Quantitative Aptitude
51. The pie chart given below shows the percentage distribution of candidates present in different sections of the 10th class of a school on Sunday.
The percentage distribution of candidates present:
Note:
- Difference between the number of candidates present in section ‘1’ and section ‘2’ is 4.
- In each of the sections ‘2’, ‘3’, ‘4’, and ‘5’, the total number of candidates (Present + Absent) is less than 51.
- No section has full attendance.
If absent candidates in section ‘1’ is 40% more than the present candidates and the ratio of absent candidates (Boys + Girls) is 5 : 3, then find the absent girl candidates in section ‘1’ is what percentage of the total present candidates in section 3?
(a) 70%
(b) 80%
(c) 90%
(d) 100%
(e) 110%
Answer: (a)
Explanation:
To find the percentage of absent girl candidates in section ‘1’ relative to the total present candidates in section ‘3’, we can follow these logical steps:
1. Identify the Section Percentages
Based on the pie chart and the legend:
- Section 1 (Horizontal Lines): 20%
- Section 2 (Diagonal Lines): 22%
- Section 3 (Vertical Lines): 15%
- Section 4 (Fine Dotted): 23%
- Section 5 (Bold Dashed): 20%
2. Find the Total Number of Present Candidates ($\mathbf{T_{p}}$)
The problem states the difference between candidates present in section ‘1’ and section ‘2’ is 4.
- Difference in percentage: $22% – 20% = 2%$
- Therefore, $2\%$ of $T_{p} = 4$
- $T_{p} = \frac{4}{0.02} = 200$
3. Calculate Candidates Present in Section 1 and Section 3
- Section 1 Present: $20\%$ of $200 = 40$
- Section 3 Present: $15\%$ of $200 = 30$
4. Determine Absent Candidates in Section 1
The absent candidates in section ‘1’ are 40% more than the present candidates.
- Absent candidates = $40 + (40\% \text{of } 40)$
- Absent candidates = $40 + 16 = 56$
5. Find Absent Girl Candidates in Section 1
The ratio of absent boys to girls is $5 : 3$. Total parts = $5 + 3 = 8$.
- Absent girls = $(\frac{3}{8}) \times 56$
- Absent girls = $3 \times 7 = 21$
6. Calculate the Final Percentage
We need to find what percentage 21 (absent girls in section 1) is of 30 (present candidates in section 3).
- Percentage = $(\frac{21}{30}) \times 100$
- Percentage = $0.7 \times 100 = 70\%$
Thus, correct answer is option (a) 70%
52. If 10 candidates who were present in section ‘1’, shifted to section ‘3’, then what is the sum of the maximum and minimum possible total (present + absent) candidates in section ‘3’?
(a) 61
(b) 71
(c) 81
(d) 91
(e) 101
Answer: (d)
Explanation:
1. Find the Total Number of Present Candidates ($\mathbf{T_p}$)
From the previous question and Constraint 1:
- Section 2 Present ($\mathbf{P_2}$): $22\%$ of $T_p$
- Section 1 Present ($\mathbf{P_1}$): $20\%$ of $T_p$
- Difference: $22% – 20% = 2%$
- $2%$ of $T_p = 4 \implies T_p = \frac{4}{0.02} = 200$
2. Calculate Original and New Present Candidates in Section 3
- Original $P_3$: $15%$ of $200 = 30$
- Shift: 10 candidates shift from Section 1 to Section 3.
- New $\mathbf{P_3 (P’_3)}$: $30 + 10 = 40$
3. Determine Max and Min Total Candidates ($\mathbf{T_3}$)
The total number of candidates in a section is the sum of present and absent candidates ($T = P + A$).
- Constraint 2: The total number of candidates in section ‘3’ must be less than 51 ($T_3 < 51$).
- Constraint 3: No section has full attendance, meaning there must be at least one absent candidate ($A_3 \ge 1$).
Finding the Maximum $\mathbf{T_3}$:
- Since $T_3 < 51$, the highest possible integer value is $T_3 = 50$.
- (In this case, $A_3 = 10$).
Finding the Minimum $\mathbf{T_3}$:
- Since there must be at least one person absent, the total must be at least one more than the number of present students.
- $T_3 \ge P’_3 + 1 = 40 + 1 = 41$.
- Thus, the minimum possible $T_3 = 41$.
4. Calculate the Sum
- Sum = $\text{Maximum } T_3 + \text{Minimum } T_3$
- Sum = $50 + 41 = 91$
Thus, correct answer is option (d) 91.
53. What is the difference between the maximum possible value of absent candidates in section ‘2’ and 3rd minimum possible value of absent candidates in section ‘5’?
(a) 6
(b) 2
(c) 3
(d) 4
(e) 5
Answer: (c)
Explanation:
To find the difference between the maximum possible number of absent candidates in section ‘2’ and the 3rd minimum possible number of absent candidates in section ‘5’, we use the data and constraints provided:
1. Re-establishing Base Data
From the previous calculations:
- Total present candidates ($T_p$) = 200.
- Section 2 Present ($P_2$): $22\%$ of $200 = 44$.
- Section 5 Present ($P_5$): $20\%$ of $200 = 40$.
2. Maximum Absent Candidates in Section ‘2’
- Constraint 2: The total number of candidates (Present + Absent) in section ‘2’ is less than 51 ($T_2 \le 50$).
- Equation: $P_2 + A_2 \le 50$
- $44 + A_2 \le 50 \implies A_2 \le 6$.
- The maximum possible value of absent candidates in section ‘2’ is 6.
3. 3rd Minimum Absent Candidates in Section ‘5’
- Constraint 2: Total candidates in section ‘5’ is less than 51 ($T_5 \le 50$).
- Constraint 3: No section has full attendance, so there must be at least one absent candidate ($A_5 \ge 1$).
- Possible values for $A_5$:
- Since $P_5 = 40$ and $T_5 \le 50$, then $40 + A_5 \le 50$, meaning $A_5$ can range from 1 to 10.
- 1st minimum: 1
- 2nd minimum: 2
- 3rd minimum: 3
- The 3rd minimum possible value of absent candidates in section ‘5’ is 3.
4. Calculate the Difference
- Difference = Maximum ($A_2$) – 3rd Minimum ($A_5$)
- Difference = $6 – 3 = 3$.
Thus, correct answer is option (c) 3
54. If the value of absent candidates in section ‘3’ is greater than ‘7’ and the total number of candidates (present + absent) in section ‘3’ is a ‘square’ number, then what is the difference between the absent candidates in section ‘3’ and total present candidates in section ‘1’ and section ‘5’ together?
(a) 65
(b) 64
(c) 63
(d) 62
(e) 61
Answer: (e)
Explanation:
1. The Core Numbers
First, we find the number of “Present” students using the fact that the difference between Section 2 (22%) and Section 1 (20%) is 4 candidates.
- $2\% = 4 \text{candidates}$
- $1\% = 2 \text{candidates}$
- Total Present ($100\%$) = 200 candidates
Now, let’s list the present students for each relevant section:
- Section 1: $20\% \text{of } 200 = \mathbf{40}$
- Section 3: $15\% \text{of } 200 = \mathbf{30}$
- Section 5: $20\% \text{of } 200 = \mathbf{40}$
2. Solve for Section 3 (The Square Number)
We need to find the number of Absent students in Section 3 ($A_3$).
- The Rule: Total students (Present + Absent) must be a Square Number and less than 51.
- We know at least 30 are present. If we add the “greater than 7” absent students, the total must be more than 37.
- The only square number between 37 and 51 is 49.
- Calculation: $49 \text{(Total)} – 30 \text{(Present)} = \mathbf{19 \text{Absent}}$.
The question asks for the difference between the Absent in Section 3 and the Present in Section 1 & 5 combined.
- Step A: Combine Section 1 and Section 5 Present: $40 + 40 = \mathbf{80}$.
- Step B: Subtract the Section 3 Absentees: $80 – 19 = \mathbf{61}$.
Thus, correct answer is option (e) 61
55. If it is given that the sum of maximum and minimum possible values of absent candidates in section ‘4’ is 4, then what is the maximum total number of candidates (Present + Absent) in section ‘4’?
(a) 49
(b) 50
(c) 51
(d) 52
(e) 53
Answer: (a)
Explanation:
The Present Candidates
From our previous steps, we know:
- The total number of present candidates across all sections is 200.
- Section 4 has 23% of those candidates.
- Present in Section 4 ($P_4$): $23\% \text{of } 200 = \mathbf{46}$.
Finding the “Absent” Numbers
The question gives us a hint about the Absent students in Section 4:
- The Minimum: Every section must have at least one person absent. So, the minimum absent is 1.
- The Rule: The problem says: $\text{Maximum Absent} + \text{Minimum Absent} = 4$.
- The Math: $\text{Maximum Absent} + 1 = 4$, which means the Maximum Absent is 3.
Calculating the Total
The “Total” is just the Present students plus the Absent students. To find the maximum total, we use the maximum number of absent students we just found:
- $\text{Total} = \text{Present} + \text{Maximum Absent}$
- $\text{Total} = 46 + 3 = \mathbf{49}$.
Thus, correct answer is option (a) 49.
56. A question is followed by 2 statements. Study the following statements carefully and state whether they are sufficient to answer the question.
A candidate appears in an examination where his marks in math is 20 marks more than what he got in English. Find his marks in Math?
Statement I: Average marks of English and Hindi is 71.5. The ratio of marks of Practical and Theory in English is 6 : 5 and in Hindi, it is 15 : 7.
Statement II: Ratio of marks obtained in the theory exam of English and Hindi is 1 : 2. The ratio of marks of English and Hindi is 5 : 8.
(a) Statement I alone is sufficient
(b) Statement II alone is sufficient
(c) Neither of the statements is sufficient
(d) Both I and II together is sufficient
(e) Either I or II is sufficient
Answer: (d)
Explanation:
To solve this, we need to find the specific value of the Math marks. We are given one initial equation:
- $Math = English + 20$
This means if we can find the value of English, we can find Math. Let’s analyze the statements:
Statement I Analysis
- Average of English ($E$) and Hindi ($H$) is 71.5: This means $E + H = 143$.
- English Ratio (Practical:Theory): $6:5$ (Total $E$ is a multiple of 11).
- Hindi Ratio (Practical:Theory): $15:7$ (Total $H$ is a multiple of 22).
- Check: We have one sum ($E + H = 143$) but two unknowns ($E$ and $H$). While the ratios tell us $E$ and $H$ must be specific multiples, there isn’t enough information to pinpoint one exact number for English.
- Result: Statement I alone is not sufficient.
Statement II Analysis
- Theory Ratio (English:Hindi): $1:2$.
- Total Marks Ratio (English:Hindi): $5:8$.
- Check: This gives us a relationship between English and Hindi, but it provides no actual numerical values (like a sum or an average). We only have ratios.
- Result: Statement II alone is not sufficient.
Combining Both Statements (I + II)
Now we use the numerical value from Statement I with the ratios from Statement II:
- From Statement I: $E + H = 143$.
- From Statement II: The ratio of $E:H$ is $5:8$.
- Calculation:
- Total parts = $5 + 8 = 13$ parts.
- $13 \text{parts} = 143$.
- $1 \text{part} = 143 \div 13 = 11$.
- English marks ($E$): $5 \text{parts} \times 11 = \mathbf{55}$.
- Find Math: Since $Math = English + 20$, then $Math = 55 + 20 = \mathbf{75}$.
By combining both statements, we found a specific numerical value for English, which allowed us to calculate the Math marks.
Thus, correct answer is option (d) Both I and II together is sufficient
57. A train can cross a car which is running with speed 20 km/h in the same direction as that of the train in 36 seconds while the train can cross a platform of length two times of the length of train in 18 seconds, then what is the length of the platform?
(a) 150m
(b) 120m
(c) 40m
(d) 80m
(e) None of these
Answer: (d)
Explanation:
To solve this, we need to find the length of the train first, which will then allow us to find the length of the platform.
1. Define Variables
- Let the length of the train be $L$.
- Since the platform is twice the length of the train, the length of the platform is $2L$.
- Let the speed of the train be $V_T$ (in m/s).
- Speed of the car = $20 \text{km/h} = 20 \times \frac{5}{18} = \frac{50}{9} \text{m/s}$.
2. Equation for the Platform
When a train crosses a platform, it covers a distance equal to the (Length of Train + Length of Platform) in 18 seconds.
- Distance $= L + 2L = 3L$
- Speed of train ($V_T$) $= \frac{\text{Distance}}{\text{Time}} = \frac{3L}{18} = \frac{L}{6} \text{m/s}$
3. Equation for the Car
When the train crosses a car moving in the same direction, we use Relative Speed ($V_T – V_{car}$). The distance covered is just the length of the train ($L$).
- Relative Speed $= V_T – \frac{50}{9}$
- Time $= 36 \text{seconds}$
- Equation: $\text{Distance} = \text{Speed} \times \text{Time}$
- $L = (V_T – \frac{50}{9}) \times 36$
4. Solve for $L$
Substitute $V_T = \frac{L}{6}$ into the equation:
- $L = (\frac{L}{6} – \frac{50}{9}) \times 36$
- $L = 36(\frac{L}{6}) – 36(\frac{50}{9})$
- $L = 6L – 200$
- $200 = 6L – L$
- $200 = 5L \implies L = 40\text{m}$
5. Find Platform Length
- Length of platform $= 2L$
- Length of platform $= 2 \times 40 = \mathbf{80\text{m}}$
Thus, correct answer is option (d) 80m
58. Direction: Answer the questions based on the information given below.
The line graph given below shows the total number of members (Males + Females) in four fitness clubs for the years 2019 and 2023.
The total number of (Males + Females) in clubs
If in club N, the percentage increase of the club members from 2019 to 2021 is x%, from 2021 to 2022 is 20%, and from 2022 to 2023 is 25%, then find the value of ‘x’? (approximately)
(a) 85.5%
(b) 96.5%
(c) 67.5%
(d) 79.5%
(e) 90.5%
Answer: (d)
Explanation:
To find the value of $x$ for club N, we work backward from the known total in 2023 to the total in 2019.
1. Extract Data from the Graph
According to the line graph for club N:
- Total members in 2019 (Dashed line) = 52
- Total members in 2023 (Solid line) = 140
2. Set Up the Calculation
We are given the growth stages from 2019 to 2023:
- 2019 to 2021: Increase of $x\%$
- 2021 to 2022: Increase of $20\%$ (Multiplier = $1.20$)
- 2022 to 2023: Increase of $25\%$ (Multiplier = $1.25$)
Let the number of members in 2021 be $Y$. We can find $Y$ by reversing the 2022 and 2023 increases:
$$140 = Y \times 1.20 \times 1.25$$
$$140 = Y \times 1.5$$
$$Y = \frac{140}{1.5} \approx 93.33$$
3. Solve for $x$
Now, we find the percentage increase ($x$) from the 2019 value (52) to the 2021 value (93.33):
- Increase amount = $93.33 – 52 = 41.33$
- Percentage increase ($x$) = $(\frac{\text{Increase}}{\text{Original Value}}) \times 100$
- $x = (\frac{41.33}{52}) \times 100$
- $x \approx 0.7948 \times 100 \approx 79.48\%$
Rounding to the nearest decimal provided in the options gives 79.5%.
Thus, correct answer is option (d) 79.5%
59. If in club N, an equal number of members joined in 2020, 2021, and 2022 and from 2019 and 2023 only 5 people left the club, then find the number of people who joined in 2020. (Assume no other club member left or joined in any given year other than the given data)
(a) 31
(b) 32
(c) 33
(d) 34
(e) 35
Answer: (a)
Explanation:
To find the number of people who joined in 2020, we can use the following steps:
1. Identify the Change in Members
According to the line graph for club N:
- Total members in 2019: 52
- Total members in 2023: 140
- Net Increase from 2019 to 2023: $140 – 52 = \mathbf{88}$
2. Account for Members who Left
The problem states that 5 people left the club between 2019 and 2023.
- If 5 people left, then the number of people who joined must be higher than the net increase to make up for those who left.
- Total Joined = Net Increase + Number of people who left
- Total Joined = $88 + 5 = \mathbf{93}$
3. Calculate Members Joined per Year
The problem states that an equal number of members joined in each of the three years: 2020, 2021, and 2022.
- Let the number of people who joined in each year be $j$.
- $3 \times j = 93$
- $j = \frac{93}{3} = \mathbf{31}$
Therefore, 31 people joined the club in 2020.
Thus, correct answer is option (a) 31
60. If the total number of male club members in club O and club M together in 2019 is 100 and in 2019, the total number of male members in club M was 75% of the total club members in club ‘N’ in 2019, then find the total number of female club members in club O in 2019 is what percent of the total number of club members in club O in 2023?
(a) 71.25%
(b) 72.50%
(c) 73.75%
(d) 74.25%
(e) 75.75%
Answer: (c)
Explanation:
Total number of male members in clubs O and M together in 2019 = 100
Male members in club M in 2019 = 75% of total members of club N in 2019
Total members of club N in 2019 = 52
Male members in club M in 2019 = 0.75 $\times$ 52 = 39
Male members in club O in 2019 = 100 – 39 = 61
Total members of club O in 2019 = 120
Female members in club O in 2019 = 120 – 61 = 59
Total members of club O in 2023 = 80
$$\frac{59}{80} \times 100 = 73.75%$$
Thus, correct answer is option (c) 73.75%
61. If in 2023, the ratio of the number of female to male club members in club ‘O’ is 9:7 respectively and only in the year 2022 few members left the club in which the number of females was 15%, then how many female members were there initially in gym ‘O’ in 2019?
(a) 55
(b) 53
(c) 54
(d) 52
(e) 51
Answer: (e)
Explanation:
Total number of members in club O in 2019 = 120
Total number of members in club O in 2023 = 80
In 2023, the ratio of female to male members in club O is $9:7$
Total parts $= 9 + 7 = 16$
Female members in club O in 2023
$$\frac{9}{16} \times 80 = 45$$
Male members in club O in 2023
$$80 – 45 = 35$$
Only in the year 2022, some members left the club and the number of females was $15%$
Total members who left the club
$$120 – 80 = 40$$
Number of female members who left in 2022
$$15% \times 40 = 6$$
Number of female members initially in club O in 2019
$$45 + 6 = 51$$
Thus, correct answer is option (e) 51
62. In 2020 in club P, the percentage increase of total members from the previous year is 25% in which males are 37.5% of the new club members who joined and the new male members are 20% more than the number of male members present in the previous year. In 2022, 20 female members joined the club. Find the ratio of the number of female members initially in 2019 and the number of female members in 2023 (Assume no other member joined or left in any year other than the given data).
(a) 61 : 79
(b) 73 : 79
(c) 71 : 79
(d) 59 : 79
(e) 67 : 79
Answer: (d)
Explanation:
Total number of members in club P in 2019 = 128
In 2020, total members increased by $25%$
Total number of members in club P in 2020
$$128 \times 1.25 = 160$$
Total number of new members who joined in 2020
$$160 – 128 = 32$$
Males are $37.5%$ of the new members who joined in 2020
Total number of new male members in 2020
$$32 \times \frac{3}{8} = 12$$
Total number of new female members in 2020
$$32 \times \frac{5}{8} = 20$$
New male members are $20%$ more than the male members present in 2019
Male members in 2019
$$\frac{12}{1.20} = 10$$
Female members in 2019
$$128 – 10 = 118$$
In 2022, $20$ female members joined the club
Female members in 2023
$$118 + 20 = 138$$
Required ratio
$$118 : 138$$
Dividing by $2$
$$59 : 79$$
Thus, correct answer is option (d) 59 : 79
63. Directions: Answer the questions based on the information given below.
There are four companies P, Q, R, and S. Data given below shows the number of products sold and unsold by these shopkeepers.
The total number of products manufactured by ‘P’ is 120, and the ratio of sold and unsold products of company Q is 3 : 1. The total number of products manufactured by company ‘Q’ and ‘S’ is 260, and the difference between the number of unsold and sold products by ‘R’ is 90. The total number of products manufactured by ‘R’ is 150, which is 12 less than that of products sold by ‘S’. The difference between the products unsold by ‘Q’ and ‘S’ is 2. The number of products sold by P is x% as that of ‘Q’.
Note: The number of products sold by each company is more than that of products unsold.
If the total number of products sold by all companies together is 432, then find the number of unsold products by company ‘S’ is how much percent more or less than that of ‘P’?
(a) 100%
(b) 80%
(c) 60%
(d) 40%
(e) 20%
Answer: (d)
Explanation:
Decoding the Base Data
Let’s break down the information for each company:
- Company R:
- Total Manufactured = 150.
- Let Sold = $a$ and Unsold = $b$.
- $a + b = 150$ and $a – b = 90$ (Since sold > unsold).
- Adding both: $2a = 240 \Rightarrow$ Sold = 120; Unsold = 30.
- Company S:
- Sold by S = (Manufactured by R) + 12 = $150 + 12 =$ 162.
- Total (Q + S) = 260 (We will use this after solving for Q).
- Company Q:
- Ratio of Sold : Unsold = $3 : 1$. Let Unsold = $z$, then Sold = $3z$.
- Total Q = $4z$.
- From the “Total Q + S” info: $4z + (\text{Sold S} + \text{Unsold S}) = 260$.
- We know Unsold Q and S differ by 2. If $z = 20$, Unsold S could be 18 or 22.
- Testing $z = 20$: Total Q = 80. Total S = $260 – 80 = 180$.
- If Total S = 180 and Sold S = 162, then Unsold S = 18.
- Check: Unsold Q (20) – Unsold S (18) = 2. This matches the condition.
- Q: Sold = 60, Unsold = 20, Total = 80.
- S: Sold = 162, Unsold = 18, Total = 180.
- Company P:
- Total Manufactured = 120.
- Sold = $x\%$ of Q’s Sold = $x\% \times 60$.
Summary Table
| Company | Sold | Unsold | Total Manufactured |
| P | $x\% \times 60$ | $120 – (x\% \times 60)$ | 120 |
| Q | 60 | 20 | 80 |
| R | 120 | 30 | 150 |
| S | 162 | 18 | 180 |
Find percent difference between Unsold S and Unsold P (given total sold = 432)
- Total Sold = Sold (P + Q + R + S) = 432.
- (x% $\times$ 60) + 60 + 120 + 162 = 432
- (x% $\times$ 60) + 342 = 432 $\Rightarrow$ x% $\times$ 60 = 90$.
- Sold P = 90.
- Unsold P = Total P – Sold P = $120 – 90 = 30$.
- Unsold S = 18.
- Difference = $30 – 18 = 12$.
- Required % (less than P) = $(\frac{12}{30}) \times$ 100 = $\textbf{40%}$.
Thus, correct answer is option (d) 40%
64. If the number of products sold by company ‘P’ is 140% more than the products unsold by company R, then find the value of ‘x’?
(a) 100%
(b) 110%
(c) 120%
(d) 130%
(e) 140%
Answer: (c)
Explanation:
Summary Table:
| Company | Sold | Unsold | Total Manufactured |
| P | $x\% \times 60$ | $120 – (x\% \times 60)$ | 120 |
| Q | 60 | 20 | 80 |
| R | 120 | 30 | 150 |
| S | 162 | 18 | 180 |
Value of x if Sold P is 140% more than Unsold R
- Unsold R = 30.
- Sold P = 30 + (140% of 30) = 30 + 42 = 72.
- We know Sold P = x% $\times$ Sold Q.
- 72 = x% $\times$ 60 $\Rightarrow$ x = ($\frac{72}{60}$) $\times$ 100 = 120%.
Thus, correct answer is option (c) 120%
65. If the number of products unsold by ‘P’ is 30, and the number of products manufactured by ‘T’ is thrice that of company ‘S’, then find the number of unsold products of company ‘T’, if the sold products of company ‘T’ is 2x.
(a) 120
(b) 100
(c) 80
(d) 60
(e) 40
Answer: (c)
66. If the difference between the number of products sold by ‘R’ and the total products manufactured by ‘S’ is ‘Y’ and the total number of products manufactured by ‘Q’ is ‘Z’, then ‘Y’ is how much percent more or less than that of ‘Z’?
(a) 15%
(b) 20%
(c) 25%
(d) 30%
(e) 35%
Answer: (c)
Explanation:
Summary Table:
| Company | Sold | Unsold | Total Manufactured |
| P | $x\% \times 60$ | $120 – (x\% \times 60)$ | 120 |
| Q | 60 | 20 | 80 |
| R | 120 | 30 | 150 |
| S | 162 | 18 | 180 |
Percent difference between Y and Z
- $Y = |\text{Sold R} – \text{Total S}| = |120 – 180| = 60$.
- $Z = \text{Total Q} = 80$.
- Difference = $80 – 60 = 20$.
- Required % (Y less than Z) = $(\frac{20}{80}) \times 100 =$ 25%.
Thus, correct answer is option (c) 25%
67. Consider the following number series and answer the question based on this.
2, 2, 4, 11, x, p
z + 5 = x – 26
Find the value of p?
(a) 387
(b) 388
(c) 389
(d) 400
(e) 401
Answer: (c)
Explanation:
Term 1 to Term 2: $2 \times 0.5 + 1 = 2$
Term 2 to Term 3: $2 \times 1 + 2 = 4$
Term 3 to Term 4: $4 \times 2 + 3 = 11$
Term 4 to Term 5 (Finding $\mathbf{x}$): Following the pattern (multiplier $2 \rightarrow 4$, addition $3 \rightarrow 4$), $11 \times 4 + 4 = 48$. Thus, $\mathbf{x = 48}$.
Term 5 to Term 6 (Finding $\mathbf{P}$): Following the pattern (multiplier $4 \rightarrow 8$, addition $4 \rightarrow 5$), $48 \times 8 + 5 = 389$. Thus, $\mathbf{P = 389}$.
Thus, correct answer is option (c) 389
68. Which of the following is true about the value of z?
(a) z is an even number
(b) z is a prime number
(c) z is a perfect square number
(d) z is greater than 90.
(e) Sum of digits of z is more than 20.
Answer: (b)
Explanation:
Step 1: Finding the value of $\mathbf{z}$
To find $z$, we first need to determine the value of $x$ from the number series in question 67. The pattern follows a doubling multiplier and an incrementing addition:
- $2 \times 0.5 + 1 = 2$
- $2 \times 1 + 2 = 4$
- $4 \times 2 + 3 = 11$
- $11 \times 4 + 4 = 48 \implies \mathbf{x = 48}$
Now, using the equation provided in question 68:
$$z + 5 = x – 2\\
z + 5 = 48 – 26\\
z + 5 = 22\\
\mathbf{z = 17}$$
Step 2: Evaluating the options for $\mathbf{z = 17}$
- (a) $z$ is an even number: False. 17 is an odd number.
- (b) $z$ is a prime number: True. 17 is a prime number as it is only divisible by 1 and itself.
- (c) $z$ is a perfect square number: False. The nearest perfect squares are 16 ($4^2$) and 25 ($5^2$).
- (d) $z$ is greater than 90: False. 17 is less than 90.
- (e) Sum of digits of $z$ is more than 20: False. The sum of the digits ($1 + 7$) is 8.
Thus, correct answer is option (b) z is a prime number
69.
Series I: 12, x, 35, 51, 70
Series II: 35, x, 73, 293, 2345
Series III: 7, 14, x, 34, 47, 62
The value of x is 23.
In which of the following series ‘x’ is not satisfied properly?
(a) Only I
(b) Only II
(c) Only III
(d) Only I and II
(e) Only II and III
Answer: (d)
Explanation:
1. Analysis of Series I
Series: $12, x, 35, 51, 70$
The pattern is based on increasing differences:
- $70 – 51 = 19$
- $51 – 35 = 16$
- $35 – x = 13$
- If we continue the pattern of subtracting 3 from the difference ($19, 16, 13, 10$):
- $x = 35 – 13 = 22$
- Check: $22 – 12 = 10$ (Matches the pattern)
- Result: In Series I, $x$ should be 22, so $x = 23$ is not satisfied.
2. Analysis of Series II
Series: $35, x, 73, 293, 2345$
The pattern involves a doubling multiplier plus one:
- $293 \times 8 + 1 = 2345$
- $73 \times 4 + 1 = 293$
- Following the pattern of multipliers ($8, 4, 2, 1$):
- $x \times 2 + 1 = 73 \implies 2x = 72 \implies x = 36$
- Check: $35 \times 1 + 1 = 36$ (Matches the pattern)
- Result: In Series II, $x$ should be 36, so $x = 23$ is not satisfied.
3. Analysis of Series III
Series: $7, 14, x, 34, 47, 62$
The pattern is based on decreasing differences:
- $62 – 47 = 15$
- $47 – 34 = 13$
- Following the pattern of odd number differences ($15, 13, 11, 9, 7$):
- $34 – x = 11 \implies x = 23$
- Check: $23 – 14 = 9$ and $14 – 7 = 7$ (Matches the pattern)
- Result: In Series III, $x = 23$ is satisfied properly.
Conclusion: Since $x = 23$ does not work for Series I and Series II, but does work for Series III, the correct answer is (d) Only I and II.
70. Directions: Answer the questions based on the information given below.
The table given below shows the data of the total funds of 4 different companies and funds invested by each company to 3 different Projects.
| Company | Total fund (Rs.) | % invested to Project ‘x’ | % donated to Project ‘y’ | % donated to Project ‘z’ |
| A | Rs. 12000 | 32% | — | 24% |
| B | — | 64% | 16% | Rs. 1905 |
| C | — | 33% | — | 27% |
| D | Rs. 13500 | — | 52% | — |
Note:
- Sum of the percentage distribution in each company is 100.
- Ratio between the total funds of company ‘A’ to company ‘C’ is 4 : 5 respectively.
- Some of the values are missing.
70. If the fund invested to project ‘y’ by company ‘A’ was distributed to an orphan child and elder person in 9 : 11 respectively, then find the difference between the fund invested to the elder person of project ‘y’ and the total fund of company ‘B’.
(a) Rs. 7115
(b) Rs. 7215
(c) Rs. 7315
(d) Rs. 7415
(e) Rs. 7515
Answer: (d)
71. If the total fund of company ‘B’ is part of their ‘actual plan’ and it is only 40% of their actual fund, then the total fund invested to project ‘x’ and project ‘y’ by company ‘B’ is what percent of their actual fund?
(a) 30%
(b) 31%
(c) 32%
(d) 33%
(e) 34%
Answer: (c)
72. If 20% of the fund invested to project ‘Y’ by the company ‘C’ was utilized in ‘women’s development’ and the rest was for ‘rural development’, then find the ratio between the fund utilized in women’s development by project ‘Y’ and fund invested by company ‘D’ to project ‘Y’?
(a) 20 : 101
(b) 20 : 103
(c) 20 : 107
(d) 20 : 111
(e) 20 : 117
Answer: (e)
73. If the fund invested by company ‘D’ in project ‘X’ and project ‘Z’ was in the ratio of 4 : 5, then find the fund invested to project ‘Z’ was what percent of the total fund of company ‘D’?
(a) 13.33%
(b) 16.66%
(c) 26.66%
(d) 33.33%
(e) 42.22%
Answer: (c)
74. If another company E has a fund that was 30% more than that of company ‘C’ and company ‘E’ invested 30%, 25% and 45% to project ‘X’, project ‘Y’, and project ‘Z’ respectively, then find the sum of the fund invested to project ‘Z’ by company ‘E’ and fund invested to project ‘X’ by company ‘B’?
(a) Rs. 11871
(b) Rs. 12871
(c) Rs. 13871
(d) Rs. 14871
(e) Rs. 15871
Answer: (d)
75. Directions: In this question, two equations numbered I and II are given. You have to solve both the equations and find out the correct option.
I. $x^2 – 7x + 12 = 0$
II. $ y^2 – 14y + 40 = 0$
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or no relationship could be established
Answer: (d)
Explanation:
Equation I: $x^2 – 7x + 12 = 0$
We need to find two numbers that multiply to $12$ and add up to $-7$.
- The numbers are $-3$ and $-4$.
- $(x – 3)(x – 4) = 0$
- Roots for $\mathbf{x}$: $3$ and $4$
Equation II: $y^2 – 14y + 40 = 0$
We need to find two numbers that multiply to $40$ and add up to $-14$.
- The numbers are $-10$ and $-4$.
- $(y – 10)(y – 4) = 0$
- Roots for $\mathbf{y}$: $10$ and $4$
Comparison Table
Now, let’s compare every possible value of $x$ with every possible value of $y$:
| x value | y value | Relationship |
| $3$ | $4$ | $x < y$ |
| $3$ | $10$ | $x < y$ |
| $4$ | $4$ | $x = y$ |
| $4$ | $10$ | $x < y$ |
Conclusion: In all cases, $x$ is either less than or equal to $y$. Therefore, the relationship is $\mathbf{x \leq y}$.
Thus, correct answer is option (d) x ≤ y
76. In the given question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I. $x^2 – 42x + 216 = 0$
II. $2y^2 + 7y + 6 = 0$
(a) x > y
(b) y ≤ x
(c) y > x
(d) x < y
(e) Either x = y or the relation between x and y cannot be established
Answer: (a)
Explanation:
Equation I: $\mathbf{x^2 – 42x + 216 = 0}$
We need to find two numbers that multiply to $216$ and add up to $-42$.
- Looking at the factors of 216: $6 \times 36 = 216$.
- Checking the sum: $-6 + (-36) = -42$.
- The equation factors as: $(x – 6)(x – 36) = 0$.
- Roots for $\mathbf{x}$: $6$ and $36$.
Equation II: $\mathbf{2y^2 + 7y + 6 = 0}$
We need two numbers that multiply to $(2 \times 6) = 12$ and add up to $7$.
- The numbers are $3$ and $4$.
- Splitting the middle term: $2y^2 + 4y + 3y + 6 = 0$.
- Factoring: $2y(y + 2) + 3(y + 2) = 0 \rightarrow (2y + 3)(y + 2) = 0$.
- Roots for $\mathbf{y}$: $-1.5$ and $-2$.
Comparison
Now, let’s compare the values of $x$ with the values of $y$:
| x value | y value | Relationship |
| $6$ | $-1.5$ | $x > y$ |
| $6$ | $-2$ | $x > y$ |
| $36$ | $-1.5$ | $x > y$ |
| $36$ | $-2$ | $x > y$ |
Conclusion: In every scenario, both roots of $x$ (which are positive) are significantly greater than both roots of $y$ (which are negative). Therefore, the relationship is x > y.
Thus, correct answer is option (a) x > y
77. Direction: In the given question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I. $x^2 – 9x + 14 = 0$
II. $y^2 + 10y + 21 = 0$
(a) x > y
(b) x < y
(c) x ≥ y
(d) x ≤ y
(e) x = y or the relationship between x and y cannot be established
Answer: (e)
78. In the given question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I. $x^2 – x – 30 = 0$
II. $y^2 + 2y – 24 = 0$
(a) x > y
(b) y ≤ x
(c) y > x
(d) x < y
(e) Either x = y or the relationship between x and y cannot be established
Answer: (e)
Explanation:
Equation I: $x^2 – x – 30 = 0$
We need two numbers that multiply to $-30$ and add up to $-1$.
- The numbers are $-6$ and $+5$.
- Factoring: $x^2 – 6x + 5x – 30 = 0$.
- $x(x – 6) + 5(x – 6) = 0$.
- $(x – 6)(x + 5) = 0$.
- Roots for $x$: $6, -5$.
Equation II: $y^2 + 2y – 24 = 0$
We need two numbers that multiply to $-24$ and add up to $+2$.
- The numbers are $+6$ and $-4$.
- Factoring: $y^2 + 6y – 4y – 24 = 0$.
- $y(y + 6) – 4(y + 6) = 0$.
- $(y + 6)(y – 4) = 0$.
- Roots for $y$: $-6, 4$.
Comparison Table:
To determine the relationship, we compare every value of $x$ with every value of $y$:
| x value | y value | Relationship |
| 6 | -6 | x > y |
| 6 | 4 | x > y |
| -5 | -6 | x > y |
| -5 | 4 | x < y |
Since $x$ is greater than $y$ in some cases ($6 > 4$) but less than $y$ in others ($-5 < 4$), no consistent relationship can be established. Whenever you find both a “$>$” and a “$<$” relationship during comparison, the answer is always “Relationship cannot be established“.
Thus, correct answer is option (e) Either x = y or the relationship between x and y cannot be established.
79. In the given question, two equations numbered I and II are given. You have to solve both the equations and mark the appropriate answer.
I. $ 2y^2 – 13y + 21 = 0$
II. $3x^2 – 16x + 21 = 0$
(a) y > x
(b) x < y
(c) x > y
(d) y ≤ x
(e) Either x = y or the relationship between x and y cannot be established
Answer: (a)
80. What approximate value should come in the place of the question mark (?) in the following question?
$14.14 \times 69.99 \times 7.96 \div 27.98 = x^2 + 23.99$
(a) 14
(b) 15
(c) 16
(d) 17
(e) 18
Answer: (c)
Explanation:
Step 1: Simplify the Equation
Round the values as follows:
- 14.14 $\approx$ 14
- 69.99 $\approx$ 70
- 7.96 $\approx$ 8
- 27.98 $\approx$ 28
- 23.99 $\approx$ 24
The equation becomes:
$$14 \times 70 \times 8 \div 28 = x^2 + 24$$
Step 2: Solve the Left Side
Perform the multiplication and division:
- First, divide 70 by 28 (or 14 by 28):
$$\frac{14}{28} = \frac{1}{2}$$ - Multiply the remaining terms:
$$\frac{1}{2} \times 70 \times 8 = 35 \times 8 = 280$$
Step 3: Solve for $x$
Now, substitute the value back into the equation:
$$280 = x^2 + 24\\
x^2 = 280 – 24\\
x^2 = 256\\
x = \sqrt{256}\\
x = 16$$
The approximate value for $x$ is 16.
Thus, correct answer is option (c) 16.
81. What approximate value should come in the place of the question mark (?) in the following question?
14.77% of 199.99 + 36.14% of 149.99 = x – 7.99% of 999.98
(a) 144
(b) 154
(c) 164
(d) 174
(e) 184
Answer: (c)
Explanation:
Step 1: Round the Values
Based on the approximation logic used in the provided solutions:
- 14.77% $\approx$ 15%
- 199.99 $\approx$ 200
- 36.14% $\approx$ 36%
- 149.99 $\approx$ 150
- 7.99% $\approx$ 8%
- 999.98 $\approx$ 1000
The equation becomes $15\% \text{of } 200 + 36\% \text{of } 150 = x – 8\% \text{of } 1000$
Step 2: Calculate Each Component
- 15% of 200:
$$0.15 \times 200 = 30$$
- 36% of 150:
$$0.36 \times 150 = 54$$
- 8% of 1000:
$$0.08 \times 1000 = 80$$
Step 3: Solve for $x$
Substitute the calculated values back into the equation:
$$30 + 54 = x – 80\\
84 = x – 80$$
To isolate $x$, add 80 to both sides:
$$x = 84 + 80\\
x = 164$$
The approximate value that should come in place of the question mark is 164.
Thus, correct answer is option (c).
82. What approximate value will come in the place of the question mark (?) in the following question?
$(7.96 \times 8.988) – (9.003 \times \sqrt{16.013}) = x – 7.789$
(a) 44
(b) 45
(c) 41
(d) 42
(e) 47
Answer: (a)
Explanation:
Step 1: Round the Values
Following the approximation logic shown in the provided solution:
- 7.96 rounds to 8.
- 8.988 rounds to 9.
- 9.003 rounds to 9.
- $\sqrt{16.013}$ rounds to $\sqrt{16}$, which is 4.
- 7.789 rounds to 8.
The simplified equation becomes:
$$(8 \times 9) – (9 \times 4) = x – 8$$
Step 2: Solve the Equation
- Calculate the products:
$\quad 8 \times 9 = 72$
$\quad 9 \times 4 = 36$ - Subtract the results:
$\quad 72 – 36 = 36$ - Final equation:
$\quad 36 = x – 8$
To isolate $x$, add 8 to both sides:
$$x = 36 + 8\\
\mathbf{x = 44}$$
The approximate value for the question mark is 44.
Thus, correct answer is option (a) 44.
83. What approximate value should come in the place of the question mark (?) in the following question?
$(35.785 – 24.76) \times (19.89 – 14.99) = x + (44.87 – 16.98)$
(a) 25
(b) 26
(c) 27
(d) 28
(e) 29
Answer: (c)
Explanation:
Step 1: Round the Values
Following the approximation logic:
- (35.785 – 24.76): Round 35.785 to 36 and 24.76 to 25.
- (19.89 – 14.99): Round 19.89 to 20 and 14.99 to 15.
- (44.87 – 16.98): Round 44.87 to 45 and 16.98 to 17.
The simplified equation becomes:
$$(36 – 25) \times (20 – 15) = x + (45 – 17)$$
Step 2: Solve the Equation
- Perform the subtractions inside the parentheses:
$\quad 36 – 25 = 11$
$\quad 20 – 15 = 5$
$\quad 45 – 17 = 28$ - Multiply the results on the left side:
$\quad 11 \times 5 = 55$ - Final simplified equation:
$\quad 55 = x + 28$
To isolate x, subtract 28 from 55:
$$x = 55 – 28\\
\mathbf{x = 27}$$
The approximate value that should come in place of the question mark is 27.
Thus, correct answer is option (c) 27.
84. The question consists of two statements numbered “I” and “II” given below it. You have to decide whether the data provided in the statements are sufficient to answer the question or not and choose the correct option accordingly.
An alloy contains silver and gold in the ratio of 9 : 14 respectively. Find the original quantity of gold in the alloy?
Statement I: If 20% of the alloy is replaced with the same quantity of gold then the ratio of the silver to gold becomes 36 : 79.
Statement II: If 75 kg of gold is added to the alloy, then the ratio of silver to gold becomes 9 : 17.
(a) The data in statement I alone is sufficient to answer, while the data in statement II alone is not sufficient to answer the question.
(b) The data in statement II alone is sufficient to answer, while the data in statement I alone is not sufficient to answer the question.
(c) The data either in statement I alone or in statement II alone is sufficient to answer the question.
(d) The data given in both statements I and II together are not sufficient to answer the question.
(e) The data in both statements I and II together are necessary to answer the question.
Answer: (b)
85. The question consists of two statements numbered “I” and “II” given below it. You have to decide whether the data provided in the statements are sufficient to answer the question or not and choose the correct option accordingly.
The ratio of the cost prices of articles ‘X’, ‘Y’ and ‘Z’ is 3 : 5 : 8. The percentage by which article ‘X’ article ‘Y’ and article ‘Z’ are marked up is 40%, 60%, and 80% respectively. Find the cost price of article ‘Z’?
Statement (I): The ratio of the selling prices of article ‘X’, article ‘Y’ and article ‘Z’ is 13 : 7 : 10
Statement (II): Total profit earned on the three articles is Rs. 120 less than the total discount given on the three articles together.
(a) The data in statement I alone is sufficient to answer, while the data in statement II alone is not sufficient to answer the question.
(b) The data in statement II alone is sufficient to answer, while the data in statement I alone is not sufficient to answer the question.
(c) The data either in statement I alone or in statement II alone is sufficient to answer the question.
(d) The data given in both statements I and II together are not sufficient to answer the question.
(e) The data in both statement I and II together are necessary to answer the question.
Answer: (d)
Explanation:
Evaluate the mathematical relationships established by each piece of data.
Initial Setup: Cost and Marked Price
The cost prices of articles ‘X’, ‘Y’, and ‘Z’ are in the ratio $3 : 5 : 8$. We can express these as:
- Cost Price (CP): $X = 3 \times 5k = 15x$, $Y = 5 \times 5k = 25x$, $Z = 8 \times 5k = 40x$
The articles are marked up by $40\%$, $60\%$, and $80\%$ respectively.
- Marked Price (MP):
- $X = 15x \times 1.4 = 21x$
- $Y = 25x \times 1.6 = 40x$
- $Z = 40x \times 1.8 = 72x$
Evaluation of Statements
Statement (I): The ratio of the selling prices of articles ‘X’, ‘Y’, and ‘Z’ is $13 : 7 : 10$.
- Let the Selling Prices (SP) be $65y$, $35y$, and $50y$.
- This statement provides a relationship between variables but does not provide any absolute numerical value (like a specific Rupee amount).
- Without a fixed value, we cannot determine the absolute cost price of ‘Z’.
- Conclusion: Statement I alone is not sufficient.
Statement (II): Total profit earned is Rs. 120 less than the total discount given.
- Profit = Total SP – Total CP = $(50y – 40x) + (35y – 25x) + (65y – 15x)$
- Discount = Total MP – Total SP = $[(72x – 50y) + (40x – 35y) + (21x – 65y)]$
- Using the condition “Profit = Discount – 120,” we can find a ratio between $y$ and $x$ ($y/x = 71/100$), but we still cannot find the independent value of $x$.
- Conclusion: Statement II alone is not sufficient.
Combined Evaluation: Even when combining both statements, we are left with a ratio between the variables $x$ and $y$ but no absolute numerical anchor to solve for a specific cost price in Rupees. Both equations derived from the statements are dependent on the same variables without a third independent piece of data to solve for them.
The data given in both statements I and II together are not sufficient to answer the question.
Thus, correct answer is option (d).
DIRECTION (Q. 86): Given question contains a statement followed by two statements are numbered as Quantity I and Quantity II. On solving these statements, we get quantities I and II respectively. Solve both quantities and choose the correct option.
86. If $a^3 = 8 $ and $b^4 = 16$
Quantity I: Value of b.
Quantity II: Value of a.
(a) Quantity I > Quantity II
(b) Quantity II > Quantity I
(c) Quantity II ≥ Quantity I
(d) Quantity I ≥ Quantity II
(e) Cannot be determined
Answer: (c)
Explanation: To determine the relationship between Quantity I and Quantity II, we first need to solve for the individual values of $a$ and $b$ based on the equations provided.
Solving for the Variables
- For $\mathbf{a}$:
- Given $a^3 = 8$.
- Taking the cube root of both sides: $a = \sqrt[3]{8}$.
- $a = 2$
- For $\mathbf{b}$:
- Given $b^4 = 16$.
- Taking the fourth root of both sides: $b = \pm\sqrt[4]{16}$.
- When an even power is involved, both positive and negative roots must be considered.
- $\mathbf{b = +2}$ or $\mathbf{-2}$
Comparing Quantities
- Quantity I (Value of $\mathbf{b}$): $+2, -2$
- Quantity II (Value of $\mathbf{a}$): $2$
Comparison Table:
| Value of a (Q-II) | Value of b (Q-I) | Relationship |
| 2 | 2 | Q-II = Q-I |
| 2 | -2 | Q-II > Q-I |
In both cases, Quantity II is either greater than or equal to Quantity I ($2 \ge 2$ and $2 \ge -2$). Therefore, the relationship is Quantity II $\mathbf{\ge}$ Quantity I.
Thus, correct answer is option (c).
87. A receives $\frac{1}{3}$ of the total profit and the remaining is divided between B and C in the ratio of 11:14. If the difference between the profit shares of C and A is Rs.300. What is B’s share?
(a) ₹2000
(b) ₹2200
(c) ₹2300
(d) ₹2400
(e) None of these
Answer: (b)
Explanation:
Step 1: Define the Total Profit
Let the total profit be represented by $x$.
Step 2: Determine Individual Shares
- A’s Share: A receives $\frac{1}{3}$ of the total profit, so A’s share = $\frac{x}{3}$.
- Remaining Profit: After A takes their share, the remaining profit is $x – \frac{x}{3} = \frac{2x}{3}$.
- B and C’s Ratio: The remaining $\frac{2x}{3}$ is divided between B and C in a ratio of $11:14$.
- B’s Share: $\frac{11}{25} \times \frac{2x}{3} = \frac{22x}{75}$.
- C’s Share: $\frac{14}{25} \times \frac{2x}{3} = \frac{28x}{75}$.
Step 3: Use the Given Difference to Solve for $x$
The problem states the difference between C’s share and A’s share is Rs. 300.
$$\text{C} – \text{A} = 300\\
\frac{28x}{75} – \frac{x}{3} = 300$$
To solve, find a common denominator (75) for the left side:
$$\frac{28x}{75} – \frac{25x}{75} = 300\\
\frac{3x}{75} = 300 \implies x = 7500$$
The total profit ($x$) is Rs. 7,500.
Step 4: Calculate B’s Share
Now, substitute the value of $x$ into the expression for B’s share:
$$\text{B’s share} = \frac{22}{75} \times 7500\\
\text{B’s share} = 22 \times 100 = \mathbf{₹2,200}$$
Thus, correct answer is option (b) ₹2,200
88. Deepak can do 2/3rd of work in 30 days. The efficiency of Rahul is 20% more than that of Deepak. Both Deepak and Rahul started working together and left the work after seven days. Puja completed the remaining work in 37 days. The efficiency of Puja is how much percent more/less than that of Deepak?
(a) 25% less
(b) 20% less
(c) 20% more
(d) 25% more
(e) None of these
Answer: (b)
Explanation:
Step 1: Determine Deepak’s Total Work Time
Deepak completes $\frac{2}{3}$ of the work in $30$ days.
- To find the time for the whole work: $\frac{30 \times 3}{2} = 45$ days.
- Let the total work be $225$ units (the LCM of the relevant numbers like $45$).
- Deepak’s efficiency = $\frac{225 \text{units}}{45 \text{days}} = 5 \text{units/day}$.
Step 2: Determine Rahul’s Efficiency
Rahul’s efficiency is $20\%$ more than Deepak’s.
- $20\%$ of $5$ is $1$, so Rahul’s efficiency = $5 + 1 = 6 \text{units/day}$.
- The combined efficiency of Rahul and Deepak is $6 + 5 = 11 \text{units/day}$.
Step 3: Calculate the Remaining Work
Rahul and Deepak worked together for $7$ days before leaving.
- Work completed in $7$ days = $11 \text{units/day} \times 7 \text{days} = 77 \text{units}$.
- Remaining work = $225 \text{units} – 77 \text{units} = 148 \text{units}$.
Step 4: Determine Puja’s Efficiency
Puja completes the remaining $148$ units of work in $37$ days.
- Puja’s efficiency = $\frac{148 \text{units}}{37 \text{days}} = 4 \text{units/day}$.
Step 5: Compare Puja and Deepak
Now we compare Puja’s efficiency ($4$) to Deepak’s efficiency ($5$).
- Difference = $5 – 4 = 1$ unit.
- Percentage decrease = $\frac{1}{5} \times 100 = 20\%$.
The efficiency of Puja is $20%$ less than that of Deepak.
Thus, correct answer is option (b).
89. A bag contains 7 red and 3 blue balls. Two balls are drawn randomly. Find the probability that one ball is red and other is blue.
(a) 21/45
(b) 23/45
(c) 29/45
(d) 19/45
(e) None of the above
Answer: (a)
Explanation:
Step 1: Total Possible Outcomes
There are a total of 10 balls (7 red + 3 blue) in the bag. We are drawing two balls randomly.
The total number of ways to draw 2 balls from 10 is calculated using combinations ($10C_2$):
$$10C_2 = \frac{10!}{2! \times (10-2)!} = \frac{10 \times 9}{2 \times 1} = 45 \text{ways}$$
Step 2: Favorable Outcomes
We want one ball to be red and the other to be blue.
- Ways to pick 1 red ball from 7: $7C_1 = 7$
- Ways to pick 1 blue ball from 3: $3C_1 = 3$
Total favorable ways = $7 \times 3 = 21$ ways.
Step 3: Calculate Probability
The probability is the ratio of favorable outcomes to the total possible outcomes:
$$\text{Required Probability} = \frac{21}{45} $$
Thus, correct answer is option (a) 21/45
90. Difference between compound interest and simple interest on Rs. 20,000 for 2 years at x% per annum is Rs. 450. Find the value of x.
(a) 15%
(b) 15 1/3%
(c) 16 2/3%
(d) 18%
(e) None of these
Answer: (a)
Explanation:
To find the value of $x$ (the interest rate), we use the standard formula for the difference between Compound Interest (CI) and Simple Interest (SI) for a period of 2 years.
The Formula
The difference between CI and SI for 2 years is given by the formula:
$$\text{Difference} = P \times \left(\frac{r}{100}\right)^2$$
Where:
- P = Principal amount = Rs. 20,000
- r = Rate of interest per annum = x%
- Difference = Rs. 450
Step-by-Step Calculation
1. Substitute the given values into the formula:
$$450 = 20000 \times \left(\frac{r}{100}\right)^2$$
2. Simplify the squared term:
$$450 = 20000 \times \frac{r^2}{10000}$$
3. Solve for $\mathbf{r^2}$:
Cancel the zeros:
$450 = 2 \times r^2$
Divide both sides by 2:
$225 = r^2$
4. Find the square root:
$$r = \sqrt{225}\\
r = \pm 15$$
Since interest rates are expressed as positive values in this context, the value of $x$ is 15%.
The rate of interest per annum is 15%. Thus, correct answer is option (a).
91. The total time taken by the boat to go ‘x’ km upstream and then return back to a certain distance is 5 hours. The ratio of the speed boat in still water to the speed of the stream is 4 : 1, respectively. If the upstream distance covered is 90 km less than the downstream distance covered and the boat can cover (3x + 60) km in still water in 6 hours, then find the value of ‘x’?
(a) 60 km
(b) 80 km
(c) 100 km
(d) 120 km
(e) 140 km
Answer: (a)
Explanation:
1. Establish Speeds: The ratio of the speed of the boat in still water ($u$) to the speed of the stream ($v$) is given as $4:1$. Let:
- Speed of boat in still water ($u$) = $4k$
- Speed of stream ($v$) = $k$
From these, we calculate the directional speeds:
- Downstream speed ($u+v$) = $4k + k = 5k$
- Upstream speed ($u-v$) = $4k – k = 3k$
2. Set Up the Time Equation
The boat travels $x$ km upstream and returns downstream. The downstream distance is $90$ km more than the upstream distance ($x + 90$ km).
- Time taken upstream = $\frac{x}{3k}$
- Time taken downstream = $\frac{x + 90}{5k}$
The total time is $5$ hours:
$$\frac{x + 90}{5k} + \frac{x}{3k} = 5\\
\frac{3(x + 90) + 5x}{15k} = 5 \implies \frac{8x + 270}{15k} = 5\\
\mathbf{8x + 270 = 75k} \quad — (Equation i)$$
3. Use the Still Water Condition
The boat covers $(3x + 60)$ km in still water ($4k$) in $6$ hours:
$$\text{Speed} = \frac{\text{Distance}}{\text{Time}} \implies 4k = \frac{3x + 60}{6}\\
\mathbf{24k = 3x + 60} \quad — (Equation ii)$$
4. Solve for $\mathbf{x}$
By dividing Equation (ii) by Equation (i), we can eliminate $k$:
$$\frac{3x + 60}{8x + 270} = \frac{24k}{75k}\\
\frac{3(x + 20)}{8x + 270} = \frac{8}{25}$$
Cross-multiplying gives:
$$75(x + 20) = 8(8x + 270)\\
75x + 1500 = 64x + 2160\\
11x = 660\\
\mathbf{x = 60}$$
The value of $x$ is 60 km. Thus, correct answer is option (a).
92. A person can cover a distance between A and B at a speed of 25 km/hr. On the return journey, he covers 200 km with 20 km/hr, and after that, he takes a different route on the return journey which increases his total distance by 45 km. If the speed of the car on the extended route is 40 km/hr. If it takes 120 min more while returning then find the distance between A and B.
(a) 215 km
(b) 250 km
(c) 275 km
(d) 295 km
(e) 315 km
Answer: (c)
Explanation:
Step 1: Set Up the Variables
Let the distance between A and B be $D$ km.
Forward Journey Time: The person covers distance $D$ at a speed of 25 km/hr.
$\text{Time}_{forward} = \frac{D}{25}$
Return Journey Time: This journey is split into two parts:
First 200 km covered at 20 km/hr.
The remaining distance on a new route. The new total distance is $(D + 45)$ km, so the remaining distance is $(D + 45 – 200)$ km, covered at 40 km/hr.
$\text{Time}_{return} = \frac{200}{20} + \frac{D + 45 – 200}{40}$
Step 2: Establish the Time Difference
The problem states the return journey takes 120 minutes (2 hours) more than the forward journey.
$$\text{Time}_{return} – \text{Time}_{forward} = 2 \text{hours}$$
$$\left(\frac{200}{20} + \frac{D – 155}{40}\right) – \frac{D}{25} = 2$$
Step 3: Solve the Equation
Simplifying the return journey components:
$\frac{200}{20} = 10$
$10 + \frac{D – 155}{40} – \frac{D}{25} = 2$
Rearranging the terms:
- $10 – 2 – \frac{155}{40} = \frac{D}{25} – \frac{D}{40}$
- $8 – \frac{155}{40} = \frac{40D – 25D}{1000}$
- $\frac{320 – 155}{40} = \frac{15D}{1000}$
- $\frac{165}{40} = \frac{15D}{1000}$
Solving for $D$:
- $15D \times 40 = 165 \times 1000$
- $600D = 165000$
- $D = \frac{165000}{600} = \mathbf{275 \text{km}}$
The distance between A and B is 275 km.
Thus, correct answer is option (c).
93. The age of the mother 8 years ago was 4 times the age of her son at that time. The age of the mother after 4 years will be $\frac{16}{7}$ times the age of the son at the time. Find the ratio between the mother’s age after 7 years and the present age of the son.
(a) 2 : 3
(b) 3 : 1
(c) 4 : 1
(d) 5 : 2
(e) 6 : 5
Answer: (b)
Explanation:
Step 1: Define Ages Based on the Past
Let the age of the son 8 years ago be x.
Since the mother’s age was 4 times the son’s age at that time, her age 8 years ago was 4x.
Step 2: Determine Ages in the Future
To find their ages 4 years from now (which is 12 years after the “8 years ago” point), we add 12 to their past ages:
- Son’s age after 4 years: $x + 12$
- Mother’s age after 4 years: $4x + 12$
Step 3: Solve for x
According to the problem, the mother’s age after 4 years will be $\frac{16}{7}$ times the son’s age at that time:
$$(4x + 12) = \frac{16}{7} \times (x + 12)$$
Multiply both sides by 7 to clear the fraction:
$$28x + 84 = 16x + 192 \\
12x = 108\\
\mathbf{x = 9}$$
Step 4: Calculate Present and Required Ages
- Present age of son: x + 8 = 9 + 8 = 17 years.
- Present age of mother: 4x + 8 = 4(9) + 8 = 36 + 8 = 44 years.
- Mother’s age after 7 years: 44 + 7 = 51 years.
Step 5: Find the Required Ratio
The ratio of the mother’s age after 7 years to the son’s present age is 51 : 17
Divide both sides by 17: 3 : 1
Thus, correct answer is option (b) 3 : 1
94. The volume of a cylinder is 4004 cm$^3$ and the diameter of the cylinder is equal to the side of a square whose perimeter is 56 cm. Find the curved surface area of the cylinder.
(a) 1122 cm$^2$
(b) 2233 cm$^2$
(c) 1144 cm$^2$
(d) 2244 cm$^2$
(e) 3311 cm$^2$
Answer: (c)
Explanation:
Step 1: Find the Radius of the Cylinder
The diameter of the cylinder is equal to the side of a square with a perimeter of $56\text{cm}$.
- Side of the Square: Since $\text{Perimeter} = 4 \times \text{side}$, then $\text{side} = \frac{56}{4} = 14\text{cm}$.
- Radius of the Cylinder: The diameter is $14\text{cm}$, so the radius ($r$) is $\frac{14}{2} = 7\text{cm}$.
Step 2: Find the Height of the Cylinder
Using the volume of the cylinder ($V = \pi r^2h = 4004\text{cm}^3$) and the radius we just found:
$$\frac{22}{7} \times 7 \times 7 \times h = 4004\\
154 \times h = 4004\\
h = \frac{4004}{154} = 26\text{cm}$$
Step 3: Calculate the Curved Surface Area (CSA)
The formula for the curved surface area of a cylinder is $2\pi rh$:
$$\text{CSA} = 2 \times \frac{22}{7} \times 7 \times 26\\
\text{CSA} = 44 \times 26 \\
\text{CSA} = 1144\text{cm}^2 $$
Thus, correct answer is option (c).
95. There are 6 consecutive odd numbers. If the product of the 1st and the 2nd number is 675, then find the average of all the numbers.
(a) 29
(b) 30
(c) 31
(d) 32
(e) 33
Answer: (b)
Explanation:
Step 1: Identify the Odd Numbers
Let the six consecutive odd numbers be (2n-3), (2n-1), (2n+1), (2n+3), (2n+5) and (2n+7).
Step 2: Solve for the First Two Numbers
The problem states that the product of the first and second numbers is $675$.
- $(2n-3) \times (2n-1) = 675$.
- Since these are consecutive odd numbers, they must be close together. We can find them by factoring $675$.
- $25 \times 27 = 675$.
Therefore, the first number is 25 and the second is 27.
Step 3: List the Sequence
Following the pattern of consecutive odd numbers, the full sequence is 25, 27, 29, 31, 33, and 35.
Step 4: Calculate the Average
The average of a set of consecutive numbers in an arithmetic progression is simply the middle value. For six numbers, it is the average of the two middle terms ($29$ and $31$):
$$\text{Average} = \frac{29 + 31}{2} = \frac{60}{2} = 30 $$
Thus, correct answer is option (b) 30.
96. The ratio of the number of boys to the number of girls is 5 : 3 and the average age of the whole class is ‘x’ and the average age of girls is ‘x + 2’ then find the average age of boys in the class.
(a) x – 1.2
(b) x + 1.25
(c) x + 2
(d) x – 2
(e) x + 2.5
Answer: (a)
Explanation:
1. Define the Quantities:
- The ratio of boys to girls is given as 5 : 3.
- Let the total number of boys = 5z.
- Let the total number of girls = 3z.
- The total number of students in the class = 5z + 3z = 8z.
2. Define the Averages:
- Average age of the whole class = x.
- Average age of girls = x + 2.
- Let the average age of boys = A.
3. Set Up the Weighted Average Equation:
The total age of all students is equal to the sum of the total age of boys and the total age of girls:
$$\text{(Total Boys} \times \text{Average Boys)} + \text{(Total Girls} \times \text{Average Girls)} = \text{Total Class} \times \text{Average Class}$$
$$(5z \times A) + (3z \times (x + 2)) = 8z \times x$$
4. Solve for A:
First, divide the entire equation by $z$ to simplify: $5A + 3(x + 2) = 8x$
Expand the terms: $5A + 3x + 6 = 8x$
Subtract $3x$ from both sides: $5A + 6 = 5x$
Isolate $5A$:
$5A = 5x – 6$
Divide by 5 to find the average age of boys ($A$):
$$A = \frac{5x}{5} – \frac{6}{5}\\
A = x – 1.2$$
The average age of the boys in the class is x – 1.2.
Thus, correct answer is option (a) x – 1.2
97. Profit earned on an article when it is sold for ₹ 624 is 20% more than the loss when it is sold for ₹ 360. If a person marked up the article by 25% on purchasing price and sold it after allowing two successive discounts of 10%. Find the selling price of the article.
(a) ₹ 534
(b) ₹ 486
(c) ₹ 537
(d) ₹ 538
(e) ₹ 539
Answer: (b)
Explanation:
Step 1: Calculate the Cost Price (CP)
- Let the cost price of the article be x.
- When sold for ₹ 624, the profit is (624 – x).
- When sold for ₹ 360, the loss is (x – 360).
- The problem states that the profit is 20% more than the loss:
$\quad 624 – x = 1.2 \times (x – 360)$. - Solving the equation:
$\quad 3120 – 5x = 6x – 2160$.
$\quad 5280 = 11x$.
$\quad \mathbf{x = 480}$ - The cost price of the article is ₹ 480.
Step 2: Calculate the Marked Price (MP)
- The article is marked up by 25% on its purchasing price.
- $\text{Marked Price} = 480 \times 1.25$.
- $\textbf{Marked Price} \mathbf{= ₹ 600}$.
Step 3: Calculate the Selling Price (SP)
- The person allows two successive discounts of 10% on the marked price.
- $\text{Selling Price} = 600 \times 0.9 \times 0.9$.
- $\text{Selling Price} = 600 \times 0.81$.
- $\textbf{Selling Price} \mathbf{= ₹ 486}$.
Thus, correct answer is option $\mathbf{(b) ₹ 486}$
98. Profit earned on selling a table for ₹ (X + 2400) is 250% more than the loss incurred on selling the table for ₹ (X + 1000). Had the table been sold for ₹ ‘X’, there would have been a loss of 70%. What is the cost price of the table?
(a) ₹1500
(b) ₹1600
(c) ₹1800
(d) ₹2000
(e) ₹2400
Answer: (d)
99. The question consists of two statements numbered “I” and “II” given below it. You have to decide whether the data provided in the statements are sufficient to answer the question or not and choose the correct option accordingly.
Find the time of the journey of Kush to reach school on time from his home given that the distance between his home and school is 250 km.
Statement I: If Kush travels with a speed of (x + 2.5) km/hr, he reaches the school from home on time.
Statement II: If Kush travels with a speed of (x – 2.5) km/hr, he will be late by 2.5 hours.
(a) The data in statement I alone is sufficient to answer, while the data in statement II alone is not sufficient to answer the question.
(b) The data in statement II alone is sufficient to answer, while the data in statement I alone is not sufficient to answer the question.
(c) The data either in statement I alone or in statement II alone is sufficient to answer the question.
(d) The data given in both statements I and II together are not sufficient to answer the question.
(e) The data in both statements I and II together are necessary to answer the question.
Answer: (e)
Explanation:
1. Analyze Statement I
- Data given: The distance is 250 km, and at a speed of $(x + 2.5)$ km/hr, Kush reaches on time.
- Formula: $\text{Time} = \frac{\text{Distance}}{\text{Speed}}$.
- Equation: $\text{Time} = \frac{250}{x + 2.5}$.
- Conclusion: This statement alone is not sufficient because we have two unknown variables ($x$ and Time) in a single equation.
2. Analyze Statement II
- Data given: At a speed of $(x – 2.5)$ km/hr, Kush is late by 2.5 hours.
- Equation: $\text{Actual Time} + 2.5 = \frac{250}{x – 2.5}$.
- Conclusion: This statement alone is also not sufficient for the same reason—one equation with two unknown variables.
3. Combine Statement I and II
By combining both equations, we can eliminate the “Time” variable to solve for $x$:
- The difference in time between the two scenarios is 2.5 hours.
- $\frac{250}{x – 2.5} – \frac{250}{x + 2.5} = 2.5$.
- After simplifying the algebraic expression, we find:
- $x^2 = 506.25$.
- $x = 22.5$ (ignoring the negative root as speed cannot be negative).
- Now, substitute $x$ back into the first equation to find the on-time journey duration:
- $\text{Time} = \frac{250}{22.5 + 2.5} = \frac{250}{25} = \mathbf{10 \text{hours}}$.
Thus, correct answer is option (e) The data in both statements I and II together are necessary to answer the question.
100. Yash spends 20% of his monthly salary on EMI. From the remaining, he spends 25% on the rent, 15% on food, 50% on groceries, and 10% on video blogging. If the difference between the amount spent on EMI and food is ₹480 then find the monthly salary of Yash.
(a) ₹5200
(b) ₹5300
(c) ₹5400
(d) ₹6000
(e) ₹5600
Answer: (d)
Explanation:
1. Establish Total Salary
Let Yash’s total monthly salary be represented by ₹100x.
2. Calculate EMI and Remaining Amount
- Amount spent on EMI: Yash spends $20\%$ of his salary on EMI.
- $20\% \text{of } 100x = \mathbf{₹20x}$.
- Remaining Amount: After the EMI, the remaining amount is:
- $100x – 20x = \mathbf{₹80x}$.
3. Calculate Food Expenditure
From the remaining $₹80x$, Yash spends $15\%$ on food:
- Amount spent on food: $15\% \text{of } 80x = \mathbf{₹12x}$.
4. Solve for $x$ using the Difference
The problem states that the difference between the amount spent on EMI ($₹20x$) and food ($₹12x$) is ₹480.
Set up the equation:
$\quad 20x – 12x = 480$
$\quad 8x = 480$
$\quad \mathbf{x = 60}$
5. Final Calculation for Monthly Salary
Now, substitute the value of $x$ back into the total salary expression:
- Monthly salary of Yash = $100 \times 60 = \mathbf{₹6000}$.
Thus, correct answer is option (d) ₹6000