Quantitative Aptitude PYQs

200 men can complete the quarter of work in 50 days

If they continue the work, time taken to complete the remaining work = 3 × 50 = 150 days

[∵ ¾ th of the work is remaining]

But they have to complete in 100 days to finish in time

Time ∝ 1/Number of employees

Let the number of extra employees required to complete the work in time be x

∴ 150/100 = (200 + x)/200

300 = 200 + x

∴ x = 100

Shortcut Trick

Length of the train = Speed of train in m/s × Time taken to cross a stationary pole.

Convert speed: 60 km/h = 60 × (5 / 18) = 50 / 3 m/s.

Length = (50 / 3) × 30 = 50 × 10 = 500 meters.

∴ The correct answer is 500 m.

Alternate Method

Given:

Speed of the train = 60 km/h

Time to cross the pole = 30 seconds

Formula Used:

Distance = Speed × Time

To convert km/h to m/s, multiply by (5 / 18)

Pole

Train Length (L) = ?

Speed = 60 km/h

Time to cross = 30 seconds

Calculations:

⇒ Speed in m/s = 60 × (5 / 18)

⇒ Speed = 300 / 18 = 50 / 3 m/s

⇒ Distance covered = Speed × Time

⇒ Length of train = (50 / 3) × 30

⇒ Length of train = 50 × 10 = 500 m

∴ The correct answer is 500 m.

Additional Information

Crossing a Point Object

When a train crosses a stationary object with negligible width like a pole, standing man, or tree, the total distance covered is equal to the length of the train itself.

Crossing an Extended Object

When a train crosses a bridge, platform, or another train, the total distance covered is the sum of their individual lengths (Distance = L₁ + L₂).

Unit Conversion Trick

To rapidly convert a speed from km/h to m/s, simply multiply by 5/18. To convert backward from m/s to km/h, multiply by 18/5.

Calculation:

Let A’s share be Rs. x

⇒ B’s share = x - 20 and C’s share = x + 20

Total amount = x + x - 20 + x + 20

⇒ 120 = 3x

⇒ x = 40

∴ B’s share = x - 20 = Rs. 20

In inverse variation M × N = constant for all values of M and N

MN = 15 × -4 = -60

-6 × A = -60

∴ A = 10

2 × B = -60

∴ B = -30

C × 60 = -60

∴ C = -1

Given :

Principal = Rs 5000

Rate₁ = 5%

Rate₂ = 5.5%

Time = 2 years

Formula used :

Simple interest = \(\frac{PRT}{100}\)        ---- (1)

Calculations :

Interest when rate is 5%, using equation (1)

⇒ Simple interest = \(\frac{5000 \times 5 \times 2}{100}\)

⇒ Rs 500

Interest when rate is 5.5%, using equation (1)

⇒ Simple interest = \(\frac{5000 \times 5.5 \times2}{100}\)

⇒ Rs 550

The required amount gained

⇒ 550 - 500

⇒ Rs 50

∴ The amount gained by him this transaction is Rs 50.

Age of son after 7 years = 17 years

⇒ Present age of son = 17 - 7 = 10 years

Average of the present ages of father and son = 25

⇒ Sum of the present ages of father and son = 25 × 2 = 50

⇒ Present age of father = 50 - 10 = 40 years

After 10 years age of father = 40 + 10 = 50 years

∴ The age of the father after 10 years will be 50 years.

5 tractors can plough 5 hectares in 5 days

5 tractors can plough 50 hectares in 50 days

1 tractor can plough 10 hectares in 50 days

∴ Number of tractors needed to plough 100 hectares = 100/ (Number of hectares 1 tractor can plough in 50 days)

⇒ 100/10 = 10

∴ 10 tractors can plough 100 hectares in 50 days

Let the initial amount invested be Rs x and the amount is compounded at 25% rate

Amount at the end of n years = Principal × (1 + Rate/100)ⁿ

10000 = x × (1 + 25/100)³

10000 = x × 1.25³

10 × 10 × 10 × 10 = x × 1.25 × 1.25 × 1.25

8 × 8 × 8 × 10 = x [∵ 10/1.25 = 8]

x = Rs. 5120

459 = 17 × 27

Second option is terminating decimal, ∴ denominator has a prime factors of 2 or 5 only

We can evaluate the first decimal and other two decimals are obtained by dividing it by 10 and 100 respectively

Let x = 0.459459459⋯

1000x = 459.459459⋯

1000x = 459 + 0.459459⋯

1000x = 459 + x

∴ 999x = 459

x = 459/999 = (17 × 27) / (37 × 27) = 17/37

∴ 0.459459459⋯ is the decimal with 37 in the denominator when represented as rational number

Given :

Reduction in rate of interest = 0.25%

Decrease in amount = Rs 64

Formula used :

Decrease in amount = \(\frac{Principal \ \times \ rate \ reduction \ in \ rate \ of \ interest \ \times \ time}{100}\)     ---(1)

Calculations :

The decrease in the income is due to the reduction in rate of interest.

Using equation (1), we get

⇒ 64 = \(\frac{Principal \ \times \ .25 \ \times \ 1}{100}\)

⇒ \(Principal = \frac{6400}{0.25}\)

⇒  Principal = Rs. 25600

∴ Original annual income is Rs 25600.

If 0 < m < 1, log₁₀ m is negative so that is the least (All other terms will be positive)

For numbers less than 1, the term with lowest power is the largest

-1 < 1 < 2

∴ m² < m < m^-1

Final order = Log₁₀m < m² < m < m^-1

Let the amount received by the widow = Rs x

∴ Amount received by daughter = 2 × x = 2x

Amount received by son = 3 × 2x = 6x

Total amount = Amount received by widow + 5 × Amount received by sons + 4 × Amount received by daughter

39000 = x + 5 × 6x + 4 × 2x

39000 = x + 30x + 8x

39000 = 39x

∴ x = 1000

Widow receives Rs. 1000

Given:

Three numbers which are co-prime to each other, are such that the product of the first two 286 and that of the last two is 770.

Concept:

(LCM × HCF) = Product of the numbers

Calculation:

Co-prime numbers have no prime factors in common so when expressed as fraction there will be no simplification

Let the numbers be A, B and C respectively

A × B = 286 and B × C = 770

A × B/ (B × C) = 286/770

A/C = 13/35

∴ A = 13 and C = 35

B = 286/A = 22

Sum of the numbers = 13 + 35 + 22 = 70

∴ The sum of three terms is 70.

Let the two digits of the age of the women be y and x (yx)

Age of the women = 10y + x

Age of the men = 10x + y

Difference in their age = 10x + y - 10y - x

⇒ 9x - 9y

Sum of their ages = 10x + y + 10y + x = 11x + 11y

Difference = 1/11 × Sum

9x - 9y = 1/11 × (11x + 11y)

9x - 9y = x + y

8x = 10y

Since x and y can only be natural numbers from 1 to 9, the only possibility is x = 5 and y = 4

Difference in their age = 9x - 9y = 45 - 36 = 9 years

For crossing the relative displacement is same and equal to the sum lengths of the trains in both cases

∴ Time ∝ 1/Relative speed

Let the speed of the goods train be x and passenger train be kx, where k is required ratio

Relative speed when moving in opposite direction = kx - (-x) = (k + 1)x [Speeds are added]

Relative speed when moving in same direction = kx - x = (k - 1)x

(Time taken when moving in same direction) / (Time taken when moving in opposite direction) = (Relative Speed in opposite direction) / (Relative Speed in same direction)

3 = (k + 1)x / [(k - 1)x]

3 = (k + 1) / (k - 1)

3k - 3 = k + 1

2k = 4

k = 2

∴ The required ratio = 2 : 1

Any number multiplied with 5 ends with 5 if it is odd

Product of all odd prime number is an odd number where one of the factor is 5

∴ The unit digit of the product is 5

Let the amount of pure copper added be ‘x’ kg

Amount of copper in 20 kg of alloy A = 2/ (2 + 3) × 20 = 8 kg

Amount of copper in 28 kg of alloy B = 3/ (3 + 4) × 28 = 12 kg

Total amount of copper in alloy C = 8 + 12 + x = 20 + x

Total amount of alloy C produced = 20 + 28 + x = 48 + x

20 + x = 6/(6 + 7) × (48 + x)

20 + x = 6/13 × (48 + x)

260 + 13x = 288 + 6x

7x = 28

∴ x = 4 kg

ax² + bx + c = x (ax + b) + c

Remainder when divided by x = c

∴ c = 3

Let P(x) be a polynomial when divided by (x - a), then the remainder is P(a)

Remainder when divided by (x - 1) = P(1)

6 = a (1) + b(1) + c

6 = a + b + 3

∴ a + b = 3

Shortcut Trick

Concept:

Consecutive numbers:

• They usually have a difference of 1 between every two numbers.
• If the terms are in  AP (i.e. common difference same), and the total number of terms is odd then the average is the same as the middle most term.

 

Calculation:

Using the above concept

Average = 5^th term = 55

Largest integer = 9^th term = 55 + 4 = 59

∴ The largest integer is 59

Alternate Method

Let the integers  are

a - 4, a - 3, a - 2, a - 1, a, a + 1, a + 2, a + 3, a + 4

Sum of 9 integers = 9a

Average = {Sum of Observations} ÷ {Total number of Observations}

Average of 9 integers = 9a/9 = 55

⇒ a = 5

⇒ Largest integer = a + 4 = 55 = 4 = 59

∴ The largest integer is 59

Formula used:

Average = Sum of observation / Number of observations

Calculation

We know that,

Average = Sum of observation / Number of observations

Sum = Average × Number of observations

Therefore, 

Total age of 15 students =(15 × 19) = 285 years

Total age of (15 + 5) i.e. 20 students = (20 × 18.5) = 370 years

Therefore, the total age of 5 new students = 370 - 285 =  85 years

Now, average of 5 new students = 85/5 = 17 year.

∴ The average age of the 5 newly admitted students is 17 years.