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QUESTION: 1

A is thrice as fast as B and is therefore able to finish a work in 40 days less than that of B. Find the time in which they can do it working together.

Solution:

Given,

Ratio of time taken by A and B = 1 : 3

∴ B takes 3 days to finish a unit work while A takes 1 day to finish the same work

The difference of time is 2 days

Given,

Difference of time is 40 days

∴ Time taken by B to complete the work = [(3/2) × 40] = 60 days

∴ A takes 20 days to finish a work

⇒ Work in 1 day by A = 1/20

⇒ Work in 1 day by B = 1/60

⇒ Work in 1 day by (A+B) = [(1/20) + (1/60)] = 4/60

∴ A and B will take days to complete the work = 15 days

QUESTION: 2

Area of the circle inscribed in a square of diagonal 4 cm is

Solution:

Let, the side of square be a

Given,

Diagonal of square (d) = 4 cm

⟹ Side of square = Diameter of inscribed circle

Using Pythagoras theorem:

⟹ d^{2} = a^{2} + a^{2}

⟹ (4)^{2} = 2a^{2}

⟹ a = 2√2cm

∴ Radius of circle = √2cm

⟹ Area of circle = π × r^{2}

∴ Area of circle = π × (√2)^{2} = 2π

QUESTION: 3

The original price of a TV set is Rs. 4,000.If the price is discounted by 10% and then raised by 5% for service contract, the price charged by the shopkeeper is

Solution:

Original price of a TV set = Rs. 4,000

∴ after 10% discount price will be = [4000 - (^{10}/_{100}) × 4000] = Rs. 3600

Given,

Increase in service contract = 5%

∴ The raised 5% service contract on new price will be = (5/100) × 3600 = Rs. 180

∴ Final selling price = 3600 + 180 = Rs. 3780

QUESTION: 4

A certain sum of money was divided between A, B and C in the ratio 1 : 3 : 5.If A received Rs. 500 the sum divided was

Solution:

Let, the distributed amount among A, B and C are x, 3x and 5x respectively

Given,

‘A’ received amount = Rs. 500

∴ x = 500

⟹ ‘B’ received amount = Rs. (500 × 3) = Rs. 1500

⟹ ‘C’ received amount = Rs. (500 × 5) = Rs. 2500

⟹ Sum of money = Rs. (500 + 1500 + 2500) = Rs. 4500

QUESTION: 5

By selling a bag at Rs. 300, profit of 10% is made. The selling price of the bag, when it is sold at 15% profit would be

Solution:

Given,

S.P_{1} = Rs.300, Profit_{1} = 10% and Profit_{2} = 15%

We know that,

Where, S.P = Selling Price, C.P = Cost Price

⟹ C.P = Rs. 272.72

⟹ Profit_{2 }= ^{Profit%}/_{100} × C.P

⟹ Profit_{2} = ^{15}/_{100} × 272.72

⟹ Profit_{2} ≈ Rs. 41

⟹ SP_{2} = C.P + Profit_{2}

⟹ SP_{2} = 272.72 + 41

∴ Selling price of the bag will be = Rs. 313.72

QUESTION: 6

The weights of two iron balls are 5 kg and 7 kg. What is the percentage weight of the 1st ball with respect to 2nd ball.

Solution:

We know that,

Percentage weight of 1st ball with respect to 2nd ball will be = ^{5}/_{7} × 100 = 71 ^{3}/_{7} %

QUESTION: 7

A Bus travels at the speed of 40 km/hr, then the distance covered by it in 5 second is

Solution:

Given,

Speed of bus = 40 km/hr = (100/9) m/s

Time = 5 s

We know that,

Distance = Speed × time

∴ Distance = (100/9) × 5 = 55.55 m

QUESTION: 8

Find the value of

Solution:

QUESTION: 9

The value of

Solution:

On solving,

QUESTION: 10

∆XYZ and ∆PQR are two similar triangles and the perimeter of ∆XYZ and ∆PQR are 160 cm and 40 cm respectively. If length of PQ = 20 cm, then length of XY is

Solution:

Let, length of XY be x

Given,

Perimeter (P_{1}) of ∆XYZ = 160 cm

Perimeter (P_{2}) of ∆PQR = 40 cm

∵ ∆XYZ and ∆PQR are similar

QUESTION: 11

If the length of a chord of a circle is equal to that of the radius of the circle, then the angle subtended, in radians, at the center of the circle by chord is

Solution:

Given,

Radius of circle = r

Chord length = radius of circle = r

Draw a perpendicular line of length x from center O

Using Pythagoras theorm,

OA^{2} = Aa^{2} + Oa^{2}

⇒ r^{2} = (^{r}/_{2})^{2} + x^{2}

⇒ x^{2} = 3r^{2}/_{4}

⇒ x = ^{√3}/_{2} r

Take any one triangle so formed

⟹ cos θ = ^{√3}/_{2}

⟹ θ = π/6

∴ The angle subtended by the chord at the center = 2 × π/_{6} = π/3

QUESTION: 12

The value of (sin^{2 }60° - cosec^{2} 45°) + (cot^{2} 30° + tan^{2 }45°)

Solution:

We know that,

sin 60° = √3/2

cosec 45° = √2

cot 30° = √3

tan 45° = 1

Putting these values in the given equation,

⟹ [(3/4) – 2] + [3 + 1] = 11/4

QUESTION: 13

The average salary of male employees in a firm was Rs. 6000 and that of females was Rs. 5600. The mean salary of all the employees was Rs. 5800. What is the % of female employees?

Solution:

Let, the number of males be ‘m’ and number of females be ‘n’

Salaries of male employees be (x_{1}, x_{2}, x_{3},…….x_{m})

Salaries of females employees be (y_{1}, y_{2}, y_{3},………y_{n})

Given,

Average salary of males = Rs.6000

⟹ m/n = 1

⟹ m = n ----(iii)

QUESTION: 14

If x +1 = √3, then the value of x^{2}+2√3

Solution:

Given,

x = √3-1

Put the value of x in given equation

⇒ x^{2} + 2√3 = (√3 - 1)^{2} + 2√3

⇒ 3 + 1 - 2√3 + 2√3 = 4

QUESTION: 15

The cube of 55 is

Solution:

We can write,

(55)^{3 }= (50 + 5)^{3}

We know that,

(a + b)^{3} = (a)^{3} + (b)^{3} + 3 × (a × b) (a + b)

⟹ (50 + 5)^{3 }= (50)^{3} + (5)^{3} + 3 × (50 × 5) (50 + 5)

⟹ (50 + 5)^{3} = (125000) + (125) + 3 × 250 × 55

∴ (50 + 5)^{3} = 166375

QUESTION: 16

In ∆PQR, ∠Q is right angle, S is the mid-point of the side PR. If PQ = 4 cm, QR = 6 cm, then the length of PS is

Solution:

Using Pythagoras's theorem,

⇒ PR^{2} = PQ^{2} + QR^{2}

⇒ PR^{2} = 4^{2} + 6^{2}

⇒ PR = √52 cm = 2√13

⇒ PS = PR/2 = √13

QUESTION: 17

The diagonals of two squares are in the ratio 5 : 7. The ratio of their area is

Solution:

Let, the diagonal of two squares be d_{1} and d_{2}. And their sides and area be a_{1} and a_{2} & A_{1} and A_{2} respectively

Given,

Using Pythagoras theorm in any one triangle

⇒ d_{1}^{2} = a_{1}^{2} + a_{1}^{2}

⇒ d_{1}^{2} = 2a_{1}^{2}

Similarly,

∴ The ratio of both squares

QUESTION: 18

The angle of elevation of a ladder leaning against a wall is 30° and the foot of the ladder is 5 m away from the wall. The length of the ladder is

Solution:

Let, the length of the ladder be ‘l’ which is QR

Use trigonometric method in given triangle PQR

⇒ cos θ = Base/Hypotenuse

⇒ cos 30° = 5/l

⇒ l = 10/√3 m

QUESTION: 19

The product of two numbers is 1800 and their HCF is 15. The numbers are

Solution:

Let, the two numbers are 15x and 15y

Given,

⇒ (15x) × (15y) = 1800

⇒ xy = 8

Value of co-primes of x and y are (1, 8)

∴ The two numbers are (1 × 15, 8 × 15) = (15, 120)

QUESTION: 20

The difference between simple and compound interests compounded annually on a certain sum of money for 2 years at 2% per annum is Rs. 3. The sum (in Rs.) is:

Solution:

Let, the sum of the money be x. CI is compound interest and SI is Simple interest

Given,

CI – SI = 3

QUESTION: 21

In a mixture of 20 litres, the ratio of milk to water is 3 : 2. Another 2 litres of water is added to the mixture. The ratio of milk to water in new mixture is

Solution:

Let, the milk be in given mixture is 3x and water be in given mixture is 2x

According to question,

⇒ 3x + 2x = 20

⇒ x = 4

∴ the milk in given mixture is 12 litres and the water be in given mixture is 8 litres

After adding 2 litres of water in the given mixture, the quantity of water will be (8 + 2) = 10 litres

∴ Ratio of milk to water in new mixture = 12/10 = 6/5 = 6 : 5

QUESTION: 22

**Direction: **The pie chart given below represents the modes of transport for 1400 officers of the Staff Selection Commission, Kolkata. Study the chart and answer the following questions.

**Question: **The number of officers who go to office by Bus is

Solution:

∵ Total number of officers = 1400

∵ Percentage of officer who go to office by Bus = 20%

∴ Number of officers who go to office by Bus = 20% of 1400 = 280

QUESTION: 23

**Direction: **The pie chart given below represents the modes of transport for 1400 officers of the Staff Selection Commission, Kolkata. Study the chart and answer the following questions.

**Question: **The number of officers who go to office by Metro Rail is

Solution:

∵ Total number of officers = 1400

∵ Percentage of officer who go to office by Metro Rail = 8%

∴ Number of officers who go to office by Metro Rail = 8% of 1400 = 112

मेट्रो रेल से कार्यालय जाने वाले अधिकारियों की संख्या क्या है?

QUESTION: 24

**Direction: **The pie chart given below represents the modes of transport for 1400 officers of the Staff Selection Commission, Kolkata. Study the chart and answer the following questions.

**Question: **Write down the difference of officers availing train and the officers availing car as mode of transport?

Solution:

∵ Total number of officers = 1400

∵ Percentage of officer using Cars = 21%

∵ Percentage of officer using trains = 36%

∴ required difference = 36% of 1400 – 21% of 1400 = 15% of 1400 = 210

QUESTION: 25

**Direction: **The pie chart given below represents the modes of transport for 1400 officers of the Staff Selection Commission, Kolkata. Study the chart and answer the following questions.

**Question: **The ratio of two - wheelers and trains being used as modes of transport is

Solution:

∵ Total number of officers = 1400

∵ Percentage of officer using two - wheelers = 15%

∵ Percentage of officer using trains = 36%

∴ Required ratio = (15% of 1400) : (36% 0f 1400) = 15 : 36 = 5 : 12

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