Mathematics and Quantitative Aptitude

Marked price = Rs 75,000
Discount = 15%

Selling price after discount: $= 75{,}000 \times \frac{85}{100} = Rs\ 63{,}750$

VAT = 15% on Rs 63,750: $= 63{,}750 \times \frac{15}{100} = Rs\ 9{,}562.50$

Amount paid by customer: $= 63{,}750 + 9{,}562.50 = Rs\ 73{,}312.50$

The customer has to pay Rs 73,312.50.

Let the borrowed sum be Rs $P$.

Simple interest for 2 years at 5%:

$I_{SI} = \frac{P \times 5 \times 2}{100} = 0.10P$

Amount to be repaid:

$A_{SI} = P + 0.10P = 1.10P$

Compound amount for 2 years at 5%:

$A_{CI} = P\left(1+\frac{5}{100}\right)^2 = P(1.05)^2 = 1.1025P$

Gain in the transaction:

$A_{CI} - A_{SI} = 20$

$1.1025P - 1.10P = 20$

$0.0025P = 20$

$P = \frac{20}{0.0025} = 8000$

Therefore, the borrowed sum is Rs 8,000.

Let the assumed mean be (A=10).

Mean: [ \bar{x}=A+\frac{\sum fd}{N} =10+\frac{30}{300} =10+0.1=10.1 ]

Standard deviation: [ \sigma=\sqrt{\frac{\sum fd^2}{N}-\left(\frac{\sum fd}{N}\right)^2} ]

[ \sigma=\sqrt{\frac{1780}{300}-\left(\frac{30}{300}\right)^2} ]

[ \sigma=\sqrt{5.9333-0.01} =\sqrt{5.9233} =2.43 ]

[ \boxed{\text{Standard deviation }(\sigma)=2.43} ]

Given:

• Tap A fills the tank in 30 hours.
• Tap B fills the tank in 15 hours.
• Both taps were opened together for some time, then A was closed.
• B alone filled the remaining tank in 12 hours.

To find: After how much time was tap A turned off?

Solution:

Work done by A in 1 hour $= \frac{1}{30}$

Work done by B in 1 hour $= \frac{1}{15}$

Work done by both taps in 1 hour:

$$\frac{1}{30}+\frac{1}{15} = \frac{1+2}{30} = \frac{3}{30} = \frac{1}{10}$$

Let tap A be turned off after $t$ hours.

Work done by both taps in $t$ hours $= \frac{t}{10}$

Work done by B alone in 12 hours:

$$12 \times \frac{1}{15} = \frac{12}{15} = \frac{4}{5}$$

Since the tank is completely filled,

$$\frac{t}{10}+\frac{4}{5}=1$$ $$\frac{t}{10}=1-\frac{4}{5}$$ $$\frac{t}{10}=\frac{1}{5}$$ $$t=2$$

Therefore, tap A was turned off after 2 hours.

Let the cost prices of the two bicycles be Rs $CP_1$ and Rs $CP_2$.

First bicycle: sold at 40% profit, SP = Rs 7140
$SP_1 = CP_1 \times \frac{140}{100}$ $7140 = CP_1 \times 1.4$ $CP_1 = \frac{7140}{1.4} = Rs\,5100$

Second bicycle: sold at 15% loss, SP = Rs 7140
$SP_2 = CP_2 \times \frac{85}{100}$ $7140 = CP_2 \times 0.85$ $CP_2 = \frac{7140}{0.85} = Rs\,8400$

Total cost price: $CP = 5100 + 8400 = Rs\,13500$

Total selling price: $SP = 7140 + 7140 = Rs\,14280$

Since $SP > CP$, there is an overall profit.

$\text{Profit} = 14280 - 13500 = Rs\,780$

$\text{Overall profit percent} = \frac{780}{13500} \times 100$ $= 5.777\ldots\% = 5\frac{7}{9}\% \approx 5.78\%$

Therefore, Arjun made an overall profit of approximately 5.78%.

Given: Total amount = Rs. 840 Ratio = 2 : 3 : 7

Steps:

Sum of ratio terms: $2 + 3 + 7 = 12$

Value of one part: $\frac{840}{12} = \text{Rs. } 70$

Shares:

First part: $2 \times 70 = \text{Rs. } 140$

Second part: $3 \times 70 = \text{Rs. } 210$

Third part: $7 \times 70 = \text{Rs. } 490$

Check: $140 + 210 + 490 = 840 \checkmark$

Answer: The amounts are Rs. 140, Rs. 210 and Rs. 490 respectively.

Given:

• Red balls = 2
• Green balls = 3
• Blue balls = 2
• Total balls = $2+3+2=7$
• Two balls are drawn at random.

To find: Probability that none of the drawn balls is blue.

Non-blue balls = Red + Green = $2+3=5$

Total ways of drawing 2 balls from 7: $n(S)={7 \choose 2}=\frac{7\times6}{2}=21$

Favourable ways of drawing 2 balls from the 5 non-blue balls: $n(E)={5 \choose 2}=\frac{5\times4}{2}=10$

Therefore, $P(E)=\frac{n(E)}{n(S)}=\frac{10}{21}$

Alternative method: $P(\text{none blue})=\frac{5}{7}\times\frac{4}{6}=\frac{20}{42}=\frac{10}{21}$

Answer: The probability that none of the balls drawn is blue is $\boxed{\frac{10}{21}}$.