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This mock test of Quantitative Aptitude - Test 9 for SSC helps you for every SSC entrance exam.
This contains 25 Multiple Choice Questions for SSC Quantitative Aptitude - Test 9 (mcq) to study with solutions a complete question bank.
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QUESTION: 1

Find the value of K for which the points A (4, 3), B(5, 4) and C(K, 6) are collinear.

Solution:

Three points A(x_{1, }y_{1}), B(x_{2, }y_{2}) and C(x_{3, }y_{3}) are collinear, if

As, A (4, 3), B (5, 4) and C (K, 6) are collinear

⇒ k - 5 = 2

⇒ k = 7

QUESTION: 2

Two circles touch each other externally at P. QR is a common tangent to the circles touching them at Q and R. The measure of ∠QPR is

Solution:

In triangle OPQ,

OP = OQ (radius of circle)

∴ ∠OQP = ∠OPQ

Let, ∠OQP = ∠OPQ = x

In triangle PCR,

CP = CR (radius of circle)

∴ ∠CPR = ∠CRP

Let, ∠CPR = ∠CRP = y

In given figure,

∠OPQ + ∠QPR + ∠RPC = 180°

⇒ x + ∠QPR + y = 180°

⇒ ∠QPR = 180° - x – y ……….eq (1)

As, we know that, radius and tangent to the circle are perpendicular to each other,

Therefore,

∠OQR and ∠CRQ are right angles

∴ ∠PQO + ∠PQR = 90°

⇒ ∠PQR = 90° - x° ………….eq (2)

∴ ∠ CRP + ∠ PRQ = 90°

⇒ ∠PRQ = 90° - y° ………..eq(3)

In triangle QPR

∠QPR + ∠PQR + ∠PRQ = 180°

⇒ 180° - x – y + 90° - x° +90° - y° = 180°

⇒ 360° - 2(x + y) = 180°

⇒ 2(x + y) = 180°

⇒ (x + y) = 90°

∠QPR = 180° - x – y

= 180° - 90°

= 90°

QUESTION: 3

The difference between interior and exterior angle of a regular polygon is 100°. Find the number of sides of polygon.

Solution:

Let out of exterior and interior angle one of the angle be x

∴ Other angle = x + 100.

Now as we know that interior angle + exterior angle = 180°

∴ x + x + 100 = 180°

⇒ 2x + 100 = 180°

⇒ 2x = 80°

⇒ x = 40°

∴ other angle = 140°

Now, since exterior angle of a regular polygon can not exceed 120°

∴ Interior angle = 140° and, Exterior angle = 40°

Now no. of sides of a regular polygon = 360 / exterior angle

∴ No. of sides = 360 / 40 = 9

QUESTION: 4

If x + y + z = 0, the value of

Solution:

We have x + y + z = 0 …(1)

We have to find the value of

Substituting values of x + y, y + z, z + x from (1)

We also know that

x^{3} + y^{3} + z^{3} - 3xyz = (x + y + z)(x^{2} + y^{2} + z^{2} - xy - yz - zx)

As, (x + y + z) = 0

Thus, x^{3} + y^{3} + z^{3} - 3xyz = 0

Shifting -3xyz to RHS

x^{3} + y^{3} + z^{3} = 3xyz

Taking -1 common from numerator as well as denominator

= 2

QUESTION: 5

If find the value of

Solution:

Given,

⇒ x^{2} + 1 = 99x

Now,

= 1/3

QUESTION: 6

A bicycle wheel makes 5000 revolutions in moving 11 km. The diameter of the wheel, in cm, is

Solution:

Let the diameter of wheel be D cm.

We know that,

⇒ Distance travelled in 1 revolution × Total no. of revolutions = Total distance travelled

Also, Distance travelled in 1 revolution = Circumference of the wheel

⇒ Circumference of wheel = π × D, where D is the diameter of the wheel

Given, Total number of revolutions = 5000

∴ π × D × 5000 = Total distance travelled

⇒ π × D × 5000 = 1100000 cm (∵ 1 km = 100000 cm)

⇒ D = 70 cm

Hence, the diameter of the wheel is 70 cm.

QUESTION: 7

If 40 men or 60 women or 80 children can do a piece of work in 6 months, then 10 men, 10 women and 10 children together do half of the work in

Solution:

Given, 40 men or 60 women of 80 children can do a piece of work in 6 months.

⇒ 40M = 60W =80C

⇒ 2M = 3W = 4C

According to question,

So, half of work will be done in number of days = x/2 = 72/13 days

⇒ Half of the work is completed in number of days

QUESTION: 8

A boat goes 10kms an hour in still water, but takes twice as much time in going the same distance against the current. The speed of the current (in km/hr) is –

Solution:

Speed of Boat in Current = Speed of Boat in still water – Speed of Current

A boat goes 10 kms an hour in still water

∴ Speed of Boat in still water = 10km/hr

Let it travels for 1 hr and covered 10 km

It takes twice as much time in going the same distance against the current

Thus, it will take 2 hr to cover the same distance of 10 km

∴ Speed of Boat against Current = 10km / 2hrs = 5km/hr

Speed of Current = 10km/hr – 5km/hr = 5km/hr

QUESTION: 9

A shopkeeper sells a pair of sunglasses at a profit of 25%. If he had bought it at 25% less and sold it for Rs. 20 less, then he would have gained 40%. The cost price of the pair of sunglasses is:

Solution:

Let the sunglasses were brought for Rs 100.

Selling price be ‘x’ .such that gain% = 25%

Selling price = Rs 100+ Rs 25 = Rs 125

Now if he had bought them at 25% less would mean

⇒ 100 – 25% of 100 = 100 – 25 = Rs 75

Hence let the cost price be Rs 75 and new SP be ‘y’ then

Gain% = 40%

⇒ (SP – CP)/CP × 100 = 40

⇒ y – 75 = 30

⇒ y = 75 + 30 = 105 Rs

Now if he sells for Previous SP – new SP = Rs. 125 – Rs 105 = Rs 20 less

Then CP = Rs 100

If he sells for Rs 20 less, then C.P.

QUESTION: 10

The average age of 8 persons in a committee is increased by 2 years when two men whose ages are 35 years and 45 years are replaced by two new men. The average age of the two new men is (in years)

Solution:

∵ Average = (Sum total of Ages of All the persons) ÷ (Total no. of persons)

Let the sum of age of 6 persons be x, age of 7th person = 35 & age of 8th Person = 45

⇒ Sum of ages of 8 person = x + 35 + 45 = x + 80

⇒ Average = A = (x + 80)/8

⇒ 8A = x + 80 .... (i)

Let the 7th & 8th person with ages 35 & 45 be replaced with persons having ages as "a" & "b" years respectively

∴ Avg of age of the 2 new men = (a + b)/2

According to question

⇒ x + a + b = 8A +16... (ii)

∴ From Eq. (i) &Eq (ii) we get

⇒ x + a + b = x + 80 + 16

⇒ a + b = 96

Dividing both sides by 2

⇒ (a + b)/2 = 48

QUESTION: 11

Two numbers are less than the third number by 20% and 28% respectively. By what per cent is the second number less than the first number?

Solution:

Let the third number be 100

Then the frist number = 100 – 20 = 80

Second number = 100 – 28 = 72

% with which second number is less than first = (first-second)/first × 100%

⇒ (80 - 72)/80 × 100 %

⇒ 800/80% = 10%

QUESTION: 12

In an examination, a student gets 20% of total marks and fails by 30 marks. Another student gets 32% of total marks which is more than the minimum pass marks by 42 marks. The pass percentage is

Solution:

Let the total marks be X

Then, marks obtained by student 1 = 0.2X

Since, student 1 fails by 30 marks after getting 20% marks

∴ According to student1, Passing marks = 0.2X + 30

Similarly, Marks obtained by student 2 = 0.32X

And according to student 2, passing marks = 0.32X - 42;

Equating eq. 1 and 2, we get

0.2X + 30 = 0.32X - 42

⇒ X = 600

∴ Passing marks = (0.2 × 600) + 30 = 150

Passing marks %

Hence, the passing marks are 25%.

QUESTION: 13

There is 100% increase to an amount in 8 years, at simple interest. Find the compound interest of Rs. 8000 after 2 years at the same rate of interest.

Solution:

Formulas to be used: -

SI = ( P × r × t ) / 100

For CI:

Where SI is Simple interest,

A is the amount at the end of time t,

P is the principal,

t is time,

r is rate

For SI, there is 100% increase to amount, thus A = 2P

⇒ SI = p

Time is 8 years.

∴ p = (p × r × t)/100

⇒ r = 100/8 = 12.5%

Now, P = 8000, t = 2years and r = 12.5%

⇒ A = 8000 × 1.125^{2}

⇒ A = Rs. 10125

CI = A – P

⇒ CI = 10125 – 8000 = Rs. 2125

QUESTION: 14

Three students A, B and C play cricket. The runs scored by A and B respectively are in the ratio 3 : 2. B’s runs to C’s runs are also in the same ratio. Together they score 342 runs. Then the runs scored by B are

Solution:

Given, runs scored by A and B respectively are in the ratio 3 : 2.

∴ A’s score

Also B’s runs to C’s runs are also in the same ratio.

∴ B’s score : C’s score = 3 : 2

Together they scored 342 runs.

∴ A’s score + B’s score + C’s score = 342

⇒ B’s score = 108

QUESTION: 15

**Direction:** The following graph shows the production of wheat flour (in 1000 tonnes) by three companies X, Y and Z over the years. Study the graph and answer the questions.

Q. The average production for five years was maximum for which company(s)?

Solution:

From the above chart, we can observe that

Total production of company X during 2000 – 2004 = (300 + 450 +250 + 500 + 400) × 1000 tonnes

= 1900 × 1000 tonnes

Average production of company X during 2000 – 2004 =

= 380 × 1000 tonnes

Total production of company Y during 2000 – 2004 = (250 + 350 + 350 + 400 + 500) × 1000 tonnes

= 1850 × 1000 tonnes

Average production of company Y during 2000 – 2004

= 370 × 1000 tonnes

Total production of company Z during 2000 – 2004 = (350 + 400 + 450 + 350 + 350) × 1000 tonnes

= 1900 × 1000 tonnes

Average production of company Y during 2000 – 2004

= 380 × 1000 tonnes

Hence we can observe that the average production for five years was maximum for X and Z

QUESTION: 16

**Direction:** The following graph shows the production of wheat flour (in 1000 tonnes) by three companies X, Y and Z over the years. Study the graph and answer the questions.

Q. What is the percentage increase in the production of company Y from 2002 to 2003?

Solution:

From the above chart, we can observe thats

Production of company Y in 2002 = 350 × 1000 tonnes

Production of company Y in 2003 = 400 × 1000 tonnes

Percentage increase

QUESTION: 17

**Direction:** The following graph shows the production of wheat flour (in 1000 tonnes) by three companies X, Y and Z over the years. Study the graph and answer the questions.

Q. What is the ratio of the average production of company X in the period 2002-2004 to the average production of company Y in the same period?

Solution:

From the above chart, we can observe that

Total production of company X during 2002 – 2004 = (250 + 500 + 400) × 1000 tonnes

= 1150 × 1000 tonnes

Average production of company X during 2002 – 2004

Total production of company Y during 2002 – 2004 = (350 + 400 + 500) × 1000 tonnes

= 1250 × 1000 tonnes

Average production of company Y during 2002 – 2004

Required ratio

= 23 : 25

QUESTION: 18

**Direction:** The following graph shows the production of wheat flour (in 1000 tonnes) by three companies X, Y and Z over the years. Study the graph and answer the questions.

Q. What is the difference between the production of company Z in 2004 and company Y in 2000 (in thousand tonnes)?

Solution:

From the above chart, we can observe that

Production of company Z in 2004 = 350 × 1000 tonnes

Production of company Y in 2000 = 250 × 1000 tonnes

Difference = (350 – 250) × 1000 tonnes

= 100 × 1000 tonnes

QUESTION: 19

If a/b = c/d= e/f = 3, then

Solution:

Given,

⇒ a/b = c/d= e/f = 3, thus

⇒ a = 3b

⇒ c = 3d

⇒ e = 3f

Putting these values in the given expression

Taking out 3^{2} common we get the given expression

= 9

Hence the value of given expression is 9.

QUESTION: 20

For what value of x is (x + 3264 × 3266) a perfect square,

Solution:

Given, (x + 3264 × 3266)

= (x + (3265 – 1) × (3265 + 1))

= (x + 3265^{2} – 1)

For this to be a perfect square, x = 1

QUESTION: 21

Which one of the following three digit numbers divides 9238 and 7091 with the same remainder in each case?

Solution:

Let the same remainder on dividing these numbers be R.

Hence, (9238 – R) & (7091 - R) both will be divisible by the required three digit number say X.

We know that if X is a factor of (9238 – R) & (7091 - R) then it will also be factor of the difference of (9238 – R) & (7091 - R) i.e.

⇒ X must also be a factor of (9238 – R – (7091 - R)) = 2147

Thus, from given 3 digit numbers only 113 completely divides 2147 and hence its factor.

Hence 113 is the required number that divides 9238 and 7091 to leave same remainder.

QUESTION: 22

The list price of a digital electronic watch is Rs. 1000. A customer gets two successive discounts on the list price, the 1st being of 15%. Calculate the 2nd discount, if the customer pays Rs. 697 for it.

Solution:

Given, Marked price = Rs. 1000

After allowing 15%,

Selling price = 1000 – 0.15 × 1000 = 0.85 × 1000 = 850

Let, A successive discount of x% is allowed,

Selling price becomes = 850 – (x/100) × 850

According to the question,

850 – (x/100) × 850 = 697

⇒ 850 (1 – x/100) = 697

⇒ 1 – x/100 = 41/50

⇒ 1 – 41/50 = x/100

⇒ 9/50 = x/100

⇒ x = 900/50 = 18%

QUESTION: 23

At an instant, the length of the shadow of a pole is √3 times the height of the pole. The angle of elevation of the Sun at that moment is

Solution:

From the diagram,

Height of the pole = AB = H

Length of the shadow = BC = L

Given, Length of the shadow is √3 times the height of the pole

∴ BC = L = H√3

From ΔABC

tan ∠ACB = Height/Base

⇒ tan ∠ACB = AB/BC

⇒ tan ∠ACB = H/H√3

⇒ tan ∠ACB = 1/√3

∵ tan 30° = 1/√3

∴ ∠ACB = 30°

∴ Angle of elevation of sun is 30°

QUESTION: 24

If for some angle θ, then the value of sin 3θ, where 2θ ≤ 90° is

Solution:

We know that

2θ = cot^{-1}(1/√3) = π/3 for 2θ ≤ 90°

⇒ θ = π/6

⇒ 3θ = π/2

Now, sin 3θ = 1

QUESTION: 25

If secθ + tanθ = 3, then the value of

Solution:

(a – b)^{2} = a^{2} + b^{2} – 2ab

Given that, secθ + tanθ = 3

⇒ secθ = 3 – tanθ

Squaring both side,

⇒ sec^{2}θ = 9 – 6×tan θ + tan^{2}θ

We know that, sec^{2}θ = 1 + tan^{2}θ

∴ 1 + tan^{2}θ = 9 – 6 × tanθ + tan^{2}θ

⇒ 6 × tanθ = 8

Now, we have to calculate

Putting the value of tan θ

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