Quantitative Aptitude PYQs

Given:

Length of the platform is 300m = 0.3 km

Speed of the train is x km/hr

Speed of a man is 6 km/hr 

Formula Used:

Time taken = Total distance/Total time

Calculation:

Let the length of train be y

Time taken by train to cross a 300m long platform = (y + 0.3)/x

⇒ (y + 0.3)/x = 30/3600

⇒ (y + 0.3)/x = 1/120

⇒ y +  0.3 = x/120

⇒ y = x/120 - 0.3

Time taken by train to cross a man walking in same direction = y/(x - 6) 

⇒ y/(x - 6) = 20/3600

On putting the value of y in the equation 

⇒ (x/120 - 0.3)/(x - 6) = 1/180

⇒ x/120 - 0.3 = (x- 6)/180

⇒ x/120 - 0.3 = x/180 - 1/30

⇒ x/120 - x/180 = 0.3 - 1/30

⇒ x/360 = 8/30

⇒ x= 96

∴ The value of x is 96 km/hr

Shortcut Trick

Length of the platform is 300m = 0.3 km

Speed of a man is 6 km/hr 

Let the length of train be y m

Speed of the train is x km/hr

According to the question,

x × 5/18 = (y + 300)/30

x × 5/18 = (y/30 + 10)      ----(1)

and

(x - 6) × 5/18 = y/20       ----(2)

on solving both we get 

⇒ x= 96

∴ The value of x is 96 km/hr

Total sales of (B1 + B2 + B3) in 2018 = 250

Total sales of (B6 + B4 + B5) in 2019 = 270

⇒ 250 : 270

⇒ 25 : 27

∴ Ratio is 25 : 27

Given:

Remainders corresponding to the divisor

Formula used:

(Divisor × Quotient) + Remainder = Dividend

Calculation:

Start from last digits

The number gives a remainder of 5 when it is divided by 7

⇒ (7x + 5)

The number gives remainder 3 when it is divided 4

⇒ [4 × (7x + 5) + 3]

Also, the number when divided by 3 gives remainder 2

⇒ [3 × {4 × (7x + 5) + 3} + 2]

⇒ [3 × (28x + 20 + 3) + 2]

⇒ 84x + 71

The number is 71 more than a multiple of 84

Hence, if the number is divided by 84

∴ The remainder will be 71

 

Let x be the number

⇒ z = 7 × 1 + 5 = 12

⇒ y = 4 × z + 3 = 51

⇒ x = 3 × y + 2 = 155

On dividing 155 by 84

∴ Remainder = 71

Important Points

If the quotient of the dividend is taken and this is used as the dividend in the next division, such as division is called Successive division. A successive division process can continue up to any number of steps until the quotient in a division became zero for the first time, i.e., the quotient in the first division is taken and divided in the second division; the quotient in the second division is taken as the dividend in the third division; the quotient in the third division is taken as the dividend in the fourth division and so on.

 

We will have to follow the rule of Successive Division (Not Normal Division). Hence, LCM Method is not followed there.

Given:

a³ + b³ = 217

(a + b) = 7

Formula Used:

(a + b)³ = a³ + b³ + 3ab (a + b)

Calculation:

Substituting the values in the formula

(7)³ = 217 + 3ab (7)

343 – 217 = 21ab

126/21 = ab

∴The value of ab is = 6

Given:

Profit ratio of three partners = 8 : 7 : 5

Period of investment of 1^st partner = 7 months

Period of investment of 2^nd partner = 8 months

Period of investment of 3^rd partner = 14 months

Concept used:

Person’s Profit = Investment × time period

Calculation:

Let the investment and time period be 7x : 8y : 14z

7x/8y = 8/7

⇒ 49x = 64y

⇒ y = 49x/64

7x/14z = 8/5

⇒ 35x = 112z

⇒ z = 35x/112

⇒ z = 5x/16

x : y : z = x : 49x/64 : 5x/16

⇒ Multiply by 64

x : y : z = 64 : 49 : 20

∴ The investment ratio is 64 : 49 : 20

Shortcut Trick We know that, 

⇒ Profit = Invested Capital × Invested Time 

Let the invested capital of all 3 be a, b, and c respectively, 

⇒ 

Now we get,

⇒ a × 7 = 8

⇒ a = 8/7

similarly, we get, b = 7/8 & c = 5/14

Now we have a = 8/7, b = 7/8,  and c = 5/14

Multiplying by 56, we get, 

⇒ 8/7(56), 7/8(56), and 5/14(56)

⇒ 64, 49, and 20

∴ The investment ratio is 64 : 49 : 20.

We have to follow the BODMAS rule.

Calculation: 

    = \([\frac{2}{3}; \div ;\frac{3}{{10}};of;\frac{4}{9}; - ;\frac{4}{5}; \times ;1\frac{1}{9}; \div ;\frac{8}{{15}};; - ;\frac{3}{4}; + \frac{3}{4}; \div \frac{1}{2}]\)

    ⇒\([\frac{2}{3}; \div ;\frac{3}{{10}}; \times ;\frac{4}{9}; - ;\frac{4}{5}; \times ;\frac{{10}}{9}; \times ;\frac{{15}}{8};; - ;\frac{3}{4}; + \frac{3}{4}; \times 2]\) 

    ⇒\([\frac{2}{3}; \div ;\frac{2}{{15}} - ;\frac{5}{3};; - ;\frac{3}{4}; + \frac{3}{2};]\)

    ⇒\([\frac{2}{3}; \times ;\frac{{15}}{2} - ;\frac{5}{3};; + ;\frac{3}{4};;]\)

    ⇒\([5; - ;\frac{{11}}{{12}};]\) = \([\frac{{49}}{{12}}]\)

∴The value of \(\frac{2}{3} \div \frac{3}{{10}}of\frac{4}{9} - \frac{4}{5} \times 1\frac{1}{9} \div \frac{8}{{15}} - \frac{3}{4} + \frac{3}{4} \div \frac{1}{2}\) is \([\frac{{49}}{{12}}]\)

Given:

Base of triangle is equal to the perimeter of square whose diagonal if 6√2

Formula used:

diagonal of square = a√2

Area of triangle = 1/2 base × height 

Perimeter of square = 4 × side 

Calculation: 

a√2 = 6√2 

a = 6 side of square 

4 × 6 = 24

24 = base of triangle according to question

now height is equal to side of square whose area is 144cm² 

b² = 144

b = 12

⇒ height of the triangle is equal to side of square

⇒ Area of triangle =1/2× 12× 24

So, Area of triangle = 144 cm²

Calculation:

Number of student obtained less than 350 = (30 + 45 + 60 + 35) = 170

Number of student obtained 400 or more = 35

Difference between number of students obtained less than 350 and number of students obtained 400 or more = (170 - 35) = 135

Percentage = 135/35 × 100 = 385.714% ≈ 385.7%

∴ The required percentage is 385.7%

Given :

A can complete a work in = 35 days

B can complete a work in = 15 days

Formula used: Work = Efficiency× Time

Calculation :

Let :

Total work = LCM of 35,15 = 105

Efficiency of A = 105/35 = 3 unit 

Efficiency of B = 105/15 = 7 unit 

According to Question they work together for 8 days, B left then A will complete the work 

⇒ (A+B)8 +AX = 105

⇒ 10 × 8 + 3X= 105×

⇒ 80 +3X = 105

⇒ 3X = 25× 3/5      ( In Question it is mentioned that A alone complete 60% of remaining work )

⇒ X = 5 days

So, A alone complete 60% of the remaining work in 5 days.

Alternate Method

 

 

 

Efficiency of  work of A and B in one days = 3 + 7

⇒ 10 

Total work done by A and B both in 8 days = 10 × 8

⇒ 80 

According to Question they work together for 8 days and B left then remaining work = 105 - 80 = 25

60% of remaining work = 25 × 60% = 15

A alone complete 60% of remaining work = 15/3 = 5

So, A alone complete 60% of the remaining work in 5 days.

Given:

AB = 6cm

BC = 10cm

AC = 8cm

AD is the median bisects BC

Concept Used:

Pythagoras theorem

AB2 + AC2 = BC2

Circumcentre theorem

Circumcentre of a right-angled triangle lies on the mid-point of the hypotenuse

AD = BD = DC

Calculation:

AB² + AC² = BC² 

6² + 8² = 10² 

Since it obeys the Pythagoras theorem

It is a right-angled triangle

Circumcentre of a right-angled triangle lies on the mid-point of the hypotenuse

AD = BD = DC

AD is a median which bisects BC into two equal parts

BD = DC = 5cm

AD = BD = DC = 5cm.

∴ The length of the median AD is 5 cm

Given:

There are two tower A and B

The angles of elevation of the tops of A and B are 30° and 60°, respectively

⇒ angle AOC = 30° , angle BOD = 60° 

OC = OD

Formula Used:

\(tan\theta = \frac{p}{b}\)

Calculation:

Let AC = h

Then, BD = H

In triangle AOC,

\(tan30^\circ = \frac{h}{OC}\)

⇒ \(\frac{1}{{\sqrt 3 }} = \frac{h}{{OC}}\)  ...........

(I) In triangle BOD,

\(tan60^\circ = \frac{H}{OD}\)

⇒ \(\frac{{\sqrt 3 }}{1} = \frac{H}{{OD}}\)   ..........

(II) By Dividing eq (I) and eq (II)

⇒ \(\frac{1}{3} = \frac{h}{H}\)

Then the ratio of height of tower B to tower A = 3 : 1

Given:

 

Formula used:

Average = (sum of all observation)/(total number of observation)

Calculation:

Total number of car sold in 2018 = 480

Total number of dealer = 6

Average = (sum of all observation)/(total number of observation)

⇒ Average = 480/6 ⇒ 80

∴ The average sales of cars in 2018 is 80

Given:

Expenditure = 1680 billions

And this is 100%.

Calculation:

The total amount of expenditure for education and Health = 6% + 4%

⇒ 10%

But 100% expenditure = 1680 billions

⇒ 1% expenditure = (1680/100) billions

⇒ 10% expenditure = 16.80 × 10 billions

⇒ 10% expenditure = 168 billions

∴ The total amount of expenditure for education and Health is 168 billions.

Shortcut Trick

Effective Simple Interest (SI) rate for 2 years = 10% × 2 = 20%

Effective Compound Interest (CI) rate for 2 years = 10% + 10% + (10 × 10) ÷ 100 = 21%

If 20% → Rs. 8,100, then 1% → 8,100 ÷ 20 = 405

⇒ CI (21%) = 405 × 21 = 8,505

∴ The correct answer is Rs. 8,505.

Alternate Method

Given:

Simple Interest (SI) = Rs. 8,100

Rate (R) = 10%

Time (T) = 2 years

Formula Used:

SI = (P × R × T) ÷ 100

CI = P × [(1 + R/100)^T − 1]

10%
10%
10%

P = 40,500

Yr 1: 4,050

Yr 2: 4,050

CI on Yr 1: 405

Total CI = 4050 + 4050 + 405 = Rs. 8,505

Calculations:

⇒ P = (SI × 100) ÷ (R × T)

⇒ P = (8100 × 100) ÷ (10 × 2)

⇒ P = 810000 ÷ 20 = 40,500

Now, finding the Compound Interest on the Principal calculated:

⇒ CI = 40500 × [(1 + 10/100)² − 1]

⇒ CI = 40500 × [(11/10)² − 1]

⇒ CI = 40500 × [121/100 − 1]

⇒ CI = 40500 × (21/100)

⇒ CI = 405 × 21 = 8,505

∴ The correct answer is Rs. 8,505.

Additional Information

Direct Relation between CI and SI for 2 years

When rate (R) and time (T = 2 years) are the same, CI can be directly found using: CI = SI × (1 + R/200).

Difference between CI and SI for 2 years

The difference between compound and simple interest on a sum P for 2 years is: Difference = P × (R/100)².

Effective Rate of Compound Interest

The total effective CI rate for 2 years can be calculated using the successive percentage formula: X + Y + (X × Y)/100.

Given:

A sum divided into A and B = Rs 27,000

Rate of simple interest = 15%

A invests for 2 years and B invests for 4 years

Formula Used:

SI = P × R × T/100

Calculation:

Let the sum of A be x

Then, the sum of B be (27,000 - x)

SI of A for 2 years = x × 2 × 15/100

⇒ 30x/100

SI of B for 4 years = (27,000 - x) × 4 × 15/100

⇒ (27,000 - x) × 60/100

⇒ 27000 × 60/100 - 60x/100

⇒ 16200 - 60x/100

According to the question:

SI of A = SI of B

⇒ 30x/100 = 16200 - 60x/100

⇒ 90x/100 = 16200

⇒ x = 18000

Sum of A = Rs 18,000

Sum of B = Rs (27,000 - 18,000) = Rs 9,000

SI of A = 30x/100 = 30 × 18000/100 = Rs 5400

SI of A = SI of B = Rs 5400

Total interest of A and B = 5400 + 5400 = Rs 10800

∴ Total interest of A and B is Rs 10800.

Given:

Perimeter of a right triangle = 60 cm

And hypotenuse = 26 cm

Formula used:

Area of right angle triangle = (1/2) × base × height

In right angle triangle, hypotenuse length² = base length² + perpendicular length²

Concept used:

Triplet of 10 is 10, 24, 26

Calculation:

We can see (10)² + (24)² = 100 + 576

⇒ 676 = (26)² = hypotenuse²

And perimeter = 10 + 24 +26

⇒ 60 = given perimeter

So our assumed triplet is correct.

Hence apart from hypotenuse, remains should be either base length or perpendicular length.

Therefore, base length= 10cm or 24cm

And perpendicular length = 24cm or 10cm

Now area of triangle = (1/2) × 10 × 24

⇒ 120 cm²

∴ The area of a right angle triangle is 120 cm².

Given:

49% of X = Y

Calculation:

49% of X = Y

⇒ Y = (49/100) × X

Now,

Y% of 50

After putting the value of Y in above

⇒ (49/100) × X × (1/100) × 50

⇒ (2450/10000) × X

⇒ 0.245 × X

For percentage, we have to multiply the above number with 100

⇒ 0.245 × X × 100

⇒ 24.5% × X

∴ Y% of 50 is 24.5% of X

Given:

Number is first increased by 40% and then decreased by 25%, again increased by 15% and then decreased by 20%. 

Calculation:

Let the number be x

⇒ \({\rm{x;}} \times \frac{{7}}{{5{\rm{;}}}} \times \frac{{3}}{{4{\rm{;}}}} \times \frac{{23}}{{20{\rm{;}}}} \times \frac{{4}}{{5}}\)

⇒ \({\rm{x;}} \times \frac{{483}}{{500{\rm{;}}}}\)

⇒ 0.966x

⇒ 0.966x is less than x

⇒ Decrease per cent in the number = \(\frac{{{\rm{x}} - 0.996{\rm{x}}}}{{100}} = 3.4{\rm{% ;}}\)

∴ Net decrease per cent in the number is 3.4%

Given:

Total Expenditure = 1680 (in billions)

Expenditure % on Defence = 12%

Expenditure % on Education = 6%

Expenditure % on Employee = 7%

Expenditure % on Subsidy = 21%

Expenditure % on Health = 4%

Expenditure % on Interest payment  = 15%

Expenditure % on proceeds to states = 34%

Expenditure % on Misc = 1%

Formula Used:

Amount of expenditure % on defence more than = \([\frac{{Defence;% ; - ;Education;% }}{{Education;% }}]\) × 100

Calculation:

Substituting the value in formula

 = \([\frac{{12; - ;6}}{6}]\) × 100 ⇒ 100%

∴Amount of expenditure  on defence is 100 % more than Education

Given:

Article purchased = 42

Article sold = 50

Formula used:

Profit% or Loss% = [(the difference between S.P and CP.)/C.P]/100

If S.P is more than C.P then it is profit 

If C.P is more than S.P then it is Loss

Calculation:

Let the C.P of one article be x 

and the S.P of one article be y

C.P for the 42 articles will be = 42x

S.P for the 50 articles will be = 50y

Now according to the question

⇒ C.P = S.P

⇒ 42x = 50y

⇒ x : y =50 : 42 ⇒ 25 : 21

Here C.P is 25 and S.P is 21. so, C.P is greater than S.P which means loss

Loss = C.P - S.P = 25 - 21 = 4

Loss% = (4/25) × 100 = 4 × 4 = 16%

∴ The loss% is 16%

 

 

We can deal in articles also

Formula Used:

Profit% or Loss% = (d/sold article) × 100

where d = difference between the no. article sold and bought

Loss = no. of articles sold is more than bought

Profit = no. of articles brought is more than sold

Calculation:

d = 50 - 42 =8

Loss% = (8/50)× 100 = 16%

∴ The loss% is 16%.