Mathematics- Top Questions & Answers pdf in Hindi

By Deepak On May 25, 2023

Mathematics- Top Questions & Answers pdf in Hindi

Hello Aspirants,

Algebra: Quadratic Formula: For a quadratic equation ax^2 + bx + c = 0, the solutions are given by x = (-b ± √(b^2 – 4ac)) / (2a).

Exponent Rules:

Product Rule: a^m * a^n = a^(m+n)

Quotient Rule: a^m / a^n = a^(m-n)

Power Rule: (a^m)^n = a^(mn)

Binomial Theorem: (a + b)^n = C(n, 0)a^nb^0 + C(n, 1)*a^(n-1)*b^1 + … + C(n, r)*a^(n-r)*b^r + … + C(n, n)a^0b^n, where C(n, r) denotes the binomial coefficient.

Calculus:

Derivative Rules:

Power Rule: d/dx(x^n) = nx^(n-1)

Product Rule: d/dx(f(x)g(x)) = f'(x)g(x) + f(x)g'(x)

Chain Rule: d/dx(f(g(x))) = f'(g(x)) * g'(x)

Fundamental Theorem of Calculus: ∫[a, b] f(x) dx = F(b) – F(a), where F(x) is the antiderivative of f(x).

Mean Value Theorem: If a function f(x) is continuous on [a, b] and differentiable on (a, b), then there exists at least one c in (a, b) such that f'(c) = (f(b) – f(a))/(b – a).

Geometry: Pythagorean Theorem: In a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides: a^2 + b^2 = c^2.

Area Formulas:

Triangle: A = (1/2) * base * height

Circle: A = π * r^2 (where r is the radius)

Rectangle: A = length * width

Trapezoid: A = (1/2) * (base1 + base2) * height

Statistics: Mean: The average of a set of numbers. It is calculated by summing all the values and dividing by the total number of values.

Median: The middle value in a set of numbers when arranged in order. If there are two middle values, the median is the average of those two values.

Standard Deviation: A measure of the amount of variation or dispersion in a set of numbers. It indicates how much the values deviate from the mean.

Probability: The likelihood of an event occurring, calculated as the ratio of favorable outcomes to total possible outcomes.

These are just a few key concepts in mathematics. Let me know if you would like more specific information or if there’s any other topic you’d like to explore!

More Mathematics Notes PDF Download

Maths Questions answers In Hindi

प्रश्न 1. log 10000 का मान है–
(a) 4 (b) 8 (c) 5 (d) 1 (Ans : a)
संकेत: log 10000 = log10 104=4

प्रश्न 2. एक लड़का 12 किमी/घण्टा की औसत गति पर स्कूल पहुँचने के लिए 20 मिनट लेता है। यदि उसे 15 मिनट में स्कूल पहुँचना है, तो औसत गति….. होनी चाहिए।
(a) 14 किमी/घण्टा (b) 15 किमी/घण्टा (c) 16 किमी/घण्टा (d) 18 किमी/घण्टा (Ans : c)
संकेत: स्कूल की दूरी =12×20/60=4 किमी
अब उसकी चाल =दूरी/समय=4/15/60
=16 किमी/घण्टा

प्रश्न 3. यदि किसी वर्ग की भुजा दोगुनी कर दी जाए, तो क्षेफल–
(a) दोगुना होता है (b) 4 गुना होता है (c) 8 गुना होता है (d) 16 गुना होता है (Ans : b)
संकेत: प्रश्नानुसार, A1=a2
A2=(2a)2=4a2
अत: स्पष्टत: यह चार गुना होगा।

प्रश्न 4. 220 का 15%=?
(a) 33 (b) 22 (c) 24 (d) 26 (Ans : a)
संकेत: 220×15/100=33

प्रश्न 5. 8888 + 888 + 88 + 8 = ?
(a) 9784 (b) 9792 (c) 9072 (d) इनमें से कोई नहीं (Ans : d)

प्रश्न 6. 450 सेबों में से 30% सड़ें हुए हैं। कितने सेब ठीक हैं?
(a) 135 (b) 140 (c) 125 (d) 315 (Ans : a)
संकेत: सड़े हुए सेबों की संख्या = 450 का 30%
= 450×30/100=135

प्रश्न 7. एक तारा पृथ्वी से लगभग 8.1×1013 किमी दूर है। मान लीजिए कि प्रकाश 3.0×105 किमी/सेकण्ड की गति से चलता है। ज्ञात कीजिए कि तारे से प्रकाश को पृथ्वी तक पहुँचने में कितना समय लगता है?
(a) 2.7×108 (b) 2.7×1011 (c) 7.5×104 (d) 7.5×103 (Ans : a)

संकेत: तारे से प्रकार को पृथ्वी एक पहुँचने में लगा समय
= दूरी/चाल = 8.1×1013/3.0×105 = 2.7×108 सेकण्ड

प्रश्न 8. एक वस्तु रु. 220 में बेचकर, नीता ने 10% लाभ प्राप्त किया। तद्नुसार वह उसे कितने में बेचे कि लाभ 30% हो जाए?
(a) रु. 220 (b) रु. 230 (c) रु. 260 (d) रु. 280 (Ans : c)
संकेत: वस्तु का क्रय मूल्य =200 × 100/110= रु. 200
30% लाभ कमाने के लिए वस्तु का विक्रय मूल्य
=200 का 130% = रु. 260

प्रश्न 9. 100 × 10 – 100 + 2000 » 100 किसके बराबर होगा?
(a) 20 (b) 920 (c) 980 (d) 1000 (Ans : b)
संकेत: 100 × 10 – 100 + 2000 » 100
= 1000 – 100 + 20 = 900 + 20 = 920

प्रश्न 10. एक व्यक्ति ने अपनी सम्पत्ति का 1/4 भाग अपनी पुत्री को दिया, ½ अपने पुत्रों को दिया और 1/5 दान कर दिया। तद्नुसार, उसने कुल कितना भाग दे दिया?
(a) 1/20 (b) 19/20 (c) 1/10 (d) 9/10 (Ans : b)
संकेत: व्यक्ति द्वारा दिया गया कुल भाग
=1/4+½+1/5=5+10+4/20=19/20

प्रश्न 11. यदि समअष्टभुज का प्रत्येक अन्त:कोण 135° हैं, तो अष्टभुज का बाह्य कोण होगा-
(a) 65° (b) 75° (c) 45° (d) 70° (Ans : c)
संकेत: अष्टभुज का प्रत्येक ब्राह्य कोण 180° – अन्त: कोण
= 180° – 135° = 45°

प्रश्न 12. एक विद्यालय में 85 लड़के व 35 लड़कियों ने सार्वजनिक परीक्षा दी। लड़कों का माध्य प्राप्त अंक 40% व लड़कियों का माध्य प्राप्त अंक 60% था, तो विद्यालय का औसत प्राप्त अंक प्रतिशत में बताइए।
(a) 50.60 (b) 54.16 (c) 45.83 (d) 48.53 (Ans : c)
संकेत: विद्यालय का औसत प्राप्तांक प्रतिशत में
= 85×40+35×60/(85+35)
= 3400+2100/120
= 550/12=45.83

प्रश्न 13. कोई धनराशि सरल ब्याज पर 20 वर्षों में दोगुनी हो जाती है। कितने वर्ष में वह चार गुनी होगी?
(a) 40 (b) 50 (c) 60 (d) 80 (Ans : c)
संकेत: साधारण ब्याज से 20 वर्ष में यदि कोई राशि दोगुनी होगी, तो 40 वर्ष में तीन गुनी एवं 60 वर्ष में चार गुनी होगी।

प्रश्न 14. किसी संख्या का 35% मिलने के लिए उस संख्या को किससे गुणा किया जाना चाहिए?
(a) 7/20 (b) 3.5 (c) √7/20 (d) 5/20 (Ans : c)
संकेत: 35%= 35/100=7/20

प्रश्न 15. 4800 का 36%×1320 का 0.2% बराबर है–
(a) 4535.52 (b) 4551.36 (c) 4561.92 (d) 4572.48 (Ans : c)
संकेत: ? = 4800 का 36/100×1320 का 0.2/100
= 1728 × 2.64
= 4561.92

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Maths Question Answer

Q 1 . एक रेलगाड़ी किसी स्टेशन पर खड़े एक व्यक्ति की 20 सेकेण्ड में पार कर जाती। यदि रेलगाड़ी की चाल 20मी./से. हो तो रेलगाड़ी की लम्बाई ज्ञात करो?
(A). 200मी.

(B). 400मी.

(C). 600मी.

(D). 120मी.

Ans. 400मी.

Q 2 . 7 व्यक्तियों का औसत आयु 22 वर्ष है यदि 1 व्यक्ति उनके साथ मिल जाता है तो उनकी औसत आयु 1 वर्ष बढ़ जाता है तो जो व्यक्ति आया उसकी आयु कितना है?
(A). 26 वर्ष

(B). 28 वर्ष

(C). 30 वर्ष

(D). 32 वर्ष

Ans. 30 वर्ष

Q 3 . 5 मित्रों की औसत आयु 40 वर्ष है उनमें से एक मित्र चला जाता है तो अब औसत आयु 38 वर्ष हो जाता है। हटा गया मित्र की आयु कितना है ?
(A). 44 वर्ष

(B). 48 वर्ष

(C). 40 वर्ष

(D). 36 वर्ष

Ans. 48 वर्ष

Q 4 . यदि कोई ट्रेन 5 मिनट में 11 किमी. चलता है तो उस ट्रेन की चाल किमी./घंटा में कितना होगा?
(A). 121 किमी./घंटा

(B). 132 किमी./घंटा

(C). 123 किमी./घंटा

(D). 125 किमी./घंटा

Ans. 132 किमी./घंटा

Q 5 . एक ट्रेन एक प्लेटफार्म जिसकी लम्बाई 120 मीटर है उसे 10 सेकेण्ड में पार कर जाती है और प्लेटफार्म पर खड़े एक व्यक्ति को 65 सेकेण्ड में पार कर जाती है तो ट्रेन की चाल किमी./घंटा में बताओ?
(A). 90 किमी./घंटा

(B). 105 किमी./घंटा

(C). 108 किमी./घंटा

(D). 111 किमी./घंटा

Ans. 108 किमी./घंटा

Q 6 . स्थान A से B की दूरी को एक कार 50 किमी./घंटा की गति से जाता है और 30 किमी./घंटा की गति से आता है तो कार का औसत चाल ज्ञात कीजिए ?
(A). 35 किमी./घंटा

(B). 37.5 किमी./घंटा

(C). 42.5 किमी./घंटा

(D). 45 किमी./घंटा

Ans. 37.5 किमी./घंटा

Q 7 . क्रमागत 7 सम संख्याओं का औसत मान 24 है तो सबसे छोटी संख्या ज्ञात कीजिए ?
(A). 18

(B). 22

(C). 16

(D). 20

Ans. 18

Q 8 . क्रमागत 7 सम संख्याओं का औसत मान 24 है तो सबसे छोटी संख्या ज्ञात कीजिए ?
(A). 18

(B). 22

(C). 16

(D). 20

Ans. 18

Q 9 . एक समूह में 10 लड़को का औसत भार 21 किलो. है अगर प्रथम 5 लड़को का औसत भार 20 किलो हो और अगले 4 लड़को का औस भार 22 किलो हो तो अन्तिम लड़के का वजन कीजिए।
(A). 20 किलो.

(B). 21 किलो.

(C). 22 किलो.

(D). 23 किलो.

Ans. 22 किलो.

Q 10 . 17 छात्रों का औसत भार 17 किलो. है गलती से 19 और 21 के स्थान पर 11 तथा 12 जोड़ दिया जाता है तो उनका सही औसत भार क्या होगा ?
(A). 15 किलो.

(B). 16.5 किलो.

(C). 17 किलो.

(D). 17.5 किलो.

Ans. 17 किलो.

Maths Questions Answers

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(x^x) when x = 2 is:

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

Solution:

x + x(x^x)

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

Answer: (a)

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 = $\frac{1.20}{1.20}$

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 = $\frac{1.20}{0.80}$

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.

Answer: (d)

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.

Answer: (d)

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.

Answer: (e)

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.

Answer: (b)

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)

Answer: (c)

8. If 10^2y = 25, then 10^-y equals:

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

Solution:

10^2y = 25

(10^y)² = 5²

10^y = 5

¹/_10^y = ¹/₅

10^-y = ¹/₅

Answer: (e)

9. The fraction ^(5x-11)/_{(2x² + 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:

Answer: (d)

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.

Answer: (c)

Maths Questions

Q.1. A sum of Rs. 25000 amounts to Rs. 31000 in 4 years at the rate of simple interest what is the rate of interest?

(a) 3%

(b) 4%

(d) 6 %

(e) None of these

Answer: [d]

Q.2. Kamla took a loan of Rs. 2400 with simple interest for as many years as the rate of interest. If she paid Rs. 864 as interest at the end of the loan period, what was the rate of interest?

(a) 3.6

(b) 6

(d) cannot determined

(e) None of these

Answer: [b]

Q.3. What is the present worth of Rs. 264 dues in 4 years at 10% simple interest per annum?

(a) 170.20

(b) 166

(d) 175.28

Answer: [C]

Q.4. A sum fetched a total simple interest of Rs. 8016.25 at the rate of 6 p.c.p.a in 5 years what is the sum?

(a) 24720.83

(b) 26730.33

(d) 26710.63

(e) None of these

Answer: [C]

Q.5. 8. Rs. 800 becomes Rs. 956 in 3 years at a certain rate of simple interest. If the rate of interest is increased by 4%, what amount will Rs 800 become in 3 years?

(a) 1020.80

(b) 1025

(d) Data inadequate

Answer: [C]

Q. 6. The product of 2 numbers is 1575 and their quotient is 9/7. Then the sum of the numbers is –

a. 74

b. 78

c. 80

d. 90

Answer: [C]

Q.7. The value of (81)3.6 * (9)2.7/ (81)4.2 * (3) is __

a. 3

b. 6

c. 9

d. 8.2

Answer: [C]

Q.8. √6+√6+√6+… is equal to –

a. 2

b. 5
c. 4
d. 3

Answer: [D]

Q.9. 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

Answer: [D]

Q.10. When (6767 +67) is divided by 68, the remainder is-

a. 1

b. 63

c. 66

d. 67

Answer: [C]

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