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This mock test of Quantitative Aptitude - Test 6 for SSC helps you for every SSC entrance exam.
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QUESTION: 1

**Directions:** The pie chart given below represents the modes of transport for 1400 employees of the staff Selection Commission, Kolkata. Study the chart and answer the questions.

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

Solution:

From the pie chart 8% of total employees go to office by metro rail.

Total employees = 1400

∴ Number of employees going to office by Metro rail = 8% of 1400 = 112

∴ 112 employees go to office by metro rail.

QUESTION: 2

**Directions:** The pie chart given below represents the modes of transport for 1400 employees of the staff Selection Commission, Kolkata. Study the chart and answer the questions.

**Question: **The number of employees who go to office by car is

Solution:

From the pie chart 21% of total employees go to office by car.

Total employees = 1400

Number of employees going to office by car = 21% of 1400 = 294

∴ 294 employees go to office by car.

QUESTION: 3

**Directions:** The pie chart given below represents the modes of transport for 1400 employees of the staff Selection Commission, Kolkata. Study the chart and answer the questions.

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

Solution:

From above question 294 employees go to office by car.

From the pie chart 15% of total employees go to office by two wheelers.

Total employees = 1400

Number of employees going to office by two wheelers = 15% of 1400 = 210

∴ 210 employees go to office by two wheelers.

∴ The ratio of two wheelers and cars being used as modes of transport is

QUESTION: 4

**Directions:** The pie chart given below represents the modes of transport for 1400 employees of the staff Selection Commission, Kolkata. Study the chart and answer the questions.

**Question: **Write down the difference: (Employees availing train - employees availing car)

Solution:

From above question 294 employees go to office by car.

From the pie chart 36% of total employees go to office by train.

Total employees = 1400

Number of employees going to office by train = 36% of 1400 = 504

∴ 504 employees go to office by train.

∴ Employees availing train - employees availing car = (504 – 294) = 210

QUESTION: 5

O is the center of circle and ∠QPS = 65°. Find the value of ∠A and ∠B respectively.

Solution:

We know that angle subtended by any two points from the circumference of the circle on the center is twice the angle subtended by those two points on any other part of the circle, so,

Points Q and S subtend an angle of 65° on the circumference so they subtend 65° × 2 = 130°.

Now, ∠ SOQ(outer) = 360° - 130° = 230°

Now similar to the previous part, ∠ SOQ(outer) is twice of angle B, so ∠ B = 230°/2 = 115°

QUESTION: 6

Find the third vertex of a triangle, if two of its vertices are at (2, 3) and (4, -2) and the centroid is at (1, 1).

Solution:

Centroid of a triangle = [(x_{1} + x_{2} + x_{3})/3 , (y_{1} + y_{2} + y_{3})/3] where (x_{1 }, y_{1}), (x_{2}, y_{2}) and (x_{3}, y_{3}) are co-ordinates of vertices of triangle

Given, two of its vertices are at (2, 3) and (4, -2) and the centroid is at (1, 1)

Let the coordinate of third vertices be (x, y)

∴ 1 = (2 + 4 + x)/3

⇒ x = - 3

And,

1 = (3 - 2 + y)/3

⇒ y = 2

So third coordinate is (-3, 2)

QUESTION: 7

The side AC of the ΔABC is produced to D, such that CD = CB. If ∠ACB = 70°, then the value of ∠ADB is

Solution:

Now we see that Δ ADB is isosceles as CD = CB and the sum of ∠ CBD and ∠ CDB is 70° (exterior angle property of a triangle). So,

∠CBD = ∠CDB (isosceles triangle)

∠CBD + ∠CDB = 70°

⇒ 2 × ∠CDB = 70°

⇒ ∠ CDB = 35°

QUESTION: 8

If x + (1/x) = 6, then find value of

Solution:

We have,

x + (1/x) = 6

on squaring both side

⇒x^{2 }+ 1/x^{2} = 36 – 2 = 34

QUESTION: 9

The angles of depression of the top and bottom of a 50 m high electric pole form the top of a building are 30° and 60° respectively. Find the height of the building.

Solution:

According to the figure,

DE is the electric pole (height = 50m), AB is the building.

∠ADC = 30° & ∠AEB = 60°

We have to find out the length of AB.

As, DC ⊥ AB & EB ⊥ AB so, DC || EB

⇒ DE = BC

Also, DE || AB so, DC = EB

Now, AD × cos30° = DC & AE × cos 60° = BE

⇒ AD × cos30° = AE × cos 60°

⇒ √3 ×AD = AE

From the figure, AB = AC + BC

⇒ AB = AC + DE [As, DE = BC]

⇒ AE × sin 60° = AD × sin 30° + 50

⇒ √3 ×AD × sin 60° = AD × sin 30° + 50

⇒ AD = 50

Now, AB = AD × sin 30° + 50

⇒ AB = 25 + 50 = 75 meter.

The height of the building = 75 meter

QUESTION: 10

If is equal to

Solution:

Formula:

(a + b)^{2} = a^{2} + b^{2} + 2ab

Squaring both sides we get,

QUESTION: 11

The simplest value of cot9° cot27° cot63° cot 81° is

Solution:

We know that cotθ = tan(90° - θ)

Cot63° = tan(90° - 63°)

Cot63° = tan27°

And cot81° = tan9°

cot9° cot27° cot63° cot 81°

= cot9° cot27° tan27° tan9°

We know that tanθcotθ = 1

= 1

QUESTION: 12

If then a^{2} – b^{2 }is equal to

Solution:

Formulas :-

Given the equation

Adding both the equation:

Since we know a^{3 }+ b^{3} + 3ab (a + b) = (a + b)^{3}

⇒ (x + 1/x)^{3} = (2a)^{3}

⇒ x + 1/x = 2a

Similarly, subtracting both the equation:

Since we know a^{3 }- b^{3} - 3ab (a - b) = (a - b)^{3}

⇒ (x - 1/x)^{3} = (-2b)^{3}

⇒ x - 1/x = -2b

Adding both the resultant equation,

⇒ x + 1/x + x – 1/x = 2a – 2b

⇒ 2x = 2a – 2b

⇒ x = a – b

Similarly, subtracting both the resultant equation,

⇒ x + 1/x - x + 1/x = 2a + 2b

⇒ 2/x = 2a + 2b

⇒ 1/x = a + b

Multiplying both the final equation:

⇒ (a + b)(a - b) = x × 1/x = 1

⇒ a^{2} – b^{2} = 1

QUESTION: 13

ABC is an equilateral triangle with side 2 cm. With A, B and C as centres and radius 1 cm three arcs are drawn. The area of the region with in the triangle bounded by three arcs is

Solution:

ABC is an equilateral triangle

Given,

AB = BC = CA = 2 cm

∠A = ∠B = ∠C = 60°

Height of equilateral triangle = (√3/2) × (side)

= (√3/2) × 2 = √3 cm

∴ Area of the triangle = (1/2) × √3 × 2 = √3 cm^{2}

Area inside the triangle inscribed by three arcs = 3 × area of one arc

= 3 × (θ/360°) × πr^{2}

= 3 × π × 1^{2} × (60/360) = π/2

∴ Required area = (√3 - π/2) cm^{2}

QUESTION: 14

ABCD is a parallelogram in which diagonals AC and BD intersect at O. If E, F, G and H are the mid points of AO, DO, CO and BO respectively, then the ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD is

Solution:

In Δ OAB,

Mid- point of OA = E

Mid- point of OB = H

Therefore,

EH|| AB

Therefore,

HE = AB/2

Similarly,

HG = BC/2

FG = CD/2

EF = AD/2

EF + HG + FG + EF = (1/2)(AB + BC + CD + AD)

⇒ Perimeter of EFGH = (1/2) perimeter of ABCD

So,

The ratio of the perimeter of the quadrilateral EFGH to the perimeter of parallelogram ABCD = 1:2

QUESTION: 15

If the roots of the equation ax^{2} + bx + c = 0 differ by 7, then which of the following expression shows the relation among a, b and c?

Solution:

Roots of equation ax^{2} + bx + c = 0 are

Difference between the roots =

Given, roots differ by 7

QUESTION: 16

A can do a piece of work in 18 days, B in 36 days and C in 72 days. All begin to do it together but A leaves after 8 days and B leaves 7 days before the completion of the work. The total number of days they worked for is

Solution:

Given,

A can do a piece of work in 18 days

∴ Part of work done by A in 1 day = 1/18

B can do a piece of work in 36days

∴ Part of work done by B in 1 day = 1/36

C can do a piece of work in 72 days

∴ Part of work done by C in 1 day = 1/72

Let,

Total number of days they work = x days

Number of days A work = 8 days

∵ B leaves7 days before the completion of the work

Number of days B work = (x – 7) days

And, Number of days C work= x days

∴ {(1/18) × 8 } + {(1/36) × (x – 7)} + {(1/72) × x} = 1

⇒ 4/9 + (x – 7)/36 + x/72 = 1

⇒ (32 + 2x -14 + x)/72 = 1

⇒ 18 + 3x = 72

⇒ 3x = 54

⇒ x = 18 days

QUESTION: 17

Form two places, 60 km apart, A and B start towards each other and meet after 6 hours. Had A travelled with 2/3 of his speed and B travelled with double of his speed, they would have met after 5 hours. The speed of A is:

Solution:

Let speed of A be a kmph and speed of B be b kmph

Then we have

⇒ a + b = 60/6

⇒ a + b = 10

or 2a + 2b = 20------[1]

also,

⇒ 2/3a + 2b = 60/5 = 12--------------[2]

[1] – [2] leads to

⇒ (4/3)a = 8

a = 6 kmph

QUESTION: 18

A container contains mixture of milk and water in the ratio of 4 : 7. 8 litres of this mixture is replaced by water. The new ratio of milk and water becomes 4 : 9. Find the initial quantity of water in the mixture.

Solution:

Given, container contains mixture of milk and water in the ratio of 4 : 7.

Let the volume of milk and water be 4a and 7a.

Now, 8 litres are removed and replaced by water.

In 1 litre of mixture has 4/11 litres of milk and 7/11 litres of water are present.

Thus in 8 litres 32/11 and 56/11 litres of milk and water are present.

∴ Volume of milk = 4a – 32/11

Volume of water = 7a – 56/11 + 8

New ratio = 4 : 9

⇒ a = 52/11

Now, initial amount of water = 7a = 364/11

QUESTION: 19

Ram and Shyam start a joint business. Ram invests 1/3 of the total sum for 9 month and Shyam gets 2/5 of the profits. For how long, the investment of Shyam was in the business?

Solution:

We know that investment × time = profit

Assume that the total sum = P & total profit is M

∴ Sum invested by Ram = ⅓P & sum invested by Shyam = ⅔P

For Ram we can say,

For Shyam we can say,

[Here T = time for which investment of Shyam was in the business]

∴ Investment of Shyam was in the business for 3 months.

QUESTION: 20

The mean of 20 items is 55. If two items 45 and 30 are removed, the new mean of the remaining items is

Solution:

Formula for mean:

Where, X = Mean, ∑X = sum of all items, N = number of items

Given, mean of 20 items is 55

⇒ two items are removed, so number of items = 18

⇒ removed items = 45 + 30 = 75

⇒ ∑X = 1100 – 75 = 1025

QUESTION: 21

The percentage of metals in a mine of lead ore is 60%. Now the percentage of silver is 3/4 % of metals and the rest is lead. If the mass of ore extracted from this mine is 8000 kg, then the mass (in kg) of lead is:

Solution:

Mass of ore extracted from mine of lead ore = 8000 kg.

Percentage of metals in a mine of lead ore = 60%

Mass of metal = 60% of 8000

= (60/100) × 8000

= 4800kg.

Percentage of silver is 3/4 % of metals and the rest is lead.

Mass of silver = 3/4 % of 4800

= (3/400) × 4800

= 36kg.

Mass of lead = (4800 – 36)kg

= 4764 kg.

QUESTION: 22

A sum becomes Rs, 10000 after 3 years and Rs, 20000 after 6 years on compound interest. The sum is

Solution:

For CI:

Where,

A is the amount at the end of time t,

P is the principal,

t is time,

r is rate

A sum becomes Rs, 10000 after 3 years and Rs, 20000 after 6 years on compound interest.

For t = 3yrs

Diving eq 3 by eq 2 we get,

QUESTION: 23

The ratio of monthly incomes of A and B is 7: 8 and their monthly expenditures in ratio of 3: 4. If each of them saves Rs. 400 per month. Find the sum of their monthly income.

Solution:

Given, ratio of monthly incomes of A and B is 7: 8.

Let the monthly incomes of A and B be 7a and 8a respectively, where a is any constant

Ratio of monthly expenditures is 3: 4

Let the monthly expenditures be 3b and 4b respectively, where b is any constant

Given, each of them saves Rs. 400

∴ 7a – 3b = 8a – 4b = 400

⇒ a = b

∴ 7a – 3a = 400

⇒ a = 100

Sum of their income = 7a + 8a = 15a

⇒ Sum of their income = Rs. 1500

QUESTION: 24

The value of (1001)^{3} is

Solution:

The value of (1001)^{3 }= 1001 × 1001 × 1001

= 1003003001

Or,

1001 = 1000 + 1

Cubing

(1001)^{3} = (1000 + 1)^{3}

We know that, (a + b)^{3} = a^{3} + b^{3 }+ 3ab (a + b)

⇒ (1001)^{3} = 1000^{3} + 1 + 3000 × (1000 + 1)

⇒ (1001)^{3 }= 1000000000 + 1 + 3003000

⇒ (1001)^{3 }= 1003003001

QUESTION: 25

For any integral of n, 3^{2n }+ 9n + 5 when divided by 3 will leave the remainder

Solution:

3^{2n }+ 9n + 5 = 3 × 3^{2n-1 }+ 9n + 3 + 2

= 3(3^{2n-1 }+ 3n + 1) + 2

As, 3(3^{2n-1 }+ 3n + 1) is completely divisible by 3.

Therefore,

Remainder is 2

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