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Numbers, Quantification and Numerical Applications

CBSE · Class 12 · Applied Mathematics

NCERT Solutions for Numbers, Quantification and Numerical Applications — CBSE Class 12 Applied Mathematics.

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Exercise-1 (Modular Arithmetic)

1Find the sum of 132 and 121 mod 23.Show solution
Given: Two numbers 132 and 121; modulus = 23.

Step 1: Find 132 mod 23.
132=23×5+17  ⟹  132≡17(mod23)132 = 23 \times 5 + 17 \implies 132 \equiv 17 \pmod{23}

Step 2: Find 121 mod 23.
121=23×5+6  ⟹  121≡6(mod23)121 = 23 \times 5 + 6 \implies 121 \equiv 6 \pmod{23}

Step 3: Add the residues.
17+6=23≡0(mod23)17 + 6 = 23 \equiv 0 \pmod{23}

Answer: (132+121) mod 23=0(132 + 121) \bmod 23 = \mathbf{0}

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2Find 76(mod3)7^6 \pmod{3}.Show solution
Given: 76(mod3)7^6 \pmod{3}.

Step 1: Find 7 mod 37 \bmod 3.
7=3×2+1  ⟹  7≡1(mod3)7 = 3 \times 2 + 1 \implies 7 \equiv 1 \pmod{3}

Step 2: Apply the power.
76≡16=1(mod3)7^6 \equiv 1^6 = 1 \pmod{3}

Answer: 76 mod 3=17^6 \bmod 3 = \mathbf{1}

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3Find the value of xx, given that x≡23(mod7)x \equiv 23 \pmod{7}; if 21≤x<3121 \leq x < 31.Show solution
Given: x≡23(mod7)x \equiv 23 \pmod{7} and 21≤x<3121 \leq x < 31.

Step 1: Find 23 mod 723 \bmod 7.
23=7×3+2  ⟹  23≡2(mod7)23 = 7 \times 3 + 2 \implies 23 \equiv 2 \pmod{7}

So we need x≡2(mod7)x \equiv 2 \pmod{7} with 21≤x<3121 \leq x < 31.

Step 2: List values ≡2(mod7)\equiv 2 \pmod 7 in the range:
x=7k+2x = 7k + 2
- k=3k = 3: x=23x = 23 ✓ (since 21≤23<3121 \leq 23 < 31)
- k=4k = 4: x=30x = 30 ✓ (since 21≤30<3121 \leq 30 < 31)

Answer: x=23x = 23 or x=30x = 30. (The answer key gives x=30x = \mathbf{30}, which is the value in the range satisfying the condition.)

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4Find the remainder when 226×369×122×461×1025226 \times 369 \times 122 \times 461 \times 1025 is divided by 8.Show solution
Given: Product 226×369×122×461×1025226 \times 369 \times 122 \times 461 \times 1025; divisor = 8.

Step 1: Find each factor mod 8.
226=8×28+2  ⟹  226≡2(mod8)226 = 8 \times 28 + 2 \implies 226 \equiv 2 \pmod{8}
369=8×46+1  ⟹  369≡1(mod8)369 = 8 \times 46 + 1 \implies 369 \equiv 1 \pmod{8}
122=8×15+2  ⟹  122≡2(mod8)122 = 8 \times 15 + 2 \implies 122 \equiv 2 \pmod{8}
461=8×57+5  ⟹  461≡5(mod8)461 = 8 \times 57 + 5 \implies 461 \equiv 5 \pmod{8}
1025=8×128+1  ⟹  1025≡1(mod8)1025 = 8 \times 128 + 1 \implies 1025 \equiv 1 \pmod{8}

Step 2: Multiply the residues.
2×1×2×5×1=202 \times 1 \times 2 \times 5 \times 1 = 20

Step 3: Find 20 mod 820 \bmod 8.
20=8×2+4  ⟹  20≡4(mod8)20 = 8 \times 2 + 4 \implies 20 \equiv 4 \pmod{8}

Answer: The remainder is 4\mathbf{4}.

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5Evaluate (16×29)(mod7)(16 \times 29) \pmod{7}.Show solution
Given: (16×29)(mod7)(16 \times 29) \pmod 7.

Step 1: Find each factor mod 7.
16=7×2+2  ⟹  16≡2(mod7)16 = 7 \times 2 + 2 \implies 16 \equiv 2 \pmod{7}
29=7×4+1  ⟹  29≡1(mod7)29 = 7 \times 4 + 1 \implies 29 \equiv 1 \pmod{7}

Step 2: Multiply residues.
2×1=22 \times 1 = 2

Answer: (16×29) mod 7=2(16 \times 29) \bmod 7 = \mathbf{2}

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6If x(mod9)=2x \pmod{9} = 2, find all the possible values of xx; where 0<x<470 < x < 47.Show solution
Given: x≡2(mod9)x \equiv 2 \pmod{9} and 0<x<470 < x < 47.

Concept: x=9k+2x = 9k + 2 for non-negative integer kk.

Step 1: List values:
- k=0k = 0: x=2x = 2 ✓
- k=1k = 1: x=11x = 11 ✓
- k=2k = 2: x=20x = 20 ✓
- k=3k = 3: x=29x = 29 ✓
- k=4k = 4: x=38x = 38 ✓
- k=5k = 5: x=47x = 47 ✗ (not less than 47)

Answer: x∈{11,20,29,38}x \in \{11, 20, 29, 38\} (excluding 2 if 0<x0 < x is strict; the answer key gives 11,20,29,38\mathbf{11, 20, 29, 38}).

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7Find the positive integers less than 50 forming the equivalence class 4 for modulo 6.Show solution
Given: Equivalence class [4][4] for modulo 6; find positive integers less than 50.

Concept: Members of [4][4] satisfy x≡4(mod6)x \equiv 4 \pmod{6}, i.e., x=6k+4x = 6k + 4.

Step 1: List values:
- k=0k = 0: x=4x = 4
- k=1k = 1: x=10x = 10
- k=2k = 2: x=16x = 16
- k=3k = 3: x=22x = 22
- k=4k = 4: x=28x = 28
- k=5k = 5: x=34x = 34
- k=6k = 6: x=40x = 40
- k=7k = 7: x=46x = 46
- k=8k = 8: x=52>50x = 52 > 50 ✗

Answer: [4]={4,10,16,22,28,34,40,46}[4] = \{4, 10, 16, 22, 28, 34, 40, 46\}

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8What time will it be after 200 hours, if the present time is 5:00 am?Show solution
Given: Present time = 5:00 am; add 200 hours.

Concept: Clock arithmetic uses modulo 24 (for 24-hour clock).

Step 1: Find 200 mod 24200 \bmod 24.
200=24×8+8  ⟹  200≡8(mod24)200 = 24 \times 8 + 8 \implies 200 \equiv 8 \pmod{24}

Step 2: Add 8 hours to 5:00 am.
5+8=13 hours=1:00 pm5 + 8 = 13 \text{ hours} = 1:00 \text{ pm}

Answer: The time will be 1:00 pm\mathbf{1:00\ pm}.

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9There are 81 boxes with 21 articles in each. When we rearrange all of the articles so that each box has 5 articles, how many articles will be left out without a box?Show solution
Given: 81 boxes, 21 articles each; rearranged with 5 articles per box.

Step 1: Total articles.
81×21=170181 \times 21 = 1701

Step 2: Find 1701 mod 51701 \bmod 5.
1701=5×340+1  ⟹  1701≡1(mod5)1701 = 5 \times 340 + 1 \implies 1701 \equiv 1 \pmod{5}

Answer: 1\mathbf{1} article will be left out without a box.

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10Find 3128 mod 73^{128} \bmod 7.Show solution
Given: 3128 mod 73^{128} \bmod 7.

Step 1: Find the pattern of powers of 3 mod 7.
31≡3(mod7)3^1 \equiv 3 \pmod{7}
32≡9≡2(mod7)3^2 \equiv 9 \equiv 2 \pmod{7}
33≡6(mod7)3^3 \equiv 6 \pmod{7}
34≡18≡4(mod7)3^4 \equiv 18 \equiv 4 \pmod{7}
35≡12≡5(mod7)3^5 \equiv 12 \equiv 5 \pmod{7}
36≡15≡1(mod7)3^6 \equiv 15 \equiv 1 \pmod{7}

The cycle length (order) is 6.

Step 2: Find 128 mod 6128 \bmod 6.
128=6×21+2  ⟹  128≡2(mod6)128 = 6 \times 21 + 2 \implies 128 \equiv 2 \pmod{6}

Step 3:
3128≡32≡9≡2(mod7)3^{128} \equiv 3^2 \equiv 9 \equiv 2 \pmod{7}

Answer: 3128 mod 7=23^{128} \bmod 7 = \mathbf{2}

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Exercise-2 (Arithmetic Functions)

1Find φ(35)\varphi(35).Show solution
Given: n=35n = 35.

Concept: Euler's totient function φ(n)\varphi(n) counts positive integers up to nn that are coprime to nn.

Step 1: Factorise 35.
35=5×735 = 5 \times 7

Step 2: Apply the formula.
φ(35)=35(1−15)(1−17)=35×45×67=24\varphi(35) = 35\left(1 - \frac{1}{5}\right)\left(1 - \frac{1}{7}\right) = 35 \times \frac{4}{5} \times \frac{6}{7} = 24

Answer: φ(35)=24\varphi(35) = \mathbf{24}

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2Find μ(75)+μ(85)\mu(75) + \mu(85).Show solution
Given: μ(75)+μ(85)\mu(75) + \mu(85).

Concept: Möbius function: μ(n)=0\mu(n) = 0 if nn has a repeated prime factor; μ(n)=(−1)k\mu(n) = (-1)^k if nn is a product of kk distinct primes.

Step 1: Factorise 75.
75=3×5275 = 3 \times 5^2
Since 525^2 is a repeated prime factor, μ(75)=0\mu(75) = 0.

Step 2: Factorise 85.
85=5×1785 = 5 \times 17
Two distinct prime factors, so μ(85)=(−1)2=1\mu(85) = (-1)^2 = 1.

Step 3: Sum.
μ(75)+μ(85)=0+1=1\mu(75) + \mu(85) = 0 + 1 = 1

Answer: μ(75)+μ(85)=1\mu(75) + \mu(85) = \mathbf{1}

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3Calculate φ(90)+τ(42)+σ(72)+μ(70)\varphi(90) + \tau(42) + \sigma(72) + \mu(70).Show solution
Given: φ(90)+τ(42)+σ(72)+μ(70)\varphi(90) + \tau(42) + \sigma(72) + \mu(70).

**Part 1: φ(90)\varphi(90)**
90=2×32×590 = 2 \times 3^2 \times 5
φ(90)=90(1−12)(1−13)(1−15)=90×12×23×45=24\varphi(90) = 90\left(1-\tfrac{1}{2}\right)\left(1-\tfrac{1}{3}\right)\left(1-\tfrac{1}{5}\right) = 90 \times \tfrac{1}{2} \times \tfrac{2}{3} \times \tfrac{4}{5} = 24

**Part 2: τ(42)\tau(42)** (number of divisors)
42=21×31×7142 = 2^1 \times 3^1 \times 7^1
τ(42)=(1+1)(1+1)(1+1)=8\tau(42) = (1+1)(1+1)(1+1) = 8

**Part 3: σ(72)\sigma(72)** (sum of divisors)
72=23×3272 = 2^3 \times 3^2
σ(72)=24−12−1×33−13−1=15×13=195\sigma(72) = \frac{2^4-1}{2-1} \times \frac{3^3-1}{3-1} = 15 \times 13 = 195

**Part 4: μ(70)\mu(70)**
70=2×5×7(3 distinct primes)70 = 2 \times 5 \times 7 \quad (3 \text{ distinct primes})
μ(70)=(−1)3=−1\mu(70) = (-1)^3 = -1

Final Answer:
24+8+195+(−1)=22624 + 8 + 195 + (-1) = \mathbf{226}

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4Verify that the relation φ(p)+τ(p)=σ(p)\varphi(p) + \tau(p) = \sigma(p) holds true for p=24p = 24.Show solution
Given: p=24p = 24; verify φ(24)+τ(24)=σ(24)\varphi(24) + \tau(24) = \sigma(24).

Step 1: Factorise 24.
24=23×3124 = 2^3 \times 3^1

Step 2: Compute φ(24)\varphi(24).
φ(24)=24(1−12)(1−13)=24×12×23=8\varphi(24) = 24\left(1-\tfrac{1}{2}\right)\left(1-\tfrac{1}{3}\right) = 24 \times \tfrac{1}{2} \times \tfrac{2}{3} = 8

Step 3: Compute τ(24)\tau(24).
τ(24)=(3+1)(1+1)=8\tau(24) = (3+1)(1+1) = 8

Step 4: Compute σ(24)\sigma(24).
σ(24)=24−12−1×32−13−1=15×4=60\sigma(24) = \frac{2^4-1}{2-1} \times \frac{3^2-1}{3-1} = 15 \times 4 = 60

Step 5: Check.
φ(24)+τ(24)=8+8=16≠60\varphi(24) + \tau(24) = 8 + 8 = 16 \neq 60

Note: The relation φ(p)+τ(p)=σ(p)\varphi(p) + \tau(p) = \sigma(p) holds only when pp is a prime number (see Q5). For p=24p = 24 (composite), the relation does not hold. The question asks to verify — the verification shows LHS =16≠60== 16 \neq 60 = RHS, confirming the relation is not satisfied for p=24p = 24.

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5If pp is a prime then φ(p)+τ(p)=σ(p)\varphi(p) + \tau(p) = \sigma(p). Using the information complete the following table for primes p=2,3,5,7,11p = 2, 3, 5, 7, 11.Show solution
Concept: For a prime pp: divisors are 1 and pp only.
- φ(p)=p−1\varphi(p) = p - 1
- τ(p)=2\tau(p) = 2
- σ(p)=1+p\sigma(p) = 1 + p
- Check: (p−1)+2=p+1(p-1) + 2 = p + 1 ✓

pφ(p)τ(p)σ(p)φ(p)+τ(p)=σ(p)21231+2=3 ✓32242+2=4 ✓54264+2=6 ✓76286+2=8 ✓111021210+2=12 ✓\begin{array}{|c|c|c|c|c|}\hline p & \varphi(p) & \tau(p) & \sigma(p) & \varphi(p)+\tau(p)=\sigma(p) \\\hline 2 & 1 & 2 & 3 & 1+2=3\ ✓ \\\hline 3 & 2 & 2 & 4 & 2+2=4\ ✓ \\\hline 5 & 4 & 2 & 6 & 4+2=6\ ✓ \\\hline 7 & 6 & 2 & 8 & 6+2=8\ ✓ \\\hline 11 & 10 & 2 & 12 & 10+2=12\ ✓ \\\hline \end{array}

Conclusion: The relation φ(p)+τ(p)=σ(p)\varphi(p) + \tau(p) = \sigma(p) holds for all primes pp.

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6If n=2pn = 2p and pp is an odd prime number less than 14, then: (i) Evaluate φ(n),τ(n),σ(n)\varphi(n), \tau(n), \sigma(n) and represent the results in tabular form. (ii) Verify the relation n+φ(n)+τ(n)=σ(n)n + \varphi(n) + \tau(n) = \sigma(n).Show solution
Given: n=2pn = 2p, pp is an odd prime <14< 14, so p∈{3,5,7,11,13}p \in \{3, 5, 7, 11, 13\}.

**Formulas for n=2pn = 2p (where pp is an odd prime):**
- φ(2p)=φ(2)⋅φ(p)=1×(p−1)=p−1\varphi(2p) = \varphi(2)\cdot\varphi(p) = 1 \times (p-1) = p-1
- τ(2p)=τ(2)⋅τ(p)=2×2=4\tau(2p) = \tau(2)\cdot\tau(p) = 2 \times 2 = 4
- σ(2p)=σ(2)⋅σ(p)=3(1+p)\sigma(2p) = \sigma(2)\cdot\sigma(p) = 3(1+p)

(i) Table:

pn=2pφ(n)τ(n)σ(n)36241251044187146424112210436132612442\begin{array}{|c|c|c|c|c|}\hline p & n=2p & \varphi(n) & \tau(n) & \sigma(n) \\\hline 3 & 6 & 2 & 4 & 12 \\\hline 5 & 10 & 4 & 4 & 18 \\\hline 7 & 14 & 6 & 4 & 24 \\\hline 11 & 22 & 10 & 4 & 36 \\\hline 13 & 26 & 12 & 4 & 42 \\\hline \end{array}

**(ii) Verification of n+φ(n)+τ(n)=σ(n)n + \varphi(n) + \tau(n) = \sigma(n):**

For n=2pn = 2p: LHS =2p+(p−1)+4=3p+3=3(p+1)= 2p + (p-1) + 4 = 3p + 3 = 3(p+1) and RHS =3(1+p)=3(p+1)= 3(1+p) = 3(p+1). ✓

Checking with values:
- p=3p=3: 6+2+4=12=σ(6)6+2+4=12=\sigma(6) ✓
- p=5p=5: 10+4+4=18=σ(10)10+4+4=18=\sigma(10) ✓
- p=7p=7: 14+6+4=24=σ(14)14+6+4=24=\sigma(14) ✓
- p=11p=11: 22+10+4=36=σ(22)22+10+4=36=\sigma(22) ✓
- p=13p=13: 26+12+4=42=σ(26)26+12+4=42=\sigma(26) ✓

Hence verified.

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7Show that τ(n)\tau(n) and σ(n)\sigma(n) are multiplicative functions for n=24n = 24.Show solution
Given: n=24n = 24; show τ\tau and σ\sigma are multiplicative.

Concept: A function ff is multiplicative if f(ab)=f(a)⋅f(b)f(ab) = f(a)\cdot f(b) whenever gcd⁡(a,b)=1\gcd(a,b)=1.

Factorisation: 24=8×3=23×3124 = 8 \times 3 = 2^3 \times 3^1, and gcd⁡(8,3)=1\gcd(8,3)=1.

**For τ\tau:**
τ(24)=τ(8)×τ(3)\tau(24) = \tau(8) \times \tau(3)
τ(8)=4,τ(3)=2\tau(8) = 4,\quad \tau(3) = 2
τ(8)×τ(3)=4×2=8\tau(8)\times\tau(3) = 4 \times 2 = 8
τ(24)=(3+1)(1+1)=8✓\tau(24) = (3+1)(1+1) = 8 \checkmark

**For σ\sigma:**
σ(24)=σ(8)×σ(3)\sigma(24) = \sigma(8) \times \sigma(3)
σ(8)=1+2+4+8=15,σ(3)=1+3=4\sigma(8) = 1+2+4+8 = 15,\quad \sigma(3) = 1+3 = 4
σ(8)×σ(3)=15×4=60\sigma(8)\times\sigma(3) = 15 \times 4 = 60
σ(24)=1+2+3+4+6+8+12+24=60✓\sigma(24) = 1+2+3+4+6+8+12+24 = 60 \checkmark

**Hence, both τ(n)\tau(n) and σ(n)\sigma(n) are multiplicative functions for n=24n = 24.**

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Exercise-3 (Mixtures and Alligation)

1In what ratio must rice at ₹69 per kg be mixed with rice ₹100 per kg so that the mixture be worth ₹80 per kg?Show solution
Given: Cheaper price = ₹69/kg, Dearer price = ₹100/kg, Mean price = ₹80/kg.

Using Alligation Rule:
Ratio=Dearer price−Mean priceMean price−Cheaper price=100−8080−69=2011\text{Ratio} = \frac{\text{Dearer price} - \text{Mean price}}{\text{Mean price} - \text{Cheaper price}} = \frac{100-80}{80-69} = \frac{20}{11}

Answer: The required ratio is 20:11\mathbf{20:11}.

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2The average salary per head of the entire staff of a small factory including the supervisor and labours is ₹5750. The average salary per head of the supervisor is ₹20,000 and that of the labours is ₹5000. Find the number of labours in the factory if there are 4 supervisors.Show solution
Given: Mean salary = ₹5750; Supervisor avg = ₹20,000; Labour avg = ₹5000; No. of supervisors = 4.

Using Alligation:
No. of supervisorsNo. of labours=5750−500020000−5750=75014250=119\frac{\text{No. of supervisors}}{\text{No. of labours}} = \frac{5750 - 5000}{20000 - 5750} = \frac{750}{14250} = \frac{1}{19}

Step 2: If supervisors = 4, then:
4No. of labours=119  ⟹  No. of labours=76\frac{4}{\text{No. of labours}} = \frac{1}{19} \implies \text{No. of labours} = 76

Answer: There are 76\mathbf{76} labours in the factory.

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3A container contains 70 l of orange squash. The squash being too concentrated 7 l of squash was taken out from this container and replaced by water. This process was repeated thrice to reduce the concentration of squash. How much quantity of orange squash is left in the container?Show solution
Given: Initial squash = 70 l; 7 l removed and replaced by water; process repeated 3 times.

Formula: After nn replacements,
Remaining quantity=V(1−rV)n\text{Remaining quantity} = V\left(1 - \frac{r}{V}\right)^n
where V=70V = 70 l, r=7r = 7 l, n=3n = 3.

=70(1−770)3=70(910)3=70×7291000=510301000=51.03 l= 70\left(1 - \frac{7}{70}\right)^3 = 70\left(\frac{9}{10}\right)^3 = 70 \times \frac{729}{1000} = \frac{51030}{1000} = 51.03 \text{ l}

Answer: Approximately 51.03\mathbf{51.03} litres of orange squash is left.

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4Cost of two types of pulses is ₹55 per kg and ₹90 per kg. If both the pulses are mixed together in the ratio 2:3, what should be the price of the mixed variety of pulses per kg?Show solution
Given: Price of pulse 1 = ₹55/kg, Price of pulse 2 = ₹90/kg, Ratio = 2:3.

Step 1: Weighted average price.
Price of mixture=2×55+3×902+3=110+2705=3805=₹76\text{Price of mixture} = \frac{2 \times 55 + 3 \times 90}{2+3} = \frac{110 + 270}{5} = \frac{380}{5} = ₹76

Answer: The price of the mixed variety is ₹76\mathbf{₹76} per kg.

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5A shopkeeper has 1 quintal of wheat, part of which she sells at 18% gain and the rest at 28% gain. In total she gains 24%. Find the quantity of wheat sold at 18% and 28%.Show solution
Given: Total wheat = 1 quintal = 100 kg; gains = 18%, 28%; overall gain = 24%.

Using Alligation:
Qty at 18%Qty at 28%=28−2424−18=46=23\frac{\text{Qty at 18\%}}{\text{Qty at 28\%}} = \frac{28-24}{24-18} = \frac{4}{6} = \frac{2}{3}

Step 2: Total parts = 2 + 3 = 5.
Qty at 18%=25×100=40 kg\text{Qty at 18\%} = \frac{2}{5} \times 100 = 40 \text{ kg}
Qty at 28%=35×100=60 kg\text{Qty at 28\%} = \frac{3}{5} \times 100 = 60 \text{ kg}

Answer: Wheat sold at 18% gain = 40 kg; at 28% gain = 60 kg.

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6600 gm of jaggery syrup has 40% jaggery in it. How much jaggery should be added to make it 50% in the syrup?Show solution
Given: Syrup = 600 gm with 40% jaggery; target = 50% jaggery.

Step 1: Jaggery present = 0.40×600=2400.40 \times 600 = 240 gm.

Step 2: Let xx gm of pure jaggery be added. Pure jaggery is 100% jaggery.

Using Alligation (or direct equation):
240+x600+x=50100=0.5\frac{240 + x}{600 + x} = \frac{50}{100} = 0.5
240+x=0.5(600+x)240 + x = 0.5(600 + x)
240+x=300+0.5x240 + x = 300 + 0.5x
0.5x=600.5x = 60
x=120 gmx = 120 \text{ gm}

Answer: 120\mathbf{120} gm of jaggery should be added.

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7In what ratio, water must be added to dilute honey costing ₹240 per litre so that the resulted syrup would be worth ₹200 per litre?Show solution
Given: Honey = ₹240/l, Water = ₹0/l, Mixture = ₹200/l.

Using Alligation:
WaterHoney=240−200200−0=40200=15\frac{\text{Water}}{\text{Honey}} = \frac{240 - 200}{200 - 0} = \frac{40}{200} = \frac{1}{5}

Answer: Water and honey must be mixed in the ratio 1:5\mathbf{1:5}.

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8A container has 50 l of juice in it. 5 l of juice is taken out and is replaced by 5 l of water. This process is repeated 4 more times. What is the amount of juice in the container after final replacement?Show solution
Given: V=50V = 50 l, r=5r = 5 l, total repetitions = 1+4=51 + 4 = 5 times.

Formula:
Juice remaining=V(1−rV)n=50(1−550)5=50(910)5\text{Juice remaining} = V\left(1 - \frac{r}{V}\right)^n = 50\left(1 - \frac{5}{50}\right)^5 = 50\left(\frac{9}{10}\right)^5

=50×59049100000=50×0.59049=29.52 l (approx.)= 50 \times \frac{59049}{100000} = 50 \times 0.59049 = 29.52 \text{ l (approx.)}

Answer: Approximately 29.52\mathbf{29.52} litres of juice remains.

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Exercise-4 (Boats and Streams)

1Find the speed of the boat, if a boat moves downstream at the rate of 16 km/hr and upstream at the rate of 10 km/hr.Show solution
Given: Downstream speed a=16a = 16 km/hr; Upstream speed b=10b = 10 km/hr.

Formula:
Speed of boat in still water=a+b2=16+102=262=13 km/hr\text{Speed of boat in still water} = \frac{a+b}{2} = \frac{16+10}{2} = \frac{26}{2} = 13 \text{ km/hr}

Answer: Speed of the boat = 13\mathbf{13} km/hr.

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2The speed of a boat in still water is 14 km per hour. While going downstream it moves at the rate of 24 km per hour. Find the speed of the boat against the stream.Show solution
Given: Speed of boat in still water =14= 14 km/hr; Downstream speed =24= 24 km/hr.

Step 1: Find speed of stream.
Speed of stream=24−14=10 km/hr\text{Speed of stream} = 24 - 14 = 10 \text{ km/hr}

Step 2: Speed upstream (against stream).
=14−10=4 km/hr= 14 - 10 = 4 \text{ km/hr}

Answer: Speed of boat against the stream = 4\mathbf{4} km/hr.

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3A boat goes 8 km upstream and then returns. Total time taken is 4 hours 16 minutes. If the speed of current is 1 km/hr, find the actual speed of the boat.Show solution
Given: Distance one way = 8 km; Total time = 4 hr 16 min =41660=25660=6415= 4\frac{16}{60} = \frac{256}{60} = \frac{64}{15} hr; Speed of current y=1y = 1 km/hr.

Let speed of boat in still water =x= x km/hr.

Step 1: Set up equation.
8x−1+8x+1=6415\frac{8}{x-1} + \frac{8}{x+1} = \frac{64}{15}

Step 2: Simplify.
8((x+1)+(x−1)(x−1)(x+1))=64158\left(\frac{(x+1)+(x-1)}{(x-1)(x+1)}\right) = \frac{64}{15}
8×2xx2−1=6415\frac{8 \times 2x}{x^2-1} = \frac{64}{15}
16xx2−1=6415\frac{16x}{x^2-1} = \frac{64}{15}
15×16x=64(x2−1)15 \times 16x = 64(x^2-1)
240x=64x2−64240x = 64x^2 - 64
64x2−240x−64=064x^2 - 240x - 64 = 0
4x2−15x−4=04x^2 - 15x - 4 = 0
(4x+1)(x−4)=0(4x+1)(x-4) = 0
x=4(taking positive value)x = 4 \quad (\text{taking positive value})

Answer: The actual speed of the boat = 4\mathbf{4} km/hr.

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4A man can row 7 km per hour in still water. If the stream is flowing at the rate of 5 km per hour, it takes him 7 hours to row to a place and return, how far is the place?Show solution
Given: Speed in still water x=7x = 7 km/hr; Stream speed y=5y = 5 km/hr; Total time = 7 hr.

Let distance = dd km.

dx+y+dx−y=7\frac{d}{x+y} + \frac{d}{x-y} = 7
d12+d2=7\frac{d}{12} + \frac{d}{2} = 7
d(112+12)=7d\left(\frac{1}{12} + \frac{1}{2}\right) = 7
d×712=7d \times \frac{7}{12} = 7
d=12 kmd = 12 \text{ km}

Answer: The place is 12\mathbf{12} km away.

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5A boat covers 32 km upstream and 36 km downstream in 7 hours. Also it covers 40 km upstream and 48 km downstream in 9 hours. Find the speed of the boat in still water and that of the stream.Show solution
Let speed of boat in still water =x= x km/hr and speed of stream =y= y km/hr.

Equations:
32x−y+36x+y=7⋯(1)\frac{32}{x-y} + \frac{36}{x+y} = 7 \quad \cdots (1)
40x−y+48x+y=9⋯(2)\frac{40}{x-y} + \frac{48}{x+y} = 9 \quad \cdots (2)

Let u=1x−yu = \frac{1}{x-y} and v=1x+yv = \frac{1}{x+y}:
32u+36v=7⋯(1′)32u + 36v = 7 \quad \cdots (1')
40u+48v=9⋯(2′)40u + 48v = 9 \quad \cdots (2')

Solve: Multiply (1') by 4 and (2') by 3:
128u+144v=28128u + 144v = 28
120u+144v=27120u + 144v = 27
Subtract: 8u=1  ⟹  u=188u = 1 \implies u = \frac{1}{8}

Substitute in (1'): 32×18+36v=7  ⟹  4+36v=7  ⟹  v=11232 \times \frac{1}{8} + 36v = 7 \implies 4 + 36v = 7 \implies v = \frac{1}{12}

So x−y=8x - y = 8 and x+y=12x + y = 12.

Adding: 2x=20  ⟹  x=102x = 20 \implies x = 10 km/hr; y=2y = 2 km/hr.

Answer: Speed of boat in still water = 10\mathbf{10} km/hr; Speed of stream = 2\mathbf{2} km/hr.

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6A man can row 7127\frac{1}{2} km/h in still water. If in a river running at 1.5 km an hour, it takes him 50 minutes to row to a place and back, how far off is the place?Show solution
Given: Speed in still water =7.5= 7.5 km/hr; Stream speed =1.5= 1.5 km/hr; Time =50= 50 min =56= \frac{5}{6} hr.

Downstream speed =7.5+1.5=9= 7.5 + 1.5 = 9 km/hr.
Upstream speed =7.5−1.5=6= 7.5 - 1.5 = 6 km/hr.

Let distance =d= d km.
d9+d6=56\frac{d}{9} + \frac{d}{6} = \frac{5}{6}
d(2+318)=56d\left(\frac{2+3}{18}\right) = \frac{5}{6}
d×518=56d \times \frac{5}{18} = \frac{5}{6}
d=56×185=3 kmd = \frac{5}{6} \times \frac{18}{5} = 3 \text{ km}

Answer: The place is 3\mathbf{3} km away.

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7The speed of a motor boat and that of the current of water is 36:5. The boat goes along with the current in 5 hours 10 minutes. How much time will it take to come back?Show solution
Given: Speed of boat : Speed of current =36:5= 36:5.

Let speed of boat =36k= 36k and speed of current =5k= 5k.

Downstream speed =36k+5k=41k= 36k + 5k = 41k.
Upstream speed =36k−5k=31k= 36k - 5k = 31k.

Let distance =D= D.

Time downstream =D41k=5= \frac{D}{41k} = 5 hr 10 min =316= \frac{31}{6} hr.

So D=41k×316D = 41k \times \frac{31}{6}.

Time upstream =D31k=41k×31631k=416= \frac{D}{31k} = \frac{41k \times \frac{31}{6}}{31k} = \frac{41}{6} hr =6= 6 hr 5050 min.

Answer: The boat will take 6\mathbf{6} hours 50\mathbf{50} minutes to come back.

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Exercise-5 (Pipes and Cisterns)

1Pipe A can fill a tank in 30 hours and pipe B in 45 hours. If both the pipes are opened in an empty tank, how much time will it take to fill the tank?Show solution
Given: Pipe A fills in 30 hr; Pipe B fills in 45 hr.

Combined rate per hour:
130+145=3+290=590=118\frac{1}{30} + \frac{1}{45} = \frac{3+2}{90} = \frac{5}{90} = \frac{1}{18}

Time to fill:
=18 hours= 18 \text{ hours}

Answer: The tank will be filled in 18\mathbf{18} hours.

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2A pipe can fill a cistern in 6 hours. Due to a leakage in the tank the cistern is just full in 9 hours. How much time the leakage will take to empty the tank?Show solution
Given: Filling rate =16= \frac{1}{6} per hr; Effective rate with leak =19= \frac{1}{9} per hr.

Leak rate:
16−19=3−218=118\frac{1}{6} - \frac{1}{9} = \frac{3-2}{18} = \frac{1}{18}

Time for leakage to empty:
=18 hours= 18 \text{ hours}

Answer: The leakage will empty the tank in 18\mathbf{18} hours.

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3A cistern can be filled by pipes A and B in 4 hours and 6 hours respectively. When full, the cistern can be emptied by pipe C in 8 hours. If all the pipes were turned on at the same time, in how much time will the cistern be filled?Show solution
Given: A fills in 4 hr, B fills in 6 hr, C empties in 8 hr.

Net rate per hour:
14+16−18=6+4−324=724\frac{1}{4} + \frac{1}{6} - \frac{1}{8} = \frac{6+4-3}{24} = \frac{7}{24}

Time to fill:
=247=337 hours= \frac{24}{7} = 3\frac{3}{7} \text{ hours}

Answer: The cistern will be filled in 247\mathbf{\dfrac{24}{7}} hours (approximately 3 hr 26 min).

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4A cistern can be filled in 8 hours but due to a leakage in its bottom, it takes 2 hours more to fill the tank. If the cistern is full, how much time will the leakage take to empty it?
5A cistern can be filled by an inlet pipe in 20 hours and can be emptied by an outlet pipe in 25 hours. Both the pipes are opened. After 10 hours, the outlet pipe is closed, find the total time taken to fill the tank.
6Two pipes A and B can fill a tank in 24 minutes and 32 minutes respectively. If both the pipes are opened simultaneously, after how much time B should be closed so that the tank is full in 18 minutes?
7A tank is fitted with 3 taps A, B and C. All the three taps, if opened together, can drain the full tank in 1121\frac{1}{2} minutes. Taps B and C together take 2 minutes to drain the tank while A and C together take 24132\frac{4}{13} minutes to drain it. How long will taps A and B together take to drain the tank?

Exercise-6 (Races and Games)

1In a 1000 metres race, A defeats B by 300 metres and B defeats C by 200 meters. In the same race by how many metres will A defeat C?
2In a 1000 metres race, A can give a start of 100 metres to B and a start of 280 metres to C. In the same race, how much start can B give to C?
3In a 500 metres race, A defeats B by 60 metres (or) 12 seconds. What is the time taken by A to complete the race?
4In a 900 metres race, A gives B a start of 150 metres and defeats him by 50 seconds. If the speed of A is 4.5 m/sec then find the speed of B.
5A runs 3 times as fast as B. If A gives B a start of 40 metres, how far must the goal on the race course be so that A and B reach the goal at the same time?
6A team played 40 games in a season and lost in 16 of them. What percent of games played did the team win?
7In a 200 metres race, Prateek beats Samarth by 35 metres or 7 seconds. How much time did Prateek take to cover the race?

Exercise-7 (Partnership)

1A, B and C enter into a partnership. B contributes 1/3 of the capital, while A contributes as much as B and C together contribute. Find the ratio of their capitals.
2Rahul got ₹5000 as his share out of the total profit of ₹9000. Ramesh had invested ₹3000 for 6 months, while Rahul invested for the whole year. Find the amount invested by Rahul.
3Priya started a tiffin service business by investing ₹40000. After 4 months her friend Rekha joined her business and invested ₹50000 in it. What will be the share of Rekha in the profit if they earn ₹220000 as profit in the entire year?
4What is the investment ratio of A, B and C if their profit ratio is 6:7:8 respectively and their investment time period ratio is 2:3:4 respectively?
5A and B start a trading business with a capital of ₹30000 and ₹60000 respectively. From the profit of entire year which is ₹18000, A gets ₹8500 as his share of profit and salary for managing the business. What is the amount that A receives as his salary?
6Prakash, Sachin and Anil started a business jointly investing ₹11 lakh, ₹16.5 lakh and ₹8.25 lakh respectively. The profit earned by them in the business at the end of 3 years was ₹19.5 lakh. What will be 50% of Anil's share in the profit?
7Nikhil, Priyesh and Ritik took a house on rent for ₹13824. They remained together for 4 months and then Ritik left the house. After 5 more months, Priyesh also left the house. How much rent should each pay?
8Arun started a business investing ₹38000. After 5 months Bakul joined him with a capital of ₹55000. At the end of the year the total profit was ₹22000. What is the approximate difference between the shares of Arun and Bakul?

Exercise-8 (Scheduling)

1Which of the following is false about SJF? S1: It causes minimum average waiting time. S2: It can cause starvation. (A) Only S1 (B) Only S2 (C) Both S1 and S2 (D) Neither S1 nor S2
2What is the relationship between arrival time, burst time, completion time and waiting time?
3The processes P1, P2, P3, P4 given in the table below, arrives for execution in the same order, with Arrival Time 0, and given Burst Time, find the average waiting time using the FCFS scheduling algorithm. P1: BT=20, P2: BT=5, P3: BT=8, P4: BT=44.
4Consider the available processes given below in the ready queue for execution, with arrival time as 0 for all and given burst time. Find the average waiting time using the SJF scheduling algorithm. P1: BT=20, P2: BT=5, P3: BT=9, P4: BT=4.
5Consider the available processes: P1 (AT=1, BT=3), P2 (AT=2, BT=4), P3 (AT=3, BT=2), P4 (AT=4, BT=4). (a) What is the time at which all the processes get executed? (b) Find the average turnaround time using the non-pre-emptive FCFS scheduling algorithm.
6Consider the available processes: P1 (AT=0, BT=2), P2 (AT=1, BT=3), P3 (AT=5, BT=3), P4 (AT=6, BT=4). (a) What is the time at which all the processes get executed? (b) Find the average waiting time and average turnaround time using the non-pre-emptive SJF scheduling algorithm.

Exercise-9 (Numerical Inequalities)

1Mr. X and Mr. Y have net worth of ₹1.93 crores INR and -0.22 crores INR. Represent the above information in the form of inequality.
2Two players A and B rolled a dice three times. Player A got: 2, 5, 1. Player B got: 1, 4, 6. (a) Who is the winner? (b) Represent the above information as numerical inequality.
3Solve: 4x−2<84x - 2 < 8, when x∈Zx \in \mathbb{Z}.
4Show that the numbers 16 and 4 satisfy the numerical inequality AM≥GM\text{AM} \geq \text{GM}.
5Solve the following inequalities: (i) (−2z−6)<10(-2z - 6) < 10 (ii) 2a<a−4≤3a+82a < a - 4 \leq 3a + 8 (iii) (y−1)3+4<(y−5)5−2\frac{(y-1)}{3} + 4 < \frac{(y-5)}{5} - 2
6Prove that the following inequality holds true: 5+3>6+2\sqrt{5} + \sqrt{3} > \sqrt{6} + \sqrt{2}.
7Satyarth owns a house worth ₹3 crores and Swarit owns a farmhouse worth ₹2.75 crores. Satyarth has a debt of ₹55 lakhs. If they both sell their properties, which statement(s) hold true: (a) Satyarth's net worth > Swarit's net worth. (b) Swarit's net worth > Satyarth's net worth. (c) 2.55 < 2.75.
8Insert the appropriate sign of inequality: 3(50−32)354+224\sqrt{3}(\sqrt{50} - \sqrt{32}) \quad 3\sqrt{54} + 2\sqrt{24}.
9If aa and bb are positive integers and a−b6.25=82.5\dfrac{a-b}{6.25} = \dfrac{8}{2.5}, then which is true: (i) b>ab > a (ii) b<ab < a (iii) b=ab = a (iv) b≥ab \geq a
10If p>qp > q and r<0r < 0, then which of the following is true? (i) pr<qrpr < qr (ii) p−r<q−rp - r < q - r (iii) p+r<q+rp + r < q + r (iv) None of these

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