Mathematical Skills Test - 8

Let the cost price of 1 kg of goods = Rs. 100

So, the selling price of 1 kg of goods = 100 ×$\frac{120}{100}$ = Rs. 120

Cost price of 900 grams of goods = Rs. 90

According to the question,

Shopkeeper charges 20% less what he normally charges

So, the new selling price = Old selling price ×$\frac{100-20}{100}$

⇒ New selling price = 120 ×$\frac{80}{100}$ = Rs. 96

So, profit = Rs. (96 - 90) = Rs. 6

So, profit %$=\frac{6}{90}\times 100$

∴ Profit% =$6\frac{2}{3}\%$

Given:

Speed of Mahesh = 20 km/hr

Speed of Rajesh = 40 km/hr

Distance between point A and B = 195 km

Speed of stream = 10 km/hr

Formula:

If the speed of boat in still water is ‘x’ km/hr and speed of stream is ‘y’ km/hr, then:

Upstream speed = Speed of boat - Speed of stream = (x – y)

Downstream speed = Speed of boat + Speed of stream = (x + y)

Calculation:

Speed of Mahesh in downstream = (20 + 10) = 30 km/hr

Distance travelled by Mahesh in 30 min

⇒ Distance travelled by Mahesh in 30 min$\left(\frac{1}{2}\mathrm{hr}\right)$ = 30 ×$\frac{1}{2}$ =15 km

Remaining distance = 195 – 15 = 180 km

Speed of Rajesh in upstream = (40 – 10) = 30 km/hr

Distance travelled by Mahesh and Rajesh together = 30 + 30 = 60 km/hr

∴ Time taken to cover this Distance

⇒ Time =$\frac{\mathrm{Distance}}{\mathrm{Speed}}$

⇒ Time =$\frac{180}{60}$ = 3 hr

∴ Distance travelled by Mahesh in 3 hrs = 30 × 3 = 90 km

Distance travelled by Mahesh in first 30 min = 15 km

∴ Total distance from point A after which they meet = 90 + 15 = 105 km

Given:

\(S=\frac{\cos x-\sin x+1}{\cos x+\sin x-1}\)

\(S=\frac{\cos x-(\sin x-1)}{\cos x+(\sin x-1)} \times \frac{\cos x-(\sin x-1)}{\cos x-(\sin x-1)}\)

\(S=\frac{[\cos x-(\sin x-1)]^{2}}{\cos ^{2} x-(\sin x-1)^{2}}\)

\(S=\frac{\cos ^{2} x+(\sin x-1)^{2}-2 \cos x(\sin x-1)}{\cos ^{2} x-\left(\sin ^{2} x-2 \sin x+1\right)}\)

\(S=\frac{\cos ^{2} x+\sin ^{2} x-2 \sin x+1-2 \cos x(\sin x-1)}{1-\sin ^{2} x-\left(\sin ^{2} x-2 \sin x+1\right)}\)

\(S=\frac{2-2 \sin x-2 \cos x(\sin x-1)}{1-\sin ^{2} x-\sin ^{2} x+2 \sin x-1}\)

\(S=\frac{2(1-\sin x)+2 \cos x(1-\sin x)}{2 \sin x-2 \sin ^{2} x}\)

\(S=\frac{(1-\sin x)(1+\cos x)}{\sin x(1-\sin x)}\)

\(S=\frac{1+\cos x}{\sin x}\)

S = cosec x + cot x

Given:

Ratio of investments of A, B and C =$\frac{1}{2}:\frac{1}{3}:\frac{1}{4}$

Total profit at the end of the year = Rs. 10400

Total profit = 10% of the total investment

Concept used:

Ratio of product of time and investment = Ratio of profit

When the investments made by all partners are for the same time period, then profit is distributed amongst them in the ratio of their investments

Calculation:

Ratio of investments =$\frac{1}{2}:\frac{1}{3}:\frac{1}{4}=6:4:3$

Let, investments of A, B and C are 6X, 4X and 3X respectively, therefore total investment = 13X

Ratio of profit = 6 ∶ 4 ∶ 3 (Using concept)

Total profit = 10400 = 10% of 13X (Given)

⇒ 10% of 13X = 10400

⇒ X =$\frac{10400}{13}$ × 10 = 8000

Investment of B = 4X = 4 × 8000 = Rs. 32000

∴ B invested Rs. 32000.

Given:

Rohit scores 30% and failed by 15.

Mohan scores 40% and passed by 35.

Let y be total marks.

According to 1st condition, passing mark = 30% of y + 15

According to 2nd condition, passing mark = 40% of y - 35

So, 40% of y - 35 = 30% of y + 15

⇒ (40% - 30%) of y = 35 + 15

⇒ 10% of y = 50

⇒ 1% of y = 5

⇒ 3% of y = 15

Passing mark = 30% of y + 15

= 30% of y + 3% of y

= 33% of y

∴ The passing mark of the examination is 33%.

Given:

Two pipes X and Y can fill the tank in “t” min

X alone takes = (t + 6) min

Y alone takes = (t + 54) min

Let the total volume of the tank be x units

X’s efficiency\(= \left[\frac{x}{(t+6)}\right]\)units/min

Time taken by both pipes X and Y to fill the tank\(= \left[\frac{x}{(t+6)} + \frac{x}{(t+54)}\right] × t = x\)

\(\Rightarrow(2 t+60) t=(t+6)(t+54)\)

\(\Rightarrow 2 t^{2}+60 t=t^{2}+60 t+324\)

\(\Rightarrow t^{2}=324\)

\(\Rightarrow t=\sqrt{324}=18\)

∴ Time taken by X and Y together to fill the tank = 18 min

Time taken by pipe 1, 2 and 3 to fill = 15 hours

Time taken by pipe 4 to empty = 40 hours

Let capacity of tank = Lcm (15 and 40) = 120 units

Efficiency of pipe 1, 2 and 3 \(= \frac{120}{15}\) = 8units/hour

Efficiency of pipe 2\(= \frac{120}{40}\)= –3units/hour

Tank filled in 1 hour = 8 + 8 + 8 = 24 units

Tank filled in next 2 hours = (8 + 8 + 8 – 3) × 2 = 42 units

Tank filled in first 3 hours = 24 + 42 = 66 units

Tank to be filled = 120 – 66 = 54 units

Time taken to fill 54 units = 54 ÷ (8 + 8 + 8) = 54 ÷ 24 =\(=2 \frac{1}{4}\)hours

Given:

First number is 35% less than third

And second number is 25% more than the third number

Formula:

Assume the values multiple of 10

Concept of ratio and proportion

Required percentage =$\frac{\mathrm{Less}/\mathrm{More}}{\mathrm{Given}\mathrm{Quantity}}\times 100$

Let the third number is 100.

First number is 35% less than third

∴ 35% of 100 = 35

⇒ First number = 100 – 35 = 65

Second number is 25% less than third

∴ 25% of 100 = 25

⇒ First number = 100 + 25 = 125

∴ First ∶ Second ∶ Third = 65 ∶ 125 ∶ 100 = 13 ∶ 25 ∶ 20

Sum of first and second = 13 + 25 = 38

Required percentage =$\frac{38-20}{20}$ × 100 = 90%

So, the required percentage is 90%.

Given:

(A + B) can do the work = 12 days

(B + C) can do the work = 15 days

(C + A) can do the work = 20 days

Calculation:

One-day work of (A + B) =$\frac{1}{12}$ days ----(i)

One-day work of (B + C) =$\frac{1}{15}$ days ----(ii)

One-day work of (C + A) =$\frac{1}{20}$ days ----(iii)

Adding equations (i), (ii) and (iii), we get

One-day work of 2(A + B + C) =$\frac{1}{12}+\frac{1}{15}+\frac{1}{20}$

⇒ 2(A + B + C) =$\frac{5+4+3}{60}$

⇒ 2(A + B + C) =$\frac{12}{60}$

⇒ A + B + C =$\frac{12}{120}=\frac{1}{10}$ days

∴ A, B and C working together will complete the work in 10 days.

Given:

Number of oranges in the basket = 3

Number of apples in the basket = 2

Number of bananas in the basket = 5

Number of outcomes = Getting either an orange or a banana

\(\therefore\)Number of outcomes\(=3+5 = 8\)

Total number of outcomes\(=3+2+5 = 10\)

We know that:

Probability of an event = \(\frac{\text{Number of outcomes that make an event}}{\text{Total number of outcomes of the experiment} (\text{when outcomes are equally likely})}\)

Therefore,

Probability \(=\frac{8}{10}\)

\(= \frac{4}{5}\)

\(\therefore\) The probability of getting an orange or a banana is \(\frac{4}{5}\).