Practice Test: Quantitative Aptitude - 4 - SSC CGL MCQ

From given data
Length of the rectangle l = (b + 2)
We know that diagonal d of a rectangle = √(l² + b²)
⇒ d = √[(b + 2)² + b²]
⇒ 5 = √[(b + 2)² + b²]
On squaring on both sides, we get
⇒ 25 = 2b² + 4b + 4
⇒ 21 = 2b² + 4b
⇒ b = -1 ± √(23/2)
∴ Length = (b + 2) = -1 ± √(23/2) + 2 = 1 ± √(23/2)

From given data
Radius of the wheel r = 3/2 = 1.5 m
⇒ Circumference of the wheel = 2πr = 2π × 1. 5 = 3π = 9. 42 m
Given that wheel makes 8 revolutions per second
⇒ Distance travelled in 1 second = 8 × 9.42 = 75.36 m
⇒ Distance travelled in one hour = 75.36 × 3600 = 271296 m = 271.296 km
∴ Speed of the train ≈ 271.30 km/hr

From given data
Let the cost price of the article be Rs. 100, then marked price = Rs. 135
Selling price = 135 – 31/5% of 135 = 135 – 8.37 = 126.63
∴ Profit percent = (126.63 – 100)/100 × 100 = 26.63 %

Let each instalment is x.
We know that Amount = p(1 + r/100)ⁿ

⇒ x = 9261
∴ Amount at each instalment = Rs. 9261

From given data
Total income earned by A and B (A + B) = 2 × 5000 = Rs. 10000  …equation 1
Total income earned by B and C (B + C) = 2 × 7250 = Rs. 14500  … equation 2
Total income earned by A and C (A + C) = 2 × 8500 = Rs. 17000  … equation 3
By adding the three equations, we get
⇒ 2(A + B + C) = 41500
⇒ A + B + C = 20750  … equation 4
Now subtract eq 2 from equation 4, we get
⇒ A = 20750 – 14500 = 6250
∴ Monthly income of A = Rs. 6250

From given data
Let the original selling price of article be Rs. x. Then, new selling price = Rs. ¾ x
Given that loss = 20%
∴ Cost price = (100/80 × 3x/4) = 15x/16
⇒ Profit = (x – 15x/16) = Rs. x/16
∴ Profit % = (profit/cost price × 100 )% = (x/16 × 16/15x × 100 ) = 6.66%

From given data
No of students failed in hindi n(H) = 30%
No of students failed in English n (E) = 25%
No of students failed in both n(H ∩ E) = 10%
⇒ Percentage of students who failed n(H U E) = n(H) + n(E) – n(H ∩ E)
⇒ n(H U E) = 30 + 25 - 10
⇒ n(H U E) = 45
Percentage of students who passed = 100 - (Percentage of students who failed) = 100 - 45 = 55%

From given data
Average speed of the two journey = 2xy/(x + y) = 2(35)(6)/(35 + 6) = 10.24 kmph
Distance travelled in 4 hours 45 min i. e, 4 ¾ hrs = 10.24 × 19/4 = 48.65 km
∴ Distance between home and temple = 48. 65/2 = 24.325 km

From the given data
⇒ 2^{x - 3} = 1/(8^{x - 4})

When bases are equal, powers must be added
⇒ 2^{4x - 15} = 2⁰
On comparing on both sides, we get
⇒ 4x - 15 = 0
⇒ 4x = 15
⇒ x = 15/4

From given data
⇒ (x^x)^1/2 = (x^3/4)^x
⇒ (x^x)^1/2 = x^{3/4 x}
⇒ x^1/2 = ¾ x
⇒ x^-1/2 = ¾
Squaring on both sides, we get
⇒ x^-1 = 9/16
⇒ 1/x = 9/16
⇒ x = 16/9

From given data,
We know that sum of two supplementary angles = 180°
⇒ (4y + 72) + (33 + y) = 180°
⇒ 5y + 105 = 180°
⇒ 5y = 75°
⇒ y = 15​°

From the given data
∠ A = 60° and∠B = π/2 radian = 90°
We know that sum of the angles in triangle = 180°
⇒ ∠A + ∠B + ∠C = 180°
⇒ ∠C = 180 - 150 = 30°
We know that 90° = 100 grad
⇒ 30° = 30 × 100/90 = 33.33 grade

From the given data
⇒ sin (180 + 45)° + sin (270)° = -sin 45° +  (-sin 90°) = - 1/√2 – 1 = ( - 1 - √ 2 )/√ 2 = -1 - 1/√2

From the given data,
⇒ Ratio of A’s work and B’s work = 2 : 1
⇒ Work done by A and B in one day = 1/16
Divide 1/16 in the ratio of 2 : 1
⇒ Work done by A in one day = (1/16 × 2/3) = 2/48 = 1/24
∴ A alone can finish the work in 24 days

From given data
a + b = 26 and a - b = 9
By adding both the equations, we get
⇒ 2a = 35
⇒ a = 35/2
Substitute a = 35/2 in a + b = 26
⇒ 35/2 + b = 26
⇒ b = 17/2
Then value of (4a²b²) = (4 (35/2 )²(17/2)²) = 88506.25

From the given data
We know that x + y = √3 and x ­- y = √2
By adding both the equations, we get
⇒ 2x = √ 3 + √ 2
⇒ x = (√ 3 + √ 2 )/2
Substitute x = (√3 + √2)/2 in x + y = √3
⇒ y = [2√3 - (√3 + √2)]/2 = (√3 –√2 )/2
Substitute x = (√3 + √2)/2 and y = (√3 - √2)/2 in (x² + y²)
⇒ (x² + y²) = [{ (√3 + √2)/2 }² + {(√3 - √2)/2 }²]
= [2(3 + 2)]/4
= 5/2

From given data

Let AB = 24 cm be a chord of circle with center O and radius OA and OB equals to 13 cm
Draw OP perpendicular to AB
⇒ AP = AB/2 = 24/2 = 12 cm
From ΔOPA, we know that
⇒ OA² = AP² + OP²
⇒ 13² = OP² + 12²
⇒ 169 - 144 = OP²
⇒ OP = 5 cm
∴ Distance between chord from centre of the circle = OP = 5 cm

From the given data
Each interior angle = 150°
∴ Exterior angle = 180° - 150° = 30°
⇒ Number of sides = 360°/exterior angle = 360°/30° = 12

From the given data
We know that 2sin²θ + 4 cos²θ = 3
⇒ 2(1 - cos²θ ) + 4 cos²θ = 3
⇒ 2 – 2cos²θ + 4 cos²θ = 3
⇒ 2 cos²θ = 3 - 2
⇒ cos²θ = 1/2
⇒ cosθ = 1/√2
⇒ θ = 45° 

From the given data
We know amount = p(1 + r/100 )ⁿ
In the first case
⇒ 2p = p(1 + r/100)³
⇒ 2 = (1 + r/100)³
In the second case
⇒ 2² = (1 + r/100)³]²
⇒ 4 = (1 + r/100)⁶
⇒ k = 4 times

From the given data

Let AB be the height of the tree = h mts
Let BC be the width of the tree = x mts and CD = 50 m
⇒ ∠ ACB = 45° and ∠ADB = 30°
From ΔABC , we get
⇒ tan 45° = AB/BC
⇒ 1 = h/x
⇒ h = x m           …eq 1
From ΔABD, we get
⇒ tan 30° = AB/BD
⇒ 1/√3 = h/(x + 50)
⇒ h = (x + 50)/√3 m         …eq 2
From eq 1 and eq 2, we get
⇒ x = (x + 50)/√ 3
⇒ x(√ 3) = x + 50
⇒ x(√ 3 - 1) = 50
⇒ x = 68. 3 m
⇒ h = x = 68. 3 m
∴ Height of the tree is 68.3 m and width of the river is 68. 3 m

From the given data
Percentage of number of students who got A grade = 42%
⇒ Number of students who got A grade = 42/100 × 20,000 = 8400

From the given data
Percentage of number of students who got A grade = 42%
⇒ Number of students who got A grade = 42/100 × 20,000 = 8400
Percentage of number of students who got B grade = 25%
⇒ Number of students who got B grade = 25/100 × 20,000 = 5000
∴ Ratio between the number of students who got A grade and B grade = 8400 : 5000 = 42 : 25

From the given data
Percentage of number of students who got C grade = 22%
⇒ Number of students who got C grade = 22/100 × 20,000 = 4400
Percentage of number of students who got D grade = 11%
⇒ Number of students who got D grade = 11/100 × 20,000 = 2200
∴ Ratio between the number of students who got C grade and D grade = 4400 : 2200 = 2 : 1

From the given data
Percentage of number of students who got A grade = 42%
⇒ Number of students who got A grade = 42/100 × 20, 000 = 8400
Percentage of number of students who got B grade = 25%
⇒ Number of students who got B grade = 25/100 × 20,000 = 5000
Difference of number of students who got A and B grades = 8400 – 5000 = 3400
Percentage of number of students who got C grade = 22%
⇒ Number of students who got C grade = 22/100 × 20,000 = 4400
Percentage of number of students who got D grade = 11%
⇒ Number of students who got D grade = 11/100 × 20,000 = 2200
Difference of number of students who got C and D grades = 4400 - 2200 = 2200
∴ Ratio between difference of number of students who got A and B grades to the difference of number of students who got C and D grades = 3400 : 2200 = 17 : 11