Hello Students,

We all know that Abhinay Sir Maths Class Notes and Book Pdf 2022 comes in every exams and math subject is important for all types of exams so today, we are sharing Abhinay Math PDF Notes Download for you. We all know that every exam has 20-25 maths questions and also we know every exam paper for limited time so improve your perpetration with Abhinay Sir Maths Class Notes and Book Pdf.

Sscresult.in is an online free notes pdf website, where we are sharing all types of Abhinay Sharma Maths Book PDF and New Syllabus, Competitive exams for UPSC, SSC, BANK, RAILWAY, LDC, NTPC and many other exams. Our Tricky mathematics PDF 2022 is very simple and easy to understand. We also cover basic subjects like Mathematics, Geography, History, General Science, Politics, etc. We also share study material including previous year question papers, current affairs, important sources, etc. for upcoming government exams.

This Maths With Abhinay Download free PDF is being provided to you free of charge, which you can download by clicking on the download button given below, and you can also DOWNLOAD some more new PDFs related to this Vedic Maths Book PDF in Hindi by going to the related notes. You can learn about all the new updates on SSCRESULT.IN by clicking on the Allow button on the screen.

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.

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 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)

= 2 + 2(4)

= 2 + 8

= 10

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

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

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.

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.

= 587,000,000,000,000 miles.

= 587 × 1012 miles.

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 %

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.

= \$ 3500 × 0.04.

= \$ 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,

Principal = \$ 2500,

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

r = 0.064

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

Solution:

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.

= 100/x 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

log (0.2)x = log 2.

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 = -0.4306….

x = -0.4 (nearest tenth)

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

10y = 5

1/10y = 1/5

10-y = 1/5

9. 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

Solution:

Partial Fraction
129Save

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.

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.

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 = 98 × 15.

49y = 1470.

y = 1470/49.

y = 30 .

Therefore, the second number is 30.