
B) 70°
C) 80°
D) 90°
Answer : D
Exp :
Here,
AC² = 12² = 144
AB² + BC² = (6√3)² + 6² = 108 + 36 = 144
∴ AC² = AB² + BC²
Here AC, AB, and BC indicate hypotenuse, perpendicular, and base respectively.
We can say that this triangle is a right-angle triangle at ∠B.
∴ The required value of ∠B = 90°
Que 44 . ABCD is a rhombus with ∠ABC = 52°. Find the value of ∠ACD.
A) 36°
B) 52°
C) 64°
D) 90°
Answer : C
Exp :
Diagonals bisect the angles and meet at point O which is the right angle.
Opposite angles are equal because the rhombus is a parallelogram.
∠ADC = 52°
∠ODC = 52/2 = 26°
∠ DOC = 90°
From ∆DOC,
∠ODC + ∠DOC + ∠OCD = 180°
=> ∠OCD = 180° – 26° – 90°
∠OCD = 64°
∴ ∠ACD = 64°
Que 45. The sum of two positive numbers is 240 and their HCF is 15. Find the number of pairs of numbers satisfying the given condition.
A) 8
B) 2
C) 4
D) 5
Answer : C
Exp :
Let two numbers be 15x and 15y respectively where x and y are co-prime.
15x + 15y = 240
=> x + y = 16
Pairs are (1, 15) ; ( 3, 13) ; (5, 11) ; (7, 9)
∴ possible number of pairs = 4
Que 46. The selling price of one article after allowing a discount of 15% on its cost price, is the same as the selling price of another article after allowing a discount of 25% on its cost price. If the sum of the cost price of both the articles is 640, then find the selling price of each article.
A) 255
B) 225
C) 195
D) 210
Answer : A
Exp :
Let, the Cost price of the two articles be P and Q respectively.
SP of 1st article = P × 85/100
SP of 2nd article = Q × 75/100
According to the question,
P × 85/100 = Q × 75/100
=> P/Q = 15/17
Now, according to the question,
(15 + 17) unit = 640
=> 1 unit = 640/32 = 20
P = 15 unit = 15 × 20 = 300
Q = 17 unit = 17 × 20 = 340
SP1 = 300 × 85/100 = 255
∴ The selling price of each article is Rs 255.
Que 47. If √x + 1/√x = 3, then find the value of x³ + 1/x³.
A) 322
B) 298
C) 308
D) 411
Answer: A
Exp :
√x + 1/√x = 3
=> x + 1/x + 2 = 3² (Squaring both sides)
=> x + 1/x = 7
=> x³ + 1/x³ + 3x × 1/x (x + 1/x) = 7³ (Cubing both sides)
=> x³ + 1/x³ + 3 × 7 = 343
∴ x³ + 1/x³ = 343 – 21 = 322
Que 48. If cos²θ/(cot²θ – cos²θ) = 3, where 0°< θ < 90°, then find the value of θ.
A) 40°
B) 45°
C) 60°
D) 75°
Answer : C
Exp :
cos²θ/(cot²θ – cos²θ) = 3
=> cos²θ = 3cot²θ – 3cos²θ
=> 4cos²θ = 3cos²θ/sin²θ
=> 4sin²θ = 3
=> sinθ = √3/2 = sin60°
θ = 60°
∴ The required value is 60°
Que 49. A and B together can do a piece of work in 12 days. A alone can do it in 18 days. In how many days B alone can do the work?
A) 24 days
B) 27 days
C) 32 days
D) 36 days
Answer : D
Exp :
LCM of 12 and 18 = 36
Efficiency of (A + B ) = 36/12 = 3
Efficiency of A = 36/18 = 2
Efficiency of B = (3 – 2) = 1
Required days = 36/1 = 36
∴ B alone complete the work in 36 days
Que 50. The base of a right prism is a square having a side of 15 cm. If its height is 8 cm then find its total surface area.
A) 750 cm²
B) 880 cm²
C) 930 cm²
D) 1000 cm²
Answer : C
Exp :
Area of base = side² = 15² = 225 cm²
Perimeter = 4 × side = 4 × 15 = 60 cm
Total surface area
= (2 × area of base ) + (perimeter × height)
= 2 × 225 + 60 × 8
= 930 cm²
∴ The total surface area is 930 cm².
Que 51. If the surface area of a sphere is 1386 cm², then its volume is
A) 4851 cm³
B) 4422 cm³
C) 4675 cm³
D) 5665 cm³
Answer : A
Exp :
Surface area = 4πr²
4πr² = 1386
=> r² = (1386/4) × (7/22)
=> r² = (9 × 49)/4
=> r = 21/2
Now volume of sphere = (4/3) πr³ = (4/3) × 22/7 × (21/2)³ = 4851 cm³
∴ The required volume of the sphere is 4851 cm³
Que 52. The numerator of a fraction is 6 less than its denominator. If the numerator is decreased by 1 and the denominator is increased by 5, then the denominator becomes 4 times the numerator. Find the fraction.
A) 3/11
B) 4/11
C) 5/11
D) 8/11
Answer : C
Exp :
Let, denominator = x ; Numerator = (x – 6)
According to the question,
(x – 6 – 1)/(x + 5) = x/4x
=> 4x – 28 = x + 5
=> 3x = 33
=> x = 11 (Denominator)
Now, (x – 6) = 11 – 6 = 5 (Numerator)
∴ The required fraction is 5/11.
Que 53. If b/a = 0.7, then find the value of (a – b)/(a + b) + 11/34
A) 1.3
B) 2.5
C) 1
D) 0.5
Answer : D
Exp :
(a – b)/(a + b) + 11/34
= (1 – b/a)/(1 + b/a) + 11/34
= (1 – 0.7)/(1 + 0.7) + 11/34
= (0.3/1.7) + 11/34
= 3/17 + 11/34
= 17/34
= 1/2
= 0.5
∴ The required value is 0.5
Que 54. If x – 3/x = 6, then find the value of (x⁴ – 27/x²)/(x² – 3x – 3).
A) 45
B) 60
C) 90
D) 120
Answer : C
Exp :
x – 3/x = 6
=> x² + 9/x² – 6 = 36 ( squaring both sides)
=> x² + 9/x² = 42
Now, (x⁴ – 27/x²)/(x² – 3x – 3)
= (x³ – 27/x³)/(x – 3/x – 3) ( Divided by x)
= (x – 3/x)(x² + 9/x² + 3)/(6 – 3)
= [6 × (42 + 3)]/3
= 90
∴ The required value is 90
Que 55. A divisor is 15 times the quotient and 3 times the remainder. If the remainder is 40, then find the dividend.
A) 1000
B) 960
C) 1140
D) 1230
Answer : A
Exp :
Remainder = 40
Divisor = 40×3 = 120
Quotient = 120/15 = 8
W know,
Dividend = Divisor × Quotient + Remainder
=> Dividend = 120 × 8 + 40 = 1000
∴ The required dividend is 1000
Que 56. A, B, and C can do a work separately in 18, 36, and 54 days respectively. They started the work together, but B and C left 5 days and 10 days respectively before the completion of the work. In how many days was the work finished?
A) 9 days
B) 13 days
C) 14 days
D) 18 days
Answer : B
Exp :
LCM of 18, 36 and 54 = 108
A’s efficiency = 108/18 = 6
B’s efficiency = 108/36 = 3
C’s efficiency = 108/54 = 2
Total efficiency = (6 + 3 + 2) = 11
B’s 5 days work = 5 × 3 = 15
C’s 10 days work = 10 × 2 = 20
Total work = (108 + 15 + 20) = 143
∴ Required days = 143/11 = 13 days
Directions (57 – 59) : Study the following pie chart and table to answer the questions.
Total number of students admitted to a university in various fields = 5000.
Solution ( 57 – 59) :
Total students in Economics = 5000 × 12/100 =600
Boys in Economics = 600 × 56/100 = 336
Total students in CS = 5000 × 15/100 = 750
Boys in CS = 750 × 44/100 = 330
Total students in IT = 5000 × 14/100 = 700
Boys in IT = 700 × 65/100 = 455
Total students in ECE = 5000 × 16/100 = 800
Boys in ECE = 800 × 72/100 = 576
Total students in EEE = 5000 × 18/100 = 900
Boys in EEE = 900 × 68/100 = 612
Total students in H.M = 5000 × 25/100 = 1250
Boys in H.M = 1250 × 80/100 = 1000
Que 57. What is the average number of boys in CS, ECE and EEE fields?
A) 450
B) 479
C) 506
D) 601
Answer : C
Exp :
Total Boys in CS, ECE and EEE = (330 + 576 + 612) = 1518
Average number of boys = 1518/3 = 506
∴ The required average number is 506
Que 58. Find the ratio of the number of boys in Economics to the number of students in Economics.
A) 25 : 14
B) 14 : 25
C) 11 : 23
D) 9 : 13
Answer : B
Exp :
Required ratio = 336 : 600 = 14 : 25
Que 59. What is the difference between the number of girls in IT and the number of girls in ECE?
A) 21
B) 30
C) 35
D) 41
Answer : A
Exp :
Girls in IT = (700 – 455) = 245
Girls in ECE = ( 800 – 576) = 224
Required difference = (245 – 224) = 21
Que 60. If sin(x + y) = cos(x – y), then find the value of cos²x.
A) 1
B) 1/2
C) 1/√2
D) 0
Answer : B
Exp :
Sin(x + y) = cos(x – y)
=> Sin(x + y) = sin (90 – x + y)
=> x + y = 90 – x + y
=> 2x = 90
=> x = 45°
Now, cos²x = Cos²45° = (1/√2)² = 1/2
∴ The required value is 1/2
Que 61. Find the value of 5 – (8 + 2√15)/4 – 1/(8 + 2√15)
A) 1
B) 5
C) 8
D) 2
Answer : A
Exp :
5 – (8 + 2√15)/4 – 1/(8+2√15)
= 5 – (8+2√15)/4 – (8 – 2√15)/{8² – (2√15)²}
= 5 – (8+2√15)/4 – (8 – 2√15)/4
= 5 – (8 +2√15 + 8 – 2√15)/4
= 5 – 16/4
= 5 – 4
= 1
∴ The required value is 1
Que 62. If cosec39° = x, then find the value of 1/cosec²51°+sin²39°+tan²51°–1/sin²51°sec²39°
A) x²
B) x² + 1
C) x² – 1
D) x + 1
Answer : C
Exp :
1/cosec²51° + sin²39° + tan²51° – 1/sin²51°sec²39°
=sin²51° + cos²51° + tan²51° – cos²39°/cos²39°
= 1 + tan²51° – 1
= tan²51°
= cot²39°
= cosec²39° – 1
= x² – 1
∴ The required value is (x² – 1)
Note :
- sin θ = cos(90° – θ)
- cos θ = 1/sec θ
- sin θ = 1/cosecθ
- tan θ = cot (90° – θ)
- cot²θ = cosec²θ – 1
Que 63. If secθ + tan θ = 3, then find the value of secθ.
A) 10/3
B) 5/3
C) 3/10
D) 3/5
Answer : B
Exp :
secθ + tan θ = 3
secθ – tanθ = 1/3
Adding these two equations, we get
2 secθ = 10/3
secθ = 5/3
∴ The required value is 5/3
Que 64. A man walks at a speed of 8 km/h. After every km, he takes a rest of 4 minutes. How much time will it take to cover a distance of 6 km?
A) 45 minutes
B) 50 minutes
C) 60 minutes
D) 65 minutes
Answer : D
Exp :
8 km covers in 60 minutes
4 minutes rest after every 1 km.
So , he covers 5 km in [(60/8) × 5 + 5 × 4] min = (300/8 + 20) min
Last 1 km covers in (60/8) × 1 = 60/8 minutes
Total time taken = (300/8 + 20 + 60/8) = 65 min
∴ The required time is 65 minutes.
Que 65. If cosθ = 5/13, then find the value of tan²θ + sec²θ.
A) 313/25
B) 25/313
C) 213/25
D) 101/15
Answer : A
Exp :
cos θ = Base/Hypotenuse = 5/13
Base(b) = 5
Hypotenuse(h) = 13
Perpendicular(p) = √(13² – 5²) = √144 = 12
Now,
tan²θ + sec²θ
= (p/b)² + (h/b)²
= (12/5)² + (13/5)²
= 144/25 + 169/25
= 313/25
∴ The required value is 313/25.
Que 66. In a triangle, ABC, AB = AC and the perimeter of ∆ABC is 8(2 + √2) cm. If the length of BC is √2 times the length of AB, then find the area of ∆ABC.
A) 8 cm²
B) 16 cm²
C) 32 cm²
D) 64 cm²
Answer : C
Exp :
Let, AB = AC = x
BC = √2x
Perimeter of ∆ABC = (AB + AC + BC) = 2x + √2x
According to the question,
2x + √2x = 8(2 + √2)
=> 2x + √2x = 2 × 8 + 8 × √2
x = 8
Semi – perimeter (S) = (8 + 8 + 8√2)/2 = 8 + 4√2
Area of the triangle ABC
= √{S(S – A)(S – B)(S – C)}
= √[(8 + 4√2)(8 + 4√2 – 8)(8 + 4√2 –8)(8 + 4√2 – 8√2)]
= √[(8 + 4√2) × 4√2 × 4√2 × (8 – 4√2)]
= 4√2 √[8² – (4√2)²]
= 4√2 √(64 – 32)
= 4√2 × √32
= 32
∴ The area of∆ABC = 32 cm²
Que 67. ABC is an equilateral triangle with a side of 12 cm and AD is the median. Find the length of GD if G is the centroid of ∆ABC.
A) √2 cm
B) 2√3 cm
C) 2√2 cm
D) 5√3 cm
Answer: B
Exp :
AD = median/height
Height of an equilateral triangle = (√3/2) × 12 = 6√3 cm
G is the centroid. So the ratio of the median AG : GD = 2 : 1
3 unit = 6√3
=> 1 unit = 6√3/3 = 2√3 cm
∴ The length of GD = 2√3 cm.
Que 68. If A’s income is 60% less than B’s income, then B’s income is what percentage more than that of A’s income.
A) 30%
B) 50%
C) 100%
D) 150%
Answer : D
Exp :
A’s income is 60% less than B’s income.
A/B = 40/100 = 2/5
Required % = (5 – 2)/2 × 100 % = 150%
∴ B’s income is 150% more than A’s income.
Que 69. What is the reflection of the point (5,–3) in the line y = 3?
A) (5, 6)
B) (5, –6)
C) (5, –9)
D) (5, 9)
Answer : D
Exp :
Since reflection is in the y-axis, so x remains unchanged.
In reflection time, y = –3 comes as y = 3, so the unit change is above 6 units. [ 3 – (–3)] = 6
So the reflection of point = [5 , 6 –(–3)] = (5, 9)
Que 70. The sum of the weights of A and B is 80 kg. 50% of A’s weight is 5/6 times of the weight of B. Find the difference between their weights.
A) 20 kg
B) 26 kg
C) 32 kg
D) 40 kg
Answer : A
Exp :
A × 50% = 5/6 × B
=> A × 1/2 = 5/6 × B
=> A/B = 5/3
Total weight of A and B together = (5 + 3) = 8 units.
Difference of weight = (5 – 3) = 2 units
According to the question,
8 units = 80
=> 2 units = (80/8) × 2 = 20 kg.
∴ The required difference between the weight is 20 kgs.
Que 71. If x + 16/x = 8, then find the value of x² + 32/x²
A) 10
B) 14
C) 18
D) 20
Answer : C
Exp :
x + 16/x = 8
=> x² + 16 – 8x = 0 (Multiplying both sides by x)
=> (x – 4)² = 0
=> x – 4 = 0
=> x = 4
Now,
x² + 32/x²
= 4² + 32/4²
= 16 + 2
= 18
∴ The required value is 18.
Que 72. The curved surface area of a cylinder is five times the area of its base. Find the ratio of the radius and height of the cylinder.
A) 3 : 5
B) 2 : 5
C) 1 : 5
D) 5 : 2
Answer : B
Exp :
The curved surface area of a cylinder = 2πrh
Area of base = πr²
According to the question,
2πrh = 5 × πr²
=> r/h = 2/5
=> r : h = 2 : 5
∴ The required ratio of the radius and height of the cylinder is 2 : 5
Que 73. If (sinθ + cosecθ)² + (cosθ + secθ)² = k + tan²θ + cot²θ, then find the value of k.
A) 1
B) 3
C) 5
D) 7
Answer : D
Exp :
(sinθ + cosecθ)² + (cosθ + secθ)² = k + tan²θ + cot²θ
=> sin²θ + cosec²θ + 2sinθcosecθ + cos²θ + sec²θ + 2cosθsecθ = k + tan²θ + cot²θ
=> 1 + 2 + 2 + cosec²θ – cot²θ + sec²θ – tan²θ = k
=> 5 + 1 + 1 = k
∴ k = 7
∴ The required value of k is 7.
Note :
- sin²θ + cos²θ = 1
- cosec²θ – cot²θ = 1
- sec²θ – tan²θ = 1
- sinθ = 1/cosecθ
- cosθ = 1/secθ
Que 74. In a two-digit number, its unit digit exceeds its tens digit by 2 and the product of the given number and the sum of its digits is equal to 460. Find the number.
A) 40
B) 44
C) 46
D) 60
Answer : C
Exp :
Option test is the best method to solve these types of questions.
Here option (C) 46, clearly shows that its unit digit exceeds its tens digit by 2.
Que 75. The ratio between the present ages of A and B is 3 : 5. If the ratio of their ages five years after becomes 13 : 20, then find the present age of B.
A) 25
B) 30
C) 35
D) 45
Answer : C
Exp :
The ratio gap should be the same.
So
3 : 5 —-(1) ×7 = 21 : 35
13 : 20 —–(2) × 2 = 26 : 40
∴ The present age of B = 35 years.
Que 76. In how much time will the simple interest on a certain sum of money be 6/5 times of the sum at 20% per annum?
A) 6 years
B) 3 years
C) 4 years
D) 8 years
Answer : A
Exp :
Let, Sum of the amount = P
Simple interest = 6P/5
S.I = PRT/100
=> 6P/5 = (P × 20 × T)/100
=> T = 6
∴ Thus the required time is 6 years.
Que 77. If x(3 – 2/x) = 3/x, then find the value of x³ – 1/x³
A) 62/27
B) 60/29
C) 1/60
D) 62/29
Answer : A
Exp :
x(3 – 2/x) = 3/x
=> 3x – 2 – 3/x = 0
=> 3(x – 1/x) = 2
=> (x – 1/x) = 2/3
Now,
= x³ – 1/x³
= (x – 1/x)³ + 3 × x × 1/x(x – 1/x)
= (2/3)³ + 3 × (2/3)
= 8/27 + 2
= 62/27
∴ The required value is 62/27
Que 78. In the given figure, measure the value of ∠A.
A) 20°
B) 30°
C) 40°
D) 60°
Answer : C
Exp :
AB = PQ
∠B = ∠Q
BC = QR
∴ ∆ABC ~ ∆PQR
∴ ∠A = ∠P
So, 2x = x + 20°
=> x = 20°
=> 2x = 40°
∴ The required value of ∠A is 40°
Que 79. A man travelled a distance of 42 km in 5 h. He travelled partly on foot at the rate of 6 km/h and partly on bicycle at the speed 10 km/h. The distance travelled on foot is –
A) 24 km
B) 18 km
C) 16 km
D) 12 km
Answer : D
Exp :
Speed = 42/5 = 8.4 km/h
Applying alligation method,
6 10
8.4
1.6 2.4
= 2 : 3
(2 + 3 =) 5 unit = 5 h
2 unit = 2 hours
Required distance = 6 × 2 = 12 km
∴ The distance travelled on foot is 12 km.
Que 80. In a ∆ABC, D is a point on BC such that AB/AC = BD/DC. If ∠B = 68° and ∠C = 52°, then the measure of ∠BAD is equal to-
A) 45°
B) 30°
C) 25°
D) 35°
Answer : B
Exp :
We know, the sum of the angles of a triangle = 180°
∠A + ∠B + ∠C = 180°
=> ∠A = 180° – 68° – 52°
=> ∠A = 60°
Since, AB/AC = BD/DC
So we can say AD is the angle bisector of ∠A.
∠BAD = 1/2 × ∠A = 60°
∴ The measure of ∠BAD is 60°
Que 81. At what rate percent per annum will a sum of Rs.15625 amount to Rs.21952 in three years, if the interest is compounded annually?
A) 12%
B) 8%
C) 6%
D) 10%
Answer : A
Exp :
Amount = P(1 + R/100)T
=> 21952 = 15625 (1 + R/100)³
=> (1 + R/100)³ = 21951/15625
=> 1 + R/100 = 28/25
=> R/100 = 28/25 – 1
=> R = (3/25) × 100
∴ R = 12%
∴ The required rate of interest is 12%
Que 82. Find the number of prime factors in the product (30)⁵ × (24)⁵
A) 15
B) 5
C) 20
D) 35
Answer : D
Exp :
Prime factors of 30 = 2 × 3 × 5
Prime factors of 24 = 2 × 2 × 2 × 3
(30)⁵ × (24)⁵
= (2 × 3 × 5)⁵ × (2 × 2 × 2 × 3)⁵
= 25+5+5+5 × 35+5 × 55
= 2²⁰ × 3¹⁰ × 5⁵
∴ Number of prime factors = (20 + 10 + 5) = 35
Que 83. The radius and height of a cylinder are in the ratio 4 : 7 and its volume 2816 cm³. Find the radius.
A) 3 cm
B) 6 cm
C) 8 cm
D) 16 cm
Answer : C
Exp :
Let, radius = 4x and height = 7x
Volume of cylinder = πr²h
According to the question,
πr²h = 2816
=> 22/7 × (4x)² × 7x = 2816
=> 16x² × 7x = 896
=> x³ = 8
∴ x = 2
∴ 4x = 8
∴ The radius is 8 cm.
Que 84. A, B and C together invest Rs.53,000 in a business. A invests Rs.5000 more than B and B invests Rs.6000 more than C. Out of a total profit of Rs.31,800 find the share of A.
A) 13800
B) 12500
C) 13000
D) 15200
Answer : A
Exp :
Let, investment of C = x
Investment of B = x + 6000
Investment of A = x + 11000
According to the question,
x + x + 6000 + x + 11000 = 53,000
=> 3x = 53000 – 17000
=> x = 36000/3
=> x = 12000
Investment of C = 12,000
Investment of B = (12,000 + 6000) = 18,000
Investment of A = (12,000 + 11000) = 23,000
The ratio of investment of A, B and C
= 23000 : 18000 : 12000
= 23 : 18 : 12
Profit share of A = (23/53) × 31800 = 13800
∴ the share of A is Rs.13800
Que 85. A dealer sold an article at a loss of 2%. Had he sold it for Rs.44 more, he would have gained 20%. Find the cost price.
A) 100
B) 120
C) 150
D) 200
Answer : D
Exp :
Loss = –2 %
gain = 20 %
overall change in SP = (20 –(–2) = 22 %
22% = 44
1% = 2
100% = 200
∴ The required cost price Rs 200
Que 86. The ratio of the height and the diameter of a right circular cone is 6 : 5 and its volume is 2200/7 cm³. What is slant height?
A) 8 cm
B) 9 cm
C) 12 cm
D) 13 cm
Answer : D
Exp :
Let, height = 6x and diameter = 5x
Radius = diameter/2 = 5x/2
Volume = (1/3)πr²h
According to the question,
(1/3) πr²h = 2200/7
=> 1/3 × 22/7 × (5x/2)² × 6x = 2200/7
=> x³ = 8
∴ x = 2
height = 6 × 2 = 12
Radius = (5/2) × 2 = 5
Slant height = √(12² + 5²) = 13 cm
∴ The slant height is 13 cm.
Que 87. A sum of Rs.1,50,000 is distributed among three persons A, B and C. So that they receive 20%, 30% and 50% respectively. A receives the same amount from another sum of money which is distributed among them so that they receive 50%, 30% and 20% respectively. Find the total amount received from both sums of money by B.
A) 56,000
B) 60,000
C) 63,000
D) 70,000
Answer : C
Exp :
A received = 150000 × (20/100) = 30,000
B received = 150000 × (30/100) = 45,000
A received same amount from another sum which is 50%.
Another sum = 30000 × 2 = 60,000
B received = 60000 × (30/100) = 18000
Total amount received by B = (45000 + 18000) = 63,000
∴ The total amount received by B is 63,000
Que 88. If x = √[–√3 + √{3 + 8√(7 + 4√3)}] , where x > o , find the value of x.
A) 1
B) √2
C) 2
D) 4
Answer : C
Exp :
x = √[–√3 + √{3 + 8√(7 + 4√3)}]
=> X = √[–√3 + √{3 + 8√(4 + 3 + 4√3)}]
=> X = √[–√3 + √{3 + 8√(2 + √3)²}]
=> X = √[–√3 + √{3 + 8(2 + √3)}]
=> X = √[–√3 + √{3 + 16 + 8√3}]
=> X = √[–√3 + √{4 + √3}²]
=> X = √[–√3 + 4 + √3]
=> X = √4
=> X = 2
∴ The required value of X is 2
Que 89. A man sells two articles at Rs.9975 each. He gains 5% on one article and losses 5% on another article. Find his overall gain or loss.
A) 20
B) 10
C) 40
D) 50
Answer : D
Exp :
SP of two articles = 9975 × 2 = 19,950
CP1 = 9975 × (100/105) = 9500
CP2 = 9975 × (100/95) = 10500
Total CP = (9500 + 10500) = 20,000
Loss = (20,000 – 19,950) = 50
∴ His overall loss = 50
Que 90. The price of a variety of a commodity is Rs.7/kg and that of another is Rs.12/kg. Find the ratio in which two varieties should be mixed so that the price of the mixture is Rs.10/kg.
A) 1 : 2
B) 1 : 3
C) 2 : 3
D) 3 : 4
Answer : C
Exp :
Applying alligation method,
7 12
10
2 3
= 2 : 3
∴ The required ratio is 2 : 3
Que 91. The average age of Kishore, his wife and their child 6 years ago was 38 years and that of his wife and their child 8 years ago was 32 years. Find the present age of Kishore.
A) 36
B) 42
C) 48
D) 52
Answer : D
Exp :
Total Present age of three persons = (38 × 3 + 6 × 3) = 132
Present ages of wife and child = (32 × 2 + 8 × 2) = 80
Present age of Kishore = (132 – 80) = 52
∴ The present age of Kishore is 52 years.
Que 92. What is to be added to 15% of 180 so that the sum is equal to 20% of 360?
A) 30
B) 40
C) 45
D) 70
Answer : C
Exp :
Let, X should be added.
X + (15/100) × 180 = (20/100) × 360
=> 27 + X = 72
=> X = 45
∴ The required number is 45
Que 93. If 2 = x + 1/[1+ {1/(5 + 1/2)}] , then find the value of x.
A) 13/15
B) 15/13
C) 3/5
D) 5/3
Answer : B
Exp :
2 = x + 1/[1+ {1/(5 + 1/2)}]
=> 2 = X + 1/[1 + 2/11]
=> 2 = X + 11/13
=> X = 2 – 11/13
=> X = 15/13
∴ The required value of X is 15/13
Que 94. Solve this: 5 – [ 96 ÷ 4 of 3 – (16 – 55 ÷ 5)}]
A) 1
B) 2
C) 4
D) 3
Answer : B
Exp :
5 – [ 96 ÷ 4 of 3 – (16 – 55 ÷ 5)]
= 5 – [ 96 ÷ 4 of 3 – (16 – 11)]
= 5 – [ 96 ÷ 4 of 3 – 5 ]
= 5 – [ 96 ÷ 12 – 5 ]
= 5 – [ 8 – 5 ]
= 5 – 3
= 2
∴ The required value is 2
Que 95. The base of a pyramid is an equilateral triangle of side 10 m. If the height of the pyramid is 40√3 m, then the volume of the pyramid is –
A) 900 m³
B) 960 m³
C) 1000 m³
D) 1200 m³
Answer : C
Exp :
Area of an equilateral triangle = √3/4 × 10² = 25√3 m²
Volume of pyramid (V) = (1/3) × area of base × height
=> V = (1/3) × 25√3 × 40√3
=> V = 1000 m³
∴ The volume of the pyramid is 1000 m³.
Que 96. A cyclic quadrilateral ABCD is such that AB = BC, AD = DC and AC, BD intersect at O and BD is angle bisector of ∠B. If ∠CAD = 46°, then the measure of ∠AOB is equal to-
A) 45°
B) 60°
C) 75°
D) 90°
Answer : D
Exp :
ABCD is a cyclic quadrilateral. So sum of its opposite angle is 180°.
∠ABC = 180° – ∠ADC = 180° – 88° = 92°
Since AB = BC
∠BAC = ∠BCA = (180° – 92°)/2 = 44°
BD is a angle bisector of B, so
∠ABO = 92°/2 = 46°
∠AOB = 180° – 46° – 44° = 90°
∴ The value is 90°
Que 97. If 1/4.263 = 0.2346, find the value of 1/0.0004263
A) 1000
B) 10000
C) 2346
D) 4263
Answer : C
Exp :
1/4.263 = 0.2346
=> 1000/4.263 = 0.2346 × 1000 (multiple by 1000)
=> 1/0.0004263 = 2346
∴ The required value is 2346
Que 98. A conical tent has to accommodate 25 people. Each person must have 4 m² of space on the ground and 80 m³ of air to breathe. Find the height of the tent.
A) 30 m
B) 40 m
C) 45 m
D) 60 m
Answer : D
Exp :
Total base area for 25 person = 25 × 4 = 100 m²
Total volume of air for 25 person = 80 × 25 = 2000 m³
Volume of conical tent (V) = (1/3) × area of base × height
=> 2000 = 1/3 × 100 × h
=> h = 60 m
∴ The height is 60 m
Que 99. If sinθ + sin²θ = 1, then find cos²θ + cos⁴θ
A) 1
B) – 1
C) 3
D) 4
Answer : A
Exp :
sinθ + sin²θ = 1
=> sinθ = 1 – sin²θ
=> sinθ = cos²θ
Now, cos²θ + cos⁴θ
= sinθ + sin²θ
= 1
∴ The required value is 1
Que 100. If α + β = 90° and α = 2β , then find 3cos²α – 2sin²β.
A) 1
B) 1/2
C) 1/4
D) 1/6
Answer : C
Exp :
α + β = 90°
=> 2β + β = 90° ( putting α = 2β)
=> β = 30°
α = 2 × 30° = 60°
3cos²α – 2sin²β
= 3cos²60° – 2sin²30°
= 3 × (1/2)² – 2 × (1/2)²
= 3/4 – 1/2
= 1/4
∴ The required value is 1/4
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