1. The sum of three numbers is 98. If the ratio of the first to the second is 2 : 3 and that of the second to the third is 5 : 8, then the second number is:

A. 20 B. 30 C. 38 D. 48

Answer: Option B
Explanation: [latex]{1}^{st}[/latex]: [latex]{2}^{nd}[/latex] → 2:3 [latex]{2}^{nd}[/latex]: [latex]{3}^{nd}[/latex] → 5:8 ————————- [latex]{1}^{st}[/latex]: [latex]{2}^{nd}[/latex]: [latex]{3}^{nd}[/latex]→10:15:24 [latex]{2}^{nd}[/latex] number = 98 x [latex]\frac{15}{49}[/latex] = 30
2. Two numbers are such as that square of one is 224 less than 8 times the square of the other. If the numbers are in the ratio of 3 : 4, they are?

A. 12, 16 B. 6, 8 C. 9, 12 D. None of these

Answer: Option B
Explanation: Numebr → 3:4 [latex]{1}^{st}[/latex] number → 3x [latex]{2}^{nd}[/latex] number → 4x 8 x 9 [latex]{x}^{2}[/latex] - 16 [latex]{x}^{2}[/latex] = 224 72 [latex]{x}^{2}[/latex] - 16 [latex]{x}^{2}[/latex] = 224 56 [latex]{x}^{2}[/latex] = 224 [latex]{x}^{2}[/latex] = 4 x = 2 Numbers are → 6.8
3. Tea worth 126 per kg and 135 per kg are mixed with a third variety in the ratio 1 : 1 : 2. If the mixture is worth 153 per kg, then the price of the third variety per kg is?

A. 169.50 B. 170 C. 175.50 D. 180

Answer: Option C
Explanation: Let tea use 1 kg, 1 kg and 2 kg Let price of [latex]{3}^{rd}[/latex] verity x [latex]\frac{126 + 135 + x + 2}{1+ 1+ 2}[/latex] = 153 261 + 2x = 153 x 4 261 + 2x = 612 2x = 351 x = 175.50
4. In a mixture of 45 litres, the ratio of milk and water is 3 : 2. How much water must be added to make the ratio 9 : 11?

A. 10 litres B. 15 litres C. 17 litres D. 20 litres

Answer: Option B
Explanation: Milk Water = 3:2x3 New, ratio → Milk:Water = 9:11 Milk:Water = 9:6 New Milk:Water = 9:11 [9 + 6r] → 45 liters 15r → 45 liters 1r → 3 liters Water added ⇒ [11 - 6r] Let price of [latex]{3}^{rd}[/latex] verity x [latex]\frac{126 + 135 + x + 2}{1+ 1+ 2}[/latex] = 5 x 3 = 15 liters
5. The ratio of the rate of flow of water in pipes varies inversely as the square of the radii of the pipes. What is the ratio of the rates of flow in two pipes of diameters 2 cm and 4 cm, respectively?

A. 1 : 2 B. 2 : 1 C. 1 : 8 D. 4 : 1

Answer: Option D
Explanation: Rate of Flow ∝ [latex]\frac{1}{{(Radius)}^{2}}[/latex] Rate of flow of first pipe = [latex]\frac{K}{{(1)}^{2}}[/latex] = K Rate of flow of Second pipe = [latex]\frac{K}{{(2)}^{2}}[/latex] = [latex]\frac{K}{4}[/latex] Ratio ⇒ K: [latex]\frac{K}{4}[/latex] = 4:1

6. Salaries of A, B and C were in the ratio 3 : 5 : 7, respectively. If their salaries were increased by 50%, 60% and 50% respectively, what will be the new ratio of the their respective new salaries?

A. 4 : 5 : 7 B. 3 : 6 : 7 C. 4 : 15 : 18 D. 9 : 16 : 21

Answer: Option D
Explanation: A:B:C → 3:5:7 New Salaries ratio: ⇒ 3 x [latex]\frac{150}{100}[/latex]: 5[latex]\frac{160}{100}[/latex]: 7[latex]\frac{150}{100}[/latex] ⇒[latex]\frac{9}{2}[/latex]:8: [latex]\frac{21}{21}[/latex] ⇒9:16:21
7. The average score of boys in an examination of a school is 71 and that of the girls is 73. The average score of the whole school in that examination is 71.8. Find the ratio of the number of boys to the number of girls that appeared in the examination.

A. 4 : 5 B. 3 : 2 C. 3 : 5 D. 5 : 2

Answer: Option B
Explanation: Let the number of boys = x number of girls = y we need to find [latex]\frac{x}{y}[/latex] The average score of boys in the examination is 71 total score of boys = 71 × x = 71x The average score of girls in the examination is 73 total score of girls = 73 × y = 73y So total mark of school = 71x + 73y ---------------(1) average score if school in examination 71.8 so total mark = 71.8 × (x+y) -----------------(2) From equations (1) and (2); 71x + 73y = 71.8(x+y) ⇒ 71x + 73y = 71.8x + 71.8y ⇒ 73y - 71.8y = 71.8x - 71x ⇒ 1.2y = 0.8x ⇒0.8x = 1.2y = [latex]\frac{x}{y}[/latex] = [latex]\frac{1.2}{0.8}[/latex] = [latex]\frac{3}{2}[/latex] = 3:2
8. Two casks of 48 L and 42 L are filled with mixtures of wine and water, the proportions in the two casks being respectively 13 : 7 and 18 : 17. If the contents of the two casks be mixed and 20 L of water is added to the whole, what will be the proportion of wine to water in the resultant solution?

A. 21 : 31 B. 12 : 13 C. 13 : 12 D. None of these

Answer: Option B
Explanation: In 48ltr cask 13+7=20 so milk = [latex]\frac{48*13}{20}[/latex] = 31.2ltr water = [latex]\frac{48*7}{20}[/latex] = 16.8ltr In 42 ltr cask 18 + 17 = 35 so milk = [latex]\frac{42*18}{35}[/latex] = 21.6ltr Water = [latex]\frac{42*17}{35}[/latex] = 20.4ltr Now both cask mixed and add 20ltr water so mixture = 48 + 42 + 20=110 Now milk = 31.2 + 21.6 = 52.8 Now water = 16.8 + 20.4 + 20=57.2 Now ratio = [latex]\frac{52.8}{57.2}[/latex] = 12:13
9. What amounts (in litres) of 90% and 97% pure acid solutions are mixed to obtain 21 L of 95% pure acid solution?

A. 6 and 15 L B. 14 and 15 L C. 12 and 15 L D. 13 and 12 L

Answer: Option A
Explanation: V1 = volume of 90% solution V2 = volume of 97% solution Volume of solution = V1 + v2 = 21 L Concentration = 95% = [latex]\frac{V1 * 90% + V2 * 97%}{21}[/latex] so 21 * 95 = 90 V1 + 97 V2 = 90 V1 + 97 (21 - V1) = 97 * 21 - 7 V1 V1 = 6 L and V2 = 15 L
10. Arvind began a business with 550 and was joined afterwards by Brij with 330. When did Brij join, if the profits at the end of the year were divided in the ratio 10 : 3?

A. After 4 months B. After 6 months C. After 4.5 months D. None of these

Answer: Option B
Explanation: [latex]\frac{550 × 12}{330 × x}[/latex] = [latex]\frac{10}{3}[/latex] x = 6 months

11. If [latex]\frac{tan θ + cot θ}{tan θ - cot θ}[/latex] = 2, (0 ≤ θ ≤ [latex]{90}^{0}[/latex]), then the value of sinθ is

A. [latex]\frac{2}{√3}[/latex] B. [latex]\frac{√3}{2}[/latex] C. [latex]\frac{1}{2}[/latex] D. 1

Answer: Option B
Explanation: Applying C and D 2[latex]\frac{tan θ }{cot θ}[/latex] = [latex]\frac{3 }{1}[/latex] [latex]\frac{{sin}^{2}θ}{{cos}^{2}θ}[/latex] = [latex]\frac{3 }{1}[/latex] [latex]{sin}^{2}θ[/latex] = 3 ( 1 - [latex]{sin}^{2}θ[/latex]) 4[latex]{sin}^{2}θ[/latex] = 3 sinθ = [latex]\frac{√3}{2}[/latex]
12. 2[latex]\frac{sin 68°}{cos 22°}[/latex] - 2[latex]\frac{cot 15°}{ 5 tan 75°}[/latex] - [latex]\frac{ 3 tan 45°.9 tan(90-70°). tan(90-50°).tan50°.tan70°}{ 5 tan 75°}[/latex]

A. -1 B. 0 C. 1 D. 2

Answer: Option C
Explanation: 2[latex]\frac{cos 22°}{ cos 22°}[/latex] - [latex]\frac{ 2 tan 75°}{ 5 tan 75°}[/latex] - [latex]\frac{ 3 tan 45°.9 tan(90-70°). tan(90-50°).tan50°.tan70°}{ 5 tan 75°}[/latex] = 2 - [latex]\frac{2}{5}[/latex] - [latex]\frac{3}{5}[/latex] = 2 - 1 = 1
13. If [latex]{cos}^{4}θ[/latex] - [latex]{sin}^{4}θ[/latex] = [latex]\frac{2}{3}[/latex], then the value of 2 [latex]{cos}^{2}θ[/latex] - 1 is

A. 0 B. 1 C. [latex]\frac{2}{3}[/latex] D. [latex]\frac{3}{2}[/latex]

Answer: Option C
Explanation: ([latex]{sin}^{2}θ[/latex] + [latex]{cos}^{2}θ[/latex])([latex]{cos}^{2}θ[/latex] - [latex]{sin}^{2}θ[/latex]) = [latex]\frac{2}{3}[/latex] 2 [latex]{cos}^{2}θ[/latex] - 1 = [latex]\frac{2}{3}[/latex]
14. If sin α sec (30° + α ) = 1 (0 < α < 30° ), then the value of sin α + cos 2α is

A. 1 B. [latex]\frac{2 + √3}{2√3}[/latex] C. 0 D. √2

Answer: Option A
Explanation: [latex]\frac{sin⁡α}{cos⁡(30°+α)}[/latex] = 1 [latex]\frac{sin⁡α}{sin⁡(90-30-α)}[/latex] [latex]\frac{sin⁡α}{sin⁡(60-α)}[/latex] sin α = sin (60 – α) α = 60 – α α = 30° sin 30° + cos 60° = [latex]\frac{1}{2}[/latex] + [latex]\frac{1}{2}[/latex] = 1
15. If [latex]\frac{(sin⁡θ+cos⁡θ)}{(sin⁡θ-cos⁡θ )}[/latex] = 3, then the value of [latex]{sin}^{4}θ[/latex] - [latex]{cos}^{4}θ[/latex] is

A. [latex]\frac{1}{5}[/latex] B. [latex]\frac{2}{5}[/latex] C. [latex]\frac{3}{5}[/latex] D. [latex]\frac{4}{5}[/latex]

Answer: Option C
Explanation: [latex]\frac{(sin⁡θ+cos⁡θ)}{(sin⁡θ-cos⁡θ )}[/latex] = 3 sin θ + cos θ = 3sin θ – 3cos θ 2sin θ = 4cos θ tan θ = 2 [latex]{sin}^{4}θ[/latex] - [latex]{cos}^{4}θ[/latex] = ([latex]{sin}^{2}θ[/latex] + [latex]{cos}^{2}θ[/latex]) ([latex]{sin}^{2}θ[/latex] - [latex]{cos}^{2}θ[/latex]) = [latex]{sin}^{2}θ[/latex] – [latex]{cos}^{2}θ[/latex] = [latex]{cos}^{2}θ[/latex] ([latex]{tan}^{2}θ[/latex] – 1) = [latex]\frac{{tan}^{2}θ - 1}{{sec}^{2}θ}[/latex] = [latex]\frac{{tan}^{2}θ - 1}{1 + {tan}^{2}θ}[/latex] = [latex]\frac{4 - 1}{1 + 4}[/latex] = [latex]\frac{3}{5}[/latex]