(3/13) of [325/(3)ˉ³] × ?² = 25 × 10⁴ × (1.5)⁴

A. 675

B. 75

C. 575

D. 625

E.

Source: Safalta.com

25Solution-

(3/13) of [325/(3)ˉ³] × ?² = 25 × 10⁴ × (1.5)⁴

=> 3 × 25 × 3³ × ?² = 25 × 10⁴ × (1.5)⁴

=> 3(¹⁺³) × 25 × ?² = 25 × 10000 × (3)⁴/(2)⁴

=> ?² = 10000/16

=> ?² = 625

=> ? = 25

Hence, the value of (?) is 25.

Q.2 What should come in the place of the question mark ‘?’ in the following question?

12 × 87 + 12 × 114 + 93 × 12 – 44 × 12 = ?

A. 2000

B. 2500

C. 3000

D. 3500

E. None of these

Solution-

12 × 87 + 12 × 114 + 93 × 12 – 44 × 12 = ?

=> 1044 + 1368 + 1116 – 528 = ?

=> 3528 – 528 = ?

=> ? = 3000

Hence, the value of ‘?’ is 3000.

Q.3 A milkman, in 50 litres mixture of milk and water keeps the ratio of milk and water in 3 : 2. One day he decided this ratio is to be 2 : 3, then find quantity of water to be further added in the mixture.

A. 25 litres

B. 30 litres

C. 40 litres

D. 41 litres

E. None of these

Solution-

In actual ratio

Quantity of milk in the mixture,

=> (50 × 3)/5 litres = 30 litres,

Quantity of water in the mixture,

=> (50 – 30)litres = 20 litres

According of question new ratio = 2: 3

Let the quantity of water added in the mixture be x liters

Then,

Milk: Water = 30: (20 + x)

=> 2/3 = 30/(20 + x)

=> 90 = 40 + 2x

=> 50 = 2x

=> x = 25 liters

Hence, the quantity of water to be added in the mixture is 25 litres.

Q.4 An amount of Rs. 400 becomes Rs. 424 in 3 years at a certain rate of simple interest, If the rate of interest increased by 8%, then find what amount will Rs. 400 becomes in 2 years?

A. Rs. 450

B. Rs. 425

C. Rs. 480

D. Data inadequate

E. None of these

Solution-

S.I. = Amount – Principle = 424 – 400 = Rs. 24

R = (24 × 100)/(400 × 3) = 2%

It is given in the question that new rate is 8% more than previous rate of interest.

New rate = 2% + 8% = 10%

New S.I. = (400 × 10 × 2)/100 = Rs. 80

New Amount = 400 + 80 = Rs. 480

Hence, the new amount is Rs. 480

Q.5 A train 300 metre long crosses a lamppost in 25 seconds. Calculate the speed of the man running in opposite direction who takes 20 seconds to cross the train.

A. 15 m/s

B. 5 m/s

C. 10 m/s

D. 2 m/s

E. 3 m/s

Solution-

Speed of train = 300/25 = 12 m/s

Let the speed of man be x m/s

Time taken to cross the train = 20s = distance of train/relative speed

Relative speed = 12 + x

20 = 300/(12 + x)

12 + x = 15

Speed of man = 15 – 12 = 3 m/s

Hence, Speed of the running man is 3 m/s.

Q.6 A vegetable vendor claims to sell his vegetables at a cost price but uses a weight of 800 gm instead of kg weight. Thus he makes a profit of:

A. 10%

B. 20%

C. 30%

D. 22%

E. 25%

Solution-

According to the formula,

Profit% = True weight – false weight/false weight × 100

=> 1000 – 800/800 × 100

=> 100/4

=> 25%

Hence, the vegetable vendor made a profit of 25%.

Direction (7-8): Study the bar graph and answer the following questions

In a bar graph, shows the number of students passed in XII class from 6 schools.

Q.7 if the fail percentage of school P is 60% then, find number of students who failed from school P is what percentage of the number of students passes from school T.

A. 80%

B. 133.33%

C. 90%

D. 105%

E. 110%

Solution-

From the given graph,

Students who passed in school P = 80

=> 40% = 80

=> 1% = 2

=> 60% = 60 × 2 = 120 (Failed students of school P)

Now, the number of students passed from school T = 90

Then, the required percentage = (120/90) × 100 = (4/3) × 100

=> 133.33%

Hence, the required percentage is 133.33%.

Q.8 If the ratio between the total number of students who passed to the total number of students who failed from all school is 7: 3, then find the total number of failed students from all schools together.

A. 210

B. 250

C. 140

D. 220

E. None of these

Solution-

Total number of passed students = 80 + 90 + 70 + 50 + 90 + 110 = 490

=> 7 units = 490

=> 1 unit = 70

Failed students = 3 units = 70 × 3 = 210

Hence, the total number of failed student from all school is 210.

Q.9 In the given question, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.

I. 5x² - 18x + 9 = 0

II. 3y² + 5y – 2 = 0

A. x > y

B. x < y

C. x y

D. x y

E. x = y or relationship between x and y cannot be established

Solution-

I. 5x² - 18x + 9 = 0

=> 5x² - 15x – 3x + 9 = 0

=> 5x(x – 3) -3(x – 3) = 0

=> (5x – 3)(x – 3) = 0

=> x = 3/5 or 3

II. 3y² + 5y – 2 = 0

=> 3y² + 6y - y – 2 = 0

=> 3y(y + 2) -1(y + 2)

=> (3y – 1)(y + 2) = 0

=> y = 1/3 or -2

Clearly x > y.

Q.10 In the given question, two equations numbered I and II are given. Solve both the equations and mark the appropriate answer.

I. 3x² - 7x + 2 = 0

II. 2y² - 11y + 15 = 0

A. x > y

B. x < y

C. x y

D. x y

E. x = y or relationship between x and y cannot be established

Solution-

I. 3x² - 7x + 2 = 0

=> 3x² - 6x - x + 2 = 0

=> 3x(x – 2) -1(x -2) = 0

=> (x – 2)(3x – 1) = 0

=> x = 2 or 1/3

II. 2y² - 11y + 15 = 0

=> 2y² - 6y – 5y + 15 = 0

=> 2y(y – 3) – 5(y – 3) = 0

=> (y – 3)(2y – 5) = 0

=> y = 3 or 5/2

Clearly x < y.