**Question 1:**

In triangle \(A B C\) , altitudes AD and BE are drawn to the corresponding bases. If \(\angle B A C=45^{\circ}\) and \(\angle A B C=\theta\), then \(\frac{A D}{B E} \text { equals }\)

1

\(\sqrt{2} \cos \theta\)

\(\cfrac{(\sin \theta+\cos \theta)}{\sqrt{2}}\)

\(\sqrt{2} \sin \theta\)

**Explanation:**

Since, Δ BEA is right angled and ∠ BAC = 450

So, Δ BEA is isosceles

AE = BE and AB = √2 BE

In Δ ADC, sin θ = AD/AB = AD/√2 BE

So, AD/BE = √2 sin θ

**Question 2:**

Consider the arithmetic progression 3, 7, 11, ... and let \(A_{n}\) denote the sum of the first n terms of this progression. Then the value of \(\frac{1}{25} \sum_{n=1}^{25} A_n\) is

404

415

455

442

**Explanation:**

A_{1} = 3

A_{2} = 3 + 7

A_{3} = 3+7+11

And so on, A_{25} = 3+7+11 +….+ 99 (since general term of the series is 4n-1)

Let N = A_{1} + A_{2} + …..A_{25} = 25(3) + 24(7) + 23(11) + …. 1(99)

The general term of the series = (26-n)(4n-1) = 105n – 4n^{2} – 26

Thus, ∑N = ∑(105n – 4n^{2} – 26) = 105(n(n+1)/2 – 4(n(n+1)(2n+1)/6 – 26n

Thus required sum = 1/25 * ( 105*25*26)/2 – 4*25*26*51/6 – 26*25)

= 105*13 – 2*26*17 – 26 = 455

**Question 3:**

In an examination, there were 75 questions. 3 marks were awarded for each correct answer, 1 mark was deducted for each wrong answer and 1 mark was awarded for each unattempted question. Rayan scored a total of 97 marks in the examination. If the number of unattempted questions was higher than the number of attempted questions, then the maximum number of correct answers that Rayan could have given in the examination is

**Explanation:**

Let number of questions correct be x, wrong be y and unattempted be z

So, x+y+z = 75 --------I

3x – y + z = 97 --------II

II + I will give

4x + 2z = 172

2x + z = 86

Now, z > x+y so min z = 38

So, max x = (86-38)/2 = 24

**Question 4:**

The number of integer solutions of the equation \(\left(x^2-10\right)^{\left(x^2-3 x-10\right)}=1\) is

**Explanation:**

\(\left(x^2-10\right)^{\left(x^2-3 x-10\right)}=1\)

Now, a^{b} = 1 if (b = 0) or (a = 1) or (a = -1 and b is even)

For, x^{2} – 3x – 10 = 0

x = 5 and -2 (two solutions)

For, x^{2} – 10 = 1

No integral solutions

For, x^{2} – 10 = -1

x^{2} = 9 => x = 3 or -3

for x= 3 or -3, x^{2} – 3x – 10 is even

Thus four integral solutions, -3, -2, 3 and 5

**Question 5:**

Suppose for all integers \(x\), there are two functions \(f\) and \(g\) such that \(f(x)+f(x-1)-1=0)\) and \(g(x)=x^2 \cdot f\left(x^2-x\right)=5\), then the value of the sum \(f(g(5))+g(f(5))\) is

**Explanation:**

f(x^{2} – x) = 5

For x = 1, f(0) = 5

Also,

f(x) + f(x-1) – 1 = 0

f(x) + f(x-1) = 1

For, x =1

f(1) + f(0) = 1

f(1) = -4

for, x = 2

f(2) + f(1) = 1

f(2) = 5

thus, f(even) = 5 and f(odd) = -4

So, f(g(5)) + g(f(5)) = f(5^{2}) + g(-4) = f(25) + g(-4) = -4+16 = 12

**Question 6:**

Two ships meet mid-ocean, and then, one ship goes south and the other ship goes west, both travelling at constant speeds. Two hours later, they are 60 km apart. If the speed of one of the ships is 6 km per hour more than the other one, then the speed, in km per hour, of the slower ship is

20

24

12

18

**Explanation:**

If speed of the slower ship is x km/hr, the faster one has speed x+6

In two hours distance travelled will be 2x and (2x+12)

The starting point and the ships’ final points will form a right angled triangle

So,

(2x)^{2} + (2x+12)^{2} = 60^{2}

4x^{2} + 4x^{2} + 48x +144 = 3600

8x^{2} + 48x – 3456 = 0

x^{2} + 6x – 432 = 0

(x+24)(x-18) = 0

x = 18

**Question 7:**

The number of integers greater than 2000 that can be formed with the digits 0, 1, 2, 3, 4, 5, using each digit at most once, is

1200

1440

1420

1480

**Explanation:**

Three cases possible

Case 1: 4 digit numbers starting with 2,3,4 or 5 without repetition = 4*5*4*3 = 240

Case 2: 5 digit numbers without repetition = 5*5*4*3*2 = 600

Case 3: 6 digit numbers without repetition = 5*5*4*3*2*1 = 600

Total = 600+600+240 = 1440

**Question 8:**

Mr. Pinto invests one-fifth of his capital at 6%, one-third at 10% and the remaining at 1%, each rate being simple interest per annum. Then, the minimum number of years required for the cumulative interest income from these investments to equal or exceed his initial capital is

**Explanation:**

If total Principal is 15 then 3 is invested at 6%, 5 at 10% and remaining 7 at 1%

SI required is atleast equal to Principal = 15

Let time be n years,

15 = 3*6*n/100 + 5*10*n/100 + 7*1*n/100

15 = 0.75n

n = 20

**Question 9:**

The average of a non-decreasing sequence of \(N\) numbers \(a_1, a_2, \ldots, a_N\) is 300 . If \(a_1\) is replaced by \(6 a_1\); the new average becomes 400 . Then, the number of possible values of \(a_1\) is

**Explanation:**

Increase in total = 6a_{1} – a_{1} = 5a_{1}

Increase in average = 100

So, 5a_{1} = 100n

a_{1 }= 20n

Since the series is non decreasing so a_{1} < Average = 300

And since increasing a_{1} increased the total so a_{1} > 0

So, 0 < 20n < 300

So, n can take 14 values and hence a_{1 }can take 14 values

**Question 10:**

If a and b are non-negative real numbers such that a + 2b = 6, then the average of the maximum and minimum possible values of (a + b) is

3.5

4.5

4

3

**Explanation:**

a+2b = 6

Max a + b when a is max and b is min so a = 6 and b = 0 => a+b = 6

Min a+b when a is min and b is max so a = 0 and b = 3 => a+b = 3

Required average = (3+6)/2 = 4.5

**Question 11:**

Working alone, the times taken by Anu, Tanu and Manu to complete any job are in the ratio 5 : 8 : 10. They accept a job which they can finish in 4 days if they all work together for 8 hours per day. However, Anu and Tanu work together for the first 6 days, working 6 hours 40 minutes per day. Then, the number of hours that Manu will take to complete the remaining job working alone is

**Explanation:**

Since time taken to finish work is in the ratio 5:8:10

So efficiency will be in the ratio 40/5:40/8:40/10 = 8:5:4

So, Anu does 8 units per hour, Tanu does 5 and Manu does 4.

Total work = 4*8*(8+5+4) = 544

Work done by Anu and Tanu = 6*6.67*(8+5) = 520

Work remaining = 544 – 520 = 24

Time taken by Manu to complete = 24/4 = 6 hours

**Question 12:**

Let \(r\) and \(c\) be real numbers. If \(r\) and \(-r\) are roots of \(5 x^3+c x^2-10 x+9=0\), then \(c\) equals

-9/2

4

-4

9/2

**Explanation:**

Since r and -r are roots therefore

5(r)^{3} + c(r)^{2} – 10(r) + 9 = 0 ------I

5(-r)^{3} + c(-r)^{2} – 10(-r) + 9 = 0 ------II

I + II gives

2cr^{2} +18 = 0

I – II gives

10r^{3} – 20 r = 0

r(10r^{2} – 20) = 0

- r = 0 or r
^{2}= 20/10 = 2

Thus 2c(2) + 18 = 0

4c = -18

c = -9/2

**Question 13:**

In an election, there were four candidates and 80% of the registered voters casted their votes. One of the candidates received 30% of the casted votes while the other three candidates received the remaining casted votes in the proportion 1 : 2 : 3. If the winner of the election received 2512 votes more than the candidate with the second highest votes, then the number of registered voters was

50240

62800

60288

40192

**Explanation:**

Let registered votes be 600x

Casted votes = 80% of 600x = 480x

One candidate gets 30% of 480x = 144x

Remaining = 480x – 144x = 336x

This is divided in the ratio 1:2:3 = 56x:112x:168x

So the diff between winner and second highest votes = 168x – 144x = 24x

Given, 24x = 2512

So, 6x = 628

So, 600x = 62800 total voters

**Question 14:**

The number of distinct integer values of n satisfying: \(\frac{4-\log _2 n}{3-\log _4 n}<0\) , is

**Explanation:**

(4 – log_{2}n)/(3 – log_{4}n) < 0

So, Case 1: (4 – log_{2}n)<0

And, (3 – log_{4}n) > 0

- log
_{2}n > 4

And, log_{4}n <3

- n > 2
^{4}and n < 4^{3} - n can take values from 17 to 63
- 47 values

Case 2: (4 – log_{2}n)>0

And, (3 – log_{4}n) < 0

- log
_{2}n < 4

And, log_{4}n >3

- n < 2
^{4}and n > 4^{3} - This is not possible

So total 47 solutions

**Question 15:**

Manu earns ₹4000 per month and wants to save an average of ₹550 per month in a year. In the first nine months, his monthly expense was ₹3500, and he foresees that, tenth month onward, his monthly expense will increase to ₹3700. In order to meet his yearly savings target, his monthly earnings, in rupees, from the tenth month onward should be

4350

4400

4300

4200

**Explanation:**

Total saving required = 550*12 = 6600

Total expenses = 9*3500 + 3*3700 = 42600

So, total earning = 42600+6600 = 49200

If earning from 10^{th} month = x

Then

9*4000 + 3x = 49200

x = 4400

**Question 16:**

On day one, there are 100 particles in a laboratory experiment. On day n, where \(n \geq 2\), one out of every n particles produces another particle. If the total number of particles in the laboratory experiment increases to 1000 on day m, then m equals

16

19

17

18

**Explanation:**

On day 2, 1 out of every 2 will produce 1 more particle, so increase of 50 and total becomes 150.

On day 3, 1 out of every 3, so 50 out of 150 will produce 1 more article, so increase of 50 and total becomes 200

This process will keep repeating and increase of 50 will happen every day.

Total increase = 1000 – 100 = 900

So number of days required after day 1 = 900/50 = 18

Thus, m = 18+1 = 19

**Question 17:**

For some natural number n, assume that (15,000)! is divisible by (n!)!. The largest possible value of n is

4

5

6

7

**Explanation:**

We know that 8! = 40320

Since 40320 > 15000

So, (8!)! cannot divide 15000!

So largest value of n = 7

**Question 18:**

There are two containers of the same volume, first container half-filled with sugar syrup and the second container half-filled with milk. Half the content of the first container is transferred to the second container, and then the half of this mixture is transferred back to the first container. Next, half the content of the first container is transferred back to the second container. Then the ratio of sugar syrup and milk in the second container is

4 : 5

6 : 5

5 : 4

5 : 6

**Explanation:**

Assuming volume of each container is 200 ltrs.

So, Container 1 has 100 L of sugar syrup and Container 2 has 100 Ltrs of milk

After first transfer,

Container 1 has 50 L of sugar syrup and Container 2 has 100 L Milk and 50 L sugar syrup

After second transfer,

Half of the second container i.e 50 L milk and 25 L sugar will get transferred to container 1

So, Container 1 has 75 L sugar syrup and 50 L milk and Conatiner 2 has 50L milk and 25 L sugar syrup.

After third transfer,

Half of the first container i.e 37.5 L sugar syrup and 25 L milk goes to second container

So, second container has 75 L milk and 62.5 Ltrs sugar syrup

Required ratio = 62.5:75 = 5:6

**Question 19:**

The length of each side of an equilateral triangle ABC is 3 cm. Let D be a point on BC such that the area of triangle ADC is half the area of triangle ABD. Then the length of AD. in cm, is

\(\sqrt{7}\)

\(\sqrt{8}\)

\(\sqrt{6}\)

\(\sqrt{5}\)

**Explanation:**

ar Δ ADC = ½ ar Δ ABD

So, ar Δ ADC = 1/3 ar Δ ABC

Since height of ΔADC and ΔABC is same

So, CD = 1/3 of BC = 1/3 of 3 = 1

In ΔADC,

Cos C = (AC^{2} + CD^{2} – AD^{2})/2.AC.CD

Cos 60 = (3^{2} + 1^{2} – AD^{2})/2*3*1

½ = (10 – AD^{2})/6

AD^{2} = 7

AD = √7

**Question 20:**

Five students, including Amit, appear for an examination in which possible marks are integers between 0 and 50, both inclusive. The average marks for all the students is 38 and exactly three students got more than 32. If no two students got the same marks and Amit got the least marks among the five students, then the difference between the highest and lowest possible marks of Amit is

20

22

21

24

**Explanation:**

Total score of 5 students = 38*5 = 190

Amit will have min score when other have max score

So, Amit’s min score = 190 – 50 – 49- 48 – 32 = 11

Amit will have max score, when others have min score

This happens when 3 students having score more than 32 have a total of 190 – 32 -31 and Amit has score 31.

So, Required difference = 31 -11 = 20

**Question 21:**

Let \(f(x)\) be a quadratic polynomial in \(x\) such that \(f(x) \geq 0\) for all real numbers \(x\). If \(f(2)=0\) and \(f(4)=6\), then \(f(-2)\) is equal to

36

12

6

24

**Explanation:**

Since, f(x) ≥ 0

And f(2) = 0

So 2 will be the only real root for f(x)

So, f(x) = a(x-2)^{2}

f(4) = a(4-2)^{2}

6 = a(4)

a = 3/2

So, f(-2) = 3/2 * (-2-2)^{2}

= 3/2 * (16) = 24

**Question 22:**

Regular polygons A and B have number of sides in the ratio 1 : 2 and interior angles in the ratio 3 : 4. Then the number of sides of B equals

**Explanation:**

Let the number of sides be x and 2x

Interior angle of A = 180 – (360/x) = (180x – 360)/x

Interior angle of B = 180 – (360/2x) = (180x – 180)/x

Given, (180x – 360)/(180x-180) = 3/4

720x – 1440 = 540x – 540

180x = 900

x = 5

So, B has 2x = 10 sides