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This mock test of Quantitative Aptitude - Test 4 for SSC helps you for every SSC entrance exam.
This contains 25 Multiple Choice Questions for SSC Quantitative Aptitude - Test 4 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

The length of a rectangle is 2 cm more than its breadth. The diagonal is 5 cm. Find the length of the rectangle?

Solution:

From given data

Length of the rectangle l = (b + 2)

We know that diagonal d of a rectangle = √(l^{2} + b^{2})

⇒ d = √[(b + 2)^{2} + b^{2}]

⇒ 5 = √[(b + 2)^{2} + b^{2}]

On squaring on both sides, we get

⇒ 25 = 2b^{2} + 4b + 4

⇒ 21 = 2b^{2} + 4b

⇒ b = -1 ± √(23/2)

∴ Length = (b + 2) = -1 ± √(23/2) + 2 = 1 ± √(23/2)

QUESTION: 2

The wheel of a railway carriage is 3 m in diameter and makes 8 revolutions per second, how fast is the train going?a. 27

Solution:

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

QUESTION: 3

A wholesaler marks his goods 35% above the cost price. If he allows a discount of 6 ^{1}/_{5}%, then his profit percent is

Solution:

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 %

QUESTION: 4

A sum of Rs. 25220 is borrowed at 5% per annum compounded annually which is paid back in 3 equal instalments. What is the amount of each instalment?

Solution:

Let each instalment is x.

We know that Amount = p(1 + r/100)^{n}

⇒ x = 9261

∴ Amount at each instalment = Rs. 9261

QUESTION: 5

The average monthly income of A and B is Rs. 5000. The average monthly income of B and C is Rs. 7250 and the average monthly income of A and C is Rs. 8500. Find the monthly income of A ?

Solution:

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

QUESTION: 6

An good is sold at a certain price. By selling it at ¾ of that price one loses 20%. Find the profit percent at its original price.

Solution:

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%

QUESTION: 7

In an examination, 30% of total students failed in Hindi, 25% failed in English and 10% failed in both. Find the percentage of those students who passed in both the subjects?

Solution:

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%

QUESTION: 8

Shiva travelled from home to temple at the rate of 35 kmph and returned at the rate of 6 kmph. If the whole journey took 4 hours 45 minutes. Find the distance between home and temple.

Solution:

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

QUESTION: 9

Find the value of x in the following expression 2^{x - 3} = 1/(8^{x - 4})

Solution:

From the given data

⇒ 2^{x - 3} = 1/(8^{x - 4})

When bases are equal, powers must be added

⇒ 2^{4x - 15} = 2^{0}

On comparing on both sides, we get

⇒ 4x - 15 = 0

⇒ 4x = 15

⇒ x = 15/4

QUESTION: 10

If then the value of x is

Solution:

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

QUESTION: 11

If (4y + 72), (33 + y) are supplementary angles, find y?

Solution:

From given data,

We know that sum of two supplementary angles = 180°

⇒ (4y + 72) + (33 + y) = 180°

⇒ 5y + 105 = 180°

⇒ 5y = 75°

⇒ y = 15°

QUESTION: 12

One angle of a triangle is 60° and other angle is π/2 radian. Find the third angle in centesimal unit?

Solution:

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

QUESTION: 13

The value of sin 225° + sin 270° is

Solution:

From the given data

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

QUESTION: 14

A can work twice as B and they together can finish a work in 16 days. In how many days will A alone finish the work?

Solution:

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

QUESTION: 15

If a + b = 26 and a - b = 9 then the value of (4a^{2}b^{2}) is,

Solution:

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 (4a^{2}b^{2}) = (4 (35/2 )^{2}(17/2)^{2}) = 88506.25

QUESTION: 16

If x + y = √3 and x - y = √2, then the value of (x^{2} + y^{2}) is

Solution:

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 (x^{2} + y^{2})

⇒ (x^{2} + y^{2}) = [{ (√3 + √2)/2 }^{2} + {(√3 - √2)/2 }^{2}]

= [2(3 + 2)]/4

= 5/2

QUESTION: 17

If the length of a chord is 24 cm which is at a certain distance from the centre of a circle of radius 13cm. Find distance between chord from centre of the circle?

Solution:

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

⇒ OA^{2} = AP^{2} + OP^{2}

⇒ 13^{2} = OP^{2} + 12^{2}

⇒ 169 - 144 = OP^{2}

⇒ OP = 5 cm

∴ Distance between chord from centre of the circle = OP = 5 cm

QUESTION: 18

Each interior angle of a regular polygon is 150°. The number of sides is

Solution:

From the given data

Each interior angle = 150°

∴ Exterior angle = 180° - 150° = 30°

⇒ Number of sides = 360°/exterior angle = 360°/30° = 12

QUESTION: 19

If 2sin^{2}θ + 4 cos^{2}θ = 3, then find the value of θ ?

Solution:

From the given data

We know that 2sin^{2}θ + 4 cos^{2}θ = 3

⇒ 2(1 - cos^{2}θ ) + 4 cos^{2}θ = 3

⇒ 2 – 2cos^{2}θ + 4 cos^{2}θ = 3

⇒ 2 cos^{2}θ = 3 - 2

⇒ cos^{2}θ = 1/2

⇒ cosθ = 1/√2

⇒ θ = 45°

QUESTION: 20

A sum of amount at r% compound interest doubles in 3 years. In 6 years it will be k times of the original principle. What is the value of k?

Solution:

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

⇒ 2^{2} = (1 + r/100)^{3}]^{2}

⇒ 4 = (1 + r/100)^{6}

⇒ k = 4 times

QUESTION: 21

A person standing on the bank of a river observes that the angle subtended by a tree on the opposite bank is 45°, when retreats 50 m from the bank he finds the angle to be 30°. find the height of the tree and breadth of the river ?

Solution:

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

QUESTION: 22

The following diagram shows the percentages of different grades of the 20000 students in the college. Observe the pie diagram and answer the questions

**Question: **Find the number of students who got A grade?

Solution:

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

QUESTION: 23

The following diagram shows the percentages of different grades of the 20000 students in the college. Observe the pie diagram and answer the questions

Question:Find the ratio between the number of students who got A grade and B grade?

Solution:

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

QUESTION: 24

The following diagram shows the percentages of different grades of the 20000 students in the college. Observe the pie diagram and answer the questions

**Question: **Find the ratio between number of students who got C grade and D grade?

Solution:

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

QUESTION: 25

The following diagram shows the percentages of different grades of the 20000 students in the college. Observe the pie diagram and answer the questions

**Question: **Find the ratio between difference of number of students who got A and B grades to the difference of the number of students who got C and D grades?

Solution:

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

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