Quantitative Aptitude - Test 9

Three points A(x_1, y₁), B(x_2, y₂) and C(x_3, y₃) are collinear, if

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

⇒ k - 5 = 2
⇒ k = 7

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°

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

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
x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx) 
As, (x + y + z) = 0 
Thus, x³ + y³ + z³ - 3xyz = 0 
Shifting -3xyz to RHS 
x³ + y³ + z³ = 3xyz
Taking -1 common from numerator as well as denominator

= 2

Given,

⇒ x² + 1 = 99x
Now,

= 1/3

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.

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 

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

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.  

∵ 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

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%

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%.

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 × 1.125²
⇒ A = Rs. 10125
CI = A – P
⇒ CI = 10125 – 8000 = Rs. 2125

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

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

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 

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

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

Given,
⇒ a/b = c/d= e/f = 3, thus
⇒ a = 3b
⇒ c = 3d
⇒ e = 3f
Putting these values in the given expression 

Taking out 3² common we get the given expression

= 9
Hence the value of given expression is 9.

Given, (x + 3264 × 3266)
= (x + (3265 – 1) × (3265 + 1))
= (x + 3265² – 1)
For this to be a perfect square, x = 1

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.

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%

 
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°

We know that
2θ = cot^-1(1/√3) = π/3 for 2θ ≤ 90°
⇒ θ = π/6
⇒ 3θ = π/2
Now, sin 3θ = 1

(a – b)² = a² + b² – 2ab
Given that, secθ + tanθ = 3
⇒ secθ = 3 – tanθ
Squaring both side,
⇒ sec²θ = 9 – 6×tan θ + tan²θ
We know that, sec²θ = 1 + tan²θ
∴ 1 + tan²θ = 9 – 6 × tanθ + tan²θ
⇒ 6 × tanθ = 8

Now, we have to calculate 
Putting the value of tan θ