Quantitative Aptitude PYQs

Calculation

Suman’s marks = 55 – 20% of 55 = 55 – 11 = 44

so, ratio = 60 : 44 = 15 : 11

Shortcut Trick

Let total work = 24 units (LCM of 8 and 24).

Efficiency of (A + B) = 24 ÷ 8 = 3 units/day.

Efficiency of B = 24 ÷ 24 = 1 unit/day.

Work done in first 2 days = 3 × 2 = 6 units. Remaining work = 24 − 6 = 18 units.

Efficiency of (B + C) = Remaining work ÷ 9 days = 18 ÷ 9 = 2 units/day.

Efficiency of C = (B + C) − B = 2 − 1 = 1 unit/day. Time for C = 24 ÷ 1 = 24 days.

∴ The correct answer is 24 days.

Alternate Method

Given: A + B can do work in 8 days. B alone can do it in 24 days. Combined work for 2 days, then B + C work for 9 days.

⇒ Efficiency of A = Efficiency of (A + B) − Efficiency of B = 3 − 1 = 2 units/day.

⇒ Work completed by A and B in 2 days = 3 × 2 = 6 units.

⇒ Remaining work = 24 − 6 = 18 units.

⇒ Time taken by B and C to finish remaining work = 9 days.

⇒ Efficiency of (B + C) = 18 ÷ 9 = 2 units/day.

⇒ Efficiency of C = Efficiency of (B + C) − Efficiency of B = 2 − 1 = 1 unit/day.

⇒ Time taken by C alone = Total Work ÷ Efficiency of C = 24 ÷ 1 = 24 days.

∴ The correct answer is 24 days.

Additional Information

Efficiency and Time Relationship

Efficiency is inversely proportional to time taken. If a person finishes work in D days, their efficiency is 1/D per day.

Combined Work Formula

If A and B work together, their combined efficiency is the sum of their individual efficiencies: E_total = E_A + E_B.

Total Work calculation

Total Work = Efficiency × Time. Using the LCM of given time periods is the most efficient way to assume total work.

Calculation∶

The series follows following pattern

4 × 1 + 1 = 5

5 × 1 - 1 = 4

4 × 2 + 1 = 9

9 × 2 - 1 = 17

17 × 3 + 1 = 52

52 × 3 - 1 = 155

∴ The value of ? is 17

Calculation∶

The series follows following pattern∶

25 – 3 = 22

53 – 25 = 28

75 – 53 = 22

103 – 75 = 28

103 + 22 = 125

∴ The value of ? is 125

The series follows the following pattern:

132 + 7 = 139

139 + 14 = 153

153 + 21 = 174

174 + 28 = 202

202 + 35 = 237

237 + 42 = 279

∴ The missing term in the series is 237.

Considering the above series,

3, 7, 14, 27, 49, 83, ?

The logic of the series can be explained as,

7 – 3 = 4

14 – 7 = 7

27 – 14 = 13

49 – 27 = 22

83 – 49 = 34

Taking double difference,

7 –  4 = 3

13 – 7 = 6

22 – 13 = 9

34 – 22 = 12

By observing the pattern of double difference,

Next double difference will be 15.

So, Next difference will be 34 + 15 = 49

⇒ ? – 83 = 49

⇒ ? = 83 + 49

⇒ ? = 132

∴ The value of ? is 132

Given:

The number series:

106, 212, 636, ?, 12720, 76320

Concept Used:

Number series following a fixed pattern.

Calculation;

The given number series follows the pattern:

106 × 2 = 212

212 × 3 = 636

636 × 4 = 2544

2544 × 5 = 12720

12720 × 6 = 76320

∴ The Missing Number in the given number series is 2544

Shortcut Trick

Rate for half-year = 10% ÷ 2 = 5%.

First deposit (₹2000) earns interest for 2 half-years: 2000 × 1.05 × 1.05 = ₹2205.

Second deposit (₹2200) earns interest for 1 half-year: 2200 × 1.05 = ₹2310.

Total Amount = 2205 + 2310 = ₹4515.

∴ The correct answer is ₹4515.

Alternate Method

Given: First deposit P₁ = ₹2000 on 1^st Jan, Second deposit P₂ = ₹2200 on 1^st July, Annual Rate R = 10% compounded half-yearly.

Formula Used: Amount A = P(1 + r/100)ⁿ, where r is the rate per period and n is the number of periods.

⇒ Half-yearly rate (r) = 10% ÷ 2 = 5%.

⇒ Amount from 1^st Jan deposit (for 2 periods) = 2000 × (1 + 5/100)²

⇒ A₁ = 2000 × (1.05) × (1.05) = 2000 × 1.1025 = ₹2205.

⇒ Amount from 1^st July deposit (for 1 period) = 2200 × (1 + 5/100)¹

⇒ A₂ = 2200 × 1.05 = ₹2310.

⇒ Total Amount = A₁ + A₂ = 2205 + 2310 = ₹4515.

∴ The correct answer is ₹4515.

Additional Information

Half-Yearly Compounding

In half-yearly compounding, the annual rate is divided by 2 and the time (number of years) is multiplied by 2.

Effective Rate of Interest

For an annual rate R compounded half-yearly, the effective annual rate is R_eff = (1 + R/200)² − 1.

Difference between SI and CI

For 2 periods (1 year compounded half-yearly), the difference is P × (r/100)², where r is the half-yearly rate.

Common Calculation

For 2021, Number of male graduates is 450.

Number of female graduate is 450 × 130 /100 = 585

Number of Male post graduate is 450 × 4 / 5 = 360

Number of female post graduate is 360 × [90/100] = 324

Similarly, we can calculate the others year values also.

Calculation

2021:

Male = 450 + 360 = 810

Female = 585 + 324 = 909

Difference = |909 – 810| = 99

2022:

Male = 360 + 240 = 600

Female = 450 + 192 = 642

Difference = |642 – 600| = 42

Required difference is 99 – 42 = 57

Common Calculation

For 2021, Number of male graduates is 450.

Number of female graduate is 450 × 130 /100 = 585

Number of Male post graduate is 450 × 4 / 5 = 360

Number of female post graduate is 360 × [90/100] = 324

Similarly, we can calculate the others year values also.

Calculation

2021:

Total PG = 360 + 324 = 684

% Female PG = (324 / 684) × 100 = 47.37%

2022:

Total PG = 240 + 192 = 432

% Female PG = (192 / 432) × 100 = 44.44%

2023:

Total PG = 400 + 280 = 680

% Female PG = (280 / 680) × 100 = 41.18%

✅ Answer: 2021

Common Calculation

For 2021, Number of male graduates is 450.

Number of female graduate is 450 × 130 /100 = 585

Number of Male post graduate is 450 × 4 / 5 = 360

Number of female post graduate is 360 × [90/100] = 324

Similarly, we can calculate the others year values also.

Calculation

2021: 450 + 585 = 1035

2022: 360 + 450 = 810

2023: 480 + 576 = 1056

Total over 3 years = 1035 + 810 + 1056 = 2901

Average = 2901 / 3 = 967

Common Calculation

For 2021, Number of male graduates is 450.

Number of female graduate is 450 × 130 /100 = 585

Number of Male post graduate is 450 × 4 / 5 = 360

Number of female post graduate is 360 × [90/100] = 324

Similarly, we can calculate the others year values also.

Calculation

2022 Male = 360 + 240 = 600

2023 Male = 480 + 400 = 880

Increase = 880 – 600 = 280

% Increase = (280 / 600) × 100 = 46.67%

✅ Answer: 46.67%

Common Calculation

For 2021, Number of male graduates is 450.

Number of female graduate is 450 × 130 /100 = 585

Number of Male post graduate is 450 × 4 / 5 = 360

Number of female post graduate is 360 × [90/100] = 324

Similarly, we can calculate the others year values also.

Calculation

2021 PG = 360 + 324 = 684

2022 PG = 240 + 192 = 432

Required Ratio = 684 : 432 = (171 : 108) = 19 : 12

Solution:

Female total:

2021: Graduates = 585, Postgraduates = 324 → Total = 909

2023: Graduates = 576, Postgraduates = 280 → Total = 856

Total female (2021 + 2023) = 909 + 856 = 1765

Male total 2022:

Graduates = 360, Postgraduates = 240 → Total = 600

Required Ratio = 1765 : 600 = 353 : 120

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

54.98% of 199.98 + 79.99% of 249.98 = ? + 64.99% of 399.99

⇒ 55% of 200 + 80% of 250 = ? + 65% of 400

⇒ 110 + 200 = ? + 260

⇒ 310 = ? + 260

∴ ? = 50

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

? = 74.99 – 2.99 × [(5.992 ÷ 6.06)  ÷ 3.03] ÷ (1/2.99) + 12.99

⇒ 75 – 3 × [(62 ÷ 6) ÷ 3] ÷ (1/3) + 13

⇒ 75 – 3 × [(36/6 ÷ 3] ÷ (1/3) + 13

⇒ 75 – 3 × [(6/3] × 3 + 13

⇒ 75 – 3 × 2 × 3 + 13

⇒ 75 – 18 + 13

⇒ 70

Follow BODMAS rule to solve this question, as per the order given below.

Step - 1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step - 2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step - 3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step - 4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

Given expression is

(19.99 × 44.99) – (69.98 × 3.13) + 359.98 = ? × 29.9

⇒ (20 × 45) – (70 × 3) + 360 = ? × 30

⇒ 900 – 210 + 360 = ? × 30

⇒ 1260 – 210 = ? × 30

⇒ 1050 = ? × 30

∴ ? = 35

Concept Used:

Follow BODMAS rule to solve this question, as per the order given below, 

​

Calculation:

2/3 of 47.99 + [6.98 + {(-15.98) + 23.98} – 6.97] = ?

⇒ 2/3 of 48 + [7 + {(-16) + 24} – 7] = ?

⇒ 2/3 of 48 + [7 + 8 – 7] = ?

⇒ 2/3 × 48 + 8 = ?

⇒ 32 + 8 = ?

⇒ 40 = ?

∴ The value of ? is 40

CALCULATIONS:

Follow the BODMAS rule to solve this question, as per the order given below.

Step-1: Parts of an equation enclosed in 'Brackets' must be solved first, and following BODMAS rule in the bracket -

Step-2: Any mathematical 'Of' or 'Exponent' must be solved next.

Step-3: Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated.

Step-4: Last but not the least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Since, we need to find out the approximate value, we can write these values to their nearest integers.

∴ Given expression can be written as:

61% of 12200 - 20% of 2210 + 3000 = 500 × ?

⇒ 7442 - 442 + 3000 = 500 × ?

⇒ 7000 + 3000 = 500 × ?

⇒ 10000 = 500 × ?

⇒ 1000 / 500 = ?

⇒ 20 = ?

∴ ? = 20.

Shortcut Trick

Ratio of Interests = (Principal_A × Rate_A × Time_A) : (Principal_B × Rate_B × Time_B)

6 : 5 = (2400 × 18 × 3) : (x × 12 × 4)

6 / 5 = (2400 × 54) / (x × 48)

x = (2400 × 54 × 5) / (48 × 6) = 50 × 9 × 5 = 2250

∴ The correct answer is ₹2250.

Alternate Method

Given:

Principal for A (P_A) = ₹2400, Rate (R_A) = 18%, Time (T_A) = 3 years

Principal for B (P_B) = ₹x, Rate (R_B) = 12%, Time (T_B) = 4 years

Ratio of SI_A : SI_B = 6 : 5

Formula Used:

Simple Interest (SI) = (P × R × T) / 100

Calculations:

⇒ SI_A = (2400 × 18 × 3) / 100 = 24 × 54 = ₹1296

⇒ Given SI_A / SI_B = 6 / 5

⇒ 1296 / SI_B = 6 / 5

⇒ SI_B = (1296 × 5) / 6 = 216 × 5 = ₹1080

⇒ SI_B = (x × 12 × 4) / 100

⇒ 1080 = (x × 48) / 100

⇒ x = (1080 × 100) / 48

⇒ x = 22.5 × 100 = 2250

∴ The correct answer is ₹2250.

Additional Information

Simple Interest Properties

Simple interest is calculated only on the Principal amount. The interest remains constant for every year if the rate is unchanged.

Amount Formula

Total Amount (A) = Principal (P) + Simple Interest (SI). It can be expressed as A = P[1 + (R × T)/100].

Ratio Method for SI

When two different principals are invested, SI₁/SI₂ = (P₁R₁T₁) / (P₂R₂T₂). This allows direct comparison without calculating individual interests.