Question 1:
If $p q r = 1$ then $\frac { 1 } { 1 + p + q ^ { - 1 } } + \frac { 1 } { 1 + q + r ^ { - 1 } } + \frac { 1 } { 1 + r + p ^ { - 1 } }$ is equivalent to

[1] p + q + r

[2] $\frac { 1 } { \mathbf { p } + \mathbf { q } + \mathbf { r } }$

[3] 1

[4] $\mathbf { p } ^ { - 1 } + \mathbf { q } ^ { - 1 } + \mathbf { r } ^ { - 1 }$

Option # 3

From the given condition pqr = 1.

Substitute the values of p, q, r at random such as $\mathrm { p } = \frac { 2 } { 3 } , \mathrm { q } = \frac { 3 } { 2 } , \mathrm { r } = 1$

Question 2:
If $\frac { a } { b + c } = \frac { b } { c + a } = \frac { c } { a + b } = r ,$ then $r$ cannot take any other value except

[1] 1/2

[2] –1

[3] 1/2 or –1

[4] –1/2 or –1

Option # 3

$\frac { a } { b + c } = \frac { b } { c + a } = \frac { c } { a + b } = r$

By option, if $r = \frac { 1 } { 2 }$

$\Rightarrow \quad 2 a - b - c = 0$

$2 b - c - a = 0$

$2 c - a - b = 0$

$\Rightarrow \quad 2 ( a + b + c ) - ( a + b + c ) - ( a + b + c ) = 0$

Similarly r = –1 is also satisfied.

Question 3:
A student gets an aggregate of 60% marks in five subject in the ratio 10 : 9 : 8 : 7 : 6. If the passing marks are 50% of the maximum marks and each subject has the same maximum marks, in how many subjects did he pass the examination?

[1] 2

[2] 3

[3] 4

[4] 5

Option # 3

Let his marks be 10, 9, 8, 7 and 6 in the five subjects. Hence, totally he has scored 40 marks. This constitutes only 60% of the total marks. Hence, total marks 40/0.6 = 66.7 or 67 approx. , which is the maximum marks in all 5 subjects.

Since the total marks in each subject is the same, hence maximum marks in each subject will be 67/5 = 13 approx. Out of this 50% is the passing marks . In other words to pass in a subject, he needs to score 6.5 marks. We can see that only in 1 subject, he scored less than this viz. 6. Hence, he passed in 4 subject.

Question 4:
The cost of diamond varies directly as the square of its weight. Once, this diamond broke into four pieces with weights in the ratio 1 : 2 : 3 : 4. When the pieces were sold, the merchant got Rs.70,000 less. Find the original price of the diamond.

[1] Rs.1.4 lakh

[2] Rs.2 lakh

[3] Rs.1 lakh

[4] Rs.2.5 lakh

Option # 3

Let the original weight of the diamond be 10x. Hence, its original price will be k(100x2 ), where k is a constant. The weights of the pieces after breaking are x, 2x, 3x and 4x. Therefore, their prices will be kx2, 4kx2 , 9kx2 and 16kx2 . So the total price of the pieces = (1 + 4 + 9 + 16) kx2 = 30kx2 .

Hence, the difference in the price of the original diamond and its pieces = 100kx 2 – 30kx 2 = 70kx 2 = 70000. Hence, kx2 =  000 and the original price = 100 kx2= 100 × 1000 = 100000 = Rs.1 lakh.

Question 5:
Three friends went for a picnic. First brought five apples and the second brought three. The third friend however brought only Rs.8. What is the share of the first friend?

[1] 8

[2] 7

[3] 1

[4] None of these

Option # 2

The number of apples = 8, so the amount eaten by each of the three is 8/3 apples therefore first friend should be paid for 5 – (8/3) and second friend should be paid for 3–(8/3) apples. They should distribute  the sum of Rs.8 in ratio 7/3 : 1/3, i.e., 7 : 1

Question 6:
Total salary of A, B & C is Rs.350. If they spend 75%, 80% & 56% of their salaries respectively their savings are as 10 : 12 : 33. Find the salary of C?

[1] 80

[2] 150

[3] 180

[4] None of These

Option # 2

A’s saving = 100 – 75 = 25% of his salary. B’s saving = 100 – 80 = 20% of his salary C’s saving = 100 – 56 = 44% of his salary 25/100 of A’s salary : 20/100 of B’s salary : 44/100 of C’s salary = 10 : 12 : 33

or 25 × A’s salary : 20 × B’s salary : 44 × C’s salary = 10 : 12 : 33

or 25 × A’s salary / 20 × B’s salary = 10/12

or A’s salary : B’s salary = 2 : 3,

B’s salary : C’s salary = 4 : 5

Thus A : B = 2 : 3, B : C = 4 : 5 Now making B common we have

A : B = 8 : 12, B : C = 12 : 15, or A : B : C = 8 : 12 : 15

Total salary = 350 Þ A’s salary = 8 / (8 + 12 + 15) × 350 = 80

B’s salary = 12 / (8 + 12 + 15) = 120, and C’s Salary = 150

Question 7:
The ratio of the age of a man and his wife is 4 : 3. After 4 years, this ratio will be 9 : 7. If at the time of the marriage, the ratio was 5:3, then how many years ago they were married?

[1] 12 years

[2] 8 years

[3] 10 years

[4] 15 years

Option # 1

Man's age $= 4 \mathrm { k } ,$ (say)

Wife's age $= 3 \mathrm { k } ,$ (say)

$\quad \frac { 4 \mathrm { k } + 4 } { 3 \mathrm { k } + 4 } = \frac { 9 } { 7 } \Rightarrow \mathrm { k } = 8$

$\therefore$ Man's age $= 32$ years

Wife's age $= 24$ years. Suppose they were married x years ago.

$\quad \frac { 32 - x } { 24 - x } = \frac { 5 } { 3 } \Rightarrow x = 12$

Question 8:
Two full tanks, one shaped like a cylinder and the other like a cone, contain jet fuel. The cylindrical tank  olds 500 litres more than the conical tank. After 200 litres of fuel has been pumped out from each tank  he cylindrical tank contains twice the amount of fuel in the conical tank. How many litres of fuel did the  cylindrical tank have when it was full?

[1] 700

[2] 1000

[3] 1100

[4] 1200

Option # 4

Work backwards from the options. If the cylinder has a capacity of 1200 litre, then the conical vessel shall have a capacity of 700 litres. Once 200 litres have been taken out from the same, the remaining holding of each of them shall be 1000 & 500.

Alternate: Let the cylinder has a capacity of X litre, then the conical vessel shall have a capacity of

(x – 500) litres.

(x – 200) = 2 (x – 700) = x = 1200.

Question 9:
The reduction in the speed of an engine is directly proportional to the square of the number of bogies attached to it. The speed of the train is 100 km/hr when there are 4 bogies and 55 kmph when there are 5 bogies. What is the maximum number of bogies that can be attached to the train so that it can move?

[1] 6

[2] 5

[3] 4

[4] None of these

Option # 2

Suppose Reduction in speed is R. Speed of the engine without any bogie = k

number of bogies attached = b, proportionality constant = c, Resultant speed = s

We have $R = c b ^ { 2 }$ and $s = k - R = k - c b ^ { 2 }$

$100 = \mathrm { k } - \mathrm { c } ( 4 ) ^ { 2 }$ or, $100 = \mathrm { k } - 16 \mathrm { c }(i)$

and $55 = k - c ( 5 ) ^ { 2 }$ or $55 = k - 25 c (ii)$

Solving (i) and (ii) we get k = 180 and c = 5. Now we have S = 180 – 5b2. If we put b = 6, S = 0

Therefore, at most we can attach 5 bogies to the engine.

Question 10:
Arvind Singh purchased a 40 seater bus. He started his services on route number 2 (from Mahu Naka to Dewas Naka with route length of 50 km). His profit (P) from the bus depends upon the number of passengers over a certain minimum number of passengers ‘n’ and upon the distance travelled by bus. His profit is Rs.3600 with 29 passengers in the bus for a journey of 36 km and Rs.6300 with 36 passengers in the bus for a journey of 42 km. What is the minimum number of passengers are required so that he will not suffer any loss.

[1] 12

[2] 20

[3] 18

[4] 15

Option # 4

The minimum number of passengers n, at which there is no loss and number of passengers travelling = m

and let the distance travelled is d, Then

$P\propto \left( m\text{ }\text{ }n \right)d$

or p = k(m – n)d? k is a constant.

When P = 3600, m = 29 and d = 36, then

3600 = k(29 – n) × 36 ...(1)

Again, when p = 6300, m = 36, d = 42, then

6300 = k(36 – n) × 42 ...(2)

Dividing equation (2) by (1)

$\frac { 6300 } { 3600 } = \frac { k ( 36 - n ) \times 42 } { k ( 29 - n ) \times 36 } \Rightarrow \frac { ( 36 - n ) } { ( 29 - n ) } = \frac { 9 } { 6 } \Rightarrow 3 n = 45 \Rightarrow n = 15$