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
\(\sqrt{2} \cos \theta\)
\(\cfrac{(\sin \theta+\cos \theta)}{\sqrt{2}}\)
\(\sqrt{2} \sin \theta\)
Option: 4 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
Option: 3 Explanation:
A1 = 3
A2 = 3 + 7
A3 = 3+7+11
And so on, A25 = 3+7+11 +….+ 99 (since general term of the series is 4n-1)
Let N = A1 + A2 + …..A25 = 25(3) + 24(7) + 23(11) + …. 1(99)
The general term of the series = (26-n)(4n-1) = 105n – 4n2 – 26
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
Option: 24 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
Now, ab = 1 if (b = 0) or (a = 1) or (a = -1 and b is even)
For, x2 – 3x – 10 = 0
x = 5 and -2 (two solutions)
For, x2 – 10 = 1
No integral solutions
For, x2 – 10 = -1
x2 = 9 => x = 3 or -3
for x= 3 or -3, x2 – 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
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
Option: 4 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 = 602
4x2 + 4x2 + 48x +144 = 3600
8x2 + 48x – 3456 = 0
x2 + 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
Option: 2 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
Option: 20 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
Option: 14 Explanation:
Increase in total = 6a1 – a1 = 5a1
Increase in average = 100
So, 5a1 = 100n
a1 = 20n
Since the series is non decreasing so a1 < Average = 300
And since increasing a1 increased the total so a1 > 0
So, 0 < 20n < 300
So, n can take 14 values and hence a1 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
Option: 2 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
Option: 6 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
Option: 1 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
2cr2 +18 = 0
I – II gives
10r3 – 20 r = 0
r(10r2 – 20) = 0
r = 0 or r2 = 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
Option: 2 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
Option: 47 Explanation:
(4 – log2n)/(3 – log4n) < 0
So, Case 1: (4 – log2n)<0
And, (3 – log4n) > 0
log2n > 4
And, log4n <3
n > 24 and n < 43
n can take values from 17 to 63
47 values
Case 2: (4 – log2n)>0
And, (3 – log4n) < 0
log2n < 4
And, log4n >3
n < 24 and n > 43
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
Option: 2 Explanation:
Total saving required = 550*12 = 6600
Total expenses = 9*3500 + 3*3700 = 42600
So, total earning = 42600+6600 = 49200
If earning from 10th 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
Option: 2 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
Option: 4 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
Option: 4 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
\(\sqrt{7}\)
\(\sqrt{8}\)
\(\sqrt{6}\)
\(\sqrt{5}\)
Option: 1 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 = (AC2 + CD2 – AD2)/2.AC.CD
Cos 60 = (32 + 12 – AD2)/2*3*1
½ = (10 – AD2)/6
AD2 = 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
Option: 1 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
Option: 4 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
Option: 10 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
