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  • Question 1
    4 / -1

    A is twice as fast as B and B is thrice as fast as C. If some distance is covered by C in 54 minutes, then in how much time will B cover?

    Solution

    Given -

    A is twice as fast as B and B is thrice as fast as C.

    C covered some distance in 54 min.

    Formula used -

    Speed = DistanceTime

    Let the time taken by B be t min.

    ⇒ A : B = 2 : 1, B : C = 3 : 1

    ⇒ A : B : C = (2 × 3) : (1 × 3) : (1 × 1)

    ⇒ A : B : C = 6 : 3 : 1 

    ⇒ Ratio of speed between A, B and C = 6 : 3 : 1

    ⇒ Ratio of time = 1613 : 1

    ⇒ Ratio of time between A, B and C = 1 : 2 : 6

    ⇒ 6 unit = 54 min

    ⇒ 1 unit = 9 min

    ⇒ 2 unit = 18 min

    ⇒ So, time taken by B = t = 18 min

    ∴ Time taken by B = 18 min.

  • Question 2
    4 / -1

    A train running at a speed of 72 km/hr takes 25 seconds to cross an electric pole. What is the length of the train?

    Solution

    Formula used:

    Distance = speed × time

    1 m/sec = 185 km /hr

    Length of train = distance

    Speed convert km/hr to m/sec

    ⇒ 72 × 518

    ⇒ Speed = 20 m/sec

    Distance = 20 × 25

    ⇒ 500 metres

    ∴ The length of the train is 500 metres.

  • Question 3
    4 / -1

    Two trains of equal length are running on parallel lines in the same direction at the rate of \(46\) km/hr and \(36\) km/hr. The faster train passed the slower train in \(72\) sec. The length of each train is:

    Solution

    Let the length of each train be \(x\) m.

    Relative speed \(= (46 – 36) = 10\) km/hr

    \(= 10 × \frac{5}{18}\)

    \(= \frac{50}{18}\) m/sec

    We know that,

    Sum of lengths of two trains \(=\) time \(×\) relative speed

    \(2x = 72 × \frac{50}{18}\)

    \(\Rightarrow x = 72 × \frac{50}{36}\)

    \(\Rightarrow x = 100\) m

  • Question 4
    4 / -1

    If \(\omega\) is a complex cube root of unit, then \(1+\omega+\omega^{2}+\ldots+\omega^{100}\) is equal to:

    Solution

    Sum of Gp series of nth terms:

    \(S_{n}=\frac{a \times\left(r^{n}-1\right)}{r-1}\)

    Where,

    a: First Term

    r: common ratio

    If \(\omega\) is a complex root of the unit. Then,

    \(\omega^{3}=1\)

    According to question

    \(1+\omega+\omega^{2}+\ldots+\omega^{100}\)

    We know that,

    \(\omega^{3}=1\)

    so, \(\left(\omega^{3}\right)^{33}=(1)^{33}\)

    \(\omega^{99}=1\)

    Now, Solving GP:

    \(=\frac{1 \times\left(\omega^{101}-1\right)}{\omega-1}\)

    \(=\frac{\omega^{99}\left(\omega^{2}-1\right)}{\omega-1}\)

    \(=\frac{\omega^{99}(\omega-1)(\omega+1)}{\omega-1}\)

    \(=\omega+1\)

  • Question 5
    4 / -1

    Which of the fractions given below, when added to 57 give 1?

    Solution

    Let the fraction be x,

     57+x=1

     1  57=x

      x =27=621 

  • Question 6
    4 / -1

    The average salary of skilled employees in a firm is Rs. \(520\) and that of the unskilled employees is Rs. \(420\). If the average salary of both the groups of employees is Rs. \(500\), then the percentage of skilled employees in the firm is:

    Solution
    Average \(=\frac{(n_1 \times x_1+n_2 \times x_2)} {(n_1+n_2)}\)
    \( 500=\frac{ (Skilled \times 520+ Unskilled \times 420)} {( Skilled + Unskilled )} \)
    \(\Rightarrow 500\) Skilled \(+500\) Unskilled \(=520\) Skilled \(+420\) Unskilled
    \(\Rightarrow 20\) Skilled \(=80\) Unskilled
    Skilled/Unskilled = \(\frac{80}{20}\)
  • Question 7
    4 / -1

    When the integers 10, 0, 5, -5, -7 are arranged in descending or ascending order, then find which of the following integer always remains in the middle of the arrangement?

    Solution

    To arrange these integers in ascending or descending order, we first place these points on the number line.

    NCERT Exemplar Class 7 Maths Integers Img 1

     

    As we know that if a point or number is located to the right of another number, then that number is larger. 

     

    Then,

    Ascending order = -7, - 5, 0, 5, 10

    Middle term = 0

    Descending order = 10, 5, 0, - 5, - 7

    Middle term = 0

    Hence, the correct option is (A)

  • Question 8
    4 / -1

    The ratio, in which tea costing Rs. 192 per kg is to be mixed with tea costing Rs. 150 per kg so that the mixed tea when sold for Rs. 194.40 per kg, gives a profit of 20%.

    Solution

    Let the CP of Mixture be Rs. x per kg. Therefore,

    x + 20% of x = SP \(\Rightarrow \frac{6x}{5}\) = 194.40

    x = Rs. 162 per kg.

    Let N kg of first tea and M kg of second tea be added.

    Now, Using Alligation, We get,

    \(\frac{N}{M} = \frac{12}{30}\) \(\Rightarrow\) N : M = 2 : 5

  • Question 9
    4 / -1

    Trimsy speaks the truth \(2\) out of \(9\) times. On selecting a card randomly from a pack of cards, she reports that it is either king or ace. Find the probability that it is actually king or ace:

    Solution

    Trimsy speaks the truth \(2\) out of \(9\) times. Probability of Truth \(=\frac{2}{9}\); Probability of lie \(=\frac{7}{9}\)

    Since, there are \(4\) ace and \(4\) king in each pack of \(52\) cards so,

    Probability of getting an king or ace \(=\frac{8}{52}\)

    Probability (getting a king or ace when, Trimsy reports that it is a king or ace) \(=\left(\frac{2}{9}\right) \times\left(\frac{8}{52}\right)\)

    \(\mathrm{P}\) (getting a king or ace when Trimsy reports that it is a king or ace) \(=\frac{4}{117}\)

  • Question 10
    4 / -1

    What is the area of a right-angled triangle, if the radius of the circumcircle is \(5 \mathrm{~cm}\) and altitude drawn to the hypotenuse is \(4 \mathrm{~cm}\)?

    Solution

    Given:

    Radius of circumcircle \(=5 \mathrm{~cm}\)

    Altitude to the hypotenuse \(=4 \mathrm{~cm}\)

    As we know,

    The hypotenuse is the diameter of a circle if an angle of a triangle is \(90^{\circ}\) which is inscribed in a circle, that triangle formed by the diameter of the circle.

    Area of triangle \(=\frac{(\mathrm{B} \times \mathrm{H})}{2}\)

    Where

    \(\mathrm B\) is the base of the triangle

    \(\mathrm{H}\) is the height of the triangle

    The hypotenuse of tringle is \(B C=2 \times\) of the radius of \(\operatorname{circles}(O C)\)

    \(\mathrm{BC}=2 \times 5\)

    \(=10 \mathrm{~cm}\)

    The altitude drawn to hypoteneus is \(A D=4 \mathrm{~cm}\)

    The area of \(\triangle \mathrm{ABC}=\frac{(10 \times 4)}{2}=20 \mathrm{~cm}^{2}\)

    \(\therefore\) The area of the right-angle triangle is \(20 \mathrm{~cm}^{2}\)

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