CDS I - Mathematics Previous Year Question Paper 2018 - CDS MCQ

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
[∵ 3/4 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

Speed of the train = 60 km/hr = 60 × 5/18 = 50/3 m/s

Distance travelled by the head of the train after it crosses the pole till the whole train crosses the pole = Length of the train
Distance = Speed × Time
50/3 × 30 = Length of the train
∴ Length of the train = 500 m

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

The extra amount he gains is due the difference in the interest rates.
Difference in the interest rate = 5.5 - 5 = 0.5%
Interest he gained = 5000 × 0.5/100 × 2 = Rs. 50 [∵ Interest = Principal × Rate/100 × Time]

Average age of father and son after 10 years = 25 + 10 = 35
Son’s age after 10 years = Son’s age after 7 years + 3
⇒ 17 + 3 = 20 years
Let the father’s age after 10 years be ‘x’
(x + 20) /2 = 35
x + 20 = 70
∴ x = 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

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

Decrease in amount = Principal × Reduction in rate of interest/100 × Time
64 = Principal × 0.25/100 × 1
∴ Principal = 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

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

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

The terms are in A.P. where the common difference is 1

For an AP, average is same as the middle most term

Middle most term = 5^th term = 55
9^th term = 5^th term + 4
⇒ 55 + 4 = 59
∴ The largest integer is 59

Let the average age of 5 students be x
Sum of ages of 15 students + Sum ages of 5 students = Sum of ages of 20 students

Sum = Average × Number of observations
15 × 19 + 5 × x = 20 × 18.5
285 + 5x = 370
5x = 85
∴ x = 17 years

Total volume = Flow rate of R × Time taken to empty

Total volume = (Flow rate of P + Flow rate of Q - Flow rate of R) × Time taken to fill if all the taps and drains are open

To use equation 1, two quantities are required
S1 gives flow rate of R = 6 litres per minute
S3 gives time taken to empty = 15 hours 20 minutes

∴ Volume of tank = 6 × (15 × 60 + 20) = 5520 litres     [∵ 1 hour = 60 minutes]

To use equation 2,
S1: gives the flow rate of R
S2 gives the total time require to fill if all taps and drains are opened.

∴ Volume = (12 + 10 - 6) × (5 × 60 + 45)
⇒ 16 × 345 = 5520 litres
Let the volume of the tank be V litres
By S3: Flow rate of Pipe R = V/ (15 × 60 + 20) = V/920

Substituting in S2
V = (12 + 10 - V/920) × (5 × 60 + 45)
V/345 = 22 - V/920
V/345 + V/920 = 22
168V/57960 + 63V/57960 = 22
231V/57960 = 22
∴ V = 5520 litres
∴ Any two statements are required to find the capacity of the tank

Let the distance between the two places be D km
Time taken = Distance/Speed

Speed in upstream = Speed of boat in still water - Speed of the stream = x - y

Speed in downstream = Speed of the boat in still water + Speed of the stream = x + y

Total time taken = Time taken upstream + Time taken downstream

z = D/ (x - y) + D/ (x + y)
z = [D(x + y) + D(x - y)] / [(x - y) (x + y)]
z = 2Dx / (x² - y²) [∵ (a - b) (a + b) = (a² - b²)]
∴ D = z(x² - y²)/2x

Let the distance between Delhi and Agra be D km

Time taken for round journey = D/60 + D/y       [∵ Time = Distance/Speed]
⇒ D(60 + y) / (60y) = 2D/Average speed
⇒ 2 × 60y = Average speed × (60 + y)
120y = 48 × (60 + y)
120y = 2880 + 48y
72y = 2880
∴ y = 40 km/h

Let the original cost price be Rs ‘x’
Profit = 32% of cost price
∴ Selling price = Cost price + Profit = 1.32x
New cost price = (100 + 20)/100 × x = 1.2x
New profit = Selling price - cost price
⇒ 1.32x - 1.2x = 0.12x
∴ Profit% = 0.12x/1.2x × 100 = 10%

Let D’s contribution be x
∴ A’s contribution = 3 × x = 3x
E’s contribution = ½ × 3x = 1.5x
B’s contribution = 1/3 × 1.5x = 0.5x
C’s contribution = 2/3 × 3x = 2x
Total contribution = 3x + 0.5x + 2x + x + 1.5x = 8x
B, C and E contribution = 0.5x + 2x + 1.5x = 4x
Difference between (A + D + E) - (B + C) = 3x + x + 1.5x - 0.5x - 2x = 3x
Share ∝ Contribution
B + C + E contribution/13500 = 4/3
∴ B + C + E contribution = 4/3 × 13500 = Rs. 18000

5¹⁷ + 5¹⁸ + 5¹⁹ + 5²⁰ = 5¹⁷ × (1 + 5 + 25 + 125)
⇒ 5¹⁷ × 156 = 5¹⁷ × 12 × 13
∴ The given expression is divisible by 13

a + b = 2c
∴ a - c = c - b
a/(a - c) + c/(b - c) = a/(a - c) - c/(a - c) [∵ b - c = -(a - c)]
⇒ (a - c)/(a - c) = 1

x^a = (y^1/a)^a = y
(x^a)^b = y^b = (z^1/b)^b = z
[(x^a)^b]^c = z^c = (x^1/c)^c = x
∴ x^abc = x
∴ abc = 1

2b = a + c
∴ b - c = a - b
x^{b - c} y^{c - a} z^{a - b} = x^{b - c} y^{c - a} z^{b - c}
⇒ (xz)^{b - c} y^{c - a}
⇒ (y²)^{b - c} y^{c - a}
⇒ (y)^{2b - c - a} = y⁰ = 1

If decimal expansion terminates, the denominator of the rational number is 2 or 5 or combinations for their powers

If the decimal expansion is non terminating but repeating, then it is a rational number with terms in prime factorization of the denominators to be numbers other than 2 or 5

If the decimal expansion is non terminating and non repeating, then it is an irrational number