1. Question
English: If f(x) = 3x - 2 and g(x) = x² + 1, find f(g(4)) + g(f(2)).
Hindi: यदि f(x) = 3x - 2 और g(x) = x² + 1 है, तो f(g(4)) + g(f(2)) ज्ञात कीजिए।
Solution:
First find:
g(4) = 4² + 1
= 16 + 1
= 17
Now:
f(g(4)) = f(17)
= 3 × 17 - 2
= 51 - 2
= 49
Next:
f(2) = 3 × 2 - 2
= 6 - 2
= 4
Now:
g(f(2)) = g(4)
= 4² + 1
= 17
Required value:
= 49 + 17
= 66
Answer: 66
2. Question
English: For positive integers, if f(x + y) = f(x) + f(y) + 2xy and f(1) = 3, find f(5).
Hindi: धनात्मक पूर्णांकों के लिए यदि f(x + y) = f(x) + f(y) + 2xy और f(1) = 3 है, तो f(5) ज्ञात कीजिए।
Solution:
Given:
f(x + y) = f(x) + f(y) + 2xy
Now:
f(2)
= f(1 + 1)
= f(1) + f(1) + 2 × 1 × 1
= 3 + 3 + 2
= 8
f(3)
= f(2 + 1)
= 8 + 3 + 2 × 2 × 1
= 11 + 4
= 15
f(4)
= f(3 + 1)
= 15 + 3 + 2 × 3 × 1
= 18 + 6
= 24
f(5)
= f(4 + 1)
= 24 + 3 + 2 × 4 × 1
= 27 + 8
= 35
Answer: 35
3. Question
English: The HCF and LCM of two numbers are 12 and 840 respectively. If their sum is 228, find the larger number.
Hindi: दो संख्याओं का HCF और LCM क्रमशः 12 और 840 है। यदि उनका योग 228 है, तो बड़ी संख्या ज्ञात कीजिए।
Solution:
Let numbers be:
12a and 12b
Since HCF = 12, a and b are co-prime.
LCM:
= 12ab
12ab = 840
ab = 70
Also:
12a + 12b = 228
a + b = 19
Numbers whose sum is 19 and product is 70:
5 and 14
Actual numbers:
60 and 168
Larger number:
= 168
Answer: 168
4. Question
English: Find the least positive number which leaves remainder 7 when divided by 12, 15 and 20, and is exactly divisible by 13.
Hindi: वह सबसे छोटी धनात्मक संख्या ज्ञात कीजिए जो 12, 15 और 20 से भाग देने पर 7 शेष छोड़े तथा 13 से पूर्णतः विभाजित हो।
Solution:
Required number:
= LCM(12,15,20) × k + 7
LCM:
12 = 2² × 3
15 = 3 × 5
20 = 2² × 5
LCM = 60
So number:
= 60k + 7
Now check divisibility by 13:
60k + 7 divisible by 13
60 mod 13 = 8
So:
8k + 7 divisible by 13
Try k = 4
8 × 4 + 7
= 39
Divisible by 13
Therefore number:
= 60 × 4 + 7
= 247
Answer: 247
5. Question
English: Simplify: (√50 + √98) / √2.
Hindi: सरलीकृत कीजिए: (√50 + √98) / √2.
Solution:
√50 = 5√2
√98 = 7√2
So numerator:
= 5√2 + 7√2
= 12√2
Now:
12√2 / √2
= 12
Answer: 12
6. Question
English: Evaluate: (64^(2/3) × 125^(1/3)) / 16^(3/4).
Hindi: मान ज्ञात कीजिए: (64^(2/3) × 125^(1/3)) / 16^(3/4).
Solution:
64^(2/3)
= (³√64)²
= 4²
= 16
125^(1/3)
= 5
16^(3/4)
= (⁴√16)³
= 2³
= 8
Therefore:
(16 × 5)/8
= 80/8
= 10
Answer: 10
7. Question
English: If f(x) = (x - 1)/(x + 1), find f(f(2)).
Hindi: यदि f(x) = (x - 1)/(x + 1) है, तो f(f(2)) ज्ञात कीजिए।
Solution:
First:
f(2)
= (2 - 1)/(2 + 1)
= 1/3
Now:
f(1/3)
= (1/3 - 1)/(1/3 + 1)
= (-2/3)/(4/3)
= -1/2
Answer: -1/2
8. Question
English: If f(x) = ax + b, f(1) = 5 and f(3) = 11, find f⁻¹(23).
Hindi: यदि f(x) = ax + b, f(1) = 5 और f(3) = 11 है, तो f⁻¹(23) ज्ञात कीजिए।
Solution:
Given:
f(x) = ax + b
f(1) = 5
a + b = 5
f(3) = 11
3a + b = 11
Subtract:
2a = 6
a = 3
Then:
b = 2
So:
f(x) = 3x + 2
Now:
f⁻¹(23)
Means:
3x + 2 = 23
3x = 21
x = 7
Answer: 7
9. Question
English: Five coins are tossed together. Find the probability of getting exactly 3 heads.
Hindi: 5 सिक्के एक साथ उछाले जाते हैं। ठीक 3 हेड आने की प्रायिकता ज्ञात कीजिए।
Solution:
Total outcomes:
= 2⁵
= 32
Ways to get exactly 3 heads:
= 5C3
= 10
Probability:
= 10/32
= 5/16
Answer: 5/16
10. Question
English: How many different arrangements can be made from the letters of the word COMMITTEE?
Hindi: COMMITTEE शब्द के अक्षरों से कितने अलग-अलग व्यवस्थापन बनाए जा सकते हैं?
Solution:
Total letters = 9
Repeated letters:
M = 2
T = 2
E = 2
Number of arrangements:
= 9! / (2! × 2! × 2!)
= 362880 / 8
= 45360
Answer: 45360
11. Question
English: Two runners run on a circular track of length 840 m at speeds 10 m/s and 8 m/s in opposite directions. After how much time will they meet for the first time?
Hindi: दो धावक 840 मीटर लंबी वृत्ताकार पथ पर 10 m/s और 8 m/s की गति से विपरीत दिशाओं में दौड़ते हैं। वे पहली बार कितने समय बाद मिलेंगे?
Solution:
Opposite direction relative speed:
= 10 + 8
= 18 m/s
Time:
= 840/18
= 140/3 seconds
= 46⅔ seconds
Answer: 46⅔ seconds
12. Question
English: 4 men can do the same work as 6 women. If 5 men and 3 women can complete a work in 12 days, then 4 men and 6 women will complete the same work in how many days?
Hindi: 4 पुरुष उतना ही कार्य कर सकते हैं जितना 6 महिलाएँ करती हैं। यदि 5 पुरुष और 3 महिलाएँ किसी कार्य को 12 दिनों में पूरा करते हैं, तो 4 पुरुष और 6 महिलाएँ वही कार्य कितने दिनों में पूरा करेंगी?
Solution:
4 men = 6 women
1 man = 3/2 women
5 men = 15/2 women
So:
5 men + 3 women
= 15/2 + 3
= 21/2 women
This completes work in 12 days.
Total work:
= 21/2 × 12
= 126 woman-days
Now:
4 men + 6 women
= 6 + 6
= 12 women
Required time:
= 126/12
= 10.5 days
Answer: 10.5 days
13. Question
English: Pipe A fills a tank in 18 hours, B fills it in 24 hours and C empties it in 36 hours. If all are opened together, find the time to fill the tank.
Hindi: पाइप A टंकी को 18 घंटे में भरता है, B 24 घंटे में भरता है और C 36 घंटे में खाली करता है। तीनों को साथ खोलने पर टंकी कितने समय में भरेगी?
Solution:
A’s 1-hour work:
= 1/18
B’s 1-hour work:
= 1/24
C’s 1-hour work:
= -1/36
Net work:
= 1/18 + 1/24 - 1/36
LCM = 72
= 4 + 3 - 2 /72
= 5/72
Time required:
= 72/5 hours
Answer: 72/5 hours
14. Question
English: The average marks of 30 students is 45. The average marks of boys is 48 and that of girls is 42. Find the number of boys.
Hindi: 30 विद्यार्थियों के औसत अंक 45 हैं। लड़कों का औसत 48 और लड़कियों का औसत 42 है। लड़कों की संख्या ज्ञात कीजिए।
Solution:
Using allegation:
Boys average = 48
Girls average = 42
Overall average = 45
Ratio of boys : girls
= (45 - 42) : (48 - 45)
= 3 : 3
= 1 : 1
Total students = 30
So boys:
= 15
Answer: 15
15. Question
English: A 30-litre mixture contains 40% acid. How much pure acid should be added to make the acid concentration 60%?
Hindi: 30 लीटर मिश्रण में 40% अम्ल है। अम्ल की मात्रा 60% करने के लिए कितना शुद्ध अम्ल मिलाया जाना चाहिए?
Solution:
Acid initially:
= 40% of 30
= 12 litres
Let pure acid added = x litres
Then:
(12 + x)/(30 + x) = 60/100
5(12 + x) = 3(30 + x)
60 + 5x = 90 + 3x
2x = 30
x = 15 litres
Answer: 15 litres
16. Question
English: An article is sold at 20% discount on marked price and still gives 25% profit. By what percentage is the marked price above the cost price?
Hindi: एक वस्तु अंकित मूल्य पर 20% छूट देकर बेची जाती है और फिर भी 25% लाभ होता है। अंकित मूल्य क्रय मूल्य से कितने प्रतिशत अधिक है?
Solution:
Let cost price = 100
Profit = 25%
So selling price:
= 125
Now selling price is after 20% discount.
That means selling price = 80% of marked price.
So:
80% of marked price = 125
Marked price:
= 125 × 100/80
= 156.25
Marked price above cost price:
= 156.25 - 100
= 56.25%
Answer: 56.25%
17. Question
English: The sides of a triangle are 20 cm, 21 cm and 29 cm. Find its inradius.
Hindi: एक त्रिभुज की भुजाएँ 20 सेमी, 21 सेमी और 29 सेमी हैं। उसकी अंतःत्रिज्या ज्ञात कीजिए।
Solution:
20² + 21² = 400 + 441
= 841
29² = 841
So it is a right triangle.
Area:
= 1/2 × 20 × 21
= 210 cm²
Semi-perimeter:
= (20 + 21 + 29)/2
= 70/2
= 35
Inradius:
= Area / semi-perimeter
= 210/35
= 6 cm
Answer: 6 cm
18. Question
English: Find the point which divides the line segment joining (2,-3) and (8,9) internally in the ratio 2 : 1.
Hindi: (2,-3) और (8,9) को जोड़ने वाले रेखाखंड को 2 : 1 के अनुपात में आंतरिक रूप से विभाजित करने वाला बिंदु ज्ञात कीजिए।
Solution:
Using section formula:
Point = [(2 × 8 + 1 × 2)/(2 + 1), (2 × 9 + 1 × -3)/(2 + 1)]
x-coordinate:
= (16 + 2)/3
= 18/3
= 6
y-coordinate:
= (18 - 3)/3
= 15/3
= 5
Answer: (6,5)
19. Question
English: The HCF and LCM of two numbers are 16 and 960 respectively. If one number is 192, find the other number.
Hindi: दो संख्याओं का HCF और LCM क्रमशः 16 और 960 है। यदि एक संख्या 192 है, तो दूसरी संख्या ज्ञात कीजिए।
Solution:
Product of two numbers = HCF × LCM
= 16 × 960
One number = 192
Other number:
= 16 × 960 / 192
= 80
Answer: 80
20. Question
English: Find the number of perfect square divisors of 2⁶ × 3⁴ × 5².
Hindi: 2⁶ × 3⁴ × 5² के पूर्ण वर्ग भाजकों की संख्या ज्ञात कीजिए।
Solution:
For a divisor to be a perfect square, powers must be even.
For 2⁶, possible even powers:
0, 2, 4, 6
= 4 choices
For 3⁴, possible even powers:
0, 2, 4
= 3 choices
For 5², possible even powers:
0, 2
= 2 choices
Total perfect square divisors:
= 4 × 3 × 2
= 24
Answer: 24
21. Question
English: If f(x) = x² - 3x + 2, find f(x + 1) - f(x).
Hindi: यदि f(x) = x² - 3x + 2 है, तो f(x + 1) - f(x) ज्ञात कीजिए।
Solution:
f(x) = x² - 3x + 2
Now:
f(x + 1)
= (x + 1)² - 3(x + 1) + 2
= x² + 2x + 1 - 3x - 3 + 2
= x² - x
Now:
f(x + 1) - f(x)
= x² - x - (x² - 3x + 2)
= x² - x - x² + 3x - 2
= 2x - 2
Answer: 2x - 2
22. Question
English: Simplify: 1/(√5 - 2) + 1/(√5 + 2).
Hindi: सरलीकृत कीजिए: 1/(√5 - 2) + 1/(√5 + 2).
Solution:
1/(√5 - 2) + 1/(√5 + 2)
Take LCM:
= [(√5 + 2) + (√5 - 2)] / [(√5 - 2)(√5 + 2)]
Numerator:
= 2√5
Denominator:
= 5 - 4
= 1
So:
= 2√5
Answer: 2√5
23. Question
English: A invests ₹8000 for 12 months, B invests ₹10000 for 9 months, and C invests for 10 months. If their profit ratio is 16 : 15 : 12, find C's investment.
Hindi: A ₹8000 को 12 महीने के लिए, B ₹10000 को 9 महीने के लिए और C 10 महीने के लिए निवेश करता है। यदि उनका लाभ अनुपात 16 : 15 : 12 है, तो C का निवेश ज्ञात कीजिए।
Solution:
Profit depends on investment × time.
A’s capital-time:
= 8000 × 12
= 96000
B’s capital-time:
= 10000 × 9
= 90000
A : B
= 96000 : 90000
= 16 : 15
So ratio matches.
Let C’s investment = x
C’s capital-time:
= 10x
Now:
96000 corresponds to 16 parts.
1 part:
= 96000/16
= 6000
C corresponds to 12 parts.
C’s capital-time:
= 12 × 6000
= 72000
So:
10x = 72000
x = 7200
Answer: ₹7200
24. Question
English: A box contains 4 red, 5 blue and 3 green balls. Two balls are drawn at random. Find the probability that both balls are of the same colour.
Hindi: एक डिब्बे में 4 लाल, 5 नीली और 3 हरी गेंदें हैं। यादृच्छया 2 गेंदें निकाली जाती हैं। दोनों गेंदों के समान रंग की होने की प्रायिकता ज्ञात कीजिए।
Solution:
Total balls:
= 4 + 5 + 3
= 12
Total ways:
= 12C2
= 66
Favourable ways:
Both red:
= 4C2
= 6
Both blue:
= 5C2
= 10
Both green:
= 3C2
= 3
Total favourable:
= 6 + 10 + 3
= 19
Probability:
= 19/66
Answer: 19/66
25. Question
English: In an AP, the 5th term is 22 and the 11th term is 46. Find the sum of the first 20 terms.
Hindi: एक AP में 5वाँ पद 22 और 11वाँ पद 46 है। पहले 20 पदों का योग ज्ञात कीजिए।
Solution:
5th term:
a + 4d = 22
11th term:
a + 10d = 46
Subtract:
6d = 24
d = 4
Now:
a + 4 × 4 = 22
a + 16 = 22
a = 6
Sum of first 20 terms:
= 20/2 × [2a + 19d]
= 10 × [2 × 6 + 19 × 4]
= 10 × [12 + 76]
= 10 × 88
= 880
Answer: 880