**Question 1:**

Let a and b be natural numbers. If a^{2} + ab + a = 14 and b^{2} + ab + b = 28, then (2a + b) equals

10

7

8

9

**Explanation:**

a^{2} + ab + a = 14 --- Eq 1

b^{2} + ab + b = 28 ------Eq 2

Eq 2 – Eq 1

b^{2 }- a^{2} + b – a = 14

(b-a)(b+a) + (b-a) = 14

(b-a)(b+a+1) = 14

Now, 14 = 1×14 or 2×7

But a < 4 since a^{2} < 14

And b < 5 since b^{2} + b < 28

So, a+b+1 < 4+5+1 = 10

So, a+b+1 = 7 and b-a = 2

Solving, b = 4 and a = 2

So, 2a+b = 2(2) + 4 = 8

**Question 2:**

For any real number x, let [x] be the largest Integer less than or equal to x. If \(\sum_{n=1}^N\left[\frac{1}{5}+\frac{n}{25}\right]=25\) then N is

**Explanation:**

1/5 + n/25 = 1

When, n = 20

So, for all values of n = 1 to 19, [1/5 + n/25] = 0

When, 1/5 +n/25 = 2

When, n = 45

So, for n = 20 to 44 i.e. 25 values [1/5 + n/25] = 1

So, N = 4

**Question 3:**

Trains A and B start traveling at the same time towards each other with constant speeds from stations X and Y, respectively. Train A reaches station Y in 10 minutes while train B takes 9 minutes to reach station X after meeting train A. Then the total time taken, in minutes, by train B to travel from station Y to station X is

6

10

15

12

**Explanation:**

Assuming that the two trains meet at Z

Now distance XZ is covered by B in 9 minutes and say it is covered by A in x minutes

Ratio of speed of A and B = x/9

Also, B will take x minutes to cover YZ since the two will travel the same time to reach the meeting point.

So, Ratio of speed of A and B = 10/(x+9)

Now, 10/(x+9) = x/9

90 = x^{2} + 9x

x^{2} + 9x – 90 = 0

(x+15)(x-6) = 0

So, x = 6

B takes 9+x = 15 minutes

**Question 4:**

A trapezium \(A B C D\) has side \(A D\) parallel to \(B C, \angle B A D=90^{\circ}, B C=3 \mathrm{~cm}\) and \(A D=8 \mathrm{~cm}\). If the perimeter of this trapezium is \(36 \mathrm{~cm}\), then its area, in sq. \(\mathrm{cm}\), is

**Explanation:**

Drop perpendicular from C to AD, intersecting AD at E.

CE II AB and CE = AB

If AB = x, CE = x and in triangle CED, CD = √(CE^{2} + ED^{2}) = √(x^{2} + 25)

Now, 8+3+x+ √(x^{2} + 25) = 36

√(x^{2} + 25) = 25 – x

(x^{2} + 25) = 625 + x^{2} – 50x

x = 12

Area = ½ (3+8)×12 = 66

**Question 5:**

For natural numbers x,y, and z, if xy + yz = 19 and yz + xz = 51, then the minimum possible value of xyz is

**Explanation:**

xy + yz = 19

y(x+z) = 19

since, x y z are natural numbers so, y = 1 and x+z = 19

Also, yz + zx = 51

So, z + zx = 51

z(1+x) = 51

z(1+19-z) = 51

z^{2} – 20z + 51 = 0

(z-3)(z-17) = 0

So, z = 17 => x = 2

Or, z = 3 => x = 16

So, Minimum xyz = 2*1*17 = 34

**Question 6:**

Let a, b, c be non-zero real numbers such that b^{2} < 4ac, and f(x) = ax^{2} + bx + c. If the set S consists of all integers m such that f(m) < 0, then the set S must necessarily be

the empty set

the set of all positive integers

the set of all integers

either the empty set or the set of all integers

**Explanation:**

**Question 7:**

A mixture contains lemon juice and sugar syrup in equal proportion. If a new mixture is created by adding this mixture and sugar syrup in the ratio 1 : 3, then the ratio of lemon juice and sugar syrup in the new mixture is

1 : 4

1 : 5

1 : 6

1 : 7

**Explanation:**

New Mixture has old mixture and sugar syrup in ratio 1:3 = 2:6

Out of this 2 of old mixture, lemon juice and sugar syrup are equal i.e. 1:1

So Total sugar syrup = 1+6 = 7 and lemon juice = 1

Reqd ratio = 1:7

**Question 8:**

The average of three integers is 13. When a natural number n is included, the average of these four integers remains an odd integer. The minimum possible value of n is

3

1

4

5

**Explanation:**

Sum of the three integers = 13*3 = 39

New total = 4x where x is odd

Smallest number greater than 39 of the for 4x where x is odd = 44

So fourth integer = 44 – 39 = 5

**Question 9:**

Amal buys 110 kg of syrup and 120 kg of juice, syrup being 20% less costly than juice, per kg. He sells 10 kg of syrup at 10% profit and 20 kg of juice at 20% profit. Mixing the remaining juice and syrup, Amal sells the mixture at ₹ 308.32 per kg and makes an overall profit of 64%. Then, Amal's cost price for syrup, in rupees per kg, is

**Explanation:**

If CP of juice per Kg = x

CP of syrup = 0.8x

So Total CP = 110(0.8x) + 120x = 208x

Now for 10 Kg syrup, SP = 1.1(0.8x) = .88x

For 20 Kg juice, SP = 1.2x

And for remaining 200 kgs, SP = 308.32

So, total SP = 10(0.88x) + 20(1.2x) + 200(308.32)

= 8.8x + 24x + 61664

So, 208x(1.64) = 32.4x + 61664

x = 200

So, Price of syrup = 0.8(200) = 160

**Question 10:**

For any natural number tt, suppose the sum of the first tt terms of an arithmetic progression is (n + 2n^{2}). If the n^{th} term of the progression is divisible by 9, then the smallest possible value of n is

4

7

9

8

**Explanation:**

Sum of n terms = n + 2n^{2}

Sum of n-1 terms = (n-1) + 2(n-1)^{2}

So, n^{th} term = n + 2n^{2} – ((n-1) + 2(n-1)^{2})

= n + 2n^{2} – n + 1 -2n^{2} +4n -2

= 4n -1

For this to be divisible by 9, smallest value of n (from options) = 7

**Question 11:**

Ankita buys 4 kg cashews, 14 kg peanuts and 6 kg almonds when the cost of 7 kg cashews is the same as that of 30 kg peanuts or 9 kg almonds. She mixes all the three nuts and marks a price for the mixture in order to make a profit of ₹1752. She sells 4 kg of the mixture at this marked price and the remaining at a 20% discount on the marked price, thus making a total profit of ₹744. Then the amount, in rupees, that she had spent in buying almonds is

1176

2520

1680

1440

**Explanation:**

Let CP of cashews, peanuts and almonds be Rs x, y, and z per kg respectively

Then, 7a = 30b = 9c = 630x

So, a = 90x, b = 21x, c = 70x

Now If MP/kg was y, then

(4+14+6)y – (4+14+6-4)(0.8y) – 4y = 1752 – 744

4y = 1008

y = 252

So, 4*90x + 14*21x + 6*70x + 1752 = (4+14+6)*252

x = 4

So, Cost of almonds = 6*70*4 = 1680

**Question 12:**

The average weight of students in a class increases by 600 gm when some new students join the class. If the average weight of the new students is 3 kg more than the average weight of the original students, then the ratio of the number of original students to the number of new students is

1 : 2

4 : 1

1 : 4

3 : 1

**Explanation:**

So Ratio of students = 2.4:0.6 = 4:1

**Question 13:**

The number of ways of distributing 20 identical balloons among 4 children such that each child gets some balloons but no child gets an odd number of balloons, is

**Explanation:**

If the students get 2a, 2b, 2c and 2d balloons each where a, b, c, d are natural numbers

Then,

2a + 2b + 2c + 2d =20

a+b+c+d = 10

Number of natural number solutions = ^{10-1}C_{4-1} = ^{9}C_{3} = 84

**Question 14:**

The largest real value of a for which the equation |x + a| + |x - 1| = 2 has an infinite number of solutions for x is

0

2

1

-1

**Explanation:**

The function |x+a| + |x-1| will have a constant value from -a to 1. For all other values of the function there will be exactly two solutions.

So, putting x = 1

|1+a| + |1-1| = 2

So, |1+a| = 2

- a = 1 or -3
- Largest value of a = 1

**Question 15:**

All the vertices of a rectangle lie on a circle of radius R. If the perimeter of the rectangle is P, then the area of the rectangle is

\(\cfrac{P^2}{16}-R^2\)

\(\cfrac{P^2}{2}-2 P R\)

\(\cfrac{P^2}{8}-\cfrac{R^2}{2}\)

\(\cfrac{P^2}{8}-2 R^2\)

**Explanation:**

If the length and breadth of the rectangle are l and b respectively,

Then, l^{2} + b^{2} = (2R)^{2}

And, 2(l+b) = P

Squaring

4(l^{2} + b^{2 }+ 2lb) = P^{2}

4R^{2} + 2lb = P^{2}/4

lb = P^{2}/8 – 2R^{2}

**Question 16:**

Let ABCD be a parallelogram such that the coordinates of its three vertices A, B, C are (1, 1), (3, 4) and (-2, 8), respectively. Then, the coordinates of the vertex D are

(-3, 4)

(0, 11)

(4, 5)

(-4, 5)

**Explanation:**

Diagonals of a parallelogram bisect each other.

So, intersection point of diagonals will be midpoint of AC = ((1-2)/2 , (1+8)/2) = (-1/2, 9/2)

This will also be midpoint of BD, so (x+3)/2 = -1/2 => x = 4

And, (y+4)/2 = 9/2 => y = 5

**Question 17:**

Let A be the largest positive integer that divides all the numbers of the form 3^{k} + 4^{k} + 5^{k} and B be the largest positive integer that divides all the numbers of the form 4^{k} + 3(4^{k}) + 4^{k}^{+2}, where k is any positive integer. Then (A + B) equals

**Explanation:**

3^{k} + 4^{k} + 5^{k} = (4-1)^{k} + 4^{k} + (4+1)^{k} = 4x when x is odd and 4x+2 when x is even

So the largest value that always divided both is 2. So, A = 2

4^{k} + 3(4^{k}) + 4^{k+2} = 4^{k} + 3(4^{k}) + 4^{2}*4^{k} = (1+3+16)(4^{k}) = 20(4^{k})

Minimum value of B = 20*4^{1} = 80

So, A+B = 82

**Question 18:**

Let \(0 \leq a \leq x \leq 100\) and \(f(x)=|x-a|+|x-100|+|x-a-50|\). Then the maximum value of \(f(x)\) becomes 100 when \(a\) is equal to

25

0

100

50

**Explanation:**

Three cases are possible

Case 1:

a+50 < 100

in this case the min value will happen at a+50 and max value will happen at a or 100

at x = a

f(x) = |x-a| + |x-100| + |x-a-50|

f(a) = 0 + 100 – a + 50 = 150 at a = 0

So, this case is not possible since max value is 100

Case 2:

a+50 = 100

a = 50

Here, max will happen at x = a

at x = a = 50

f(x) = |x-a| + |x-100| + |x-a-50|

f(a) = 0 + 50 + 50 = 100

This is possible

Case 3:

a + 50 > 100

Here the max will again happen at x = a and min will happen at x =100

at x = a

f(x) = |x-a| + |x-100| + |x-a-50|

f(a) = 0 + 100 – a + 50 = 150 at a = 0

which is not possible

So, a = 50

**Question 19:**

Pinky is standing in a queue at a ticket counter. Suppose the ratio of the number of persons standing ahead of Pinky to the number of persons standing behind her in the queue is 3 : 5. If the total number of persons in the queue is less than 300, then the maximum possible number of persons standing ahead of Pinky is

**Explanation:**

Number of people in front of Pinky = 3x

Behind Pinky = 5x

Total = 3x+5x+1 = 8x+1 <300

x<299/8

Max x = 37

So max value of 3x = 3*37 = 111

**Question 20:**

In a class of 100 students, 73 like coffee, 80 like tea and 52 like lemonade. It may be possible that some students do not like any of these three drinks. Then the difference between the maximum and minimum possible number of students who like all the three drinks is

52

53

48

47

**Explanation:**

If a, b, c, d are the number of students who like exactly one, two, three and zero drinks then

a+b+c+d = 100

a + 2b + 3c = 73+80+52 = 205

b+2c – d = 105

c <105/2

Max c = 52

For min c,

a=d=0 => b+c=100

Min c = 105 – 100 = 5

So, required difference = 52 – 5 = 47

**Question 21:**

Alex invested his savings in two parts. The simple interest earned on the first part at 15% per annum for 4 years is the same as the simple interest earned on the second part at 12% per annum for 3 years. Then, the percentage of his savings invested in the first part is

37.5%

60%

62.5%

40%

**Explanation:**

If investment in each part is x and y

x*15*4/100 = y*12*3/100

5x = 3y

So, if x = 3, y = 5

Required ratio = 3/(3+5) * 100 = 37.5%

**Question 22:**

In a village, the ratio of number of males to females is 5 : 4. The ratio of number of literate males to literate females is 2 : 3. The ratio of the number of illiterate males to illiterate females is 4 : 3. If 3600 males in the village are literate, then the total number of females in the village is

**Explanation:**

Literate males = 3600

Literate females = 3/2 * 3600 = 5400

If illiterate males = 4y => illiterate females = 3y

Then,

(3600+4y)/(5400+3y) = 5/4

14400 + 16y = 27000 + 15y

y = 12600

So, Number of females = 3(12600) + 5400 = 43200