Quantitative Aptitude PYQs

Given:

Dimensions of a solid metallic cube = 9 cm

Dimensions of a solid metallic cuboid = 5 cm, 13 cm, 31 cm

Rate of Polishing a new cube = Rs. 10 per cm²

Formula Used:

Total Surface area of a cube = 6(side)²

Volume of cube = (side)³

Volume of cuboid = (length × breadth × height)

Amount to polish the cube = Rate × Total surface area of cube

Calculation:

Volume of the new cube = Volume of solid metallic cube + Volume of solid metallic cuboid

Volume of solid metallic cube = (9 cm)³ = 729 cm³

Volume of solid metallic cuboid = (5 cm × 13 cm × 31 cm) = 2015 cm³

⇒ Volume of the new cube = (729 cm³ + 2015 cm³) = 2744 cm³ 

⇒ (side)³ = 2744 cm³ 

⇒ side of new cube = 14 cm  

Total surface area of the new cube = 6(14 cm)² = 6 × 196 cm² = 1176 cm²

Amount to polish the cube = Rs. 10 per cm² × 1176 cm² = Rs. 11,760

∴ Cost to polish the new cube is Rs. 11,760

Given: 

Number of cars having speed between (40-50)kmph = 50 

Number of cars having speed between (50-60)kmph = 70 

Number of cars having speed between (60-70)kmph = 85 

Number of cars having speed between (70-80)kmph = 90 

Number of cars having speed between (80-90)kmph = 60 

Number of cars having speed between (90-100)kmph = 45 

Calculations:

Total cars on highway = 50 + 70 + 85 + 90 + 60 + 45 = 400

Cars running with the speed of 90 km/hr and above = 45

Percentage of cars running with the speed of 90km/hr and above = (45/400) × 100 = 11.25%

∴ 11.25% cars were running with the speed of 90 km/hr and above.

Given:

R = 17%

Difference between compound interest and simple interest for 2 years = Rs. 433.50

Formula used:

Difference between compound interest and simple interest for 2 years = P × R/100 × R/100

Compound interest = P × (1 + R/100)^T - P

Calculation:

Difference between compound interest and simple interest for 2 years = P × R/100 × R/100

⇒ 433.50 = P × 17/100 × 17/100

⇒ 43350 × 100 = P × 289

⇒ P = 4335000/289 = 15000

Compound interest = P × (1+R/100)T - P

⇒ 15000 × (1 + 17/100)² - 15000

⇒ 15000 × [(117/100 × 117/100) – 1]

⇒ 15000 × (13689 – 10000)/10000

⇒ (15 × 3689)/10

⇒ Rs. 5533.5

∴ Compound interest for two years is Rs. 5533.5 

Alternate Method

2 years Effective rate for S.I. = RT 

⇒ 17 × 2 = 34% 

2 years Effective rate for C.I = 2R + R²/100 

⇒ 2 × 17 + 17²/100

⇒ 34 + 2.89 = 36.89% 

Difference in rates = 36.89% – 34% = 2.89% 

Now, 2.89% = 433.50 

Here we have to find C.I so = (433.50/2.89) × 36.89 = 5533.5 

∴ The compound interest for two years is Rs. 5533.5

Given:

Radius of solid sphere = 11 cm

Radius of small solid sphere = 2 cm 

Formula Used:

Volume of a sphere = (4/3)πr³

Number of small sphere = (Volume of solid sphere)/(Volume of small solid sphere)

Where, r = Radius of the sphere 

Calculation:

Volume of the solid sphere with radius 11 cm = (4/3) × π × 11³

Volume of the small solid sphere with radius 2 cm = (4/3) × π × 2³

Number of small solid sphere = (4/3) × π × 11³/(4/3) × π × 2³ = 1331/8 = 166.37 ≈ 166

∴ Maximum 166 of such spheres can be made.

Given:

The monthly salary of a person = Rs. 75,000

E was 60% of the income, T was 20% of E, and C was 15% of T

When his salary got raised by 40% he maintained the percentage level of E, but T

became 30% of E and C became 20% of T.

Calculation :

Family expense (E) = 60% of income

⇒ 60% of 75000 = (60 × 75000)/100 = 45000

Taxes (T) = 20% of (E)

⇒ 20% of 45000 = (20 × 45000)/100 = 9000

Charity (C) = 15% of (T)

⇒ 15% of 9000 = (15 × 9000)/100 = 1350

Total saving = 75000 – (45000 + 9000 + 1350) = 75000 – 55350 = 19650

New salary = 75000 + 40% of 75000

⇒ 75000 + (40 × 75000)/100 = 75000 + 30000 = 105000

Family expense (E) = 60% of income

⇒ 60% of 105000 = (60 × 105000)/100 = 63000

Taxes (T) = 30% of (E)

⇒ (30 × 63000)/100 = 18900

Charity (C) = 20% of (T)

= (20 × 18900)/100

= 3780

Total saving = 105000 – (63000 + 18900 + 3780) = 19320

Ratio of the savings of his earlier salary to that of his present salary = 19650 ∶ 19320

= 655 ∶ 644

∴ Ratio of the savings of his earlier salary to that of his present salary is 655 ∶ 644.

Given:

 

Calculation:

Number of Cars Between the Speed Interval of (60 - 70) Km/h = 85

Number of Cars between the Speed Interval of (70 - 80) Km/h = 90

Difference between the Two Speed Intervals = 90 – 85 = 5

Less Percentage = (Difference/Number of Cars between Interval (70 - 80)Km/h) × 100

⇒ Less Percentage = (5/90) × 100

⇒ 50/9 = 5.6%

∴ The Number of Cars between speed intervals (60 - 70) Km/h is 5.6% less than the Number of Cars between speed intervals (70 - 80) Km/h

Total number of girls from A,B and C = 500 + 550 + 650 = 1700

Total number of boys from all school = 600 + 450 + 750 + 700 + 800 = 3300

⇒ 1700/3300 = 17/33

∴ The required ratio is 17 ∶ 33

Given:

Total numbers of employee = 450

Concept used :

Central angle = [(Value)/(sum of the total values)] × 360° 

Calculation:

Number of employees in department E = 20% of 450

⇒ Required number = (20 × 450)/ 100

⇒ 90

The central angle of the employees in department E = (90/450) × 360°

⇒ 72° 

∴ The central angle of the employees in department E is 72°

Given:

The cistern gets filled together (A + B) in 6 hours.

Concept used:

Effeciency = 1/(Total time)

Calculation:

Let the time taken by B be X hr.

Work done by B in 1 hr. = 1/X

Now, time taken by A = (X – 5) hr.

So, work done by in 1 hr. = 1/(X – 5) hr.

Again, work done by (A +B) in 1 hr = 1/6

According to the question:

(1/X) + 1/(X – 5) = 1/6

⇒(X – 5 + X)/{X × (X – 5)} = 1/6

⇒ 6(2X – 5) = X² – 5X

⇒ 12X - 30 = X2 – 5X

⇒ X2 – 5X – 12X + 30 = 0

⇒ X² – 17X + 30 = 0

⇒ X² – 15X – 2X + 30 = 0

⇒ X (X – 15) – 2 (X – 15) = 0

⇒ (X – 2) (X – 15) = 0

⇒ X – 2 = 0

⇒ X = 2 (B can never fill the tank in 2 hours because if it fills it in 2 hours that means time taken by A in negative and it is not possible)

Either,

X – 15 = 0

⇒ X = 15

⇒ A = 15 – 5 = 10 hr.

∴ Time taken by A is 10 hr.

Concept used:

Average = (Total sum)/(Total numbers)

Calculations:

Total number of girls and boys in school

⇒ 600 +500 + 450 + 550 + 750 + 650 + 700 + 600 + 800 + 650 = 6250

⇒ Average = 6250/5 = 1250

∴ The required average is 1250

Given:

Average Price of Three Chairs = Rs. 14,014

Ratio of the Prices of three Chairs = 3 : 4 : 7

Formula Used:

Average = Sum of Prices of Articles/ Number of Articles

Concept Used:

Sum of Ratios will be equal to the Sum of Prices of All Chairs

Calculation:

14,014 = Sum of Prices of Chairs/3

⇒ Sum of Prices Of chairs = 14,014 × 3 = 42,042

Sum of Prices of Chairs is also equal to the sum of Ratios given

i.e. 3 + 4 + 7 = 14 Units

⇒ 14 Units = Rs. 42,042

⇒ 1 Unit = 42,042/14 = Rs. 3003

Since, we have to find the Value of Highest Price Chair i.e. 7 Units

⇒ 7 Units = 3003 × 7 = Rs. 21021

∴ The Highest price of Chair is Rs. 21,021

Given:

\(\frac{0.325 \ \times \ 0.325\ + \ 0.175 \ \times\ 0.175\ + \ 25\ \times \ 0.00455}{5 \ \times \ 0.0065 \ \times \ 3.25\ - \ 7\ \times \ 0.175\ \times\ 0.025}\ + \ \frac{0.5}{1.5}\)

Formula Used:

(a + b)² = a² + b² + 2ab

(a² – b²) = (a + b) (a - b)

Concept Used:

We need to identify the numbers to fit them into the above-stated formula.

Calculation:

Let a = 0.325 and b = 0.175

In the numerator,

25 × 0.00455 = 2 × 0.325 × 0.175 = 2ab

In the denominator,

5 × 0.0065 × 3.25 = (0.325)² = a²

7 × 0.175 × 0.025 = (0.175)² = b²

On Comparing Formulas with a given equation we get,

\(⇒ \frac{{{{\left( {0.325\ + \ 0.175} \right)}^2}}}{{\left( {0.325\ +\ 0.175} \right)\left( {0.325\ -\ 0.175} \right)}} + \frac{{0.5}}{{1.5}} \)

⇒ \(\frac{(0.325 + 0.175)}{(0.325 - 0.175)} + \frac{0.5}{1.5}\)

\(= \frac{{0.5}}{{0.15}} + \frac{{0.5}}{{1.5}} \)

= (5.5)/(1.5)

= 11/3

∴ The simplified Value of this Expression is 11/3

Given:

x2 - 3x + 1 = 0

Formula used:

x³ + 1/x³ = (x + 1/x)³ – 3(x + 1/x)

Calculation:

x2 – 3x + 1 = 0

⇒ x² + 1 = 3x

⇒ x + 1/x = 3

⇒ x3 + 1/x3 = (x + 1/x)3 – 3(x + 1/x)

⇒ x3 + 1/x³ = 27 – 9

⇒ x3 + 1/x³ = 18

Required value = (x⁴ + 1/x²) ÷ (x² + 1)

⇒ x(x³ + 1/x³) ÷ x(x + 1/x)

⇒ (x3 + 1/x3) ÷ (x + 1/x)

⇒ 18 ÷ 3 = 6

∴ The correct answer is 6

Given:

Average weight of A, B and C = 70 kg

Average weight of A, B, C and D = 60 kg

Weight of E = Weight of D + 5 kg

Average Weight of B, C, D and E = 59 kg

Formula Used:

Average = (Sum of Values)/(Total Number) 

Calculation:

Sum of weight of A, B and C = 3 × 70 kg = 210 kg

Sum of weight of A, B, C and D = 4 × 60 kg = 240 kg

Weight of D = (Sum of weight of A, B, C and D) – (Sum of weight of A, B, and C)

⇒ Weight of D =  (240 – 210) = 30 kg

Weight of E = 30 kg + 5 kg = 35 kg

Sum of weight of B, C, D and E = 4 × 59 kg = 236 kg

Weight of A = (Sum of weight of A, B, C and D + Weight of E) – (Sum of weight of B, C, D and E)

⇒ Weight of A = 240 kg + 35 kg – 236 kg = 39 kg

Average weight of A, D and E = (Sum of weight of A, D and E)/3 = (39 kg + 30 kg + 35 kg)/3

⇒ Average weight of A, D and E = (104 kg)/3 = 34.66 kg ≈ 35 kg

∴ Average weight of A, D and E is 35 kg.

Given:

Man spends 75% of his income.

Income increases by 28% 

Expenditure increases by 20%

Concept:

Saving = Income – Expenditure 

Solution:

Let the Income of a man = 100

Expenditure = 75% of 100

⇒ (75/100) × 100 = 75

Saving = Income – Expenditure 

Initial Saving = 100 – 75 = 25      ----(1)

Now as per the question,

His income increased by 28%

⇒ His new income = Old income + 28% of Old income

⇒ His new income = (128/100) × 100 = 128

Similarly, his expenditure increased by 20%

⇒ His new expenditure = (120/100) × 75

⇒ His new expenditure = 90

Saving_(new) = Income_(new) – Expenditure_(new)

⇒ Saving(new) = 128 - 90 = 38      ----(2)

Now, % increase in savings = {(38 – 25)/25} × 100      ----(from 1 and 2)

⇒ 13 × 4 = 52%

∴ The % increase in saving is 52%

Shortcut Trick

 

∴ The % increase in saving = (38 - 25) ×  100/25 = 13 × 4 = 52%

Given:

A and B can do a piece of work in 36 days,

B and C can do the same work in 60 days,

A and C can do the same work in 45 days.

Concept used:

Efficiency = 1/(Total time)

Calculation:

(A + B)’s 1 day work = 1/36      ----(i) 

(B + C)’s 1 day work = 1/60      ----(ii) 

(C + A)’s 1 day work = 1/45      ----(iii)

From equations (i), (ii) and (iii),

2(A + B + C)’s 1 day work = 1/36 + 1/60 + 1/45

⇒ (5 + 3 + 4)/180 = 1/15

⇒ (A + B + C)’s 1 day work = 1/(15 × 2) = 1/30      ----(iv)

From equations (iii) and (iv) we get

B’s 1 day work = 1/30 – 1/45 = (3 – 2)/90 = 1/90

Time taken by B alone to complete the work = 90

∴ B alone can complete the same work in 90 days

 Alternate Method

Let us find the LCM of  36, 60 and 45

LCM of 36, 60 and 45 = 180

Let the total work be 180 units. 

A + B can do = 180/36 = 5 units/day      ----(i)

B + C can do = 180/60 = 3 units/day      ----(ii)

C + A can do = 180/45 = 4 units/day      ----(iii) 

Adding the above equations, we get 

2(A + B + C) = 5 + 3 + 4 

⇒ A + B + C = 12/2 = 6      ----(iv)

Subtracting equation (iii) from (iv) 

B = 6 - 4 = 2units/day 

⇒ Time taken by B = 180/2 = 90days

∴ B alone can complete the same work in 90 days

Given:

A man sold an article at a gain of 10% and if he had bought the article at 20% less and sold it for Rs. 1000 more, he would have made a profit of 40%.

Formula used:

S.P = C.P × (1 + P%/100)

P = S.P – C.P

P% = (P/C.P) × 100

Where,

S.P = Selling price 

C.P = Cost price 

P = Profit

Calculation:

Let CP be x

Then, SP = 110x/100 = 11x/10

New CP = 80% of x = 80x/100 = 4x/5

Gain = 40%

New SP = 140% of 4x/5 = (140/100) × 4x/5 = 28x/25

According to the question,

(28x/25) – (11x/10) = 1000

⇒ (56x – 55x)/50 = 1000

⇒ x/50 = 1000

⇒ x = Rs. 50,000

Hence, CP = Rs. 50,000

And SP = (11/10) × 50000 = Rs. 55000

∴ The earlier selling price of the article is Rs. 55000

Given:

Pipes A, B and C can fill an empty tank in 30/7 hours

A and B are filling pipes and C is an emptying pipe.

Time taken by A and B together to fill the tank = 15 hours 

Time taken by C to empty the tank alone = 12 hours 

Concept:

Efficiency = 1/(Total time)

Calculation: 

let time taken by B be X hr.

Work done by B in 1 hr = 1/X hr.

Work done by  (A + B – C) in 1 hr = 7/30

According to the question:

⇒ (1/15) + (1/X) – (1/12) = 7/30

⇒ 1/X = (7/30) + (1/12) – (1/15)

⇒ 1/X = (14 + 5 – 4)/60

⇒ 1/X = 15/60

⇒ 1/X = 1/4

⇒ X = 4

∴ Time taken by B is 4 hr.

Alternate Method 

 

A + B + C = 14 (Total Efficiency)

⇒ 4 + B + (-5) = 14

⇒ B = 15

Total work done by B = 60/15 = 4hr

∴ Time taken by B is 4 hr.

Given:

x + y + z = 17

xy + yz + zx = 111

xyz = 171 

Concept used:

1.) (x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + zx)

2.) x3 + y3 + z3 – 3xyz = (x + y + z)[x2 + y2 + z2​ – (xy + yz + zx)]

Calculations:

(x + y + z)² = x² + y² + z² + 2(xy + yz + zx)

⇒ 17² = x2 + y2 + z² + 2 × 111

⇒ x² + y² + z²  = 289 – 222

⇒ x2 + y2 + z2  = 67

Using algebraic identity

x3 + y3 + z3 – 3xyz = (x + y + z)[x2 + y2 + z2​ – (xy + yz + zx)]

⇒ x3 + y3 + z3 – 3 × 171 = 17 × (67 – 111)

⇒ x3 + y3 + z3 – 513 = -748

⇒ x3 + y3 + z3 = -748 + 513

⇒ x3 + y3 + z3 = -235

\(\sqrt[3]{{(x^3+y^3+z^3+xyz)}}\) = ∛(-235 + 171)

⇒ -4

∴ The correct answer is -4

Given:

\(;\left( {5\frac{1}{{4;}} ÷ ;\frac{{3;}}{7}of;\frac{1}{2}} \right) ÷ \left( {5\frac{1}{9} - 7\frac{7}{{8;}} ÷ ;9\frac{9}{{20}}} \right) × \frac{{11}}{{21}} - \left( {5 ÷ 2;of\frac{1}{2}} \right)\)

Concept used:

Follow the BODMAS rule according to the table given below:

Calculation:

\(;\left( {5\frac{1}{{4;}} ÷ ;\frac{{3;}}{7}of;\frac{1}{2}} \right) ÷ \left( {5\frac{1}{9} - 7\frac{7}{{8;}} ÷ ;9\frac{9}{{20}}} \right) × \frac{{11}}{{21}} - \left( {5 ÷ 2;of\frac{1}{2}} \right)\)

⇒ [(21/4) ÷ (3/14)] ÷ [(46/9) – (63/8) ÷ (189/20)] × 11/21 – (5 ÷ 1)

⇒ [(21/4) × (14/3)] ÷ [(46/9) – (63/8) × (20/189)] × 11/21 – 5

⇒ (49/2) ÷ [(46/9 – 5/6)] × (11/21) – 5

⇒ (49/2) ÷ [(92 – 15)/18] × (11/21) – 5

⇒ (49/2) ÷ (77/18) × (11/21) – 5

⇒ (49/2 × 18/77 × 11/21) – 5

⇒ 3 – 5

⇒ –2

∴ The value of ? is –2.