CDS I - Mathematics Previous Year Question Paper 2016 - CDS MCQ

4x² – 16x + λ/4 = 0
⇒ 16x² – 64x + λ = 0
Sum of root = α + β = 64/16 = 4
Product of root = αβ = λ/16
Given 1 < α < 2 and 2 < β < 3
⇒ Value of αβ should be between 2 and 6
But we need to consider that α + β = 4
⇒ 3 < αβ < 4
Putting αβ = λ/16
⇒ 3 < λ/16 < 4
⇒ 48 < λ < 64
∴ λ can take values from 49 to 63 i.e. 15 no. of values.

Multiplying x in first part, y in second part and z in third part.

Putting the given values

5P9 + 3R7 + 2Q8 = 1114

We can see that, the sum of last digits i.e. (5 + 3 + 2) = 10 and 1 is added as carry

And 2 is added as carry in the sum of middle digits of the numbers (P + Q + R)

⇒ (P + Q + R + 2) will be 11 only.
⇒ P + Q + R = 9

For maximum value of Q, values of P and R should be minimum
If P = R = 0 then Q = 9
∴ Maximum value of Q = 9

We can observe that: The difference in divisor and reminder is same in every case.
(2 – 1) = (3 – 2) = (4 – 3) = (5 – 4) = (6 – 5) = 1
LCM of 2, 3, 4, 5 and 6 = 60
⇒ Required number = 60 – 1 = 59
Set A = [(60 – 1), (120 – 1), (180 – 1), ….] = [59, 119, 179, …..]
Between 0 and 100, only 59 is such integer.
∴ Only one integer between 0 and 100 belongs to the set A.

Let the number be x
⇒ 8x – x/8 = 2016
⇒ 64x – x = 2016 × 8
⇒ 63x = 16128
⇒ x = 256
∴ The number is 256.

x² + ax + b = 0
For real roots, necessary condition is:
a² – 4b ≥ 0
⇒ a² ≥ 4b
And (a, b) ϵϵ {1, 2, 3, 4}
Possible pairs of a & b according to the above conditions:
(2, 1), (3, 1), (4, 1), (3, 2), (4, 2), (4, 3), (4, 4)
∴ There are 7 possible pairs of a and b.

Suppose the numbers are ‘a’ and ‘b’ with a > b
a² + b² = 208 and a² = 18b
Solving the equations
⇒ b² + 18b = 208
⇒ b² + 18b – 208 = 0
⇒ (b + 26)(b – 8) = 0
⇒ b = 8, b = -26
If b = 8
a² = 18b = 18 × 8 = 144
⇒ a = 12
∴ Difference of the larger and the smaller number = 12 – 8 = 4

A = {7, 8, 9, 10, 11, 12} and B = {7, 10, 14, 15}
(A – B) = {8, 9, 11, 12}
(B – A) = {14, 15}
∴ Number of elements in (A – B) and (B – A) are 4 and 2 respectively.

Using hit and trial method:
1) 4.65 × 10 = 46.5
2) 4.65 × 20 = 93
3) 4.65 × 21 = 97.65
4) 4.65 × 25 = 116.25
∴ He will be able to save an exact number of rupees in 20 days.

Number of rounds to be completed = 4/0.25 = 16
Since the ratio of speeds of A and B is 5 : 4
⇒ When A covers 5 rounds, B will complete 4 rounds and they will meet.
⇒ A will pass B after 5^th, 10^th & 15^th round.
∴ The winner passes the other thrice.

Observing all the options
1. 15/1600 = 0.009375
2. 23/8 = 2.875
3. 35/50 = 0.7
4. 17/6 = 2.8333333……
∴ Here we can see that 17/6 has non-terminating and repeating decimal expression.

25x² – 15x + 2
⇒ 25x² – 10x – 5x + 2
⇒ 5x(5x – 2) – 1(5x – 2)
⇒ (5x – 1)(5x – 2)
⇒ x = 1/5 and 2/5
⇒ α and β will be equal to 1/5 and 2/5.
If zeros are (2α)^-1 and (2β)^-1
Then the zeros of that polynomial are 5/2 and 5/4
∴ Polynomial = (x – 5/2)(x – 5/4) = (x² – 15x/4 + 25/8) = (8x² – 30x + 25)

[If a^{(n – 1)} is divided by ‘n’ where ‘a’ and ‘n’ are co-prime numbers then remainder will be 1]
Hence here also ‘2’ and ‘101’ are co-prime numbers.
∴ Remainder = 1

Suppose total number of workers = x
⇒ Number of women workers = x/3
⇒ Number of men workers = 2x/3
⇒ Number of married women = (x/3) × ½ = x/6
⇒ Number of married men = (2x/3) × 3/4 = x/2
∴ Ratio of married women to married men = (x/6) : (x/2) = 1 : 3

B alone can do the work in 30 hours and A can do 50% more work than B in same time;
⇒ Time taken by A to complete the work alone = 30/1.5 = 20 hours
Suppose total work = 60 units (LCM of 20 & 30)

⇒ Efficiency of A = 60/20 = 3
⇒ Efficiency of B = 60/30 = 2
⇒ Unit of work done by B in 12 hours = 12 × 2 = 24 units
⇒ Remaining work = 60 – 24 = 36 units
∴ Time taken by (A + B) to complete remaining work = 36/5 = 7.2 hours

Suppose the distance be ‘x’ km and speed be ‘n’ km/minutes
⇒ Increased speed = 1.2n
⇒ x/n – x/1.2n = 20
⇒ (0.2x/1.2n) = 20
⇒ x/n = 120 minutes

Here ‘x/n’ is the time taken to cover the distance with the original speed.

∴ Train will take 120 minutes to cover the same distance with the original speed.

Given Cost of 2.5 kg rice is Rs. 125
⇒ Cost of 1 kg rice = 125/2.5 = Rs. 50
⇒ Cost of 4 kg pulses = cost of 9 kg rice = 50 × 9 = Rs. 450
⇒ Cost of 1 kg pulses = 450/4 = Rs. 112.5
⇒ Cost of 1.5 kg tea = Cost of 14 kg pulses = 112.5 × 14 = Rs. 1575
⇒ Cost of 1 kg tea = 1575/1.5 = Rs. 1050
⇒ Cost of 5 kg nuts = Cost of 2 kg tea = 1050 × 2 = Rs. 2100
∴ Cost of 11 kg nuts = (11/5) × 2100 = Rs. 4620

Ratio in which the profit will be shared = Investment × Time for which investment was made

⇒ (1 × 12) : (2 × 11) : (3 × 10) : (4 × 9) : (5 × 8) : (6 × 7) : (7 × 6) : (8 × 5) : (9 × 4) : (10 × 3) : (11 × 2) : (12 × 1)

⇒ 12 : 22 : 30 : 36 : 40 : 42 : 42 : 40 : 36 : 30 : 22 : 12

Here we can see that shares of both F and G in the profit will be maximum.

x² + px + q = 0
If a and b are the roots of the equation
⇒ Sum of roots = (p + q) = -p     ----1
⇒ Product of roots = pq = q     ----2
On solving equation 1 and 2, we get
p = 1 and q = -2
∴ p = 1 only

1729 = 1728 + 1 = 12³ + 1³ and 1729 = 1000 + 729 = 10³ + 9³

∴ It can be written as the sum of cubes of two positive integers in two ways only.

Let the cost price of the item = Rs. x
⇒ Increased cost price = marked price = Rs. 1.2x
⇒ Overall selling price = 1.2x × 0.9 = Rs. 1.08x
Profit = Selling price - Cost price = 1.08x - x = Rs. 0.08x
∴ Percentage gain = 0.08x/x × 100 = 8%

HCF of p and q is 1
1. Both p and q may be prime
Suppose p = 2 and q = 3
⇒ HCF of p and q = 1
⇒ Statement 1 is correct

2. One number may be prime, and the other is composite.
Suppose p = 3 and q = 4
⇒ HCF of p and q = 1
⇒ Statement 2 is correct

3. Both the numbers may be composite
Suppose p = 9 and q = 10
⇒ HCF of p and q = 1
⇒ Statement 3 is correct.
∴ All three statements are correct.

n^th term of this series can be written as


Sum of the series 
Sum of the series 
∴ Sum of the series 

Downstream speed = 20/2 = 10 km/hr
Upstream speed = 4/2 = 2 km/hr
⇒ Speed of current = (Downstream speed - Upstream speed)/2
∴ Speed of the current = (10 - 2)/2 = 4 km/hr

Since x and y are two odd integers, their squares will be odd always and sum of two odd numbers is always even.

⇒ Statement 1 is correct.
Suppose x = 1 and y = 3
⇒ x² + y² = 1 + 9 = 10
It is even but not divisible by 4.
∴ Only statement 1 is correct.

2^{x + 2} × 27^{x/ (x - 1)} = 9
⇒ 2^{x + 2} × 3^{3x/ (x - 1)} = 3²
⇒ 2^{x + 2} = 3^{[2 - {(3x/x - 1)}]}
⇒ 2^x+2 = 3^{[(x + 2) / (x - 1)]}

Taking log
⇒ (x + 2)log2 = [(x + 2)/(x - 1)]log3
⇒ x - 1 = (log3/log2)
⇒ x = 1 + (log3/log2)
⇒ x = 1 - (log2/log3)

And
(x + 2) = 0 ⇒ x = -2
∴ Roots of the equation = -2, 1 - (log2/log3)

2¹²² + 4⁶² + 8⁴² + 4⁶⁴ + 2¹³⁰
⇒ 2¹²² + 2¹²⁴ + 2¹²⁶ + 2¹²⁸ + 2¹³⁰
⇒ 2¹²² (1 + 4 + 16 + 64 + 256)
⇒ 2¹²² × 341
Since 341 is divisible by 11
∴ 2¹²² + 4⁶² + 8⁴² + 4⁶⁴ + 2¹³⁰ is divisible by 11 only

Let the maximum marks for the examination be ‘n’.

Candidate scoring x% marks in an examination fails by ‘a’ marks but another candidate who scores y% marks gets ‘b’ marks more than the minimum pass marks;

∴ (n × x)/100 + a = (n × y)/100 – b
⇒ (a + b) = n/100 × (y – x)