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Q1. At IIM Bangalore, 60% of
the students are boys and the rest are girls. Further 15% of the boys and 7.5%
of the girls are getting a fee waiver. If the number of those getting a fee
waiver is 900, find the total number of students getting 50% concession, if it
is given that 50% of those not getting a fee waiver are eligible to get half
fee concession?

(a) 3600

(b) 2800

(c) 3200

(d) 3300

(e) None of these

Q2. A person bought some
candies and sold 80% of them at the price he paid for total candies. He sold
the remaining candies at 10% profit. If the total profit is 1386 Rs. then find
the cost price of the all candies.

(a) 6300

(b) 6100

(c) 5700

(d) 6500

(e) 7500

Q3. S₁ is a series of five
consecutive multiple of three, whose sum is 180 and S₂ is the series of four
consecutive multiple of four whose second smallest number is 13 more than
second highest number of S₁ series. Find the average of smallest number of S₁
series and highest number of S₂ series.

(a) 51

(b) 49

(c) 47

(d) 45

(e) 43

Q4. ‘2n’ years ago ratio of
Amit’s age to Inder’s age is 5 : 4. ‘n’ years ago ratio of Inder’s age to
Satish’s age is 9 : 7. Difference between present age of Amit to Satish’s
present age is 12 years. Find the sum of present ages of all three if ratio of
Amit’s age to Satish’s age after ‘n’ year will be 13 : 9.

(a) 81

(b) 84

(c) 87

(d) 90

(e) Cannot be determined

Q5. A student finds the average of five two digits
numbers. If One number is reversed and the average is taken again then the
average increase by 5.4. If all five digits are consecutive multiple of four,
then find the number which is reversed?

(a) 58

(b) 36

(c) 74

(d) 48

(e) None of these

**Directions (6-10):**In each of these questions, two equations (I) and (II) are given. You have to solve both the equations and give answer,

(a)
if x<y

(b)
if x≤y

(c)
if x = y, or no relation can be established between x and y

(d)
if x>y

(e)
if x≥𝑦

Q6.
I. 9x

^{2}= 1
II.
4y

^{2}+11y−3 = 0
Q7.
I. x

^{2}−5 =0
II.
4y

^{2}−24y+35 = 0
Q8.
I. x

^{2}−5x−14=0
II. y

^{2 }+7y+10 = 0
Q9.
I. 5x+7y=−43

II.
9x−17y=41

Q10.
I. 2x

^{2}− (4+√13)x+2√13=0
II. 10y

^{2}− (18+5√13)y+9√13=0**Solutions**

S1. Ans.(d)

Sol. Let total students are x 15100×60100x+15200×40100𝑥=900 9100x+3x100=900
12x=900×100 x=7500

Total students getting fee
waiver =675+225=900

Required no. of students =(7500−900)/2=3300

S2. Ans.(a)

Sol.

Let total candies be 100 and
price per candy be x

∴ total C.P = 100x

Total S.P = 100𝑥+110/100 ×20𝑥

ATQ, 1386=(100𝑥+22𝑥)−100𝑥

∴ required
total c.p = 100× 1386/22=6300

S3. Ans.(d)

Sol.

Let S₁ is a series consists 3x
– 6, 3x – 3, 3x, 3x + 3, 3x + 6

ATQ,

3x – 6 + 3x – 3 + 3x + 3x + 3 +
3x + 6 = 180

⇒ x = 12

S₁ series = 30, 33, 36, 39, 42

Second smallest no. of S₂ = 39
+ 13 = 52

S₂ series = 48, 52, 56, 60

Required average= (30+60)/2=90/2=45

S4. Ans.(d)

Sol.

Let, Present age of Amit, Inder
and Satish be x, y and z respectively.

ATQ,

(x –2n)/(y –2n) = 5/4

⇒ 4x
– 8n = 5y – 10n

⇒ 5y
– 4x = 2n …(i)

(y –n)/(z –n)= 9/7

⇒ 7y – 7n = 9z – 9n

⇒ 2n = 9z – 7y …(ii)

Equating (i) & (ii)

5y – 4x = 9z – 7y

12y = 9z + 4x

And, x – z = 12

(x+n)/(z+n)= 13/9

⇒ 9x + 9n = 13z + 13n

⇒ 4n = 9x – 13z

9x – 13z = 18z – 14y

9x + 14y = 31z

Now x = 12 + z

9 (12 + z) + 14y = 31z

⇒ 22z – 14y = 108

or 11z – 7y = 54 …(iii)

and 10y – 8x = 9x – 13z

17x = 10y + 13z

17 (12 + z) = 10y + 13z

204 + 17z = 10y + 13z

4z + 204 = 10y …(iv)

On solving (iii) and
(iv)

y = 30, z = 24, x =36

Required sum = 30 + 24 +
36 = 90

S5. Ans.(b)

Sol.

Let the number which is
reversed is “10a+b”

If it is reversed then
the number becomes “10b+a”

ATQ,

10𝑏+𝑎−10𝑎−𝑏=5×5.4

⇒9𝑏−9𝑎=27

⇒𝑏−𝑎=3

Numbers can be 14, 25, 36, 47,
58 and 69 but all the five digits are multiple of 4

⇒ The number should be 36

S6.
Ans.(c)

Sol.
𝑥=1/3 −1/3

4𝑦

^{2}+12𝑦−𝑦−3=0
4𝑦(𝑦+3)−1(𝑦+3)=0

𝑦=14,−3

So no relation can be established

S7.
Ans.(a)

Sol.
𝑥=√5,−√5

4𝑦

^{2}−14𝑦−10𝑦+35=0
2𝑦(2𝑦−7)−5(2𝑦−7)=0

𝑦=5/2, 7/2

𝑥<𝑦

S8. Ans.(e)

Sol. 𝑥

^{2}−7𝑥+2𝑥−14=0
𝑥(𝑥−7)+2(𝑥−7)=0

𝑥=7,−2

𝑦

^{2}+5𝑦+2𝑦+10=0
𝑦=−2,−5

𝑥≥y

S9. Ans.(d)

Sol. Solving these equations.

𝑥=−3,𝑦=−4

𝑥>𝑦

S10. Ans.(e)

Sol. 2𝑥

^{2}−4𝑥−√13𝑥+2√13=0
2𝑥(𝑥−2)−√13(𝑥−2)=0

𝑥=√13/2,2

10𝑦

^{2}−18𝑦−5√13𝑦+9√13=0
2𝑦(5𝑦−9)−√13(5𝑦−9)=0

𝑦=√13/2,95

𝑥≥𝑦