Q. 1
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
After a 22 percent deduction. A's net salary is Rs. 1,600. A's gross salary is nearly
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Step-by-Step Explanation & Concept Rationale
Let the gross salary be $G$.
A $22\%$ deduction leaves $100\% - 22\% = 78\%$ of the gross salary.
\[ 0.78 \times G = 1600 \]
\[ G = \frac{1600}{0.78} \approx \text{Rs. } 2,051.28 \]
Therefore, the gross salary is nearly Rs. 2,051.
A $22\%$ deduction leaves $100\% - 22\% = 78\%$ of the gross salary.
\[ 0.78 \times G = 1600 \]
\[ G = \frac{1600}{0.78} \approx \text{Rs. } 2,051.28 \]
Therefore, the gross salary is nearly Rs. 2,051.
Q. 2
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Aslam purchased a Chair for Rs. 2,000 and sold it to Arshad at a loss of 10 percent Arshad sold it to Akbar at a loss of 10 percent while Akbar sold it to Qumar at a gain of 10 percent. The amount Qumar paid for it would be Rs......
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Step-by-Step Explanation & Concept Rationale
We can track the value of the chair through each successive transaction:
Aslam sells to Arshad at a 10% loss:
Price paid by Arshad = \(2000 \times (1 - 0.10) = 2000 \times 0.90 = \mathbf{\text{Rs. 1,800}}\)
Arshad sells to Akbar at a 10% loss:
Price paid by Akbar = \(1800 \times (1 - 0.10) = 1800 \times 0.90 = \mathbf{\text{Rs. 1,620}}\)
Akbar sells to Qumar at a 10% gain:
Price paid by Qumar = \(1620 \times (1 + 0.10) = 1620 \times 1.10 = \mathbf{\text{Rs. 1,782}}\)
Aslam sells to Arshad at a 10% loss:
Price paid by Arshad = \(2000 \times (1 - 0.10) = 2000 \times 0.90 = \mathbf{\text{Rs. 1,800}}\)
Arshad sells to Akbar at a 10% loss:
Price paid by Akbar = \(1800 \times (1 - 0.10) = 1800 \times 0.90 = \mathbf{\text{Rs. 1,620}}\)
Akbar sells to Qumar at a 10% gain:
Price paid by Qumar = \(1620 \times (1 + 0.10) = 1620 \times 1.10 = \mathbf{\text{Rs. 1,782}}\)
Q. 3
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Ali bought a electric fan priced at Rs. 2,000. He was given two successive discounts of 10 and 5 percent. If he had to pay 10 percent sales tax, the net amount he paid was.
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Step-by-Step Explanation & Concept Rationale
Apply the first discount (10%):
Discount amount = \(2,000 \times 0.10 = \text{Rs. 200}\)
Price after 1st discount = \(2,000 - 200 = \text{Rs. 1,800}\)
Apply the second discount (5%) to the reduced price:
Discount amount = \(1,800 \times 0.05 = \text{Rs. 90}\)
Price after 2nd discount = \(1,800 - 90 = \text{Rs. 1,710}\)
Add the 10% sales tax to the final discounted price:
Sales tax amount = \(1,710 \times 0.10 = \text{Rs. 171}\)
Net amount paid = \(1,710 + 171 = \mathbf{\text{Rs. 1,881}}\)
Discount amount = \(2,000 \times 0.10 = \text{Rs. 200}\)
Price after 1st discount = \(2,000 - 200 = \text{Rs. 1,800}\)
Apply the second discount (5%) to the reduced price:
Discount amount = \(1,800 \times 0.05 = \text{Rs. 90}\)
Price after 2nd discount = \(1,800 - 90 = \text{Rs. 1,710}\)
Add the 10% sales tax to the final discounted price:
Sales tax amount = \(1,710 \times 0.10 = \text{Rs. 171}\)
Net amount paid = \(1,710 + 171 = \mathbf{\text{Rs. 1,881}}\)
Q. 4
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
A man sells a radio and a mixer for Rs. 350 each. On one he gains 10 percent and on the other loses 10 percent. Thus on the whole, he
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Step-by-Step Explanation & Concept Rationale
When two items are sold at the same price—one at a gain of \(x\%\) and the other at a loss of \(x\%\), the overall transaction always results in a loss. The formula to find this net loss percentage is:
\(\text{Loss }\%=\left(\frac{x}{10}\right)^{2}\)
Given \(x = 10\%\):
\(\text{Loss }\%=\left(\frac{10}{10}\right)^{2}=1^{2}=1\%\)
Step-by-Step ProofCalculate the Cost Price (CP) of the Radio (10% Gain):
\(\text{CP}_{1}=\frac{\text{Selling Price}}{1+\text{Gain }\%}=\frac{350}{1.10}=\text{Rs. }318.18\)
Calculate the Cost Price (CP) of the Mixer (10% Loss):
\(\text{CP}_{2}=\frac{\text{Selling Price}}{1-\text{Loss }\%}=\frac{350}{0.90}=\text{Rs. }388.89\)
Find the Totals:Total Cost Price:
\(\text{Rs. } 318.18 + \text{Rs. } 388.89 = \text{Rs. } 707.07\)
Total Selling Price:
\(\text{Rs. } 350 + \text{Rs. } 350 = \text{Rs. } 700.00\)
Calculate Net Loss Percentage:
\(\text{Net Loss}=\text{Rs. }707.07-\text{Rs. }700.00=\text{Rs. }7.07\)
\(\text{Loss }\%=\left(\frac{7.07}{707.07}\right)\times 100=1\%\)
\(\text{Loss }\%=\left(\frac{x}{10}\right)^{2}\)
Given \(x = 10\%\):
\(\text{Loss }\%=\left(\frac{10}{10}\right)^{2}=1^{2}=1\%\)
Step-by-Step ProofCalculate the Cost Price (CP) of the Radio (10% Gain):
\(\text{CP}_{1}=\frac{\text{Selling Price}}{1+\text{Gain }\%}=\frac{350}{1.10}=\text{Rs. }318.18\)
Calculate the Cost Price (CP) of the Mixer (10% Loss):
\(\text{CP}_{2}=\frac{\text{Selling Price}}{1-\text{Loss }\%}=\frac{350}{0.90}=\text{Rs. }388.89\)
Find the Totals:Total Cost Price:
\(\text{Rs. } 318.18 + \text{Rs. } 388.89 = \text{Rs. } 707.07\)
Total Selling Price:
\(\text{Rs. } 350 + \text{Rs. } 350 = \text{Rs. } 700.00\)
Calculate Net Loss Percentage:
\(\text{Net Loss}=\text{Rs. }707.07-\text{Rs. }700.00=\text{Rs. }7.07\)
\(\text{Loss }\%=\left(\frac{7.07}{707.07}\right)\times 100=1\%\)
Q. 5
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Three successive discounts of 10 percent are equivalent to a single discount of ___ percent.
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Step-by-Step Explanation & Concept Rationale
Assume the initial price is 100.
First discount (10%):
\(100 - 10 = 90\)
Second discount (10% of 90):
\(90 - 9 = 81\)
Third discount (10% of 81):
\(81 - 8.1 = 72.9\)
Total equivalent discount:
\(100 - 72.9 = 27.1\)
First discount (10%):
\(100 - 10 = 90\)
Second discount (10% of 90):
\(90 - 9 = 81\)
Third discount (10% of 81):
\(81 - 8.1 = 72.9\)
Total equivalent discount:
\(100 - 72.9 = 27.1\)
Q. 6
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
A dealer marks his goods 20 percent above his cost price. If he gives a discount of 10 percent on his marked price, the profit he earns on his good is --- percent.
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Step-by-Step Explanation & Concept Rationale
Let CP=100
MP=100+20=120
SP=$120\times\frac{(100-10)}{100}=108$.
Profit = 8%.
MP=100+20=120
SP=$120\times\frac{(100-10)}{100}=108$.
Profit = 8%.
Q. 7
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
A man buys a scooter for Rs. 7,000 and sells it for Rs. 11,500. The percentage profit is nearly ------ percent.
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Step-by-Step Explanation & Concept Rationale
Find the profit amount: \(\text{Selling Price} - \text{Cost Price} = 11,500 - 7,000 =\) Rs. 4,500.
Calculate the profit percentage:
\(\text{Profit }\%=\left(\frac{\text{Profit}}{\text{Cost Price}}\right)\times 100\)
\(\text{Profit }\%=\left(\frac{4,500}{7,000}\right)\times 100\approx 64.28\%\)
When rounded to the nearest tenth, it gives 64.3%.
Calculate the profit percentage:
\(\text{Profit }\%=\left(\frac{\text{Profit}}{\text{Cost Price}}\right)\times 100\)
\(\text{Profit }\%=\left(\frac{4,500}{7,000}\right)\times 100\approx 64.28\%\)
When rounded to the nearest tenth, it gives 64.3%.
Q. 8
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
If Rs. 1200 yields Rs. 594 as interest for six years at simple interest, the interest rate would be nearly ------ percent.
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Step-by-Step Explanation & Concept Rationale
Simple Interest Formula: \(SI = \frac{P \times R \times T}{100}\)
Given values:
Principal (\(P\)) = Rs. 1200
Simple Interest (\(SI\)) = Rs. 594
Time (\(T\)) = 6 years
Rearrange the formula to solve for Rate (\(R\)):
\(R=\frac{SI\times 100}{P\times T}\)
\(R=\frac{594\times 100}{1200\times 6}\)
\(R=\frac{59400}{7200}=8.25\%\)
Given values:
Principal (\(P\)) = Rs. 1200
Simple Interest (\(SI\)) = Rs. 594
Time (\(T\)) = 6 years
Rearrange the formula to solve for Rate (\(R\)):
\(R=\frac{SI\times 100}{P\times T}\)
\(R=\frac{594\times 100}{1200\times 6}\)
\(R=\frac{59400}{7200}=8.25\%\)
Q. 9
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Rs. 2,000 at 5 percent compound interest after four years will become nearly.
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Step-by-Step Explanation & Concept Rationale
Compound Interest Formula: \(A = P \left(1 + \frac{R}{100}\right)^T\)
Given values:
Principal (\(P\)) = Rs. 2,000
Rate (\(R\)) = 5%
Time (\(T\)) = 4 years
Calculate the amount:
\(A=2000\times \left(1+0.05\right)^{4}\)
\(A=2000\times (1.05)^{4}\)
\(A=2000\times 1.2155\approx \text{Rs. }2431.01\)
When rounded to the closest option, it gives Rs. 2,430.
Given values:
Principal (\(P\)) = Rs. 2,000
Rate (\(R\)) = 5%
Time (\(T\)) = 4 years
Calculate the amount:
\(A=2000\times \left(1+0.05\right)^{4}\)
\(A=2000\times (1.05)^{4}\)
\(A=2000\times 1.2155\approx \text{Rs. }2431.01\)
When rounded to the closest option, it gives Rs. 2,430.
Q. 10
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
-- -- -- becomes Rs. 2,500 in four years at $6\frac{1}{4}$ percent, simple interest.
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Step-by-Step Explanation & Concept Rationale
Convert the rate to a decimal: \(6\frac{1}{4}\% = 6.25\%\) per year.
Calculate total interest percentage over 4 years:
\(6.25\% \times 4 \text{ years} = 25\%\).
Set up the amount formula:
The total amount (\(A\)) equals the Principal (\(P\)) plus \(25\%\) of the Principal.
\(A=P+0.25P=1.25P\)
Solve for Principal (\(P\)):
\(2500=1.25P\)
\(P=\frac{2500}{1.25}=2000\)
Calculate total interest percentage over 4 years:
\(6.25\% \times 4 \text{ years} = 25\%\).
Set up the amount formula:
The total amount (\(A\)) equals the Principal (\(P\)) plus \(25\%\) of the Principal.
\(A=P+0.25P=1.25P\)
Solve for Principal (\(P\)):
\(2500=1.25P\)
\(P=\frac{2500}{1.25}=2000\)
Q. 11
Quantitative Aptitude Test
Difficulty: Medium
(1 Mark)
Divide Rs. 793 into three parts such that their amounts 2, 3 and 4 years may be equal, the rate of interest being 5%.
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Step-by-Step Explanation & Concept Rationale
Understand Simple Interest Amount formula:
The formula for the total amount (\(A\)) after simple interest is applied is:
\(A=P\left(1+\frac{R\times T}{100}\right)\)
Where \(P\) is the principal part, \(R\) is the rate (5%), and \(T\) is the time.
Set up the amounts for the three parts (\(P_1, P_2, P_3\)):
For 2 years:
\(A_1 = P_1 \left(1 + \frac{5 \times 2}{100}\right) = P_1 \left(\frac{110}{100}\right)\)
For 3 years:
\(A_2 = P_2 \left(1 + \frac{5 \times 3}{100}\right) = P_2 \left(\frac{115}{100}\right)\)
For 4 years:
\(A_3 = P_3 \left(1 + \frac{5 \times 4}{100}\right) = P_3 \left(\frac{120}{100}\right)\)
Equate the amounts:Since the amounts are given to be equal (\(A_1 = A_2 = A_3\)):
\(110\cdot P_{1}=115\cdot P_{2}=120\cdot P_{3}\)
Find the ratio of the parts:
To find the ratio \(P_1 : P_2 : P_3\), we take the reciprocal of the coefficients:
\(P_{1}:P_{2}:P_{3}=\frac{1}{110}:\frac{1}{115}:\frac{1}{120}\)
The formula for the total amount (\(A\)) after simple interest is applied is:
\(A=P\left(1+\frac{R\times T}{100}\right)\)
Where \(P\) is the principal part, \(R\) is the rate (5%), and \(T\) is the time.
Set up the amounts for the three parts (\(P_1, P_2, P_3\)):
For 2 years:
\(A_1 = P_1 \left(1 + \frac{5 \times 2}{100}\right) = P_1 \left(\frac{110}{100}\right)\)
For 3 years:
\(A_2 = P_2 \left(1 + \frac{5 \times 3}{100}\right) = P_2 \left(\frac{115}{100}\right)\)
For 4 years:
\(A_3 = P_3 \left(1 + \frac{5 \times 4}{100}\right) = P_3 \left(\frac{120}{100}\right)\)
Equate the amounts:Since the amounts are given to be equal (\(A_1 = A_2 = A_3\)):
\(110\cdot P_{1}=115\cdot P_{2}=120\cdot P_{3}\)
Find the ratio of the parts:
To find the ratio \(P_1 : P_2 : P_3\), we take the reciprocal of the coefficients:
\(P_{1}:P_{2}:P_{3}=\frac{1}{110}:\frac{1}{115}:\frac{1}{120}\)
Q. 12
Quantitative Aptitude Test
Difficulty: Hard
(1 Mark)
Between two stations the first, second and third class fares were fixed at first in the ratio 8:6:3 but afterwards the first class fare were reduced by 1/6 and the second class by 1/12. In a year, the number of first, second and third class passengers were respectively as 9:12:26 and the money taken at the booking offices was Rs. 1088. How much was paid by the first class passengers.
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Step-by-Step Explanation & Concept Rationale
Find the new fare ratios:
First Class:
Reduced by \(\frac{1}{6}\) \(\rightarrow 8 \times \left(1 - \frac{1}{6}\right) = 8 \times \frac{5}{6} = \frac{20}{3}\)
Second Class: Reduced by \(\frac{1}{12}\) \(\rightarrow 6 \times \left(1 - \frac{1}{12}\right) = 6 \times \frac{11}{12} = \frac{11}{2}\)
Third Class: Remains unchanged \(\rightarrow \mathbf{3}\)
Calculate the ratio of revenue collected (\(\text{New Fare} \times \text{Number of Passengers}\)):
First Class: \(\frac{20}{3} \times 9 = \mathbf{60}\)
Second Class: \(\frac{11}{2} \times 12 = \mathbf{66}\)
Third Class: \(3 \times 26 = \mathbf{78}\)
The revenue collection ratio for First : Second : Third class is 60 : 66 : 78.
Find total units of revenue:
\(60+66+78=\mathbf{204}\text{ units}\)
Calculate money paid by First Class passengers:
\(\text{First Class Share}=\frac{60}{204}\times 1088\)
\(\text{First Class Share}=\mathbf{Rs.320}\)
First Class:
Reduced by \(\frac{1}{6}\) \(\rightarrow 8 \times \left(1 - \frac{1}{6}\right) = 8 \times \frac{5}{6} = \frac{20}{3}\)
Second Class: Reduced by \(\frac{1}{12}\) \(\rightarrow 6 \times \left(1 - \frac{1}{12}\right) = 6 \times \frac{11}{12} = \frac{11}{2}\)
Third Class: Remains unchanged \(\rightarrow \mathbf{3}\)
Calculate the ratio of revenue collected (\(\text{New Fare} \times \text{Number of Passengers}\)):
First Class: \(\frac{20}{3} \times 9 = \mathbf{60}\)
Second Class: \(\frac{11}{2} \times 12 = \mathbf{66}\)
Third Class: \(3 \times 26 = \mathbf{78}\)
The revenue collection ratio for First : Second : Third class is 60 : 66 : 78.
Find total units of revenue:
\(60+66+78=\mathbf{204}\text{ units}\)
Calculate money paid by First Class passengers:
\(\text{First Class Share}=\frac{60}{204}\times 1088\)
\(\text{First Class Share}=\mathbf{Rs.320}\)
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