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### Vedic Maths Book CONTENT – ### Maths Questions for Competitive Exams with Solutions

1. Mrs. Rodger got a weekly raise of \$145. If she gets paid every other week, write an integer describing how the raise will affect her paycheck.  (Vedic Maths Book PDF Free Download)

Solution:

Let the 1st paycheck be x (integer).

Mrs. Rodger got a weekly raise of \$ 145.

So after completing the 1st week she will get \$ (x+145).

Similarly after completing the 2nd week she will get \$ (x + 145) + \$ 145.

= \$ (x + 145 + 145)

= \$ (x + 290)

So in this way end of every week her salary will increase by \$ 145.

2. The fraction (5x-11)/(2×2 + x – 6) was obtained by adding the two fractions A/(x + 2) and B/(2x – 3). The values of A and B must be, respectively:

(a) 5x, -11, (b) -11, 5x, (c) -1, 3, (d) 3, -1, (e) 5, -11  (Vedic Maths Book PDF Free Download)

Solution: 3. Mr. Jones sold two pipes at \$1.20 each. Based on the cost, his profit one was 20% and his loss on the other was 20%. On the sale of the pipes, he:

(a) broke even, (b) lost 4 cents, (c) gained 4 cents, (d) lost 10 cents, (e) gained 10 cents

Solution:

Selling price of the first pipe = \$1.20  (Vedic Maths Book PDF Free Download)

Profit = 20%

Let’s try to find the cost price of the first pipe

CP = Selling price – Profit

CP = 1.20 – 20% of CP

CP = 1.20 – 0.20CP

CP + 0.20CP = 1.20

1.20CP = 1.20

CP = 1.201.20
CP = \$ 1

Selling price of the Second pipe = \$1.20

Loss = 20%

Let’s try to find the cost price of the second pipe  (Vedic Maths Book PDF Free Download)

CP = Selling price + Loss

CP = 1.20 + 20% of CP

CP = 1.20 + 0.20CP

CP – 0.20CP = 1.20

0.80CP = 1.20

CP = 1.200.80
CP = \$1.50

Therefore, total cost price of the two pipes = \$1.00 + \$1.50 = \$2.50

And total selling price of the two pipes = \$1.20 + \$1.20 = \$2.40

Loss = \$2.50 – \$2.40 = \$0.10

Therefore, Mr. Jones loss 10 cents.  (Vedic Maths Book PDF Free Download)

4. The distance light travels in one year is approximately 5,870,000,000,000 miles. The distance light travels in 100 years is:

(a) 587 × 108 miles, (b) 587 × 1010 miles, (c) 587 × 10-10 miles, (d) 587 × 1012 miles, (e) 587 × 10-12 miles

Solution:

The distance of the light travels in 100 years is:

5,870,000,000,000 × 100 miles.

= 587 × 1012 miles.

5. What is |-26|?

(a) -26,

(b) 26,

(c) 0,

(d) 1

Solution:

|-26|

6. Multiply: (x – 4)(x + 5)

(a) x2 + 5x – 20,

(b) x2 – 4x – 20,

(c) x2 – x – 20,

(d) x2 + x – 20.

Solution:

(x – 4)(x + 5).

= x(x + 5) -4(x + 5).

= x2 + 5x – 4x – 20.

= x2 + x – 20.

7. Factor: 5×2 – 15x – 20.

(a) 5(x-4)(x+1),

(c) -5(x+4)(x-1),

(d) 5(x+4)(x+1).

Solution:

5×2 – 15x – 20.

= 5(x2 – 3x – 4).

= 5(x2 – 4x + x – 4).

= 5{x(x – 4) +1(x – 4)}.  (Vedic Maths Book PDF Free Download)

= 5(x-4)(x+1).

8. If 102y = 25, then 10-y equals:

(a) -1/5, (b) 1/625, (c) 1/50, (d) 1/25, (e) 1/5

Solution:

102y = 25

(10y)2 = 52

1/10y = 1/5

10-y = 1/5

9. The value of x + x(xx) when x = 2 is:

(a) 10, (b) 16, (c) 18, (d) 36, (e) 64

Solution:

x + x(xx)

Put the value of x = 2 in the above expression we get,

2 + 2(22)

= 2 + 2(2 × 2)  (Vedic Maths Book PDF Free Download)

= 2 + 2(4)

= 2 + 8

= 10

10. The sum of three numbers is 98. The ratio of the first to the second is 2/3, and the ratio of the second to the third is 5/8. The second number is:

(a) 15, (b) 20, (c) 30, (d) 32, (e) 33

Solution:

Let the three numbers be x, y and z.  (Vedic Maths Book PDF Free Download)

Sum of the numbers is 98.

x + y + z = 98………………(i)

The ratio of the first to the second is 2/3.

x/y = 2/3.

x = 2/3 × y.

x = 2y/3.

The ratio of the second to the third is 5/8.  (Vedic Maths Book PDF Free Download)

y/z = 5/8.

z/y = 8/5.

z = 8/5 × y.

z = 8y/5.

Put the value of x = 2y/3 and z = 8y/5 in (i).

2y/3 + y + 8y/5 = 98

49y/15 = 98.

49y = 1470.

y = 1470/49.

y = 30 .

Therefore, the second number is 30.  (Vedic Maths Book PDF Free Download)

### Maths Questions for Competitive Exams –

1. Which is greater than 4?

(a) 5,

(c) -1/2,

(d) -25.

Solution:

5 greater than 4.

(a) -1,

(b) -1/2,

(c) 0,

(d) 3.

Solution:

The smallest number is -1.

3. Combine terms: 12a + 26b -4b – 16a.

(a) 4a + 22b,

(b) -28a + 30b,

(c) -4a + 22b,

(d) 28a + 30b.

Solution:

12a + 26b -4b – 16a.

= 12a – 16a + 26b – 4b.

= -4a + 22b.

4. Simplify: (4 – 5) – (13 – 18 + 2).  (Vedic Maths Book PDF Free Download)

(a) -1,

(b) –2,

(c) 1,

(d) 2.

Solution:

(4 – 5) – (13 – 18 + 2).

= -1-(13+2-18).

= -1-(15-18).

= -1-(-3).

= -1+3.

5. What is |-26|?

(a) -26,

(b) 26,

(d) 1

Solution:

|-26|

= 26.

5. A man has \$ 10,000 to invest. He invests \$ 4000 at 5 % and \$ 3500 at 4 %. In order to have a yearly income of \$ 500, he must invest the remainder at:

(a) 6 % , (b) 6.1 %, (c) 6.2 %, (d) 6.3 %, (e) 6.4 %  (Vedic Maths Book PDF Free Download)

Solution:

Income from \$ 4000 at 5 % in one year = \$ 4000 of 5 %.

= \$ 4000 × 5/100.

= \$ 4000 × 0.05.

= \$ 200.

Income from \$ 3500 at 4 % in one year = \$ 3500 of 4 %.

= \$ 3500 × 4/100.

= \$ 140.

Total income from 4000 at 5 % and 3500 at 4 % = \$ 200 + \$ 140 = \$ 340.

Remaining income amount in order to have a yearly income of \$ 500 = \$ 500 – \$ 340.

= \$ 160.

Total invested amount = \$ 4000 + \$ 3500 = \$7500.

Remaining invest amount = \$ 10000 – \$ 7500 = \$ 2500.

We know that, Interest = Principal × Rate × Time

Interest = \$ 160,

Rate = r [we need to find the value of r],

Time = 1 year.

160 = 2500 × r × 1.

160 = 2500r

160/2500 = 2500r/2500 [divide both sides by 2500]

0.064 = r

Change it to a percent by moving the decimal to the right two places r = 6.4 %

Therefore, he invested the remaining amount \$ 2500 at 6.4 % in order to get \$ 500 income every year.

6. Jones covered a distance of 50 miles on his first trip. On a later trip he traveled 300 miles while going three times as fast. His new time compared with the old time was:

(a) three times as much, (b) twice as much, (c) the same, (d) half as much, (e) a third as much

Let speed of the 1st trip x miles / hr. and speed of the 2nd trip 3x / hr.

We know that

Speed = Distance/Time.

Or, Time = Distance/Speed.

So, times taken to covered a distance of 50 miles on his first trip = 50/x hr.

And times taken to covered a distance of 300 miles on his later trip = 300/3x hr.

So we can clearly see that his new time compared with the old time was: twice as much.

7. If (0.2)x = 2 and log 2 = 0.3010, then the value of x to the nearest tenth is:

(a) -10.0, (b) -0.5, (c) -0.4, (d) -0.2, (e) 10.0

Solution:

(0.2)x = 2.

Taking log on both sides

x log (0.2) = 0.3010, [since log 2 = 0.3010].

x log (2/10) = 0.3010.

x [log 2 – log 10] = 0.3010.

x [log 2 – 1] = 0.3010,[since log 10=1].

x [0.3010 -1] = 0.3010, [since log 2 = 0.3010].

x[-0.699] = 0.3010.

x = 0.3010/-0.699.

x = -0.4 (nearest tenth)

6. Multiply: (x – 4)(x + 5)

(a) x2 + 5x – 20,

(b) x2 – 4x – 20,

(c) x2 – x – 20,

(d) x2 + x – 20.  (Vedic Maths Book PDF Free Download)

Solution:

(x – 4)(x + 5).

= x(x + 5) -4(x + 5).

= x2 + 5x – 4x – 20.

= x2 + x – 20.

7. Factor: 5×2 – 15x – 20.

(a) 5(x-4)(x+1),

(b) -2(x-4)(x+5),

(c) -5(x+4)(x-1),

(d) 5(x+4)(x+1).

Solution:

5×2 – 15x – 20.

= 5(x2 – 3x – 4).

= 5(x2 – 4x + x – 4).  (Vedic Maths Book PDF Free Download)

= 5{x(x – 4) +1(x – 4)}.

= 5(x-4)(x+1).

8. Factor: 3y(x – 3) -2(x – 3).

(a) (x – 3)(x – 3),

(b) (x – 3)2,

(c) (x – 3)(3y – 2),

(d) 3y(x – 3).

3y(x – 3) -2(x – 3).

= (x – 3)(3y – 2).

9. Solve for x: 2x – y = (3/4)x + 6.

(a) (y + 6)/5,

(b) 4(y + 6)/5,

(d) 4(y – 6)/5.

Solution:

2x – y = (3/4)x + 6.

or, 2x – (3/4)x = y + 6.

or, (8x -3x)/4 = y + 6.

or, 5x/4 = y + 6.

or, 5x = 4(y + 6).  (Vedic Maths Book PDF Free Download)

or, 5x = 4y + 24.

or, x = (4y + 24)/5.

Therefore, x = 4(y + 6)/5.

10. Simplify:(4×2 – 2x) – (-5×2 – 8x).

Solution:

(4×2 – 2x) – (-5×2 – 8x)

= 4×2 – 2x + 5×2 + 8x.

= 4×2 + 5×2 – 2x + 8x.  (Vedic Maths Book PDF Free Download)

= 9×2 + 6x.

= 3x(3x + 2).

11. Find the value of 3 + 2 • (8 – 3)

(a) 25,

(c) 17,

(d) 24,

(e) 15.

Solution:

3 + 2 • (8 – 3)

= 3 + 2 (5)

= 3 + 2 × 5

= 3 + 10

12. Rice weighing 33/4 pounds was divided equally and placed in 4 containers. How many ounces of rice were in each?

Solution:

33/4 ÷ 4 pounds.

= (4 × 3 + 3)/4 ÷ 4 pounds.

= 15/4 × 1/4 pounds.

= 15/16 pounds.

Now we know that, 1 pound = 16 ounces.

Therefore, 15/16 pounds = 15/16 × 16 ounces.

= 15 ounces.

Solution:

16w3 – u4w3.

= w3(16 – u4).

= w3(42 – ((u2)2).

= w3(4 + u2)(4 – u2).

= w3(4 + u2)(22 – u2).

= w3(4 + u2)(2 + u)(2 – u).

Answer: w3(4 + u2)(2 + u)(2 – u).

14. Factor: 3x4y3 – 48y3.

Solution:

3x4y3– 48y3.

= 3y3(x4 – 16).

= 3y3[(x2)2 – 42].

= 3y3(x2 + 4)(x2 – 4).

= 3y3(x2 + 4)(x2 – 22).

= 3y3(x2 + 4)(x + 2)(x -2).

Answer: 3y3(x2 + 4)(x + 2)(x -2)

15. What is the radius of a circle that has a circumference of 3.14 meters?

Solution:

Circumference of a circle = 2πr.

Given, circumference = 3.14 meters.

Therefore,

2πr = Circumference of a circle

or, 2πr = 3.14.

or, 2 × 3.14r = 3.14,[Putting the value of pi (π) = 3.14].

or, r = 3.14/6.28.

or, r = 0.5.

16. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Ans: (6767+ 67) = 67(6766 + 166)

As 66 is an even number

We know 6766 is an even number

We know (6766 + 1) is perfectly divisible by (67 + 1)

i.e. 68

We know The remainder = 66

17. In a division sum, the divisor is 4 times the quotient and twice the remainder. If a and b are respectively the divisor and the dividend then-

a. 4b-a2 /a =3

b. 4b-2a /a2 =2

d. A(a+2)/b = 4

Ans : As divisor is a, and dividend is b.

We know Quotient = a/4

And Remainder = a/2

We know b = a * a/4 + a/2

We know 4b = a2 + 2a

We know a(a+2)/b = 4

18. If 738 A6A is divisible by 11, then the value of A is-

a. 6

b. 3

c. 9

d. 1

Ans : As 738A6 A is divisible by 11.

We know A + A + 3 = 6 + 8 + 7

We know A = 9

19. The east positive integer that should be subtracted from 3011 * 3012 so that the difference is perfect square is-

a. 3009

b. 3010

c. 3011

d. 3012

Ans : We know that 3011 * 3012 = 3011 (3011 + 1)

We know Required least number = 3011

20. P, Q, R are employed to do a work for Rs. 5750. P and Q together finished 19/23 of work and Q and R together finished 8/23 of work. Wage of Q, in rupees, is-

a. 2850

b. 3750

c. 2750

d. 1000

Ans : Work done by Q = 19/23 + 8/23 – 1

We know Wage of Q = 4/23 * 5750

= Rs. 1000

21. A can do a piece of work in 24 day, B in 32 days and C in 64 days. All begin to do it together, but A leaves after 6 days and B leaves 6 days before the completion of the work. How many days did the work last?

a. 15

b. 20

d. 30

Ans : Work done by A = 6/24 = ¼

Work done by B = (x-6)/32
(where x is no. of days in which work is competed)

We know that ¼ + x – 6/32 + x/64 = 1
We know 16 + 24 – 12 + x /64 = 1
We know 3x + 4 = 64
We know x = 60/3 = 20 days

22. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

b. 78

c. 80

d. 90

Ans : Let the numbers be x and y .

We know xy = 1575

And x/y = 9/7

We know y2 = 1225

We know y = 35 and x = 45

We know The sum of the numbers = 45+35

= 80

23. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

c. 9

d. 8.2

Ans : (81)3.6 * (9)2.7/(81)4.2 * (3) = (3)14.4 * (3)5.4/(3)16.8 * (3)

= 314.4+5.4-16.8-1

= 32

24. √6+√6+√6+… is equal to –

a. 2
b. 5
c. 4
d. 3

Ans : Let √6+√6+√6+…. be x.
We know x =√6+x
We know x2 = 6+x
We know x2 – x – 6 = 0
We know (x-3) (x+2) = 0

25. The sum of the squares of two natural consecutive odd numbers is 394. The sum of the numbers is –

a. 24

b. 32

c. 40

d. 28

Ans : Let the consecutive odd numbers be x and (x+2)

We know that x2 + (x + 2)2 = 394

We know   x2 + x2 + 4x + 4 = 394

We know   2 x2 +4x – 390 = 0

We know   x2 + 2x – 195 = 0

We know  (x +15) (x-13) = 0

We know   x = 13

We know    Required sum = 13 +15 =28   (Vedic Maths Book PDF Free Download)

##### TEST YOURSELF | क्या आप यह प्रश्न कर सकते हैं | Unsolved Questions:

1. A sports field is 300 feet long. Write a formula that gives the length of x sports fields in feet. Then use this formula to determine the number of sports fields in 720 feet.

2. A recipe calls for 2 1/2 cups and I want to make 1 1/2 recipes. How many cups do I need?

3. Fahrenheit temperature F is a linear function of Celsius temperature C. The ordered pair (0, 32) is an ordered pair of this function because 0°C is equivalent to 32°F, the freezing point of water. The ordered pair (100, 212) is also an ordered pair of this function because 100°C is equivalent to 212° F, the boiling point of water.  (Vedic Maths Book PDF Free Download)

4. Mario answered 30% of the questions correctly. The test contained a total of 80 questions. How many questions did Mario answer correctly? 