Mathematics Mock Test - 8 - CDS MCQ

For number of zeroes we must count number of 2 and 5 in prime numbers below 100.
We have just 1 such pair of 2 and 5.
Hence we have only 1 zero.

Let the common remainder be x.

32506 – x is divisible by n.

34041 – x is divisible by n.

Difference of (32506 – x) and (34041 – x) = (32506 – x) – (34041 – x)

⇒ 32506 – x – 34041 + x

⇒ 32506 – 34041

⇒ 1535 

Factors of 1535 = 1 × 5 × 307 × 1535

3-digit number = 307

⇒ n = 307

∴ The value of n is 307.

• There are several rational numbers between 4 and 5. The numbers are between 16/4 and 20/4. 
• Therefore, the answer is c, that is, 17 / 4, 18 / 4, 19 / 4.

Let the number of girls be 2x and number of boys be x.

Girls getting admission = 0.6x

Boys getting admission = 0.45x

Number of students not getting admission = 3x – 0.6x -0.45x = 1.95x

Percentage = (1.95x/3x) * 100 = 65%

Assuming the maximum marks =100a, then Meena got 40a

After increasing her score by 50%, she will get 40a(1+50/100)=60a

Passing score = 60a+35

Post review score after 20% increase = 60a*1.2=72a

=>Hence, 60a+35+7=72a

=>12a=42   =>a=3.5

=> maximum marks = 350 and passing marks = 210+35=245

=> Passing percentage = 245*100/350 = 70

(85410 + 36885 + 24705) = 147000; 147000/1600 = 91.875

When we multiply each number by 9, the average would also get multiplied by 9.

Hence, the new average = 18 X 9 = 162.

David's Weight = (4 x 70) – (3 x 74) = 280 – 222 = 58
Eric’s weight = 58 + 3 = 61

Now, we know that:
Ross + Joey + Chandler + David = 4 x 70 = 280
Joey + Chandler + David + Eric = 75 x 4 = 300.

Hence, Ross’s weight is 20 kg less than Eric’s weight. Ross = 61 - 20 = 41 kg. 

A and B complete a work in = 15 days
⇒ One day's work of (A + B) = 1/ 15

B complete the work in = 20 days
⇒ One day's work of B = 1/20

⇒ A's one day's work = 1/15 − 1/20 = (4−3)/6 = 1/60

Thus, A can complete the work in = 60 days.

So Option A is correct

Let distance between the two places = d km
Let total time taken by faster horse = t hr
⇒ Total time taken by slower horse = (t + 5) hr,

Therefore,
speed of the faster horse = d/t km/hr
speed of the slower horse = d/(t + 5) km/hr 
The two horses meet each other in 3 hour 20 min i.e. in 3(1/3) hr = 10/3 hr
In this time, total distance travelled by both the horses together is d. 

∴ d/(t+5) * 10/3 + d/t * 10/3 = d
⇒ 10/(3(t+5)) + 10/3t = 1
⇒ 10t + 10(t+5) = 3t(t+5)
⇒ 20t + 50 = 3t² + 15t
⇒ 3t² − 5t − 50 = 0
⇒ 3t² + 10t − 15t − 50 = 0
⇒ t(3t + 10) − 5(3t + 10) = 0
⇒ (3t + 10)(t − 5) = 0
⇒ t = 5 (ignoring -ve value) 

Thus, Total time taken by slower horse = 5 + 5 = 10 hr

So Option B is correct

The discriminant of the quadratic equation is (-12)² - 4(3)(10) i.e., 24.
As this is positive but not a perfect square, the roots are irrational and unequal.

Let the roots of the quadratic equation be x and 3x.
Sum of roots = -(-12) = 12

⇒ x + 3x = 4x = 12
⇒ x = 3

Product of the roots: 3x² = 3(3)² = 27.

We have |n - 60| < |n - 100| < |n - 20|
Now, the difference inside the modulus signified the distance of n from 60, 100, and 20 on the number line.

This means that when the absolute difference from a number is larger, n would be further away from that number.

The absolute difference of n and 100 is less than that of the absolute difference between n and 20.

Hence, n cannot be ≤ 60, as then it would be closer to 20 than 100. Thus we have the condition that n>60.

The absolute difference of n and 60 is less than that of the absolute difference between n and 100.

Hence, n cannot be ≥ 80, as then it would be closer to 100 than 60.

Thus we have the condition that n<80.

The number which satisfies the conditions are 61, 62, 63, 64……79. Thus, a total of 19 numbers.

Alternatively

as per the given condition: |n - 60| < |n - 100| < |n - 20|

Dividing the range of n into 4 segments. (n < 20, 20<n<60, 60<n<100, n > 100)

1) For n < 20.

|n-20| = 20-n, |n-60| = 60- n, |n-100| = 100-n

considering the inequality part: |n - 100| < n - 20|

100 -n < 20 -n,

No value of n satisfies this condition.

2) For 20 < n < 60.

|n-20| = n-20, |n-60| = 60- n, |n-100| = 100-n.

60- n < 100 – n and 100 – n < n – 20

For 100 -n < n – 20.

120 < 2n and n > 60. But for the considered range n is less than 60.

3) For 60 < n < 100

|n-20| = n-20, |n-60| = n-60, |n-100| = 100-n

n-60 < 100-n and 100-n < n-20.

For the first part 2n < 160 and for the second part 120 < 2n.

n takes values from 61 …………….79.

A total of 19 values

4) For n > 100

|n-20| = n-20, |n-60| = n-60, |n-100| = n-100

n-60 < n – 100.

No value of n in the given range satisfies the given inequality.

Hence a total of 19 values satisfy the inequality.

2 ≤ |x – 1| × |y + 3| ≤ 5
The product of two positive number lies between 2 and 5.
As x is a negative integer, the minimum value of |x – 1| will be 2 and the maximum value of |x – 1| will be 5 as per the question.

When, |x – 1| = 2, |y + 3| can be either 1 or 2
So, for x =  -1, y can be – 4 or – 2 or – 5 or -1.
Thus, we get 4 pairs of (x, y)

When |x – 1| = 3, |y + 3| can be 1 only
So, for x = – 2, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x – 1| = 4, |y + 3| can be 1 only
So, for x = – 3, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

When |x – 1| = 5, |y + 3| can be 1 only
So, for x = – 4, y can be -4 or -2
Thus, we get 2 pairs of the values of (x, y)

Therefore, we get a total of 10 pairs of the values of (x, y)
Hence, option A is the correct answer.

Cos x – Sin x = √2 Sin x
⇒ Cos x = Sin x + √2 Sin x



⇒ Sin x = (√2 - 1) Cos x
⇒ Sin x = √2 Cos x - Cos x
⇒ Sin x + Cos x = √2 Cos x

Cos x – Sin x = √2 Sin x 

=> Cos x = Sin x + √2 Sin x 
=> Cos x = Sin x + √2 Sin x 
=> Sin x = Cosx/(√2+1) * Cos x 
=> Sin x = (√2−1)/(√2−1) * 1/(√2+1) * Cos x
=> Sin x = (√2−1)/((√2)²−(1)²)* Cos x
=> Sin x = (√2 - 1) Cos x
=> Sin x = √2 Cos x – Cos x
=> Sin x + Cos x = √2 Cos x

Hence, the correct answer is Option A.

Given numbers are 1.08 , 0.36 and 0.90
G.C.D. i.e. H.C.F of 108, 36 and 90 is 18
Therefore, H.C.F of given numbers = 0.18            

Sales tax = 150 / 5 = 30
∵ SP contains Rs. 30 component of sales tax.
Of the remainder (150 – 30 = 120) 1/3rd is the profit.
Thus, the profit = 120 / 3 = 40
Hence, Cost price = 120 – 40 = 80

So option B is correct

S.P. at 10% profit = Rs.(110y/100) = 11y/10
New C.P. of article = 85y/100 = 17y/20
S.P. = Rs.(17y/20 * 110/100)
New S.P. of article = Rs. 187/200

According to Question,
11y/10 - 187y/200 = 33
33y/200 = 33
y = 200

So option B is correct

S.I. for 1 year = Rs. (854 - 815) = Rs. 39.

S.I. for 3 years = Rs.(39 * 3) = Rs. 117.

 Principal = Rs. (815 - 117) = Rs. 698.

We can get SI of 3 years = 12005 - 9800 = 2205

SI for 5 years = (2205/3)*5 = 3675 [so that we can get principal amount after deducting SI]

Principal = 9800 - 3675 = 6125 

So Rate = (100*3675)/(6125*5) = 12%

The correct option is A 

• Selected year will be a non leap year with a probability 3/4
• Selected year will be a leap year with a probability 1/4
• A selected leap year will have 53 Mondays with probability 2/7
• A selected non leap year will have 53 Mondays with probability 1/7
• E→ Event that randomly selected year contains 53 Mondays

P(E) =  (3/4 × 1/7) + (1/4 × 2/7)
P(Leap Year/ E) = (2/28) / (5/28) = 2/5 

Total number of balls = 4 + 5 + 6 = 15

Let S be the sample space.

• n(S) = Total number of ways of drawing 3 balls out of 15 = ¹⁵C₃

Let E = Event of drawing 3 balls, all of them are yellow.

• n(E) = Number of ways of drawing 3 balls from the total 5 = ⁵C₃
  (∵ there are 5 yellow balls in the total balls)

[∵ ⁿC_r = ⁿC_(n-r). So ⁵C₃ = ⁵C₂. Applying this for the ease of calculation]

log₁₀ 80 = log₁₀ (8 x 10)
⇒ log₁₀ 8 + log₁₀ 10
⇒ log₁₀(2³) + 1
⇒ 3 log₁₀ 2 + 1
⇒ (3 x 0.3010) + 1
⇒ 1.9030

Explanation:

The quadratic equation whose roots are reciprocal of 2x² + 5x + 3 = 0 can be obtained by replacing x by 1/x.
Hence, 2(1/x)² + 5(1/x) + 3 = 0
=> 3x² + 5x + 2 = 0

Explanation:

I. (a + 4)² = 0 => a = -4
II.(b - 3)(b - 1) = 0
=> b = 1, 3 => a < b